Expert: Sherry Wallin - 7/3/2009

Question
1) 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?

2)Three types of pencils,J,K,and L, cost $0.05, $0.10, and $0.25 each, respectively.If a box of 32 of these pencils costs a total of $3.40 and there are twice as many K pencils as L pencils in the box, how many J pencils are there in the box?

Answer
Hi Shruti~
   Let x be the number of 5 cent (.05x) and 25 cent stamps (.25x). Since Kim bought twice as many 10 cent stamps as 5 cent stamps we know that she bought 2x ten-cent stamps as well .1(2x) = .2x.
This gives you an equation of .05x +.25x + .2x < = 2.65 or
.50x < = 2.65 -> x < = 5.3 so it must be that x = 5. So now you know you have 5 nickel stamps, 10 dime stamps and 5 quarter stamps for a total cost of $2.50 which means you can buy 15 one-cent stamps.

    Let x = the number of L pencils so there are 2x K pencils. Since you know there are 32 pencils you know that the number of J pencils is 32 -(x + 2x). You also know that the cost is $3.40 so
.05(32-3x) (.1)2x + .25x = 3.40 -> .3x + 1.60 = 3.40 -> .3x = 1.80 ->
x = 6, thus the number of L pencils is 6 and the number of K pencils is 12 and then number of J pencils is 32 - 18 = 14 J pencils.

Math Prof