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QUESTION: 1

Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for **"HCF and LCM"** under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.

**Q.** Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Solution:

Given numbers are 43, 91 and 183.

Subtract smallest number from both the highest numbers.

we have three cases: 183 > 43; 183 > 91 and 91 > 43

183 – 43 = 140

183 – 91 = 92 and

91 – 43 = 48

Now, we have three new numbers: 140, 48 and 92.

Find HCF of 140, 48 and 92 using prime factorization method, we get

HCF (140, 48 and 92) = 4

**The highest number that divide 183, 91 and 43 and leave the same remainder is 4.**

QUESTION: 2

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

Solution:

QUESTION: 3

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

Solution:

QUESTION: 4

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

Solution:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

QUESTION: 5

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

Solution:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

Required number (9999 - 399) = 9600.

QUESTION: 6

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

Solution:

Let the numbers be 37*a* and 37*b*.

Then, 37*a* x 37*b* = 4107

*ab* = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) *i.e.,* (37, 111).

Greater number = 111.

QUESTION: 7

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

Solution:

QUESTION: 8

The Greatest Common Divisor of 1.08, 0.36 and 0.9 is:

Solution:

Given numbers are 1.08 , 0.36 and 0.90

H.C.F of 108, 36 and 90 is 18 [ ∵∵ G.C.D is nothing but H.C.F]

Therefore, H.C.F of given numbers = 0.18

QUESTION: 9

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

Solution:

Let the numbers 13*a* and 13*b*.

Then, 13*a* x 13*b* = 2028

⇒ *ab* = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers *a* and *b* are said to be **coprime** or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

QUESTION: 10

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Solution:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90*k* + 4, which is multiple of 7.

Least value of *k* for which (90*k* + 4) is divisible by 7 is *k* = 4.

∴ Required number = (90 x 4) + 4 = 364.

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