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The weight of every type A widget is the same, the weight of 04 Jan 2010, 12:12
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35% (medium)

Question Stats:

78% (02:41) correct 22% (03:05) wrong based on 259 sessions

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The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

A. 12/13
B. 21/40
C. 2/3
D. 77/96
E. 10/7

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8A:3B 5B:7C = 40A:15B 15B:21C Therefore (40 +15)/(15+21).....clearly not the correct answer, but I'm not sure why.
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The weight of every type A widget is the same, the weight of 04 Jan 2010, 15:33
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Here's an inexact approach:

8A = 3B and 5B = 7C

Therefore, 40A = 15B = 21C.

The term 40A, for example, means that we have 40 widgets weighing A pounds each. Right now we only have the number of units that provide for an equal weight of all three widget types. We must solve for a weight before we can continue.

40A = 15B
B = (40/15)A
B = 2.75A

If A = 1 then B = 2.75. Now we just have to find C.

40A = 21C
C = (40/21)A and 40/21 is about equal to 2. So let's say that C = 2

We now have:
(1+2.75)/(2+2.75) = 3.75/4.75

This should be a hair less than 4/5. Looking through the answers, the only choice close to this is D.) 77/96
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The weight of every type A widget is the same, the weight of 04 Jan 2010, 16:40
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40A = 15B = 21C.....
now convert each in value of a common constant say A...
SO B=40/15=8A/3 , C=40A/21....
A+B/C+B=(40A/15+A)/(40A/15+40A/21)=(55A/15)/(480A/105)=(55A*105)/(480A*15)=77/96
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The weight of every type A widget is the same, the weight of 05 Jan 2010, 06:28
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As it is often said don't do more than needed. So in this case as well we don't need to find the complex relations between A, B & C. The question gives us the relation between A & B, B & C and it asks us the ratio for A+B and B+C.
So my approach is:

8A = 3B => B = 8/3 A --- (1)
Hence A+B = A + 8/3 A = 11/3 A.

Similarly, 5B=7C => C=5/7 B.
Hence B+C = B + 5/7 B = 12/7 B

Finally, A+B : B+C = (11/3 A) / (12/7 B)
= 77/36 * A/B
= 77/36 * 3/8 From (1)
= 77/96. Ans (d)
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The weight of every type A widget is the same, the weight of 05 Jan 2010, 12:31
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Thanks for all the responses. I'm failing to understand why we need to convert further i.e why does the 77/36 need to be multiplied?

Thanks again for the help.
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The weight of every type A widget is the same, the weight of 05 Jan 2010, 13:18
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You can use this method, if this is clear:

8A = 3B -> A = 3/8 B
A + B = 3/8 B + B = 11/8 B

5B = 7C -> C = 5/7 B
B + C = B + 5/7B = 12/7 B

Therefore:
\(\frac{A+B}{B+C} = \frac{11/8 B}{12/7 B} = \frac{11*7}{12*8} = \frac{77}{96}\)

Hope its clear..

Cheers!
JT
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The weight of every type A widget is the same, the weight of 05 Jan 2010, 19:55
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Question gives A/B=3/8
C/B=5/7

LCM of 8 & 7 = 56
A/B=3*7/8*7
C/B=5*8/7*8

A+B/B+C=21+56/40+56
=77/96.
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The weight of every type A widget is the same, the weight of 10 Jun 2013, 22:56
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thanks
The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

a) 12/13
b) 21/40
c) 2/3
d)77/96
e)10/7


8A:3B 5B:7C = 40A:15B 15B:21C Therefore (40 +15)/(15+21).....clearly not the correct answer, but I'm not sure why.

Solution: Bunuel shed light if I am wrong................

weight of 1 type A widget lets say Wa so for B is Wb and for c is Wc

Given that 8Wa=3Wb and 5Wb=7Wc

not we can get from the above Wa/Wb=3/8 and Wb/Wc=7/8 no equate both we got Wa/Wb=21/56 and Wb/Wc=56/40

question asked us Wa+Wb/Wb+Wc= 21+56/56+40=77/96 ans....


Rgds
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The weight of every type A widget is the same, the weight of 11 Jun 2013, 01:24
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prasannajeet
thanks
The weight of every type A widget is the same, the weight of every type B widget is the same, and the weight of every type C widget is the same. If the weight of 8 type A widgets is equal to the weight of 3 type B widgets, and the weight of 5 type B widgets is equal to the weight of 7 type C widgets. What is the ratio of the total weight of 1 type A widget and 1 type B widget, to the total weight of 1 type B widget and 1 type C widget?

a) 12/13
b) 21/40
c) 2/3
d)77/96
e)10/7


8A:3B 5B:7C = 40A:15B 15B:21C Therefore (40 +15)/(15+21).....clearly not the correct answer, but I'm not sure why.

Solution: Bunuel shed light if I am wrong................

weight of 1 type A widget lets say Wa so for B is Wb and for c is Wc

Given that 8Wa=3Wb and 5Wb=7Wc

not we can get from the above Wa/Wb=3/8 and Wb/Wc=7/8 no equate both we got Wa/Wb=21/56 and Wb/Wc=56/40

question asked us Wa+Wb/Wb+Wc= 21+56/56+40=77/96 ans....


Rgds
Prasannajeet

Your solution is correct.

A:B = 3:8 = 21:56
B:C = 7:5 = 56:40

(A + B):(B + C) = (21 + 56):(56 + 40) = 77:96

Answer: D.
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The weight of every type A widget is the same, the weight of 08 Aug 2014, 03:49
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We are looking for: \(\frac{A+B}{B+C}\)

The question gives us:
\(3B = 8A\) --> \(\frac{3}{8}B = A\)
and
\(5B = 7C\) --> \(\frac{5}{7}B = C\)

Now set in the equation we need to solve:

\(\frac{\frac{3}{8}B + B}{\frac{5}{7}B + B} = \frac{\frac{11}{8}B}{\frac{12}{7}B} = \frac{11B*7B}{8B*12B} = \frac{77}{96}\)

Answer is D.
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The weight of every type A widget is the same, the weight of 13 Aug 2014, 03:21
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\(\frac{a}{b} = \frac{3}{8}\)

Applying componendo / dividendo

\(\frac{a+b}{b} = \frac{3+8}{8} = \frac{11}{8}\) .............. (1)

\(\frac{c}{b} = \frac{5}{7}\)

Applying componendo / dividendo

\(\frac{c+b}{b} = \frac{5+7}{7} = \frac{12}{7}\) .................... (2)

Divide (1) by (2)

\(\frac{\frac{a+b}{b}}{\frac{c+b}{b}} = \frac{\frac{11}{8}}{\frac{12}{7}} = \frac{11*7}{12*8} = \frac{77}{96}\)
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The weight of every type A widget is the same, the weight of 21 Jan 2021, 18:16
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Not a difficult problem, but a basic math mistake costs precious time...this GMAT is so unforgiving.

Start by expressing everything in terms of one common variable

A = 3B /8 and B = 7C/5

Here's the question: A + B / B + C = ?

Substitute...

[3B / 8 + 7C / 5] / 7C/5 + C
(3(7C + 5)/8) + (7C / 5) / 7C/5 + C
= 77/96

Answer is D.
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