32
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

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.

Q:

Ratio of water and milk in a container is 2 : 3. If 40 liter mixture removed from the container and same quantity of milk is added to it then the ratio of water to milk becomes 1 : 4. Find the initial quantity of mixture ?

 A) 75 lit B) 80 lit C) 85 lit D) 90 lit

Explanation:

From the given data,

let the initial quantity of the mixture = 5x

Then,

Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.

10 954
Q:

A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, the profit % ?

 A) 22.5% B) 25% C) 20% D) 33.33%

Explanation:

As water costs free, water sold at cost price of milk gives the profit.

Required profit % = 5/20 x 100 = 5 x 5 = 25%.

5 842
Q:

Shiva purchased 280 kg of Rice at the rate of 15.60/kg and mixed it with 120 kg of rice purchased at the rate of 14.40/kg. He wants to earn a profit of Rs. 10.45 per kg by selling it. What should be the selling price of the mix per kg?

 A) Rs. 22.18 B) Rs. 25.69 C) Rs. 26.94 D) Rs. 27.54

Explanation:

Rate of rice of quantity 280 kg = Rs. 15.60/kg

Rate of rice of quantity 120 kg = Rs. 14.40/kg

He want to earn a profit of Rs. 10.45/kg

Rate of Mix to sell to get profit of 10.45 =

8 1418
Q:

Two bottles contains mixture of milk and water. First bottle contains 64% milk and second bottle contains 26% water. In what ratio these two mixtures are mixed so that new mixture contains 68% milk?

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

Explanation:

% of milk in first bottle = 64%

% of milk in second bottle = 100 - 26 = 74%

Now, ATQ

64%            74%

68%

6                    4

Hence, by using allegation method,

Required ratio = 3 : 2

8 2211
Q:

In what ratio must wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg so that the mixture be worth Rs. 3/kg?

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

Explanation:

Given rate of wheat at cheap = Rs. 2.90/kg

Rate of wheat at cost = Rs. 3.20/kg

Mixture rate = Rs. 3/kg

Ratio of mixture =

2.90                              3.20

3

(3.20 - 3 = 0.20)            (3 - 2.90 = 0.10)

0.20 : 0.10 = 2:1

Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.

12 1767
Q:

An alloy contains gold and silver in the ratio 5 : 8 and another alloy contains gold and silver in the ratio 5 : 3. If equal amount of both the alloys are melted together, then the ratio of gold and silver in the resulting alloy is ?

 A) 113/108 B) 105/103 C) 108/115 D) 103/113

Explanation:

As given equal amounts of alloys are melted, let it be 1 kg.

Required ratio of gold and silver =

Hence, ratio of gold and silver in the resulting alloy = 105/103.

10 2919
Q:

A tin a mixture of two liquids A and B in the proportion 4 : 1. If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5. How much of the liquid B was there in the tin? What quantity does the tin hold?

 A) 58 l B) 65 l C) 50 l D) 62 l

Explanation:

Let the tin contain 5x litres of liquids

=> 5(4x - 36) = 2(x + 36)

=> 20x - 180 = 2x + 72

=> x = 14 litres

Hence, the initial quantity of mixture = 70l

Quantity of liquid B

= 50 litres.

16 3862
Q:

An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?

 A) 45 gm B) 50 gm C) 55 gm D) 60 gm

Explanation:

Initial quantity of copper = = 40 g

And that of Bronze = 50 - 40 = 10 g

Let 'p' gm of copper is added to the mixture

=> = 40 + p

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.