profit or loss

[rrobiinn]

if Jon buys equal number of two types of mangoes at the rate of 9 mangoes per dollar and 11 mangoes per dollar, and then sell those at the rate of 10 mangoes per dollar, then what will be his profit or loss?

I can solve it using shortcut method, but now want to learn about how its done actually in details.

[Tommy Wallach]

Hey Robin,

On a question like this, you'd likely be able to use logic/answer choices to get at shortcuts. Often the GMAT gives questions where the shortcuts are so efficient that you shouldn't do the math. However, the math here should be pretty easy, if you know your fraction/percent equivalents.

9 mangoes per dollar = 11.11 cents per mango
11 mangoes per dollar = 9.09 cents per mango
10 mangoes per dollar = 10 cents per mango

If we imagine 2 and 2 mangoes, that would be 22.22 + 18.18 = 40.4 cents
If we sell those four mangoes for 10 cents each = 40 cents

Loss of .4 cents, or in percents, approximately 1 percent.

Hope that helps!

-t

P.S. If you were implying that we should do this without plugging in numbers, that would be totally crazy. This isn't a shortcut, but the real way to do the question. The algebra to solve for percent change would look like this:

(11.11x + 9.09x - 10(2x))/(11.11x + 9.09x) * 100

In other words: Difference/Original * 100 = (Cost of purchase - Cost of Sale)/(Cost of purchase) * 100

The x's will immediately cancel out, so you're left doing what I did, as if you'd plugged in 1 instead of 2.

Hope that helps!

-t

[The Iceman]

if Jon buys equal number of two types of mangoes at the rate of 9 mangoes per dollar and 11 mangoes per dollar, and then sell those at the rate of 10 mangoes per dollar, then what will be his profit or loss?

I can solve it using shortcut method, but now want to learn about how its done actually in details.

Since you need to find profit/loss % and the ratio will be independent of the number of mangoes that you consider, you can assume a nice number here. For example lcm(9,11) or 99 mangoes at each of those rates.

Hence, Profit % = (19.8-(11+9))*100/20 = 1% loss

Note: Even if you assume x mangoes for each of those rates, x will be eliminated

Profit % = x(2/10 - (1/11+ 1/9))*100/[x(1/11+1/9)] = -2*99*100/990*20 = 1% loss

[Tommy Wallach]

Heyo,

Iceman is absolutely right. The answer is 1%. My explanation has been edited (I originally wrote 9.9 instead of 9.09, as I meant). I still prefer plugging in a small number to a big number, as long as you know what 1/11th and 1/9th look like as decimals. : )

-t

[GMATGuruNY]

if Jon buys equal number of two types of mangoes at the rate of 9 mangoes per dollar and 11 mangoes per dollar, and then sell those at the rate of 10 mangoes per dollar, then what will be his profit or loss?

I can solve it using shortcut method, but now want to learn about how its done actually in details.

To make the math easy, let the number of mangos purchased at each rate be equal to the LCM of 9, 11 and 10:
9*11*10 = 990.

Case 1: 990 mangos purchased at 9 mangos per dollar
Total cost = 990/9 = $110.
The selling rate is 10 mangos per dollar.
Thus:
Total revenue = 990/10 = $99.

Case 2: 990 mangos purchased at 11 mangos per dollar
Total cost = 990/11 = $90.
The selling rate is 10 mangos per dollar.
Thus:
Total revenue = 990/10 = $99.

Cost in case 1 + cost in case 2 = 110+90 = 200.
Revenue in case 1 + revenue in case 2 = 99+99 = 198.
Amount lost = 200-198 = 2.
Percent lost = loss/cost = 2/200 = 1/100 = 1%.

[Lifetron]

What's the shortcut ?

[Tommy Wallach]

You could conceivably know that this is the same as increasing by ten percent and decreasing by ten percent, which always results in a loss of ten percent (regardless of the order of increase and decrease).

T