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Expert replies
Source: — Problem Solving |

by perwinsharma » Mon Jul 28, 2014 4:12 am
So we're trying to find ALL consumer installment credit, let's call that Q.

So at the end of the year we know that 36% of Q comes from automobile installment credit. So we need to figure out the number for automobile installment credit, and then we can solve for Q. We're told that 57 billion is 1/3 of the automobile installment credit amount. So the full amount of automobile installment credit (in billions) is: 57*3 (since 1/3 of the amount is 57 billion, we multiply by 3 to find the full amount. This gives 171 (billion).

So going back to the stem we know that 36% of outstanding consumer credit (which we are calling Q) comes from automobile credit. So 171 is 36% of what number?

.36*Q=171

Q=171/.36

Q=475
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by GMATinsight » Mon Jul 28, 2014 5:15 am
abhasjha wrote:Image
Automobile Installment Credit = (36/100) x all Outstanding Credit

Given $57 Billion = (1/3) x Automobile Installment Credit

i.e.$57 Billion = (1/3) x (36/100) x all Outstanding Credit

i.e. all Outstanding Credit = (100/36) x (3/1) x 57 = (25/3) x 57 = $475 Billion

Answer: Option C
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by GMATGuruNY » Mon Jul 28, 2014 11:27 am
At the end of year X, automobile installment credit accounted for 36% of all outstanding consumer installment credit. At that time automobile finance companies extended $57 billion of credit, or 1/3 of the automobile installment credit. How many billion dollars of consumer installment credit was outstanding at that time?

A. 62
B. 171
C. 475
D. 513
E. 684
An alternate approach is to convert the percentages to RATIOS.

Let A = automobile installment credit, C = all outstanding consumer installment credit, and E = extended credit.

Automobile installment credit accounted for 36% of all outstanding consumer installment credit.
Thus:
A:C = 36:100 = 9:25.

Finance companies extended 1/3 of the automobile installment credit.
In other words, $1 of credit was extended for every $3 of automobile installment credit.
Thus:
E:A = 1:3 = 3:9.

Combining the two ratios, we get:
E:A:C = 3:9:25
E:C = 3:25.

Since the actual value of E = 57, and 3*19 = 57, the multiplier for the ratio above is 19:
E:C = (3*19) : (25*19) = 57:475.
Thus, C = 475.

The correct answer is C.
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by Brent@GMATPrepNow » Mon Jul 28, 2014 2:49 pm
At the end of year X, automobile installment credit accounted for 36% of all outstanding consumer installment credit. At that time automobile finance companies extended $57 billion of credit, or 1/3 of the automobile installment credit. How many billion dollars of consumer installment credit was outstanding at that time?

A. 62
B. 171
C. 475
D. 513
E. 684
This questions illustrates the importance of ALWAYS checking the answer choices BEFORE getting bogged down with time-consuming calculations. For more on this important strategy, you can watch our free video: https://www.gmatprepnow.com/module/gener ... es?id=1111

IMPORTANT: Notice that the answer choices are reasonably spread apart. This means we may be able to use some ESTIMATION to find the answer quickly.

$57 billion of credit equals 1/3 of the automobile installment credit
So, the automobile installment credit = (3)($57 billion) = $171 billion

Automobile installment credit accounted for 36% of all outstanding consumer installment credit
IMPORTANT ESTIMATION CONCEPTS:
If x is 25% of y, then y = 4x
If x is 33 1/3% of y, then y = 3x
If x is 50% of y, then y = 2x
Since 36% is a little more than 33 1/3% we can conclude the following: if x is 36% of y, then y = a little less than 3x


The question tells us that automobile installment credit (worth $171 billion) accounts 36% of all outstanding consumer installment credit.
In other words, $171 billion is 36% of all outstanding consumer installment credit.
So, all outstanding consumer installment credit = little less than 3($171)
3(171) = 513, so the correct answer is a little less than $513 billion.
Answer choice C is a little less than $513 billion, so it must be the correct answer.

Cheers,
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by phanikpk » Mon Jul 28, 2014 4:19 pm
1/3 of AIC= 57 Mn
AIC= 57*3= 171 Mn
AIC= 0.36 of CIC

So, CIC= AIC*100/36= 171*100/36= 171*25/9= 25*19= 475
Option C
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by theCodeToGMAT » Mon Jul 28, 2014 8:17 pm
total = A
end of X = 0.36 * A
57 = 1/3 * 0.36 * A
57/0.12 = A
5700/12 = A
just below 500

So, {C}
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