1. Total number of men, women and children working in a factory is 18. They earn Rs. 4000 in a day. If the sum of the wages of all men, all women and all children is in ratio of 18 : 10 : 12 and if the wages of an individual man, woman and child is in ratio 6 : 5 : 3, then how much a woman earn in a day?

a) Rs. 120

b) Rs. 150

c) Rs. 250

d) Rs. 4000

Explanation: Ratio of number of men, women and children,

$$\eqalign{ & = \frac{{18}}{6}:\frac{{10}}{5}:\frac{{12}}{3} \cr & = 3:2:4 \cr} $$

Total (Men + Women + Children) = 18

3X + 2X + 4X = 18

9X = 18

X = 2

Hence, number of women = 2X = 2 × 2 = 4

Share of all women = $$\frac{{10 \times 4000}}{{40}}$$ = Rs. 1000 [18 + 10 + 12 = 40]

Share of each woman = $$\frac{{1000}}{4}$$ = Rs. 250

2. A and B are two alloys in which ratios of gold and copper are 5 : 3 and 5 : 11 respectively. If these equally amount of two alloys are melted and made alloy C. What will be the ratio of gold and copper in alloy C?

a) 25 : 23

b) 33 : 25

c) 15 : 17

d) 17 : 15

Explanation : Ratio of Gold and Copper in Alloy A = 5 : 3

Ratio of Gold and Copper in Alloy B = 5 : 11

Amount of Gold in Alloy A = $$\frac{5}{8}$$

Amount of Gold In Alloy B = $$\frac{5}{{16}}$$

Amount of Copper in A = $$\frac{3}{8}$$

Amount of Copper in B = $$\frac{{11}}{{16}}$$

Amount of Gold In C,

= (Amount of gold in A + Amount of gold in B) = $$\frac{5}{8}$$ + $$\frac{5}{{16}}$$ = $$\frac{{10 + 5}}{{16}}$$ = $$\frac{{15}}{{16}}$$

Amount of Copper in C,

= Amount of Copper in A + Amount of Copper in B = $$\frac{3}{8}$$ + $$\frac{{11}}{{16}}$$ = $$\frac{{17}}{{16}}$$

Ratio of Gold and Copper in C,

$$ = \frac{{15}}{{16}}:\frac{{17}}{{16}} = 15:17$$

3. A bag contains an equal number of one rupee, 50 paise and 25 paise coins. If the total value is Rs. 35, how many coins of each type are there?

a) 15

b)18

c) 20

d) 27

Explanation: Let X coins of each type of there

Total Value = Rs. 35

X + $$\frac{{\text{X}}}{2}$$ + $$\frac{{\text{X}}}{4}$$ = 35

4X + 2X + X = 140

7X = 140

X = 20

4. A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket?

a) 21 liters

b) 23 liters

c) 24 liters

d) 27 liters

Explanation: Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:

$$\eqalign{ & = \left[ {7x - {\frac{7}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {7x - {\frac{{21}}{4}} } \right]\,{\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & = \left[ {5x - {\frac{5}{{12}}} \times 9} \right]\, \text{litres} \cr & = \left[ {5x - {\frac{{15}}{4}} } \right]\,{\text{litres}} \cr & \therefore \,{\text{ratio becomes}}, \cr & \frac{{ {7x - {\frac{{21}}{4}} } }}{{ {\left(5x - {\frac{{15}}{4}}\right)+9 } }} = \frac{7}{9} \cr & \frac{{ {28x - 21} }}{{ {20x + 21} }} = \frac{7}{9} \cr & {252x - 189} = 140x + 147 \cr & 112x = 336 \cr & x = 3 \cr & {\text{So the can contained}}, \cr & = 7 \times x \cr & = 7 \times 3 \cr & = 21\,{\text{litres of A initially}}{\text{.}} \cr} $$

5. A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B, then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket?:

a) 25 litres

b) 15 litres

c) 18 litres

d) None of these

Explanation: Let bucket contains 5x and 3x of liquids A and B respectively.

When 16 litres of mixture are drawn off, quantity of A in mixture left:

$$\eqalign{ & {5x - {\frac{5}{8}} \times 16} = {5x - 10} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & {3x - {\frac{3}{8}} \times 16} = {3x - 6} \cr & {\text{Now the ratio becomes,}} \cr & \frac{{ {5x - 10} }}{{ {3x - 6} }} = \frac{3}{5} \cr & 25x - 50 = 9x - 18 \cr & 16x = 32 \cr & x = 2 \cr & {\text{Quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr} $$

6. The salaries of A, B and C are in the ratio 1 : 3 : 4. If the salaries are increased by 5%, 10% and 15% respectively, then the increased salaries will be in the ratio

a) 20 : 66 : 95

b) 21 : 66 : 96

c) 21 : 66 : 92

d) 19 : 66 : 92

Explanation: Let A's Salary = Rs. 100

Then, B's Salary = Rs. 300

And, C's Salary = Rs. 400

Salary has given in 1 : 3 : 4 ratio

Now,

5% increase in A's Salary,

A's new Salary = (100 + 5% of 100) = Rs. 105

B's Salary increases by 10%, Then,

B's new Salary = (300 + 10% of 300) = Rs. 330

C's Salary increases by 15%,

C's new Salary = (400 + 15% of 400) = Rs. 460

Then, ratio of increased Salary,

A : B : C = 105 : 330 : 460 = 21 : 66 : 92

7.If A : B = 2 : 3 and B : C = 4 : 5 then A : B : C is

a) 2 : 3 : 5

b) 5 : 4 : 6

c) 8 : 12 : 15

d) 6 : 4 : 5

Explanation:

$$\eqalign{ & \frac{{\text{A}}}{{\text{B}}} = \frac{2}{3} \cr & \frac{{\text{B}}}{{\text{C}}} = \frac{4}{5} \cr} $$

A : B : C = 2 × 4 : 3 × 4 : 3 × 5 = 8 : 12 : 15

8.If two times A is equal to three times of B and also equal to four times of C, then A : B : C is

a) 2 : 3 : 4

b) 3 : 4 : 2

c) 4 : 6 : 3

d) 6 : 4 : 3

Explanation:

$$\eqalign{ & 2A = 3B \cr & Or,\,B = \left( {\frac{2}{3}} \right)A;\,{\text{and}} \cr & 2A = 4C \cr & Or,\,C = \left( {\frac{1}{2}} \right)A; \cr & {\text{Hence}}, \cr & A:B:C = A:\frac{{2A}}{3}:\frac{A}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:\frac{2}{3}:\frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6:4:3 \cr} $$

9.In a school having roll strength 286, the ratio of boys and girls is 8 : 5. If 22 more girls get admitted into the school, the ratio of boys and girls becomes

a) 12 : 7

b) 10 : 7

c) 8 : 7

d) 4 : 3

Explanation: Boys : girls = 8 : 5 (let the boys = 8x, girl = 5x)

Total strength = 286

8x + 5x = 286

13x = 286

x = $$\frac{{286}}{{13}}$$ = 22

Boys = 176 and girls = 110

22 more girls get admitted then number of girls become

(5x + 22) = 110 + 22 = 132

New ratio of boys and girls = 176 : 132 = 4 : 3.

10. Two numbers are in ratio 4 : 5 and their LCM is 180. The smaller number is

a) 9

b) 15

c) 36

d) 45

Explanation: Let two numbers be 4x and 5x

Their LCM = 180 and HCF = x

Now,

1

^{st}number × 2

^{nd}number = LCM × HCF

4x × 5x = 180 × x

20x = 180

x = 9

The smaller number = 4 × 9 = 36