This set of Aptitude Questions and Answers (MCQs) focuses on “LCM – Set 2”.
1. Find the largest 5-digit number which is divisible by 15, 18, 24 and 27.
a) 99630
b) 99360
c) 99650
d) 99450
View Answer
Explanation: LCM of 15, 18, 24 and 27 is 1080.
The largest 5-digit number is 99999. On dividing it by 1080, we get 639 as remainder.
Hence, the largest 5-digit number which is divisible by 15, 18, 24 and 27 is 99999-639 = 99360.
2. Find the smallest 5-digit number which is divisible by 36, 75 and 108.
a) 10800
b) 11000
c) 10600
d) 10200
View Answer
Explanation: LCM of 36, 75 and 108 is 2700.
The smallest 5-digit number is 10000. On dividing it by 2700, we get 1900 as remainder.
Therefore, the smallest 5-digit number which is divisible by 36, 75 and 108 is 10000+(2700-1900) = 10800.
3. Find the largest number which when subtracted from 5000, the remainder is divisible by 28, 44 and 84.
a) 4480
b) 4256
c) 4076
d) 4128
View Answer
Explanation: LCM of 28, 44 and 84 is 924.
Therefore, the largest number to be subtracted is 5000-924 = 4076.
4. Find the least number which when divided by 8, 9, 12 and 15 leaves the same remainder 5 in each case.
a) 325
b) 415
c) 485
d) 365
View Answer
Explanation: LCM of 8, 9, 12 and 15 is 360.
Hence, the required number is 360+5 = 365.
5. Find the largest 4-digit number which when divided by 16, 20, 25 and 48, leaves the same remainder 15 in each case.
a) 9825
b) 9615
c) 9565
d) 9655
View Answer
Explanation: LCM of 16, 20, 25 and 48 is 1200.
The largest 4-digit number is 9999. On dividing it by 1200, we get 399 as remainder.
The largest 4-digit which is divisible by 16, 20, 25 and 48 is 9999-399 = 9600.
Therefore, the required number is 9600+15 = 9615.
6. Find the least number which when divided by 18, 25, 30 and 54 leaves remainder 15, 22, 27 and 51 respectively.
a) 1347
b) 1538
c) 1298
d) 1435
View Answer
Explanation: Here, 18-15=3, 25-22=3, 30-27=3 and 54-51=3.
LCM of 18, 25, 30 and 54 is 1350.
Therefore, the required number is 1350-3= 1347.
7. What is the least number when divided by 3, 5, 8, 9 and 15 leaves a remainder of 2 in each case, but when divided by 19 leaves no remainder?
a) 842
b) 862
c) 792
d) 722
View Answer
Explanation: LCM of 3, 5, 8, 9 and 15 is 360.
The required number is of the form 360*k+2.
The least value of k for which 360*k+2 is divisible by 19 is 2.
Hence, the required number is 360*2+2 = 722.
8. A pendulum strikes 6 times in 3 seconds and 7 times in 5 seconds. If both pendulums start striking at the same time, how many clear strikes can be listened in 2 minutes?
a) 378
b) 383
c) 408
d) 413
View Answer
Explanation: First pendulum strikes once in 3/6 = 1/2 seconds. Second pendulum strikes once in 5/7 seconds.
LCM of 1/2 and 5/7 = \(\frac{LCM \, of \, 1 \, and \, 5}{HCF \, of \, 2 and \, 7}\)=5.
So, they strike together after every 5 seconds.
And they strike together for \(\Big(\frac{120}{5}+1\Big)\) = 25 times in 2 minutes.
Therefore, total number of clear strikes heard is=\(\Big(\frac{120}{\frac{1}{2}}+\frac{120}{\frac{5}{7}}\Big)\)-25=240+168-25=383.
9. What is the third term in the sequence of numbers that leave remainders of 4, 5 and 6 when divided by 6, 7 and 8 respectively?
a) 502
b) 526
c) 488
d) 460
View Answer
Explanation: Here, 6-4=2, 7-5=2 and 8-6=2.
LCM of 6, 7 and 8 is 168.
Therefore, the third term in the sequence of numbers is 168*3-2 = 502.
10. Three wheels can complete 25, 21 and 16 revolutions per minute respectively. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?
a) 60 minutes
b) 45 seconds
c) 180 seconds
d) 60 seconds
View Answer
Explanation: For one complete revolution, the first, second and third wheel take 60/25, 60/21, 60/16 seconds i.e., 12/5, 20/7, 15/4 seconds respectively.
Therefore, the time taken for all the red spots to touch the ground simultaneously = LCM \(\Big(\frac{12}{5},\frac{20}{7},\frac{15}{4}\Big)=\frac{LCM \, of \, 12, \, 20 \, and \, 15}{HCF \, of \, 5, \, 7 \, and \, 4}\)=60 seconds=1 minute.
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