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A street vendor sells two newspapers, one for 25 cents and t
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02 Nov 2018, 09:06
02 Nov 2018, 09:06
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A street vendor sells two newspapers, one for 25 cents and the other for 40 cents. If she sold 100 newspapers for $28.00, how many newspapers did she sell at 25 cents?
A. 80
B. 60
C. 50
D. 40
E. 20
A. 80
B. 60
C. 50
D. 40
E. 20
A street vendor sells two newspapers, one for 25 cents and t
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02 Oct 2019, 11:01
02 Oct 2019, 11:01
Expert Reply
A street vendor sells two newspapers, one for 25 cents and the other for 40 cents. If she sold 100 newspapers for $28.00, how many newspapers did she sell at 25 cents?
A. 80
B. 60
C. 50
D. 40
E. 20
Kudos for the right answer and explanation
A. 80
B. 60
C. 50
D. 40
E. 20
Kudos for the right answer and explanation
MyGuru Representative
Joined: 09 Apr 2020
Posts: 35
Given Kudos: 0
Re: A street vendor sells two newspapers, one for 25 cents and t
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13 Apr 2020, 19:49
13 Apr 2020, 19:49
1
Expert Reply
Carcass wrote:
A street vendor sells two newspapers, one for 25 cents and the other for 40 cents. If she sold 100 newspapers for $28.00, how many newspapers did she sell at 25 cents?
A. 80
B. 60
C. 50
D. 40
E. 20
A. 80
B. 60
C. 50
D. 40
E. 20
This problem is fairly straightforward, but there are a few strategies for solving it. Let's take a look at two of them.
1) Weighted Average Approach
Since 100 newspapers sell for $28.00, that means that the average cost per paper is $0.28. That's pretty close to $0.25, suggesting that far more than half the papers will be the cheaper one.
Rule out C, D, and E, since those are half or less of the 100 total papers sold. A is probably going to be the right answer, since the vast majority of the papers will be the $0.25 one, but let's check:
80*0.25 + 20*0.4 = 20 + 8 = 28. Answer: A
2. Linear System--Substitution
If you prefer an algebraic approach, just treat this as a linear system of equations. Set up your first equation to represent the total number of papers (100), and the second to represent the total cost ($28).
x+y=100
0.25x + 0.4y = 28
We're solving for x, so we want to eliminate y. Rewrite y in terms of x:
y = 100 - x
And substitute:
0.25x + 0.4(100 - x) = 28
0.25x + 40 - .4x = 28
Subtract 0.25 x from both sides:
40 = .15x + 28
12 = .15x
(You don't need to use the calculator to divide the decimal here--just recognize that 12/15 = 4/5 = .8, and move the decimal point two places to the right.
x = 80
Answer: A
