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In the formula above, a, b, c, and d are positive numbers.

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carcass
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In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   Updated on: 12 Sep 2013, 09:37
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x = 8(a/c) + 16(b/c)

In the formula above, a, b, c, and d are positive numbers. If c = a + b and b > a, which of the following could be the value of x?

(A) 8
(B) 10
(C) 12
(D) 15
(E) 21
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D

Originally posted by carcass on 12 Sep 2013, 09:32.
Last edited by Bunuel on 12 Sep 2013, 09:37, edited 1 time in total.
Moved to PS forum.
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Bunuel
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   12 Sep 2013, 09:44
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carcass wrote:
x = 8(a/c) + 16(b/c)

In the formula above, a, b, c, and d are positive numbers. If c = a + b and b > a, which of the following could be the value of x?

(A) 8
(B) 10
(C) 12
(D) 15
(E) 21


\(\frac{8a}{c} + \frac{16b}{c}=\frac{8a+16b}{c}=\frac{8a+16b}{a+b}\).

You can notice that it's a weighted average formula for \(a+b\) items, where the weight of each of the items of \(a\) is 8 and the weight of each of the items of \(b\) is 16. Now, since b > a, then the average must be closer to 16, then to 8. Only D fits.

Answer: D.
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   08 Sep 2014, 03:50
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carcass wrote:
x = 8(a/c) + 16(b/c)

In the formula above, a, b, c, and d are positive numbers. If c = a + b and b > a, which of the following could be the value of x?

(A) 8
(B) 10
(C) 12
(D) 15
(E) 21


x = 8a/c + 16b/c
x= 8(a+b)/c + 8b/c
now substituting c = a + b in the above equation
x = 8 + 8b/(a+b)
The ratio b/(a+b) is smaller than 1 thus answer choice E is eliminated... and as a < b ... thus the ratio b/(a+b) is larger than 1/2 or 0.5... thus
8*b/(a+b) is larger than 4... thus the only answer choice left is D
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Bunuel
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   12 Sep 2013, 09:45
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Bunuel wrote:
carcass wrote:
x = 8(a/c) + 16(b/c)

In the formula above, a, b, c, and d are positive numbers. If c = a + b and b > a, which of the following could be the value of x?

(A) 8
(B) 10
(C) 12
(D) 15
(E) 21


\(\frac{8a}{c} + \frac{16b}{c}=\frac{8a+16b}{c}=\frac{8a+16b}{a+b}\).

You can notice that it's a weighted average formula for \(a+b\) items, where the weight of each of the items of \(a\) is 8 and the weight of each of the items of \(b\) is 16. Now, since b > a, then the average must be closer to 16, then to 8. Only D fits.

Answer: D.


Similar question to practice from OG:
Quote:
If x, y, and k are positive numbers such that (x/(x+y))(10) + (y/(x+y))(20) = k and if x < y, which of the following could be the value of k?

A. 10
B. 12
C. 15
D. 18
E. 30


Discussed here: if-x-y-and-k-are-positive-numbers-such-that-x-x-y-128231.html
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In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   15 Sep 2015, 13:30
x = 8(a/c) + 16(b/c)

just by looking at the options .. we should be able to understand that 8 or its multiple i.e 16 should be divisible by C..why ? Because none of the options are in p/q form.

let us say c =4 which gives us a=1 and b=3

the equation thus becomes x=8(1/4) + 16(3/4) =2+12=14 --close enough :)
the other value c can take is 8
which gives us possible values for (a,b) as (1,7) (2,6) (3,5)

take 1,7

x=8(1/8)+16(7/8)=15

That's how I managed to do.
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   11 Jan 2016, 19:17
oh man..I did the long way..
8a+8b/a+b = 8
now, 8b/a+b =??
we can eliminate A right away..
tried several numbers..then got to:
a=1
b=7
a+b=8
8b=56.
56/8=7.
7+8=15.

D
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   27 May 2019, 14:45
CounterSniper wrote:
x = 8(a/c) + 16(b/c)

just by looking at the options .. we should be able to understand that 8 or its multiple i.e 16 should be divisible by C..why ? Because none of the options are in p/q form.

let us say c =4 which gives us a=1 and b=3

the equation thus becomes x=8(1/4) + 16(3/4) =2+12=14 --close enough :)
the other value c can take is 8
which gives us possible values for (a,b) as (1,7) (2,6) (3,5)

take 1,7

x=8(1/8)+16(7/8)=15

That's how I managed to do.


I did exactly the same way! Can someone clarify if this method is OK to use for future such questions? Feels like guess work and lucked out with this one.
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   28 May 2019, 18:06
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carcass wrote:
x = 8(a/c) + 16(b/c)

In the formula above, a, b, c, and d are positive numbers. If c = a + b and b > a, which of the following could be the value of x?

(A) 8
(B) 10
(C) 12
(D) 15
(E) 21



Simplifying, we have:

x = (8a + 16b)/c

x = (8a + 16b)/(a + b)

We see that we have a weighted average, and since b is greater than a, then x must be closer to 16 and also between 8 and 16, so x could be 15.

Answer: D
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   15 May 2021, 10:28
x = 8(a/c) + 16(b/c)

or X = \(\frac{(8a +16b)}{(a+b)}\)

Since b>a so, X will be close to 16.

So, I think D. :)
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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink] New post   31 Aug 2023, 04:58
x = (8a + 16b)/(a + b)
x = 8(a + 2b)/(a + b)

So (a+b) must be 8, by putting that value ans is 15, where a=1, b=7

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Re: In the formula above, a, b, c, and d are positive numbers. [#permalink]
31 Aug 2023, 04:58
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