ELIMINATION METHOD WORD PROBLEMS WORKSHEET

Problem 1 :

Two numbers add upto 20 and they differ by 6. What are the two numbers?

Problem 2 :

Henry bought bought rulers and erasers in a total of 45. The price of each ruler is $5 and that of an eraser is $3. If he had paid a total of $185 for his purchase, what is the price of a ruler and an eraser?

Problem 3 :

A park charges $10 for adults and $5 for kids. How many many adults tickets and kids tickets were sold, if a total of 548 tickets were sold for a total of $3750?

Problem 4 :

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1, it becomes ½ if we only add 1 to the denominator. What is the fraction?

Problem 5 :

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Problem 6 :

In a three digit number, the middle digit is zero and the other two digits add upto 11. If 297 be subtracted from it, the digits are reversed. Find the number.

Problem 7 :

Of two positive numbers, three times the larger number is equal to four times the smaller number. If the sum of one-half of the larger number and two-third of the smaller number is equal to 8, find the two numbers.

Problem 8 :

In a business, John has invested money in stock and bonds. He earns 5% of the money invested in stock and 7% in bonds. If his total investment is $25000, find the money invested in stock and bonds. 

Answers

1. Answer :

Let x and y be the two numbers.

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

x - y = 6 ----(2)

(1) + (2) :

Divide both sides by 2.

x = 13

Substitute x = 13 into (1).

13 + y = 20

Subtract 13 from both sides.

y = 7

Therefore, the numbers are 13 and 7.

2. Answer :

Let r and e be the costs of a ruler and an eraser respectively.

r + e = 45 ----(1)

5r + 3e = 185 ----(2)

3(1) - (2) :

Divide both sides by -2.

r = 25

Substitute r = 25 into (1).

25 + e = 45

Subtract 25 from both sides.

e = 20

Therefore,

number of rulers = 25

number of erasers = 20

3. Answer :

Let a and k be number of adult and kids tickets respectively.

a + k = 548 ----(1)

10a + 5k = 3750 ----(2)

5(1) - (2) :

Divide both sides by -5.

a = 202

Substitute a = 202 into (1).

202 + k = 548

Subtract 202 from both sides.

k = 346

Therefore,

number of adult tickets = 202

number of kids tickets = 346

4. Answer :

Let ˣ⁄y be the required fraction.

Given : When 1 is added to the numerator and 1 is subtracted from the denominator, the fraction reduces to 1.

⁽ˣ ⁺ ¹⁾⁄₍y – ₁₎ = 1

 x + 1 = y - 1

 x – y = -2 ----(1)

Given : When only 1 is added to the denominator, the fraction becomes ½.

ˣ⁄₍y ₊ ₁₎ = ½

 2x = y + 1

 2x – y = 1 ----(2)

(2) - (1) :

Substitute x = 3 into (1).

3 - y = -2

Subtract 3 from both sides.

-y = -5

y = 5

ˣ⁄y = ⅗

Therefore, the fraction is ⅗.

5. Answer :

Let x and y be the present ages of Nuri and Sonu respectively.

Five years ago, Nuri was thrice as old as Sonu.

x – 5 = 3(y – 5)

x – 5 = 3y – 15

x – 3y = -10 ----(1)

Ten years later, Nuri will be twice as old as Sonu.

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x – 2y = 10 ----(2)

(2) - (1) :

Substitute y = 20 into (2).

x - 2(20) = 10

x - 40 = 10

Add 40 to both sides.

x = 50

Therefore,

the present age of Nuri = 50 years

the present age of Sonu = 20 years

6. Answer :

Let x0y be the required three-digit number.

x + y = 11 ----(1)

If 297 be subtracted from it, the digits are reversed.

x0y - 297 = y0x

100(x) + 10(0) + 1(y) - 297 = 100(y) + 10(0) + 1(x)

100x + 0 + y - 297 = 100y + 0 + x

100x + y - 297 = 100y + x

99x - 99y = 297

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

(1) + (2) :

Divide both sides by 2.

x = 7

Substitute x = 7 into (1).

7 + y = 11

Subtract 7 from both sides.

y = 4

x0y = 704

Therefore, the three digit number is 704.

7. Answer :

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

Given : Three times the larger number is equal to four times the smaller number.

3x = 4y

3x - 4y = 0 ----(1)

Given : The sum of one-half of the larger number and two-third of the smaller number is equal to 8.

(½)x + (⅔)y = 8

Least common multiple of (2, 3) = 6.

Multiply both sides by 6 to get rid of the denominators 2 and 3.

3x + 4y = 48 ----(2)

(1) + (2) :

Divide both sides by 6.

x = 8

Substitute x = 8 into (1).

3(8) - 4y = 0

24 - 4y = 0

Subtract 24 from both sides.

-4y = -24

Divide both sides by -4.

y = 6

Therefore, the two numbers are 8 and 6.

8. Answer :

John has invested money in stock and bonds. He earns 5% of the money invested in stock and 7% in bonds. If the total investment is $25000 and the total earning is $1550, find the money invested in stock and bonds. 

Let x and y be the amounts of money invested stock and bonds respectively.

x + y = 25000 ----(1)

Given : John earns 5% of the money invested in stock and 7% in bonds.

5% of x + 7% of y = 1550

0.05x + 0.07y = 1550

Multiply both sides by 100.

5x + 7y = 155000 ----(2)

7(1) - (2) :

Divide both sides by 2.

x = 10000

Substitute x = 10000 into (1).

10000 + y = 25000

Subtract 10000 from both sides.

y = 15000

Therefore,

money invested in stock = $10,000

money invested in bonds = $15,000

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.