A vessel is filled to its capacity with pure milk. Ten litres are withdrawn from it and replaced by water. This procedure is repeated again. The vessel now has 32 litres of milk. Find the capacity of the vessel (in litres).
Lets assume the capacity of the vessel = x litres.
x (1- 10/x)^2 = 32
x^2 +100-20x = 32x
x^2 +100 - 52x= 0
x^2 - 50x - 2x + 100 = 0
=> (x-2) (x-50) = 0
=> x=2 & x=50
As 10 litres is withdrawn so vessel capacity will be 50 litres.
Workspace
NA
SHSTTON
26
Solv. Corr.
38
Solv. In. Corr.
64
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1 M:19 S
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32 / 46
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In a mixture of petrol and kerosene, petrol is only 99 litres. If this same quantity of petrol would be presented in another mixture of petrol and kerosene where total volume would be 198 litres less than the actual mixture then the concentration of petrol In the present mixture would have been 13.33% point less than that. What is the concentration of petrol in actual mixture?
Lets assume amount of petrol = x
So, (x*100)/198=13.33
x=26.3934
Now same amount of petrol is in 99 liters also.
so the percentage in actual mixture is
(26.3934*100)/99 =26.66%
Workspace
NA
SHSTTON
55
Solv. Corr.
85
Solv. In. Corr.
140
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A juice container was 4/5 full. Ravinder poured 4 glasses from it and then poured two glasses back. The container is 3/4 full now. How many glasses can be filled if the entire container of juice is emptied?
Lets assume the quantity of tank = x
x/7 + 22 = x/5
=> x/5-x/7 = 22
=> 2x/35 = 22 => x = 385
Workspace
NA
SHSTTON
15
Solv. Corr.
10
Solv. In. Corr.
25
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You need a 18% acid solution for a certain test, but your supplier only ships a 13% solution and a 43% solution. You need 120 lts of the 18% acid solution. the 13% solution costs Rs 82 per ltr for the first 67 ltrs, and Rs 66 per ltr for any amount in access of 67 ltrs. What is the cost of the 13% solution you should buy?
Let us assume we need "a" liters of 13% acid solution and "b" liters of 43% acid solution. Now
=>18 = (a * 13 + b * 43)/(a + b)
=>a/b = 5/1
So we need 100 liters of 13% acid solution, and 20 liters of 18% acid solution.
Final cost = 82 * 67 + 66 * 33 = 7672
Workspace
NA
SHSTTON
11
Solv. Corr.
6
Solv. In. Corr.
17
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Three containers A, B and C are having mixtures of milk and water in the ratio of 1:5, 3:5, 5:7 respectively. If the capacities of the containers are in the ratio 5:4:5, find the ratio of milk to water, if all the three containers are mixed together.
Assume that there are 500,400 and 500 litres respectively in the 3 containers.
Then, we have, 83.33, 150 and 208.33 litres of milk in each of the three containers.
Thus, the total milk is 441.66 litres. Hence, the amount of water in the mixture is
1400 - 441.66 = 958.33liters.
Hence, the ratio of milk to water is
441.66 : 958.33 => 53:115 (using division by .3333)
The calculation thought process should be
(441*2+2) : (958*3+1) = 1325 : 2875.
Dividing by 25 => 53:115.
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NA
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21
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17
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38
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On a certain assembly line, the rejection rate for Hyundai i10s production was 4 percent, for Hyundai i20s production 8 percent and for the 2 cars combined 7 percent. What was the ratio of Hyundais i10 production?
How many liters of a 90% of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid?
concentration is given which is wrt 100, hence we can take x lt of 90% and (30-x)of 75%
x*90 + (30-x)*75 = 30*78 hence the ans is 6 ltr.
Workspace
NA
SHSTTON
13
Solv. Corr.
12
Solv. In. Corr.
25
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Eesha bought two varities of rice costing Rs. 50 per kg and Rs. 60 per kg and mixed them in some ratio. Eesha sold that mixture at Rs. 70 per kg making a profit of 20 %. What was the ratio of the micxture?
In 30kg solution sugar is 2%
So finding out the quantity of sugar = 0.02*30= 0.6kgs
Now let x kgs of sugar be added to make 10% concentration
So (0.6+x)/(30+x) = 10/100
Solving we get x= 2.6kgs.
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