**Answer**:Option A

**Explanation**:

Total money = 20 + 40 = 60

In this question the total money is given so we need to multiply the value of the coins in the ratio.

10 × 0.10 x + 8 × 0.25x + 9 × 0.50x = 60

x + 2x + 4.5x = 60

7.5 x = 60

x = 8

Number of 50 paise coins = 9 × 8 = 72

**2. Three numbers A, B and C are in the ratio of 12 : 15 : 25. If sum of these numbers is 312, ratio between the difference of B and A and the difference of C and B is –**

**Answer**:Option C

**Explanation**:

Let the number A, B and C be 12a , 15a and 25a respectively

Then from the question 12a + 15a + 25a = 312

⇒ a = \(\frac{312}{52}\) = 6

⇒ required ratio = \(\frac{15 × 6 – 12 × 6}{25 × 6 – 15 × 6}\)

= \(\frac{3}{10}\) = 3 : 10

**3. The ratio of Nicotine to Heroine in four types of drugs of equal quantity is 2 : 3, 3 : 7, 4: 11 and 11 :9 respectively. The four drugs are mixed together. What is the ratio of Nicotine to Heroine after mixing?**

**Answer**:Option C

**Explanation**:

Let the quantity of each drugs be ‘a’

Then the quantity of nicotine after mixing is

(\(\frac{2}{5}\) + \(\frac{3}{10}\) + \(\frac{4}{15}\) + \(\frac{11}{20}\)) a

= (\(\frac{2 × 12 + 3 × 6 + 4 × 4 + 11 × 3}{60}\)) a

= (\(\frac{91}{60}\)) a

and quantity of heroine after mixing is

(\(\frac{3}{5}\) + \(\frac{7}{10}\) + \(\frac{11}{15}\) + \(\frac{9}{20}\)) a

= (\(\frac{3 × 12 + 7 × 6 +11× 4 + 9 × 3}{60}\)) a

= (\(\frac{149}{60}\)) a

⇒ the required ratio of nicotine to heroin after mixing

= \(\frac{\frac{91}{60}a}{\frac{149}{60}a}\)

= 91 : 149

**4. Out of three positive numbers, the ratio of the first and the second numbers is 3 : 4 that of the second and the third numbers is 5 : 6 if the product of the second and the third numbers is 4320. What is the sum of three numbers?**

**Answer**:Option A

**Explanation**:

Ratio of first and second numbers = 3 : 4

The ratio of second and third numbers = 5: 6

Let the second number = 5x, third number = 6x

Product of second and third numbers = 4320

5x × 6x = 4320

\({x}^{2}\) = 144

x = 12

2nd number = 60, 3rd number = 72,

1st number = 60 × \(\frac{3}{4}\) = 45

Sum of three numbers = 60 + 72 + 45 = 177

**5. The cost price of 2 shirts and 3 jeans is Rs. 2200 and the cost price of 2 jeans and 4 shirts is Rs. 2400. Find the ratio between the cost price of the jeans and the shirt.**

**Answer**:Option B

**Explanation**:

Let the cost price of one shirt = x Rs., the cost price of one jeans = y Rs.

According to the question,

2x + 3y = 2200 ….1

2y + 4x = 2400

2x + y = 1200 ….2

After solving these 2 equation,

x = 350 Rs. , y = 500 Rs.

Ratio = 500 : 350 = 10 : 7

**Answer**:Option C

**Explanation**:

Let the Numbers be 5x, 6x and 7x

∴ \({5x}^{2}\) + \({6x}^{2}\) + \({7x}^{2}\) = 990

⇒ 25\({x}^{2}\) + 36\({x}^{2}\) + 49\({x}^{2}\) = 990

⇒ 110\({x}^{2}\) = 990

⇒ \({x}^{2}\) = 9

⇒ x = 3

∴ Numbers = 15, 18 and 21.

**2. The ratio of Sita’s, Riya’s and Kunal’s monthly income is 84 : 76 : 89. If Riya’s annual income is Rs. 4,56,000, what is the sum of Sita’s and Kunal’s annual incomes? (In some cases monthly income is used while in other annual income is used.)**

**Answer**:Option D

**Explanation**:

Ratio of monthly income and ratio of annual income will be the same,

ie. 84 : 76 : 89

Applying the rule of proportion, we get

Riya’s income in the ratio: Riya’s total income:: Sum of Sita’s and Kunal’s income in ratio: Sum of Sita’s and Kunal’s income in value

76 : 456000 : : (84 + 89) : x

∴ Sum of Sita’s and Kunal’s annual income in value

= \(\frac{456000}{76}\) × (84 + 89) = Rs. 1038000

**3. Two numbers are in ratio of 21 : 26. If 8 is added in each, the new numbers are in ratio of 5 : 6. Find the ratio of numbers, if 6 is subtracted from each number? **

**Answer**:Option A

**Explanation**:

Let the numbers be 21x and 26x.

(\(\frac{21x + 8}{26x + 8}\)) = \(\frac{5}{6}\)

6(21x + 8) = 5(26x + 8)

126x + 48 = 130x + 40

x = 2

So, numbers will be 42 and 52.

If 6 is subtracted, then numbers will be 36 and 46.

Required ratio = 36 : 46. i.e. 18 : 23

**4. If A varies directly as B and inversely as C and A = 6, when B = 2 and C= 3, what is the value of A when B = 8 and C = 6?**

**Answer**:Option A

**Explanation**:

Let the constant be x,

so putting the first scenario in equation

A = x × \(\frac{B}{C}\) , we get:

6 = x × \(\frac{2}{3}\) or x = 9

We can find out A in the second scenario by putting the value of x as 9

A = 9 × \(\frac{8}{6}\) or A = 12

**5. The income of Asaram, Satpal and Rahim in the ratio of 12 : 9 : 7 and their spendings are in the ratio 15 : 9 : 8. If Asaram saves 25% of his income. What is the ratio of the savings of Asaram, Satpal and Rahim?**

**Answer**:Option A

**Explanation**:

Income = Expenditure + Saving

Asaram: 12x = 15y + 3x (3x = 25% of 12x)

Satpal: 9x = 9y + (9x – 9y)

Rahim: 7x = 8y + (7x – 8y)

Therefore, 12x – 3x = 15y

⇒ \(\frac{x}{y}\) = \(\frac{5}{3}\)

⇒ y = \(\frac{3x}{5}\)

Therefore, savings = (income-expenditure)

Asaram = 12x – 9x = 3x

Satpal = 9x – 9y = 9x – \(\frac{27x}{5}\) = \(\frac{18x}{5}\)

Rahim = 7x – 8y = 7x – \(\frac{24x}{5}\) = \(\frac{11x}{5}\)

i.e., the ratio of savings of Asaram : Satpal : Rahim

= 3x : \(\frac{18}{5}\)x : \(\frac{11}{5}\)x

= 15 : 18 : 11

**Answer**:Option D

**Explanation**:

Amount of zinc = (\(\frac{100 × 13}{20}\)) kg = 65kg.

**2. The ratio of number of men and women in a factory of 720 workers is 7 : 5. How many more women should be joined to make the ratio 1 : 1? **

**Answer**:Option C

**Explanation**:

Number of men = (\(\frac{720 × 7}{12}\)) = 420.

Number of women = (720 – 420) = 300.

∴ Number of women to be joined = (420 – 300) = 120.

**3. What is the ratio whose terms differ by 40 and the measure of which is \(\frac{2}{7}\) ?**

**Answer**:Option B

**Explanation**:

Let the ratio be x : (x + 40).

Then, \(\frac{x}{(x + 40)}\) = \(\frac{2}{7}\) ⇔ 7x = 2x + 80 ⇔ x =16.

∴ Required ratio is 16 : 56.

**4. Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B is**

**Answer**:Option A

**Explanation**:

From the equation, we get

5% of A + 4% of B = \(\frac{2}{3}\) (6% of A + 8% of B)

⇒ 15% of A + 12% of B = 12% of A + 16% of B

⇒ [15 – 12] % of A = [16 – 12] % of B

⇒ 3 % of A = 4% of B

⇒ \(\frac{A}{B}\) = \(\frac{4}{3}\)

Therefore, A : B = 4 : 3.

**5. In a certain school, the ratio of boys to girls is 7 : 5. If there are 2400 students in the school, then how many girls are there?**

**Answer**:Option D

**Explanation**:

Let the number of boys and girls are 7x and 5x, respectively.

Given, the total number of students = 2400

⇒ 7x + 5x = 2400 ⇒ 12x = 2400

∴ x = 200

∴ Required number of girls = 5x = 5 × 200 = 1000.