Class 10 Maths Chapter 1 Real Numbers Practice Paper 4

Q1. The exponent of 7 in the prime factorization of 4410 is
(a) 2
(b) 3
(c) 4
(d) 5

Answer: Answer: (a) 2
Solution: 4410 = 2 × 2205 = 2 × 3² × 5 × 7²
Therefore, the exponent of 7 is 2.

Q2. If two positive integers p and q are written as p = a³b² and q = a²b⁴, where a and b are prime numbers,
then the LCM (p, q) is:
(a) ab
(b) a²b³
(c) a³b⁴
(d) a⁴b⁴

Answer: Answer: (c) a³b⁴
Solution: p = a³b², q = a²b⁴
LCM = Product of highest powers of all prime factors = a³b⁴

Q3. The HCF and the LCM of 18, 24, 30 respectively are
(a) 6, 360
(b) 18, 240
(c) 6, 240
(d) 240, 6

Answer: Answer: (a) 6, 360
Solution: 18 = 2 × 3², 24 = 2³ × 3, 30 = 2 × 3 × 5
HCF = 2 × 3 = 6, LCM = 2³ × 3² × 5 = 360

Q4. If the HCF of 78 and 143 is expressible in the form 78m - 143, then the value of m is
(a) 2
(b) 3
(c) 4
(d) 5

Answer: Answer: (a) 2 Solution: Using Euclidean algorithm:
143 = 78 × 1 + 65
78 = 65 × 1 + 13
65 = 13 × 5 + 0
HCF = 13 = 78 × 2 - 143 × 1, so m = 2

Q5. Meera has 48 cm long red and 72 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) 8 cm as it is the HCF of 48 and 72
(b) 8 cm as it is the LCM of 48 and 72
(c) 24 cm as it is the LCM of 48 and 72
(d) 24 cm as it is the HCF of 48 and 72

Answer: Answer: (d) 24 cm as it is the HCF of 48 and 72
Solution: To cut ribbons into equal pieces, we need HCF(48, 72) = 24 cm

Q6. The largest number which divides 85 and 136 leaving remainders 7 and 10 respectively is
(a) 18
(b) 26
(c) 78
(d) 126

Answer: Answer: (b) 26
Solution: Required number = HCF(85-7, 136-10) = HCF(78, 126) = 26

Q7. If 5460 = 2m × 3ⁿ × 5k × 7p, then the value of m + n + k + p is
(a) 5
(b) 6
(c) 7
(d) 8

Answer: Answer: (a) 5
Solution: 5460 = 2² × 3 × 5 × 7²
So m = 2, n = 1, k = 1, p = 2
m + n + k + p = 2 + 1 + 1 + 2 = 6. Wait, let me recalculate:
5460 = 4 × 1365 = 4 × 3 × 455 = 4 × 3 × 5 × 91 = 4 × 3 × 5 × 7 × 13
Actually, 5460 = 2² × 3 × 5 × 7 × 13, so the answer should be (a) 5.

Q8. If p = 2² × 5, q = 2³ × 3 × 7, r = 3ⁿ × 7 and LCM (p, q, r) = 2³ × 3² × 5 × 7, then n is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer: Answer: (b) 2
Solution: p = 2² × 5¹, q = 2³ × 3¹ × 7¹, r = 3ⁿ × 7¹
LCM = 2³ × 3² × 5¹ × 7¹
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 6240 and their HCF is 13, then their LCM is 480.
Reason (R): HCF is always a factor of LCM.
(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: (b) Both A and R are true but R is not the correct explanation of A
Solution: Product = HCF × LCM, so LCM = 6240/13 = 480. Both statements are true.
However, R doesn't explain why the calculation in A is correct.

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): 8ⁿ ends with the digit zero, where n is natural number.
Reason (R): Any number ends with digit zero, if its prime factor is of the form 2m × 5ⁿ, where m, n are natural 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. Questions 11 to 14 carry 2 marks each.

Answer: Answer: (d) A is false but R is true
Solution: 8ⁿ = (2³)ⁿ = 2³ⁿ. This never contains factor 5, so cannot end in 0.
Statement R is correct about numbers ending in 0.

Q11. Explain why 3 × 5 × 7 + 7 and 7 × 11 × 13 + 11 × 7 are composite numbers.

Answer: Solution:
First number: 3 × 5 × 7 + 7 = 105 + 7 = 112 = 2⁴ × 7 (composite)
Second number: 7 × 11 × 13 + 11 × 7 = 1001 + 77 = 1078 = 2 × 7² × 11 (composite)
Both numbers have factors other than 1 and themselves.

Q12. Two numbers are in the ratio 3 : 4 and their LCM is 240. What is the HCF of these numbers?

Answer: Solution:
Let the numbers be 3x and 4x where x is their HCF.
LCM = (3x × 4x)/x = 12x
Given: 12x = 240, so x = 20
Therefore, HCF = 20

Q13. Show that any number of the form 9ⁿ, where n ∈ N can never end with digit 0. (2017)

Answer: Solution:
9ⁿ = (3²)ⁿ = 3²ⁿ
For a number to end with 0, it must be divisible by 10 = 2 × 5
Since 9ⁿ contains only factor 3 (no factors of 2 or 5), it can never end with 0.

Q14. The LCM of two numbers is 8 times their HCF. The sum of LCM and HCF is 540. Find the HCF of the two numbers. Questions 15 to 17 carry 3 marks each.

Answer: Solution:
Let HCF = h, then LCM = 8h
Given: h + 8h = 540
9h = 540
h = 60
Therefore, HCF = 60

Q15. Prove that √5 is an irrational number. (2023)

Answer: Solution:
Proof by contradiction:
Assume √5 is rational, so √5 = p/q where p, q are integers with no common factors.
Squaring: 5 = p²/q², so 5q² = p²
This means p² is divisible by 5, so p is divisible by 5.
Let p = 5k, then 5q² = 25k², so q² = 5k²
This means q is also divisible by 5. But this contradicts our assumption that p and q have no common factors.
Therefore, √5 is irrational.

Q16. 4 Bells toll together at 8.00 am. They toll after 6, 9, 10 and 15 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(6, 9, 10, 15)
LCM = 2 × 3² × 5 = 90 seconds = 1.5 minutes
In 3 hours = 180 minutes
Number of times = 180/1.5 = 120
So they will toll together 120 times after 8:00 AM.

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

Answer: Solution:
Proof by contradiction:
Assume 3 + 4√5 is rational = r
Then 4√5 = r - 3 (rational)
So √5 = (r - 3)/4 (rational)
But this contradicts the given fact that √5 is irrational.
Therefore, 3 + 4√5 is irrational.

Q18. (a) Find the largest possible positive integer that divides 138, 175 and 268 leaving remainder 6, 7 and 10 respectively. (3)
(b) An army contingent of 756 soldiers is to march behind an army band of 42 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(138-6, 175-7, 268-10) = HCF(132, 168, 258)
132 = 2² × 3 × 11, 168 = 2³ × 3 × 7, 258 = 2 × 3 × 43
HCF = 2 × 3 = 6
(b) Maximum columns = HCF(756, 42)
756 = 2² × 3³ × 7, 42 = 2 × 3 × 7
HCF = 2 × 3 × 7 = 42 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 75 cm, 90 cm and 105 cm respectively.
(i) What is the HCF of 75 and 105? (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(75, 105): 75 = 3 × 5², 105 = 3 × 5 × 7
HCF = 3 × 5 = 15
(ii) Total distance = LCM(75, 90, 105)
LCM = 2 × 3² × 5² × 7 = 1260
Sum of exponents = 1 + 2 + 2 + 1 = 6
(iii) Minimum distance = LCM(75, 90, 105) = 1260 cm = 12.6 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 9 m 60 cm, 7 m 20 cm and 5 m 40 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 (960 and 720) = 120, then find LCM (960 and 720). (1)

Answer: Solution:
(i) Convert to cm: 960 cm, 720 cm, 540 cm
HCF(960, 720, 540) = 120 cm
(ii) LCM(960, 720, 540) = 25920 cm
(iii) Using HCF × LCM = Product of numbers
LCM = (960 × 720)/120 = 5760
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