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A.18 ltr

B.24 ltr

C.32 ltr

D.42 ltr

2.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.169.50

B.170

C.175.50

D.200

3.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

4.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

B.42

C.54

D.63

5.A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

A) 10 liters

B) 20 liters

C) 30 liters

D) 40 liters

6.Milk and water are mixed in a vessel A as 4:1 and in vessel B as 3:2. For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C.

A) 1:9

B) 9:1

C) 3:7

D)7:3

7.From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:

A) 45L

B) 36.45L

C) 40.5L

D) 42.5L

A) 27 %

B) 26 %

C) 29 %

D) 21 %

9.A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk. What is the % of milk in the mixture?

A) 71.02%

B) 76.92%

C) 63.22%

D) 86.42%

10.A mixture of 70 litres of Fruit Juice and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12.5% water ?

A) 2 lit

B) 4 lit

C) 1 lit

D) 3 lit

SOLUTIONS:

1.

2.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

130.50 X

153

(X-153). 22.50

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

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

3.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−21/4)7x-7/12×9=7x-214 litres.

Quantity of B in mixture left = (5x−512×9)=(5x−15/4)5x-5/12×9=5x-154litres.

(7x−21/4)/(5x−15/4)+9=7/97

x-2145x-154+9=79

⇒28x−2120x+21=79

⇒28x-2120x+21=79

⇒x=3⇒x=3

4.

Answer & ExplanationAnswer: 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 =[x(1−8x)4]litres

∴[x(1−8x)4x] = 1681

⇒[1−8x]4=(23)4

9 7

8.40

1.40 0.60

⇒x=24

5.Answer: A) 10 liters

Explanation:

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

6.

Answer & ExplanationAnswer: D) 7:3

Explanation:

Ratio of Milk and water in a vessel A is 4 : 1

Ratio of Milk and water in a vessel B is 3 : 2

Ratio of only milk in vessel A = 4 : 5

Ratio of only milk in vessel B = 3 : 5

Let 'x' be the quantity of milk in vessel C

Now as equal quantities are taken out from both vessels A & B

=> 4/5 : 3/5

x

3/5-x x - 4/5

=> 3/5−x / x−4/5 = 1/1 (equal quantities)

=> x = 7/10

Therefore, quantity of milk in vessel C = 7

=> Water quantity = 10 - 7 = 3

Hence the ratio of milk & water 7:3

7.

Answer & ExplanationAnswer: B) 36.45L

Explanation:

General Formula:

Final or reduced concentration = initial concentration x (1−amount being replaced in each operation/ total amount)^n

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time

Therefore, 50×(1−550)^3

= 36.45L

8.

Answer: A) 27 %

Explanation:

Explanation:

Let the percentage of benzene = X

(30 - X)/(X- 25) = 6/4 = 3/2

=> 5X = 135

X = 27

(30 - X)/(X- 25) = 6/4 = 3/2

=> 5X = 135

X = 27

So, required percentage of benzene = 27 %

9.Answer: B) 76.92%

Explanation:
Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

10.Answer: A) 2 lit

Explanation:
Quantity of fruit juice in the mixture = 70 - [70 x (10/100) ]= 63 litres.

After adding water, juice would form 87.5% of the mixture.

Hence, if quantity of mixture after adding x liters of water, [(87.5) /100 ]*x = 63 => x = 72

Hence 72 - 70 = 2 litres of water must be added