Question

3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he:

(a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents

Answer 1

Let The Cost Price for the first pipe be "x" Dollars

Selling Price = x + 20% of x 

1.20 = x + 20x / 100

1.20 = x + x/5

1.20 = (5x + x)/5

1.20 * 5 = 6x

6 = 6x

x = 1 Dollar

So ,

Cost Price of The Pipe in Which Mr. Jones Got Profit of 20% = 1 Dollar

Let The Cost Price Of The Second Pipe be "y" dollar

Selling Price of second Pipe = y - 20% of y

1.20 = y - 20y/100

1.20 = y - y/5

1.20 = (5y - y)/5

1.20 * 5 =4y

6 = 4y

y = 6/4

y = 3/2

y = 1.5 Dollars 

Total Cost Price Which Mr Jones Paid = 1 + 1.5

                                                                 = 2.5 Dollars

Total Selling Price Which Mr Jones Received = 1.2 + 1.2

                                                                             = 2.4 Dollar

Mr. Jones Lost 10 cents

so,

ANSWER = (d) = lost 10 cents 

Answer 2

Answer:

Step-by-step explanation:

Solution:

Selling price of the first pipe = $1.20

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price - Profit

CP = 1.20 - 20% of CP

CP = 1.20 - 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = 1.201.20

CP = $ 1

Selling price of the Second pipe = $1.20

Loss = 20%

Let’s try to find the cost price of the second pipe

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP - 0.20CP = 1.20

0.80CP = 1.20

CP = 1.200.80

CP = $1.50

Therefore, total cost price of the two pipes = $1.00 + $1.50 = $2.50

And total selling price of the two pipes = $1.20 + $1.20 = $2.40

Loss = $2.50 – $2.40 = $0.10

Therefore, Mr. Jones loss 10 cents.

Answer:  (d)