Quiz-1

Total amount of sugar = 1000 kg

He sells 1st part of sugar at 8 % profit and the rest (2nd) at 18 % profit.

He gains mean profit on whole sugar is 14 %.

By the rule of allegation, we have:

Profit on 1st part Profit on 2nd part

8% 18%

Mean Profit

14%

(18 – 14 = 4%) (14 – 8 = 6%)

Ration of 1st and 2nd parts = 4: 6 = 2: 3

⇒ Quantity of 2nd part = 3/ (2 + 3) x 1000 kg

= 600 kg.

So the correct option is (2).

Let quantity of solution sold be 1 litre

When the solution is sold for 297, the profit percent = 10%

Hence CP of of solution = SP × 100/(100+P%)

⇒CP = 297 × 100/(100+10)

⇒CP = 270

Ratio of quantities of brand A and brand B is 3:7

Let their costs be 2c and 3c.

⇒ 0.3×2c + 0.7×3c = 270

⇒ 2.7c = 270

⇒ c = 100

Hence cost of Brand B = 3c = 300

We have SP = 75 per litre

Profit Per cent = 37.5%

Hence

⇒ CP = SP × 100/(100 + 37.5)

⇒ CP = 7500/137.5 = 54.5454

Hence, to make the CP from 60 to 54.54,

Hence,

Required ratio is

= 54.54 : 60 - 54.54

= 54.54 : 5.454

= 10:1

% of Liquid in Solution A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Litres

Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Litres

Volume of Water in Solution B = (70/100) × 10 = 7 Litres

If Solution A is mixed with Solution B

⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100

% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

We need to mix rice varieties costing Rs 9.3 and Rs 10.8 per kg to make the mixture at Rs 10 /kg

To do so,

We need to add these varieties in the ratio

⇒10.8 - 10 : 10 - 9.3

= 0.8 : 0.7

= 8 : 7

Net Volume of the New Mixture = 6 + 4 = 10 Litres

Net Volume of the Solution in the New Mixture = (20/100) × 6 + (60/100) × 4

Net Volume of the Solution in the New Mixture = (6 / 5) + (12 / 5) = (18 / 5) = 3.6 Litres

% of Solution in the Mixture = (Volume of Solution/Net Volume) × 100

% of Solution in the Mixture = (3.6 / 10) × 100 = 36%

Hence Option (A)

Let the quantity of tea a = x

Let the quantity of tea b = y

Let the quantity of tea c = z

Given : y = z ...... (i)

x + y + z = 100 ....... (ii)

Cost of Mixture = (Sum of Cost of individual items in a mixture / total quantity of mixture)

90 = (95x + 100y + 70z) / (x + y + z)

90x + 90y + 90z = 95x + 100y + 70z

5x + 10y = 20z
5x = 20y - 10y = 10y............................... (from Eq. (i), y =z )

x = 2y .................... (iii)

From Equation (i), (ii) & (iii)

x + y + z = 100

2y + y + y = 100

4y = 100

y = 25

Hence from equation (iii) x = 50

Hence from equation (i) z =25

Hence Option (b)

Amount of alcohol in the initial mixture = 20% of 15 litres

= 0.20 × 15 = 3 litres

Amount of water in initial mixture = 15 – 3 = 12 litres

Given 3 litres of water is added

⇒ Total volume of mixture = 15 + 3 = 18 litres.

Total amount of water in new mixture = 12 + 3 = 15 litres.

Percentage of alcohol in new mixture = (3/18) × 100

= 16.67 % = 17% (Rounded to nearest integer)

Total Volume of solution = 15 litres
Alcohol Content in solution = 20% = (20/100) × 15 = 3 litres
Water added to previous Solution = 5 litres
⇒ New Volume = Old Volume + Volume Added = 15 + 5 = 20 litres
Alcohol Content = same as before = 3 litres
% Alcohol Content = (Volume of Alcohol in Solution / Total Volume of Solution) × 100
⇒ % Alcohol Content = (3 / 20) × 100 = 15%

We have been given two brands at Rs 9 and Rs 4 per kg.

Hence we have

Hence, these 2 brands should be added in the ratio 3 : 2

Hence, Let quantity of Brand A and Brand B be 3m and 2m.

Then,

We have

⇒ 3m+2m = 40

⇒ 5m = 40

⇒ m = 8

Hence, quantity of brand A = 3m

= 3 × 8

= 24 kg