# Quantitative Aptitude Quiz for UPSI(Profit and Loss): 05 April 2021

Updated Mon, 05 Apr 2021 08:29 PM IST

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1. How much pure alcohol must be added to 400 ml of a solution containing 16% of alcohol to change the concentration of alcohol in the mixture to 40%

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 ∶ 9

7. 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 ∶ 5

8. 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

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