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If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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22 Nov 2013, 09:19
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If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true? I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\) (A) I only (B) I and II (C) II and III (D) II only (E) I, II and III I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks.
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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22 Nov 2013, 13:39
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BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Don't do algebra, just think it out. I) A+B, both of which are less than one. They can't be greater than 2. II) B is clearly larger than A (look at numerators & denominators), so when you add C to it you're almost at 2. A is the smallest number here, think of them as percentages if that helps, and when you subtract that from B you still have some left over. THen add C in, which is almost 1, on its own, you can see it'st rue. III) Remember when you square a fraction you get a smaller number, so you have a smaller number than what those two fractions started out as. Some fairly quick guestimating about what they equal (again, it's easier perhaps to think of them as being percentages) and you can see it's less than 1



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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19 Jan 2014, 11:33
Hi, I used fraction to solve this, but it takes time. I want to know whether we can solve it using other than Fraction.
Last edited by goodyear2013 on 21 Jan 2014, 09:42, edited 1 time in total.



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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19 Jan 2014, 16:46
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BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Hi, To solve this the best approach will be  write the numbers in other forms either in fraction or some other forms.(Fraction will take loger time) I tried this  \(a = \frac{28}{31}\) It can be written as, \(a = \frac{(313)}{31}\) \(a = 1 \frac{3}{31}\) similar B and C can be written as \(b =1\frac{1}{30}\) and \(c =1\frac{2}{33}\). Now 1) a+b >2 13/31+11/30 = 23/311/30 (we are trying to subtract something from 2 and it can not be greater than 2) So Not True.2) b + c  a > 1 11/30 +1  2/33  1+ 3/31 = 2 + 3/31  1/30  2/33.( which will be always greater than 1) So True. 3) a+ b^2  c^2 > 1 a+(b+c)(bc) a+ (2 1/302/33)(2/331/30) (Here you need to calculate 2/33 == 0.0606 or 1/30 = 0.0333) 13/31 + (2 1/302/33)(2/331/30) after + sign term is less than 3/31. So you can say it whole is less than 1.
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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20 Jan 2014, 11:01
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bhatiamanu05 wrote: BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Hi, To solve this the best approach will be  write the numbers in other forms either in fraction or some other forms.(Fraction will take loger time) I tried this  \(a = \frac{28}{31}\) It can be written as, \(a = \frac{(313)}{31}\) \(a = 1 \frac{3}{31}\) similar B and C can be written as \(b =1\frac{1}{30}\) and \(c =1\frac{2}{33}\). Now 1) a+b >2 13/31+11/30 = 23/311/30 (we are trying to subtract something from 2 and it can not be greater than 2) So Not True.2) b + c  a > 1 11/30 +1  2/33  1+ 3/31 = 2 + 3/31  1/30  2/33.( which will be always greater than 1) So True.
3) a+ b^2  c^2 > 1 a+(b+c)(bc) a+ (2 1/302/33)(2/331/30) (Here you need to calculate 2/33 == 0.0606 or 1/30 = 0.0333) 13/31 + (2 1/302/33)(2/331/30) after + sign term is less than 3/31. So you can say it whole is less than 1. You have a mistake in the red marked equation. \(... = 1 + 3/31  1/30  2/33\) what makes it less clear that it is always greater than 1. Though I would du an educated guess here... 1. is quite easy to assess. 2. and 3. are quite complicated. But maybe this might help you: when x < y:\(\frac{x}{y} < \frac{x+1}{y+1} < \frac{x+2}{y+2}\) What you also could do is, ask yourself which one is the biggest fraction. Therefore write the fractions next to each other and do a little thinking. \(\frac{28}{31}  \frac{29}{30}  \frac{31}{33}\) To check which is bigger, multiply the numerator of the first with the denominator of the second and vice verse. The fraction of the numerator which "creates" the bigger value is the bigger fraction. \(28*30 < 29*31\) and \(29*31 < 30*33\) thereforce \(a < b < c\)
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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04 Apr 2014, 22:16
Option \(D\). Option \(A,B\) and \(E\) can be eliminated by simple observation:In the first statement \(a<1\) and \(b<1\) so their sum cannot be greater than \(2\). Now we don't even have to consider statement II since it's in both \(C\) and \(D\). As for third statement, \(b<1\) So \(b^2<b\)
And \(b>c\) (we can find this out by making the denominators in \(b\) and \(c=330\)) When difference of their squares is taken,it'll be so less than even \(bc\) that after we add \(a\) it won't be greater than \(1\). Consider:\(b^2=(319/330)^2\) and \(c^2=(310/330)^2\) \(b^2c^2=(9*629)/(330)^2\) which is very very small. [ \(a\) is approx \(0.94\).So we need \(0.06\) to be added to \(a\) to make it equal to \(1\).]



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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12 Jun 2015, 23:40
BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Fast Approach (1.16 min): I. <1 + <1 = <2 hence eliminate As we eliminated I, we eliminate choices A, B, and E. So we only have to decide between C and D. C: II and III D: II only. Hence, we should only check III. As if III is true, then we pick C. If III is false, then we automatically are left with D. Check III: By simple comparison of fractions, you can immediately infer that a < c < b. Estimate: Assume a = 0.97, c = 0.98 , b= 0.99 III: 0.97 + (0.99)^2  (0.98)^2 < 1? even if you do not square 0.99, and 0.98, 0.97+(0.990.98) = 0.97 + (.01) < 1. Hence, squaring them would make the number even smaller. So FALSE. Hence, we are left with choice D.



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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francoimps wrote: BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Fast Approach (1.16 min): I. <1 + <1 = <2 hence eliminate As we eliminated I, we eliminate choices A, B, and E. So we only have to decide between C and D. C: II and III D: II only. Hence, we should only check III. As if III is true, then we pick C. If III is false, then we automatically are left with D. Check III: By simple comparison of fractions, you can immediately infer that a < c < b. Estimate: Assume a = 0.97, c = 0.98 , b= 0.99 III: 0.97 + (0.99)^2  (0.98)^2 < 1? even if you do not square 0.99, and 0.98, 0.97+(0.990.98) = 0.97 + (.01) < 1. Hence, squaring them would make the number even smaller. So FALSE. Hence, we are left with choice D.Hi Francoimps, Almost everything you have done is correct except one step (Highlighted below) which I am skeptical about being correct all the time Assume a = 0.97, c = 0.98 , b= 0.99Also one step that you have written By simple comparison of fractions, you can immediately infer that a < c < b. Estimate:In my opinion this SIMPLE COMPARISON OF FRACTION is not so simple for many and perhaps this consumes some time of the test takers So I am giving a little clarity here on comparison of such fractions Given: \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\) Method 1: Cross multiplicationComparing \(a = \frac{28}{31}\) by writing it on left side and \(b =\frac{29}{30}\) by writing it on Right side Cross multiply by multiplying the denominators of fractions into the numerator of other fraction to be compared with i.e. writing on left side \(28*30\) and writing on Right side \(29*31\) Left side is smaller than right side i.e. a (written on left) will be smaller than b (written on right side) Similarly compare b with c and know that a < c < b Method 2: Algebraic Understandinga is smaller than b is very simply visible as the numerator and denominator both as equally less by 1 than numerator and denominator of b Rule: if u/v<1 and x>0, then u/v < (u+x)/(v+x) Compare \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\) From b to c, Numerator increased by 2 from 29 to 31, i.e. less than 10% increase From b to c, Denominator increased by 3 from 30 to 33, i.e. Greater than 10% increase since numerator has increased by lesser percentage and denominator has increased by greater percentage therefore new fraction (i.e. c) will be smaller than b Compare \(a = \frac{28}{31}\), and \(c =\frac{31}{33}\)From a to c, Numerator increased by 3 from 28 to 31, i.e. Greater than 10% increase From a to c, Denominator increased by 2 from 31 to 33, i.e. Less than 10% increase since numerator has increased by Greater percentage and denominator has increased by Lesser percentage therefore new fraction (i.e. c) will be Greater than a Hence, a < c < b
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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02 Jan 2016, 02:06
BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Let 28 = 1 ; 29 = 2 ; 29 = 3 ; ... ; 33 = 6 => a = 0.25 , b = 0.66 , and c = 0.66 by this only II is true. hence D. Bunuel is this method correct? i guess this would give approximate answers but i assume it can be used.
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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13 Mar 2016, 22:44
Hi,
Here is how I avoided long calculation.
It is easy to see that (I) is false, so I will skip that.
For (II), let's see, we have: \(\frac{29}{30}+\frac{31}{33}\frac{28}{31}\).
Let's rearange a bit to avoid big numbers: \(\frac{31}{33}+(\frac{29}{30}\frac{28}{31})\)
Now, if \((\frac{29}{30}\frac{28}{31})>\frac{2}{33}\), then (II) is true. Otherwise, it's false.
\(\frac{29}{30}\frac{28}{31}=(1\frac{1}{30})(1\frac{3}{31})=\frac{1}{30}+\frac{3}{31}=\frac{1}{30}+(\frac{1}{31}+\frac{2}{31})\)
As the difference between \(\frac{1}{30}\) and \(\frac{1}{31}\) is negligible and very close to zero, let's ignore that. Because \(\frac{2}{31}\) is clearly greater than \(\frac{2}{33}\), (II) is true.
(III) is a bit tricky, but we can employ the same approach.
If \((\frac{29}{30})^2(\frac{31}{33})^2>\frac{3}{31}\), then (III) is true. Otherwise, it's false.
\((\frac{29}{30})^2(\frac{31}{33})^2=(1\frac{1}{30})^2(1\frac{2}{33})^2=\frac{2}{30}+\frac{1}{900}+\frac{4}{33}+\frac{4}{(33^2)}\)
Rearange a bit: \(\frac{2}{30}+\frac{4}{(33^2)}+\frac{1}{900}+\frac{4}{33}\)
The first two fraction are clearly less than \(\frac{3}{31}\) (since they are negative).
Next, \(\frac{1}{900}<\frac{3}{31}\).
\(\frac{4}{33}=\frac{3}{33}+\frac{1}{33}\).
Now, pay attention: \(\frac{3}{33}<\frac{3}{31}\)
AND \(\frac{1}{33}<\frac{2}{30}\). Hence, \(\frac{1}{33}\frac{2}{30}<0\)
So, all in all, \(\frac{2}{30}\frac{4}{(33^2)}+\frac{1}{900}+\frac{4}{33}<\frac{3}{31}\), meaning that (III) is false.



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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13 Mar 2016, 23:47
BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. In this type of problems one should bee able to eliminate choices without any calculation. I is false and requires simple calculation. If i is false we can eliminate A,B and E without looking in to each. Among C and D , II is common. So we do not have to check for II. We need to check only for III
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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05 Dec 2017, 00:01
This is how I did it. a=28/31=.90x (where x is the third number after the decimal  you don't need to find out x or go further) b=29/30=.90x (where x is the third number after the decimal  you don't need to find out x or go further) c=31/33=.92x (where x is the third number after the decimal  you don't need to find out x or go further) Now let's look at the options: I. a+b>2 This is false, so eliminate all options where it says that I is true. We are left with options C and D at this point. Now go for the option is not common in C and D, so we choose to work with III. III. a+b^2c^2>1 From our mathematical operations at the beginning, we have found out that that c>b, so b^2c^2 must yield a negative value. Hence, the result simplifies to: 0.90x+(negative value)>1 This is false. Thus, we are left with option D only. This elimination method is the fastest. Hope it helps. Feel free to reach out in case of any query.
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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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13 Dec 2017, 21:15
BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. Don't go in the trap of making denominators common. Go for Decimal approach. If you are good in maths and can do faster calculation, decimal will give you answer in less than 2.5 minutes. If you are struggling with decimal I will suggest to watch some Vedic Maths videos as it will help you in faster calculation



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If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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15 Dec 2017, 14:11
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Statement 1 cannot be true. Something less than 1 plus something less than 1 = something less than 2.
Statement2 Add a to both sides so we can compare if b+c > 1+a
We can rewrite the two sides in terms of how much less than 2 each is.
\(b+c=21/302/33\) \(1+a=23/31\)
so we can compare rewrite the inequality as \(\frac{1}{30}\frac{2}{33}>? \frac{3}{31}\) add terms to both sides so we can compare \(\frac{3}{31} >? \frac{1}{30}+\frac{2}{33}\) multiply everything by 31 \(3>? 31/30+62/33\) \(3>? 1+2+1/304/33\) \(4/33>? 1/30\) \(4(30) > 33(1\)) Since that statement is clearly true we can conclude statement 2 is true.
Statement 3
\(a+b^2>? 1+c^2\) \((28/31)+(29/30)^2>? 1+(93/99)^2\) \((28/31)+(841/900)>?1.88\) 28/31 is very close to .9 and 855/900 is .95 we can say the left side= .9+.93 and the right side =1.88 The right side is bigger so statement 3 is not true.
The test fails so statement 3 cannot be true.



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Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true? [#permalink]
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26 Dec 2017, 05:31
SravnaTestPrep wrote: BabySmurf wrote: If \(a = \frac{28}{31}\), \(b =\frac{29}{30}\) and \(c =\frac{31}{33}\). Which of the following is true?
I. \(a+b > 2\) II. \(b+ca > 1\) III. \(a + b^2  c^2 > 1\)
(A) I only (B) I and II (C) II and III (D) II only (E) I, II and III
I didn't understand really how to do this in 23 minutes. Is there any quick way to solve this or similar problems? Even creating common denominators for this to figure out II. took me a long time. The only statement I could evaluate quickly was I. as all fractions are smaller than 1 and therefore the sum of a+b is obviously not larger than 2. But I struggled with II. and III. Numerators and denominators are all "around 30". So maybe the trick lies in doing something with that information, or is that on purpose as a distraction? Thanks. In this type of problems one should bee able to eliminate choices without any calculation. I is false and requires simple calculation. If i is false we can eliminate A,B and E without looking in to each. Among C and D , II is common. So we do not have to check for II. We need to check only for III clearly b is greater than c because 29/30 > 31/33 ... so sqB  Sq C is never negative




Re: If a = 28/31, b=29/30 and c=31/33. Which of the following is true?
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