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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
Answers
Answer:
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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
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- Therefore, Mr. Jones loss 10 cents.
- 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%.
↦ S.P of the first pipe = $1.20
- Profit = 20%.
Now, we have to find the C.P of the first pipe.
↦ CP = SP - Profit
↦ CP = 1.20 - 20% of CP
↦ CP = 1.20 - 0.20CP
↦ CP + 0.20CP = 1.20
↦ 1.20CP = 1.20
↦ CP =
↦ CP = $ 1
↦ S.P of the Second pipe = $1.20
- Loss = 20%.
Now, we have to find the C.P of the second pipe.
↦ CP = SP + Loss
↦ CP = 1.20 + 20% of CP
↦ CP = 1.20 + 0.20CP
↦ CP - 0.20CP = 1.20
↦ 0.80CP = 1.20
↦ CP =
↦ CP = $1.50
Hence, total CP of the two pipes =$ 1.00 + $1.50 =$ 2.50
And total SP of the two pipes =$1.20 + $1.20 = $2.40
Loss = $ 2.50 - $ 2.40 = $ 0.10
- Thus, Mr. Jones loss 10 cents.