Crack IBPS PO : Mixture and Allegation Day 6
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Mixture and Alligation
No.of Questions: 10
Time : 10 minutes
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Question 1 of 10
1. Question
1 pointsCategory: AptitudeA bottle is fully filled with a mixture of which 5 parts are water and 10 parts of another liquid . What fraction of the mixture must be taken and replaced with water. So, that there is equal quantity of water and other liquid in the bottle?
Correct
(a)
Let the bottle contain total 15 parts,
Let x parts of the mixture be taken off
So,
10 10x/15=55x/15+x
x=15/4
So, part of mixture replaced with = 15/(4×15)=1/4
Incorrect
(a)
Let the bottle contain total 15 parts,
Let x parts of the mixture be taken off
So,
10 10x/15=55x/15+x
x=15/4
So, part of mixture replaced with = 15/(4×15)=1/4

Question 2 of 10
2. Question
1 pointsCategory: AptitudeA student gets 40%marks in physics and 60% mark in maths. If this overall percentage is 48% and the maximum marks for maths is 100%, whatis the maximum mark for physics?
Correct
(b)
Since maximum marks for maths is given as 100, this maximum marks for physics should be 150.
Incorrect
(b)
Since maximum marks for maths is given as 100, this maximum marks for physics should be 150.

Question 3 of 10
3. Question
1 pointsCategory: AptitudeA waiter stole wine from a butt of sherry which contained 30% spirit and replaced it with wine containing 6% spirit. The butt was of 12% strength. How much of the butt did he steal?
Correct
(d)
Now only ¼ of the sherrywith 30% spirit is left. Hence he has stolen ¾ of the butt.
Incorrect
(d)
Now only ¼ of the sherrywith 30% spirit is left. Hence he has stolen ¾ of the butt.

Question 4 of 10
4. Question
1 pointsCategory: AptitudeA merchant sold 400 kg of rice, part of which he sells at 10% gain and the rest at 18% gain. He gains 15% on the whole. Find the quantity sold at 18%.
Correct
(c)
The ratio of 10% gain to 18% gain is to be at 3:5. The quantity of 18% gain =
Incorrect
(c)
The ratio of 10% gain to 18% gain is to be at 3:5. The quantity of 18% gain =

Question 5 of 10
5. Question
1 pointsCategory: AptitudeIn three vessels the respective ratios of milk and water are 4:3, 1:2 and 2:3 in the three respective vessels. If all the three vessels are poured into a single large vessel, find the proportion of milk and water in the mixture.
Correct
(e)
Incorrect
(e)

Question 6 of 10
6. Question
1 pointsCategory: AptitudeTwo glasses filled with mixture of fruit juice and water in the proportion of 3:2 and 2:5 respectively are emptied into a third glass. What is this proportion of fruit juice and water in the third glass?
Correct
(d)
As per formula,
or 31:39
Incorrect
(d)
As per formula,
or 31:39

Question 7 of 10
7. Question
1 pointsCategory: AptitudeIn 1kg mixture of sand and cement, 20% is cement. How much sand is to be added so that the proportion of cement becomes 10%
Correct
(a)
There are 200 grams of cement in 1 kg (1,000 grams) of the mixture.
If we add x grams of sand, the total amount would be (1,000 + x) grams of mixture and we need 200 grams of cement to be 10% of that:0.1(1,000 + x) = 200
x = 1,000.
Incorrect
(a)
There are 200 grams of cement in 1 kg (1,000 grams) of the mixture.
If we add x grams of sand, the total amount would be (1,000 + x) grams of mixture and we need 200 grams of cement to be 10% of that:0.1(1,000 + x) = 200
x = 1,000.

Question 8 of 10
8. Question
1 pointsCategory: AptitudeA certain quantity of milk is mixed with 10 litres of water and the resultant mixture is worth Rs.12/ per litre. If the cost of pure milk is Rs.20/ per lit, then how much milk is there in the mixture?
Correct
(b)
Mixture = Milk + Water
Cost of pure milk = Rs.20 per litre
Cost of water = Rs.0
Mean price of mixture = Rs.12
The required ratio is 2:3 =>10:15
Given quantity of water is 10 litres
Milk = 10 ×1.5 = 15 litres
Validation total water 10 + total milk: 15
Mixture > 25
Mixture cost = 25 × 12 = 30
Original cost = 15 × 20 = 300
Incorrect
(b)
Mixture = Milk + Water
Cost of pure milk = Rs.20 per litre
Cost of water = Rs.0
Mean price of mixture = Rs.12
The required ratio is 2:3 =>10:15
Given quantity of water is 10 litres
Milk = 10 ×1.5 = 15 litres
Validation total water 10 + total milk: 15
Mixture > 25
Mixture cost = 25 × 12 = 30
Original cost = 15 × 20 = 300

Question 9 of 10
9. Question
1 pointsCategory: AptitudeIn what ratio must water be added to milk gain 20% by selling the mixture at cost.
Correct
(c)
Let the cost price of milk be Rs.10/
Then S.P of mixture: Rs.10/
It is given that the S.P. is with a profit of 20%.
That is if Rs.100 is CP > S.P is 120
or if SP is 120 > CP 100
If SP of mixture 10 > CP 100/120×10 = 25/3
Now allegation formula
The ratio of water to milk is 1:5,
Validation: Let no take litre of water, and then it is 5 litres of milk.
Original cost of milk: 5 10 = Rs. 50
Cost of mixture = 6 10 = Rs. 60
(1 water + 5 milk)
Profit 60 – 50 = 10
=> % 10/50×100 = 20%
Incorrect
(c)
Let the cost price of milk be Rs.10/
Then S.P of mixture: Rs.10/
It is given that the S.P. is with a profit of 20%.
That is if Rs.100 is CP > S.P is 120
or if SP is 120 > CP 100
If SP of mixture 10 > CP 100/120×10 = 25/3
Now allegation formula
The ratio of water to milk is 1:5,
Validation: Let no take litre of water, and then it is 5 litres of milk.
Original cost of milk: 5 10 = Rs. 50
Cost of mixture = 6 10 = Rs. 60
(1 water + 5 milk)
Profit 60 – 50 = 10
=> % 10/50×100 = 20%

Question 10 of 10
10. Question
1 pointsCategory: AptitudeIn what ratio must a businessman mix two types of rice costing Rs.36 and Rs.48 per kg. So that by selling this mixture at Rs.44 per kg he gains 10% profit.
Correct
(a)
The S.P of the mixture is Rs.44 which includes a profit of 10%.
Find the C.P of the mixture: of Rs.100 in CP > S.P is 110 (10% Profit)
110 is SP > CP 100
44 is SP >100/110×44 = 40 (C.P)
By allegation method,
I type & II type should be mixed in the ratio 2:1.
Validation: Cost of 2 kg of I type = 36 2 = 72
Cost of 1 kg of II type = 48 1 = 48
120
S.P of 3kgs @ 44 per kg = 132
Profit % = 12/120×100=10%
Incorrect
(a)
The S.P of the mixture is Rs.44 which includes a profit of 10%.
Find the C.P of the mixture: of Rs.100 in CP > S.P is 110 (10% Profit)
110 is SP > CP 100
44 is SP >100/110×44 = 40 (C.P)
By allegation method,
I type & II type should be mixed in the ratio 2:1.
Validation: Cost of 2 kg of I type = 36 2 = 72
Cost of 1 kg of II type = 48 1 = 48
120
S.P of 3kgs @ 44 per kg = 132
Profit % = 12/120×100=10%
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