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Q:

A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

A) 57	B) 67
C) 77	D) 87

Answer & Explanation Answer: C) 77

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = $\sqrt{5929}$ = 77.

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Filed Under: Square Roots and Cube Roots

Q:

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

A) 1677	B) 1683
C) 2523	D) 3363

Answer & Explanation Answer: B) 1683

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

Required number = (840 x 2 + 3) = 1683.

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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:

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: B) 2

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

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Filed Under: HCF and LCM

Q:

Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.

A) 1:3	B) 2:3
C) 3:4	D) 4:5

Answer & Explanation Answer: B) 2:3

Explanation:

By the rule of alligation:
Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind

Required ratio = 60 : 90 = 2 : 3

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Filed Under: Alligation or Mixture

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

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