 # Previous Year LCM and HCF Questions for SSC CGL/MTS Exams

Dear maths lovers, Let your practice begins in minuteness but ends in magnificence.It is impossible to study maths properly by just reading and listening. So, practise, practise & more practise. For that, we are providing here Quant Quiz of 15 questions on LCM and HCF, in accordance with the syllabus of SSC CGL.We have also provided Study Notes on all the topics.
Q1. The LCM of two numbers is 1920 and their HCF is 16. If one of the numbers is 128, find the other number.
(a) 204
(b) 240
(c) 260
(d) 320

Q2. The HCF and product of two numbers are 15 and 6300 respectively. The number of possible pairs of the numbers is:
(a) 4
(b) 3
(c) 2
(d) 1
Q3. The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 125. If one of the numbers is 100, then the other number is:
(a) 5
(b) 25
(c) 100
(d) 125

Q4. The HCF and LCM of two numbers are 13 and 455 respectively. If one of the numbers lies between 75 and 125, then, that numbers is:
(a) 78
(b) 91
(c) 104
(d) 117
Q5. The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is:
(a) 450
(b) 454
(c) 540
(d) 544
Q6. Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as a remainder:
(a) 99999
(b) 99958
(c) 99960
(d) 99962
Q7. The least number which when divided by 16, 18, 20 and 25 leaves 4 as a remainder in each case but when divided by 7 leaves no remainder is:
(a) 17004
(b) 18000
(c) 18002
(d) 18004
Q8. What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder?
(a) 312
(b) 962
(c) 1562
(d) 1586
Q9. The traffic lights at three different road crossings change after 24 seconds, 36 seconds and 54 seconds respectively. If they all change simultaneously at 10:15:00 AM, then at what time will they again change simultaneously?
(a) 10:16:54 AM
(b) 10:18:36 AM
(c) 10:17:02 AM
(d) 10:22:12 AM
Q10. From a point on a circular track 5 km long A, B and C started running in the same direction at the same time with the speed of 2(1/2)  km per hour, 3 km per hour and 2 km per hour respectively. Then on the starting point, all three will meet again after:

(a) 30 hours

(b) 6 hours
(c) 10 hours
(d) 15 hours
Q11. Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder in each case. Then, the sum of the digits in N is:
(a) 4
(b) 5
(c) 6
(d) 8
Q12. Three sets of English, Mathematics and Science books containing 336, 240, 96 books respectively have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. The total number of stacks will be:
(a) 14
(b) 21
(c) 22
(d) 48
Q13. Three numbers are in the ratio 2: 3: 4 and their H.C.F. is 12. The L.C.M. of the numbers is:
(a) 144
(b) 192
(c) 96
(d) 72
Q14. The H.C.F and L.C.M. of two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1: 4, then the larger of the two numbers is:
(a) 12
(b) 108
(c) 48
(d) 84
Q15. The number between 3000 and 4000 which is exactly divisible by 30, 36 and 80 is:
(a) 3625
(b) 3250
(c) 3500
(d) 3600
Solutions

S2. Ans.(c)
Sol. Let the numbers be 15x and 15y
So, 15x × 15y = 6300
× y = 28
Possible pairs    = (1, 28) √
= (2, 14) ×
= (4, 7) √
Two possible pairs.

S5. Ans.(d)
Sol. LCM of 15, 12, 20, 54 = 540
Then number = 540 + 4 = 544
[4 being remainder]
S6. Ans.(d)
Sol. The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12 = 120
After dividing 99999 by 120, we get 39 as remainder
99999 – 39 = 99960
99960 is the greatest five digit number divisible by the given divisors.
In order to get 2 as a remainder in each case, we will simply add 2 to 99960.
∴ Greatest number = 99960 + 2 = 99962
S7. Ans.(d)
Sol. LCM of 16, 18, 20 and 25 = 3600
∴ Required number = 3600K + 4 which is exactly divisible by 7 for certain value of K.
When K = 5,
Number = 3600 × 5 + 4
= 18004 which is exactly divisible by 7.
S8. Ans.(b)
Sol. LCM of 3, 5, 6, 8, 10 and 12 = 120
∴ Required number
= 120x + 2, which is exactly divisible by 13.
120x + 2 = 13 × 9x + 3x + 2
Clearly 3x + 2 should be divisible by 13.
For x = 8, 3x + 2 is divisible by 13.
∴ Required number = 120x + 2 = 120 × 8 + 2
= 960 + 2 = 962
S9. Ans.(b)
Sol. LCM of 24, 36 and 54 seconds
= 216 seconds
= 3 minutes 36 seconds
∴ Required time = 10:15:00 + 3 minutes 36 seconds
= 10:18:36 a.m.

S11. Ans.(a)
Sol. The greatest number N = HCF of (1305 – x), (4665 – x) and (6905 – x), where x is the remainder
= HCF of (4665 – 1305), (6905 – 4665) and (6905 – 1305)
= HCF of 3360, 2240 and 5600
∴ N = 1120
Sum of digits = 1 + 1 + 2 + 0 = 4

S13. Ans.(a)
Sol. Let the numbers be 2x, 3x and 4x respectively.
∴ HCF = x = 12
∴ Numbers are: 2 × 12 = 24, 3 × 12 = 36, 4 × 12 = 48
LCM of 24, 36, 48
= 2 × 2 × 2 × 3 × 3 × 2 = 144
S14. Ans.(d)
Sol. HCF of numbers = 21
∴ Numbers = 21x and 21y
Where x and y are prime to each other.
Ratio of numbers = 1 : 4
∴ Larger number = 21 × 4 = 84
S15. Ans.(d)
Sol. Firstly, we find the LCM of 30, 36 and 80 = 720
∴ Required number = Multiple of 720
= 720 × 5 = 3600
because 3000 < 3600 < 4000 Join India's largest learning destination

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