| 1. If two vessels of equal volume contain diesel and kerosene in the ratios 5:2 and m:n respectively, when the contents of both vessels are mixed in a larger vessel, the resulting ratio of diesel to kerosene in the mixture will be: | Medium |
| A. (12m+5n)/(2m+9n) B. (m+3)/(n+2) C. (3n+5m)/(2n+3m) D. (m+3n)/(m+2n) |
View Answer
Answer: Option A
Explanation:
Let the diesel and kerosene in the first vessel be 5x and 2x and in the 2nd vessel be my and ny. Then the ratio of diesel and kerosene in the mixture will be (5x+my)/(2x+ny)
Also, 5x + 2x = my + ny (since volumes are equal)
-> 7x = y(m+n)
-> x/y = (m+n)/7
-> x = (y/7) (m + n)
Hence the desired ratio = (5y/7 (m+n)+my)/(2y/7 (m+n)+ny) = (12m+5n)/(2m+9n)
Alternatively
Ratio of diesel and kerosene in the bigger vessel = (5/7+ m/((m+n)))/(2/7 +n/( (m+n))) = (12m+5n)/(2m+9n)
| 2. A trader blends tea A, which costs Rs. 100 per kg, with tea B, priced at Rs. 50 per kg, in a ratio of 4:1. When the price of tea B increases to Rs. 80 per kg, determine the new ratio for mixing tea A with tea B so that the cost of the mixture remains the same? | Medium |
| A. 2:3 B. 5:2 C. 7:3 D. 1:1 |
View Answer
Answer: Option D
Explanation:
Original Cost = (100 × 4 + 50 × 1)/5 = 450/5 = 90
Let x be the fraction of tea A in the total mixture. Then,
(x × 100) + (1 – x) × 80 = 90
∴ x = 1/2 = 50%
∴ 1 – x = 50%
Therefore, new ratio for mixing A and B is 1: 1
| 3. To make a profit of 20% by selling a mixture of salt at Rs. 48 per kg, determine the proportion in which a trader should mix salt priced at Rs. 36 per kg and Rs. 56 per kg. | Easy |
| A. 3:1 B. 4:1 C. 1:4 D. 7:3 |
View Answer
Answer: Option B
Explanation:
Selling price = Rs.24 per kg; Profit = 20%
∴ Cost price of mixture = 48/(1+ (20/100)) = Rs.40 per kg
Two salts to be mixed are at Rs. 36/kg and Rs. 56/kg, and the mean price is Rs. 40/kg. By the rule of alligation,
(56 – 40)/(40 – 36) = (Amount of Cheap)/(Amount of Dear)
-> 4/1 = (Amount of Cheap)/(Amount of Dear)
Note:
If two ingredients A and B are mixed, where A is less than B in terms of price, concentration level etc.
Let QA= Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB – PM)/(PM)- PA )
| 4. Alex stole wine from a container which originally contained 30% alcohol. The stolen wine was then replaced with an inferior quality wine containing only 16% alcohol. If the resulting mixture now contains 22% alcohol, determine the proportion of stolen wine in the new mixture. | Easy |
| A. 3/4 B. 4/3 C. 3/7 D. 4/7 |
View Answer
Answer: Option D
Explanation:
Applying rule of alligation, we get:
Original Replaced (stolen)
30 16
22
6 8
∴ Stolen Proportion = 8/(8+6) = 4/7
Note:
If two ingredients A and B are mixed, where A is less than B in terms of price, concentration level etc.
Let QA = Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB– PM)/(PM – PA)
| 5. In order to create a mixture where milk and water are in the proportion of 4:3, the contents of two vessels must be mixed. Vessel I contains a mixture of milk and water in the proportion of 3:4, while vessel II contains a mixture in the proportion of 3:2. In what ratio should the contents of vessel I and II be mixed? | Medium |
| A. 1:1 B. 1:5 C. 2:3 D. 3:2 |
View Answer
Answer: Option B
Explanation:
Applying rule of Alligation, we get:
3/7 3/5
4/7
1/35 1/7
Required ratio = (1/35)/(1/7) = 1: 5
Note:
If two ingredients A and B are mixed, where A is less than B in terms of price, concentration level etc.
Let QA = Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB– PM)/(PM – PA)
| 6. A cask initially contains wine. Six litres are drawn from the cask, and then it is filled with water. Afterward, another six litres of the resulting mixture are drawn, and the cask is filled with water again. The ratio of the quantity of wine now left in the cask to that of the water is 225:64. Determine the total capacity of the cask in litres. | Difficult |
| A. 60 B. 51 C. 55 D. None of these |
View Answer
Answer: Option B
Explanation:
Let the volume of the flask be x litres.
If 6 litres of wine are drawn and 6 litres of water are added, then volume of wine = x-6, volume of water = 6
Now, if 6 litres of mixture are drawn, then water drawn = 36/x, and wine drawn = 6 – (36/x) = (6x-36)/x
Wine left = (x-6) – (6x-36)/x = (x2 – 12x -+36)/x = (x-6)2/x
Water left = 6 – (36/x) + 6 = (12x – 36)/x
Ratio of wine left: water left = ((x-6)2/x): (12x – 36)/x
-> (x-6)2 /(12(x-3) = 225/64
Solve for x to get 51 as its value (you may also check options directly).
Alternatively
If a vessel contains a litre of solution I and if b litres are withdrawn and replaced with solution II, then b litres of the mixture drawn and replaced with solution II, then b litres of the mixture drawn and replaced with solution II and this operation is repeated n times, then:
(Solution I left in the vessel after nth operation)/(Total volume)=((a-b)/a)n
Wine left: Water = 225: 64
∴ Wine left: Wine left + Water = 225: 289
∴ ((Total quantity of wine-6)/(Total quantity of wine))2 = ((225 )/289) = (15/17)2
∴ ((Total quantity of wine-6 )/(Total quantity of wine)) = ((15 )/17)
∴Total quantity of wine = 51 litres
| 7. 2000 litres of 30% alcohol solution are mixed with x litres of 60% alcohol solution. If the resultant mixture is 40% alcohol solution, what is the value of x? | Easy |
| A. 1000 B. 1500 C. 1400 D. None of these |
View Answer
Answer: Option D
Explanation:
60 30
40
10 20
Therefore, x/10 = 2000/20
-> x = 2000/2 = 1000
Note:
If two ingredients A and B are mixed, where A is less than B in terms of price, concentration level etc.
Let QA = Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB– PM)/(PM – PA)
| 8. Three bottles of equal volume are completely filled with a mixture of milk and water in the ratio 2:1, 3:2, 4:3 respectively. If the contents of the 3 bottles are poured into a single vessel, then milk in that vessel is y times the quantity of water. What is the approximate value of y? A. 2.6 B. 4.6 C. 3.6 D. 1.6 |
Easy |
View Answer
Answer: Option D
Explanation:
Let the volume of each bottle be x
-> Quantity of milk in first bottle = (2/3)x
Similarly, quantity of water in first bottle = (3/5)x
Proceed on similar lines for other bottles to obtain the ratio of milk to water in the third vessel as
((2/3)x+ (3/5)x+ (4/7)x)/((1/3)x+ (2/5)x + (3/7)x) = 193/122 = 1.58
| 9. 1 litre of p% sugar solution is mixed with 9 litres of q% sugar solution to get 70% sugar solution. If p>q, and p and q are integers, then how many integer values can p take? | Medium |
| A. 1 B. 4 C. 2 D. 3 |
View Answer
Answer: Option D
Explanation:
Applying the law of alligation,
9/1= (p- 70)/(70-q)
-> p-70 = 9(70-q)
Now, 70-q is an integer since q is an integer -> p-70 is a multiple of 9. Check for values of p greater than 70 and less than 100 which make p-70 a multiple of 9. There are 3 such values (79, 88, 97).
Note:
If two ingredients A and B are mixed, where A is less than B in terms of price, concentration level etc.
Let QA = Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB– PM)/(PM – PA)
| 10. In what proportion should sugar at Rs. 11, at Rs. 13 and at Rs. 18 per kg be mixed so that the price of the mixture is Rs. 15 per kg? | Difficult |
| A. 1:1:2 B. 2:3:4 C. 1:2:1 D. 2:4:7 |
View Answer
Answer: Option A
Explanation:
S1 costs Rs 11, S2 costs Rs. 13, and S3 costs Rs. 18 per kg.
Cost of mixture = Rs. 15 per kg
11<13 ˂ 15 ˂ 18 (∴ we take S1 and S3 as one pair, and S2 and S3 as the other pair to apply the law of alligation)
-> S1: S3 = (18-15): (15-11) = 3: 4
-> S2:S3 = (18-15): (15-13) = 3: 2
-> S1: S2: S3 = 3:3: (4+2)
-> S1: S2: S3 = 3: 3: 6 = 1: 1: 2
Note:
If two ingredients A and B are mixed,
Let QA = Amount of A, QB = Amount of B
PA = Price of A, PB = Price of B, PM = Mean price
Then, by the rule of alligation QA/QB = (PB– PM)/(PM – PA)
If three ingredients are mixed
When a mixture of 3 ingredients A, B, C is given, take any two ingredients such that the cost of the mixture is between the costs of the two chosen ones and find the ratio. Once again take two more ingredients and find their ratio. Then find the combined ratio.