Class 10 Maths Chapter 1 Real Numbers Practice Paper 5

Q1. The exponent of 3 in the prime factorization of 5832 is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: Answer: (a) 3
Solution: 5832 = 8 × 729 = 2³ × 3⁶
Therefore, the exponent of 3 is 6. Answer should be corrected to (d) 6.

Q2. If two positive integers m and n are written as m = p²q³ and n = p³q², where p and q are prime numbers,
then the LCM (m, n) is:
(a) pq
(b) p²q²
(c) p³q³
(d) p⁴q⁴

Answer: Answer: (c) p³q³
Solution: m = p²q³, n = p³q²
LCM = Product of highest powers of all prime factors = p³q³

Q3. The HCF and the LCM of 16, 20, 24 respectively are
(a) 4, 240
(b) 8, 480
(c) 4, 480
(d) 480, 4

Answer: Answer: (a) 4, 240
Solution: 16 = 2⁴, 20 = 2² × 5, 24 = 2³ × 3
HCF = 2² = 4, LCM = 2⁴ × 3 × 5 = 240

Q4. If the HCF of 91 and 156 is expressible in the form 91m - 156, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2 Solution: Using Euclidean algorithm:
156 = 91 × 1 + 65
91 = 65 × 1 + 26
65 = 26 × 2 + 13
26 = 13 × 2 + 0
HCF = 13 = 91 × 2 - 156 × 1, so m = 2

Q5. Ravi has 56 cm long green and 96 cm long yellow ribbon. He cuts each ribbon into pieces such that all pieces are of equal length. What is the length of each piece?
(a) 8 cm as it is the HCF of 56 and 96
(b) 8 cm as it is the LCM of 56 and 96
(c) 28 cm as it is the LCM of 56 and 96
(d) 28 cm as it is the HCF of 56 and 96

Answer: Answer: (a) 8 cm as it is the HCF of 56 and 96
Solution: To cut ribbons into equal pieces, we need HCF(56, 96) = 8 cm

Q6. The largest number which divides 98 and 147 leaving remainders 6 and 9 respectively is
(a) 23
(b) 46
(c) 92
(d) 138

Answer: Answer: (b) 46
Solution: Required number = HCF(98-6, 147-9) = HCF(92, 138) = 46

Q7. If 7560 = 2m × 3ⁿ × 5k × 7p, then the value of m + n + k + p is
(a) 6
(b) 7
(c) 8
(d) 9

Answer: Answer: (b) 7
Solution: 7560 = 2³ × 3³ × 5 × 7
So m = 3, n = 3, k = 1, p = 1
m + n + k + p = 3 + 3 + 1 + 1 = 8. Answer should be (c) 8.

Q8. If a = 3² × 7, b = 3³ × 5 × 11, c = 5ⁿ × 11 and LCM (a, b, c) = 3³ × 5² × 7 × 11, then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2
Solution: a = 3² × 7¹, b = 3³ × 5¹ × 11¹, c = 5ⁿ × 11¹
LCM = 3³ × 5² × 7¹ × 11¹
For this to be true, n must be 2.

Q9. In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
Assertion (A): If product of two numbers is 7280 and their HCF is 14, then their LCM is 520.
Reason (R): HCF × LCM = Product of two numbers.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Answer: Answer: (a) Both A and R are true and R is the correct explanation of A
Solution: Product = HCF × LCM, so LCM = 7280/14 = 520. Both statements are true.
R correctly explains the formula used in A.

Q10. In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
Assertion (A): 12ⁿ ends with the digit zero, where n is natural number.
Reason (R): Any number ends with digit zero, if its prime factorization contains both 2 and 5 as factors.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true. Questions 11 to 14 carry 2 marks each.

Answer: Answer: (d) A is false but R is true
Solution: 12ⁿ = (2² × 3)ⁿ = 2²ⁿ × 3ⁿ. This never contains factor 5, so cannot end in 0.
Statement R is correct about numbers ending in 0.

Q11. Explain why 5 × 7 × 11 + 11 and 11 × 13 × 17 + 13 × 11 are composite numbers.

Answer: Solution:
First number: 5 × 7 × 11 + 11 = 385 + 11 = 396 = 2² × 3² × 11 (composite)
Second number: 11 × 13 × 17 + 13 × 11 = 2431 + 143 = 2574 = 2 × 3 × 11 × 39 (composite)
Both numbers have factors other than 1 and themselves.

Q12. Two numbers are in the ratio 4 : 5 and their LCM is 320. What is the HCF of these numbers?

Answer: Solution:
Let the numbers be 4x and 5x where x is their HCF.
LCM = (4x × 5x)/x = 20x
Given: 20x = 320, so x = 16
Therefore, HCF = 16

Q13. Show that any number of the form 15ⁿ, where n ∈ N can never end with digit 0. (2017)

Answer: Solution:
15ⁿ = (3 × 5)ⁿ = 3ⁿ × 5ⁿ
For a number to end with 0, it must be divisible by 10 = 2 × 5
Since 15ⁿ contains factor 5 but no factor of 2, it can never end with 0.

Q14. The LCM of two numbers is 7 times their HCF. The sum of LCM and HCF is 640. Find the HCF of the two numbers. Questions 15 to 17 carry 3 marks each.

Answer: Solution:
Let HCF = h, then LCM = 7h
Given: h + 7h = 640
8h = 640
h = 80
Therefore, HCF = 80

Q15. Prove that √7 is an irrational number. (2023)

Answer: Solution:
Proof by contradiction:
Assume √7 is rational, so √7 = p/q where p, q are integers with no common factors.
Squaring: 7 = p²/q², so 7q² = p²
This means p² is divisible by 7, so p is divisible by 7.
Let p = 7k, then 7q² = 49k², so q² = 7k²
This means q is also divisible by 7. But this contradicts our assumption that p and q have no common factors.
Therefore, √7 is irrational.

Q16. 4 Bells toll together at 7.00 am. They toll after 8, 12, 15 and 18 seconds respectively. How many times will they toll together again in the next 3 hours?

Answer: Solution: The bells will toll together at intervals equal to LCM(8, 12, 15, 18)
LCM = 2³ × 3² × 5 = 360 seconds = 6 minutes
In 3 hours = 180 minutes
Number of times = 180/6 = 30
So they will toll together 30 times after 7:00 AM.

Q17. Given that √7 is irrational, prove that 4 + 3√7 is irrational. (CBSE Sample Paper 2022)
Questions 18 carry 5 marks.

Answer: Solution:
Proof by contradiction:
Assume 4 + 3√7 is rational = r
Then 3√7 = r - 4 (rational)
So √7 = (r - 4)/3 (rational)
But this contradicts the given fact that √7 is irrational.
Therefore, 4 + 3√7 is irrational.

Q18. (a) Find the largest possible positive integer that divides 156, 208 and 286 leaving remainder 8, 10 and 12 respectively. (3)
(b) An army contingent of 864 soldiers is to march behind an army band of 48 members in a Republic Day parade. The two groups are to march in the same number of columns. What is the maximum number of columns they can march? (2) Questions 19 to 20 carry 4 marks each.

Answer: Solution:
(a) Required number = HCF(156-8, 208-10, 286-12) = HCF(148, 198, 274)
148 = 2² × 37, 198 = 2 × 3² × 11, 274 = 2 × 137
HCF = 2
(b) Maximum columns = HCF(864, 48)
864 = 2⁵ × 3³, 48 = 2⁴ × 3
HCF = 2⁴ × 3 = 48 columns

Q19. A morning walk may help improve your mental clarity and ability to focus throughout the day. A recent study found that amongst older adults, those who started their days with a morning walk improved their cognitive function, compared to those who remained sedentary. Walking may also help you think more creatively. In a morning walk three students step off together, their steps measure 84 cm, 96 cm and 108 cm respectively.
(i) What is the HCF of 84 and 108? (1)
(ii) Find the sum of exponents of the prime factors of total distance. (1)
(iii) What is the minimum distance each should walk so that he can cover the distance in complete steps? (2)

Answer: Solution:
(i) HCF(84, 108): 84 = 2² × 3 × 7, 108 = 2² × 3³
HCF = 2² × 3 = 12
(ii) Total distance = LCM(84, 96, 108)
LCM = 2⁵ × 3³ × 7 = 6048
Sum of exponents = 5 + 3 + 1 = 9
(iii) Minimum distance = LCM(84, 96, 108) = 6048 cm = 60.48 m

Q20. A family room is an informal, all purpose room in a house. The family room is designed to be a place where family and guests gather for group recreation like talking, reading, watching TV and other family activities. The length, breadth and height of a room are 10 m 80 cm, 8 m 40 cm and 6 m 60 cm.
(i) Determine the longest rod which can measure the three dimensions of the room exactly. (2)
(ii) What is LCM of the given three measurements? (1)
(iii) If the HCF (1080 and 840) = 120, then find LCM (1080 and 840). (1)

Answer: Solution:
(i) Convert to cm: 1080 cm, 840 cm, 660 cm
HCF(1080, 840, 660) = 60 cm
(ii) LCM(1080, 840, 660) = 30240 cm
(iii) Using HCF × LCM = Product of numbers
LCM = (1080 × 840)/120 = 7560
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