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profit and loss help!!!

by anuu » Fri May 06, 2011 9:36 pm
Hi Guys,

I came across the below problem in an online practice test...can anybody pls provide suggestions to solve it?

If the cost of each book is reduced by $10, a person can buy 30 more books for $1800. Find the original price of the book.

My approach is :

let the original cost of book be x$
if 'n' books are bought then the cost is nx$

now, If the cost of each book is reduced by $10, a person can buy 30 more books for $1800

i.e (x-10)(n+30) = 1800

I'm stuck here..any suggestions pls?

Regards,
Anu
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by prakhag » Fri May 06, 2011 10:43 pm
Given:

n.x = 1800 - (1)

(x-10)(n+30) = 1800 - (2)

Expand eqn (2):

nx-10n+30x-300 = 1800

Use equation (1) to subsitute nx=1800 and x=1800/n;

You will get x=30.
Last edited by prakhag on Sat May 07, 2011 10:00 am, edited 1 time in total.
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by aravindan_v » Fri May 06, 2011 10:53 pm
Hi Anu,

I am assuming that the cost 1800 is a constant. ie, the cost of n books at x price is 1800. Also, the cost of n+30 books at x-10 price is 1800.

from the above statements we can get two equations,
1. xn = 1800
2. (x-10)(n+30) = 1800

from this we can arrive to the original price x as 30.

Cheers,
Aravindan
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by manpsingh87 » Fri May 06, 2011 11:21 pm
anuu wrote:Hi Guys,

I came across the below problem in an online practice test...can anybody pls provide suggestions to solve it?

If the cost of each book is reduced by $10, a person can buy 30 more books for $1800. Find the original price of the book.

My approach is :

let the original cost of book be x$
if 'n' books are bought then the cost is nx$

now, If the cost of each book is reduced by $10, a person can buy 30 more books for $1800

i.e (x-10)(n+30) = 1800

I'm stuck here..any suggestions pls?

Regards,
Anu
let the cost of the book be x; therefore as per given information we have;
1800/(x-10)-1800/x=30;
1800(x-x+10)=30(x^2-10x);
60*10=x^2-10x;
x^2-10x-600;
(x-30)(x+20)=0;
hence x=30;
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by GMATGuruNY » Sat May 07, 2011 3:09 am
At a certain store, the selling price of each book is the same. If the selling price of each book were to be reduced by $10, 30 more books could be purchased for $1800 than could be purchased at the original price. What is the original price of each book?

A. $20
B. $30
C. $40
D. $50
E. $60
No need to set up a quadratic equation. We can plug in the answers, which represent the original price of each book.

Answer choice C: $40 per book
Number that can be purchased at $40 each = 1800/40 = 45.
Number that can be purchased at $30 each = 1800/30 = 60.
Difference = 60-45 = 15.
Doesn't work.
Since the difference needs to be larger, the original cost must be smaller, so that more books can be purchased at each price.
Eliminate C, D and E.

Answer choice B: $30 per book.
Number that can be purchased at $30 each = 1800/30 = 60.
Number that can be purchased at $20 each = 1800/20 = 90.
Difference = 90-60 = 30. Success!

The correct answer is B.
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by mk101 » Sat May 07, 2011 12:05 pm
anuu wrote:Hi Guys,

I came across the below problem in an online practice test...can anybody pls provide suggestions to solve it?

If the cost of each book is reduced by $10, a person can buy 30 more books for $1800. Find the original price of the book.

My approach is :

let the original cost of book be x$
if 'n' books are bought then the cost is nx$

now, If the cost of each book is reduced by $10, a person can buy 30 more books for $1800

i.e (x-10)(n+30) = 1800

I'm stuck here..any suggestions pls?

Regards,
Anu

Look for factors of 1800 that differ by 30 - (10, 40) (20,50) (30, 60) (60,90) - these are the only possible pairs of original number of books bought and the number of books bought later on.

Divide 1800 by each pair and look for that pair which gives you a difference of 10

60 and 90 fit our requirement - 1800/60 = 30 and 1800 =20
the above values satisfy the conditions
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