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A. 100/6 %
B. 50/6 %
C. 100/3 %
D. 50/3 %
E. None of the above
Sol. Original Mixture = x L
In (x + 1) Mixture, quantity of milk = (x + 1)* (25/100) = (x + 1)/4
one litre of milk is added to the new mixture
[((x + 1)/4 )+ 1 ]/ x + 2 = 40%
x = 3 ; quantity of milk = (3 + 1)/4 = 1L
percentage of milk in the original mixture = 1/3 * 100 = 100/3 %.
2. 60 kg of a certain variety of Sugar at Rs.32 per kg is mixed with 48 kg of another variety of sugar and the mixture is sold at the average price of Rs.28 per kg. If there be no profit or no loss due to the new selling price, then what is the price of second variety of Sugar?
A. Rs.25
B. Rs.23
C. Rs.29
D. Rs.27
E. None of the above
Sol. Total CP of first variety = 60 * 32 = 1920
Total CP of second variety = 48 * x = 48x
SP of Mixture = 1920 + (108 * 28) = 3024
1920 + 48x = 3024 => x = 23.
3. A vessel is filled with 120 litres of Chemical solution, Acid “A”. Some quantity of Acid “A” was taken out and replaced with 23 litres of Acid “B” in such a way that the resultant ratio of the quantity of Acid “A” to Acid “B” is 4:1. Again 23 litres of the mixture was taken out and replaced with 28 litre of Acid “B”. What is the ratio of the Acid “A” to Acid “B” in the resultant mixture?
A. 43 : 29
B. 46 : 23
C. 47 : 21
D. 46 : 29
E. None of the above
Sol. In 23 litre mixture, Quantity of Acid “B” = 23 * 1/5 = 4.6 litre
Acid “A” in the mixture = 23 – 4.6 = 18.4 litre
120 – x / 23 = 4 / 1
x = 28
Ratio = 92-18.4 : 18.4 + 28
Ratio = 46 : 29.
4. 18 litres of Petrol was added to a vessel containing 80 litres of Kerosene. 49 litres of the resultant mixture was taken out and some more quantity of petrol and kerosene was added to the vessel in the ratio 2:1. If the respective ratio of kerosene and petrol in the vessel was 4:1, what was the quantity of kerosene added in the vessel?
A. 1 litre
B. 2 litre
C. 5 litre
D. 3 litre
E. None of the above
Sol. Total quantity of the mixture = 18+80 = 98 litre
quantity of petrol remaining = 18/2 = 9
quantity of kerosene remaining = 80/2 = 40
(40 + 2x) / (9 + x) = 4 / 1
x = 2
Quantity of kerosene added in the vessel = 2x = 4 litre.
5. A vessel which contains a mixture of acid and water in ratio 13:4. 25.5 litres of mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of acid is added to the mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture in the vessel before the replacement in litres?
A. 58 litre
B. 68 litre
C. 78 litre
D. 48 litre
E. None of the above.
Sol. Quantity of Acid = 13x
Quantity of water = 4x
Total = 17x
Resultant Mixture = 17x – 25.5 + 2.5 + 5 = 17x – 18
Resultant water = 4x – 25.5 * (4/17) + 2.5 = 4x – 3.5
Resultant mixture contains 25% water
(17x – 18)*25/100 = 4x – 3.5
x = 4
Initial quantity = 17*4 = 68.
6. A jar contains a mixture of two liquids A and B in the ratio 3:1 . When 15 ltrs of this mixture is drawn off and replaced with 9 ltrs of B,then ratio between A and B becomes 3:4. How many ltrs of liq. A was initially present in the jar ?
A. 20
B. 20.50
C. 20.25
D. 20.75
E. None of these
Sol. Let assume A = 3x ltrs and B = x ltrs.
Now 15 ltrs is drawn and 9 ltrs of B is added,
A = 3x - (3/4)*15 = (12x-45)/4
B = x - (1/4)*15 + 9 = (4x+21)/4
Now this ratio is equal to 3:4 , thus,
3/4 = (12x-45)/(4x+21)
x = 6.75
Thus A = 3x = 3*6.75 = 20.25 ltrs
7. Two equal glasses filled with mixture of alcohol and water in the ratio 2:1 and 1:1 respectively emptied in third glass. Find the ratio of alcohol and water in the third glass ?
A. 5:3
B. 7:3
C. 7:5
D. 8:5
E. None of these
Sol. For 1st glass,
Lets assume, alcohol = 2x and water = x
For 2nd glass,
Alcohol = y and water = y
Now as both glasses are equal, thus,
3x =2y
x = 2y/3
Now for 3rd glass,
Alcohol = 2x + y and water = x + y
Ratio of this will be,
(2x+y)/(x+y)
Put x = 2y/3 and solve,
= 7/5
Thus alcohol:water is 7:5.
8. A mixture of 15 ltrs contains two liquids A and B in the ratio 2:1. 5 ltrs is drawn and replaced with 5 ltrs of B, if the same process is repeated again, then find the amount of A at the end of it ?
A. 10
B. 12
C. 15
D. 16
E. None of these
Sol. Initial amount of A = (2/3)*15 = 10
Initial amount of B = (1/3)*15 = 5
Amount of A left after 2 withdrawls,
= 10*[(10-5)/5]^2
= 10 ltrs.
9. 400 gm spirit solution has 30 % spirit in it , what is the ratio of spirit should be added to make it 80 % in the solution ?
A. 2:5
B. 4:3
C. 5:2
D. 2:7
E. None of these
Sol. Using alligation rule,
= (100-80)/(80-30)
= 20 : 50.
10. Box A contains wheat worth Rs.30 per kg and box B contains wheat worth Rs.40 per kg.If both A and B are mixed in the the ratio 4:7 then the price of mixture per kg is ?
A. 36.36
B. 35.80
C. 42.50
D. 31.30
E. None of these
Sol. Lets assume the price of mixture per kg is 'x'
Using the alligation rule,
(40-x)/(x-30) = 4/7
Solving, x = 36.6
11. In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?
A. 11:77:7
B. 11:45:7
C. 25:45:8
D. 27:23:6
E. None of these
Sol. Using the alligation rule,
= (174-141)/(141-121)
= 33 : 21
= 11 : 7.
12. A vessel contains 63 litres of a mixture of milk and water. The ratio of milk to water is 3:4. If 14 litres of mixture is taken out from that vessel and then 6 litres of water added to it,What will be the percentage of milk in the final mixture ?
A. 40%
B. 39%
C. 38%
D. 37%
E. None of these
Sol. Milk : 63 × (3/7) = 27lit
Water : 63 × (4/7) =36lit
Milk in 14 lit mixture = 14 × (3/7) = 6lit
New mixture = (63 – 14 + 6) = 55 lit
Milk in new mix = 27 – 6 = 21
% of milk in the mix = (21/55) × 100 = 38.18% = 38%.
13. Sum of Rs.96 was shared among 48 boys and girls,each girl receive Rs.2.40 and boy receive Rs.1.80.How much amount did boys
A. 26
B. 27
C. 32
D. 30
E. None of these
Sol. Using the alligation rule,
= 0.2 : 0.4
= 1 : 2
No. of boys = (2/3)*48 = 32.
14. A dairy man pays Rs.6.40 per liter of milk. He adds water and sells the mixture at Rs.8 per liter thereby making 37.5% profit. Find the ratio of the water to milk received by the customers?
A. 1:15
B. 1:20
C. 1:12
D. 1:10
E. None of these
Sol. Suppose milk x and water y liters
Reqd. ratio of water and milk= y:x
CP of x liters milk=Rs.6.4x
SP of x liters milk=Rs.8(x+y)
64=8(x+y)*100/137.5
x:y=1:10.
15. In a 80 litre mixture of milk and water, the % of water is only 25%. The milkman gave 15 litres of this mixture to a customer and then added 15 litres of water to the remaining mixture. What is the % of milk in the final mixture ?
A. 61%
B. 57%
C. 49%
D. 63%
E. None of these
Sol. Water = 80*25/100 = 20litre
Milk = 80-20 = 60litre
18litre mixture
Milk = 15*60/80 = 11.25litres
Remaining quantity of milk = 60-11.25 = 48.75litres
% of milk in the mixture = 48.75*100/80 = 60.93 = 61%.
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