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Q:

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

A) 1008	B) 1015
C) 1022	D) 1032

Answer & Explanation Answer: B) 1015

Explanation:

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

= 1008 + 7

= 1015

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Q:

Find the Greatest Number that will devide 43, 91 and 183 so as to leave the same remainder in each case

A) 4	B) 7
C) 9	D) 13

Answer & Explanation Answer: A) 4

Explanation:

Required Number = H.C.F of (91- 43), (183- 91) and (183-43)

= H.C.F of 48, 92, and 140 = 4

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Q:

The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is:

A) 91	B) 910
C) 1001	D) 1911

Answer & Explanation Answer: A) 91

Explanation:

Required number of students = H.C.F of 1001 and 910 = 91

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Q:

L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2	B) -1
C) 1	D) 2

Answer & Explanation Answer: A) -2

Explanation:

H. C. F of two prime numbers is 1. Product of numbers = 1 x 161 = 161.

Let the numbers be a and b . Then , ab= 161.

Now, co-primes with product 161 are (1, 161) and (7, 23).

Since x and y are prime numbers and x >y , we have x=23 and y=7.

Therefore, 3y-x = (3 x 7)-23 = -2

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Q:

The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125 , then that number is

A) 77	B) 88
C) 99	D) 110

Answer & Explanation Answer: A) 77

Explanation:

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35

Now, co-primes with product 35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35) and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is (55,77)

Hence , required number = 77

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Q:

If the sum of two numbers is 55 and the H.C.F and L.C.M of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to

A) 55/601	B) 601/55
C) 11/120	D) 120/11

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b . Then, a+b =55 and ab = 5 x 120 = 600.

Therefore, Required sum = $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b} = \frac{55}{600} = \frac{11}{120}$

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Q:

The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

A) 10	B) 46
C) 70	D) 90

Answer & Explanation Answer: A) 10

Explanation:

Let the numbers be x and (100-x).

Then, $x (100 - x) = 5 * 495$

=> $x^{2} - 100 x + 2475 = 0$

=> (x-55) (x-45) = 0

=> x = 55 or x = 45

The numbers are 45 and 55

Required difference = (55-45) = 10

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Q:

The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

A) 12	B) 48
C) 84	D) 108

Answer & Explanation Answer: C) 84

Explanation:

Let the numbers be x and 4x. Then, $x \times 4 x = 84 \times 21 \Leftrightarrow x^{2} = \frac{84 \times 21}{4} \Leftrightarrow x = 21$

Hence Larger Number = 4x = 84

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