Problem Solving

is there a specific formula for a question like this?

Kevin earns a 15% commission on his first $50,000 in sales each year, and a 20% commission on sales in excess of $50,000. If Kevin's total commission last year was $12,500, what was the total amount of his sales?

Earning is made up of 2 parts.

(1) 15% of the first 50,000 >> (15/100)(50,000) >> 15*500 >> 7,500

(2) 20% of the sales in excess of 50,000. If S is the total sales, the amount in excess of 50,000 would be S-50,000. So this part of earning is 20% of S-50,000 >> (20/100)(S-50,000) >> (S/5 - 10,000)

Total earning is 7,500 + (S/5 - 10,000) if he sold more than $50,000

We know that his total earnings were $12,500. To find S, the total amount of his sales just solve for it: 12,500 = 7,500 + (S/5 - 10,000)

-Patrick

ohjoon2 wrote:is there a specific formula for a question like this?

Kevin earns a 15% commission on his first $50,000 in sales each year, and a 20% commission on sales in excess of $50,000. If Kevin's total commission last year was $12,500, what was the total amount of his sales?

Hi,

this question would be super-fast to solve using backsolving; if you post the answer choices (which you should always do, since the answers on a multiple choice exam often shape your attack on the question), I'd be happy to show you how to do so.

15/100(50000)+20X/100=12500

;X=25000

25000+50000=75000

Can someone tell me if this is correct?

artistocrat wrote:15/100(50000)+20X/100=12500

;X=25000

25000+50000=75000

Can someone tell me if this is correct?

yea thats right!

Preet

First calculate what is 15% of 50000 = 15/100 * 50000 = 7500
so leftover commission = 12500 - 7500 = 5000
and this 5000 commission comes from 20% so that amount = 100/20 * 5000 = 25000
so total amount of his sales = 50000 + 25000 = 75000

so its a 2 step problem and may be solved in 30 sec.
step 1. 15/100 * 50000 = 7500(upto 10 sec.), 12500 - 7500 = 5000(upto 5 sec.)
step 2. 100/20 * 5000 = 25000(upto 10 sec.), 50000 + 25000 = 75000(upto 5 sec.)


add upto 30 sec. for reading the problem too..



btw back solving would be even faster than this..as stuart said..

thanks all for the help.

p.s.
stuart-it was a sample question that i found online, there wasn't any answer choices. i just wanted to know the formula.