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For positive integers a, b, and c, a < b < c < 100. Which of the following has the greatest value?
A. a/100
B. (a+b)/(100+b)
C. (a+c)/(100+c)
D. (a+b+c)/(100+b+c)
E. The answer cannot be determined from the information provided.
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A. a/100
B. (a+b)/(100+b)
C. (a+c)/(100+c)
D. (a+b+c)/(100+b+c)
E. The answer cannot be determined from the information provided.
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For any proper fraction \(\frac{x}{y}\) (x<y and both x and y are positive)
\(\frac{x}{y}\)<\(\frac{x+1}{y+1}\)<\(\frac{x+2}{y+2}\)<...<\(\frac{x+n}{y+n}\)
where n is a positive integer. The higher the value of n, the more greater \(\frac{x+n}{y+n}\) is than the original fraction.
Example
\(\frac{x}{y}\)<\(\frac{x+5}{y+5}\)<\(\frac{x+50}{y+50}\)
Now we need to arrange
\(\frac{a}{100}\), \(\frac{a+b}{100+b}\), \(\frac{a+c}{100+c}\) and \(\frac{a+(b+c)}{100+(b+c)}\)
Since 0<b<c<b+c then the order would be
\(\frac{a}{100}\)<\(\frac{a+b}{100+b}\)<\(\frac{a+c}{100+c}\)<\(\frac{a+(b+c)}{100+(b+c)}\)
D has the greatest value. Hence Answer D
\(\frac{x}{y}\)<\(\frac{x+1}{y+1}\)<\(\frac{x+2}{y+2}\)<...<\(\frac{x+n}{y+n}\)
where n is a positive integer. The higher the value of n, the more greater \(\frac{x+n}{y+n}\) is than the original fraction.
Example
\(\frac{x}{y}\)<\(\frac{x+5}{y+5}\)<\(\frac{x+50}{y+50}\)
Now we need to arrange
\(\frac{a}{100}\), \(\frac{a+b}{100+b}\), \(\frac{a+c}{100+c}\) and \(\frac{a+(b+c)}{100+(b+c)}\)
Since 0<b<c<b+c then the order would be
\(\frac{a}{100}\)<\(\frac{a+b}{100+b}\)<\(\frac{a+c}{100+c}\)<\(\frac{a+(b+c)}{100+(b+c)}\)
D has the greatest value. Hence Answer D
General Discussion
Posts: 398
SHORTCUT:let us assume a, b, c be 25, 50, 75 respectively
A. 25/100 =0.25
B. 75/150 = 0.5
C. 100/175 = 0.57
D. 150/225 = 0.66
Option D
A. 25/100 =0.25
B. 75/150 = 0.5
C. 100/175 = 0.57
D. 150/225 = 0.66
Option D
Agree... D... also it allows all the variables to feature in the option... which is necessary to ascertain the answer in this case
Sent from my GT-I9060I using GMAT Club Forum mobile app
Sent from my GT-I9060I using GMAT Club Forum mobile app
Bunuel
a, b, c = Pos INTs
a < b < c < 100
Let's test numbers, Let try a =1, b =2, & b =3;
so we have
A. a/100 -------------------> 1/100
B. (a+b)/(100+b) ---------> 3/102
C. (a+c)/(100+c) ---------> 4/103
D. (a+b+c)/(100+b+c) ---> 5/105
E. The answer cannot be determined from the information provided.[/quote]
Let's try another integer values: a =97, b =98, c =99
A) 97/100 B) 195/198 C) 196/199 D) 294/297
Since option D has the highest numerator and highest denominator, and the denominator of each option is greater than the numerator by the same value, option D has the greatest value.
Now, assuming at this stage you need to compare each option against the other and you have to deal with big values, here's a shortcut i just found (hope, I'm right
):For instance, let's check which is bigger, A or B: A) 97/100 vs. B) 195/198 ----> A) 198*97 vs B) 195*100,
just compare the positive differences of the multiplying figures and the option with the lower difference is likely to be bigger.
That is, since 195 - 100 = 97 is less than 199 - 97 = 102, option B is bigger (check!)
For B & C: B) 195/198 vs C) 196/199 ----> B) 199*195 vs. C) 198*196 ----> 199 - 195 = 4 > 198 - 196 = 2, option C is bigger
For C & D: C) 196/199 vs. D) 294/297 ----> C) 297*196 vs. 294*199 ----> 297 - 196 = 201 > 294 - 199 = 195, option D is bigger
Answer: D
picking smart numbers here is great: try 25, 50, 75 and be careful with calculations
D is the answer
D is the answer
Hi Bunuel,
Thanks for this question!
Although I did get the correct answer ( I used two examples of a, b and c to verify: 1,2,3 and 25,50, 75). And since option D was higher in both the cases, I assumed D would be correct.
However, how can we be so sure that E isn't the correct option? I wanted to understand the question better so that if a similar question comes up during the exam - I can be more certain of why I am eliminating one of the options.
Thanks in advance!
Thanks for this question!
Although I did get the correct answer ( I used two examples of a, b and c to verify: 1,2,3 and 25,50, 75). And since option D was higher in both the cases, I assumed D would be correct.
However, how can we be so sure that E isn't the correct option? I wanted to understand the question better so that if a similar question comes up during the exam - I can be more certain of why I am eliminating one of the options.
Thanks in advance!
Posts: 8
poojamathur21
I think I know the answer.
1/n < 2/(n+1) < 3/(n+2) etc. with n >1
It could be said in a logic form but I don't know how to put it. Still waiting for Bunuel's explanation.
MaverickTone
Yup, you're right. I think I got it. Bunuel, please confirm.
If we have a fraction say a / b, any number (x>0) when added to both the numerator and denominator makes the fraction bigger, and when subtracted makes the fraction smaller.
=> a-x/b-x < a/b < a+x/b+x
Posted from my mobile device
poojamathur21
Here is a POST by Magoosh which discusses this issue in detail.
Hope it helps.
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poojamathur21
It depends on whether the fraction a/b is a proper fraction (a<b) or its an improper fraction (a>b)
If a/b is a proper fraction, then adding any positive number x (x>0) to both numerator and denominator Increases the value of the fraction
OR a/b < (a+x)/(b+x)
But if c/d is say an improper fraction (c>d), then adding any positive number x (x>0) to both numerator and denominator Decreases the value of the fraction.
OR c/d > (c+x)/(d+x)
So lets consider a proper fraction a/100 (a<100). If we add a positive number b to both numerator and denominator, its value will increase
Thus a/100 < (a+b)/(100+b)
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Bunuel
since b > a
so, a/100 < (a+b)/(100+b)
since c > b
so, (a+b)/(100+b) < (a+c)/(100+c)
since b+c > c
so, (a+c)/(100+c) < (a+b+c)/(100+b+c)
So , a/100 < (a+b)/(100+b) < (a+c)/(100+c) < (a+b+c)/(100+b+c)
Answer D..
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Posts: 49
Bunuel
Logical Thinking
I took 3 numbers < 100 for a,b and c such as 97,98 and 99.
I use a numerator and denominator rule (mentioned in Manhattan guide )to solve this question without calculation. To be frank I couldn't think of solid solution.
Rule : Adding the same number to both the numerator and denominator brings the fraction close to 1, regardless of the fraction value.
since, a<b<c< 100. The fraction < 1.
Now scanning through the answer options, A to D are combination of a, b and c.
Using the above rule I know that D is closed to 1. Hence D option is largest among A to D. Therefor marked D.
To understand this approach please refer to Manhattan guide
Ans: D
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Bunuel
Solution:
We can let a = 1, b = 2 and c = 3, we see that the value of the answer choices would be:
A) 1/100
B) 3/102
C) 4/103
D) 6/105
Since the denominators are really about the same, we see that choice D has the largest value. (Notice that if all the denominators are 100, their values would be 0.01, 0.03, 0.04, and 0.06, respectively.)
Now if we let a = 97, b = 98 and c = 99, the value of the answer choices would be:
A) 97/100
B) 195/198
C) 196/199
D) 294/297
Although each answer choice has a value approximately equal to 1, we see that each numerator is 3 less than its denominator. In that case, the fraction with the largest denominator would be the one that has the largest value. So again, choice D has the largest value.
Answer: D
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when difference between numerator and denominator is same, the larger denominator usually has the higher value.
Like 5/6<6/7<7/8<8/9
Here clearly the same number is being added to numerator and denominator and thus the largest denominator wins which is option D
Like 5/6<6/7<7/8<8/9
Here clearly the same number is being added to numerator and denominator and thus the largest denominator wins which is option D
Bunuel
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