Q1. Can a triangle exist with side lengths 5 cm, 7 cm, and 13 cm? Verify using the triangle inequality.
Answer: To determine if a triangle exists with side lengths 5 cm, 7 cm, and 13 cm, we apply the triangle inequality: the sum of any two sides must be greater than the third side.
– Check 1: 5 + 7 = 12 < 13 (not satisfied).
– Check 2: 5 + 13 = 18 > 7 (satisfied).
– Check 3: 7 + 13 = 20 > 5 (satisfied).
– Since one condition (5 + 7 > 13) is not satisfied, a triangle cannot exist with these side lengths.
Q2. Check if a triangle exists with side lengths 6 m, 8 m, and 12 m.
Answer: Apply the triangle inequality:
Check 1: 6 + 8 = 14 > 12 (satisfied).
– Check 2: 6 + 12 = 18 > 8 (satisfied).
– Check 3: 8 + 12 = 20 > 6 (satisfied).
– All conditions are satisfied, so a triangle exists with these side lengths.
Q3: For a triangle with side lengths 3 cm, 4 cm, and 6 cm, confirm if it satisfies the triangle inequality.
Answer: The triangle inequality states that each side must be less than the sum of the other two:
– Check 1: 3 + 4 = 7 > 6 (satisfied).
– Check 2: 3 + 6 = 9 > 4 (satisfied).
– Check 3: 4 + 6 = 10 > 3 (satisfied).
Since all conditions are met, the side lengths 3 cm, 4 cm, and 6 cm satisfy the triangle inequality, confirming a triangle can exist.
Q4: Does a triangle with side lengths 2 mm, 9 mm, and 15 mm satisfy the triangle inequality?
Answer: Apply the triangle inequality:
– Check 1: 2 + 9 = 11 < 15 (not satisfied).
– Check 2: 2 + 15 = 17 > 9 (satisfied).
– Check 3: 9 + 15 = 24 > 2 (satisfied).
Since one condition fails (2 + 9 > 15), the side lengths do not satisfy the triangle inequality, and no triangle exists.
Q5: In triangle PQR, if angle P = 40 degrees and angle Q = 80 degrees, find the measure of angle R.
Answer: The angle sum property states that the sum of angles in a triangle is 180 degrees.
Angle P + Angle Q + Angle R = 180 degrees.
40 + 80 + Angle R = 180.
120 + Angle R = 180.
Angle R = 180 – 120 = 60 degrees.
Thus, Angle R = 60 degrees.
Q6: In triangle XYZ, Angle X = 55 degrees and Angle Y = 65 degrees. Find the exterior angle at vertex Z.
Answer: First, find Angle Z using the angle sum property:
Angle X + Angle Y + Angle Z = 180 degrees.
55 + 65 + Angle Z = 180.
120 + Angle Z = 180.
Angle Z = 60 degrees.
The exterior angle at Z is supplementary to Angle Z (they form a straight angle):
Q7: A spider is at one corner of a rectangular box and needs to reach the opposite corner by walking on the surface. If the box dimensions are 10 cm, 12 cm, and 15 cm, calculate the length of the shortest path.
Ans: The shortest path is found by unfolding the box. Possible nets include:
Q8: In triangle LMN, if angle L = 50 degrees and angle M = 70 degrees, find the exterior angle at vertex N and determine its relationship with angles L and M.
Ans: First, find angle N using the angle sum property: Angle L + Angle M + Angle N = 180 degrees. Substituting, 50 + 70 + Angle N = 180. This simplifies to 120 + Angle N = 180, so Angle N = 60 degrees. The exterior angle at N is supplementary to Angle N: Exterior angle = 180 – 60 = 120 degrees. To find the relationship, note that the exterior angle at N equals the sum of the opposite interior angles (L and M): 50 + 70 = 120 degrees. Thus, the exterior angle at N is 120 degrees, and it equals the sum of angles L and M.
Q9: In triangle XYZ, the exterior angle at vertex X is 130 degrees, and angle Y = 55 degrees. Find angle Z and analyze whether the triangle is acute-angled, right-angled, or obtuse-angled based on its angles.
Ans: To determine angle Z in triangle XYZ and classify the triangle, we proceed as follows: Calculate Angle X
The exterior angle at X is supplementary to angle X: Angle X = 180 – Exterior angle at X.
Substitute: Angle X = 180 – 130 = 50 degrees.
Calculate Angle Z
Use the angle sum property: Angle X + Angle Y + Angle Z = 180 degrees.
Substitute: 50 + 55 + Angle Z = 180.
Simplify: 105 + Angle Z = 180.
Solve: Angle Z = 180 – 105 = 75 degrees.
Classify the Triangle
List angles: Angle X = 50 degrees, Angle Y = 55 degrees, Angle Z = 75 degrees.
Definitions: Acute-angled: all angles < 90 degrees. Right-angled: one angle = 90 degrees. Obtuse-angled: one angle > 90 degrees.
Check: 50, 55, 75 are all less than 90 degrees.
Conclusion: The triangle is acute-angled.
Therefore, angle Z is 75 degrees, and the triangle is acute-angled.
Q10: In triangle UVW, if angle U = 45 degrees and angle V = 60 degrees, find the third angle W and determine the ratio of angles U, V, and W.
Ans: To find angle W in triangle UVW and the ratio of angles U, V, and W, we proceed as follows: Calculate Angle W
Use the angle sum property: Angle U + Angle V + Angle W = 180 degrees.
Substitute: 45 + 60 + Angle W = 180.
Simplify: 105 + Angle W = 180.
Solve: Angle W = 180 – 105 = 75 degrees.
Determine Ratio of Angles
List angles: Angle U = 45 degrees, Angle V = 60 degrees, Angle W = 75 degrees.
Express ratio: 45 : 60 : 75.
Find the greatest common divisor: 45, 60, 75 are divisible by 15.
Q1: Priya is experimenting with sums of odd numbers. What happens to the parity of the sum when she adds (a) 3 odd numbers (b) 4 odd numbers?
Sol: For 3 odd numbers (e.g., 1 + 3 + 5 = 9), the sum is odd because each odd number contributes one unpaired unit, and 3 unpaired units remain unpaired.
For 4 odd numbers (e.g., 1 + 3 + 5 + 7 = 16), the sum is even because the 4 unpaired units can be paired. (a) Odd, (b) Even
Q2: Anjali has 5 boxes and number cards with odd numbers (1, 3, 5, 7, 9, …). She needs to place one card in each box so they sum to 20. Is this possible?
Sol: The sum of 5 odd numbers is always odd (since each odd number adds an unpaired unit, and 5 unpaired units remain odd). Since 20 is even, it’s impossible. No, it’s not possible.
Q3: Rohan observes the Virahãnka-Fibonacci sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … What is the parity of the 7th term (21)?
Sol: The 7th term is 21. Since 21 ÷ 2 = 10 remainder 1, it’s odd.
Q4: In the cryptarithm P + P + P = QP, where P and Q are digits, and QP is a two-digit number, find P and Q.
Sol: Let P be a digit (0–9). Then, 3P = QP (a two-digit number with units digit P). This implies 3P = 10Q + P. Subtract P: 2P = 10Q. Thus, P = 5Q. Since P and Q are digits, Q = 1, P = 5 (as 5 × 1 = 5). Check: 5 + 5 + 5 = 15, and QP = 15 (Q = 1, P = 5).
Q5: Meena wants to write 6 as a sum of 1s and 2s in all possible ways. How many ways can she do this?
Sol: This follows the Virahãnka-Fibonacci sequence, where the number of ways to write n as a sum of 1s and 2s is the (n+1)th term. For n = 6, the 7th term is 13 (sequence: 1, 2, 3, 5, 8, 13, 21, …). Hence there are 13 ways
Q6: Siblings Vikram and Nisha, born one year apart, claim the sum of their ages is 45. Is this possible?
Sol: Their ages are consecutive (e.g., n and n+1). The sum n + (n+1) = 2n + 1 is always odd (since 2n is even, adding 1 makes it odd). Since 45 is odd, it’s possible. Example: 22 + 23 = 45. Yes, it’s possible.
Q7: What is the 50th odd number in the sequence 1, 3, 5, 7, …?
Sol: The nth odd number is given by the formula 2n – 1. For n = 50, 2 × 50 – 1 = 100 – 1 = 99.
Q8: Arjun has a 15 × 22 grid. What is the parity of the number of small squares in this grid?
Sol: The number of squares is 15 × 22. Parity depends on the factors: 15 is odd, 22 is even. Odd × Even = Even (e.g., 15 × 22 = 330, which is even). Hence, Even
Q1: What is the measure of each angle formed when two perpendicular lines intersect?
Ans: When two lines are perpendicular, they intersect at right angles, each measuring 90°.
This is because when two lines cross and are perpendicular, they form 4 equal angles at the point of intersection.
A full circle around a point measures 360°.
So,
360°÷4=90°
Therefore, each of the four angles is 90°, making the lines perpendicular.
Q2: In a transversal intersecting two parallel lines, if one corresponding angle is 65°, what is the measure of the other corresponding angle in the pair?
Ans: When a transversal intersects two parallel lines, corresponding angles are always equal.
So, if one corresponding angle is 65°, then the other corresponding angle will also be:
65°
Q3:A transversal intersects two lines, forming an interior angle on the same side of the transversal that sums to 180°. Prove that the two lines are parallel, and find the measures of the alternate interior angles if one is 75°.
Ans: Step 1: Proving the lines are parallel
When a transversal cuts two lines and the interior angles on the same side of the transversal add up to 180°, it means the two lines are parallel.
This is based on the Converse of the Consecutive Interior Angles Theorem, which states:
If the interior angles on the same side of a transversal are supplementary (sum = 180°), then the lines are parallel.
So, since the given interior angles add up to 180°, the lines are parallel.
Step 2: Finding the alternate interior angles
We are told that one alternate interior angle is 75°.
In parallel lines cut by a transversal, alternate interior angles are equal.
So, the other alternate interior angle is also:
75°
Q4: If a transversal intersects two parallel lines and one exterior angle is 130°, what is the measure of its corresponding exterior angle?
Ans: When a transversal intersects two parallel lines, corresponding angles, including exterior angles, are equal. If one exterior angle is 130°, its corresponding exterior angle is also 130°.
Q5: Fold a square paper horizontally twice to create three creases. Describe the relationship between these creases and the vertical edges of the paper, and explain why.
Ans: When a square paper is folded horizontally twice, it creates three horizontal creases.
These creases are:
Parallel to each other, and
Parallel to the top and bottom (horizontal) edges of the square paper.
Since the folds are made horizontally, they do not tilt or slant, and they run in the same direction as the horizontal edges.
The vertical edges of the paper are perpendicular to these creases. That means the creases and the vertical edges intersect at right angles (90°).
Q8: If a number machine follows the rule “3a – b”, find the output when a = 5 and b = 2.
Ans: Output = 3 x 5 – 2 = 15 – 2 = 13.
Q9: A snail climbs “u” cm during the day and slips back “d” cm at night. Write an expression for its net progress after 5 days and 5 nights.
Ans: Net progress = 5 x (u – d).
Q10:A train stops for 2 minutes at each of 3 stations. If the travel time between stations is “t” minutes, write an expression for the total journey time.
Ans: Total time = Travel time + Stop time = 3 x t + 6 minutes.
Q11: Simplify 4(x + y) – y.
Ans: Distribute and combine like terms: 4x + 4y – y = 4x + 3y.
Q12:If 10y – 3 and 10(y – 3) are two expressions, compare their values when y = 4.
Ans: 10y – 3 = 10 x 4 – 3 = 37. 10(y – 3) = 10 x (4 – 3) = 10. They are not equal.
Q1: Rewrite the expression 45 − (12 + 8) by removing the brackets and explain the changes in the signs of the terms inside.
Ans: When removing the brackets preceded by a negative sign, the signs of the terms inside change. Thus, 45 − (12 + 8) becomes 45 − 12 − 8. On solving , 45 – 12 – 8 = 25.
Q2: Identify the terms in the expression 7 × 3 + 10 − 5 and evaluate its value.
Ans: The terms are 7 × 3, 10, and −5. Evaluating the expression:
7 × 3 = 21, then 21 + 10 = 31, and finally 31 − 5 = 26.
Thus, the value of the expression is 26.
Q3: Compare the expressions 25 + 13 and 24 + 14 without calculating their values. Explain your reasoning.
Ans: The first expression 25 + 13 can be written as (24 + 1) + (14 − 1), which simplifies to 24 + 14. Thus, both expressions are equal.
Q4: Write an expression for the total cost if 5 notebooks cost ₹12 each and 3 pens cost ₹8 each. Also, identify the terms.
Ans: The expression is 5 × 12 + 3 × 8. The terms are 5 × 12 and 3 × 8. The total cost is 60 + 24 = ₹84.
Q5: Simplify the expression 100 − (30 − 10) by removing the brackets and justify the sign changes.
Ans: Removing the brackets gives 100 − 30 + 10. The sign of 30 becomes negative, and −10 becomes positive. Hence the equation becomes, 100 – 30 + 10 = 110 – 30 = 80. The value is 80.
Q6: A shopkeeper sells apples at ₹20 per kg and oranges at ₹15 per kg. A customer buys 3 kg of apples and 2 kg of oranges. (a) Write an expression for the total cost. (b) Compare the two expressions and verify if they yield the same value.
Ans: (a) The expression for the total cost is 3 × 20 + 2 × 15. (b) Let’s write two different expressions:
First expression: 20 × 3 + 15 × 2
Second expression (actual total): ₹60 + ₹30
Now check if they are equal:
20 × 3 = 60
15 × 2 = 30
So, 60 + 30 = ₹90
Both expressions give the same result.
Q7: Consider the expressions 50 − (15 + 5) and 50 − 15 − 5. (a) Evaluate both expressions. (b) Explain why the second expression is derived from the first by removing brackets. (c) What happens if the brackets are preceded by a positive sign, such as 50 + (15 − 5)?
Ans: (a) Both expressions evaluate to 30. As 50 − (15 + 5) = 50 – 15 – 5 = 50 – 20 = 30 and 50 − 15 − 5 = 50 – 20 = 30 (b) When brackets are removed, the negative sign before the brackets changes the signs of the terms inside. Hence, 50 − (15 + 5) = 50 − 15 − 5. (c) If the brackets are preceded by a positive sign, the signs inside remain unchanged: 50 + (15 − 5) = 50 + 15 − 5 = 60.
Q8: A train has 8 compartments with 12 seats each and 5 compartments with 15 seats each. (a) Write an expression to find the total number of seats. (b) Use the distributive property to show that both expressions yield the same result.
Ans:
(a) The expressions are:
8 × 12 + 5 × 15
(b) The distributive property does not apply here directly. The correct evaluation is:
8 × 12 = 96 and 5 × 15 = 75, so total seats = 96 + 75 = 171.
Q9: A snail climbs 4 cm up a pole during the day and slips back 2 cm at night. The pole is 12 cm high. (a) Write an expression to represent the snail’s progress after 3 days. (b) Use the expression to find how many days it will take for the snail to reach the top.
Ans: (a) After 3 days, the snail’s progress is 3 × (4 − 2) = 6 cm. (b) On the 5th day, the snail reaches 4 × (4 − 2) + 4 = 12 cm (since it doesn’t slip back after reaching the top).
Q10: Riya is saving money to buy a toy that costs ₹250. She saves ₹20 every week from her pocket money. (a) Write an expression to represent how much she saves after 5 weeks. (b) How many weeks will it take for her to save enough money to buy the toy?
Ans:
(a) Amount saved in 5 weeks = 5 × 20 = ₹100 (b) To find the number of weeks to save ₹250: Let the number of weeks be w. 20 × w = 250 w = 250 ÷ 20 = 12.5 weeks So, she will need 13 weeks to save enough money.
Ans: 12,34,56,789 Explanation: In the Indian system we group digits as crore, lakh, thousand, and hundreds. Here, “twelve crore” gives 12 in the crore place, “ thirty-four lakh” gives 34 in the lakh place, “fifty-six thousand” gives 56 in the thousand place, and “seven hundred eighty-nine” fills the last three digits .
Q2: Convert the number 9,45,32,106 into words in the Indian system.
Ans: Nine crore forty-five lakh thirty-two thousand one hundred six
Q3: Using the digits 0, 1, 3, 5, 6, 8, 9, form the greatest and the smallest seven-digit numbers.
Ans:Greatest number:
To form the greatest number, we arrange the digits in descending order:
9, 8, 6, 5, 3, 1, 0 So, the greatest seven-digit number is: 9,865,310
Smallest number:
To form the smallest number, we arrange the digits in ascending order, but a number cannot start with 0. So, we start with the smallest non-zero digit, which is 1, and then arrange the remaining digits in ascending order:
1, 0, 3, 5, 6, 8, 9 So, the smallest seven-digit number is: 10,35,689
Q4: Express 7,56,83,291 in the international system and write it in words.
Ans: 75,683,291; seventy-five million six hundred eighty-three thousand two hundred ninety-one Explanation: In the international system we group digits by threes from the right: 75 | 683 | 291. Reading: 75 million 683 thousand 291; in words: seventy-five million six hundred eighty-three thousand two hundred ninety-one.
Q5: Identify the digit in the ten-thousands place of 4,28,75,160 and state its place value.
Ans: Digit = 7; place value = 7 × 10,000 = 70,000 Explanation: In 4,28,75,160 the digits from right are units (0), tens (6), hundreds (1), thousands (5), ten-thousands (7). Thus the ten-thousands digit is 7, whose value is 70,000. .
Q6: Compare using “>” or “<”: 2,34,56,789 ___ 2,43,56,789.
Ans: 2,34,56,789 < 2,43,56,789 Explanation: Compare place by place from the left: Both have 2 crore, but in the lakh place 34 < 43, so 2,34,56,789 is less than 2,43,56,789. .
Q7: Arrange in descending order:
• 12,34,56,789
• 12,345,678
• 1,23,45,678
• 1,234,567
Ans: To arrange the numbers in descending order, we compare them based on the number of digits and place value.
Let’s write the numbers in standard international format for better comparison:
Q8: Which is greater, 0.345 million or 34.5 lakh, and by how much?
Ans: 34.5 lakh is greater by 3,105,000 Explanation: 0.345 million = 345,000. 34.5 lakh = 3,450,000. Subtracting: 3,450,000 − 345,000 = 3,105,000. So 34.5 lakh exceeds 0.345 million by 3,105,000.
Q9: True or False: 1 billion is equal to 100 crore in the Indian system.
Ans: True; 1 billion = 1,000 million = 100 crore Explanation: Since 1 million = 10 lakh, 1 billion = 1,000 million = 10,000 lakh = 100 crore.
Q10: Round off 6,24,894 to the nearest thousand.
Ans: 6,25,000 Explanation: Check the hundreds digit (8) in 894 is greater than 5, so thousands place (24 thousand) rounds up to 25 thousand. Other digits become zeros: 6,25,000. .
Q11: Round off 8,23,49,161 to the nearest ten lakh.
Ans: Step 1: Identify the ten lakh place. In 8,23,49,161, the ten lakh digit is 2 (in 23 lakh).
Step 2: Look at the digit in the lakh place, which is 3.
Since 3 < 5, we do not increase the ten lakh digit.
Replace all digits after the ten lakh place with zeros.
So, 8,23,49,161 rounded to the nearest ten lakh is:
8,20,00,000
Q12: Round off 7,68,429 to the nearest ten thousand.
Ans: 7,70,000 Explanation: Check the thousands digit (8) in 8,429 is greater than 5, so ten-thousands place (6 ten-thousands) rounds up to 7 ten-thousands. Result: 7,70,000.
Q13: How many lakhs make a billion?
Ans: 10,000 Explanation:
We know that:
1 billion = 1,000,000,000
1 lakh = 1,00,000
Now, to find how many lakhs make 1 billion, we divide:
1,000,000,000 ÷ 1,00,000 = 10,000
Q14: A delivery van travels 56,789 km in Year 1 and 67,890 km in Year 2. What is the total distance covered in two years?
Ans: 1,24,679 Km Explanation: 56,789 + 67,890 = 1,24,679 km
Q15: A museum receives a donation of ₹ 3.27 crore. Express this amount in rupees and in the international system (with words).
Ans: ₹ 32,700,000; thirty-two million seven hundred thousand
Explanation: Since 1 crore = 10,000,000, 3.27 crore = 3.27 × 10,000,000 = 32,700,000 rupees. In international grouping that is 32 | 700 | 000 → 32 million 700 thousand, read as thirty-two million seven hundred thousand.
