# Word Problem Set 2

## Problem:

Alvin bought 4 magazines at \$3.60 each and a storybook for \$8.60. Lionel spent \$7 more than Alvin. He bought comics at \$1.50. How much did Alvin spend? How many copies of comics did Lionel buy?

### Given:

Let:
A – Alvin spent = 4 magazines (\$3.60/magazine) + 1 storybook (\$8.60) eq. 1
L – Lionel spent = no. of comics (\$1.50/comics) eq. 2
Take note, Lionel spent \$7 more than Alvin, meaning, L = A + \$7 eq. 3

(a) Alvin spent (A) ?
(b) Copies of comics Lionel bought?

### Solution:

(a) Using eq. 1, we get:
A = 4 magazines (\$3.60/magazine) + 1 storybook (\$8.60)
A = \$14.40 + \$8.60
A = \$23.00
(b) Using the value of A in eq. 3, we get:
L = A + \$7 = \$23.00 + \$7.00 = \$30.00
Using eq. 2,
L = no. of comics (\$1.50/comics)
\$30.00 = no. of comics (\$1.50/comics)
\$30.00/\$1.50 = no. of comics (\$1.50/comics) /\$1.50
20 = no. of comics

## Problem:

Lydia and her sister each saved half of their pocket money every month. Lydia took 12 months more than her sister to save P3,600. Her sister saved P750 per month. How much was Lydia’s pocket money each month?

### Given:

x – Lydia’s savings each month = ½ (Lydia’s pocket money each month (L) ) eq. 1
y – Sister’s savings each month = ½ (Sister’s pocket money each month(S)) = P750/month eq. 2
take note, both were able to save P3,600 over a period (months)
we know that, Lydia’s months = Sister’s months + 12 months eq. 3

Lydia’s pocket money (L)?

### Solution:

Solve for sister’s months to reach total savings of P3,600:
Sister’s months = total savings / y = P3,600 / P750 = 4.8 months
Using eq. 3,
Lydia’s months = Sister’s months + 12 months
Lydia’s months = 4.8 months + 12 months
Lydia’s months = 16.8 months
Solve for Lydia’s savings each month:
Lydia’s months = total savings / x
16.8 months = P3,600 / x
x = P3,600/16.8 months = P214.29 each month
Solve for Lydia’s pocket money each month using eq. 1
x = ½ (L)
P214.29 = ½ (L)
2 (P214.29) = ½ (L) (2)
P428.58 = L

## Problem:

Mr. Jones packed 14 tarts in each of his 25 plastic boxes and had 15 tarts left on a tray. He decided to re-pack all the tarts into paper boxes such that each box contained 12 tarts, and put the remaining tarts aside. If he sold each box for \$12.90, how much did he get for selling all the boxes he had packed?

### Given:

Initial packing: 14 tarts each plastic box for 25 plastic boxes with 15 tarts left
Repacked: 12 tarts each paper box
Sold: \$12.90 each paper box

Total sale?

### Solution:

(a) Solve for number of tarts baked:
Number of tarts baked = tarts initially packed + tarts left eq. 1
Tarts initially packed = (tarts in each plastic box ) x (number of plastic boxes)
Tarts initially packed = (14) x (25) = 350 tarts
Number of tarts baked = tarts initially packed + tarts left
Number of tarts baked = 350tarts + 15tarts = 365 tarts
(b) Solve for the number of repacked tarts:
Number of repacked tarts = (number of tarts baked) / (number of tarts each paper box)
Number of repacked tarts = (365 tarts) / ( 12 tarts each paper box)
Number of repacked tarts = 30 boxes with 5 tarts left
(c) Solve for the total sale
Total sale = (number of repacked tarts ) x (sale price per paper box)
Total sale = (30 boxes) x (\$12.90)
Total sale = \$387

## Problem:

A fruit vendor bought 200 apples for \$80. He found that some were bad and threw them away. He sold the remaining apples at \$0.60 each and collected a total of \$105. How many apples did he throw away?

### Given:

Total apples bought = 200 apples @ \$80
Let:
x – number of good apples
y – number of bad apples
Total apples = (number of good apples) + (number of bad apples)
Total apples = x + y eq. 1
Take note, good apples were sold at \$0.60 each with total sale of \$105

### Solution:

Number of good apples sold = (total sale) / (sale price per good apple)
x = (\$105) / (\$0.60)
x = 175 good apples
using eq. 1,
total apples = x + y
200 apples = 175 apples + y
200 – 175 = y
25 apples = y

## Problem:

Mary and Janet had 420 stamps altogether. Mary gave 180 of her stamps to a friend and Janet lost 90 stamps. Mary was left with 4 times the number of stamps Janet had left. How many stamps had Janet left?

### Given:

Let:
x – Mary’s stamps left
y – Janet’s stamps left
we know initially, that, Mary’s stamps + Janet’s stamps = 420 stamps eq. 1
take note, Mary gave 180 stamps & Janet lost 90 stamps
take note, Mary’s stamps left is 4 times the number of Janet’s stamps left, hence:
x = y (4) = 4y eq. 2

Janet’s stamps left (y)?

### Solution:

Using eq. 1,
(Mary’s stamps) + (Janet’s stamps) = 420 stamps
(a) Solve for Mary’s stamps:
Mary’s stamps = x + (stamps she gave)
Mary’s stamps = x + 180 eq. 3
(b) Solve for Janet’s stamps:
Janet’s stamps = y + (stamps she lost)
Janet’s stamps = y + 90 eq.4
Using eq. 3 & 4, eq. 1 becomes:
(Mary’s stamps) + (Janet’s stamps) = 420 stamps
(x + 180) + (y + 90) = 420 eq. 5
Using eq. 2 in eq. 5,
((4y) + 180) + (y + 90) = 420
4y + 180 + y + 90 = 420
4y + y = 420 – 180 – 90
5y = 150
5y/5 = 150/5
y = 30 stamps

## Problem:

A shopkeeper bought a total of 40 paper and plastic lanterns for \$248. He sold 8 plastic lanterns and had equal number of paper and plastic lanterns left. How many plastic lanterns did he buy?

### Given:

Let:
x – number of paper lanterns bought
y – number of plastic lanterns bought
we know that, x + y = 40 eq. 1
take note, total cost of 40 lanterns is \$248
take note, after he sold 8 plastic lanterns, the number of plastic lanterns left = number of paper lanterns left

plastic lanterns bought (y) ?

### Solution:

Number of paper lanterns = (number of paper lanterns sold) + (number of paper lanterns left)
Take note, there were no sold paper lanterns, so, number of paper lanterns left is equal to the number of paper lanterns bought:
Number of paper lanterns = (0) + x
Number of paper lanterns = x eq. 2
Number of plastic lanterns = (number of plastic lanterns sold) + (number of plastic lanterns left)
Number of plastic lanterns = (8) + (number of plastic lanterns left) eq. 3
We know from eq. 1 that:
Number of paper lanterns + number of plastic lanterns = 40
x + y = 40
or
x + (8 + (number of plastic lanterns left)) = 40
REMEMBER, the number of plastic lanterns left = number of paper lanterns left = x
x + (8 + x) = 40
x + x = 40 – 8
2x = 32
2x/2 = 32/2
x = 16 paper lanterns
using eq. 1,
x + y = 40
16 + y = 40
y = 40 -16
y = 24 plastic lanterns