# 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

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

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

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

2x  =  y + 6

Substitute (y - 1) for x.

2(y - 1)  =  y + 6

2y - 2  =  y + 6

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

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

WORD PROBLEMS

Word problems on simple equations

Word problems on linear equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

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

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

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

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

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

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

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6 