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605-655 (Medium)|   Word Problems|         
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At a certain photoprocessing shop, the first standard-size print of a 08 May 2022, 02:33
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45% (medium)

Question Stats:

71% (02:00) correct 29% (02:09) wrong based on 1428 sessions

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At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the same negative costs $1. What is the total cost, in dollars, of y standard-size prints of each of x different negatives?

A. 4xy + (x - 1)y

B. 4xy + (y - 1)x

C. 4x + xy

D. 4x + xy - 1

E. 4x + x(y - 1)


Source: Skills Insight
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E
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At a certain photoprocessing shop, the first standard-size print of a 26 Oct 2022, 09:50
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summerindecember
Hi stne, I don't understand the question.

Why is "(y−1)1"? Why do we need to add "(y−1)1" to "4"? Why do we need to subtract 1 from y?
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At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the same negative costs $1. What is the total cost, in dollars, of y standard-size prints of each of x different negatives?

A. 4xy + (x - 1)y

B. 4xy + (y - 1)x

C. 4x + xy

D. 4x + xy - 1

E. 4x + x(y - 1)

The first print of a negative costs $4, and each additional print of the same negative costs $1.

One print of a negative costs $4;
Two prints of the negative cost $4 (for the first print) plus $1 (for the second print) = $5
Three prints of the negative cost $4 (for the first print) plus $1 (for the second print) + $1 (for the third print) = $6
...

So, the first print costs $4 and all subsequent prints of the same negative cost $1 each.

Now, y prints of a negatives will cost $4 for the first print, and $1 for each of the the remaining prints of the negative, so $1 for each of the remaining (y-1) prints. So, the total of $4*1 + $1*(y - 1). For example:

    10 prints of a negatives will cost $4 for the first print, and $1 for each of the the remaining 10 -1 = 9 prints of that negative: $4*1 + $1*(10 - 1).

Now, if we have x different negatives, then to print y prints of each, we'd need x*(4 + (y - 1)) = 4x + x(y - 1). For example:

    Say we have 3 different negatives and want to print 10 prints of each. 10 prints of one negative will cost 4 + (10 - 1) dollars and 10 prints of 3 different negatives will cost 3*(4 + (10 - 1)).

Answer: E.

Hope it's clear.

P.S. This is an official question, so I suggest to study it carefully.
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At a certain photoprocessing shop, the first standard-size print of a 16 May 2022, 22:40
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Bunuel
At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the same negative costs $1. What is the total cost, in dollars, of y standard-size prints of each of x different negatives?

A. 4xy + (x - 1)y

B. 4xy + (y - 1)x

C. 4x + xy

D. 4x + xy - 1

E. 4x + x(y - 1)


Source: Skills Insight
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For one negative the cost for \(y\) prints :

\(4 +(y-1)1\)

For \(x\) negatives the cost for \(y\) prints:

\(=x(4 +(y-1)1)\)

\(=4x+x(y-1)\)

Ans E
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At a certain photoprocessing shop, the first standard-size print of a 26 Oct 2022, 03:22
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Hi stne, I don't understand the question.

Why is "(y−1)1"? Why do we need to add "(y−1)1" to "4"? Why do we need to subtract 1 from y?
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At a certain photoprocessing shop, the first standard-size print of a 03 Sep 2023, 03:09
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Bunuel
summerindecember
Hi stne, I don't understand the question.

Why is "(y−1)1"? Why do we need to add "(y−1)1" to "4"? Why do we need to subtract 1 from y?

At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the same negative costs $1. What is the total cost, in dollars, of y standard-size prints of each of x different negatives?

A. 4xy + (x - 1)y

B. 4xy + (y - 1)x

C. 4x + xy

D. 4x + xy - 1

E. 4x + x(y - 1)

The first print of a negative costs $4, and each additional print of the same negative costs $1.

One print of a negative costs $4;
Two prints of the negative cost $4 (for the first print) plus $1 (for the second print) = $5
Three prints of the negative cost $4 (for the first print) plus $1 (for the second print) + $1 (for the third print) = $6
...

So, the first print costs $4 and all subsequent prints of the same negative cost $1 each.

Now, y prints of a negatives will cost $4 for the first print, and $1 for each of the the remaining prints of the negative, so $1 for each of the remaining (y-1) prints. So, the total of $4*1 + $1*(y - 1). For example:

    10 prints of a negatives will cost $4 for the first print, and $1 for each of the the remaining 10 -1 = 9 prints of that negative: $4*1 + $1*(10 - 1).

Now, if we have x different negatives, then to print y prints of each, we'd need x*(4 + (y - 1)) = 4x + x(y - 1). For example:

    Say we have 3 different negatives and want to print 10 prints of each. 10 prints of one negative will cost 4 + (10 - 1) dollars and 10 prints of 3 different negatives will cost 3*(4 + (10 - 1)).

Answer: E.

Hope it's clear.

P.S. This is an official question, so I suggest to study it carefully.
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the way you are explaining is amazing.....hats off you....it took hours for me to crack....but when i read your explanation i felt is is very easy .....thank you so much ...
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At a certain photoprocessing shop, the first standard-size print of a 27 Jan 2025, 20:09
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The wording of the question is ambiguous and may lead to confusion, I'm surprised this is an official Q.

I understood this Q as if all X negatives were different for each y as the Q states: "y standard-size prints of each of x different negatives". So if all X negatives are different, then the solution would be 4xy.

For example, if we need to print 2 standard-size prints of each of 3 different negatives, the cost would be = $4*3 + $4*3 = $4*3*2 (or 4*x*y).

The answer shown above requires an assumption that all Ys are the same.

Right?
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At a certain photoprocessing shop, the first standard-size print of a 28 Jan 2025, 01:18
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LongTermWarrior
The wording of the question is ambiguous and may lead to confusion, I'm surprised this is an official Q.

I understood this Q as if all X negatives were different for each y as the Q states: "y standard-size prints of each of x different negatives". So if all X negatives are different, then the solution would be 4xy.

For example, if we need to print 2 standard-size prints of each of 3 different negatives, the cost would be = $4*3 + $4*3 = $4*3*2 (or 4*x*y).

The answer shown above requires an assumption that all Ys are the same.

Right?
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y's are number of prints for each of x different negatives; the wording of question seems clear enough.

Taking your example,
Let's say if I have 3 different negatives => x = 3
I want 2 prints of each => y = 2

For 1 negative, the cost will come to => 4*(1) + 1*(2-1) = $5
For 3 negatives, the cost will come to => 3 * 5 = $15

4x + x(y-1) = 12 + 3(2-1) = $15
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At a certain photoprocessing shop, the first standard-size print of a 15 Jun 2026, 20:03
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This question illustrates a frequent question type on the GMAT: the “flat fee plus” problem (other common examples include taxi fares and phone plans). The key with this type of question is to properly account for the flat fee portion in the “plus” calculation. If you aren’t careful, it’s easy to double-count the item that is covered by the flat fee.

To ensure that you count properly, picking numbers is often a smart strategy on these questions. Here, it’s easiest to pick small integers (aside from the number 1).

Let’s pick x = 2 and y = 3. So we’re getting 3 prints each of 2 different negatives. The first print is $4 and the additional 2 prints are each $1, so the 3 prints for each negative cost $6. Since you have 2 negatives, the total cost is $12.

Now check the answer choices using x = 2 and y = 3 to see what matches $12:

(A) 4xy + (x - 1)y --> 4xy is already 4(2)(3) = $24, so choice A is too big.
(B) 4xy + (y - 1)x --> Also too big for the same reason.
(C) 4x + xy --> 4(2) + (2)(3) = $14
(D) 4x + xy – 1 --> 4(2) + (2)(3) – 1 = $13
(E) 4x + x(y - 1) --> 4(2) + 2(3 - 1) = $12

Only choice E matches, so it is correct.

You could also solve using algebraic translation. The first print for each negative costs $4. Then, we subtract 1 from y to leave us with y – 1 additional prints, each of which costs $1, so the additional prints for each negative cost $(y – 1).

Thus, the cost of all the prints for each negative is 4 + (y – 1). You might be tempted to combine the 4 and the -1, but noting that 4 is in all the answer choices, let’s leave it alone.

Finally, to find out the total cost we multiply the number of negatives x times the cost per negative 4 + (y – 1):

x(4 + (y – 1))

Then distribute the x:

4x + x(y – 1)

This is answer choice E.

Note the trap answer C, which would result if you forgot to subtract 1 from y to account for the fact that the first print was already paid for in the $4 fee. Picking numbers makes this more obvious (without taking a lot of time to calculate), so for most students will be preferable here.

Either way you solve, it’s critical to account for the initial portion (the first print) in the plus calculation (the additional prints), advice that applies to all “flat fee plus” questions.
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At a certain photoprocessing shop, the first standard-size print of a 16 Jun 2026, 06:55
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Bunuel
At a certain photoprocessing shop, the first standard-size print of a negative costs $4, and each additional print of the same negative costs $1. What is the total cost, in dollars, of y standard-size prints of each of x different negatives?

A. 4xy + (x - 1)y

B. 4xy + (y - 1)x

C. 4x + xy

D. 4x + xy - 1

E. 4x + x(y - 1)


Source: Skills Insight
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Plug in simple values.

x = 1, y = 2. Total cost of 2 prints of 1 negative will be $4 + $1 =$5.00

A. 4xy + (x - 1)y = 8 + ..Non-negative.. (Eliminate)

B. 4xy + (y - 1)x = 8 + ..Non-negative.. (Eliminate)

C. 4x + xy = 4 + 2 (Eliminate)

D. 4x + xy - 1 = 4 + 2 - 1 = 5 (Possible)

E. 4x + x(y - 1) = 4 + 1 = 5 (Possible)


x = 2, y = 1. Total cost of 1 print of 2 negatives will be $4 + $4 =$8.00

D. 4x + xy - 1 = 8 + 1 (Eliminate)

E. 4x + x(y - 1) = 8 (Correct Answer)

Answer (E)
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At a certain photoprocessing shop, the first standard-size print of a 16 Jun 2026, 14:53
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summerindecember
Hi stne, I don't understand the question.

Why is "(y−1)1"? Why do we need to add "(y−1)1" to "4"? Why do we need to subtract 1 from y?
Show more
Good question! The question indicates that y represents the number of standard-sized prints. The first print costs $4, so let’s start with that. If more prints are ordered, each of them is $1. So we’ll need to add the $4 cost of the first print to the cost of additional prints: $1 times * # of additional prints.

The # of additional prints sold for $1 is going to be the total amount of prints (y), except that we won’t count the first print, because that was already accounted for. So the # of additional prints is y-1 (the total minus the first $4 print).

So to find the cost of y prints of a single photo, we add $4 + $1 * (y -1), which equals 4 + 1(y-1).


For more practice problems, check out ManhattanPrep’s Free GMAT QBank and Starter Kit.


Best,
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