Class 10 Maths Chapter 1 Real Numbers Practice Paper 7

Q1. The exponent of 5 in the prime factorization of 9375 is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: Answer: (c) 5
Solution: 9375 = 3 × 3125 = 3 × 5⁵
Therefore, the exponent of 5 is 5.

Q2. If two positive integers u and v are written as u = r³s and v = rs⁴, where r and s are prime numbers, then
the LCM (u, v) is:
(a) rs
(b) r²s²
(c) r³s⁴
(d) r⁴s⁵

Answer: Answer: (c) r³s⁴
Solution: u = r³s¹, v = r¹s⁴
LCM = Product of highest powers of all prime factors = r³s⁴

Q3. The HCF and the LCM of 36, 48, 60 respectively are
(a) 12, 720
(b) 6, 1440
(c) 12, 1440
(d) 720, 12

Answer: Answer: (a) 12, 720
Solution: 36 = 2² × 3², 48 = 2⁴ × 3, 60 = 2² × 3 × 5
HCF = 2² × 3 = 12, LCM = 2⁴ × 3² × 5 = 720

Q4. If the HCF of 119 and 187 is expressible in the form 119m - 187, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2 Solution: Using Euclidean algorithm:
187 = 119 × 1 + 68
119 = 68 × 1 + 51
68 = 51 × 1 + 17
51 = 17 × 3 + 0
HCF = 17 = 119 × 2 - 187 × 1, so m = 2

Q5. Kiran has 72 cm long purple and 120 cm long orange ribbon. She cuts each ribbon into pieces such that all pieces are of equal length. What is the length of each piece?
(a) 12 cm as it is the HCF of 72 and 120
(b) 12 cm as it is the LCM of 72 and 120
(c) 24 cm as it is the LCM of 72 and 120
(d) 24 cm as it is the HCF of 72 and 120

Answer: Answer: (d) 24 cm as it is the HCF of 72 and 120 Solution: To cut ribbons into equal pieces, we need HCF(72, 120)
72 = 2³ × 3², 120 = 2³ × 3 × 5
HCF = 2³ × 3 = 24 cm

Q6. The largest number which divides 126 and 189 leaving remainders 9 and 15 respectively is
(a) 29
(b) 58
(c) 117
(d) 174

Answer: Answer: (b) 58
Solution: Required number = HCF(126-9, 189-15) = HCF(117, 174) = 58

Q7. If 9240 = 2m × 3ⁿ × 5k × 7p × 11q, then the value of m + n + k + p + q is
(a) 6
(b) 7
(c) 8
(d) 9

Answer: Answer: (b) 7
Solution: 9240 = 2³ × 3 × 5 × 7 × 11
So m = 3, n = 1, k = 1, p = 1, q = 1
m + n + k + p + q = 3 + 1 + 1 + 1 + 1 = 7

Q8. If a = 5² × 11, b = 5³ × 7 × 13, c = 7ⁿ × 13² and LCM (a, b, c) = 5³ × 7² × 11 × 13², then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2
Solution: a = 5² × 11¹, b = 5³ × 7¹ × 13¹, c = 7ⁿ × 13²
LCM = 5³ × 7² × 11¹ × 13²
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 9504 and their HCF is 18, then their LCM is 528.
Reason (R): The product of HCF and LCM of two numbers is equal to the product of the 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 = 9504/18 = 528. 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): 24ⁿ ends with the digit zero, where n is natural number.
Reason (R): For a number to end with zero, it must have both 2 and 5 as prime 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: 24ⁿ = (2³ × 3)ⁿ = 2³ⁿ × 3ⁿ. This never contains factor 5, so cannot end in 0.
Statement R is correct about numbers ending in 0.

Q11. Explain why 11 × 13 × 17 + 17 and 17 × 19 × 23 + 19 × 17 are composite numbers.

Answer: Solution:
First number: 11 × 13 × 17 + 17 = 2431 + 17 = 2448 = 2⁴ × 3² × 17 (composite)
Second number: 17 × 19 × 23 + 19 × 17 = 7429 + 323 = 7752 = 2³ × 3 × 17 × 19 (composite)
Both numbers have factors other than 1 and themselves.

Q12. Two numbers are in the ratio 6 : 8 and their LCM is 504. What is the HCF of these numbers?

Answer: Solution:
Let the numbers be 6x and 8x where x is their HCF.
LCM = (6x × 8x)/2x = 24x (since HCF(6,8) = 2)
Given: 24x = 504, so x = 21
Therefore, HCF = 21

Q13. Show that any number of the form 27ⁿ, where n ∈ N can never end with digit 0. (2017)

Answer: Solution:
27ⁿ = (3³)ⁿ = 3³ⁿ
For a number to end with 0, it must be divisible by 10 = 2 × 5
Since 27ⁿ contains only factor 3 (no factors of 2 or 5), it can never end with 0.

Q14. The LCM of two numbers is 5 times their HCF. The sum of LCM and HCF is 720. Find the HCF of the two numbers. Questions 15 to 17 carry 3 marks each.

Answer: Solution:
Let HCF = h, then LCM = 5h
Given: h + 5h = 720
6h = 720
h = 120
Therefore, HCF = 120

Q15. Prove that √13 is an irrational number. (2023)

Answer: Solution:
Proof by contradiction:
Assume √13 is rational, so √13 = p/q where p, q are integers with no common factors.
Squaring: 13 = p²/q², so 13q² = p²
This means p² is divisible by 13, so p is divisible by 13.
Let p = 13k, then 13q² = 169k², so q² = 13k²
This means q is also divisible by 13. But this contradicts our assumption that p and q have no common factors.
Therefore, √13 is irrational.

Q16. 4 Bells toll together at 5.00 am. They toll after 10, 14, 18 and 21 seconds respectively. How many times will they toll together again in the next 4 hours?

Answer: Solution: The bells will toll together at intervals equal to LCM(10, 14, 18, 21)
LCM = 2 × 3² × 5 × 7 = 630 seconds = 10.5 minutes
In 4 hours = 240 minutes
Number of times = 240/10.5 ≈ 22.86
So they will toll together 22 times after 5:00 AM.

Q17. Given that √13 is irrational, prove that 7 + 6√13 is irrational. (CBSE Sample Paper 2022)
Questions 18 carry 5 marks.

Answer: Solution:
Proof by contradiction:
Assume 7 + 6√13 is rational = r
Then 6√13 = r - 7 (rational)
So √13 = (r - 7)/6 (rational)
But this contradicts the given fact that √13 is irrational.
Therefore, 7 + 6√13 is irrational.

Q18. (a) Find the largest possible positive integer that divides 182, 247 and 338 leaving remainder 11, 13 and 17 respectively. (3)
(b) An army contingent of 1008 soldiers is to march behind an army band of 63 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(182-11, 247-13, 338-17) = HCF(171, 234, 321)
171 = 3² × 19, 234 = 2 × 3² × 13, 321 = 3 × 107
HCF = 3
(b) Maximum columns = HCF(1008, 63)
1008 = 2⁴ × 3² × 7, 63 = 3² × 7
HCF = 3² × 7 = 63 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 92 cm, 115 cm and 138 cm respectively.
(i) What is the HCF of 92 and 138? (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(92, 138): 92 = 2² × 23, 138 = 2 × 3 × 23
HCF = 2 × 23 = 46
(ii) Total distance = LCM(92, 115, 138)
LCM = 2² × 3 × 5 × 23 = 1380
Sum of exponents = 2 + 1 + 1 + 1 = 5
(iii) Minimum distance = LCM(92, 115, 138) = 1380 cm = 13.8 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 14 m 40 cm, 11 m 20 cm and 8 m 80 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 (1440 and 1120) = 160, then find LCM (1440 and 1120). (1)

Answer: Solution:
(i) Convert to cm: 1440 cm, 1120 cm, 880 cm
HCF(1440, 1120, 880) = 80 cm
(ii) LCM(1440, 1120, 880) = 31680 cm
(iii) Using HCF × LCM = Product of numbers
LCM = (1440 × 1120)/160 = 10080
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