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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.

The required sum = $\frac{1}{a} + \frac{1}{b}$ = $\frac{a + b}{a b}$ = $\frac{55}{600}$ = $\frac{11}{120}$

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Filed Under: HCF and LCM

Q:

Which of the following has the most number of divisors?

A) 99	B) 101
C) 176	D) 182

Answer & Explanation Answer: C) 176

Explanation:

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

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Filed Under: HCF and LCM

Q:

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A) 123	B) 127
C) 235	D) 305

Answer & Explanation Answer: B) 127

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

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Filed Under: HCF and LCM

Q:

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

Answer & Explanation Answer: D) 364

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4 = 364.

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Filed Under: HCF and LCM

Q:

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A) 4	B) 10
C) 15	D) 16

Answer & Explanation Answer: D) 16

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes,they will together (30/2)+1=16 times

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Filed Under: HCF and LCM

Q:

A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A) 1/3	B) 1/4
C) 1/5	D) 1/7

Answer & Explanation Answer: C) 1/5

Explanation:

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = $(3 - \frac{3 x}{8} + x)$ litres.

Quantity of syrup in new mixture = $(5 - \frac{5 x}{8})$ litres.

$(3 - \frac{3 x}{8} + x) = (5 - \frac{5 x}{8})$

=> 5x + 24 = 40 - 5x

=> 10x = 16 => x = 8/5

So, part of the mixture replaced = $(\frac{8}{5} \times \frac{1}{8})$ = 1/5.

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Filed Under: Alligation or Mixture

Q:

Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?

A) 4:3	B) 3:4
C) 5:6	D) 7:9

Answer & Explanation Answer: D) 7:9

Explanation:

Let the C.P. of spirit be Re. 1 litre.
Spirit in 1 litre mix. of A = 5/7 litre, C.P. of 1 litre mix.
in A = Re. 5/7
Spirit in 1 litre mix. of B = 7/13 litre, C.P. of 1 litre mix.
in B = Re. 7/13
Spirit in 1 litre mix. of C = 8/13 litre, Mean price = Re. 8/13.

By the rule of alligation, we have:
Cost of 1 litre mixture in A Cost of 1 litre mixture in B