LCM and HCF

Q1.

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

279	283
308	318

Answer:

Option C

Solution:

Other number			= (11×7700 / 275)
			= 308.

Wrong Answer

Q2.

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

101	107
109	111

Answer:

Option D

Solution:

Therefore Greater number = 111.
Let the numbers be 37a and 37 b. Then, 37a ×37b = 4107
<=> ab = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 ×1, 37 ×3) i.e , (1 , 111)

Wrong Answer

Q3.

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

4	7
9	3

Answer:	Option A

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

Wrong Answer

Q4.

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:

276	229
322	345

Answer:	Option C

Solution:
Clearly, the numbers are (23 x 13) and (23 x 14). .'. Larger number = (23 x 14) = 322.

Wrong Answer

Q5.

The H.C.F of 204, 1190 and 1445 is:

17	18
19	21

Answer:

Option A

Solution:

204 = 2² x 3 x 17; 1190 = 2 x 5 x 7 x 17; 1445 = 5 x 17²

.'.	H.C.F. = 17

Wrong Answer

Written

Sunday, November 16, 2014