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Kritisood
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 21 Jun 2020, 10:46
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Difficulty:

55% (hard)

Question Stats:

73% (03:20) correct 27% (03:26) wrong based on 421 sessions

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A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200

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Is there a way to do this w/o the cumbersome calculations?
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chetan2u
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 21 Jun 2020, 22:51
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Kritisood
A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200

Is there a way to do this w/o the cumbersome calculations?
Show more

Of course if you have time, do the calculations and get to your answer.

Say you are short of time and are looking for the best possible guess..

1) First point - profit or loss
there is 40% profit on some and 140*0.8 or 12% on others so clearly profit...Eliminate A and B

2) Second point - The answer should be a multiple of 3.
We have 10000+2000 or 12000 being multiplied by different quantities in each case, so answer has to be a multiple of 3....Eliminate E, as 1+8+3+2=14, not a multiple of 3

3) Compare remaining choices - 24,000 is very small compared to 139,200
Now, the initial 20 tractors give a profit of 40% so 20*12000*0.4=96000.....Eliminate C - 24000, as answer >96000

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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 21 Jun 2020, 14:43
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Kritisood
A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200
Show more

Consider the following case:
Cost per car = 100
For 40% profit to be earned the first month, the selling price in the first month = $140
For the selling price to decrease by 20% the following month, the selling price in the following month = 140 - (20% of 140) = 140-28 = $112
The blue value indicates that the profit in the second month = 12%

In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month.
Implication:
Of every 5 cars, 2 earn a profit of 40% in the first month, while 3 earn a profit of 12% in the following month.
Average profit percentage per 5 cars \(= \frac{(2*40) + (3*12)}{5}\)%= \(\frac{116}{5}\)% \(= \frac{116}{500}\)

A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work.
Total cost = 50(10000+2000) = 50*12,000

Since the cars earn an average profit of \(\frac{116}{5}\)%, we get:
Overall profit \(= \frac{116}{500} * 50*12000 = 116*1200 = 139,200\)

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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 21 Jun 2020, 20:58
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Average profit
= (20×40+30×12)÷50
=23.2
Profit
=12000×50×0.232
=139200

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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 10 Oct 2020, 12:35
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Kritisood
A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200

Consider the following case:
Cost per car = 100
For 40% profit to be earned the first month, the selling price in the first month = $140
For the selling price to decrease by 20% the following month, the selling price in the following month = 140 - (20% of 140) = 140-28 = $112
The blue value indicates that the profit in the second month = 12%

In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month.
Implication:
Of every 5 cars, 2 earn a profit of 40% in the first month, while 3 earn a profit of 12% in the following month.
Average profit percentage per 5 cars \(= \frac{(2*40) + (3*12)}{5}\)%= \(\frac{116}{5}\)% \(= \frac{116}{500}\)

A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work.
Total cost = 50(10000+2000) = 50*12,000

Since the cars earn an average profit of \(\frac{116}{5}\)%, we get:
Overall profit \(= \frac{116}{500} * 50*12000 = 116*1200 = 139,200\)

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I wanted to ask is this a GMAT level question? Since it involves a lot of calculation and is also time consuming.
Please respond
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 11 Oct 2020, 11:06
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Kritisood
A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200

Consider the following case:
Cost per car = 100
For 40% profit to be earned the first month, the selling price in the first month = $140
For the selling price to decrease by 20% the following month, the selling price in the following month = 140 - (20% of 140) = 140-28 = $112
The blue value indicates that the profit in the second month = 12%

In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month.
Implication:
Of every 5 cars, 2 earn a profit of 40% in the first month, while 3 earn a profit of 12% in the following month.
Average profit percentage per 5 cars \(= \frac{(2*40) + (3*12)}{5}\)%= \(\frac{116}{5}\)% \(= \frac{116}{500}\)

A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work.
Total cost = 50(10000+2000) = 50*12,000

Since the cars earn an average profit of \(\frac{116}{5}\)%, we get:
Overall profit \(= \frac{116}{500} * 50*12000 = 116*1200 = 139,200\)

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D
Show more

GMATGuruNY,

What am I doing wrong?
Total Cost =
50*10,000= 500,000
2000*50= 100,000
600,000

Total sales = first month 20*14000= 280,000
Second month = 30 * 11,200 =336,000

So Total incoming 280000+ 336000= 616,000
Hence profit 616000-600000= 16000 .
???
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 11 Oct 2020, 13:37
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stne
What am I doing wrong?
Total Cost =
50*10,000= 500,000
2000*50= 100,000
600,000

Total sales = first month 20*14000= 280,000
Second month = 30 * 11,200 =336,000
???
Show more

The colored values assume a cost per tractor of $10,000 and thus are incorrect.
Since the dealer spends $2,000 to repair each $10,000 tractor, his total cost per tractor = $12,000.
Red value, corrected:
Selling price for the 20 tractors that earn a 40% profit = 12,000 * 1.4 = $16,800.
Blue value, corrected:
Selling price for the remaining 30 tractors after a 20% discount = 16,800 * 0.8 = $13,440.
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 11 Oct 2020, 13:47
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stne
What am I doing wrong?
Total Cost =
50*10,000= 500,000
2000*50= 100,000
600,000

Total sales = first month 20*14000= 280,000
Second month = 30 * 11,200 =336,000
???

The colored values assume a cost per tractor of $10,000 and thus are incorrect.
Since the dealer spends $2,000 to repair each $10,000 tractor, his total cost per tractor = $12,000.
Red value, corrected:
Selling price for the 20 tractors that earn a 40% profit = 12,000 * 1.4 = $16,800.
Blue value, corrected:
Selling price for the remaining 30 tractors after a 20% discount = 16,800 * 0.8 = $13,440.
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Thank you Mitch.
Makes sense.
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 23 Dec 2025, 20:36
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I just had one doubt, isn't profit % usually based on selling price and mark up is on cost?
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Bunuel KarishmaB egmat
@GMATGuruNY
Isn't profit % on selling price and not on cost?
shouldn't we do 12,000/60% instead?

markup is the term used for % on cost..
Please guide & correct me


Kritisood
A tractor dealer purchased 50 old tractors, each at $10,000 and spent $2000 on each of them for repair work. In the first month, he sold 40% of the tractors, each, at a profit of 40% and sold the remaining tractors in the next month, each at a price 20% less than the selling price of each tractor in the previous month. What was the overall profit or loss for the dealer?

A. Loss of $24,000
B. Loss of $139,200
C. Profit of $24,000
D. Profit of $139,200
E. Profit of $183,200

Show Spoiler
Is there a way to do this w/o the cumbersome calculations?
Show more
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A tractor dealer purchased 50 old tractors, each at $10,000 and spent 24 Dec 2025, 00:56
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You are confusing the concepts.

Profit% is on Cost.
Margin% is profit on selling price.
Markup is the increase in price on cost which is same as profit if no discount is given at the point of sale.

rak08
Bunuel KarishmaB egmat
@GMATGuruNY
Isn't profit % on selling price and not on cost?
shouldn't we do 12,000/60% instead?

markup is the term used for % on cost..
Please guide & correct me



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