ps set: section 23

This topic has expert replies
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Fri Jun 01, 2007 11:02 pm

ps set: section 23

by thumpin_termis » Sat Jul 07, 2007 8:52 pm
17. Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent stamps, what is the least number of 1-cent stamps she could have bought ?
(A) 5
(B) 10
(C) 15
(D) 20
(E) 25

.
.
.
.
.
.
OA is C

Master | Next Rank: 500 Posts
Posts: 468
Joined: Sat Mar 03, 2007 10:17 pm
Thanked: 5 times

by moneyman » Sat Jul 07, 2007 9:36 pm
My approach to this problem is by picking numbers..

We have four denominations
1c
5c
10c
25c

Lets pick a number for the number of 5 cents...Let the number be 5.

Then according to the question we will have 10 10cents and 5 25cents and this amounts to $2.50 and we are said that the total is $2.65.Therefore the number of 1cent coins would be 15($2.65-$2.50)..Answer C

Try picking other numbers greater than 5 and you will find that it does not ahere to the question.
Maxx

Senior | Next Rank: 100 Posts
Posts: 69
Joined: Wed Apr 25, 2007 1:38 am
Followed by:1 members

by discreet » Sat Jul 07, 2007 9:56 pm
Here is How I approached the problem :

Let x denote the no. of 5 cent stamps and y denote the no. of 1 cent stamps.

so,Lets form an equation from the given condition

=> 5x+ 25x + 2x(10) + y = 265

Finally,we have 50x + y = 265

Thumb rule that can be used for such type of questions : When you are asked to minimize a quantity,you must maximize all others and vice-versa.

Here,we are asked the minimum no. of 1 cent stamps

so,Maximum possible value of x is 5 giving us 250.So,the remaining has to be 1 cent stamp => 15.

Master | Next Rank: 500 Posts
Posts: 468
Joined: Sat Mar 03, 2007 10:17 pm
Thanked: 5 times

by moneyman » Sat Jul 07, 2007 10:01 pm
Dats a great approach discreet..
Maxx