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
Source: safalta
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. 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