SUBSTITUTION WORD PROBLEMS WORKSHEET

Problem 1 :

One number is greater than thrice the other number by 2. If four times the smaller number exceeds the greater by 5, find the numbers.

Problem 2 :

The ratio of income of two persons is 1 : 2 and the ratio of their expenditure is 3 : 7. If the first person saves $2000 and the second person saves $3000, find the expenditure of each person.

Problem 3 :

Three chairs and two tables cost $700 and five chairs and six tables cost $1100. Find the cost of each chair.

Problem 4 :

Five years ago, a man was four times as old as his son. Five year hence, the man will be two and half times as old as his son. Find their present ages.

Problem 5 :

A father's age is three times the sum of the age of his two son. After 5 years, his age will be twice the sum of the ages of his two sons. Find the present age of the father.

Problem 6 :

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

Solutions

Problem 1 :

One number is greater than thrice the other number by 2. If four times the smaller number exceeds the greater by 5, find the numbers.

Solution :

Let x be the smaller number and y be the greater number.

Given : One number is greater than thrice the other number by 2.

Then, we have

y = 3x + 2 -----(1)

Given : Four times the smaller number exceeds the greater by 5.

Then, we have

4x = y + 5

Substitute (3x + 2) for y.

4x = 3x + 2 + 5

4x = 3x + 7

Subtract 3x from each side.

x = 7

Substitute 7 for x in (1).

(1)-----> y = 3(7) + 2

y = 21 + 2

y = 23

So, the numbers are 7 and 23.

Problem 2 :

The ratio of income of two persons is 1 : 2 and the ratio of their expenditure is 3 : 7. If the first person saves $2000 and the second person saves $3000, find the expenditure of each person.

Solution :

From the ratio of income 1 : 2, the incomes of the two persons can be assumed as x and 2x.

From the ratio of expenditure 3 : 7, the expenditures of the two persons can be assumed as 3y and 7y.

Savings = Income - Expenditure

The first person saves $3000.

Then, we have

x - 3y = 3000

Add 2y to each side.

x = 3000 + 3y -----(1)

The second person saves $7000.

Then, we have

2x - 7y = 7000

Substitute (3000 + 3y) for x.

(1)-----> 2(2000 + 3y) - 7y = 3000

4000 + 6y - 7y = 3000

4000 - y = 3000

Subtract 1000 from each side.

-y = -1000

Multiply each side by -1.

y = 1000

3y = 3(1000) = 3000

7y = 7(1000) = 7000

So, the expenditures of the two persons are $3000 and $7000.

Problem 3 :

Three chairs and two tables cost $700 and five chairs and six tables cost $1100. Find the cost of each chair.

Solution :

Let x and y be the costs of each chair and table respectively.

Then, we have

3x + 2y = 700 -----(1)

5x + 6y = 1700 -----(2)

Multiply (1) by 3.

(1) ⋅ 3 -----> 9x + 6y = 2100

Subtract 9x from each side.

6y = 2100 - 9x

Substitute (2100 - 9x) for 6y in (2).

(2)-----> 5x + 2100 - 9x = 1700

2100 - 4x = 1700

Subtract 2100 from each side.

-4x = -400

Divide each side ny -4.

x = 100

So, the cost of each chair is $100.

Problem 4 :

Five years ago, a man was four times as old as his son. Five year hence, the man will be two and half times as old as his son. Find their present ages.

Solution :

Let 'x' and 'y' be the present ages of the man and his son respectively.

Five years ago :

Age of the man = x - 5

Age of the son = y - 5

Given : Five years ago, the man was four times as old as his son

x - 5 = 4(y - 5)

x - 5 = 4y - 20

Add 5 to each side.

x = 4y - 15 -----(1)

Five years hence :

Age of the man = x + 5

Age of the son = y + 5

Given : Five years hence, the man will be two and half times as old as his son.

x + 5 = 2.5(y + 5)

x + 5 = 2.5y + 12.5

Subtract 5 from each side.

x = 2.5y + 7.5 -----(2)

From (1) and (2),

4y - 15 = 2.5y + 7.5

Subtract 2.5y from each side.

1.5y - 15 = 7.5

Add 15 to each side.

1.5y = 22.5

Divide each side by 1.5.

y = 15

Substitute 15 for y in (1).

(1)-----> x = 4(15) - 15

x = 60 - 15

x = 45

So, the present ages of the man and son are 45 years and 15 years respectively.

Problem 5 :

A father's age is three times the sum of the age of his two son. After 5 years, his age will be twice the sum of the ages of his two sons. Find the present age of the father.

Solution :

Let x be the present age of the father and y be the sum of the present ages of the two sons.

At present :

x = 3y -----(1)

After 5 years :

x + 5 = 2(y + 5 + 5)

x + 5 = 2(y + 10)

x + 5 = 2y + 20

Subtract 5 from each side.

x = 2y + 15

Substitute 3y for x.

3y = 2y + 15

Subtract 2y from each side.

y = 15

Substitute 15 for y in (1).

(1)-----> x = 3(15)

x = 45

So, the present age of the father is 45 years.

Problem 6 :

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

Solution :

Let x/y be the required fraction.

Given : If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1.

(x + 2) / (y + 1) = 1

Multiply each side by (y + 1).

x + 2 = 1(y + 1)

x + 2 = y + 1

Subtract 2 from each side.

x = y - 1 -----(1)

Given : If the numerator is decreased by 4 and the denominator by 2, it becomes 1/2.

(x - 4) / (y - 2) = 1 / 2

Multiply each side by 2(y - 2).