# WORD PROBLEMS USING ELIMINATION METHOD

Problem 1 :

Find the value of two numbers if their sum is 12 and their difference is 4.

Solution :

Let x and y be the two numbers.

Given : Their sum is 12.

x + y = 12 ----(1)

Given : Their difference is 4.

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

(1) + (2) :

Divide both sides by 2.

x = 8

Substitute x = 8 into (1).

8 + y = 12

Subtract 12 from both sides.

y = 4

Therefore, the numbers are 8 and 4.

Problem 2 :

Kevin bought pen and pencils in a total of 50. The cost of each pen is \$2 and that of a pencil is \$1.50. If he had paid a total of \$85 for his purchase, find the cost of a pen and a pencil.

Solution :

Let x and y be the costs of a pen and a pencil respectively.

Given : Kevin bought pen and pencils in a total of 50.

x + y = 50 ----(1)

Given : The cost of each pen is \$2, a pencil is \$1.50 and Kevin paid a total of \$85 for his purchase.

2x + 1.5y = 85 ----(2)

1.5(1) - (2) :

Divide both sides by 0.5.

x = 20

Substitute x = 20 into (1).

20 + y = 50

Subtract 20 from both sides.

y = 30

Therefore,

number of pens = 20

number of pencils = 30

Problem 3 :

The sum of two numbers is 7. The difference between 5 times the larger and 3 times the smaller is equal 11. Find the numbers.

Solution :

Let x and y be the two numbers such that x > y.

Given : The sum of two numbers is 7.

x + y = 7 ----(1)

Given : The difference between 5 times the larger and 3 times the smaller is equal 11.

5x - 3y = 11 ----(2)

3(1) + (2) :

Divide both sides by 8.

x = 4

Substitute x = 4 into (1).

4 + y = 7

Subtract 3 from both sides.

y = 3

Therefore, the two numbers are 4 and 3.

Problem 4 :

Chase and Sara went to the candy store. Chase bought 5 pieces of fudge and 3 pieces of bubble gum for a total of \$5.70. Sara bought 2 pieces of fudge and 10 pieces of bubble gum for a total of \$3.60. Find the cost of 1 piece of fudge and 1 piece of bubble gum?

Solution :

Let f and g be the costs of 1 piece of fudge and 1 piece of bubble.

From the given information,

5f + 3b = 5.7 ----(1)

2f + 10b = 3.6 ----(2)

2(1) - 5(2) :

Divide both sides by -44.

x = 0.15

Substitute x = 0.15 into (2).

2(0.15) + 10b = 3.6

0.3 + 10b = 3.6

Subtract 0.3 from both sides.

10b = 3.3

Divide both sides by 10.

b = 0.33

Therefore,

cost of 1 piece of fudge = \$0.15

cost of 1 piece of bubble gum = \$0.33

Problem 5 :

Daily earnings of Oliver and Henry are in the ratio 3 : 4 and their expenditures are in the ratio 5 : 7. If each saves \$50 per day, find the daily earnings of each.

Solution :

From the earnings ratio 3 : 4,

earnings of Oliver = 3x

earnings of Henry = 4x

From the expenditures ratio 5 : 7,

expenditure of Oliver = 5y

expenditure of Henry = 7y

The relationship between income, expenditure and savings :

Income - Expenditure = Savings

Then, we have

3x - 5y = 50 ----(1)

4x - 7y = 50 ----(2)

7(1) - 5(2) :

Therefore,

earnings of Oliver = 3(100) = \$300

earnings of Henry = 4(100) = \$400

Problem 6 :

The sum of the present ages of a father and his son is 45 years. 10 years hence, the difference between their ages is 25 years. Find the present age of the father and son.

Solution :

Let f and s be the present ages of father and son.

Given : The sum of the present ages 45.

f + s = 45 ----(1)

Given : 10 years hence, the difference between the ages is 25.

(f + 10) - (s + 10) = 25

f + 10 - s - 10 = 25

f - s = 25 ----(2)

Divide both sides by 2.

f = 35

Substitute f = 35 into (1).

35 + s = 45

Subtract 35 from both sides.

s = 10

Therefore,

present age of the father = 35 years

present age of the son = 10 years

Problem 7 :

In a three digit number, the middle digit is zero and sum of the other two digits is 9. If 99 is added to it, the digits are reversed. Find the three digit number.

Solution :

Let x0y be the required three digit number.

x + y = 9 ----(1)

Given : When 99 is added to it, the digits are reversed.

x0y + 99 = y0x

100(x) + 10(0) + 1(y) + 99 = 100(y) + 10(0) + 1(x)

100x + 0 + y + 99 = 100y + 0 + x

100x + y + 99 = 100y + x

99x - 99y = -99

Divide both sides by 99.

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

(1) + (2) :

Divide both sides by 2.

x = 4

Substitute x = 4 into (1).

4 + y = 9

Subtract 4 from both sides.

y = 5

x0y = 405

Therefore, the three digit number is 405.

Problem 8 :

Two numbers are such that twice the greater number exceeds twice the smaller one by 18. One-third of the smaller number and one-fifth of the greater number are together 21. Find the two numbers.

Solution :

Let and y be the two numbers such that x > y.

Given : Twice the greater number exceeds twice the smaller one by 18.

2x - 2y = 18

Divide both sides by 2.

x - y = 9 ----(1)

Given : One-third of the smaller number and one-fifth of the greater number are together 21.

5y  + 3x = 315

3x + 5y = 315 ----(2)

5(1) + (2) :

Divide both sides by 8.

x = 45

Substitute x = 45 into (1).

45 - y = 9

Subtract 45 from both sides.

-y = -36

y = 36

Therefore, the two numbers are 45 and 36.

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