SUBSITUTION METHOD WORD PROBLEMS

Problem 1 :

Find two integers such that three times the smaller is 33 more than twice the larger and twice the smaller plus five times the larger equals 250.

Solution :

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

3x = 33 + 2y --->(1)

2x + 5y = 250 --->(2)

From (1)

x = (33 + 2y)/3

x = 11 + (2y/3)

2[11+ (2y/3)] + 5y = 250

22 + (4y/3) + 5y = 250

22 + (19y/3) = 250

19y/3 = 250 - 22

19y/3 = 228

y = 228 ⋅ (3/19)

y = 36

By applying y = 36 in (2)

2x + 5(36) = 250

2x + 180 = 250

2x = 250 - 180

2x = 70

x = 35

So, two numbers are 35 and 36.

Problem 2 :

Two hammers and a screwdriver cost a total of $34. A hammer and 3 screwdrivers cost a total of $32. Find the price of each type of tool.

Solution :

Let x and y be the cost of one hammer and screwdriver respectively.

2x + y = 34 ---> (1)

x + 3y = 32 --- > (2)

From (1), y = 34 - 2x

x + 3(34 - 2x) = 32

x + 102 - 6x = 32

-5x = 32 - 102

-5x = -70

x = 70/5

x = 14

Applying x = 14 in (2), we get

14 + 3y = 32

3y = 32 - 14

3y = 18

y = 18/3

y = 6

Thus, hammer cost $14 and screwdriver cost $6.

Problem 3 :

Four adults and three children go to a theatre for $74, where as two adults and five children are charged $58. Find the price of an adult’s ticket and a child’s ticket.

Solution :

Let a = adults and c = children

4a + 3c = $74 --- > (1)

2a + 5c = $58 --- > (2)

From (1)

4a = 74 - 3c

a = (74 - 3c) / 4 --- > (3)

From (2)

2a = 58 - 5c

a = (58 - 5c) / 2 --- > (4)

Solving simultaneously,

(58 - 5c) / 2 = (74 - 3c) / 4

Cross multiply,

4(58 - 5c) = 2(74 - 3c)

232 - 20c = 148 - 6c

232 - 148 = -6c + 20c

84 = 14c

c = 84/14

c = 6

Each child’s ticket cost is $6.

Then the adult’s ticket cost = (74 - 3c) / 4

= (74 - 3(6)) / 4

= (74 - 18) / 4

= 56/4

= $14

So, adult’s ticket cost is $14.

Problem 4 :

Three blankets and a sheet cost me $190. Two sheets and a blanket cost a total of $100. Find the cost of one blanket and one sheet.

Solution :

Let b = blankets and s = sheet

3b + 1s = 190 --- > (1)

1b + 2s = 100 --- > (2)

From (1)

s = 190 - 3b

Substitute s = 190 - 3b in (2)

1b + 2(190 - 3b) = 100

1b + 380 - 6b = 100

-5b = 100 - 380

-5b = -280

b = 280/5

b = 56

By applying b = 56 in (1)

3(56) + s = 190

168 + s = 190

s = 190 - 168

s = 22

So, blanket cost is $56 and sheet cost is $22.                   

Problem 5 :

Seven peaches plus eight nashi cost me $273 altogether. However, three peaches and one nashi cost a total of $66. Find the cost of one piece of each fruit.

Solution :

Let p and n be the cost of one peaches and nashi respectively.

7p + 8n = 273 --- > (1)

3p + 1n = 66 --- > (2)

From (2)

n = 66 - 3p

Substitute n = 66 - 3p in (1)

7p + 8(66 - 3p) = 273

7p + 528 - 24p = 273

-17p = 273 - 528

-17p = -255

p = 255/17

p = 15

By applying p = 15 in (2)

3(15) + n = 66

45 + n = 66

n = 66 - 45

n = 21

So, peach cost is Ұ15 and nashi cost is Ұ21.

Problem 6 :

A purse contains $3.75 in 5 cent and 20 cent coins. If there are 33 coins altogether, how many of each type of coin are in the purse?

Solution :

Let n = number of 5 cent coins

m = number of 20 cent coins

n + m = 33 --- > (1)

5n + 20m = 375 (cents) --- > (2)

From (1)

m = 33 - n

Substitute m = 33 - n in (2)

5n + 20(33 - n) = 375

5n + 660 - 20n = 375

-15n = 375 - 660

-15n = -285

n = 19

By applying n = 19 in (1)

19 + m = 33

m = 33 - 19

m = 14

So, 19 five cents and 14 twenty cents.