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). 

2(x - 4)  =  1(y - 2)

2x - 8  =  y - 2

Add 8 to each side. 

2x  =  y + 6

Substitute (y - 1) for x.

2(y - 1)  =  y + 6

2y - 2  =  y + 6

Add 2 to each side. 

2y  =  y + 8

Subtract y from each side. 

y  =  8

Substitute 8 for y in (1).

(1)-----> x  =  8 - 1

x  =  7

Therefore, we have

x / y  =  7 / 8

So, the required fraction is 7/8.

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