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Q:

A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is

A) 400 kg	B) 560 kg
C) 600 kg	D) 640 kg

Answer & Explanation Answer: C) 600 kg

Explanation:

By the rule of alligation:

Profit of first part Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3.

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.

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Filed Under: Alligation or Mixture

Q:

How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?

A) 36 Kg	B) 42 Kg
C) 54 Kg	D) 63 Kg

Answer & Explanation Answer: D) 63 Kg

Explanation:

By the rule of alligation:

C.P. of 1 kg sugar of 1st kind C.P. of 1 kg sugar of 2nd kind

Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3.

Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind.

Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.

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Filed Under: Alligation or Mixture

Q:

The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is

A) Rs. 19.50	B) Rs. 19
C) Rs. 18	D) Rs. 18.50

Answer & Explanation Answer: C) Rs. 18

Explanation:

Let the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have :

Cost of 1 kg of type 1 rice Cost of 1 kg of type 2 rice

$∴$ (20-x)/(x-15) = 2/3

$\Rightarrow$ 60 - 3x = 2x - 30

$\Rightarrow$ x = 18.

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Filed Under: Alligation or Mixture

Q:

An observer 1.6 m tall is $20 \sqrt{3}$ away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:

A) 21.6 m	B) 23.2 m
C) 24.72 m	D) None of these

Answer & Explanation Answer: A) 21.6 m

Explanation:

Draw BE // CD

Then, CE = AB = 1.6 m,

BE = AC =

$\frac{D E}{\tan 30} = > D E = \tan 30 x 20 \sqrt{3} = > D E = \frac{1}{\sqrt{3}} x 20 \sqrt{3} = 20$

Therefore, CD = CE + DE = (1.6 + 20) m = 21.6 m.

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Filed Under: Height and Distance
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Q:

A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour?

A) 3.6	B) 7.2
C) 8.4	D) 10

Answer & Explanation Answer: B) 7.2

Explanation:

Speed = $(\frac{600}{5 \times 60}) m / s e c$ = 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr

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Filed Under: Time and Distance

Q:

Two trains starting at the same time from two stations 200 km apart and going in opposite directions cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds?

A) 9 : 20	B) 11 : 9
C) 11 : 20	D) None of these

Answer & Explanation Answer: B) 11 : 9

Explanation:

In the same time, they cover 110 km and 90 km respectively.

Therefore, Ratio of their speeds = 110 : 90 = 11 : 9

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Filed Under: Time and Distance

Q:

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minutes?

A) 100 m	B) 150 m
C) 190 m	D) 200 m

Answer & Explanation Answer: A) 100 m

Explanation:

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr

Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

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Filed Under: Time and Distance

Q:

The distance between two cities A and B is 330 Km. A train starts from A at 8 a.m. and travel towards B at 60 km/hr. Another train starts from B at 9 a.m and travels towards A at 75 Km/hr. At what time do they meet?

A) 10 a.m	B) 10.30 a.m
C) 11 a.m	D) 11.30 a.m

Answer & Explanation Answer: C) 11 a.m

Explanation:

Suppose they meet x hrs after 8 a.m

then,

[Distance moved by first in x hrs] + [Distance moved by second in (x-1) hrs] = 330.

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