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Q:

Find the odd man out?

396, 462, 572, 427, 671, 264.

A) 671	B) 462
C) 427	D) 264

Answer & Explanation Answer: C) 427

Explanation:

Here the given series is 396, 462, 572, 427, 671, 264.

In all the terms, the middle digit is the sum of first and third digit except 427.

So the Odd number in the given series is 427.

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Filed Under: Odd Man Out
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Q:

In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?

A) 50	B) 40
C) 30	D) 20

Answer & Explanation Answer: C) 30

Explanation:

There is a meal for 200 children. 150 children have taken the meal.

Remaining meal is to be catered to 50 children.

Now, 200 children = 120 men.

50 children = $(\frac{120}{200}) 50$ =30men

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Filed Under: Chain Rule

Q:

A fort had provision of food for 150 men for 45 days. After 10 days, 25 men left the fort. The number of days for which the remaining food will last, is:

A) 29/5	B) 37/4
C) 42	D) 54

Answer & Explanation Answer: C) 42

Explanation:

After 10 days : 150 men had food for 35 days.

Suppose 125 men had food for x days.

Now, Less men, More days (Indirect Proportion)

125 : 150 :: 35 : x => 125 x x = 150 x 35

=>

=> x = 42.

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Filed Under: Quantitative Aptitude - Arithmetic Ability

Q:

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:

A) 173 m	B) 200 m
C) 273 m	D) 300 m

Answer & Explanation Answer: C) 273 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, $∠ A C B = 30 ° a n d ∠ A D B = 45 °$

$\frac{A B}{A C} = \tan 30 ° = \frac{1}{\sqrt{3}} = > A C = A B * \sqrt{3} = 100 \sqrt{3 m}$

$\frac{A B}{A D} = \tan 45 ° = 1 = > A D = A B = 100 m$

CD=(AC+AD)= $(100 \sqrt{3} + 100) m = 100 (\sqrt{3} + 1) = 100 * 2.73 = 273 m$

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Filed Under: Height and Distance

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:

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A) 26 minutes and 18 seconds	B) 42 minutes and 36 seconds
C) 45 minutes	D) 46 minutes and 12 seconds

Answer & Explanation Answer: D) 46 minutes and 12 seconds

Explanation:

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

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Filed Under: Quantitative Aptitude - Arithmetic Ability

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:

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

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