For positive integers a, b, and c, a

Question

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.

tapabrata · Accepted Answer

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

KrishnakumarKA1 · Answer

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

Sumo23 · Answer

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

duahsolo · Answer

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. 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

SeregaP · Answer

picking smart numbers here is great: try 25, 50, 75 and be careful with calculations

D is the answer

PMZ21 · 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!

MaverickTone · Answer

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.

PMZ21 · Answer

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

Bunuel · Answer

Here is a  POST  by Magoosh which discusses this issue in detail.

Hope it helps.

amanvermagmat · Answer

It depends on whether the fraction a/b is a proper fraction (ab) 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)

shashankism · Answer

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..

singhabhijeet · Answer

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

ScottTargetTestPrep · Answer

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

Adit_ · Answer

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

For positive integers a, b, and c, a < b < c < 100. Which of the follo

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