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01. Gold is 19 times as heavy as water and copper 9 times. In what ratio should these metals be mixed that the mixture may be 15 times as heavy as water?
A. 1:2 B. 2:1
C. 2:3 D. 3 : 2

Answer and Explanation

Answer: 3 : 2

Explanation:
Quantity of Metal Gold=19, Copper = 9;
Resultant = 15
Proportion 6 : 4
Gold : Copper = 6 : 4 = 3 : 2

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02. A mixture has 100 kg of sugar, part of which he sells at 7 p.c. profit and the rest at 17 p.c. profit. He gains 10 p.c. on the whole. Find how much he sold at 7% profit?
A. 70 kg and 30 kg B. 30 kg and 70 kg
C. 15 kg and 43 kg D. 25 kg and 30 kg

Answer and Explanation

Answer: 30 kg and 70 kg

Explanation:
If the cost price of each kind of sugar be Rs. 100 then the selling price of one kind is Rs .107 (7% profit),Selling price of the other kind is Rs . 117 (17% profit).
Selling price of the mixture = Rs.110
SP of Each R1=117,R2=107
Mixture = 110
Proportion 3:7
Let the quantity of each in 100 kg be 3x & 7x
3x + 7x = 100=>x=10.The quantity of each in 100 kg is 30kg and 70 kg

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03. What quantity of sugar costing Rs 6.10 per kg must be mixed with 126 kg of sugar priced at Rs. 2.85 per kg, so that 20% may be gained by selling the mixture at Rs 4.80 per kg?
A. 49 kg B. 59 kg
C. 69 kg D. 68 kg

Answer and Explanation

Answer: 69 kg

Explanation:
R1=Sugar costing Rs 2.85 per kg
R2= Sugar costing Rs 6.10 per kg
RM=Average cost of mixture
We know that the mixture is being sold at a profit of 20%
ie, cost price of mixture=(100/120)*4.80=Rs 4

RM = Rs 4 = 400 paise

R1=285 paise
R2=610 paise

N1=210
N2=115

N1/N2 = 210/115 = 42/23

If quantity of sugar at 285 paise per kg is 42 kg,
Then sugar at 610 paise per kg = 23 kg

∴ if sugar at 285 paise per kg is 126 kg,
Then sugar at 610 paise per kg=69 kg

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04. 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 the liquid A was contained by the can initially?
A. 10 B. 21
C. 20 D. 25

Answer and Explanation

Answer: 21

Explanation:

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05. Vessel ‘A’ contains 150 litres of mixture of milk and water in the ratio of 3:1. Vessel ‘B’ contains 200 litres of the mixture in the ratio of 5:2. If the mixture from both the vessels are poured into vessel ‘C’, find the ratio of the resulting mixture.
A. 15 : 7 B. 8 : 3
C. 15 : 2 D. 143 : 53

Answer and Explanation

Answer: 143 : 53

Explanation:
As vessel A contains 150 litres of mixture of milk and water in the ratio 3:1, the quantity of milk is 3*150/4 litres, and that of water is 150/4 litres.
Similarly, as vessel B contains 200 litres of mixture of milk and water in the ratio 5:2, the quantity of milk is 5*200/7, and that of water is 2*200/7 litres.
After mixing the total quantity is 350 litres.
The quantity of milk is {(3*150/4) + (5*200/7)} = 7150/28
The quantity of water is {(150/4) + (2*200/7)} = 2650/28
The ratio of the resulting mixture = 7150/2650 = 143 : 53

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06. To gain profit, a milkman adds 1.5 litre of water to a certain amount of milk. He sells the adulterated milk at Rs.28 per litre and pure milk costs Rs.36 per litre. Find the amount of milk.
A. 5.25 B. 6
C. 5.5 D. 5

Answer and Explanation

Answer: 5.25

Explanation:
Let the amount of milk be x.
36x = 28(x + 1.5)
36x = 28x + 42
8x = 42
x = 5.25 litres

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07. A vessel has 500 ml of pure milk. From these vessels 50 ml of the milk is removed and 50 ml of water poured into the vessel. If this operation is repeated 2 more times, what is the percentage of milk in the vessel at the end?
A. 24.30% B. 81%
C. 72.9% D. 75%

Answer and Explanation

Answer: 72.9%

Explanation:
We know [(a – b) / a]n = c.
Where a = 500 ml, b = 50ml, n = 3
Proportion of the milk = [(500 – 50) / 500]3
= (450/500)3
= (9/10)3
= 0.729
Percentage of milk in the vessel = 100c = 72.9%

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08. A lump of two metals weighing 18 grams is worth Rs 74 but if their weights be interchanged, it would be worth Rs 60.10. If the price of the god is Rs 7.20 per gram, find the weight of the other metal in the mixture.
A. Rs 0.15/gm B. Rs 1.25/gm
C. Rs 0.25/gm D. Rs 1.33/gm

Answer and Explanation

Answer: Rs 0.25/gm

Explanation:
Let the weight of the gold be x gm
Weight of other metal = (18 - x) gm
Let the price of the other metal be Rs y / gm
7.20x + y(18 – x) = 84
7.20(18 – x) + yx =60.10
Solving these tow equations simultaneously , x =10,y = 0.25
Quality of the other metal = 18 – 10=8 gm
Price of other metal = Rs 0.25/gm

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09. A mixture of 70 liters of wine and water contains 10 % of water. How much water must be added to make water 37 % of the resulting mixture?
A. 30 liters B. 31 liters
C. 32 liters D. 35 liters

Answer and Explanation

Answer: 30 liters

Explanation:
The mixture contains (10/100)70 = 7 liters of water
It contains (70 - 7) = 63 liters of wine
Let x liters of water be added
(7 + x)/(70 + x)=37/100
X = 30 liters

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10. How much sugar costing Rs. 10 per kg be mixed with 80 kg of sugar costing Rs. 4 per kg so that the resultant mixture costs Rs. 5 per kg?
A. 15kg B. 15.5kg
C. 16kg D. 17kg

Answer and Explanation

Answer: 16kg

Explanation:
Price of the cheaper item = 4
Price of the costlier item = 10
Mean price = 5

Quantity of Costlier/ Quantity of Cheaper = 1/5
= Quantity of costlier = Quantity of the cheaper / 5
= Quantity of costlier = 80 / 5 = 16 kg.

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11. The ratio between the length and the breadth of a rectangular field is 3 : 2. If only the length is increased by 5 m, the new area of the field will be 2,600 sq metres. What is breadth of the rectangular field?
A. 40 m B. 60 m
C. 65 m D. Cannot be determined

Answer and Explanation

Answer: 40 m

Explanation:
Let length = 3x m and breadth = 2x m
Then, (3x + 5) * 2x = 2600 => 6x2 + 10x – 2600 = 0
3x2 + 5x – 1300 = 0 (3x + 65) (x – 20) = 0 , x = 20
Breadth = 2x = 40 m

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