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# H.C.F and L.C.M Questions

Home > Quantitative Aptitude > H.C.F and L.C.M > General Questions
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The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275, then the other is:

A308

B208

C318

D283

Explanation:

lcm*hcf = product of 2 nos

11*7700=275*?

ans =308

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Let 13 and 273 are the HCF and LCM of two numbers respectively, and if one of them is less than 140 and greater than 60 then what will be that number?

A121

B91

C75

D131

Explanation:

Solution: Product of two number = Product of their HCF and LCM

pq = 13*273

pq = 3549.

Now, co-primes with product 3549 are (1,3549), (21,169), (39,91) and (13,273)

Acc. to the condition given in the question only one satisfy them that is (39,91)

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Sudhir goes to the market once every 64 days and Sushil goes to the same market once every 72 days. They met each other one day. How many days later will they meet each other again?

A567

B576

C765

DNone of these

Explanation:

Find the LCM of 64,72
i.e 2*2*2*8*9=576

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Sum of squares of two numbers is 2754, their HCF is 9, LCM is 135, Find the numbers

A45 and 36

B45 and 27

C54 and 27

DNone of these

Explanation:

Product of two no. = H.C.F*L.C.M
So,x*y=135*9=1215 -----(1)
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184
So,x+y=72 ----------- (2)
By solving eq. (1) & (2)
nos. are 45 and 27

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Printer A prints 8192 character per min and printer B prints 13862 character per min four character are equal to one word. Printer A starts at 7:15 am while Printer B starts at 7:29 am then at what time both will have same no. of words printed.

A8:49:14 AM

B7:49:14 AM

C9:49:14 AM

D7:49:14 PM

Explanation:

Four character are equal to one word
8192/4*(t+14)=13862/4*t
8192*(t+14)=13862*t yields the same result
t=20.2272 minutes = about 20 min 14 sec
7:49:14 am

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If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?

A0

B2

C3

D5

Explanation:

Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a

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find the number between 100 to 400 which is divisible by either 2,3,5,7..

A120

B100

C205

D210

Explanation:

LCM of 2,3,5,7=210

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A teacher can divide her class into groups into groups of 5,13 and 17. What is the smallest possible strength of the class?

A835

B940

C1105

D1220

Explanation:

For smallest possible class strength, we consider LCM of the given numbers. So LCM (5, 13, 17) = 1105.

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HCF of 2472,1284 and a third number 'n'is 12.If their LCM is 8*9*5*103*107.then the number 'n'is..

A2^2*3^2*5^1

B2^2*3^2*7^1

C2^2*3^2*8103

DNone of these

Explanation:

2472 = 2^3×3×103
1284 = 2^2×3×107
HCF = 2^2×3
LCM = 2^3×3^2×5×103×107
HCF of the numbers is the highest number which divides all the numbers. So N should be a multiple of 2^2×3
LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.
So N = 2^2×3^2×5^1

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The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A1677

B1683

C1898

D3363

ENone of these

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 * 2 + 3) = 1683.

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