Class 10 Maths Chapter 1 Real Numbers Practice Paper 9

Q1. The exponent of 11 in the prime factorization of 7436 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2
Solution: 7436 = 4 × 1859 = 4 × 11 × 169 = 2² × 11 × 13²
Therefore, the exponent of 11 is 1. Answer should be (a) 1.

Q2. If two positive integers w and z are written as w = t²u³ and z = t⁴u, where t and u are prime numbers, then
the LCM (w, z) is:
(a) tu
(b) t²u²
(c) t⁴u³
(d) t⁶u⁴

Answer: Answer: (c) t⁴u³
Solution: w = t²u³, z = t⁴u¹
LCM = Product of highest powers of all prime factors = t⁴u³

Q3. The HCF and the LCM of 45, 54, 72 respectively are
(a) 9, 1080
(b) 18, 2160
(c) 9, 2160
(d) 1080, 9

Answer: Answer: (a) 9, 1080
Solution: 45 = 3² × 5, 54 = 2 × 3³, 72 = 2³ × 3²
HCF = 3² = 9, LCM = 2³ × 3³ × 5 = 1080

Q4. If the HCF of 133 and 203 is expressible in the form 133m - 203, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2 Solution: Using Euclidean algorithm:
203 = 133 × 1 + 70
133 = 70 × 1 + 63
70 = 63 × 1 + 7
63 = 7 × 9 + 0
HCF = 7 = 133 × 2 - 203 × 1, so m = 2

Q5. Shreya has 84 cm long silver and 140 cm long golden ribbon. She cuts each ribbon into pieces such that all pieces are of equal length. What is the length of each piece?
(a) 14 cm as it is the HCF of 84 and 140
(b) 14 cm as it is the LCM of 84 and 140
(c) 28 cm as it is the LCM of 84 and 140
(d) 28 cm as it is the HCF of 84 and 140

Answer: Answer: (d) 28 cm as it is the HCF of 84 and 140 Solution: To cut ribbons into equal pieces, we need HCF(84, 140)
84 = 2² × 3 × 7, 140 = 2² × 5 × 7
HCF = 2² × 7 = 28 cm

Q6. The largest number which divides 154 and 231 leaving remainders 10 and 15 respectively is
(a) 36
(b) 72
(c) 144
(d) 216

Answer: Answer: (b) 72
Solution: Required number = HCF(154-10, 231-15) = HCF(144, 216) = 72

Q7. If 10584 = 2m × 3ⁿ × 7k × 13p, then the value of m + n + k + p is
(a) 6
(b) 7
(c) 8
(d) 9

Answer: Answer: (b) 7
Solution: 10584 = 2³ × 3² × 7 × 21 = 2³ × 3² × 7²
Actually, let me recalculate: 10584 = 8 × 1323 = 8 × 3² × 147 = 8 × 9 × 3 × 49 = 2³ × 3³ × 7²
So m = 3, n = 3, k = 2, p = 0 (no factor of 13)
There seems to be an error in the question as 10584 doesn't contain 13 as a factor.

Q8. If p = 7² × 13, q = 7 × 11² × 17, r = 11ⁿ × 17³ and LCM (p, q, r) = 7² × 11³ × 13 × 17³, then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (c) 3
Solution: p = 7² × 13¹, q = 7¹ × 11² × 17¹, r = 11ⁿ × 17³
LCM = 7² × 11³ × 13¹ × 17³
For this to be true, n must be 3.

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 11220 and their HCF is 22, then their LCM is 510.
Reason (R): For any two positive integers a and b, HCF(a,b) × LCM(a,b) = a × b.
(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 = 11220/22 = 510. 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): 30ⁿ ends with the digit zero, where n is natural number.
Reason (R): A number ends with zero if it is divisible by both 2 and 5.
(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: (a) Both A and R are true and R is the correct explanation of A
Solution: 30ⁿ = (2 × 3 × 5)ⁿ = 2ⁿ × 3ⁿ × 5ⁿ. This contains both factors 2 and 5, so ends in 0.
Both statements are true and R explains A.

Q11. Explain why 13 × 17 × 19 + 19 and 19 × 23 × 29 + 23 × 19 are composite numbers.

Answer: Solution:
First number: 13 × 17 × 19 + 19 = 4199 + 19 = 4218 = 2 × 3 × 19 × 37 (composite)
Second number: 19 × 23 × 29 + 23 × 19 = 12673 + 437 = 13110 = 2 × 3 × 5 × 19 × 23 (composite)
Both numbers have factors other than 1 and themselves.

Q12. Two numbers are in the ratio 7 : 9 and their LCM is 630. What is the HCF of these numbers?

Answer: Solution:
Let the numbers be 7x and 9x where x is their HCF.
LCM = (7x × 9x)/x = 63x (since HCF(7,9) = 1)
Given: 63x = 630, so x = 10
Therefore, HCF = 10

Q13. Show that any number of the form 33ⁿ, where n ∈ N can never end with digit 0. (2017)

Answer: Solution:
33ⁿ = (3 × 11)ⁿ = 3ⁿ × 11ⁿ
For a number to end with 0, it must be divisible by 10 = 2 × 5
Since 33ⁿ contains only factors 3 and 11 (no factors of 2 or 5), it can never end with 0.

Q14. The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 800. Find the HCF of the two numbers. Questions 15 to 17 carry 3 marks each.

Answer: Solution:
Let HCF = h, then LCM = 4h
Given: h + 4h = 800
5h = 800
h = 160
Therefore, HCF = 160

Q15. Prove that √17 is an irrational number. (2023)

Answer: Solution:
Proof by contradiction:
Assume √17 is rational, so √17 = p/q where p, q are integers with no common factors.
Squaring: 17 = p²/q², so 17q² = p²
This means p² is divisible by 17, so p is divisible by 17.
Let p = 17k, then 17q² = 289k², so q² = 17k²
This means q is also divisible by 17. But this contradicts our assumption that p and q have no common factors.
Therefore, √17 is irrational.

Q16. 4 Bells toll together at 4.00 am. They toll after 12, 15, 20 and 24 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(12, 15, 20, 24)
LCM = 2³ × 3 × 5 = 120 seconds = 2 minutes
In 4 hours = 240 minutes
Number of times = 240/2 = 120
So they will toll together 120 times after 4:00 AM.

Q17. Given that √17 is irrational, prove that 8 + 7√17 is irrational. (CBSE Sample Paper 2022)
Questions 18 carry 5 marks.

Answer: Solution:
Proof by contradiction:
Assume 8 + 7√17 is rational = r
Then 7√17 = r - 8 (rational)
So √17 = (r - 8)/7 (rational)
But this contradicts the given fact that √17 is irrational.
Therefore, 8 + 7√17 is irrational.

Q18. (a) Find the largest possible positive integer that divides 195, 273 and 364 leaving remainder 13, 15 and 19 respectively. (3)
(b) An army contingent of 1134 soldiers is to march behind an army band of 72 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(195-13, 273-15, 364-19) = HCF(182, 258, 345)
182 = 2 × 7 × 13, 258 = 2 × 3 × 43, 345 = 3 × 5 × 23
HCF = 1
(b) Maximum columns = HCF(1134, 72)
1134 = 2 × 3⁴ × 7, 72 = 2³ × 3²
HCF = 2 × 3² = 18 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 95 cm, 114 cm and 133 cm respectively.
(i) What is the HCF of 95 and 133? (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(95, 133): 95 = 5 × 19, 133 = 7 × 19
HCF = 19
(ii) Total distance = LCM(95, 114, 133)
LCM = 2 × 3 × 5 × 7 × 19 = 3990
Sum of exponents = 1 + 1 + 1 + 1 + 1 = 5
(iii) Minimum distance = LCM(95, 114, 133) = 3990 cm = 39.9 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 15 m 60 cm, 12 m 40 cm and 9 m 30 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 (1560 and 1240) = 40, then find LCM (1560 and 1240). (1)

Answer: Solution:
(i) Convert to cm: 1560 cm, 1240 cm, 930 cm
HCF(1560, 1240, 930) = 310 cm
(ii) LCM(1560, 1240, 930) = 18600 cm
(iii) Using HCF × LCM = Product of numbers
LCM = (1560 × 1240)/40 = 48360
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