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.