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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:

Tea worth of Rs. 135/kg & Rs. 126/kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be____?

A) Rs. 169.50 B) Rs.1700
C) Rs. 175.50 D) Rs. 180
 
Answer & Explanation Answer: C) Rs. 175.50

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

 

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind 

 

 

  

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

 Hence, price of the third variety = Rs.175.50 per kg.

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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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Q:

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

A) 18 litres B) 24 litres
C) 32 litres D) 42 litres
 
Answer & Explanation Answer: B) 24 litres

Explanation:

Let the quantity of the wine in the cask originally be x litres

 
Then, quantity of wine left in cask after 4 operations =x1-8x4litres

 

x1-8x4x = 1681  

 

 1-8x4=234 

 

x=24

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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 

 60 - 3x = 2x - 30

  x = 18.

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Q:

A can contains a mixture of two liquids A and B in the ratio  7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

A) 10 B) 20
C) 21 D) 25
 
Answer & Explanation Answer: C) 21

Explanation:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

 

Quantity of A in mixture left =7x-712×9=7x-214 litres.

 

Quantity of B in mixture left =  5x-512×9=5x-154litres.

 

7x-2145x-154+9=79

 

28x-2120x+21=79

 

x=3

 So, the can contained 21 litres of A.

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Q:

The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is

A) 15 cm B) 25 cm
C) 35 cm D) 42 cm
 
Answer & Explanation Answer: C) 35 cm

Explanation:

Required Length = H.C.F of 700 cm, 385 cm and 1295 c

                             = 35 cm.

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Filed Under: HCF and LCM

Q:

An observer 1.6 m tall is 203away 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 =

 DEtan30=> DE = tan30 x 203=> DE = 13x203 = 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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