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Word Problems

by BTGmoderatorRO » Sat Nov 11, 2017 7:45 am
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.
I am having a big problem in interpreting this question. Can someone break this down pls???? <i class="em em-triumph"></i>
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by ErikaPrepScholar » Mon Nov 13, 2017 4:48 am
We know from the question that Kim bought one 25-cent stamp for every 5-cent stamp. We also know that Kim bought two 10-cent stamps for every 5-cent stamp. If she buys another 5, 10, or 25 cent stamp, she must buy more of the others to keep the ratios the same. This means that Kim purchased the stamps in 50-cent groups: one 5-cent stamp, one 25-cent stamp, and two 10-cent stamps - if she buys one 5, 10, or 25 cent stamp, she must spend a total of 50-cents to keep the ratios the same.

This means that Kim can buy up to $2.50 worth of stamps with only 5, 10, and 25-cent stamps. Since she bought a total of $2.65 worth of postage stamps, she must have bought $0.15 worth of 1-cent stamps, or 15 stamps.

Using logic like this is probably the fastest way to solve the problem, but we can also create equations. Letting the number of 1-cent stamps = A, 5-cent = B, 10-cent = C, and 25-cent = D, the problem tells us:

B = D
C = 2B
0.01A + 0.05B + 0.1C + 0.25D = 2.65

Combining those equations gives us:

0.01A + 0.05B + 0.1(2B) + 0.25(B) = 2.65
0.01A + 0.5B = 2.65

This tells us that for every 5-cent stamp, Kim must spend 50-cents. Then we can plug in options to minimize A.
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by Scott@TargetTestPrep » Tue Jan 09, 2018 10:36 am
Roland2rule wrote: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
We can create the following variables:

a = number of 1-cent stamps

b = number of 5-cent stamps

c = number of 10-cent stamps

d = number of 25-cent stamps

Thus:

2.65 = 0.01a + 0.05b + 0.10c + 0.25d

and

b = d

and

c = 2b

Thus, we have:

2.65 = 0.01a + 0.05b + 0.10(2b) + 0.25b

2.65 = 0.01a + 0.50b

265 = a + 50b

265 - 50b = a

We want the value of b to be as large as possible and still maintain that the value of 265 - 50b is positive and that value will be the least value of a. We see that if b = 5, then a = 15 (if b > 5, then 265 - 5b, or a, will be negative). Thus, the least value of a is 15.

Answer: C

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by [email protected] » Tue Jan 09, 2018 12:01 pm
Hi Roland2rule,

We're told that Kim bought a total of $2.65 worth of postage stamps in four denominations: she bought an EQUAL number of 5-cent and 25-cent stamps and TWICE as many 10-cent stamps as 5-cent stamps. We're asked for the least number of 1-cent stamps that she could have bought.

This question can be approached with a great 'brute force' approach based on the 'relationships' among the types of stamps that are purchased. Since Kim bought an EQUAL number of 5-cent and 25-cent stamps AND she bought TWICE as many 10-cent stamps as 5-cent stamps, we have a relationship among those 3 types of stamps (a ratio of 1:1:2):

IF... she bought one 5-cent, one 25-cent and two 10-cent stamps, then the total spent is 50 cents
IF... she bought two 5-cent, two 25-cent and four 10-cent stamps, then the total spent is 100 cents
Etc.
Notice that the pattern is some multiple of 50-cents....

With a total of $2.65, you could have 5 of those 50-cent groups (and 5x50 = $2.50), leaving the remaining 15 cents for 1-cent stamps... meaning that the minimum number of 1-cent stamps purchased = 15.

Final Answer: C

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