In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg?
A) 1:3
B) 3:1
C) 1:2
D) 2:1
Answer & Explanation
Answer:
D) 2:1
Explanation:
1st Variety RS.18/kg 2nd Variety Rs.24/kg
Mixture Rs.20/kg
the ratio of QDP/QCP = 4/2 = 2:1
Q.2:
In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?
A) 3 : 7
B) 5 : 7
C) 7 : 3
D) 7 : 5
Answer & Explanation
Answer:
C) 7 : 3
Explanation:
Cost of 1 kg Pulses of 1st Kind RS.15 Cost of 1 kg Pulses of 2nd Kind Rs.20
Mixture Rs.16 (Mean price)
Hence Quantity of cheaper (QCP):Quantity of Dearer (QDP) = Cp of Dearer quantity -mean price:mean price - Cp of Cheaper quantity = 20-16.5:16.5 - 15=3.5:1.5
the ratio of QCP/QDP = 3.5/1.5 = 7:3
Q.3:
The ratio in which the price at Rs.7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs.6.30 a kg is:
A) 1:3
B) 2:3
C) 3:4
D) 4:5
Answer & Explanation
Answer:
B) 2:3
Explanation:
Cost of 1 kg of dearer rice RS.7.5 Cost of 1 kg cheaper rice Rs.5.7
Mixture Rs.6.3 (Mean price)
Hence Quantity of cheaper (QCP):Quantity of Dearer (QDP) = Cp of Dearer quantity -mean price:mean price - Cp of Cheaper quantity = 6.3 - 5.7 : 7.2-6.3 = 0.6/0.9
the ratio of QCP/QDP = 0.6/0.9 = 2:3
Q.4:
In what ratio must a grocer mix teas worth Rs.60 a kg and Rs.65 a kg.So that by selling the mixture at Rs. 68.20 a kg, He may gain 10%?
A) 3:2
B) 3:4
C) 3:5
D) 4:5
Answer & Explanation
Answer:
A) 3:2
Explanation: [ According to trick Explained in the Video ]
S.P of 1kg mix = RS.68.20
Gain = 10% C.P of 1kg mix =Rs.(100/110 x 68.20) =Rs.62 (mean price)
(Cheaper Tea) : (Dearer Tea) = 65-62:62-60 =3 : 2
Q.5:
How many kg of Sugar at 50 P per kg must a man mix with 25 kg of sugar at 34P per kg so that by selling the mixture at 44P per kg he gains 10% on the outlay?
A) 10 kg
B) 15 kg
C) 20 kg
D) 18 kg
Answer & Explanation
Answer:
B) 15 kg
Explanation: [ According to trick Explained in the Video ]
1st price RS.50 p /kg 2nd price Rs.25 p /kg
Cost price of mixture = (44/110) x 100 = 40 (mixture)
Mixture Rs.40/kg
Hence Quantity of cheaper (QCP):Quantity of Dearer (QDP) = Cp of Dearer quantity -mean price:mean price - Cp of Cheaper quantity = 40-36:50 - 40=6:10
the ratio of QDP/QCP = 6/10 = 3:5
∴ Sugar of 50 paisa per kg to be mixed = (3/5) x 25 = 15kg
Q.6:
How much sugar at Rs. 9.5 a kg should be added to 17 kg of tea at Rs. 20 a kg so that the mixture be worth Rs. 13 a kg.?
A) 11 kg
B) 17 kg
C) 21 kg
D) 34 kg
Answer & Explanation
Answer:
D) 34 kg
Explanation:
Ratio in which tea and sugar should be mixed
= 20 – 13 : 13 – 9.5 = 7 : 3.5 ⇒ 7 : 3.5 ⇒ 2:1 .
Let x be quantity at 9.5/kg. ∴ 2 : 1 = x : 17,hence x = 34 kg.
Q.7:
How many kg of custard powder costing Rs. 42 per kg must be mixed with 16 kg of custard powder costing Rs. 60 per kg so that 20 % may be gained by selling the mixture at Rs. 60 per kg?
A) 11 kg
B) 14 kg
C) 12 kg
D) 20 kg
Answer & Explanation
Answer:
D) 20 kg
Explanation:
SP = 60. Gain= 20%.
CP = (100/120) x 60. ∴ CP = 50.
Ratio between the 2 varieties of custard powder = 60 – 50 : 50 –42 =
10 : 8. ∴ if x is the required quantity then 10 : 8 = x : 16 ⇒ x =20 kg.
Q.8:
How much chicory at Rs. 5 a kg should be added to 20 kg of coffee at Rs. 12 a kg so that the mixture be worth Rs. 7.50 a kg.?
A) 21 kg
B) 15 kg
C) 36 kg
D) 42 kg
Answer & Explanation
Answer:
C) 36 kg
Explanation: [ According to trick Explained in the Video ]
Ratio in which coffee and chicory should be mixed
= 12 – 7.5 : 7.5 – 5 = 4.5 : 2.5 ⇒ 9 : 5.
Let x be quantity at 4/kg. ∴ 9 : 5 = x : 20 ⇒ x = 36kg
Q.9:
Cost of two types of pulses is Rs.15 and Rs, 20 per kg, respectively. If both the pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg?
A) Rs. 22 per kg
B) Rs. 30 per kg
C) Rs. 10 per kg
D) Rs. 18 per kg
Answer & Explanation
Answer:
D) Rs. 18 per kg
Explanation: [ According to trick Explained in the Video ]
Let the cost of mixed variety of pulse be Rs. x
As per the alligation rule,
2:3 = (20-x) : (x-15)
⇒ 2x+3x = 60+30
⇒ 5x = 90
⇒ x = 18
Q.10:
A dealer has 1000 kg sugar and he sells a part of it at 8% profit and the rest of it at 18% profit. The overall profit he earns is 14%. What is the quantity which is sold at 18% profit?
A) 250 kg
B) 600 kg
C) 620 kg
D) 400 kg
Answer & Explanation
Answer:
B) 600 kg
Explanation:
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (18-14) : (14-8) = 4:6 = 2:3
Quantity of sugar sold at 18% profit = 3/5 × 1000 = 600kg
A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?
A) 7 liters
B) 15 liters
C) 10 liters
D) 9 liters
Answer & Explanation
Answer:
C) 10 liters
Explanation: [ According to trick Explained in the Video ]
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.
Q.2:
How many liters of oil at Rs.40 per liter should be mixed with 240 liters of a second variety of oil at Rs.60 per liter so as to get a mixture whose cost is Rs.52 per liter?
A) 120 liters
B) 180 liters
C) 110 liters
D) 160 liters
Answer & Explanation
Answer:
D) 160 liters
Explanation:
1st Variety RS.40/ltr 2nd Variety Rs.60/ltr
Mixture Rs.52/kg
Hence Quantity of cheaper (QCP):Quantity of Dearer (QDP) = Cp of Dearer quantity -mean price:mean price - Cp of Cheaper quantity = 60-52:52 - 40=8:12
the ratio of QDP/QCP = 8/22 = 4:3
Q.3:
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?
A) 40
B) 44
C) 48
D) 52
Answer & Explanation
Answer:
B) 44
Explanation:
Let the capacity of the can be T litres.
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
Hence, this day was SundayAfter adding milk, quantity of milk in the mixture = 6/11 T.
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44.
Q.4:
A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A) 26.34 Litre
B) 27.36 Litre
C) 28 Litre
D) 29.16 Litre
Answer & Explanation
Answer:
D) 29.16 Litre
Explanation: [ According to trick Explained in the Video ]
Amount of milk left after 3 operation = [40(1-4/40)^3] liters
(40 * 9/10 *9/10 * 9/10) =29.16 liters
Q.5:
A mixture of 70 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12 1/2% water?
A) 2
B) 8
C) 4
D) 5
Answer & Explanation
Answer:
A) 2
Explanation: [ According to trick Explained in the Video ]
Quantity of milk in the mixture = 90/100 (70) = 63 litres.
After adding water, milk would form 87 1/2% of the mixture.
Hence, if quantity of mixture after adding x liters of water,
(87 1/2) / 100 x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added
Q.6:
Five litres of wine is removed from a cask full of wine and is replaced with water. Five litres of this mixture is then removed and replaced with water. If the ratio of wine to water in the cask is now 16 : 9, how much wine did the cask hold ?
A) 25 litres
B) 50 litres
C) 100 litres
D) 150 litres
Answer & Explanation
Answer:
A) 25 litres
Explanation: [ According to trick Explained in the Video ]
Let the cask holds x liters of wine. 5 liters of wine is replaced with water.
This operation is done 2 times. ∴ [(x - 5) / x]^2 = 16 / (16+9)
⇒ [(x - 5) / x]^2 = 16 / 25
⇒ [(x - 5) / x] = 4/5 ⇒ x = 25 liters
Q.7:
In a mixture of milk and water, there is only 26% water. After replacing the mixture with 7 liters of pure milk , the percentage of milk in the mixture become 76%. The quantity of mixture is:
A) 65 liters
B) 91 liters
C) 38 liters
D) None of these
Answer & Explanation
Answer:
B) 91 liters
rather than finding odd days between both the dates i will use the trick and directly find the day on Oct 10, 2001 because it will save time.
Explanation: [ According to trick Explained in the Video ]
Milk water
74% 26%
76% 24%
Left amount = Initial amount(1-replaced amount/total amount)
24 = 26( 1- 7:k)
k= 91 liters
Q.8:
Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q, the ratio of alcohol and water in the resulting mixture is?
A) 16 : 5
B) 14 : 5
C) 16 : 7
D) 19 : 5
Answer & Explanation
Answer:
D) 19 : 5
Explanation: [ According to trick Explained in the Video ]
Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.
Q.9:
In what ratio must water be mixed with milk to gain 16 2/3 % by selling the mixture at cost price?
A) 1:6
B) 2:3
C)4:3
D) 6:1
Answer & Explanation
Answer:
A) 1:6
Explanation: [ According to trick Explained in the Video ]
Let the C.P of the milk be Re. 1 per litre
Then S.P of 1 litre of mix =Re 1, Gain% =50/3% C.P of 1 Litre of Mix =Rs(100 x 3/350 x 1) =Re 6/7
C.P of 1 liter of water Rs.0 ,c.p of 1 liter of milk Rs.1
Ratio of water and milk =1/7 = 6/7 = 1: 6
Q.10:
How much water must be added to a bucket which contains 40 liters of milk at the cost price of Rs.3.50 per liter so that the cost of milk reduces to Rs.2 per liter?
A) 25 liters
B) 28 litres
C) 30 liters
D) 35 liters
Answer & Explanation
Answer:
B) 28 litres
Explanation: [ According to trick Explained in the Video ]
Total cost price =Rs(40 x 7/2) = Rs.140
Cost per litre =Rs.2,
Total quantity = 140/2 = 70 Litres.
