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Mixture A had 15% cement by weight

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Mixture A had 15% cement by weight [#permalink] New post   30 Sep 2021, 04:34
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A mixture A had 15% cement by weight. One-fourth of the weight of mixture A was then taken from mixture B, which was 40% cement by weight and added to mixture A. What is the percentage increase in cement in the new mixture?

A. 2
B. 3
C. 4
D. 5
E. 6
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D
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Mixture A had 15% cement by weight [#permalink] New post   05 Oct 2021, 03:07
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koala vishnukavili

The stem is convoluted hence we find it hard.

Let's say mixture \(A = 60kg\) & \(B = 100kg\)

Cement is \(15\)% by weight in mixture \(A\) so cement \(= 9kg\)

Read the big words:
One-fourth of the weight of mixture A was then taken from mixture B, which was 40% cement by weight and added to mixture A.

So, \(15kg\) (\(\frac{1}{4}\)th of \(A\)) is taken from mixture \(B\). So \(15kg\) of mixture \(B\) will have \(6kg\) cement.

Total weight of \(A = 60 + 15 = 75kg\)

in A, the weight of cement \(= 9 + 6 = 15kg\)

New %age = \(\frac{15}{75}*100\)% \(= 20\)%

%age increase \(= 5\)%

Answer D
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Re: Mixture A had 15% cement by weight [#permalink] New post   30 Sep 2021, 05:13
One-fourth of the weight of mixture A was then taken from mixture B, which was 40% cement by weight and added to mixture A
This line is very confusing. Can someone please provide the solution to this?
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Re: Mixture A had 15% cement by weight [#permalink] New post   04 Oct 2021, 03:32
This is what I understood from the question:

Let the total weight of Mixture A be = 200.

Given, "A mixture A had 15% cement by weight" => 15/100(200) => 30. -> (equation 1)

We were given, "1/4th of the weight of the Mixture A is from Mixture B"

Which can be, Mixture B = 1/4(200) => 50 -> (equation 2), which is the weight of Mixture B that is given to Mixture A.

Now, also given: "1/4th of the weight of the Mixture A is from Mixture B(equation 2), which was 40% cement by weight and added to mixture A".

So, from the above statement, we can get the exact weight of cement. Which is => 40% of (equation 2) => 40/100(50) => 20. -> (equation 3)

Now, The percent increase from the previous cement(equation 1) to (equation 3).

Which is, ( (30 - 20)/20 ) x 100 => ( 10/20 ) x 100 => 0.5 x 100 => 50. I ended up here!

Don't know what to do next! Any help is much appreciated

Thanks in advance
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Re: Mixture A had 15% cement by weight [#permalink] New post   05 Oct 2021, 01:26
vishnukavili wrote:
This is what I understood from the question:

Let the total weight of Mixture A be = 200.

Given, "A mixture A had 15% cement by weight" => 15/100(200) => 30. -> (equation 1)

We were given, "1/4th of the weight of the Mixture A is from Mixture B"

Which can be, Mixture B = 1/4(200) => 50 -> (equation 2), which is the weight of Mixture B that is given to Mixture A.

Now, also given: "1/4th of the weight of the Mixture A is from Mixture B(equation 2), which was 40% cement by weight and added to mixture A".

So, from the above statement, we can get the exact weight of cement. Which is => 40% of (equation 2) => 40/100(50) => 20. -> (equation 3)

Now, The percent increase from the previous cement(equation 1) to (equation 3).

Which is, ( (30 - 20)/20 ) x 100 => ( 10/20 ) x 100 => 0.5 x 100 => 50. I ended up here!

Don't know what to do next! Any help is much appreciated

Thanks in advance


But according to your solution there is no increase rather it's a decrease in your answer.
i.e. From 30 you got 20
So there's a 33% decrease
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Re: Mixture A had 15% cement by weight [#permalink] New post   05 Oct 2021, 01:47
koala wrote:
vishnukavili wrote:
This is what I understood from the question:

Let the total weight of Mixture A be = 200.

Given, "A mixture A had 15% cement by weight" => 15/100(200) => 30. -> (equation 1)

We were given, "1/4th of the weight of the Mixture A is from Mixture B"

Which can be, Mixture B = 1/4(200) => 50 -> (equation 2), which is the weight of Mixture B that is given to Mixture A.

Now, also given: "1/4th of the weight of the Mixture A is from Mixture B(equation 2), which was 40% cement by weight and added to mixture A".

So, from the above statement, we can get the exact weight of cement. Which is => 40% of (equation 2) => 40/100(50) => 20. -> (equation 3)

Now, The percent increase from the previous cement(equation 1) to (equation 3).

Which is, ( (30 - 20)/20 ) x 100 => ( 10/20 ) x 100 => 0.5 x 100 => 50. I ended up here!

Don't know what to do next! Any help is much appreciated

Thanks in advance


But according to your solution there is no increase rather it's a decrease in your answer.
i.e. From 30 you got 20
So there's a 33% decrease



Yes yess..😅 i'm actually stuck.. is it just me or the question is really tricky ?

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Re: Mixture A had 15% cement by weight [#permalink] New post   05 Oct 2021, 04:06
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rx10 wrote:
koala vishnukavili

The stem is convoluted hence we find it hard.

Let's say mixture \(A = 60kg\) & \(B = 100kg\)

Cement is \(15\)% by weight in mixture \(A\) so cement \(= 9kg\)

Read the big words:
One-fourth of the weight of mixture A was then taken from mixture B, which was 40% cement by weight and added to mixture A.

So, \(15kg\) (\(\frac{1}{4}\)th of \(A\)) is taken from mixture \(B\). So \(15kg\) of mixture \(B\) will have \(6kg\) cement.

Total weight of \(A = 60 + 15 = 75kg\)

in A, the weight of cement \(= 9 + 6 = 15kg\)

New %age = \(\frac{15}{75}*100\)% \(= 20\)%

%age increase \(= 5\)%

Answer D

Great explanation . However, the stem is a little bit odd
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Re: Mixture A had 15% cement by weight [#permalink] New post   05 Oct 2021, 04:15
Thank you rx10! Completely clear.

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