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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (03:32) correct
39%
(03:01)
wrong
based on 62
sessions
History
Date
Time
Result
Not Attempted Yet
Peter buys some pens, pencils and erasers using a certain amount of money. The total price of 1 pencil and 1 eraser is equal to the price of 1 pen. He bought thrice as many pens as pencils and half as many pencils as erasers. If he had exchanged the number of pens and erasers, he would have saved $ 20. How much money (in $) did he spend on buying pencils?
A. 10
B. 15
C. 20
D. 25
E. 30
Source : Wizako
A. 10
B. 15
C. 20
D. 25
E. 30
Source : Wizako
01 Sep 2022, 00:58
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Let pens = x, let pencils = y and let erasers = z
The total price of 1 pencil and 1 eraser is equal to the price of 1 pen:
\(x = z + y\)
He bought thrice as many pens as pencils and half as many pencils as erasers:
Tells us that the ratio of \(x:y:z\) = \(3:1:2\). Let's assume that these are the exact amounts that Peter bought.
If he had exchanged the number of pens and erasers, he would have saved $ 20:
Actual amount spent: \(3x + y + 2z = a\)
Using \(x = z + y\) we can rewrite the above as \(3(z + y) + y + 2z = a\)
\(4y + 5z = a\)
Amount spent if he had exchanged the number of pens and pencils: \(2x + y + 3z = a - 20\)
Using \(x = z + y\) we can rewrite the above as \(2(z + y) + y + 3z = a - 20\)
\(3y + 5z = a - 20\)
Subtracting \(3y + 5z = a - 20\) from \(4y + 5z = a\) will give \(y = 20\)
Answer C
The total price of 1 pencil and 1 eraser is equal to the price of 1 pen:
\(x = z + y\)
He bought thrice as many pens as pencils and half as many pencils as erasers:
Tells us that the ratio of \(x:y:z\) = \(3:1:2\). Let's assume that these are the exact amounts that Peter bought.
If he had exchanged the number of pens and erasers, he would have saved $ 20:
Actual amount spent: \(3x + y + 2z = a\)
Using \(x = z + y\) we can rewrite the above as \(3(z + y) + y + 2z = a\)
\(4y + 5z = a\)
Amount spent if he had exchanged the number of pens and pencils: \(2x + y + 3z = a - 20\)
Using \(x = z + y\) we can rewrite the above as \(2(z + y) + y + 3z = a - 20\)
\(3y + 5z = a - 20\)
Subtracting \(3y + 5z = a - 20\) from \(4y + 5z = a\) will give \(y = 20\)
Answer C
01 Sep 2022, 01:37
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I found it useful to use different notations for quantities and prices
Let's say upper case letters for the first and lower case letters for the latter
As a consequence we have that the total amount spent by Peter is
Pp + Nn + Ee
Where
P: number of pens bought
p: price of a single pen
N: number of pencils bought
n: price of a single pencil
E: number of erasers bought
e: price of a single eraser
Now, knowing that
p = n + e
And that
P = 3N
And
E = 2N
We can substitute in the above obtaining
3N * (n + e) + Nn + 2Ne = N * (4n + 5e)
Had Peter exchanged the number of Pens and Eraser, he would have spent
2N * (n + e) + Nn + 3Ne = N * (3n + 5e)
Which would have meant a saving of $ 20,00
As a consequence we have that
N * (4n + 5e) - N * (3n + 5e) = Nn = $ 20,00
Now the solution is simply there
Nn = $ 20,00
That is, the amount spent by Peter in pencils is twenty dollars
IMO, the answer is "C"
Posted from my mobile device
Let's say upper case letters for the first and lower case letters for the latter
As a consequence we have that the total amount spent by Peter is
Pp + Nn + Ee
Where
P: number of pens bought
p: price of a single pen
N: number of pencils bought
n: price of a single pencil
E: number of erasers bought
e: price of a single eraser
Now, knowing that
p = n + e
And that
P = 3N
And
E = 2N
We can substitute in the above obtaining
3N * (n + e) + Nn + 2Ne = N * (4n + 5e)
Had Peter exchanged the number of Pens and Eraser, he would have spent
2N * (n + e) + Nn + 3Ne = N * (3n + 5e)
Which would have meant a saving of $ 20,00
As a consequence we have that
N * (4n + 5e) - N * (3n + 5e) = Nn = $ 20,00
Now the solution is simply there
Nn = $ 20,00
That is, the amount spent by Peter in pencils is twenty dollars
IMO, the answer is "C"
Posted from my mobile device
