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- Question 1 of 14
1. Question
The H.C.F. and L.C.M. of two numbers are 50 and 250 respectively. If the first number is divided by 2, the quotient is 50. The second number is:
Hint
First number = (50 × 2) = 100.
Second number
- Question 2 of 14
2. Question
The sum of two numbers is 2000 and their L.C.M. is 21879. The two numbers are:
Hint
Let the numbers be x and (2000 – x).
Then, their L.C.M. = x(2000 – x).
So, x(2000 – x) = 21879
⇒ x² – 2000x + 21879 = 0
⇒ (x –1989)(x – 11) = 0
⇒ x = 1989 or 11.
Hence, the numbers are 1989 and 11.
- Question 3 of 14
3. Question
The H.C.F. of two numbers is 8. Which one of the following can never be their L.C.M.?
Hint
H.C.F. of two numbers divides their L.C.M. exactly. 8 is not a factor of 60.
- Question 4 of 14
4. Question
The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.?
Hint
Since H.C.F. is always a factor of L.C.M., we cannot have three numbers with H.C.F. 35 and L.C.M. 120.
- Question 5 of 14
5. Question
Which of the following fractions is the largest?
Hint
L.C.M. of 8, 16, 40 and 80 = 80.
So, 7/8 is the largest.
- Question 6 of 14
6. Question
Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12 and 16 will leave the same remainder 4?
Hint
Suppose least number be x
1856 – x = n(LCM of 7, 12, 16) + 4
⇒ 1856 -x = n (336) + 4
we should take n = 5 so that n(336) is nearest to 1856 and n (336)< 1856
1856 – x = 1680 + 4 = 1684
x = 1856 – 1684 = 172
- Question 7 of 14
7. Question
Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.
Hint
Required number
= H.C.F of (148 – 4), (246 – 6) and (623 – 11)
= H.C.F of 144, 240 and 612 = 12
- Question 8 of 14
8. Question
Monica, Veronica and Rachael begin to jog around a circular stadium. They complete their revolutions in 48 seconds, 64 seconds and 72 seconds respectively. After how many seconds will they be together at the starting point?
Hint
Required time = LCM of 48, 64 and 72
LCM = 2 × 2 × 2 × 2 × 3 × 4 × 3 = 576 seconds.
- Question 9 of 14
9. Question
Three girls start jogging from the same point around a circular track and each one completes one round in 24 seconds, 36 seconds and 48 seconds respectively. After how much time will they meet at one point?
Hint
Required time
= L.C.M of 24, 36 and 48
= 144 seconds
= 2 minutes 24 seconds
- Question 10 of 14
10. Question
Three friends A, B and C start running around a circular stadium and complete a single round in 24, 36 and 30 seconds respectively. After how many minutes will they meet again at the starting point?
Hint
Required time
= L.C.M of 24, 36 and 30
= 360 seconds
= 6 minutes
- Question 11 of 14
11. Question
Swapnil, Aakash and Vinay begin to jog around a circular stadium. They complete their revolutions in 36 seconds, 48 seconds and 42 seconds respectively. After how many seconds will they be together at the starting point?
Hint
They will be together at the starting point after the
L.C.M of 36, 48 and 42 = 1008 seconds
- Question 12 of 14
12. Question
The circumferences of the fore and hind-wheels of a carriage are 2⅖ and 3³⁄₇ respectively. A chalk mark is put on the point of contact of each wheel with the ground at any given moment. How far will the carriage have travelled so that its chalk marks may be again on the ground at the same time?
Hint
Required distance
= L.C.M of
Hence, carriage will travelled 24m so that its chalk marks may be again on the ground at the same time.
- Question 13 of 14
13. Question
Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again.
Hint
L.C.M of 18, 24 & 32 = 288
Hence they would chime after every 288 min.
⇒ 4 hrs 48 min
- Question 14 of 14
14. Question
HCF of 3240, 3600 and a third number is 36 and their LCM is 2⁴ × 3⁵ × 5² × 7². The third number is :
Hint
Let the third number be x.
Product of numbers = LCM × HCF
Therefore,
⇒
⇒
⇒
=
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