Quant Question of the Day Chat

Bunuel, Is there a way to retake the incorrect questions from the GC mocks as a customized test ? Afaik GC mock questions don’t show up in the error log.

Got the option under the custom test tab.

Hi. Can someone please explain how answer to DS 2 - E?

Can you post the link pls?

Check this new Arithmetic question from GMAT Club:

https://gmatclub.com/forum/topic-427904.html
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Answer to DS 2 is option C. Pls refer to this link: how-old-is-ferdinand-1-ferdinand-and-kris-are-the-same-age-427885.html

Can you explain please how you guys solved this one

Data Sufficiency Butler: March 2024
March 29	DS 1	DS 2

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Problem Solving Butler: February 2023
March 29	PS 1	PS 2

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The underlying concept is that if you add a constant to a bunch of numbers(each number in that bunch) , the SD of the new bunch of numbers doesn’t change.
Also if you multiply each number in that bunch by a constant, the SD of the new bunch = old SD * mod of the constant.

So essentially y is a bunch of numbers generated by adding a constant of 2 (which has no effect on the SD) and then multiplying the numbers by a constant.
So the new SD will be 1*5

for finding max/min of x-y for 2 given ranges can we find it by subtracting the 4 extreme values straight and diagonally like we do to find for xy’s max and min?

Check this easy Arithmetic question:

https://gmatclub.com/forum/topic-427955.html
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One more fresh Arithmetic problem to practice:

https://gmatclub.com/forum/topic-427957.html
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Try this new tricky Arithmetic question:

https://gmatclub.com/forum/if-k-is-an-i ... 27961.html
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can someone explain how to do this?




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Discussed here: at-a-certain-photoprocessing-shop-the-first-standard-size-print-of-a-389858.html
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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.­
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can anyone help with this euler path

as a sequence of vertices in the order they are visited

Man this seems like a scam

It’s tyring to open https://gmatclub.com/chat no my phone