Mathematics for Class 5 ( Maths Mela ) Solutions

Child TV Reporter

Samaira and Kabir decided to act like reporters and collect information from their friends.
They collected data from 35 friends and recorded it in a notebook as shown below.

Ans: 

Look at the table and answer the following questions.

Q1: How many children watch TV for more than half an hour?
Ans: Number of children who watch TV for more than half an hour =13 + 7 + 3 + 3 = 26 children.

Q2: How many children watch TV for less than two hours?
Ans: Number of children who watch TV for less than two hours = 9 + 13 + 7 = 29 children.

Q3: The number of children who watch TV for more than two hours is _________.
Ans: Number of children who watch TV for more than two hours is 3.

Q4: More children watch TV for two hours than half an hour. (True/False)
Ans: Given statement is false, since only 3 children watch TV for two hours, but 9 children watch TV for half an hour.

 Yes, watching TV for too long can cause eye strain and tiredness.

Stock-Taking in a Shop

Joseph Uncle takes stock of the play items (toys, board games, and sports items) in his store a week before the summer break. He tries to record the items in his shop using a pictograph. He notices that there are too many items of each kind in his shop and it is not easy to make a picture for every item
Dipesh, one of his helpers, suggested using one picture (icon) for every 5 items of each kind.
His pictograph is shown below.
Dipesh used a scale while recording the items in the pictograph. A scale helps record a large number of things using fewer icons.
Now, answer the following questions based on the above pictograph.

Q1: How many toys does Joseph Uncle have?
Ans: Joseph Uncle has 8 × 5 = 40 toys

Q2: How many board games does Joseph Uncle have?
Ans: Joseph Uncle has 10 × 5 = 50 board games.

Q3: How many total play items does Joseph Uncle have?
Ans: Joseph Uncle has 12 × 5 = 60 sports items.
Thus, total play items Joseph Uncle has = 40 + 50 + 60 = 150.

Q: Is there any other scale that you can use to make the pictograph?
Ans: Yes, we can use another scale to make the pictograph simpler.

Two-wheelers on the Road

Deepti noted down the number of two-wheelers passing her house in one hour on three different days.
She used one icon to show 3 two-wheelers.

Observe the pictograph and answer the following questions.

Q1: Which day had the most two-wheelers passing her house?
Ans: Friday

Q2: How many total two-wheelers did she record over three days?
Ans: Two-wheeler recorded on Monday
= 8 × 3 = 24 two-wheelers

Two-wheeler recorded on Wednesday
= 4 × 3 = 12 two-wheelers

Two-wheeler recorded on Friday
= 12 × 3 = 36 two-wheelers

Total two-wheelers Deepti recorded over three days = 24 + 12 + 36 = 72 two-wheelers

Q3: How many fewer two-wheelers were seen on Wednesday than on Monday?
Ans: On Monday, Deepti saw 24 two-wheelers
On Wednesday, Deepti saw 12 two-wheelers
Difference = 24 – 12 = 12 two-wheelers
∴ 12 fewer two-wheelers were seen on Wednesday than on Monday.

Q4: How many more two-wheelers were seen on Friday than on Wednesday?
Ans:
On Wednesday, Deepti saw 12 two-wheelers.
On Friday, Deepti saw 36 two-wheelers.
Difference = 36 – 12 = 24 two-wheelers.
24 more two-wheelers were seen on Friday than on Wednesday.

Q5: Choose any other scale and represent the same data using a pictograph in your notebook.
Ans: Here 1 = 4 two-wheelers

Recording a Day

One day Raman and Sheela decided to record how they spent their day.
Raman recorded his daily routine in the table below.
Sheela recorded her routine in the following manner.
Observe Raman’s and Sheela’s routines and answer the following questions.

Q1: Whose daily routine shows more time spent on sleeping?
Ans: Raman spent 9 hours on sleeping and Sheela spent 8 hours on sleeping.
So, Raman’s daily routine shows more time spent on sleeping.

Q2: Who spends more hours in the school?
Ans: Raman spends 6 hours at school and Sheela spends 7 hours at school.
So, Sheela spends more hours in school.

Q3: How many more hours does Sheela spend studying compared to Raman?
Ans: Sheela spends 4 hours and Raman spends 2 hours in studying.
Difference = 4 – 2 = 2 hours
Sheela spends 2 more hours studying compared to Raman.

Q4: Is there any activity on which they spend the same amount of time? If yes, name the activity.
Ans: No, there is no activity on which they spend the same amount of time.

Q5: Based on their data, whose routine do you think is more balanced? Why?
Ans: Sheela’s routine appears to be more balanced with a good allocation for sleep, studying, and other activities.

Day in My Life

Q: Record your daily routine (24 hours) in hours and minutes, as necessary. Note the time spent on activities like sleeping, studying, playing, eating, and others.

 Make a bar graph of the time you spend on different activities in the space given below.
Ans:

Whose Index Finger is Longer?

Cut long paper strips from waste paper. Give one strip (each of the same width) to each of your friends. Now, put the paper strip on your index finger and tear off the extra strip extending above your fingers.
Paste these paper strips along the horizontal line in the given bar graph.

Write the answers to the following questions based on your graph.
1. Whose index finger is the longest?

2. The length of the longest index finger is _________ cm.

3. The smallest index finger is _________

4.  It belongs to _________.
Ans: Do it yourself.

Food Wastage in the School Canteen

Rani stays in a residential school. Her school’s dining hall displays the amount of food wasted and the number of children the food could have fed. Given below is the data collected over the weekdays for different food items.

Q: Rani was shocked to see the data. What do you think about food wastage? How can we reduce the wastage of food? What can we do with the leftover food?
Ans: The food wastage is not only a waste of resources and money, but also disrespectful to the efforts of farmers and cooks.
The wastage of food can be reduced by:
(i) Preparing food according to the number of people eating.
(ii) Serving food in smaller portions.
(iii) Storing leftover food properly in the refrigerator.
(iv) Teaching children and adults the value of food.
The leftover food can be:
(i) Shared with neighbours, relatives, and friends.
(ii) Donated to NGOs for distribution to the needy.
(iii) Used creatively to make new dishes.

Observe the above graph and answer the following questions.
Q1: Which food item had the highest amount of wastage? _____________
Ans: Khichdi had the highest amount of wastage i.e., of 8 kg.

Q2: Which food item had the least amount of wastage? _______________
Ans: Idli-Sambhar had the least amount of wastage i.e., of 4 kg.

Q3: How much total food wastage was recorded in these days? _________
Ans: Total food wastage recorded in Monday to Friday = 5 + 6 + 8 + 7 + 4 = 30 kg

Q4: If 1 kg of food waste can feed 3 children, how many children could have been fed with the total food wasted?
Ans: 1 kg of food waste can feed 3 children.
30 kg of food waste can feed 3 × 30
= 90 children.

Q5: ________ day had less food wastage than day.
Ans: Friday had less food wastage than Wednesday.

Q6: If the same food items are to be repeated next week, can you predict which food item is likely to be wasted the most?
Ans: Khichdi had the highest wastage of 8 kg this week. So, Khichdi is most likely to be wasted the most again next week.

Page No. 187

True or False

Q: Observe the above picture carefully. Based on your observation, find out which of the following statements are true or false.

Ans:

Q: Do you see why 12 is a multiple of 1, 2, 3, 4, 6, and 12?
Ans:
2 × 6 = 12, 3 × 4 = 12, 12 × 1 = 12

Let Us Do

Q1: Make different arrays for the following numbers. Identify the factors in each case.
(a) 10
Ans:

Factors of 10 are 1, 2, 5, and 10.

(b) 14
Ans:

Factors of 14 are 1, 2, 7, and 14.

(c) 13
Ans:

Factors of 13 are 1 and 13 only. So, 13 is a prime number.

(d) 20
Ans:

Factors of 20 are 1, 2, 4, 5, 10, and 20.

(e) 25
Ans:

Factors of 25 are 1, 5, and 25.

(f) 32
Ans:

Factors of 32 are 1, 2, 4, 8, 16, and 32.

(g) 37

Ans:

Factors of 37 are 1 and 37 only. So, 37 is a prime number.

(h) 46
Ans:

Factors of 46 are 1, 2, 23 and 46.

(i) 54
Ans:

Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54.

Q: What do you notice about the common multiples of 3 and 4? Discuss in class.
Ans: Do it yourself.

Page No. 166-170

Let Us Do

Q1: Find 5 common multiples of the following pairs of numbers.
(a) 2 and 3
Ans:
2 and 3
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, …
Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, …
 5 common multiples of 2 and 3 are: 6, 12, 18, 24, and 30.
(b) 5 and 8
Ans:

5 and 8
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, …
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, …
 5 common multiples of 5 and 8 are: 40, 80, 120, 160, and 200.

(c) 2 and 4
Ans:
Multiples of 2:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, …
Multiples of 4:
4, 8, 12, 16, 20, 24, 28, 32, …
 5 common multiples of 2 and 4 are:
4, 8, 12, 16, and 20.

(d) 3 and 9
Ans:
Multiples of 3:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, …….
Multiples of 9:
9, 18, 27, 36, 45, 54 …
5 common multiples of 3 and 9 are:
9, 18, 27, 36, and 45.

(e) 5 and 10
Ans:
Multiples of 5:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …
Multiples of 10:
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, …
5 common multiples of 5 and 10 are:
10, 20, 30, 40, and 50.

(f) 9 and 12
Ans:
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, …
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144, 156, 168, 180, 192, …
5 common multiples of 9 and 12 are: 36, 72, 108, 144, and 180.

(g) 8 and 12
Ans:
Multiples of 8:
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, …
Multiples of 12:
12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ………..
5 common multiples of 8 and 12 are:
24, 48, 72, 96, and 120.

(h) 6 and 8
Ans:
Multiples of 6:
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, …
Multiples of 8:
8,16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, …
 5 common multiples of 6 and 8 are: 24, 48, 72, 96 and 120.

(i) 6 and 9
Ans:
Multiples of 6:
6, 12, 18, 24, 30, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, …
Multiples of 9:
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, …
5 common multiples of 6 and 9 are: 18, 36, 54, 72 and 90.

Q: What do you notice about the common multiples of different pairs of numbers? Discuss in class.
Ans:
Do it yourself.

Q2: Food is available at the end of a cobbled road. Robby, the rabbit, takes a jump of 4 each time. Deeku, the deer, takes a jump of 6 each time. They both start at 0. Will both Robby and Deeku reach the food? Who will reach first? How do you know? Explain your answer.
Ans:

Robby (Rabbit): Jumps 4 steps at a time. This means Robby will land on multiples of 4 (0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64).
Deeku (Deer): Jumps 6 steps at a time. This means Deeku will land on multiples of 6 (0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60).
Robby: Yes, 64 is a multiple of 4 (4 x 16 = 64). Therefore, Robby will land exactly on 64.
Deeku: No, 64 is not a multiple of 6. (60 is 6 × 10, and 66 is 6 × 11. 64 falls in between). So, Deeku will not land exactly on 64.
Since Robby can land on 64 and Deeku cannot, Robby will be the one to reach the food.

Q:  Mowgli’s friends live along the trail on the marked places below. Which of his friends will he be able to visit, if he jumps by 2 steps starting from 0?

Ans: He will meet the ant, the frog, the bird, the bear, and the rabbit.

Q: Did Mowgli meet the ant, frog, bird and the rabbit? Notice their positions— 4, 12, 14, and 50. 2 is a common factor of these numbers.
Ans: Yes, Mowgli met the ant, the frog, the bird, and the rabbit.

Q: Which of his friends will he be able to meet if he jumps by 3 steps?
Ans: If Mowgli jumps by 3 steps, at 9 he met the spider, at 12 he met frog, at 21 he met snake, at 30 he met bear, at 39 he met deer, and at 57 he met with monkey as 3 is a common factor of the numbers 9, 12, 21, 30, 39, and 57.

Q: Which numbers will he touch if he jumps by 5 steps?
Ans: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55.
Q: 5 is a common factor of the numbers___________.
Ans: 5 is a common factor of the numbers: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, …
Q: Which numbers will he touch if he jumps by 10 steps?
Ans: 10, 20, 30, 40,50.
Q: 10 is a common factor of the numbers_______.
Ans:10 is a common factor of the numbers: 10, 20, 30, 40, 50, 60, …
Q4: Let us find some common factors of the numbers 24 and 36. Note that all jumps in the following questions start from 0.
(a) Can we jump by 2 steps at a time to reach both 24 and 36? Yes/No. 2 is/is not a common factor of 24 and 36. ,
(b) Can we jump by 3 steps at a time to reach both 24 and 36? Yes/No. 3 is/is not a common factor of 24 and 36.
(c) Can we jump by 4 steps at a time to reach both 24 and 36? Yes/No. 4 is/is not a common factor of 24 and 36.
(d) What other jumps can we take to reach both 24 and 36?
(e) How many common factors can you find for 24 and 36? List them.
(f) What about jumping by 1 step each time to reach both 24 and 36?
Ans:
(a) Yes, we can jump by 2 steps at a time to reach both 24 and 36. 2 is a common factor of 24 and 36.
(b) Yes, we can jump by 3 step at a time to reach both 24 and 36. 3 is a common factor of 24 and 36.
(c) Yes, we can jump by 4 steps at a time to reach both 24 and 36. 4 is a common factor of 24 and 36.
(d) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Common factors = 1, 2, 3, 4, 6, 12.
So we can also jump by 6 steps and 12 steps.
(e) The common factors of 24 and 36 are: 1, 2, 3, 4, 6, 12. They are six in number.
(f) Every number is a multiple of 1. So, 24 and 36 are also reached by 1-step jumps.
Q5: What are the common factors of 12 and 13?
Ans: Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 13 = 1, 13
So, the common factor of 12 and 13 is 1.
Q6:  Find which of the following numbers can be reached by jumps of 4 steps?
4 is the common factor of the numbers _____________________________.
Ans:
4 divides 16, 36 and 48 exactly.
So, numbers that can be reached by jumps of 4 steps are 16, 36 and 48.
4 is the common factor of the numbers  16, 36, and 48.
Q7:  Find the common factors of the following pairs of numbers.
(a) 12 and 16
Ans:
Factors of 12: (1), (2), 3, (4), 6, and 12.
Factors of 16: (1), (2), (4), 8, and 16.
Common factors of 12 and 16 are: 1, 2 and 4.

(b) 8 and 12
Ans:
Factors of 8: (1), (2), (4), and 8.
Factors of 12: (1), (2), 3, (4), and 12.
Common factors of 8 and 12 are: 1, 2, and 4

(c) 4 and 16
Ans:
Factors of 4: (1), (2), and (4).
Factors of 16: (1), (2), (4), 8, and 16. Common factors of 4 and 16 are: 1, 2 and 4

(d) 2 and 9
Ans:
Factors of 2: (1), and 2 Factors of 9: (1), 3, and 9
Common factor of 2 and 9 is only 1.

(e) 3 and 5
Ans:
Factors of 3: (1) and 3.
Factors of 5: (1) and 5.
Common factor of 3 and 5 is only 1.

(f ) 12 and 15
Ans:
Factors of 12: (1), (2), 3, 4, 6 and 12.
Factors of 15: (1), (3), 5 and 15.
Common factors of 12 and 15 are: 1 and 3.

(g) 20 and 5
Ans:
Factors of 20: (1) 2, 4, (5) 10 and 20.
Factors of 5: (1) and (5).
Common factors of 20 and 5 are: 1 and 5.

(h) 9 and 21
Ans:
Factors of 9: (1), (3) and 9.
Factors of 21: (1), (3), 7 and 21.
Common factors of 9 and 21 are: 1 and 3.

(i) 6 and 27
Ans:
Factors of 6: (1), 2, (3) and 6.
Factors of 27: (1), (3), 9 and 27.
Common factors of 6 and 27 are: 1 and 3.

Q: What do you notice about the common factors of different pairs of numbers? Discuss in class.
Ans: Do it yourself.

Q8:  State whether the following statements are true (T) or false (F). .
(a) Factors of even numbers must be even.
Ans: Example: Factors of 12 = 1, 2, 3, 4, 6, 12, which includes the odd numbers 1 and 3.
False.

(b) Multiples of odd numbers cannot be even.
Ans: Example: Multiples of 3 = 3, 6, 9, 12, which includes even numbers 6, 12.
False.

(c) Factors of odd numbers cannot be even.
Ans: Example: Factors of 15 = 1, 3, 5, 15 → all odd.
Example: Factors of 9 = 1, 3, 9 → all odd.
True.

(d) One of the common multiples of two consecutive numbers is their product.
Ans: Example: Consecutive numbers 4 and 5
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40….
Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60….
Common multiples: 20, 40, 60…
True.

(e) The only common factor of any two consecutive numbers is 1.
Ans: Example: Consecutive numbers 6 and 7
Factors of 6: 1, 2, 3, 6
Factors of 7: 1, 7
Common factors: 1
True.

(f) 0 cannot be a factor of any number.
Ans: True

Q9: Sher Khan, the tiger, goes hunting every 3rd day. Bagheera, the panther, goes hunting every 5th day. If both of them start on the same day, on which days will they be hunting together?
Ans: Sher Khan hunts every 3rd day → 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 …
Bagheera hunts every 5th day → 5, 10, 15, 20, 25, 30, 35, 40, 45, 50 …
The days on which they will be hunting together are the common multiples of 3 and 5.
Common multiples = 15, 30, 45 …..
So, Sher Khan and Bagheera will hunt together on every 15th day, 30th day, 45th day, and so on.

Q10:  (a) In the trail shown earlier, Sher Khan’s house is on number 25 and that of Baloo the bear is on number 30. Mowgli wants to meet his friend Baloo the bear but wants to avoid Sher Khan’s house. How long (in steps) could each jump be?
(b) What number of jumps (in steps) he could choose so that he can meet both Kaa, the snake, at 21 and Akela, the wolf, at 35?
Ans:
(a) To meet Baloo the bear and avoid Sher Khan, Mowgli must choose a jump length that is a factor of 30 but not a factor of 25.
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30.
Factors of 25: 1, 5, 25.
Common factors: 1, 5.
Removing common factors from the factors of 30, we get 2, 3, 6, 10, 15, and 30.
So, possible jump lengths could be 2, 3, 6, 10, 15, and 30.

(b) To meet Kaa the snake and Akela the wolf, Mowgli must choose a jump length that is a common factor of both 21 and 35.
Factors of 21: 1, 3, 7, 21.
Factors of 35: 1, 5, 7, 35.
Common factors: 1 and 7.
So, possible jump lengths could be 1 and 7.

Q11: Sort the following numbers into those that are
(a) divisible by 2 only
(b) divisible by 5 only
(c) divisible by 10 only
(d) divisible by 2, 5, and 10.
Ans:
(a) 22, 30, 38, 40, 56, 62, 66, 78, 84, 90.
(b) 25, 30, 40, 45, 55, 75, 90, 95.
(c) 30, 40, 90.
(d) 30, 40, 90.

Page No. 155

Raghav practices yoga in the morning.

Let us find out

Q1: At what time did Raghav start practising Yoga?
Ans:
Raghav started practicing Yoga at 05:35 a.m.

Q2: At what time did he finish?
Ans:
Raghav finished at 05:55 a.m.

Q3: How much time did he spend practising Yoga?
Ans:
He started at 05:35 a.m. and
finished at 05:55 a.m.

Since 55 – 35 = 20So, he spent 20 minutes practicing Yoga.

Q4: Find the time elapsed between the given time periods. Share your strategies.
(а) 01:15 p.m. to 01:42 p.m.?
Ans:
Time elapsed between 01:15 p.m. to 01: 42 p.m. = 27 minutes

(b) 03:18 p.m. to 08:18 p.m.?
Ans:
Time elapsed between 03:18 p.m. to 08:18 p.m. = 5 hours

(c) 09:15 a.m. to 11:30 a.m.?
Ans:
Time elapsed between 09:15 a.m. to 11:30 a.m.
= 2 hours 15 minutes.

Q5: The table below shows the time taken by 3 children to paint a picture.

(а) Who took the longest time?
(b) Who took the least time?
Ans:
(a) Since, 2 hours 10 minutes > 1 hour 35 minutes > 1 hour 20 minutes
So, Rani took the longest time.

(b) Raghav took the least time.

Page No. 156-158

Q6: Fill in the blanks by writing the time in the appropriate format.
Ans: 

Q7:
Ans: 

School Race

Akira, Sunita, and Mary are participating in a 200 m-walking race.
Q: Do you notice the use of a new unit, ‘seconds’ in the picture?
1 min = 60 seconds
In situations like a race, ‘seconds’ help us observe small differences in time taken by participants. Each participant took 1 minute, but how much more?
Identify the child who won the race. How much time did the child take?
Ans: The child who won the race is encircled.
She took 1 minute 55 seconds.

Let us Do

Estimate whether you would take seconds or minutes to complete the following activities. Tick the appropriate cell.
Ans:

List events other than the ones listed above that you can do or that can happen in less than a minute or in a few seconds.
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
_______________________________________
Ans:
A lightning strike.
A balloon bursting.
Sending a text message.
Tearing a paper.
Opening or closing a door.
Standing up from a chair.
Giving a handshake or high-five.
Receiving a phone notification.
Unlocking a phone with a fingerprint or face scan.
Sneezing or coughing.

Estimate and verify how many of these you can do in a minute.
Ans:
Do it yourself.

Let Us Find

Q1: Find out the number of times you can skip the rope in 10 seconds.
Ans: 7 – 8 times

Q2: How long does it take you to write the word FRIEND?
Ans: 2 seconds

Q3: How long does it take you to run 100m?
Ans: 20 – 25 seconds

Page No. 159 -160

Let Us Do

Observe the clocks and fill in the blanks.

Q1: Rani took __________ sec. to get out of her bed.
Ans:
Rani took 30 sec. to get out of her bed.

Q2: Raghav took ________ sec. to move from his room to the kitchen.
Ans:
Raghav took 35 sec. to move from his room to the kitchen.

Q3: Ritu took ________ sec. to pick up a piece of paper from the floor.
Ans:
Ritu took 15 sec. to pick up a piece of paper from the floor.

Q4: Raghu took ________ sec. to wash his spoon and plate.
Ans:
Raghu took 60 sec. to wash his spoon and plate.

Q: Draw the missing seconds hand on the clocks on the right side.
Ans: 

Page No. 161 -163

Let Us Do

Q1: Use the double number line below to complete the conversions.
Ans: 

(b) Do as instructed. You can use the above double number line for the following  conversions.
Ans:
(i) 1 hr 10 min = 60 min + 10 min = 70 min
So, 1 hour 10 minutes = 70 minutes.

(ii) 2 hr 45 min = 120 min + 45 min = 165 min
So, 2 hours 45 minutes = 165 minutes,

(iii) 3 hr 15 min = 3 × 60 + 15 min = 180 min + 15 min = 195 min
So, 3 hours 15 minutes = 195 minutes.

(iv) 4 hr 20 min = 4 times 60 + 20 min
= 240 min + 20 min = 260 min
So, 4 hours 20 minutes = 260 minutes.

(v) 75 min = 60 min + 15 min = 1 hr 15 min
So, 75 minutes = 1 hour 15 minutes.

(vi) 150 min = 120 min + 30 min = 2 hr 30 min
So, 150 minutes = 2 hours 30 minutes.

(vii) 220 min = 120 min + 60 min + 40 min = 2 hr + 1 hr + 40 min = 3 hr 40 min
So, 220 minutes = 3 hours 40 minutes.

(viii) 390 min = 240 min + 120 min + 30 min = 4 hr + 2 hr + 30 min = 6 hr + 30 min
So, 390 minutes = 6 hours 30 minutes

(c) Fill in the blanks in the double number line to complete the conversions.

Ans: 

Ans: (i) 320 sec = 300 sec + 20 sec = 5 min + 20 sec

So, 320 sec = 5 min 20 sec

(ii) 225 sec = 120 sec + 60 sec + 45 sec
= 2 min + 1 min + 45 sec
So, 225 sec = 3 min 45 sec

(iii) 700 sec = 600 sec + 60 sec + 40 sec
= 10 min + 1 min + 40 sec
So, 700 sec = 11 min 40 sec

(iv) 1000 sec = 900 sec + 60 sec + 40 sec
= 15 min + 1 min + 40 sec
So, 1000 sec = 16 min 40 sec

(v) 10 min 13 sec = 600 sec + 13 sec
= 613 sec
So, 10 min 13 sec = 613 sec

(vi) 4 min 8 sec = 2 min + 2 min + 8 sec
= 120 sec + 120 sec + 8 sec = 248 sec
So, 4 min 8 sec = 248 sec

(vii) 15 min 40 sec = 900 sec + 40 sec
= 940 sec
So, 15 min 40 sec = 940 sec

Q2: Raghav studies Mathematics, English, Hindi, and the World Around Us subjects. He took 50 minutes to study each of the subjects. Find the total time taken in hours and minutes. Share your strategies with your class.
Ans: Time taken to study each subject = 50 minutes.
Total time taken to study all subjects = 4 × 50 minutes
= 200 minutes
= 180 minutes + 20 minutes
= 3 hours + 20 minutes
= 3 hours 20 minutes.
Therefore, the total time taken is 3 hours and 20 minutes.

Q3: Raghu left his house at 08:00 hours and arrived at his Nana ji’s house at 09:05 hours. How long did he take to reach his house?
Ans: Raghu left his house at 08:00 hours.
Raghu arrived at his Nanaji’s house at 09:05 hours.
From 08:00 to 09:00 → 1 hour
From 09:00 to 09:05 → 5 minutes
So, Raghu took 1 hour and 5 minutes to reach his Nana ji’s house.

Q4: Jyoti went to play at 06:15 PM. She came back after 1 hour 45 minutes. At what time did she reach home?
Ans: Jyoti left at 06:15 PM.
She stayed out for 1 hour and 45 minutes.

Add 1 hour → 07:15 PM
Add 45 minutes → 08:00 PM

Jyoti reached home at 08:00 PM.

Q5: Ragini took 1 hour 10 minutes to do her homework. She finished it at 09:40 PM. What time did she start?
Ans: Ragini finished at 09:40 PM.
Time taken = 1 hour 10 minutes.

09:40 – 1 hour = 08:40 PM
08:40 – 10 minutes = 08:30 PM

Ragini started her homework at 08:30 PM. 

Q6: A group of children left for a picnic at 08:30 AM. They returned after 4 hours and 10 minutes. At what time did they return?
Ans: They left at 08:30 AM.
Time spent = 4 hours 10 minutes.

Add 4 hours → 12:30 PM
Add 10 minutes → 12:40 PM

The children returned at 12:40 PM.

Q7: Raji started her homework at 06:00 PM. She finished her homework in 1 hour 30 minutes. At what time did she finish?
Ans: Raji started at 06:00 PM.
Time taken = 1 hour 30 minutes.

Add 1 hour → 07:00 PM
Add 30 minutes → 07:30 PM

Raji finished her homework at 07:30 PM.

Q8: Alya goes out to play at 05:30 PM and comes back after 1 hour 10 minutes. At what time does she come back?
Ans: Alya left at 05:30 PM.
Time spent = 1 hour 10 minutes.

Add 1 hour → 06:30 PM
Add 10 minutes → 06:40 PM

Alya came back at 06:40 PM. 

Q9: If the lunch break of a school starts at 12:30 PM and ends in 35 minutes, what time will lunch end?
Ans: Lunch starts at 12:30 PM.
Duration = 35 minutes.

Add 30 minutes → 01:00 PM
Add 5 minutes → 01:05 PM

Therefore, lunch will end at 01:05 PM.

Q10: It is 08:35 PM right now. What time will it be after 8 hours and 25 minutes?
Ans: Current time = 08:35 PM

08:35 PM + 8 hours = 04:35 AM (next day)
04:35 AM + 25 minutes = 05:00 AM

The time after 8 hours and 25 minutes will be 05:00 AM (next day).

Page No. 142

Let Us Do

Q1: Find the perimeter of the following shapes. All sides of the following shapes are equal.

Ans: Perimeter of a shape
= Sum of lengths of all sides 
= (4 + 4 + 4 + 4 + 4) cm
= 20 cm

Perimeter = sum of all six equal sides.
Each side = 5 cm, number of sides = 6.
Perimeter = 5 + 5 + 5 + 5 + 5 + 5 cm
= 30 cm

Q2: Draw two rectangles each having the following perimeters.
(a) 26 cm
Ans:

(b) 18 cm
Ans:

Page No. 143

Preetha and Adrit’s grandmother is making a rug with square patches. The picture below shows the rug. 
How many patches have they used to make this?

Ans:
They have used 90 square patches to make the rug.
This number can be found by counting the number of patches in one row and multiplying by the number of rows, or by counting all patches directly.

Q: Preetha and Adrit are trying to cover their table with different shapes. Preetha covered it with triangles and circles. Adrit covered it with squares and rectangles.

They found that __________, __________ and __________ shapes cover the top of the table without gaps and overlaps. __________ shape leaves gaps.
__________ triangles cover Table 1.
__________ squares cover Table 3.
__________ rectangles cover Table 4
Ans:

They found that triangle, square and rectangle shapes cover the top of the table without gaps and overlaps. Circle shape leaves gaps.
20 triangles cover Table 1.
8 squares cover Table 3.
12 rectangles cover Table 4.

Q: To find the area of a region, we usually fill it with shapes that tile (no gaps or overlaps), like squares, rectangles and triangles.
Do circles tile? Can we use them to cover a region?
The area of Table 1 is __________ triangle units.
The area of Table 3 is __________ square units.
The area of Table 4 is __________ rectangle units
Ans:
No, circles do not tile. We cannot use them to cover a region because circles leave gaps between them when placed side by side.
The area of Table 1 is 20 triangle units.
The area of Table 3 is 8 square units.
The area of Table 4 is 12 rectangle units.

Q: Now, try to cover the top of your table without gaps and overlaps with the following objects of same size.
(a) Notebooks
(b) Lunch boxes
(c) Pencil boxes
(d) Maths textbooks
Which of the above objects covered the region completely?
Ans:
Do it yourself.

Page No. 144

Let Us Do

Preetha is playing with tiles. She covers her desk with different shapes as shown below.

Look at the different tiles on her desk and answer how many of the following shapes will cover the desk.
(a) Green triangles _______
(b) Red triangles _______
(c) Blue squares _______
Ans:

In the given picture, the desk is made up of 54 unit squares.

(a) Two green triangles make one full square. Therefore,
Total number of green triangles = 54 × 2 = 108 green triangles.

(b) If one red triangle equals 9/2 unit squares (that is 4.5 square units), then
Total area to cover = 54 unit squares.
Number of red triangles required = 54 ÷ 4.5 = 12 red triangles.

Therefore, 12 red triangles will cover the desk.

(c) One blue square = 1 unit square.
Therefore, the number of blue squares required = 54 blue squares.

Page No. 145-148

Let Us Do

Q1: Compare the areas of the two gardens given below on the square grid. Share your observations.

Area of Garden A = _____ cm square
Area of Garden B = _____ cm square
Ans:
Garden A is covered by 10 unit squares, each of side 1 cm.
Therefore, the area of Garden A = 10 square cm.
Garden B is covered by 12 unit squares, each of side 1 cm.
Therefore, the area of Garden B = 12 square cm.
Observation: Garden B has a larger area than Garden A by 2 square cm.

Q2: Trace your palm on the square grid given below and find the approximate area of your palm. Compare the area of your palm with your friend’s palm. Who has a bigger palm?

Ans:

Do it yourself.

Q3: Collect leaves of different kinds. Put them on a square grid and find their area.
(a) Name the leaf with the largest area.
(b) Name the leaf with the smallest area.
Ans:
Do it yourself.

Q4: The following mats are made of square patches of equal size. How many square patches will be required to cover each mat? Would both mats require an equal or different number of patches? Trace and cut out a small square of the size given below and find the area.

Ans:

The number of square patches required to cover the mat (a) = 10 square patches.
The number of square patches required to cover the mat (b) = 12 square patches.
Therefore, both mats require a different number of patches.

Q: Trisha makes these two rectangles. She says, “I increased the area of my rectangle, and the perimeter increased. Do you think this is always true?

Ans: 

Take a rectangle with length = 12 cm and breadth = 2 cm.
Area = 12 × 2 = 24 square cm.
Perimeter = 2(12 + 2) = 2 × 14 = 28 cm.
Now change the rectangle to length = 7 cm and breadth = 4 cm.
Area = 7 × 4 = 28 square cm.
Perimeter = 2(7 + 4) = 2 × 11 = 22 cm.
Here, area increased from 24 square cm to 28 square cm, but the perimeter decreased from 28 cm to 22 cm.
Therefore, Trisha’s claim is not always true. Increasing the area does not always increase the perimeter; both measures depend on the shape’s dimensions.

Let Us Explore

Q1: Tick the shapes with the same area. Find the perimeters of these shapes. What do you notice? Discuss.

Ans: 

(a) Area = 12 square cm.
Perimeter = 6 cm + 2 cm + 6 cm + 2 cm = 16 cm.
(b) Area = 12 square cm.
Perimeter = 2 cm + 1 cm + 1 cm + 5 cm + 2 cm + 5 cm + 1 cm + 1 cm = 18 cm.
(c) Area = 12 square cm.
Perimeter = 3 cm + 4 cm + 3 cm + 4 cm = 14 cm.
(d) Area = 12 square cm.
Perimeter = 4 cm + 3 cm + 4 cm + 3 cm = 14 cm.
(e) Area = 8 square cm.
Perimeter = 8 cm + 1 cm + 8 cm + 1 cm = 18 cm.
(f) Area = 12 square cm.
Perimeter = 12 cm + 1 cm + 12 cm + 1 cm = 26 cm.
Therefore, shapes (a), (b), (c), (d) and (f) each have area 12 square cm, while (e) has area 8 square cm.
We notice that shapes with the same area can have different perimeters.

Q2: Tick the shapes with the same perimeter. Find the areas of these shapes. What do you notice? Discuss.

Ans: (a) Perimeter = 3 cm + 3 cm + 1 cm + 1 cm + 2 cm + 2 cm = 12 cm.
Area = 7 square cm.

(b) Perimeter = 3 cm + 1 cm + 1 cm + 2 cm + 2 cm + 3 cm = 12 cm.
Area = 7 square cm.

(c) Perimeter = 1 cm + 2 cm + 2 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 1 cm = 16 cm.
Area = 7 square cm.

(d) Perimeter = 1 cm + 4 cm + 2 cm + 1 cm + 3 cm + 5 cm = 16 cm.
Area = 7 square cm.

(e) Perimeter = 7 cm + 1 cm + 7 cm + 1 cm = 16 cm.
Area = 7 square cm.
Thus, these shapes show that different shapes can have the same area even if their perimeters are different. All these shapes have area 7 square cm.

Page No. 149-150

Let Us Do

Q1: Draw different shapes having the same area as the given shape. Write the perimeter of each shape. What do you notice? Discuss.

Ans: 

(a) Area = 18 square cm.
Perimeter = 9 cm + 2 cm + 9 cm + 2 cm = 22 cm.

(b) Area = 18 square cm.
Perimeter = 6 cm + 3 cm + 6 cm + 3 cm = 18 cm.

(c) Area = 18 square cm.
Perimeter = 3 cm + 1 cm + 2 cm + 3 cm + 5 cm + 4 cm = 18 cm.

(d) Area = 18 square cm.
Perimeter = 2 cm + 5 cm + 1 cm + 2 cm + 4 cm + 2 cm + 1 cm + 5 cm = 22 cm.

(e) Area = 18 square cm.
Perimeter = 1 cm + 1 cm + 5 cm + 1 cm + 1 cm + 4 cm + 1 cm + 1 cm + 5 cm + 1 cm + 1 cm + 4 cm = 26 cm.
Thus, shapes having the same area can have different perimeters.

Q2:  Is the area of shape (a) less than the area of shape (b) given below? Discuss.

Preetha and Adrit’s grandmother is making another square patchwork. She arranges the patches as shown below. Can you guess how many patches she will need? How did you find it?

Ans:
Area of shape (a) = 9 unit squares.
Area of shape (b) = 6 full unit squares + 6 half unit squares = 6 + (6 × 1/2) = 6 + 3 = 9 unit squares.
So, area of shape (a) = area of shape (b). Both have equal area.

Page No. 151-153

Let Us Do

Q1: Find the area of your classroom floor in square meters. Take the help of your teacher to measure the length and breadth of the floor. What is the perimeter of the classroom floor?
Ans: Do it yourself.

Q2: Find the area and perimeter of the following shapes.

Ans:

(a) Side of the square = 6 cm.
Area = Length × Length
= 6 × 6
= 36 square cm.
Perimeter = 4 × Length
= 4 × 6
= 24 cm.

(b) Length = 4 cm, Breadth = 7 cm.
Area of a rectangle = Length × Breadth
= 4 × 7
= 28 square cm.
Perimeter of a rectangle = 2(Length + Breadth)
= 2 × (4 + 7)
= 2 × 11
= 22 cm.

(c) Length = 12 cm, Breadth = 4 cm.
Area of a rectangle = Length × Breadth
= 12 × 4
= 48 square cm.
Perimeter of a rectangle = 2(Length + Breadth)
= 2 × (12 + 4)
= 2 × 16
= 32 cm.

(d) Side of the square = 3 cm.
Area of a square = Side × Side
= 3 × 3
= 9 square cm.
Perimeter of a square = 4 × Side
= 4 × 3
= 12 cm.

(e) Length = 6 cm, Breadth = 5 cm.
Area of a rectangle = Length × Breadth
= 6 × 5
= 30 square cm.
Perimeter of a rectangle = 2(Length + Breadth)
= 2 × (6 + 5)
= 2 × 11
= 22 cm.

Q3: Find the area and perimeter of the following objects. Use a scale or measuring tape to find the length and the breadth of each of the objects.

Ans: Do it yourself.

Q4: Find the area of a rectangular field whose length is 42 m and breadth is 34 m.
Ans:
Length = 42 m, Breadth = 34 m.
Area of a rectangle = Length × Breadth
= 42 × 34
= 1,428 square m.
Therefore, the area of the rectangular field is 1,428 square m.

Q5: The area of a rectangular garden is 64 square m and its length is 16 m. What is its breadth?
Ans:
Area = 64 square m, Length = 16 m.
Area = Length × Breadth
64 = 16 × Breadth
Breadth = 64 ÷ 16 = 4 m.
Therefore, the breadth of the rectangular garden is 4 m.

Q6: Find the area of the following figure with the dimensions as marked in the figure.

Ans:
Length = 32 cm.
Breadth = (6 + 12) cm = 18 cm.
Area of the rectangular figure = Length × Breadth
= 32 × 18
= 576 square cm.

Page No. 136

Alphabet Cutout

Q1: Which of the following alphabet cutouts can be made by just drawing half (1/2) or quarter (1/4) of the letter? You can do it by drawing lines of symmetry on the letters.

Which of the letters have a horizontal line of symmetry? _________________ 
Which of the letters have a vertical line of symmetry? ____________________ 
Which letters have both vertical and horizontal lines of symmetry?________

Ans: (i) Alphabets made by drawing 1/2 of the letter: E, T, and V.

Explanation: a letter that has a single line of symmetry can be made by drawing one half and reflecting it. For example, E here is shown with a horizontal line of symmetry, while T and V have vertical lines of symmetry. 

(ii) Alphabets made by drawing 1/4 of the letter: O and X.

(iii) Letters with horizontal line of symmetry:

Explanation: For (ii), letters like O and X have both vertical and horizontal lines of symmetry, so one quarter of these letters is enough to make the whole letter by repeating the quarter piece.

(iv) Letters with vertical line of symmetry:

(v) Letters with both vertical and horizontal lines of symmetry:

Let Us Do

Q1: Use lines of symmetry to make paper cutouts of diya, boat, and other designs. Look along the border of the page to find the pictures.
Ans: Do it yourself. Hint: Fold the paper along the line of symmetry and cut out the shape on one side only. When you unfold the paper you will get a complete symmetrical design.

Page No. 137-138

Let Us Make a Windmill Firki

Q1: Observe the dot in the firki. Does the firki look the same after 1/4,1/2,3/4 and a full turn?
Ans: Yes, the firki looks the same after 1/4, 1/2, 3/4, and a full turn.

Explanation: The firki has four identical blades placed around the centre. Because the blades are identical, turning the firki by 1/4 turn (90°) moves one blade to the place of the next; the pattern lines up exactly after 1/4, 1/2, 3/4 and full turns. 

Q2: Observe the letters below. Do they look the same when turned? Dots have been marked on the letters to keep track of the orientation of letters. You may also cut out the letters and fix the centre point of the letter by a nail or use tracing paper to check if the letter looks the same when turned.

Ans:

Explanation: For the letters, fix the centre point and rotate to check which letters match their original position; the pictures above show the results.

Let Us Do

Q1: Find symmetry in the digits.

Which digit(s) have reflection symmetry? ___________________________ 
Which digit(s) have rotational symmetry? ___________________________ 
Which digit(s) have both rotational and reflection symmetries? ________

 Now, let us look at the following numbers: || , |00| Do these have (a) rotational symmetry, (b) reflection symmetry or (c) both symmetries? Give examples of 2-, 3-, and 4-digit numbers which have rotational symmetry, reflection symmetry, or both.Ans: (i) Digits with reflection symmetry = 1, 8, 0.

(ii) Digits with rotational symmetry = 1, 8, 0.

(iii) Digits with both rotational and reflection symmetry = 1, 8 and 0.

Q5: Now, let us look at the following numbers: ||, |00|
Do these have
(a) rotational symmetry,
(b) reflection symmetry or
(c) both symmetries?
Ans:
||, |00| have both rotational and reflection symmetry.

Q6: Give examples of 2-, 3-, and 4-digit numbers which have rotational symmetry, reflection symmetry, or both.
Ans: 

(i) Numbers with Reflection Symmetry:
• 2-digit: 11, 88
• 3-digit: 181, 303, 808
• 4-digit: 1001, 1331, 8008

(ii) Numbers with Rotational Symmetry:
• 2-digit: 80, 88, 00
• 3-digit: 808, 880, 800
• 4-digit: 8008, 8888, 0880

(iii) Numbers with Both Reflection & Rotational Symmetry:
• 2-digit: 88, 00
• 3-digit: 808, 888
• 4-digit: 8008, 8888, 0880

Explanation: Reflection symmetry (also called mirror symmetry) means a vertical or horizontal line can divide the digit so that one side is the mirror image of the other. Rotational symmetry means the digit looks the same after turning by 180° (half turn) or some other fraction. The examples above are numbers that remain the same when reflected or rotated using common digit shapes; you can check each example by folding a printed number along the line of symmetry or by rotating a cutout.

Page No. 139-140

Making Designs

Q1: (a) Does the design have rotational symmetry?
Yes/No.

Ans: No, the design doesn’t have rotational symmetry. 

Explanation: When the figure is turned it does not match its original position at fixed turns, so it is not rotationally symmetric.

(b) Try to change the design by adding some shape(s) so that the new design looks the same after a 1/2 turn. Draw the new design 2 in your notebook.
Ans: 

Explanation: Add symmetrical elements opposite each other so that after a half turn (180°) each part moves to the position of an identical part.

(c) Now try to modify or add more shapes so that the new design looks the same after turn.
Draw the new design in your notebook.
Ans: 

Explanation: To get higher order rotational symmetry, repeat the same motif around the centre at equal angles so the design matches after the desired turn.

(d) Do the new designs have reflection symmetry? If yes, draw the lines of symmetry.
Ans: Yes, the new designs have reflection symmetry.

Explanation: The added shapes are placed so that a vertical or horizontal line divides the design into two mirror-image halves; the picture above shows the lines of symmetry.

Let Us Think

Q1: Does this design look the same after 1/2 turn?
Ans: Yes, the design looks the same after 1/2 turn. 

Explanation: The design has two identical halves opposite each other; a half turn interchanges these halves so the design matches its original position.

Q2: Does this design look the same after a 1/4 turn?
Ans: No, the design does not look the same after a 1/4 turn. 

Explanation: The design does not repeat every 90°, so a quarter turn changes the positions of parts and they do not match the original.

Colour the square given in the adjoining figure using two colours so that the design looks the same after every 1/4 turn.
How many times does this shape look the same during a full turn?

Q3: Do these designs have reflection symmetry also? Draw the line(s) of symmetry.

Ans:

The image looks the same after every 1/4 turn.
During a full turn, it looks the same four times. The shaded design does not have reflection symmetry.

Explanation: A design that repeats every 1/4 turn will match its original position four times during a full 360° rotation. If no mirror line divides the shaded design into identical halves, it does not have reflection symmetry.

Let Us Do

Q1: Cut out squares and equilateral triangles with the same side length. These are provided at the end of the book.
Make different symmetrical designs by using these two shapes.

Ans: (Sample)

Hint: Arrange the shapes around a centre to make rotational patterns, or place them as mirror images to make reflection-symmetric designs.

Q2: Does this shape have reflection symmetry? If yes, draw its line(s) of symmetry.

Ans: Yes, the shape has reflection symmetry. Its lines of symmetry are drawn as shown in the picture below.

Explanation: A line drawn through the centre divides the shape into two mirror-image halves; the picture shows one or more such lines.

Q3: Does it have rotational symmetry? If yes, at which turn?
Ans: The shape has rotational symmetry of 1/2 turn also. Explanation: A half turn (180°) maps the shape onto itself because two opposite parts swap places and coincide.

Q4: Does it have both symmetries?
Ans: Yes, it has both reflection and rotational symmetry.

Let Us Explore

Q1: Below are images of wooden blocks and a part of their prints. Match each block to its correct print by drawing a line. One is done for you.

Ans:

Page No. 141

Q. Observe the pattern made by the wooden block below. We get the final print by using the block 4 times.

Ans: 

Design B looks the same after every 1/4 turn.
The design has rotational symmetry.

Explanation: The block is repeated four times around a centre so the print repeats every 90° turn.

Let us Do

Observe the shapes given on the border. Which of the shapes have reflection symmetry? Put a (✔) mark on them. Put a * on the shapes that have rotational symmetry.

Ans: 

Explanation: Mark shapes that can be folded along a line to give matching halves with ✔. Put * on shapes that match themselves after a rotation by the specified angle.

Project Work

Q1: Create symmetrical patterns and designs using vegetable blocks. Some are shown below.

Ans: Do it yourself. Suggestion: Cut vegetables into simple shapes, stamp them with ink or paint, and arrange the prints around a centre or along a mirror line to make symmetrical designs. Photograph or paste your patterns into your project notebook

Page No. 119

Q: Observe the following array of coconuts. Write two division facts using the given multiplication fact.

Ans: 35 ÷ 1 = 35

Q: Write the appropriate multiplication fact for the array shown below. Write two division facts that follow from the multiplication fact.

Ans:

Page No. 120-121

Let us Play

Identify the numbers that can fill the circles such that the numbers in the squares are the products or the quotients of the numbers in the circles.

Ans: 

(Answer may vary)

Let Us Do

Q1: Solve the following multiplication problems. Write two division statements in each case.

Ans:

Q2: Solve the following division problems. Notice the patterns and discuss in class.

Ans: 

Patterns in Division and Place Value

Ans: 

Q: Now fill the place value chart.

Ans: 

We observed that when dividend is increasing 10 times, the quotient is also increasing 10 times.

Page No. 122

Let Us Do

Q1: Sabina cycles 160 km in 20 days and same distance each day. How many kilometres does she cycle each day?
Ans:
Total distance = 160 km.
Total days = 20.
Distance per day = 160 ÷ 20 = 8 km.
Sabina cycles 8 km each day.

Q2: How many notes of ₹ 100 does Seema need to carry if she wants to buy coconuts worth ₹ 4200?
Ans:

Total cost of coconuts = ₹4200
Denomination value = ₹100 notes
Number of notes needed = 4200 ÷ 100 = 42.
Seema needs 42 notes of ₹100.

Q3: The owner of an electric store has decided to distribute? 5500 equally amongst 5 of his employees as a Diwali gift. What amount will each employee get?
What will happen if he distributes the same amount of money among 10 employees? Will each employee get more or less? How much money would he have to distribute if everyone must get the same amount as earlier?
Ans: Total money = ₹5500.
Number of employees = 5.
Amount per employee = 5500 ÷ 5 = 1100.
Each employee will get ₹1100.

(i) Distribution among 10 employees:
Total money = ₹5500
Number of employees = 10
Amount per employee = 5500 ÷ 10 = ₹550.
So, each employee would now get ₹550, which is less than ₹1100.

(ii) If each of the 10 employees must get ₹1100, total money required = ₹1100 × 10 = ₹11,000.
So, he would need ₹11,000.

Q4: Place the numbers 1 to 8 in the following boxes so that all the four operations, division, multiplication, addition and subtraction are correct. No number must be repeated.

Ans: 

Q5: Fill in the blanks
(a) _____ ÷ 18 = 100.
(b) _____ ÷ 10 = 610.
(c) _____ ÷ 100 = 72.
(d) _____ ÷ 100 = 10.
(e) 870 ÷ _____ = 87.
(f) _____ ÷ 100 = 70.
(g) 200 ÷ _____ = 2.
(h) 130 ÷ _____ = 13.
Ans:
(a) 1800 ÷ 18 = 100.
(b) 6100 ÷ 10 = 610.
(c) 7200 ÷ 100 = 72.
(d) 1000 ÷ 100 = 10.
(e) 870 ÷ 10 = 87.
(f) 7000 ÷ 100 = 70.
(g) 200 ÷ 100 = 2.
(h) 130 ÷ 10 = 13.

Page No. 123

Try It!

Ans: 

Page No. 124

Let Us Solve

Solve the following problems using the strategies used in the previous question.
(a) 256 ÷ 4
Ans:

(b) 545 ÷ 5
Ans:

(c) 147 ÷ 7
Ans:

(d) 1212 ÷ 6
Ans:

(e) 648 ÷ 12
Ans:

(f) 9648 ÷ 48
Ans:

(g) 775 ÷ 25
Ans:

(h) 796 ÷ 4
Ans:

Page No. 125

Let Us Learn to Divide

Q: 726 ÷ 4
Ans:

No, we cannot write 200 instead of 100 as 200 × 4 = 800 > 726

Q: 902 ÷ 16
Ans:

No, we cannot use a tens digit larger than 5 (60 × 16 = 960 > 902).

Q: Is 726 = 4 × 181? Yes/No.
So, 726 = 4 × 181 + _______.
Ans:
4 × 181 = 724.
So, the correct answer is No.

726 = 4 × 181 + 2

Q: Is 902 = 16 × 56? Yes/No.
So, 902 = 16 × 56 + _____.
Ans:
16 × 56 = 896.
So, the correct answer is No.
902 = 16 × 56 + 6

Page No. 126-128

Let Us Solve

Solve the following word problems

Q1: Rani is planning to host a party. She estimates that 250 guests will attend. She plans to serve one samosa to each guest. Samosas are available in packs of 6 or 8. Which pack should Rani buy? Explain your answer.
Ans: Number of guests = Number of samosa = 250

So, 41 packs of 6 samosas and 4 are left in packet. 31 packs of 8 same left in the packet. Rani Should buy packs of 6 samosas.

Q2: 342 students from a school are going on a trip to the Science Park. Each bus can carry a maximum of 41 students. How many buses does the school need to arrange?
Ans: Total students = 342
No. of students in 1 bus = 41
No. of buses required = 342 ÷ 41

Division shows 41 × 8 = 328, remainder 14. Since 14 students still need seats, one more bus is needed.
Therefore, total buses required = 8 + 1 = 9 buses.

Answer: 9 buses.

Q3: Sofia has only ₹ 50 and ₹ 20 notes. She needs to pay ₹ 520 using these notes. How many ₹ 50 and ₹ 20 notes does she need to make ₹ 520? Find out the different possible combinations.
Ans:
Let us check some possibilities by using division.

So, 10 × ₹ 50 + 1 × ₹ 20 = ₹ 520

So, 8 × ₹ 50 + 6 × ₹ 20 = ₹ 520

So, 6 × ₹ 50 + 11 × ₹ 20 = ₹ 520
So, Sofia can pay using ten ₹ 50 notes and one ₹ 20 note, eight ₹ 50 notes and six ₹ 20 notes, or six ₹ 50 notes and eleven ₹ 20 notes.
There can be other possible combinations also.

Q4: Three friends decide to split the money spent on their picnic equally. They buy snacks and sweets for ₹ 157, juice and fruits for ₹ 124 and pulav and paratha for ₹ 136. How much should each person pay to share the cost equally?
Ans: Money spent by three friends-
Snacks and sweets = ₹ 157
Juice and fruits = ₹ 124
Pulav and paratha = ₹ 136
Total = ₹ 157 + ₹ 124 + ₹ 136 = ₹ 417
Now each friend spends = ₹ 417 ÷ 3

Therefore, each person needs to pay ₹ 139.

Q5: Identify the remainder, if any. Check if N = D × Q + R.
(a) 887 ÷ 3
Ans:

Here, N = 887, D = 3, Q = 295, R = 2
Since, 3 × 295 + 2 = 885 + 2 = 887
Therefore, N = D × Q + R

(b) 283 ÷ 8
Ans:

Here, N = 283, D = 8, Q = 35, R = 3
Since, 8 × 35 + 3 = 280 + 3 = 283
Therefore, N = D × Q + R

(c) 745 ÷ 5
Ans:

Here, N = 745, D = 5, Q = 149, R = 0
Since, 5 × 149 + 0 = 745 + 0 = 745
Therefore, N = D × Q + R

(d) 767 ÷ 26
Ans:

Here, N = 767, D = 26, Q = 29, R = 13
Since, 26 × 29 + 13 = 754 + 13 = 767
Therefore, N = D × Q + R

(e) 530 ÷ 41
Ans:

Here, N = 530, D = 41, Q = 12, R = 38
Since, 41 × 12 + 38 = 492 + 38 = 530
Therefore, N = D × Q + R

(f) 888 ÷ 67
Ans:

Here, N = 888, D = 67, Q = 13, R = 17
Since, 67 × 13 + 17 = 871 + 17 = 888
Therefore, N = D × Q + R

Kalpavruksha Coconut Oil

Q1: In a particular year, Susie and Sunitha used 4376 coconuts for extracting coconut oil. They can extract 1 L of oil from 8 coconuts. What quantity of oil were they able to extract?

Ans:  They would get 4376 ÷  8 litres of coconut oil.

4376 ÷ 8 = 547. They extracted 547 L of oil in the year.

Both methods are valid and can be effective depending on the students’ understanding and preferences.

How much will they earn if they sell the oil at ₹175 for 1 l? 

Ans: They will earn ₹ 547 × 175.

547 × 175 = 95,725. Therefore, they will earn ₹ 95,725.

Q2: Coconut husk is used for making coir. Coir is a natural fibre used in gardening, farming, boat-making, and making decorative items. Susie and Sunitha’s farm sells coconut husk at ₹23 per kilogram. They earned ₹  9913 from the sale of husk in May. What quantity of husk did they sell in May?

Ans: The quantity of husk sold in May is 9913 ÷ 23 kg.

9913 ÷ 23 = 431. Susie and Sunitha’s farm sold 431 kg of coconut husk in May.

Q3: In the hot summer months, tender coconuts are sold for ₹ 35. Ibrahim earns ₹ 8890 in a week. How many tender coconuts did he sell? The number of tender coconuts sold by Ibrahim is 8890 ÷ 35.

Q: Ibrahim sold ______ tender coconuts.
Ans:

Ibrahim sold 254 tender coconuts.

Q: Ibrahim had bought the tender coconuts for ₹ 20 each. How much extra money did he earn by selling the coconuts at ₹ 35?
Ans:
The cost of 254 coconuts at ₹ 20 each = 254 × ₹ 20 = ₹ 5,080.
He earned ₹ 8,890 from the sale.
The extra amount he earned is ₹ 8,890 – ₹ 5,080 = ₹ 3,810.

Page No. 130-131

Let Us Divide

(a) 7,032 ÷ 6

Ans:

(b) 3,005 ÷ 5

Ans:

3005 = 3 Thousands + 0 hundreds + 0 tens + 5 ones.
3 Thousands ÷ 5 → not possible without regrouping
Regroup 3 thousand into hundreds.
30 hundreds
30 hundreds ÷ 5 = 6 hundreds
0 tens ÷ 5 = 0 tens
5 ones ÷ 5 = 1 ones

(c) 2,874 ÷ 14

Ans:

(d) 9,805 ÷ 32

Ans:

Let Us Do

Q1: Find the missing numbers such that there is no remainder. Remember, there could be more than one solution.

Ans: Do it Yourself!

I am a 3-digit number.

• If you divide me by 5, you get 42.
• If you multiply me by 2, you get 420. 

What number am I?
Ans: 3-digit number
42 × 5 = 210
210 × 2 = 420
I am 210.

Page No. 132-133

Let Us Solve

Q1: A theatre company can accommodate 45 people during one show.
(a) A total of 475 people bought tickets for a puppet show. How many shows are needed to seat all the people who bought tickets?
(b) There are 2 shows in a day. How many days will be needed to accommodate all the people?
Ans:
(a) The number of shows required for 475 people

= 475 ÷ 45

Ten shows will accommodate 450 people and one more show will accommodate for the remaining 25 people.
Therefore, 11 shows are needed to seat all the people who bought tickets.

(b) From part (a) number of shows required is 11.
If there are 2 shows in a day, then in 5 days there will be 10 shows, and on 6th day the 11th show will be held.
Thus, 6 days are needed to accommodate all the people.

Q2: Naina bought 5 kg of ice cream as a birthday treat for her 23 friends. 400 g ice cream was left after everyone had an equal share. How much ice cream did each of her friends eat?
Ans: Total ice cream shared = 5000 g – 400 g = 4600 g
Share of each of her friend = 4600 ÷ 23 = 200
Therefore, each of her friends ate 200 g of ice cream.

Q3: Megha packs 15 packets of ragi-oats biscuits for a 4-day group trip. Each packet contains 8 biscuits. There are 6 people in the group. If distributed evenly, how many biscuits can one person have each day.
Ans: Total number of biscuits in 15 packets = 15 × 8 = 120
Total number of biscuits distributed per day = 120 ÷ 4 = 30
Number of biscuits one person can have per day = 30 ÷ 6 = 5
Therefore, each person can have 5 biscuits each day.

Q4: Solve the following and identify the remainder, if any. Check whether N = D × Q + R in each case.
(a) 9,045 ÷ 5
Ans:

Here, N = 9045, D = 5, Q = 1809, R = 0
Since, 9045 = 5 × 1809 + 0
Therefore, N = D × Q + R

(b) 1,034 ÷ 4
Ans:

Here, N = 1034, D = 4, Q = 258, R = 2
Since, 1034 = 4 × 258 + 2
Therefore, N = D × Q + R

(c) 2,504 ÷ 7
Ans:

Here, N = 2504, D = 7, Q = 357, R = 5
Since, 2504 = 7 × 357 + 5
Therefore, N = D × Q + R

(d) 8,900 ÷ 15
Ans:

Here, N = 8900, D = 15, Q = 593, R = 5
Since, 8900 = 15 × 593 + 5
Therefore, N = D × Q + R

(e) 9,876 ÷ 32
Ans:

Here, N = 9876, D = 32, Q = 308, R = 20
Since, 9876 = 32 × 308 + 20
Therefore, N = D × Q + R

(f) 7,506 ÷ 24
Ans:

Here, N = 7506, D = 24, Q = 312, R = 18
Since, 7506 = 24 × 312 + 18
Therefore, N = D × Q + R

Q5: Find the solutions for part A. Observe the relations between the quotient, divisor and dividend and use it to answer parts B and C.

Ans:

Q6: A company in Mumbai organises cycle rallies from Mumbai to Panjim, Goa, every year. They aim to cover 576 km in 12 days.
(а) How much distance should they cycle every day, to cover the distance evenly?
(b) After reaching Ratnagiri, they rest for 1 day. How much distance should they cycle each day to reach Goa in 4 days? Assume that they cover the distance evenly.

Ans:

(a) The distance they should cycle every day = 576 ÷ 12 = 48 km

(b) Distance from Ratnagiri to Goa = 232 km
The distance they should cycle each day to reach Goa in 4 days = 232 ÷ 4 = 58 km.

Q7: Given below are a few problems. You may need some additional information to solve these. Identify the missing information. Write the missing information and find the answer.
(a) A fruit vendor sells 6 baskets of mangoes. Each basket contains 12 mangoes. How much did the vendor earn in total?
(b) A school has 8 classrooms, and each classroom has an equal number of desks. How many desks are there in each classroom?
(c) Rahul buys 5 cricket bats for his team. The total bill is ₹ 3500. How much does one bat cost?
(d) A restaurant serves 125 plates of idlis in a day. The total earnings from selling all the idli plates is ₹ 6250. How many idlis are there in each plate?
Ans:
(a) Missing information: The cost of each mango.
Let each mango cost = ₹ 10
The cost of each basket = 12 × 10 = ₹ 120
Therefore, the cost of 6 baskets = 6 × 120 = ₹ 720
Thus, the vendor earns ₹ 720 in total.
(Answer may vary)

(b) Missing information: Total number of desks in the school.
Let the total number of desks in the school be 160.
Therefore, the total number of desks in each classroom = 160 ÷ 8 = 20
(Answer may vary)

(c) No information is missing.
The cost of one bat = 3500 ÷ 5 = ₹ 700

(d) Missing information: The cost of one idli. Let the cost of each idli is ₹25. The cost of one plate of idlis
= 6250 ÷ 125 = ₹ 50
Therefore, the number of idlis in one plate = 50 ÷ 25 = 2
(Answer may vary)

Q8: To make one bookshelf, a carpenter needs the following things 4 long wooden panels 8 short wooden panels 16 small clips 4 large clips 32 screws
The carpenter has a stock of 264 long wooden panels, 306 short wooden panels, 2400 small clips, 120 large clips, and 2800 screws. How many bookshelves can the carpenter make? Discuss your thoughts.

Ans:

Since 264 ÷ 4 = 66
Therefore, the long wooden panels are sufficient for 66 bookshelves.
306 ÷ 8 = 38 and remainder 2.

So, the short wooden panels are sufficient 38 bookshelves.
2400 ÷ 16 = 150 and remainder 0.

So, the small clips are sufficient for 150 bookshelves.
120 ÷ 4 = 30

So, large clips are sufficient for 30 bookshelves.
2800 ÷ 32 = 87 and remainder 16.

So, screws are sufficient for 87 bookshelves. Based on the above data the carpenter can make 30 bookshelves.

Page No. 134

Vegetable Market

Q: Munshi Lai has a big farm in Bihar. Every Saturday, he sells the vegetables from his farm at Sundar Sabzi Mandi. Munshi ji maintains a detailed record of the quantity of vegetables he sends to the Mandi and the cost of each vegetable. The following table shows his record book on one Saturday.
His naughty grandson has erased some numbers from his record book. Help Munshi Lai complete the table.

Ans:

Let Us Solve

Q: Divide the following. Try dividing using place values, whenever you can. Identify the remainder, if any, and check whether N = D × Q + R.
1. 506 ÷ 5
Ans:

N = 506, D = 5, Q = 101, R = 1
5 × 101 + 1 = 505 + 1 = 506
Therefore, N = D × Q + R

2. 918 ÷ 8
Ans:

N = 918, D = 8, Q = 114, R = 6
8 × 114 + 6 = 912 + 6 = 918
Therefore, N = D × Q + R

3. 8,126 ÷ 7
Ans:

N = 8126, D = 7, Q = 1160, R = 6
7 × 1160 + 6 = 8120 + 6 = 8126
Therefore, N = D × Q + R

4. 9,324 ÷ 4
Ans:

N = 9324, D = 4, Q = 2331, R = 0
Therefore, N = D × Q + R

5. 876 ÷ 6
Ans:

N = 876, D = 6, Q = 146, R = 0
Therefore, N = D × Q + R

6. 7,008 ÷ 3
Ans:

N = 7008, D = 3, Q = 2336, R = 0
Therefore, N = D × Q + R

7. 934 ÷ 12
Ans:

N = 934, D = 12, Q = 77, R = 10
12 × 77 + 10 = 924 + 10 = 934
Therefore, N = D × Q + R

8. 829 ÷ 23
Ans:

N = 829, D = 23, Q = 36, R = 1
23 × 36 + 1 = 828 + 1 = 829
Therefore, N = D × Q + R

9. 705 ÷ 18
Ans:

N = 705, D = 18, Q = 39, R = 3
18 × 39 + 3 = 702 + 3 = 705
Therefore, N = D × Q + R

10. 8,704 ÷ 32
Ans:

N = 8704, D = 32, Q = 272, R = 0
Therefore, N = D × Q + R

11. 6,790 ÷ 45
Ans:

N = 6790, D = 45, Q = 150, R = 40
45 × 150 + 40 = 6750 + 40 = 6790
Therefore, N = D × Q + R

12. 5,074 ÷ 21
Ans:

N = 5074, D = 21, Q = 241, R = 13
21 × 241 + 13 = 5074
Therefore, N = D × Q + R

Page No. 135

Mathematical Statements

Q1: Find out whether the following statements are True (T) or False (F). A true sentence is one where both sides of the ‘=’ sign have the same value.
(a) 8 × 9 = 70 + 2
(b) 20 – 6 = 7 × 3
(c) 48 – 3 = 4 × 4
(d) 89 – 9 = 90 + 0
(e) 25 + 10 = 45 – 10
Ans:
(a) 8 × 9 = 72 and 70 + 2 = 72
Therefore, 8 × 9 = 70 + 2 is True.

(b) 20 – 6 = 14 and 7 × 3 = 21
Since 14 and 21 are not equal.
Therefore, 20 – 6 = 7 × 3 is False.

(c) 48 – 3 = 45 and 4 × 4 = 16
Note: 48 – 3 = 45, which is not equal to 16. There appears to be a typographical error in the original statement. If the intended statement was 48 – 32 = 4 × 4, then it would be true. As given,
48 – 3 = 45 and 4 × 4 = 16, so the equality is False.

(d) 89 – 9 = 80 and 90 + 0 = 90
Since, 80 and 90 are not equal.
Therefore, 89 – 9 = 90 + 0 is False.

(e) 25 + 10 = 35 and 45 – 10 = 35
Therefore, 25 + 10 = 45 – 10 is True.

Q2:  Complete the following statements such that they are true.
(a) 7 × 6 = _____ + 17
(b) 87 + 6 = _____ × 31
(c) 63 + _____ = 74 – 4
(d) _____ ÷ 9 = 16 ÷ 2
Ans:
(a) 7 × 6 = 42 and 25 + 17 = 42
Therefore, 7 × 6 = 25 + 17

(b) 87 + 6 = 93 and 3 × 31 = 93
Therefore, 87 + 6 = 3 × 31

(c) 74 – 4 = 70 and 63 + 7 = 70
Therefore, 63 + 7 = 74 – 4

(d) 16 ÷ 2 = 8 and 72 ÷ 9 = 8
Therefore, 72 ÷ 9 = 16 ÷ 2

Q3:  Think about the following statements and find examples as suggested below.
(a) “When two odd numbers are added, the sum is even.”
Find 5 examples for the above statement. Can you find an example to show that the statement can be false?
(b) “Multiplying a number by 2 can give an odd number.”
Give some example for this statement. Can you find any?
(c) “Halving a number always leads to an even number.”
Give 3 examples for the statement. Can you find 3 examples when this is not true?
Ans:

(a) This statement is always true. Examples of odd + odd = even:

• 3 + 5 = 8
• 7 + 9 = 16
• 11 + 13 = 24
• 15 + 17 = 32
• 21 + 23 = 44

There are no counterexamples; the sum of two odd numbers is always even.

(b) This statement is false. Multiplying any whole number by 2 always gives an even number. Examples:

• 2 × 1 = 2 (even)
• 2 × 3 = 6 (even)
• 2 × 5 = 10 (even)

There are no examples where multiplying an integer by 2 gives an odd number.

(c)This statement is sometimes true and sometimes false. Examples where halving gives an even number:

• 8 ÷ 2 = 4 (even)
• 12 ÷ 2 = 6 (even)
• 20 ÷ 2 = 10 (even)

Examples where halving does not give an even number (it gives an odd number):

• 6 ÷ 2 = 3 (odd)
• 10 ÷ 2 = 5 (odd)
• 14 ÷ 2 = 7 (odd)

So halving a number sometimes leads to an even number and sometimes to an odd number.

Q4: Tick in the appropriate cell for the following statements.

Ans:

Page No. 104

Check! Check!

Q: Anu has recorded the weights of the items in her house. Check if she has recorded them correctly by putting a tick against them if they look correct.
Ans:

Let Us Do

Q: Read the scales. Write the correct weight in the space given below.
Ans:
From parts (a) to (d), each small division on the scale represents 100 g, and in parts (e) and (f), each small division on the scale represents 10 g.

Page No. 105

Different Units but Same Measure

Q: Match the bags that have the same weights. You can use the double number line given below.

Ans:

Page No. 106

Let Us Find

Q1: Shamim and Rehan observed someone buying sugar weighing 5 kg 50 g. They thought of the quantity in grams. How much is it?
Ans:
1 Kg = 1000g
5 kg = 5,000 g
5 kg 50 g = 5,000 g + 50 g = 5050 g
So, Shamim is right.
Q2: Complete the conversions by filling in the blanks. You can use the double number line given below on which some numbers have been marked. 

Ans: 

(a) 7 kg 67 g = ___________ g
Ans:

7 kg = 7,000 g

7 kg 67 g = 7,000 g + 67 g = 7,067g

7 kg 67g = 7,067 g

(b) 3 kg 300 g = ___________ g
Ans:
3 kg = 3,000 g
3 kg 300 g = 3,000 g + 300 g
= 3,300 g
3 kg 300 g = 3,300 g

(c) 8 kg 69 g = ___________ g
Ans:
8 kg = 8,000 g
8 kg 69 g = 8,000 g + 69 g
= 8,069 g
8 kg 69 g = 8,069 g

(d) 10,760 g = _____ kg ____ g
Ans:
10,760 g = 10,000 g + 760 g
= 10 kg + 760 g
10,760 g = 10 kg 760 g

(e) 4,080 g = _____ kg ____ g
Ans:
4,080 g = 4,000 g + 80 g
= 4 kg + 80 g
4,080 g = 4 kg 80 g

(f) 12,042 g = _____ kg ____ g
Ans:
12,042 g = 12,000 g + 42 g
= 12 kg + 42 g
12,042 g = 12 kg 42 g

Page No. 107 

Comparison between Different Weights

Q1:  Harpreet’s family planned a picnic over the weekend. Her mother and father packed different food items to take along. The following is the list of fruits they carried.

Among the fruits they carried, which one has the
(a) highest weight?
(b) least weight?
(c) Arrange the items in descending order of their weight.
Ans:
(a) Weight of the watermelon = 3 kg = 3,000 g
Weight of the pineapple = 1 kg 750 g
= 1,750 g
Weight of the apples = 1 kg 250 g
= 1,250 g
Weight of the mangoes = 2 kg = 2,000 g
Among these weights, 3,000g is the highest, which is the weight of the watermelon.
Therefore, among the fruits they carried, watermelon has the highest weight.

(b) Among the given weights, 1,250g is the least, which is the weight of the apples. Therefore, among the fruits they carried, apples have the least weight.

(c) The decreasing order of weights is:
3,000 g > 2,000 g > 1,750 g > 1,250 g
Therefore, the items in decreasing order of their weights are: Watermelon, Mangoes, Pineapple, Apples.

Q2:  Compare the weights using <, =, > signs.
(a) 1 kg 600 g ____ 1,700 g
Ans:
1 kg 600 g = 1,600 g < 1,700 g
1 kg 600 g < 1,700 g

(b) 1 kg 600 g ____ 1 kg 60 g
Ans:
In 1 kg 600 g and 1 kg 60 g
1 kg = 1 kg but 600 g > 60 g
1 kg 600 g > 1 kg 60 g

(c) 10 kg 35 g ____ 10035 g
Ans:
10 kg 35 g = 10,035 g

(d) 1 kg 600 g ____ 2 kg 500 g
Ans:
In 1 kg 600 g and 2 kg 500 g 1 kg < 2 kg
1 kg 600 g < 2 kg 500 g

(e) 5 kg 50 g ____ 4 kg 500 g
Ans:

5 kg 50 g = 5,050 g

4 kg 500 g = 4,500 g

5,050 g > 4,500 g

So, 5 kg 50 g > 4 kg 500 g

(f) 900 g + 7,000 g ____ 7 kg + 900 g
Ans:
900 g + 7,000 g = 900 g + 7 kg
900 g + 7,000 g = 7 kg + 900 g

Page No. 108-109

Let Us Find

Q1:  If a sugar sachet weighs 5 g, how much will it be in milligrams?
Ans:
1 g = 1,000 mg
5 g = 5 × 1,000 = 5,000 mg
Therefore, the sugar sachet weighs 5,000 milligrams.

Q2:  Complete the double number line below appropriately.
Ans:

Q3:  An ornament weighs 4 g 100 mg. What will be the weight in milligrams?
Ans:

Q4: A goldsmith has made an ornament weighing 10 g 500 mg. What will its weight be in milligrams?
Ans:
1g = 1000 milligram
10 g = 10,000 mg
10 g 500 mg = 10,500 mg
Therefore, the weight of the ornament is 10,500 mg.

Q5:  Compare the weights using <, =, > signs.
(a) 20 g ___________ 200 mg
Ans:
20 g = 20,000 mg
20,000 mg > 200 mg
20 g > 200 mg

(b) 16 g 50 mg ___________ 50 g 16 mg
Ans:
16 g < 50 g
16 g 50 mg < 50 g 16 mg

(c) 2,010 mg ___________ 2 g 100 mg
Ans:
2,010 mg = 2,000 mg + 10 mg
= 2 g 10 mg
Now, 2 g = 2 g but 10 mg < 100 mg
2,010 mg < 2 g 100 mg

(d) 9,000 mg ___________ 90 g
Ans:
9,000 mg = 9 g < 90 g
9,000 mg < 90 g

(e) 5,000 g ___________ 7,500 g
Ans:
5,000 g < 7,500 g

(f) 800 mg + 88 mg ___________ 880 mg + 8 mg
Ans:
800 mg + 88 mg = 888 mg,
880 mg + 8 mg = 888 mg
800 mg + 88 mg = 880 mg + 8 mg

Q6: Observe the pictures given below and fill in the blanks.
Ans:

Q7: Answer the following questions.
(a) 5,000 kg = ____ quintals = ____ tonne
(b) 9,000 kg = ____ quintals
(c) ____ kg = 8 tonnes
Ans:
(a) 100 kg = 1 quintal
5,000 kg = 5,000/100 quintals
= 50 quintals
Also, 10 quintals = 1 tonne
50 quintal = 50/10 tonnes
= 5 tonne
Therefore, 5,000 kg = 50 quintals = 5 tonne

(b) 9,000 kg = 9,000/100 quintals
= 90 quintals

(c) 1 tonne = 1,000 kg
8 tonnes = 8 × 1,000 kg = 8,000 kg
Therefore, 8,000 kg = 8 tonnes.

King’s Weight

Q: In a kingdom, the king donates wheat grains equal to 10 times his weight on his birthday.
(a) If he donates 800 kg of wheat grain this birthday, what is his current weight?
Ans:
The king donates 800 kg of wheat grains on his birthday. It means 10 times the weight of the king this birthday is 800 kg. Therefore, 10 x king’s weight = 800 kg
So, king’s weight = 800/10 kg
= 80 kg

(b) If he had donated 780 kg of wheat grain on his last birthday, what was his weight last year?
Ans:
The king donated 780 kg of wheat grains on his last birthday. It means 10 times the king’s weight last year = 780 kg.
So king’s weight last year = 780/10 kg
= 78 kg

(c) How much weight did he gain in a year until this birthday?
Ans:
Weight gain = Weight of the king on this birthday – Weight of the king on last birthday
= 80 kg – 78 kg
= 2 kg

Page No. 111

Let Us Do

Q1: A restaurant owner uses 5 kg 200 g, 8 kg 900 g, and 12 kg 600 g of onions over 3 days. What is the total weight of onions used by the restaurant owner in 3 days?
Ans: Uses of onions on 1st day = 5 Kg 200 g
Uses of onions on 2nd day = 8 kg 900 g
Uses of onions on 3rd day = 12 kg 600 g
Total weight of the onions used in 3 days =

So, the total weight of the onions used by the restaurant owner in 3 days is 26 kg 700 g.

Q2: Aarav is helping his grandfather at the fruit stall. He lifts two baskets of apples weighing 2 kg 100 g and 3 kg 950 g. What is the total weight of apples he lifted?
Ans: First basket weight = 2 kg 100 g
Second basket weight = 3 kg 950 g
Total weight of two basket =

Therefore, the total weight of the apples Aarav lifted is 6 kg 50 g.

Q3: 4 kg 500 g of sand is used from a sack weighing 10 kg. How much sand is left in the sack?
Ans: Weight of the used sand = 4 kg 500 g
Total sand weight = 10 kg
Remaining sand weight =

Therefore, 5 kg 500 g sand is left in the sack.

Q4: A rice sack weighs 9 kg 750 g. After some rice is used, it weighs 3 kg 700 g. How much rice was used?
Ans: Total weight of rice sack = 9 kg 750 g
Sack weight after used rice = 3 kg 700 g
Weight of used rice =

Therefore, 6 kg 50 g rice was used.

Q5: A delivery truck delivered 17 kg 900 g of supplies in the morning and 12 kg 700 g in the afternoon. How much total supplies did it deliver?
Ans: Weight of the delivered supplies in the morning = 17 kg 900 g
Weight of the delivered supplies in the afternoon = 12 kg 700 g
Weight of the total supplies delivered =

Therefore, the total supplies delivered by the delivery truck is 30 kg 600 g.

Q6: A box of books weighs 14 kg 750 g. After removing some books, the weight of the box is 10 kg 500 g. What is the weight of the books removed?
Ans: Weight of a box of books = 14 kg 750 g
Weight of a box after some books removed = 10 kg 500 g
Weight of the books removed =

Therefore, the weight of the books removed is 4 kg 250 g.

Q7: In a community kitchen of a Gurdwara, 65 kg of flour was purchased on one day. Out of this, 42 kg 275 g flour was used for preparing langar. The next day, an additional 52 kg 500 g of flour was bought. What is the total quantity of flour now available in the kitchen store?
Ans: Weight of flour purchased = 65 kg
Weight of flour used in langar = 42 kg 275 g
Weight of flour purchased next day = 52 kg 500 g
Flour purchased + additional flour bought:

Flour now available in the kitchen store:

Therefore, the total flour available in the kitchen is 75 kg 225 g.

Page No. 112-114

Let Us Do

Q1: The cost of some grocery items is given in the following table. Find the total cost of each item.
Ans:
Rice: Cost of 12 kg = 12 × ₹ 60 = ₹ 720
Cost of lkg (1,000 g) = ₹ 60
Cost of 500 g = ₹ 30
Therefore, cost of 12 kg 500 g = ₹ 720 + ₹ 30
= ₹ 750

Flour: Cost of 7 kg = 7 × ₹ 40 = ₹ 280
Cost of 1 kg (250 g + 250 g + 250 g + 250 g)
= ₹ 40 (10 + 10 + 10 + 10)
Cost of 250 g = ₹ 10
Therefore, cost of 7 kg 250 g = ₹ 280 + ₹ 10
= ₹ 290

Sugar: Cost of 5 kg = 5 × ₹ 45 = ₹ 225
Chana dal: Cost of 3 kg = 3 × ₹ 70 = ₹ 210
Cost of 1 kg (1000 g) = ₹ 70
Cost of 100 g = ₹ 7
Cost of 600 g = 6 × ₹ 7 = ₹ 42
Therefore, cost of 3 kg 600 g = ₹ 210 + ₹ 42
= ₹ 252

Besan: Cost of 4 kg = 4 × ₹ 60 = ₹ 240

Jaggery: Cost of 1 kg (1000 g) = ₹ 50
Cost of 100 g = ₹ 5
Cost of 400 g = 4 ×₹ 5 = ₹ 20
Cost of 1 kg 400 g = ₹ 50 + ₹ 20 = ₹ 70
Hence, the complete table is:

Q2: 4 people need 500 g rice for a meal. How much rice will be needed for 8 people if they eat similar quantity of rice?
Ans:
For 4 people, rice needed = 500 g
For 1 person, rice needed = 500/4 = 125 g
So, for 8 people, rice needed = 8 × 125 = 1000 g = 1 kg
Therefore, 8 people will need 1 kg of rice.

Q3: 5 kg of tomatoes cost ₹ 73. How much will 10 kg of tomatoes cost?
Ans:
5 kg × 2 = 10 kg
₹ 73 × 2 = ₹ 146
10 kg of tomatoes will cost ₹ 146.

Q4: Nitesh is a scrap dealer. How much would he have paid for
(а) 16 kg of old newspaper, if he paid ₹ 8 for every 1 kg of newspaper?
Ans:
1 kg × 16 = 16 kg
₹ 8 × 16 = ₹ 128
Nitesh would have paid ₹ 128.

(b) 20 kg iron, if he paid ₹ 200 for every 10 kg of iron?
Ans:
10 kg × 2 = 20 kg
₹ 200 × 2 = ₹ 400
Nitesh would have paid ₹ 400.

(c) 10 kg plastic, if he paid ₹ 30 for 5 kg of plastic?
Make double number lines for answering (b) and (c).
Ans:
5 kg × 2 = 10 kg
₹ 30 × 2 = ₹ 60
Nitesh would have paid ₹ 60.

Measuring Capacity

Q1: You must have seen tea being prepared at your home. How much water and milk do we need to make 2 cups of tea?
Ans: We need 200 mL of water and 200 mL of milk to make 2 cups of tea.

Q: Do we need 1 litre of water to make 2 cups of tea?
Ans: No, we do not need 1L of water to make 2 cups of tea. It is too much.

Q: Is 500 mL of water enough for 2 cups of tea?
Ans: No, 500 mL of water is also too much for 2 cups of tea.

Q2: A bucket can hold a maximum of 20 mL of water. Is this statement correct? Which unit should be used in such a situation?
Ans:The statement is incorrect. We should use litres instead of mL in this situation because mL is a very small unit in the context of capacity of a bucket.

Big to Small, Small to Big

Q1: Ramiz brings a 500 mL water bottle to school. He drinks two bottles at school. How much water does he drink at school?
Ramiz drinks ____ mL + ____ mL = ____ mL.
Ramiz drinks ____ l of water in a day.
Ans: Capacity of 1 water bottle = 500 mL
So, the capacity of 2 water bottle = 500 mL + 500 mL = 1000 mL
Ramiz drinks 1000 mL/1 litre of water in a day.

Q2: Muskaan drinks 3 L of water in a day. How many times would she need to refill a 500 mL water bottle?
Ans: Capacity of water bottle = 500 mL
Total consumption of water in a day by Muskaan = 3 litres
Number of refills needed = 3,000 mL ÷ 500 mL = 6 times
Muskaan drinks 3,000 mL of water in a day.

Q3: Write the total capacity of the following containers in each blank.
Ans: First Jug: 1 litre + 500 mL + 100 mL = 1 litre 600 mL
Second Jug: 1 litre + 500 mL = 1 litre 500 mL
Third Jug: 100 mL + 100 mL + 500 mL = 700 mL
Fourth Jug: 1 litre + 1 litre + 100 mL = 2 litres 100 mL

Different Units but Same Measure

The Milkman’s Delivery

 Khayal chacha delivers fresh cow milk to homes.   Bhalerao’s family orders 2 L of milk everyday. 

Q: This family has a vessel marked in mL only. What mark will you see in the vessel corresponding to 2 L?
Ans:

Q: Khayal chacha delivers the following amounts of milk each week to different families.
Ans:

Page No. 115-116

Let Us Think

Q1: Mary and Daisy filled their bottle with 11 400 mL of water. They wondered about the capacity of the bottle in mL. How much is it?
Who do you think is correct and why?
Ans:
1 L = 1000 mL
1L 400 mL = 1000 mL + 400 mL
= 1400 mL
Therefore, Mary is correct.

Q2: Convert and fill in the blanks appropriately. You can use the double number line given earlier.
(a) 3 L 8 mL = _____mL
(b) 9 L 90 mL = _____ mL
(c) 14,075 mL = ____L ____mL
(d) 8 L 86 mL = ____ mL
(e) 12,200 mL = ____L_____mL
(f) 18,350 mL = ____L ____mL
Ans:
(a) 3 L = 3,000 mL
3 L 8 mL = 3,008 mL

(b) 9 L = 9,000 mL
9 L 90 mL = 9,090 mL

(c) 14,075 mL = 14,000 mL + 75 mL
= 14L + 75 mL
14,075 mL = 14 L 75 mL

(d) 8L = 8,000 mL
8 L 86 mL = 8,086 mL

(e) 12,200 mL = 12,000 mL + 200 mL
= 12 L + 200 mL
12,200 mL = 12 L 200 mL

(f) 18,350 mL = 18,000 mL + 350 mL
= 18 L + 350 mL
18,350 mL = 18 L 350 mL

Let Us Compare

1. Kiran owns a petrol pump. She records the details of the sales of petrol in a day

2. 
(a) How much more fuel is bought for buses than for trucks?
(b) What is the total quantity of fuel filled from the petrol pump on that day?
Ans:

(a) Fuel bought for buses = 1,800 L
Fuel bought for trucks = 1,500L
Difference = 1,800 – 1,500 = 300
300 L more fuel is bought for buses than for trucks.

(b) The total quantity of fuel filled from the petrol pump on that day is:
1,500 L + 1800 L + 500 L + 96 L + 125 L
= 4,021 L

3. Compare the following quantities using the signs <, =, >.
(a) 5 L 600 mL ___________ 5,400 mL
Ans:
5 L 600 mL = 5000 mL + 600 mL
= 5,600 mL > 5,400 mL
5 L 600 mL > 5,400 mL

(b) 10 L 100 mL ___________ 1 L 600 mL
Ans:
10 L > 1 L
10 L 100 mL > 1 L 600 mL

(c) 190 mL + 800 mL ___________ 800 mL +109 mL
Ans:
800 mL = 800 mL and
190 mL > 109 mL
190 mL + 800 mL > 800 mL + 109 mL

(d) 3 l 600 mL ___________ 3,600 mL
Ans:
3 l 600 mL = 3000 mL + 600 mL
= 3600 mL
31 600 mL = 3600 mL

(e) 4 l 50 mL ___________ 4 l 500 mL
Ans:
4 L = 4L and 50 mL < 500 mL
4L 50 mL < 4L 500 mL

4. Sam and Tina fill petrol in their bikes. Tina bought 2 L 500 mL of petrol. Sam bought 2 L 800 mL more petrol than Tina. How much petrol did Sam buy?

Ans: Tina’s petrol = 2 L 500 mL = 2500 mL
Extra petrol Sam bought = 2 L 800 mL = 2800 mL
Sam’s petrol = 2500 mL + 2800 mL = 5300 mL
In litres: 5300 mL = 5L 300 mL
So, Sam bought 5L 300 mL of petrol.

Page No. 118

Let Us Solve

Q1: Riya is filling water bottles for a picnic. She fills one 2 L bottle and four 500 mL bottles. Her friend, Aarav fills three 750 mL bottles. Who filled more water, Riya or Aarav? How much more?
Ans:
Water filled in four 500 mL bottles is 4 × 500 mL = 2,000 mL = 2L
Total amount of water filled by Riya = 2L + 2L = 4L
Water filled by Aarav = 3 × 750 mL = 2,250 mL
= 2 L 250 mL
4L > 2L 250 mL
So, Riya filled more water.

Therefore, Riya filled 1L 750 mL more water than Aarav.

Q2: A bottle of milk is poured equally into 8 glasses, leaving 120 mL of milk in the bottle.
(a) If each glass has a capacity of 360 mL, what is the total capacity of 8 glasses?
(b) How much milk was there in the bottle initially?
(c) If 1 l of milk costs ₹ 40, how much will 3 l milk cost?
Ans:
(a) Total capacity of 8 glasses
= 8 × 360 mL = 2,880 mL = 2L 880 mL

(b) The amount of milk in the bottle initially
= the total amount of milk poured in the glasses + the amount of milk left
= 2,880 mL + 120 mL
= 3,000 mL
= 3 L

(c) Cost of 1 l milk is = ₹ 40
Cost of 3 l milk = 40 × 3 = ₹120

Q3: A juice vendor has a 5 l container of orange juice. Each glass has a capacity 250 mL.
(a) How many full glasses can he serve before the container becomes empty?
Ans:
1l = 1000 mL = 500 mL + 500 mL
= 250 mL + 250 mL + 250 mL + 250 mL
So, 4 full glasses can be served in 1L.

1 l × 5 = 5 l
4 glasses × 5 = 20 glasses
Therefore, he can serve 20 full glasses before the container becomes empty.

(b) If he has already served 10 glasses, how much juice is left?
Ans:
Total juice = 5000 mL
Juice in 10 glasses = 10 × 250 mL = 2500 mL
Juice left = 5000 mL – 2500 mL = 2L 500 mL

(c) If 250 mL of juice is sold at ₹25, how much will he earn by selling 5 L juice?
Ans:
We know that 20 × 250 mL = 5000 mL(5 l)

250 mL × 20 = 5000 mL
₹ 25 × 20 = ₹ 500
Therefore, he will earn ₹ 500 by selling 5 l juice.

Q4: In a factory, 8 L 400 mL of oil needs to be equally poured into 7 containers for storage. How much oil will each container hold?
Ans: Given:

• Total oil = 8 L 400 mL
• Number of containers = 7

We know that, 1 L = 1000 mL

So, 8 L = 8000 mL

⇒ 8 L 400 mL = 8000 + 400 = 8400 mL

​8400 mL ÷ 7 = 1200 mL

Now, Convert back to litres and millilitres: 1200 mL = 1 L 200 mL

Q5: If one container can hold 1 L 75 mL of buttermilk, how much buttermilk will be there in 8 such containers?
Ans:
1 container x 8 = 8 containers
1L 75 mL × 8 = 1L × 8 + 75mL × 8
= 8 L + 600 mL = 8 L 600 mL

Therefore, 8 L 600 mL buttermilk will be there in 8 such containers.

Page No. 92-93

Weaving Mats

You may have seen woven baskets of different kinds. If you look closely, you will notice different weaving patterns on each basket.
We will try weaving some mats with paper strips.

Q1: Let us make paper mats.
You will need —A coloured paper (30 cm long and 20 cm wide) and eight paper strips of two different colours (3 cm wide and longer than 20 cm).
Ans:
(a) Take a coloured paper 30 cm long and 20 cm wide.
(b) Fold the coloured paper in half along the longer side.
(c) Draw vertical lines at equal distances from the folded edge and cut slits leaving a gap of 3 cm at the top.
(d) Carefully unfold the paper. There will be no cuts in the paper at the top and the bottom.
(e) Now cut 8 paper strips of 3 cm width in 2 colours and of length slightly longer than 20 cm.
(f) Take one colour strip and weave it across the slits going one under and one over, and again one under and one over. Repeat it for the first row.
(g) Take one more strip of another colour and weave it across the slits going 1 over and one under, and again one over and one under. Repeat it for the second row.
(h) Weave all the strips in the same alternating pattern. Neatly fold any extra strip ends behind the mat. Your mat is ready!

Q2: Can you work out the steps for any of these designs and weave the pattern?
Write the steps of the pattern in your notebook for each row until it starts repeating.

Ans:
For Image 1:
Row 1: 1 under (do not repeat), 3 over, 3 under, 3 over, … (repeat).
Row 2: 3 over, 3 under, 3 over, 3 under, 3 over, … (repeat).
Row 3: 2 over (do not repeat), 3 under, 3 over, 3 under, 3 over, … (repeat).
Row 4: 2 under (do not repeat), 3 over, 3 under, 3 over, … (repeat).
Row 5: 3 under, 3 over, 3 under, 3 over, … (repeat).
Row 6:1 over (do not repeat), 3 under, 3 over, 3 under, … (repeat).

For Image 2:
Row 1: 1 under, 3 over, 1 under, 3 over, 1 under, … (repeat).
Row 2: 2 under (do not repeat), 1 over, 3 under, 1 over, 3 under, 1 over, … (repeat).
Row 3: 1 over, 3 under, 1 over, 3 under, 1 over, … (repeat).
Row 4: 2 over (do not repeat), 1 under, 3 over, 1 under, … (repeat).
Row 5: 1 under, 3 over, 1 under, 3 over, 1 under, … (repeat).
Row 6: 2 under (do not repeat), 1 over, 3 under, 1 over, 3 under, … (repeat)

Let Us Try

Draw the following pattern on a grid paper. Part of it is done for you.
Now, complete the rest of the grid to get the full design.
Ans:
Do it yourself.

Page No. 94-99

Find Out

Q: Can squares (a regular 4-sided shape) fit together around a point without any gap or overlap? Try it out using cutouts of squares (a sample square is given at the end of the book). How many squares did you need?

Ans: 

Yes, squares can fit around a point without leaving any gap or overlap.
The total angle around a point is 360°, and the corner angle of a square is 90°.
360° ÷ 90° = 4.
So, exactly four squares fit around a point.

Q: Can five squares fit together around a point without any gaps or overlaps? Why or why not?
Ans:
No, five squares cannot fit together around a point without gaps or overlaps. This is because the angle at each corner of a square is 90° and 5 × 90°= 450°.
Since 450° is greater than 360°, five squares will overlap and cannot fit neatly around a point.

Q: Can regular hexagons (6-sided shapes with equal sides) fit together around a point without any gaps or overlaps? Try and see (a sample hexagon is given at the end of the book). How many fit together at a point?
Ans: 
Yes, regular hexagons can fit together around a point without any gaps or overlaps.
The angle at each corner of a regular hexagon is 120°, and the total angle around a point is 360°.
360° ÷ 120° = 3.
So, 3 hexagons fit together perfectly around a point.

Q: Here is a tessellating pattern with more than one shape.
What shapes have been used in this pattern?
Ans:
Equilateral triangles and regular hexagons used in the above pattern.

Q: Continue the pattern given below and colour it appropriately.
Ans:
Do it yourself.

Q: Do regular octagons fit together without any gaps or overlaps? Try drawing the same and check.
Ans:
No, regular octagons do not fit together perfectly without any gaps or overlaps to form a tessellation.

Q: Look at the pattern given below. What shapes are coming together at the marked points? Are the same set of shapes coming together at these points? Continue the pattern and colour it appropriately.
Ans:
At each marked point, two octagons and one square come together.
Yes, the same set of shapes (two octagons and one square) meet at all the red-marked points.

Q: Here is a tiling pattern made using two different shapes-squares and triangles. Are the triangles equilateral? Why or why not?

What shapes are coming together at the marked points?
Are the same set of shapes coming together at these points? Continue the pattern and colour it appropriately.

Ans: Since the sides of the squares used in the tiling pattern are equal, the sides of the triangles used to fill in the gaps between the squares must be equal. Thus, the triangles are equilateral.

At each marked point, two squares and one equilateral triangle are coming together.

Yes, the same set of shapes are coming together at all marked points.

Q: What geometrical shapes can you make by fitting 2 of these triangles together? Trace the shapes you created.Ans:

Q1: How many different types of triangles can you make?
Ans: 4, isosceles triangle, scalene triangle, equilateral triangle and right triangle.

Q2: Is it possible to make a triangle where all three sides are equal (equilateral triangle)?
Ans:
Yes, it is possible to make a triangle where all three sides are equal. Such a triangle is called an equilateral triangle.

Q3: Is it possible to make a triangle where all three sides are unequal?
Ans: Yes, it is possible to make a triangle where all three sides are unequal. Such a triangle is called a scalene triangle.

Q4: How many different 4-sided shapes (quadrilaterals) can you make?
Here are three possible shapes.
Have you made a shape like the one shown on the right?
Ans:

A rectangle, a kite, and a parallelogram are the three 4-sided shapes that can be made.

Q5: Measure the sides of each of these two quadrilaterals A and B. What do you notice?
Are there any pairs of sides that are equal? Which pairs are equal—adjacent or opposite?
Ans: On measuring, it is found that the opposite sides of the quadrilaterals A and B are equal.

Q6: In the grid given below, draw two different kites and parallelograms each.
Ans:
Do it yourself.

Q7: Now, use 3 triangles from the rhombus to form shapes. How many sides do each one of them have?
Using 3 triangular pieces of the rhombus, try creating a (a) 3-sided shape, (6) 4-sided shape, and (c) 5-sided shape.
Ans:
Do it yourself.

Q8: Which of these shapes can be made with all 4 pieces? Try and find out.
(a) Square
(b) Rectangle
(c) Triangle
(d) Pentagon (5-sided)
(e) Hexagon (6-sided)
(f) Octagon (8-sided)
Ans:
(a) Square
If you divide a rhombus into four triangles using its diagonals, you get four identical right triangles. With these four pieces:
(a) Square: Not possible (unless the rhombus was already a square).
(b) Rectangle: Possible. Two triangles can form a rectangle, and combining two rectangles gives a larger rectangle.
(c) Triangle: Not possible—four pieces cannot be arranged into a perfect triangle.
(d) Pentagon (5-sided): Cannot be formed.
(e) Hexagon (6-sided): Possible, by arranging the four triangles with their hypotenuses facing outward.
(f) Octagon (8-sided): Not possible with just four pieces.

Tangram

Q: Look at the tangram set given at the end of your textbook. Cut out all the shapes. Name them.
(a) How are they same or different from each other?
(b) What do you notice about the angles of each of the shapes?
(c) What do you notice about the sides of each of the shapes?Ans: 

(a) Shape 1, 2, 3, 5 and 6 are triangles. Out of these shapes 2 and 3 are equal and shapes 1 and 5 are equal. Shape 4 is a square and shape 7 is a parallelogram.

(b) In shape 2, 3 and 6 only two angles are equal where as in shape 1, 5 and 4 all angles are equal. In shape 7 opposite angles are equal.

(c) In shape 2, 3 and 6 only two sides are equal where as in shape 1, 5 and 4 all sides are equal. In shape 7 opposite sides are equal.

Page No. 100

Which Shape Am I?

Q: Match the statements with appropriate shapes. Do some of them describe more than one shape?
Ans:

Kites

Make your own kite shape.
(a) Start with a square piece of paper.
(b) Take one corner of the paper and fold it towards the opposite corner, creating a sharp crease along the diagonal.
(c) Open and fold the corner A inwards, aligning the edge with the crease you just made.
(d) Repeat on the other side, folding the other corner B inwards to align with the crease at the centre.
You have a kite shape!
Q: What shapes do you see in the kite?
Ans:
Three right-angled triangle in which two are of same size.

Page No. 101

Play with Circles

Do you remember a circle?
 (a) Draw a circle with a compass and mark its centre.
 (b) Draw its diameter. Mark the endpoints of the diameter. 
 (c) Draw another diameter of the circle and mark the endpoints. 
(d) Now join the four points. 

What shape is formed? Check the sides of the quadrilateral and the angles obtained. 

Ans: The opposite sides and angles are equal. All angles are right angles.

Try with a different pair of diameters. 

What do you notice about the shape that is formed? 

Ans: Everytime we get a rectangle.

Is it possible to create a 4-sided shape other than a rectangle through this process?

Ans: We can create a rectangle only with this process.

Page No. 102

Cube Connections

Q1: Here are three views of a cube. Can you draw them on the net in the correct order?
Ans:

Q2: Here are some big solid cube frames. How many small cubes have been removed from each cube?

Ans:

(a) The full cube has 3 × 3 × 3 = 27 small cubes.

After removing the small cubes only the big cube frame has 20 small cubes. So, total removed cubes = 27-20 = 7 cubes.

(b) The full cube has 4 × 4 × 4 = 64 small cubes.
After removing the small cubes the big cube frame has 32 small cubes.
So, total removed cubes = 64 – 32
= 32 cubes.

(c) The full cube has 5 × 5 × 5 = 125 small cubes.
After removing the small cubes the big cube frame has 44 small cubes.
So, total removed cubes = 125 – 44
= 81 cubes.

Q3:  Nisha has glued 27 small cubes together to make a large solid cube. She paints the large cube red. How many of the original small cubes have—
(a) three faces painted red?
(b) two faces painted red?
(c) one face painted red?
(d) no faces painted red?
Ans:
Nisha has a large solid cube made from 27 small cubes. Since 3 × 3 × 3 = 27, the large cube is a 3 × 3 × 3 cube.
She paints the entire large cube red.
(a) The large solid cube has 8 corner cubes, and each corner small cube is painted red.
So, 8 small cubes are three faces painted red.
(b) Since a cube has 12 edges or sides, and 1 cube in the middle of each edge is painted red.
So, there are 12 small cubes with two faces painted red.
(c) These are the small cubes located in the center of each face of the large cube. A cube has 6 faces. Each face of the large cube is a 3 × 3 square of small cubes. The center small cube of each 3 × 3 face has only one face exposed to the outside, or painted red. So, there are 6 small cubes with one face red.
(d) For a 3 × 3 × 3 = 27 cube, if we remove the outer layer of cubes, we are left with an inner cube. This means there is only 1 small cube right in the very center of the large cube that has no faces painted red.

Puzzle

Tanu arranged 7 shapes in a line. She used 2 squares, 2 triangles, 1 circle, 1 hexagon, and 1 rectangle.
Find her arrangement using the following clues:
(a) The square is between the circle and the rectangle.
(b) The rectangle is between the square and the triangle.
(c) The two triangles are next to the square.
(d) The hexagon is to the right of the triangle
(e) The circle is to the left of the square.
Ans: The arrangement is: Triangle, Circle, Square, Rectangle, Triangle, Hexagon, Square.

Page No. 103

Icosahedron and Dodecahedron

Q: What do these names mean? Once you count their faces, you will know.
Ans: Icosahedron: It is a geometric solid with 20 faces, typically shaped as equilateral triangles in the case of a regular icosahedron.
Dodecahedron: It is a three-dimensional shape having twelve plane faces, in particular a regular solid figure with twelve equal pentagonal faces.

Q: What shapes do you see in an icosahedron and a dodecahedron?
Ans:
Icosahedron: Equilateral triangles, Dodecahedron: Regular pentagons

Q: Do all the faces look the same?
Ans:
Icosahedron: Yes, Dodecahedron: Yes

Q: How many faces meet at a vertex (point)?
Ans:
Icosahedron: 5 faces (Equilateral triangles), Dodecahedron: 3 faces (Regular pentagons)

Q: Do the same number of faces meet at each vertex?
Ans:
Icosahedron: Yes, Dodecahedron: Yes

Q: How many edges do you see?
Ans: The edge is the line where two faces meet.
Icosahedron: It has 30 edges.
Dodecahedron: It has 30 edges.

Q: How did you count them such that you do not miss out any edge or count an edge twice?
Ans: For the Icosahedron: We can see that each of the 20 triangles has 3 sides,
so 20 x 3 = 60.
But each edge is shared by 2 triangles. So, we counted every edge twice. The real number is 60 ÷ 2 = 30 edges.
For the Dodecahedron: Each of the 12 pentagons has 5 sides,
so 12 x 5 = 60.
Again, each edge is shared by 2 pentagons.
So, 60 ÷ 2 = 30 edges.

Q: Can you think of any other solid shapes that have faces that look the same?
Ans:
Yes, there are a few other types of shapes where all faces are identical. These are known as platonic solids. Besides the icosahedron and dodecahedron, the other platonic solids are:

• Tetrahedron: 4 faces, all are equilateral triangles.
• Cube: 6 faces, all are squares.
• Octahedron: 8 faces, all are equilateral triangles.

Do the same number of faces meet at each common vertex?
Ans: Yes.

You can also build some 3-D shapes using straws or ice-cream sticks and clay or play dough. Which shapes did you make?

Ans: With straws and clay, we can make:
Cubes and cuboids (like a box).
Pyramids with a square base (square pyramid) or a triangular base (tetrahedron).
Triangular prisms (like a tent shape).

Page No. 70-71

Let Us Think

Q1: The given shapes stand for numbers between 1 and 24. The same shape denotes the same number across all problems. Find the numbers hiding in all the shapes.

Ans: This can be solved by using a trial-and-error method while keeping the conditions consistent across all diagrams.
One valid set of values for the shapes is shown below. These values are chosen so that each equation formed by the shapes is true when substituted:

Q2: Place the digits 2, 5, and 3 appropriately to get a product close to 100. Share your reasoning in class.

Ans: If we place the digits to form 53 × 2 = 106, the product is 106. Rounding 106 to the nearest ten gives 110, which is closer to 100 than the other possible products made from the digits 2, 5 and 3. 
Thus, 53 × 2 is a good choice to get a product near 100.

Q3: A dairy has packed butter milk pouches in the following manner. Find the number of pouches kept in each arrangement. One is done for you.

Ans: 

Q4: Which number am I?
I am a two-digit number, help of the following clues.
(а) I am greater than 8.
(b) I am not a multiple of 4.
(c) I am a multiple of 9.
(d) I am an odd number.
(e) I am not a multiple of 11.
(f) I am less than 50.
(g) My ones digit is even.
(h) My tens digit is odd.

There’s a contradiction here. If it’s an odd number (clue d), its ones digit must be odd (1, 3, 5, 7, or 9). But clue (g) says its ones digit is even (0, 2, 4, 6, or 8).
A number cannot be both odd and have an even ones digit at the same time.

Ans: No, all clues were used consistently to find the number.
Clue (d) says “odd number,” while clue (g) says the ones digit is even; these two clues conflict with one another because an odd number cannot have an even ones digit. Hence, the clues are inconsistent.

Clues that would directly help narrow down the number:

• (c) Multiple of 9 – This limits options significantly
• (f) Less than 50 – Combined with being two-digit, gives range 10-49
• (d) Odd number OR (g) One’s digit is even – One of these (whichever is correct)
• (h) Tens digit is odd – Further narrows the options

Clues that might be redundant once we narrow it down:

• (a) Greater than 8 – This is already satisfied by being a two-digit number (all two-digit numbers are ≥10)
• (b) Not a multiple of 4 – This would already be satisfied if the number is odd (clue d), since odd numbers can never be multiples of 4
• (e) Not a multiple of 11 – This might eliminate one specific option, but may not be necessary

Q5: Make your own numbers.
Choose any two numbers and one operation from the grid. Try to make all the numbers between 0 and 20. For example 2 can be formed as 4 – 2. Could you make all the numbers?

Ans: One example given is 36 – 25 = 11. (Answers will vary; try different pairs and operations to obtain numbers from 0 to 20.) 

Q: Which numbers could you not make? Is it possible to make these numbers using three numbers? You can use two operations, if needed. Which numbers between 0-20 can you get in more than one way?
Ans:
Try this as a class activity: list the numbers you could not form using two numbers and one operation, then see if adding a third number or a second operation helps. Record numbers that have more than one representation.

Page No. 72-74

Order of Numbers in Multiplication

Q: Daljeet Kaur runs a milk processing unit. She has arranged the butter packets in the following ways. Find the number of butter packets in each case. What pattern do you notice (or observe)? Discuss in class.

Ans: 

The pattern observed is that the product of two numbers remains the same when we interchange their positions; that is, multiplication is commutative: a × b = b × a.

What is 9 × 0? 0 × 9?
Ans: 9 × 0 = 0 and 0 × 9 = 0. Multiplying any number by zero gives zero.

Patterns in Multiplication by 10s and 100s

Q1: Let us revise multiplication by 10s and 100s.
(a) 4 × 10 = _____
(b) 20 × 10 = _____
(c) 10 × 40 = _____
(d) 10 × 10 = 100
(e) 20 × 50 = ______
(f) 80 × 10 = ______
(g) 3 × 100 = 100 × 3 = 300
(h) 8 × 100 = _____ = ______
(i) 10 × 100 = _____ = ______
Ans:
(a) 4 × 10 = 40
(b) 20 × 10 = 200
(c) 10 × 40 = 400
(d) 10 × 10 = 100
(e) 20 × 50 = 1000
(f) 80 × 10 = 800
(g) 3 × 100 = 100 × 3 = 300
(h) 8 × 100 = 100 × 8 = 800
(i) 10 × 100 = 100 × 10 = 1000

Q2: Find the answers to the following questions. Fill in the table below and describe the pattern. Discuss in class.

Ans: 

Q: Let us fill in the table and observe the patterns.

Ans: 

Page No. 75-76

Doubling and Halving

Q: Butter packets are arranged in the following ways. Let us find some strategies to calculate the total number of packets.

Sol: 

(c) Solve the following problems like the previous ones.

Ans:

Q: This halving and doubling strategy works well when we have to multiply with numbers like 5 and 25. Discuss why?

Ans: Halving one factor and doubling the other does not change the product because (a ÷ 2) × (b × 2) = a × b. This is useful to simplify calculations when one factor becomes easier to multiply after halving or doubling.

Why it works well for 5 and 25

1. For ×5:
5=102
So, for any even N,

i.e. halve N and then append a zero (×10).
Example: 38×5=(19)×10=190.

2. For ×25:
25=1004
So, for any N divisible by 4,

i.e. halve twice (divide by 4) and then append two zeros (×100).
Example: 84×25=(21)×100=2100

(d) Find the product by halving and doubling either the multiplier or the multiplicand.
(1) 5 × 18
(2) 50 × 28
(3) 15 × 22
(4) 25 × 12
(5) 12 × 45
(6) 16 × 45
Ans: 

(e) Give 5 examples of multiplication problems where halving and doubling will help in finding the product easily. Find the products as well.
Ans:

Nearest Multiple

(a) 4 × 19

(b) 14 × 21

(c) Give 5 examples of problems where you can use the nearest multiple to find the product easily. Find the products as well.
Ans:
(1) 5 × 31 = 5 × 30 + 5 = 150 + 5 = 155
(2) 7 × 29 = 7 × 30 – 7 = 210 – 7 = 203
(3) 12 × 49 = 12 × 50 – 12 = 600 – 12 = 588
(4) 8 × 101 = 8 × 100 + 8 = 800 + 8 = 808
(5) 16 × 99 = 16 × 100 – 16 = 1600 – 16 = 1584
(Answer may vary)

(d) Find the products of the following numbers by finding the nearest multiple.
(1) 7 × 52
(2) 12 × 28
(3) 75 × 31
(4) 99 × 15
(5) 8 × 25
(6) 22 × 42
Ans:
(1) 7 × 52 = 7 × 50 + 7 × 2 = 350 + 14 = 364
(2) 12 × 28 = 12 × 30 – 12 × 2 = 360 – 24 = 336
(3) 75 × 31 = 75 × 30 + 75 = 2250 + 75 = 2325
(4) 99 × 15 = 100 × 15 – 15 = 1500 – 15 = 1485
(5) 8 × 25 = 10 × 25 – 2 × 25 = 250 – 50 = 200
(6) 22 × 42 = 22 × 40 + 22 × 2 = 880 + 44 = 924

Page No. 77-78

Let Us Solve

Use strategies flexibly to answer the following questions. Discuss your thoughts in class.

Q1: A school has an auditorium with 35 rows, with 42 seats in each row. How many people can sit in this auditorium?
Ans:
Number of rows = 35
Number of seats in each row = 42
Total number of seats in all = 35 × 42 = 35 × 40 + 35 × 2
= 1400 + 70
= 1470
Thus, 1470 people can sit in the auditorium.

Q2:  Priya jogs 4 kilometres every day. How many kilometres will she jog in 31 days?
Ans:
Priya jogs every day = 4 km
Number of kilometres she will jog in 31 days = 4 × 31 = 4 × 30 + 4 = 120 + 4 = 124 km
Thus, Priya will jog 124 km in 31 days.

Q3: A school has received 36 boxes of books with 48 books in each box. How many total books did the school receive in the boxes?
Ans:
Number of books in each box = 48
Number of boxes received by the school = 36
Total number of books received by the school = 36 × 48 = 36 × 50 – 36 × 2
= 1800 – 72
= 1728 books
Thus, the school received 1728 books in all.

Q4: Priya uses 16 metres of cloth to make 4 kurtas. How much cloth would she need to make 8 kurtas?
Ans:
Cloth needed to make 4 kurtas = 16 m
Cloth needed to make 1 kurta = 16 ÷ 4 = 4 m
Cloth needed to make 8 kurtas = 8 × 4 = 32 m.

Q5: Gollappa has 29 cows on his farm. Each cow produces 5 litres of milk per day. How many litres of milk do the cows produce in total, each day?
Ans:
Number of cows on Gollappa’s farm = 29
Milk produced by each cow = 5 litres
Total quantity of milk produced in his farm = 5 × 29 = 5 × 30 – 5 = 150 – 5 = 145 litres
Thus, 145 litres of milk are produced each day.

Q6: Maska Cow Farm has 297 cows. Each cow requires 18 kg of fodder per day. How much total fodder is needed to feed 297 cows every day?
Ans:
Total number of cows at Maska Cow Farm = 297
Fodder required by each cow = 18 kg per day
Total amount of fodder required for 297 cows = 18 × 297 = 18 × 300 – 18 × 3 = 5400 – 54
= 5346 kg
Thus, 5346 kg of fodder will be required for 297 cows every day.

Page No. 79-80

Let Us Multiply

(a) 32 × 8

Ans: 

(b) 69 × 45

Ans: 

Let Us Do

Q: Solve the following problems like Nida did.
(a) 

Ans: 

(b) 83 × 9

Ans: 

(c) 67 × 28

Ans: 

(d) 53 × 37

Ans: 

Q2: Solve the following problems like Kanti did.
(a) 94 × 5
Ans: 

(b) 49 × 6
Ans:

(c) 37 × 53
Ans:

(d) 28 × 79
Ans:

Q3: Solve the following problems like John.
(a) 86 × 3
Ans:

(b) 72 × 7
Ans:

(c) 94 × 36
Ans:

(d) 66 × 22
Ans:

Q4: Solve the following problems:
(a) A movie theatre has 8 rows of seats, and each row has 12 seats. If half the seats are filled, how many people are watching the movie? If 3 more rows get filled, how many total people will be there?

Ans:

Number of rows of seats in the movie theatre = 8

Number of seats in each row = 12

Since, half seats are filled, that is 4 rows out of 8 are filled.

So, number of people who watched the movie = 4 × 12
Thus, 48 people watched the movie.
Now, 3 more rows get filled.
Therefore, number of people who will watch the movie = 7 × 12
Thus, 84 people will be there, if 3 more rows get filled.

(b) In a test match between India and West Indies, the Indian team hit twenty- four 4s and eighteen 6s across the two innings. How many runs were scored in 4s and 6s each? 234 runs were made by running between the wickets. If 23 runs were extras, how many runs were scored by Indian team in the two innings?
Ans:
In the test match, number of 4s hit by Indian team = 24
Therefore, runs scored by Indian team by hitting 4s = 24 × 4 = 96 runs 
In the test match, number of 6s hit by Indian team = 18

Therefore, runs scored by Indian team by hitting 6s = 18 × 6 = 108 runs

Runs scored by running between the wickets = 234 runs

Runs scored by extras = 23 runs

Therefore, total runs scored by the Indian team in two innings = 96 + 108 + 234 + 23 = 461 runs

(c) Anjali buys 15 bulbs and 12 tube lights from Sudha Electricals. Each bulb costs ₹25 and each tube light costs ₹34. How much money should Anjali give to the shopkeeper?
Ans:
Cost of 15 bulbs = 15 × ₹ 25 = ₹ 375

Cost of 12 tube lights = 12 × ₹ 34 = ₹ 408

Total cost of 15 bulbs and 12 tube lights = ₹ 375 + ₹ 408 = ₹ 783
Thus, Anjali will give ₹ 783 to the shopkeeper to buy 15 bulbs and 12 tube lights.

(d) A shopkeeper sold 28 bags of rice. Each bag costs ?350. How much money did he earn by selling rice bags?
Ans:
The cost of each rice bag = ₹ 350
Since, a shopkeeper sold 28 bags of rice. 

Therefore, total cost of 28 bags of rice = 350 × 28 = ₹ 9800
Thus, the shopkeeper earned ₹ 9800 by selling 28 bags of rice.

(e) A school library has 86 shelves and each shelf has 162 books. Find the number of books in the library.
Ans:
Number of shelves in the library = 86 (80 + 6)
Number of books in each shelf = 162 (100 + 60 + 2)

Therefore, the total number of books in the library = 162 × 86 = 13932

Page No. 83-84

Let Us Do

Q1: Solve the following problems like Nida did.
a) 548 × 6
Ans:

b) 682 × 3
Ans:

(c) 324 × 18
Ans: 

(d) 507 × 23
Ans:

(e) 190 × 65
Ans:

Q2: Solve the following problems like John.
(a) 123 × 84
Ans:

(b) 368 × 32
Ans:

(c) 159 × 324
Ans:

(d) 239 × 401

Ans:

(e) 592 × 5

Ans:

(f) 101 × 22
Ans:

Q3: Let us solve a few questions like Milli’s father.

Ans:

Now use Mili’s father’s method to solve the following questions.

(a) 807 × 5
Ans:

(b) 143 × 28
Ans:

(c) 309 × 9
Ans:

(d) 450 × 38
Ans:

(e) 584 × 23
Ans:

(f) 302 × 13
Ans:

(g) 604 × 54
Ans:

(h) 112 × 23
Ans:

(i) 237 × 19
Ans:

Page No. 85

Check, Check!

Check if the following children’s solutions are correct. If correct, explain why the solution is correct. If it is incorrect, then identify the error and correct the solution.
(a) Asma’s solutions for 46 × 59

Ans:

Asma broke the multiplication into parts correctly and added the partial products. Her final result is correct. Only the order of the partial products shown is reversed, but addition is commutative so the final sum remains the same.

(b) Pankaj’s solution for 203 × 54

Ans:
Pankaj made a place-value error: he used 20 instead of 200 and 5 instead of 50 when forming partial products. This gives an incorrect final product. The correct method is to use the actual place values (200 and 50) when making the partial products.

(c) Lado’s solution for 38 × 150

Ans:
She split 38 as 30 + 8 and 150 as 100 + 50 and then multiplied each part correctly. Her decomposition and final sum of partial products are correct.

(d) Kira’s solution for 193 × 272

Ans:
Kira broke the first number correctly, but she used 7 instead of 70 in one partial product for the second number. That is a place-value mistake. The correct partial products must use 70 (tens) and 200 (hundreds).

(e) Asher’s solution

Ans:
Asher made an error identifying place values in 323: the digits 3 and 2 stand for hundreds, tens and ones as 300, 20 and 3. Correctly using these place values, 626 × 323 = 202,198.

Page No. 86-90

Let Us Do

Q1: Identify the problems that have the same answer as the one given at the top of each box. Do not calculate.

Ans:

Q2: Find easy ways of solving these problems.
(a) 16 × 25
Ans:

(b) 12 × 125
Ans:

(c) 24 × 250
Ans:

(d) 36 × 25
Ans:

(e) 28 × 75
Ans:
28 × 75 = 30 × 75 – 150
= 2250 – 150
= 2100

(f) 300 × 15
Ans:
300 × 15 = 4500

(g) 50 × 78
Ans:

(h) 199 × 63
Ans:
199 × 63 = 200 × 63 – 63
= 12600 – 63
= 12537

(i) 128 × 35
Ans:
128 × 35 = 130 × 35 – 70
= 4550 – 70
= 4480

Q3: Write 5 other examples for which you can find easy ways of getting products.
Ans:
(a) 201 × 19
= 200 × 19 + 19
= 3800 + 19
= 3819

(b) 149 × 25
= 150 × 25 – 25
= 3750 – 25
= 3725

(c) 25 × 78

(d) 28 × 18
= 30 × 18 – 18 × 2
= 540 – 36
= 504

(e) 998 × 4
= 1000 × 4 – 8
= 4000 – 8
= 3992
(Answer may vary)

Q4: Find the answers to the following questions based on the given information.
(a) 17 × 23 = 391
(b) 17 × 24 = _______
(c) 17 × 22 = _______
(d) 16 × 23 = _______
(e) 8 × 9 = 72
(f) 18 × 9 = _______
(g) 28 × 9 = _______
(h) 108 × 9 = _______
(i) 18 × 23 = _______

Ans:
(a) 17 × 23 = 391
(b) 17 × 24 = 408
(c) 17 × 22 = 374
(d) 16 × 23 = 368
(e) 8 × 9 = 72
(f) 18 × 9 = 162
(g) 28 × 9 = 252
(h) 108 × 9 = 972
To find 17 × 24, how much is to be added to 17 × 23 _____ 17 or 23?

Ans:
17 × 24 = 17 × 23 + (17) = 391 + (17)
To find 18 × 23, how much is to be added to 17 × 23 _____ 17 or 23?

Ans:

18 × 23 = 17 × 23 + (23)

= 391 + (23)

= 414

Let Us Think

Q1: Find the possible values of the coloured boxes in each of the following problems. The same colour indicates the same number in a problem. Some problems can have more than one answer.

Ans:

Q2: Estimate the products on the left and match them to the numbers given on the right.

Ans:

The King’s Reward

One day, a king decided to reward three of his most talented ministers. The king called them to his court and said, “You all have served my empire with great dedication. As a reward, I give you three choices of gold.

Which of the rewards would you have chosen?
Ans:
Do it yourself.

After a week, the 3 ministers were surprised at the final amount of gold coins. Guess who received the most gold coins? Calculate how much gold
Ans:
Minister-1
5 × 1 = 5 → Day 1
5 × 2 = 10 → Day 2
10 × 2 = 20 → Day 3
20 × 2 = 40 → Day 4
40 × 2 = 80 → Day 5
80 × 2 = 160 → Day 6
160 × 2 = 320 → Day 7

Minister-2
3 × 1 = 3 → Day 1
3 × 3 = 9 → Day 2
9 × 3 = 27 → Day 3
27 × 3 = 81 → Day 4
81 × 3 = 243 → Day 5
243 × 3 = 729 → Day 6
729 × 3 = 2187 → Day 7

Minister-3
1 × 1 = 1 → Day 1
1 × 5 = 5 → Day 2
5 × 5 = 25 → Day 3
25 × 5 = 125 → Day 4
125 × 5 = 625 → Day 5
625 × 5 = 3125 → Day 6
3125 × 5 = 15625 → Day 7
Minister 1 received 320 gold coins, minister 2 received 2187 coins and minister 3 received 15625 coins.
Hence, minister 3 received most number of gold coins.

Multiplication Patterns

Q1: Notice how the multiplier, multiplicand, and products are changing in each of the following. What is the relationship of the new product with the original product?- Solve (a) completely, and then predict the answers for the rest.
(a) 16 × 44 = 704
(1) 8 × 88 = 704
(2) 8 × 22 = 176
(3) 16 × 22 = ______
(4) 32 × 44 = ______
Ans:
(a) 16 × 44 = 704
(1) 8 × 88 = 704
(2) 8 × 22 = 176
(3) 16 × 22 = 352
(4) 32 × 44 = 1408

b) 12 × 32 = 384
(1) 6 × 16 = ______
(2) 24 × 16 = ______
(3) 24 × 64 = ______
(4) 12 × 16 = ______
Ans:
12 × 32 = 384
(1) 6 × 16 = 96
(2) 24 × 16 = 384
(3) 24 × 64 = 1536
(4) 12 × 16 = 192

Q2: Observe and complete the given patterns.

Ans: 

Here are some numbers.

Remember number pairs from Grade 4? Any two adjacent numbers in a row or a column are number pairs. Can you identify the pair whose product is the smallest and another pair whose product is the largest? Do you need to find every product or can you find this by looking at the numbers?

Ans:

Use reasoning to identify the smallest and largest products: the smallest product will come from the smallest adjacent numbers and the largest product from the largest adjacent numbers. (Do this activity by inspection and a few multiplications as needed.)

Page No. 91

Let Us Solve

Q1: Mala went to a book exhibition and bought 18 books. The shop was selling 3 books for ₹ 150. After buying the books, she still had ₹ 20 left. How much money did Mala have at the beginning?
Ans:
The number of books bought by Mala = 18
Since, the cost of 3 books is ₹ 150.
So, the cost of 1 book = ₹ 150 ÷ 3 = ₹ 50
Therefore, the total cost of 18 books = 18 × 50 = (20 × 50) – (2 × 50) = 1000 – 100 = 900
So, the cost of 18 books = ₹ 900
Since, Mala still had ₹ 20 left, the amount she had at the beginning = ₹ 900 + ₹ 20 = ₹ 920.

Q2: A village sports club organises a women’s football tournament. The club earned money by selling match tickets and charging fees for team participation.
They sold 57 tickets for ₹ 115 each.
They had 3 teams joining the tournament, with each team paying a participation fee of ₹ 1,599.
The teams paid ₹ 1,750 in total rent the football ground and ₹ 1,129 for food and water.
(a) How much money did the club collect in total from ticket sales and team participation fees?
(b) What were the total expenses on renting the ground and food and water?
Ans:
Money collected by selling tickets = ₹ (57 × 115) = ₹ (60 × 115 – 3 × 115) = ₹ (6900 – 345) = ₹ 6555
Deposit of participation fees by the 3 teams = ₹ (3 × 1599) = ₹ (3 × 1600 – 3) = ₹ 4800 – 3 = ₹ 4797

Rent paid for the football ground by the teams in total = ₹ 1750
Cost of food and water for the teams = ₹ 1129

(a) The total money collected by the club from the ticket sales and team participation fees = ₹ 6555 + ₹ 4797 = ₹ 11,352

(b) Total expenses on renting the ground and food and water = ₹ 1750 + ₹ 1129 = ₹ 2879

Q3: Ananya is watching Republic Day celebrations on city’s public ground. There are 12 rows of students sitting in front of her and 17 rows behind her. There are 18 students to her right and 22 students to her left.
(a) How many rows of students are there in total?
(b) How many students are there in Ananya’s row?
(c) What is the total number of students on the ground?
Ans:
(a) Number of rows of students in the ground = 12 + 1 + 17 = 30 rows (include Ananya’s row).
(b) Number of students in Ananya’s row = 18 + 1 + 22 = 41 students (include Ananya).
(c) Total number of students on the ground = 30 × 41 = 1230

Q4: Multiply.
(a) 67 × 78
(b) 34 × 56
(c) 45 × 263
(d) 86 × 542
(e) 432 × 107
(f) 310 × 120
Ans:
(a) 67 × 78 = 67 × 80 – 2 × 67
= 5360 – 134
= 5226

(b) 34 × 56 = 30 × 56 + 4 × 56
= 1680 + 224
= 1904

(c) 45 × 263 = 40 × 263 + 5 × 263
= 10,520 + 1315
= 11,835

(d) 86 × 542 = 86 × (540 + 2)
= 86 × 540 + 2 × 86
= 46,440 + 172
= 46,612

(e) 432 × 107 = 432 × 100 + 432 × 7
= 43,200 + 3024
= 46,224

(f) 310 × 120 = 300 × 120 + 10 × 120
= 36,000 + 1200
= 37,200

Q5: If 67 × 67 = 4489, without multiplication find 67 × 68.
Ans:
Since, 67 × 67 = 4489
∴ 67 × 68 = 67 × 67 + 67 = 4489 + 67 = 4556

Q6: If 99 × 100 = 9900, without multiplication find 99 × 99.
Ans:
Since, 99 × 100 = 9900
∴ 99 × 99 = 99 × 100 – 99 = 9900 – 99 = 9801

Table of contents
Page 42
Page 43
Page 44
Page 45

View More

Page 42

Q. In each of the following, there are two groups of numbers. Look carefully at the numbers in each group and their sums. Interchange pairs of numbers between the two groups to make their sums equal. Try to do this using the least number of moves. You could write each number on a small piece of paper.

Ans:

Swap (2 ↔ 3) → both sums become 20. (1 move)

Swap (5 ↔ 9) → both sums become 43. (1 move)

Swap (11 ↔ 13) and (15 ↔ 17) → both sums become 72. (2 moves)

Swap (77 ↔ 81) and (78 ↔ 82) → both sums become 322. (2 moves)

Explanation: In each case we swap numbers so that the difference between the two group sums is removed. A single swap changes each group by the difference between the two swapped numbers; choose swaps so that the net change equalises the sums with the fewest moves.

Page 43

Fuel Arithmetic

Q1. A lorry has 28 litres of fuel in its tank. An additional 75 litres is filled. What is the total quantity of fuel in the lorry? The total quantity of fuel in the tank is 28 L + 75 L.

28 L + 75 L = 103 L. So the lorry has 103 litres of fuel after filling.

Let us try one more.

Q2. Find the sum of 49 and 89.

Ans:

49 + 89 = 138. You can add tens first (40 + 80 = 120) and then units (9 + 9 = 18); 120 + 18 = 138.

Let Us Solve

Q. Add the following numbers. Wherever possible, find easier ways to add pairs of numbers.
1. 15 + 79
2. 46 + 99
3. 38 + 35
4. 5 + 89
5. 76 + 28
6. 69 + 20
Ans:
1. 

15 + 79 = 94. (Add 15 + 80 – 1 = 95 – 1 = 94.)

2. 

46 + 99 = 145. (46 + 100 – 1 = 146 – 1 = 145.)

3. 

38 + 35 = 73. (30 + 30 = 60 and 8 + 5 = 13; 60 + 13 = 73.)

4. 

5 + 89 = 94. (5 + 90 – 1 = 95 – 1 = 94.)

5. 

76 + 28 = 104. (76 + 24 = 100, plus 4 more = 104; or 76 + 20 + 8 = 104.)

6. 

69 + 20 = 89. (Add tens and units separately: 60 + 20 = 80 and 9 + 0 = 9; total 89.)

Page 44

Relationship Between Addition and Subtraction

Q1. Find the relationship between the numbers in the given statements and fill in the blanks appropriately. 
(a)  If 46 + 21 = 67, then,
67 – 21 = _______.
67 – 46 = _______.
(b)  If 198 – 98 = 100, then,
100 + _______ = 198.
198 – _______ = 98.
(c) If 189 + 98 = 287, then,
287 – 98 = _______.
287 – 189 = _______.
(d) If 872 – 672 = 200, then,
200 + _______ = 872.
872 – _______ = 672.
Ans:
(a) If 46 + 21 = 67, then,
67 – 21 = 46.
67 – 46 = 21.

(b) If 198 – 98 = 100, then,
100 + 98 = 198.
198 – 100 = 98.

(c) If 189 + 98 = 287, then,
287 – 98 = 189.
287 – 189 = 98.

(d) If 872 – 672 = 200, then,
200 + 672 = 872.
872 – 200 = 672.

Explanation: These examples show that addition and subtraction are inverse operations. From an addition sentence A + B = C, we get two subtraction sentences C – A = B and C – B = A.

Q2. In each of the following, write the subtraction and addition sentences that follow from the given sentence.

Ans:
(a) If 78 + 164 = 242, then
242 − 164 = 78
242 − 78 = 164

(b) If 462 + 839 = 1301, then
1301 − 839 = 462
1301 − 462 = 839

(c) If 921 − 137 = 784, then
784 + 137 = 921
921 – 784 = 137

(d) If 824 − 234 = 590, then
824 – 590 = 234
590 + 234 = 824

Explanation: Each pair shows how a given addition or subtraction can be turned into the related subtraction and addition sentences by reversing the operation.

Page 45

Let Us Solve

Q1. What is the difference between 82 and 37?

Ans:

Difference = 82 – 37 = 45.

Check your answer. Is 37 + ____ = 82?
Ans:Yes, 37 + 45 = 82.

2. 57 – 11 = ______________

Ans: 57 – 11 = 46.

3. 23 – 19 = ______________

Ans: 23 – 19 = 4.

4. 49 – 21 = ______________

Ans: 49 – 21 = 28.

5. 56 – 18 = ______________

Ans: 56 – 18 = 38.

6. 93 – 35 = ______________

Ans: 93 – 35 = 58.

7. 84 – 23 = ______________

Ans: 84 – 23 = 61.

8. 70 – 43 = ______________

Ans: 70 – 43 = 27.

9. 65 – 47 = ______________

Ans: 65 – 47 = 18.

Sums of Consecutive Numbers

Numbers that follow one another in order without skipping any number are called consecutive numbers. Here are some examples –

Q1. In each of the boxes above, state whether the sums are even or odd. Explain why this is happening.
Ans: (i) Sum of 2 consecutive numbers: Always odd.
Reason: Each pair has one even and one odd number; even + odd = odd.

(ii) Sum of 3 consecutive numbers: Always divisible by 3. The parity depends on the starting number; it can be even or odd but the sum is always a multiple of 3 because the three remainders modulo 3 add to 0.

(iii) Sum of 4 consecutive numbers: Always even.
Reason: There are two even and two odd numbers; even + even = even and odd + odd = even, so total is even.

Q2. What is the difference between the two successive sums in each box? Is it the same throughout?
Ans: (i) Sum of 2 consecutive numbers: 
Differences:
5 – 3 = 2
7 – 5 = 2
9 – 7 = 2.
The difference is always 2.
(ii) Sum of 3 consecutive numbers: 
Differences:
9 – 6 = 3
12 – 9 = 3
15 – 12 = 3
The difference is always 3.
(iii) Sum of 4 consecutive numbers:
Differences:
14 – 10 = 4
18 – 14 = 4
22 – 18 = 4
The difference is always 4.
So, the difference between successive sums is the same throughout each box because each successive group shifts every number by 1, increasing the total by the count of numbers.

Q3. What will be the difference between two successive sums for:
(a) 5 consecutive numbers 
(b) 6 consecutive numbers
Ans:
(a) Sum of 5 consecutive numbers:
1 + 2 + 3 + 4 + 5 = 15
2 + 3 + 4 + 5 + 6 = 20
3 + 4 + 5 + 6 + 7 = 25
Difference: 20 – 15 = 5, 25 – 20 = 5.
The difference is 5 because each new group adds 1 to each of the five numbers.
(b) Sum of 6 consecutive numbers:
1 + 2 + 3 + 4 + 5 + 6 = 21
2 + 3 + 4 + 5 + 6 + 7 = 27
3 + 4 + 5 + 6 + 7 + 8 = 33
Difference: 27 – 21 = 6, 33 – 27 = 6.
The difference is 6 for the same reason: each of the six numbers increases by 1.

Page 46

Let us see some more interesting patterns in sums.

Notice how the sums of 3, 4, and 5 consecutive numbers are related to the numbers being added. 

Use your understanding to find the following sums without adding the numbers directly.
(a) 67 + 68 + 69   
(b) 24 + 25 + 26+ 27 
(c) 48 + 49 + 50 + 51 + 52
(d) 237 + 238 + 239 + 240 + 241 + 242
Sol: 

Expanded solutions:

(a) 67 + 68 + 69 = 3 × 68 = 204 (middle number × 3).

(b) 24 + 25 + 26 + 27 = 4 × 25.5 = 102 (average × count).

(c) 48 + 49 + 50 + 51 + 52 = 5 × 50 = 250 (middle number × 5).

(d) 237 + 238 + 239 + 240 + 241 + 242 = 6 × 239.5 = 1,437 (average × 6).

Tip: For a sequence of consecutive numbers, multiply the average (middle value or average of two middle values) by the number of terms to get the sum quickly.

Page 48

Let Us Solve

Q1. Find the following sums. Try not to write TTh, Th, H, T, and O at the top. Align the digits carefully.
(a)  238 + 367
(b) 1,234 + 12,345
(c) 12 + 123
(d) 46,120 + 12,890
(e) 878 + 8,789
(f) 1,749 + 17,490
Ans:
(a) 238 + 367

= 605.

(b) 1,234 + 12,345

 = 13,579.

(c) 12 + 123

 = 135.

(d) 46,120 + 12,890

 = 59,010.

(e) 878 + 8,789

 = 9,667.

(f) 1,749 + 17,490

 = 19,239.

Method note: Align digits by place value (units under units, tens under tens, etc.) and add from right to left carrying where needed.

Q2. The great Indian road trip!
Nazrana and her friends planned a road trip across India, starting from Delhi. They first drove to Mumbai, then Goa, then Hyderabad, and finally Puri.
Look at the distances marked on the map and help them find the total distance travelled.

Ans: In the given map, the distances between the cities are as follows:
Delhi to Mumbai = 1600 km.
Mumbai to Goa = 590 km
Goa to Hyderabad = 670 km
Hyderabad to Puri = 1055 km
Total distance = 1600 + 590 + 670 + 1055 = 3915 km.
The total distance travelled by Nazrana and her friends is 3,915 km.

Q3. Find 2 numbers among 5,205, 6,220, 7,095, 8,455, and 4,840 whose sum is closest to the following.

(а) 10,000
Ans: 5,205 + 4,840 = 10,045.
This sum is the closest to 10,000 among all pairs; the difference is 45.

(b) 15,000
Ans: 6,220 + 8,455 = 14,675.
The difference from 15,000 is 325; this is the closest possible pair.

(c) 13,000
Ans: 8,455 + 4,840 = 13,295.
The difference from 13,000 is 295; this is the closest among pairs.

(d) 16,000
Ans: 7,095 + 8,455 = 15,550.
The difference from 16,000 is 450; this is the nearest pair available.

Pages 50-52

Q1. Subtract the following. Try not to write TTh, Th, H, T, and O at the top. Align the digits carefully.
(a) 4,578 – 2,222
Ans:

 4,578 – 2,222 = 2,356.

(b) 15,324- 11,780
Ans:

15,324 – 11,780 = 3,544.

(c) 5,423 – 423
Ans:

5,423 – 423 = 5,000.

(d) 123 – 12
Ans:

123 – 12 = 111.

(e) 77,777 – 777
Ans:

77,777 – 777 = 77,000.

(f) 826 – 752
Ans:

826 – 752 = 74.

Q2. Mary’s train journey to Delhi.
Mary is on a train journey. She starts from Kolkata with ₹12,540.
She spends ₹3,275 on food and other expenses during her trip to Varanasi. In Varanasi, her uncle gives her a gift worth ₹4,900. She then travels to Delhi, spending ₹2,645 on the train ticket. She spends ₹1,275 on souvenirs in Delhi. How much money is Mary left with at the end of the Delhi trip?

Ans: Mary starts her journey from Kolkata with ₹12,540.
She spent ₹3,275 on food and other expenses to reach Varanasi.
In Varanasi, her uncle gave her ₹4,900 as a gift.
She then spent ₹2,645 on a train ticket to Delhi.
In Delhi, she bought souvenirs worth ₹1,275.
Total money she had after receiving gift = ₹12,540 + ₹4,900 = ₹17,440.
Total expenses during the journey = ₹3,275 + ₹2,645 + ₹1,275 = ₹7,195.
Money left with Mary = ₹17,440 – ₹7,195 = ₹10,245.
Thus, Mary had ₹10,245 left after completing her Delhi trip.

Q3. Members of a school council have raised ₹70,500. They plan to set up a Maths Lab with some games and models worth ₹39,785, buy library books worth ₹9,545, and purchase sports equipment worth ₹19,548. 
(a) Estimate whether the school council has raised enough money to make the purchases. Share your thoughts in class. 
(b) Check your estimate with calculations.
Ans: (a) Estimated amount required for Maths Lab with games and models (₹39,785) ≈ ₹40,000.
Estimated amount required for library books (₹9,545) ≈ ₹10,000.
Estimated amount required for sports equipment (₹19,548) ≈ ₹20,000.
Total estimated amount = ₹40,000 + ₹10,000 + ₹20,000 = ₹70,000.
Amount raised = ₹70,500.
So, the school council has enough money according to the estimate.

(b) Now, calculating the actual total cost:
₹39,785 + ₹9,545 + ₹19,548 = ₹68,878.
Money raised by the school council = ₹70,500.
Balance amount = ₹70,500 – ₹68,878 = ₹1,622.
Therefore, the school council will be left with ₹1,622 after all the purchases.

Q4. A truck can carry 8,250 kg of goods. A factory loads 3,675 kg of cement and 2,850 kg of steel on it.
(a) What is the total weight loaded onto the truck?
(b) How much more weight can the truck carry before reaching its maximum capacity?
Ans: (a) Weight of cement = 3,675 kg.
Weight of steel = 2,850 kg.
Total weight of cement and steel = 3,675 kg + 2,850 kg = 6,525 kg.
So, total weight loaded onto the truck = 6,525 kg.

(b) Maximum capacity of truck = 8,250 kg.
Remaining capacity = 8,250 kg – 6,525 kg = 1,725 kg.
The truck can carry 1,725 kg more before reaching its maximum capacity.

Quick Sums and Differences

Sukanta likes the numbers 10, 100, 1,000, and 10,000. He wants to figure out what number he should add to a given number such that the sum is 100 or 1,000. Help him fill in the blanks with an appropriate number.
59 + _____ = 100
Try this method for the number 59.
Ans: 

Method: Subtract the given number from the target. For 59 to reach 100, 100 – 59 = 41. So 59 + 41 = 100.

Now, use this method to solve the following.
877 + ________ = 1,000 and 666 + ________ = 1,000
4,103 + ________ = 10,000 and 5,555 + ________ = 10,000

Ans:

877 + 123 = 1,000.

666 + 334 = 1,000.

4,103 + 5,897 = 10,000.

5,555 + 4,445 = 10,000.

Will this method work if the units digit is 0? What do you think? What other methods can you use to find the missing number to fill in the blanks? Share your thoughts in the class.

(a) 180 + ________ = 1,000

Ans:

180 + 820 = 1,000.

(b) 760 + ________ = 1,000

Ans:

760 + 240 = 1,000.

(c) 400 + ________ = 1,000

Ans:

400 + 600 = 1,000.

Namita likes the number 9. She wants to subtract 9 or 99 from any number. Find a way to quickly subtract 9 or 99 from any number.

(a) 67 – 9 = _____

Ans:

67 – 9 = 58.

(b) 83 – 9 = _____

Ans:

83 – 9 = 74.

(c) 144 – 9 = _____

Ans: 

144 – 9 = 135.

(d) 187 – 99 = _____

Ans:

187 – 99 = 88.

(e) 247 – 99 = _____

Ans:

247 – 99 = 148.

(f) 763 – 99 = _____

Ans:

763 – 99 = 664.

Rule: To subtract 9 or 99 from any number: subtract 10 or 100 first, then add 1 to the result. For example, to do 67 – 9, compute 67 – 10 = 57 and then add 1 to get 58.

Namita wonders if she can get 9 or 99 as the answer to any subtraction problem. Find a way to get the desired answer.
(a) 32 – _____ =9
(b) 66 – _____ =9
(c) 877 – _____ = 99
(d) 666 – _____ = 99
Ans: This is reverse subtraction. 
We are given the starting number and the difference and must find the number that was subtracted. 
Do: Missing number = Starting number – Difference.
(a) 32 – _____ = 9
_____ = 32 – 9 = 23.
(b) 66 – _____ = 9
_____ = 66 – 9 = 57.
(c) 877 – _____ = 99
_____ = 877 – 99 = 778.
(d) 666 – _____ = 99
_____ = 666 – 99 = 567.