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

A. 4

B. 7

C. 9

D. 13

Ans:A

Explanation: Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)= H.C.F. of 48, 92 and 140 = 4.

- 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

A. 276

B. 299

C. 322

D. 345

Ans:C

Explanation: Clearly, the numbers are (23 x 13) and (23 x 14).

Larger number = (23 x 14) = 322.

- 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:

A. 4

B. 5

C. 6

D. 8

Ans:A

Explanation: 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

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

A. 9000

B. 9400

C. 9600

D. 9800

Ans:C

Explanation: 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.

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

A. 101

B. 107

C. 111

D. 185

Ans:C

Explanation: 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.

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

A. 40

B. 80

C. 120

D. 200

Ans:A

Explanation: Let the numbers be 3*x*, 4*x* and 5*x*.

Then, their L.C.M. = 60*x*.

So, 60*x* = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

- The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

A. 12

B. 16

C. 24

D. 48

Ans:D

Explanation: Let the numbers be 3*x* and 4*x*. Then, their H.C.F. = *x*. So, *x* = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

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

A. 74

B. 94

C. 184

D. 364

Ans:D

Explanation: 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.

- 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:

A. 279

B. 283

C. 308

D. 318

Ans:C

Explanation:(11*7700)/275=308

- The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

A. 1008

B. 1015

C. 1022

D. 1032

Ans:C

Explanation: Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7 = 1015

- Four bells commence tolling together and toll at intervals of 6,7, 8 and 9 seconds respectively. In 2hours how many times do they toll together?

A. 15

B. 16

C. 50

D. None

Ans:A

Explanation: lcm of 6,7,8,9 is 504

in 2hrs( 7200 seconds),it will ring together 7200/504=14.2==>14

initially 1 time

total= 14+1=15 times

12. Two number , both greater than 27,have hcf 27 and lcm 162 .The sum of the number is

A. 189

B. 162

C. 135

D. 81

Ans:C

Explanation: H.C.F*l.C.M= product of two no.

162*27=4374

now find out the factors of 4374

ANd find out the suitavle factors.I.e 54*81=4374 which have h.cf=27 and L.c.m-162.

therefore 54+81=135 Ans(c)