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

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,
\(a = \frac{(31-3)}{31}\)
\(a = 1- \frac{3}{31}\)
similar B and C can be written as
\(b =1-\frac{1}{30}\) and \(c =1-\frac{2}{33}\).

Now 1) a+b >2

1-3/31+1-1/30 = 2-3/31-1/30 (we are trying to subtract something from 2 and it can not be greater than 2) So Not True.

2) b + c - a > 1

1-1/30 +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)(b-c)

a+ (2- 1/30-2/33)(2/33-1/30) (Here you need to calculate 2/33 == 0.0606 or 1/30 = 0.0333)

1-3/31 + (2- 1/30-2/33)(2/33-1/30) after + sign term is less than 3/31.

So you can say it whole is less than 1.

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

Hi, I used fraction to solve this, but it takes time.
I want to know whether we can solve it using other than Fraction.

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 \(a < b < c\)

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 \(b-c\) that after we add \(a\) it won't be greater than \(1\).
Consider:\(b^2=(319/330)^2\) and \(c^2=(310/330)^2\)
\(b^2-c^2=(9*629)/(330)^2\) 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 \(1\).]

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+(0.99-0.98) = 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= 0.99

Also 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 multiplication

Comparing \(a = \frac{28}{31}\) 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 Understanding

a 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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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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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.

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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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^2-c^2>1
From our mathematical operations at the beginning, we have found out that that c>b, so b^2-c^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.

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

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.

\(b+c=2-1/30-2/33\)
\(1+a=2-3/31\)

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
\(3>? 31/30+62/33\)
\(3>? 1+2+1/30-4/33\)
\(4/33>? 1/30\)
\(4(30) > 33(1\))
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.

clearly b is greater than c because 29/30 > 31/33 ... so sqB - Sq C is never negative

Hi

Well alternate approach for this question could be if we know the theory of numerator inc and decrease.

Imagine XY cordinate plane. positive X axis Numerator Increases , positive Y axis denominator increases, negative x axis numerator decreases and negative y axis denominator decreases.

Let Numerator be N and Denominator be D

So in I quadrant Numerator Inc and Denominator Increases
II quadrant Numerator Decreases and Denominator Increases
III quadrant Numerator decreases and Denominator decreases
IV quadrant Numerator Inc and Denominator Decreases.

Now Just remember II and IV quadrants .
II Quadrant if while comparing fractions numerator decreases and Denominator Increases the fraction decreases.
IV Quadrant if while comparing fractions numerator increases and Denominator decreases the fraction increases.
So in I quadrant N Inc by greater % than D then fraction inc , or if D inc by greater % than N fraction dec
in III quad if N decreases by greater % than D then fraction decreases, or if D decreases by greater % than N the fraction inc.


So back to this question lets arrange denominators in increasing order

So \(\frac {29}{30}\) , \(\frac {28}{31}\) , \(\frac {31}{33}\)

Notice in these two fractions \(\frac {29}{30}\) , \(\frac {28}{31}\) numerator is decreasing and denominator is increasing so it must be
that \(\frac {29}{30}\) > \(\frac {28}{31}\)

Now comparing fractions \(\frac {28}{31}\) , \(\frac {31}{33}\) we have Numerator increases and denominator increases but here Numerator increases by greater% than denominator so it must be \(\frac {28}{31}\) < \(\frac {31}{33}\)

If you compare \(\frac {29}{30}\) & \(\frac {31}{33}\) here N and D both increase but D increases by greater % than Numerator So fraction dec
so we have

So \(\frac {28}{31}\) <\(\frac {31}{33}\) < \(\frac {29}{30}\)

So a<c<b
Now \(\frac {29}{30}\) is slightly less than 1. lest say b= .98, then other two(a and c ) are less than .98

I. a+b>2a+b>2

Even if a is .97 then also a+b will not be greater than 2.
II. b+c−a>1
.98+0.96- .93 > 1
So true .


III. a+b2−c2>1
Even if b and c are 1 then also the exp doesnt hold good becuase a <1
So not true.

Hence the option D

Probus
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Given:
1. \(a = \frac{28}{31}\)
2. \(b =\frac{29}{30}\)
3. \(c =\frac{31}{33}\)

Asked: Which of the following is true?

Approximations:
\(a = \frac{28}{31}= \frac{27}{30} approx\)
\(b = \frac{29}{30}\)
\(c = \frac{31}{33} = \frac{28}{30} approx\)

I. \(a+b > 2\)
a + b = 27/30 + 29/30 = 56/30 < 2
NOT TRUE

II. \(b+c-a > 1\)
b + c - a = 29/30 + 28/30 - 27/30 = 30/30 = 1
Since approximation does not confirm result
b + c - a = 29/30 + 31/33 - 28/31 = 1 - 1/30 + 1 - 2/33 - (1 - 3/31) = 1 + 3/31 - 1/30 - 2/33 = 1 + 2.9/30 - 1/30 - 1.8/33 = 1 + .1/30 > 1
TRUE

III. \(a + b^2 - c^2 > 1\)
a + (b+c)(b-c) = 27/30 + 57/30 * 1/30 = 27/30 + 19/300 = 28.9/30 < 1
NOT TRUE

IMO D
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Hi, it took me a while too before I came to the conclusion, but here's how we can solve it under 2-3minutes:

Convert the fractions into decimals to save time on the calculations later.

So by this logic,
\(a = \frac{28}{31} = 0.90\)
\(b =\frac{29}{30} = 0.96\)
\(c =\frac{31}{33} = 0.93\)

Now, they've given us 3 equations from which we need to determine which one is true.

Substitute the values of \(a\) & \(b\) in
I. \(a+b > 2\)
\(0.90 + 0.96 = 1.86 < 2\)

Therefore, answer is not Option A. Since we were able to prove \(1.86 < 2\), we can eliminate Options B & E also. (Because we were able to determine that I. \(a+b > 2\) is false)

This leaves us with only 2 options - C & D.

Let's solve III. \(a + b^2 - c^2 > 1\) to determine if III can be true or not. If III is false, we can eliminate Option C & leaving Option D as the answer.

III. \(a + b^2 - c^2 > 1\)
\(0.90 + (0.96)^2 - (0.93)^2\)
\(0.90 + (0.96 + 0.93)(0.96 - 0.93)\) (Since \((a^2 - b^2) = (a+b)(a-b)\))
\(0.90 + (1.89)(0.03)\)

You can either leave the sum here as we can see that \((1.89)(0.03)\) results to 4 decimal places after 0 & adding 0.90 will make no difference as the result will still be less than 1.

However, for better understanding, I have solved this further for you.

\(0.90 + (1.89)(0.03)\) ====> \(0.90 + 0.0567\) ====> \(0.9567 < 1\)

Clearly, III. \(a + b^2 - c^2 > 1\) is also false, leaving us with only Option D.

Therefore, Option D is the right answer!

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