a) 150 ml
b) 160 ml
c) 170 ml
d) 180 ml
Solution quantity = 400ml
Let the quantity of pure alcohol to be added in 400ml be A ml.
alcohol in 400ml solution = 16 × 400/100 = 64ml.
Then,
⇒ 400 × 16/100 + A = (400 + A) × 40/100
⇒ 64 + A = 160 + 2A/5
⇒ 3A/5 = 96
⇒ A = 96 × 5/3
⇒ A = 160
2. In a particular type of alloy the ratio of iron to carbon is 2 : 3.
The amount of iron that should be added to 15 kg of this material to make the ratio of the contents1 : 1 is
a) 3 kg
b) 4 kh
c) 2 kg
d) 5 kg
Given, iron : carbon = 2 : 3
In 15 kg of this alloy, we have
amount of iron =2/5×15 = 6kg
amount of carbon =3/5×15 = 9kg
Hence, the amount of iron to be added to this mixture to make the ratio of the contents 1:1 is 3 kg of iron.
3. In what ratio should Assam Tea costing Rs.
300 per kg be mixed with Darjeeling Tea costing Rs.
400 per kg, so that by selling the mixture at Rs.
Source: safalta
408 per kg there is gain of 20%.a) 1 : 2
b) 3 : 2
c) 2 : 3
d) 4 : 3
Selling price of mixture is Rs. 408 per kg
Profit% = (selling price/Cost price – 1) × 100
⇒ 20 = (408/Cost price – 1) × 100
⇒ 0.2 + 1 = 408/Cost price
⇒ Cost price of mixture = 408/1.2 = Rs.340 per kg
∴ Ratio of Assam Tea to the Darjeeling Tea = (400 - 340) : (340 - 300) = 60 : 40 = 3 : 2
4. A mixture is made by mixing alcohol and water in the ratio 9 ∶ 7.
If ‘x’ litres of alcohol and ‘3x’ litres of water is mixed in 80 litres of mixture, the new ratio becomes 13 ∶ 14.
Find the quantity of new mixture.
a) 72l
b) 144l
c)108l
d) 120l
Sum of first ratio = 9 + 7 = 16
Quantity of alcohol in 80 litres of mixture = 9/16 × 80 = 45 litres
Quantity of water in 80 litres of mixture = 7/16 × 80 = 35 litres
When ‘x’ litres of alcohol and ‘3x’ litres of water is added,
Quantity of alcohol in new mixture = 45 + x
Quantity of water in new mixture = 35 + 3x
But, ratio of new mixture = 13 ∶ 14
⇒ (45 + x)/(35 + 3x) = 13/14
⇒ 14(45 + x) = 13(35 + 3x)
⇒ 630 + 14x = 455 + 39x
⇒ 39x – 14x = 630 – 455
⇒ x = 175/25 = 7
∴ Total quantity of new mixture = 45 + x + 35 + 3x = 80 + 4(7) = 80 + 28 = 108 litres
5. 20 L of a mixture contains alcohol and water in the ratio 2 : 3.
If 4 L of water is mixed in it, the percentage of alcohol in the new mixture will be
a) 25%
b) 33.33%
c) 50%
d) 66.66 %
Amount of alcohol in the mixture = 2/5 × 20 = 8 L
And amount of water in the mixture = 20 – 8 = 12 L
Given, 4 L of water is added. Hence,
Amount of water in the new mixture = 16 L
Hence, the percentage of alcohol in the new mixture = 8/24×100 = 33.33%
6. The ratio of milk and water in three samples is 1 ∶ 3, 3 ∶ 5 and 11 ∶ 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?a) 7 : 9
b) 9 : 7
c) 4 : 5
d) 5 : 4
Let ‘x’ quantity of each of three mixtures to be mixed
Quantity of milk in the new mixture = (1/4)x + (3/8)x + (11/16)x = 21x/16
Quantity of water in the new mixture = (3/4)x + (5/8)x + (5/16)x = 27x/16
∴ Required ratio = (21x/16) ∶ (27x/16) = 7 ∶ 97. From a homogeneous mixture of salt and sugar containing 4 parts of salt and 5 parts of sugar, 1/4th part of the mixture is drawn off and replaced with salt. The ratio of salt and sugar in the new mixture is?
a) 4 : 5
b) 7 : 5
c) 5 : 7
d) 5 : 4
Let amount of the mixture of salt and sugar is 1 kg
Mixture contains 4 parts of salt and 5 parts of sugar
⇒ Amount of salt in the mixture = 4/9 kg
⇒ Amount of sugar in the mixture = 5/9 kg
After replacing 1/4th part with salt,
⇒ Amount of salt in the new mixture = (4/9 – 4/9 × ¼ + 1/4) kg = (16 – 4 + 9) /36 kg = 7/12 kg
⇒ Amount of sugar in the new mixture = (5/9 – 5/9 × 1/4) kg = 5/12 kg
∴ Required ratio = 7/12 ∶ 5/12 = 7 ∶ 58. A mug contained 90 liter alcohol. From this mug 9 liter of alcohol was taken out and replaced by water. This process was repeated once more times. How much alcohol is now contained in the mug?
a) 81 l
b) 72 l
c) 72.9 l
d) 71.9 l
Amount of alcohol after the 2nd operation
⇒ a × {1 – (b/a)}n
⇒ 90 × {1 - (9/90)}2
⇒ 90 × {(90 - 9)/90}2
⇒ 90 × (81/90) × (81/90)
⇒ 72.9 litres
9. A vendor sells potatoes at a cost price, but he mixes some rotten potatoes and thereby gains 25%.
The percentage of rotten potatoes in the mixture is?
a) 15%
b) 20%
c) 25 %
d) 30%
Let cost price of 1 kg of pure potato in Rs. be ‘x’
Let quantity of pure potato in the mixture be ‘m’ kg
⇒ Cost price of 1 kg of the mixture in Rs. = mx
∵ selling price of the mixture = cost price of pure potatoes
⇒ (Selling price – Cost price) /Cost price × 100% = 25%
⇒ (x – mx) /mx = 0.25
⇒ 1 – m = 0.25m
⇒ m = 0.80
⇒ Amount of rotten potatoes in the mixture = (1 – 0.80) kg = 0.20 kg
∴ Required percentage = 0.20/1 × 100% = 20%
10. 6 kg sugar costing Rs. 10/kg is mixed with 4 kg sugar costing Rs. 15/kg. What is the average cost of the mixture per kilogram?a) 12/kg
b) 15/kg
c) 16/kg
d) 18/kg
Quantity of sugar = 6 kg
Price of sugar per kg = Rs. 10
∴ Total cost = 10 × 6 = Rs. 60
Another quantity of sugar = 4 kg
Price of sugar per kg = Rs. 15
∴ Total cost = 15 × 4 = Rs. 60
⇒ Total cost of 10 kg of sugar = 60 + 60 = Rs. 120
⇒ Cost per kilogram = 120/10 = Rs. 12 per kilogram
