#### Data Interpretation

• ( 1 ) One quality of rice at Rs. 8.40 per kg is mixed with another quality at a certain rate in the ratio 5 : 6. If the mixture so formed be worth Rs. 10 per kg, what is the rate per kg of the second quality of rice?

• 1) Rs. 12.17
• 2) Rs. 10.60
• 3) Rs. 10.80
• 4) Rs. 15
• Discussion in forum
Answer : 1) Rs. 12.17
Solution : Let the rate of second quality be Rs x per Kg.
C.P of 1Kg rice of 1st kind = 840p
C.P of 1 Kg rice of 2nd kind = 100x p
Mean price = 1000p
By rule of alligation we have required ratio 5 : 6 So we get required ratio, (x-10) : 2.6 :: 5 : 6
=> x = 12.17 per Kg

discussion

Answer : 1) Rs. 12.17

• ( 2 ) A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?

• 1) 1 kg
• 2) 3 kgs
• 3) 5 kgs
• 4) 6 kgs
• Discussion in forum
Answer : 3) 5 kgs
Solution : If the selling price of mixture is Rs.30/kg and the merchant makes a profit of 20%,
then the cost price of the mixture =30/1.2 = Rs.25/kg.
We need to find out the ratio in which the three varieties are mixed to obtain a mixture costing Rs.25 /kg.
Let variety A cost Rs.20/kg, variety B cost Rs.24 / kg and variety C cost Rs.30/kg.
The mean desired price falls between B and C.
Step 1: Find out the ratio Qa:Qc using alligation rule.
Qa/Qc = (30-25)/(25-20) = 1/1
Be warned if we do not simplify the ratio and let if remain Qa/Qc = 5/5, it changes the final answer, which will be the COREECT answer as well. See at the end of this answer for more details.
Step 2: Find out the ratio Qb:Qc using alligation rule.
Qb/Qc = (30-25)/(25-24) = 5/1
Step 3: Qc is found by adding the value of Qc in step 1 and step 2 =1+1=2
Therefore, the required ratio =1:5:2
If there are 2 kgs of the third variety in the mixture, then there will be 5 kgsof the second variety in the mixture.

discussion

Answer : 3) 5 kgs

• ( 3 ) From a cask of milk containing 30 litres, 6 litres are drawn out and the cask is filled up with water.If the same process is repeated a second, then a third time, what will be the number of litres of milk left in the cask?

• 1) 5.12 litres
• 2) 12 litres
• 3) 14.38 litres
• 4) 15.36 litres
• Discussion in forum
Answer : 4) 15.36 litres
Solution : The problem can be solved by traditional method but it is cumbersome process to do that. The problem is simple if its solution is simpler. Hence we will go for a simpler solution for this kind of problem.
There is a short cut method to find the Quantity of milk left after nth operation.
It is given by [(x - y)/x]n of the whole quantity, where x is initial quantity of milk in the cask y is the quantity of milk withdrawn in each process and n is the number of process.
Hence from the above rule it can be say that Quantity of milk left after the 3rd operation = [(30 - 6)/30]3 * 30 = 15.36 liters.

discussion

Answer : 4) 15.36 litres

• ( 4 ) In a can, there is a mixture of milk and water in the ratio 4 : 5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6 : 5. Find the capacity of the can?

• 1) 40
• 2) 44
• 3) 48
• 4) 52
• Discussion in forum
Answer : 2) 44
Solution : Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.

discussion

Answer : 2) 44

• ( 5 ) A can contains a mixture of two liquids. A and B in proportion 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially ?

• 1) 10
• 2) 20
• 3) 21
• 4) 25
• Discussion in forum
Answer : 3) 21
Solution : Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A in mixture left = (7x - (7/12) x 9) litres = (7x - 21/4) litres.
Quantity of B in mixture left = (5x - (5/12) x 9) litres = (5x - 15/4) litres.
(7x - 21/4)/[(5x - 15/4) + 9] = 7/9
=> (28x - 21)/(20x + 21) = 7/9
=> 252x - 189 = 140x + 147
=> 112x = 336
=> x = 3.
So, the can contained 21 litres of A

discussion

Answer : 3) 21

• ( 6 ) The ratio of milk and water in 66 litres. of adulterated milk is 5 : 1. Water is added to it to make the ratio 5 : 3. The quantity of water added is :

• 1) 22 litres.
• 2) 24.750 litres.
• 3) 16.500 litres.
• 4) 20 litres.
• Discussion in forum
Answer : 1) 22 litres.
Solution : concentration of water in mixture1 = 1/6 (since the ratio of milk and water = 5:1) ...(1)
concentration of water in pure milk= 1 ...(2)
Now the above mentioned items are mixed to form mixture2 where milk and water ratio = 5 : 3
=> concentration of water in mixture2 = 3/8
By rule of alligation, => Quantity of mixture1 : Quantity of water
= 5/8 : 5/24 = 1/8:1/24 = 1:1/3
Given that Quantity of mixture1 = 66 litres
=> 66 : Quantity of water =1:1/3
=> Quantity of water = 66*1/3=22 litres.

discussion

Answer : 1) 22 litres.

• ( 7 ) A trader has 50 kg of pulses, part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4%. How much quantity is sold at 14% profit and that at 6% loss?

• 1) 5 kg, 45 kg
• 2) 15 kg, 35 kg
• 3) 10kg, 40 kg
• 4) None of these
• Discussion in forum
Answer : 1) 5 kg, 45 kg
Solution : Hence, ratio of quantity sold at 14% profit and 6% loss,
= 2:18 = 1:9
Hence, pulses sold at 14% profit,
= (50*1)/10 = 5 kg
At 6% Wheat sold at loss = 45 kg.

discussion

Answer : 1) 5 kg, 45 kg

• ( 8 ) A mixture contains wine and water in the ratio 3:2 and another mixture contains them in the ratio 4:5. How many litres of latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water?

• 1) 27/5 litres
• 2) 27/3 litres
• 3) 9/2 litres
• 4) 15/4 litres
• Discussion in forum
Answer : 1) 27/5 litres
Solution : Mixture(M1) contains wine and water in ratio = 3 : 2
For 1 litre Mixture(M1),
Wine1 = 3/5 litres
Water1 = 2/5 litres
Mixture(M2) contains wine and water in ratio = 4 : 5
For 1 litre Mixture(M2),
Wine2 = 4/9 litres
Water2 = 5/9 litres
Now 3 litres of M1 must be mixed with 'X' litres of M2 to get equal quantities of Wine and Water
For 3 litres Mixture(M1),
Wine1 = 9/5 litres
Water1 = 6/5 litres
For X litres Mixture (M2), Wine2 = 4X/9
Water2 = 5X/9
Total quantity of Wine (Wine1 + Wine2) = Total Quantity of Water (Water1 + Water2)
==> 9/5 + 4X/9 = 6/5 + 5X/9
==> X = 27/5 litres
Hence 27/5 litres of M2 must be mixed with 3 itres of M1 to get equal quantities of wine and water.

discussion

Answer : 1) 27/5 litres

• ( 9 ) In a zoo, there are Rabbits and Pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there?

• 1) 90
• 2) 100
• 3) 110
• 4) 120
• Discussion in forum
Answer : 3) 110
Solution : Heads Count = 200.
Legs count = 580.
Average Legs count for per head = 580/200 = 29/10. Rabbits : Pigeons = (9/10) : (11/10)= 9 : 11.
Number of Pigeons = (200 *11)/20 = 110.

discussion

Answer : 3) 110

• ( 10 ) A lump of two metals weighing 18 g is worth Rs. 87 but if their weight is interchanged, it would be worth Rs. 78.60. If the price of one metal be Rs. 6.70 per gram, find the weight of the other metal in the mixture.

• 1) 8g
• 2) 12g
• 3) 15g
• 4) 18g
• Discussion in forum
Answer : 1) 8g
Solution : Cost of (18 g of 1st metal + 18 g of 2nd metal) = Rs. 165.60
Cost of (1 g of 1st metal + 1 g metal of 2nd metal) = Rs. 9.20
Hence cost of 1 g of 2nd metal,
= 9.20 - 6.70
= Rs. 2.5
Mean price = Rs. 87/18 Now, quantity of 1st metal /quantity of 2nd metal = 14/6:56/30 = 5:4
Quantity of 2nd metal = (18*4)/9 = 8 g.

discussion

Answer : 1) 8g