Mathematics Mock test on LCM HCF

Mathematics Quiz

1. The LCM and HCF of two numbers are 84 and 21, respectively. If the ratio of two numbers is 1 : 4, then the larger of the two numbers is:

(a) 21
(b) 48
(c) 84
(d) 108

Correct Answer: (c). Let the numbers be x and 4x. Using the formula LCM × HCF = Product of Numbers, we get x × 4x = 84 × 21 → x = 21. Hence, the larger number is 4x = 84.

2. The LCM of two numbers is 4800 and their HCF is 160. If one of the numbers is 480, then the other number is:

(a) 16
(b) 16000
(c) 160
(d) 1600

Correct Answer: (d). Using the formula LCM × HCF = Product of Numbers, we get 4800 × 160 = 480 × Other Number → Other Number = 1600.

3. Three numbers are in the ratio 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F is:

(a) 40
(b) 80
(c) 120
(d) 200

Correct Answer: (a). Let the numbers be 3x, 4x, and 5x. Their LCM is 60x = 2400 → x = 40. Hence, HCF = x = 40.

4. The HCF and LCM of two numbers are 11 and 385 respectively. If one number lies between 75 and 125, then that number is:

(a) 77
(b) 88
(c) 99
(d) 110

Correct Answer: (a). Using the formula LCM × HCF = Product of Numbers, we get 11 × 385 = x × y → xy = 4235. Since one number lies between 75 and 125, it must be 77 (as 77 × 55 = 4235).

5. Let ‘K’ be the greatest number that will divide 1305, 4665, and 6905, leaving the same remainder 25 in each case. Then the sum of the digits of ‘K’ is:

(a) 7
(b) 5
(c) 6
(d) 8

Correct Answer: (a). Subtracting the remainder from each number gives 1305 - 25 = 1280, 4665 - 25 = 4640, 6905 - 25 = 6880. The HCF of these numbers is 160. Sum of digits of 160 = 1 + 6 + 0 = 7.

6. The least number, which when divided by 48, 60, 72, 108, 140 leaves 38, 50, 62, 98, and 130 remainders respectively, is:

(a) 11115
(b) 15110
(c) 15120
(d) 15210

Correct Answer: (b). The difference between divisors and remainders is constant (10). Find the LCM of 48, 60, 72, 108, 140 and subtract 10. LCM = 15120 → Required number = 15120 - 10 = 15110.

7. HCF of first 200 prime numbers which are of the form 10p + 1 is:

(a) 10
(b) 7
(c) 6
(d) None of these

Correct Answer: (d). Prime numbers of the form 10p + 1 are co-prime to each other. Hence, their HCF is 1.

8. The LCM of 15/2, 4/3, 9/27 is:

(a) 1/54
(b) 10/27
(c) 20/3
(d) None of these

Correct Answer: (c). LCM of numerators = LCM(15, 4, 9) = 180. HCF of denominators = HCF(2, 3, 27) = 1. Hence, LCM = 180/1 = 20/3.

9. If HCF(a, b) = 12 and a × b = 1800, then LCM(a, b) =:

(a) 900
(b) 150
(c) 90
(d) 3600

Correct Answer: (b). Using the formula LCM × HCF = Product of Numbers, we get LCM × 12 = 1800 → LCM = 150.

10. There are 264 girls and 408 boys in a school. These children are to be divided into groups of equal number of boys and girls. The maximum number of boys or girls in each group will be:

(a) 11
(b) 17
(c) 24
(d) 36

Correct Answer: (c). The HCF of 264 and 408 is 24. Hence, the maximum number of boys or girls in each group is 24.

11. Three bells begin tolling at the same time and continue to do so at intervals of 21, 28, and 30 seconds respectively. The bells will toll together again after:

(a) 7 seconds
(b) 420 seconds
(c) 630 seconds
(d) 1764 seconds

Correct Answer: (b). The LCM of 21, 28, and 30 is 420. Hence, the bells will toll together after 420 seconds.

12. The ratio of two numbers is 3 : 4, their HCF is 4. Their LCM is:

(a) 12
(b) 16
(c) 24
(d) 48

Correct Answer: (d). Let the numbers be 3x and 4x. HCF = x = 4. Hence, the numbers are 12 and 16. LCM = 48.

13. Product of two co-prime numbers is 117. Their LCM should be:

(a) 1
(b) 117
(c) Equal to their HCF
(d) 0

Correct Answer: (b). For co-prime numbers, HCF = 1. Since LCM × HCF = Product of Numbers, LCM = 117.

14. Which of the following pairs of fractions adds up to a number more than 5?

(a) 5/3, 3/4
(b) 7/3, 11/5
(c) 11/4, 8/3
(d) 13/5, 11/6

Correct Answer: (c). Adding 11/4 and 8/3 gives (33 + 32)/12 = 65/12 ≈ 5.42, which is greater than 5.

15. The length and breadth of a rectangular field are 55 m and 45 m respectively. The length of the largest rod (in m) that can measure the length and breadth of the field exactly is:

(a) 11 m
(b) 9 m
(c) 5 m
(d) 10 m

Correct Answer: (c). The HCF of 55 and 45 is 5. Hence, the largest rod that can measure both dimensions exactly is 5 m.

16. One pendulum ticks 57 times in 58 seconds and another 608 times in 609 seconds. If they started simultaneously, find the time after which they will tick together.

(a) 211/19 s
(b) 1217/19 s
(c) 1218/19 s
(d) 1018/19 s

Correct Answer: (c). The time intervals for the pendulums are 58/57 and 609/608 seconds. The LCM of these intervals is 1218/19 seconds.

17. Four runners started running simultaneously from a point on a circular track. They took 200 sec, 300 sec, 360 sec, and 450 sec to complete one round. After how much time do they meet at the starting point for the first time?

(a) 1800 sec
(b) 3600 sec
(c) 2400 sec
(d) 4800 sec

Correct Answer: (a). The LCM of 200, 300, 360, and 450 is 1800 seconds. Hence, they meet after 1800 seconds.

18. The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of three digits.

(a) 161
(b) 171
(c) 181
(d) 191

Correct Answer: (a). The difference between 11284 and 7655 is 3629. The three-digit divisor of 3629 is 161.

19. Three bells toll at intervals of 9, 12, and 15 minutes respectively. All the three begin to toll at 8 a.m. At what time will they toll together again?

(a) 8:45 a.m.
(b) 10:30 a.m.
(c) 11:00 a.m.
(d) 1:30 p.m.

Correct Answer: (c). The LCM of 9, 12, and 15 is 180 minutes (3 hours). Adding 3 hours to 8 a.m. gives 11:00 a.m.

20. Four bells begin to toll together and toll respectively at intervals of 6, 5, 7, 10, and 12 seconds. How many times will they toll together in one hour excluding the one at the start?

(a) 7 times
(b) 8 times
(c) 9 times
(d) 11 times

Correct Answer: (a). The LCM of 6, 5, 7, 10, and 12 is 420 seconds (7 minutes). In one hour (3600 seconds), they toll together 3600 ÷ 420 = 8 times, excluding the initial toll.