# Arithmetic Aptitude :: H.C.F and L.C.M

Home > Arithmetic Aptitude > H.C.F and L.C.M > General Questions

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The G.C.D. of 1.08, 0.36 and 0.9 is:

A0.03

B0.9

C0.18

D0.108

| | | Asked In nagarroHCL TechnologiesAccenture1 |

Explanation:

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)
So H.C.F of given numbers = 0.18

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There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

A80

B82

C85

D87

| | | Asked In nagarro |

Explanation:

Given that numbers are co primes,
and two products have the middle number in common.

=> Middle number = H.C.F. of 551 and 1073 = 29

so first number is = 551/29 = 19
=> Third number = 1073/29 = 37

Therefore, sum of these numbers is = (19 + 29 + 37) = 85

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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

| | | Asked In InfosysGlobal Edge |

Explanation:

lcm*hcf = product of 2 nos

11*7700=275*?

ans =308

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Let p and q be two prime numbers such that p is greater than q. If 319 is their LCM then the difference of thrice of q and p is:

A1

B0

C4

D9

| | | Asked In IBMGlobal Edge |

Explanation:

Solution: We know that HCF of two prime numbers is 1.

Product of HCF and LCM = 1 x 319 = 319.

Remember that, Product of two number = Product of their HCF and LCM

pq = 319

Now, co-primes with product 319 are (1,319) and (29,11)

Since p > q, p = 29 and q = 11

Then 3q - p = 3 - 29 = 4.

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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

| | | Asked In IBM |

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

| | | Asked In Wipro |

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

| | | Asked In WiproGlobal Edge |

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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Find the largest 4-digit number, which gives the remainder 7 and 13 when divided by 11 and 17?

A9900

B9907

C9807

DNone of these

| | | Asked In WiproGlobal Edge |

Explanation:

LCM of 11 and 17 = 187.
When divided by 11 remainder 7,so difference 4.
When devided by 17 remainder 13,so difference 4.
Largest no exactly devide by 11 & 17=9911
The no's = 9911-4 = 9907.

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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

| | | Asked In IBM |

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

| | | Asked In IBM |

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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