**Problem 1 :**

The average age of three boys is 25 years and their ages are in the proportion 3 : 5 : 7. Find the age of the youngest boy.

**Solution :**

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x + 5x + 7x) / 3 = 25

15x = 75

x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

So, the age of the youngest boy is 15 years.

**Problem 2 :**

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7 : 6, find his new weight.

**Solution :**

**Given :** Original weight of John = 56.7 kg.

He is going to reduce his weight in the ratio 7:6.

We can use the following hint to find his new weight, after it is reduced in the ratio 7 : 6.

His new weight is

= (6 ⋅ 56.7) / 7

= 48.6 kg.

So, John's new weight is 48.6 kg.

**Problem 3 :**

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2 : 3.

**Solution :**

Sum of the terms in the given ratio is

= 3 + 5

= 8

So, no. of boys in the school is

= 720 ⋅ (3/8)

= 270

No. of girls in the school is

= 720 ⋅ (5/8)

= 450

**Given :** Number of new girls admitted in the school is 18.

Let x be the no. of new boys admitted in the school.

After the above new admissions,

No. of boys in the school = 270 + x

No. of girls in the school = 450 + 18 = 468

**Given :** The ratio after the new admission is 2 : 3.

Then, we have

(270 + x) : 468 = 2 : 3

Use cross product rule.

3(270 + x) = 468 ⋅ 2

810 + 3x = 936

3x = 126

x = 42

So, the number of new boys admitted in the school is 42.

**Problem 4 :**

The monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenditures are in the ratio 7 : 9. If each saves $50 per month, find the monthly income of the second person.

**Solution :**

From the given ratio of incomes ( 4 : 5 ),

Income of the 1st person = 4x

Income of the 2nd person = 5x

(Expenditure = Income - Savings)

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

So, we have

(4x - 50) : (5x - 50) = 7 : 9

Use cross product rule.

9(4x - 50) = 7(5x - 50)

36x - 450 = 35x - 350

x = 100

Then, the income of the second person is

= 5x

= 5(100)

= 500.

So, income of the second person is $500.

**Problem 5 :**

The ratio of the prices of two houses was 16 : 23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11 : 20. Find the original price of the first house.

**Solution :**

From the given ratio 16 : 23,

Original price of the 1st house = 16x

Original price of the 2nd house = 23x

After increment in prices,

Price of the 1st house = 16x + 10% of 16x

= 16x + 1.6x

= 17.6x

Price of the 2nd house = 23x + 477

After increment in prices, the ratio of prices becomes 11:20.

Then, we have

17.6x : (23x + 477) = 11 : 20

Use cross product rule.

20(17.6x) = 11(23x + 477)

352x = 253x + 5247

99x = 5247

x = 53

Then, original price of the first house is

= 16x

= 16(53)

= 848

So, original price of the first house is $848.

Apart from the word problems given above, if you need more word problems on ratio and proportion, please click the following links.

**Ratio and Proportion Word Problems - 1**

**Ratio and Proportion Word Problems - 2**

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

You can also visit the following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**