Class 10 Maths Chapter 1 Real Numbers Practice Paper 6

Q1. The exponent of 2 in the prime factorization of 6048 is
(a) 4
(b) 5
(c) 6
(d) 7

Answer: Answer: (a) 4
Solution: 6048 = 16 × 378 = 2⁴ × 2 × 189 = 2⁵ × 3³ × 7
Therefore, the exponent of 2 is 5. Answer should be (b) 5.

Q2. If two positive integers x and y are written as x = a⁴b² and y = a²b⁵, where a and b are prime numbers,
then the LCM (x, y) is:
(a) a²b²
(b) a³b³
(c) a⁴b⁵
(d) a⁶b⁷

Answer: Answer: (c) a⁴b⁵
Solution: x = a⁴b², y = a²b⁵
LCM = Product of highest powers of all prime factors = a⁴b⁵

Q3. The HCF and the LCM of 14, 28, 35 respectively are
(a) 7, 140
(b) 14, 280
(c) 7, 280
(d) 280, 7

Answer: Answer: (a) 7, 140
Solution: 14 = 2 × 7, 28 = 2² × 7, 35 = 5 × 7
HCF = 7, LCM = 2² × 5 × 7 = 140

Q4. If the HCF of 104 and 169 is expressible in the form 104m - 169, then the value of m is
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2 Solution: Using Euclidean algorithm:
169 = 104 × 1 + 65
104 = 65 × 1 + 39
65 = 39 × 1 + 26
39 = 26 × 1 + 13
26 = 13 × 2 + 0
HCF = 13 = 104 × 2 - 169 × 1, so m = 2

Q5. Priya has 63 cm long red and 108 cm long blue ribbon. She cuts each ribbon into pieces such that all pieces are of equal length. What is the length of each piece?
(a) 9 cm as it is the HCF of 63 and 108
(b) 9 cm as it is the LCM of 63 and 108
(c) 27 cm as it is the LCM of 63 and 108
(d) 27 cm as it is the HCF of 63 and 108

Answer: Answer: (d) 27 cm as it is the HCF of 63 and 108 Solution: To cut ribbons into equal pieces, we need HCF(63, 108)
63 = 3² × 7, 108 = 2² × 3³
HCF = 3² = 9 cm. Answer should be (a) 9 cm.

Q6. The largest number which divides 112 and 168 leaving remainders 8 and 12 respectively is
(a) 26
(b) 52
(c) 104
(d) 156

Answer: Answer: (b) 52
Solution: Required number = HCF(112-8, 168-12) = HCF(104, 156) = 52

Q7. If 8190 = 2m × 3ⁿ × 5k × 7p, then the value of m + n + k + p is
(a) 5
(b) 6
(c) 7
(d) 8

Answer: Answer: (b) 6
Solution: 8190 = 2 × 3² × 5 × 7 × 13
So m = 1, n = 2, k = 1, p = 1
But this doesn't match the given form. Let me recalculate:
8190 = 2 × 4095 = 2 × 3² × 455 = 2 × 3² × 5 × 91 = 2 × 3² × 5 × 7 × 13
Since 13 is not in the given form, there's an error in the question.

Q8. If p = 2⁴ × 3, q = 2² × 5² × 7, r = 3ⁿ × 7² and LCF (p, q, r) = 2⁴ × 3² × 5² × 7², then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2
Solution: p = 2⁴ × 3¹, q = 2² × 5² × 7¹, r = 3ⁿ × 7²
LCM = 2⁴ × 3² × 5² × 7²
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 8736 and their HCF is 16, then their LCM is 546.
Reason (R): For any two positive integers, HCF × LCM = 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 = 8736/16 = 546. 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): 18ⁿ ends with the digit zero, where n is natural number.
Reason (R): A number ends with digit zero if and only if it is divisible by 10.
(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: 18ⁿ = (2 × 3²)ⁿ = 2ⁿ × 3²ⁿ. This never contains factor 5, so cannot end in 0.
Statement R is correct about numbers ending in 0.

Q11. Explain why 7 × 11 × 13 + 13 and 13 × 17 × 19 + 17 × 13 are composite numbers.

Answer: Solution:
First number: 7 × 11 × 13 + 13 = 1001 + 13 = 1014 = 2 × 3 × 13² (composite)
Second number: 13 × 17 × 19 + 17 × 13 = 4199 + 221 = 4420 = 2² × 5 × 13 × 17 (composite)
Both numbers have factors other than 1 and themselves.

Q12. Two numbers are in the ratio 5 : 7 and their LCM is 420. What is the HCF of these numbers?

Answer: Solution:
Let the numbers be 5x and 7x where x is their HCF.
LCM = (5x × 7x)/x = 35x
Given: 35x = 420, so x = 12
Therefore, HCF = 12

Q13. Show that any number of the form 21ⁿ, where n ∈ N can never end with digit 0. (2017)

Answer: Solution:
21ⁿ = (3 × 7)ⁿ = 3ⁿ × 7ⁿ
For a number to end with 0, it must be divisible by 10 = 2 × 5
Since 21ⁿ contains only factors 3 and 7 (no factors of 2 or 5), it can never end with 0.

Q14. The LCM of two numbers is 6 times their HCF. The sum of LCM and HCF is 700. Find the HCF of the two numbers. Questions 15 to 17 carry 3 marks each.

Answer: Solution:
Let HCF = h, then LCM = 6h
Given: h + 6h = 700
7h = 700
h = 100
Therefore, HCF = 100

Q15. Prove that √11 is an irrational number. (2023)

Answer: Solution:
Proof by contradiction:
Assume √11 is rational, so √11 = p/q where p, q are integers with no common factors.
Squaring: 11 = p²/q², so 11q² = p²
This means p² is divisible by 11, so p is divisible by 11.
Let p = 11k, then 11q² = 121k², so q² = 11k²
This means q is also divisible by 11. But this contradicts our assumption that p and q have no common factors.
Therefore, √11 is irrational.

Q16. 4 Bells toll together at 6.00 am. They toll after 9, 12, 16 and 20 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(9, 12, 16, 20)
LCM = 2⁴ × 3² × 5 = 720 seconds = 12 minutes
In 4 hours = 240 minutes
Number of times = 240/12 = 20
So they will toll together 20 times after 6:00 AM.

Q17. Given that √11 is irrational, prove that 6 + 5√11 is irrational. (CBSE Sample Paper 2022)
Questions 18 carry 5 marks.

Answer: Solution:
Proof by contradiction:
Assume 6 + 5√11 is rational = r
Then 5√11 = r - 6 (rational)
So √11 = (r - 6)/5 (rational)
But this contradicts the given fact that √11 is irrational.
Therefore, 6 + 5√11 is irrational.

Q18. (a) Find the largest possible positive integer that divides 169, 234 and 325 leaving remainder 7, 9 and 13 respectively. (3)
(b) An army contingent of 945 soldiers is to march behind an army band of 54 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(169-7, 234-9, 325-13) = HCF(162, 225, 312)
162 = 2 × 3⁴, 225 = 3² × 5², 312 = 2³ × 3 × 13
HCF = 3
(b) Maximum columns = HCF(945, 54)
945 = 3³ × 5 × 7, 54 = 2 × 3³
HCF = 3³ = 27 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 88 cm, 99 cm and 121 cm respectively.
(i) What is the HCF of 88 and 121? (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(88, 121): 88 = 2³ × 11, 121 = 11²
HCF = 11
(ii) Total distance = LCM(88, 99, 121)
LCM = 2³ × 3² × 11² = 8712
Sum of exponents = 3 + 2 + 2 = 7
(iii) Minimum distance = LCM(88, 99, 121) = 8712 cm = 87.12 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 12 m 60 cm, 9 m 80 cm and 7 m 70 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 (1260 and 980) = 140, then find LCM (1260 and 980). (1)

Answer: Solution:
(i) Convert to cm: 1260 cm, 980 cm, 770 cm
HCF(1260, 980, 770) = 70 cm
(ii) LCM(1260, 980, 770) = 27720 cm
(iii) Using HCF × LCM = Product of numbers
LCM = (1260 × 980)/140 = 8820
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