GRE- Quantitative Comparisons-I

Funda for Quantitative Comparisons:

   

   Observation is the key here. We need not find the absolute values in most of the cases as relative comparison of the given values is enough. So, don’t rush for huge simplifications or calculations. Be smart and use common sense to come out with the best answer options. If you practice this type of math by keeping this in mind, you will crack it.

Now we will solve some Quantitative comparison problems related to numbers and fractions:

Each question consists of two quantities, one in column A and one in column B. You are to compare the two quantities and choose

            A if the quantity in column A is greater;

            B if the quantity in column B is greater;

            C if the quantities are equal;

            D if the relationship cannot be determined from the information given;

Q1)      Column A        Column B

                5/7                     13/49

Solution:

As the denominators are 7 and 49, let us make them equal by multiplying numerator and denominator of 5/7 with 7,

(5*7)/(7*7) = 35/49

Now the two given values : 35/49 and 13/49

Since 35>13, first value is greater than the second hence answer is option “A”

Q2)      Column A        Column B

              4/5 - 4/7          4/7 - 2/5

Solution:

We can observe that the denominators of fractions in each case are equal ie., 5 and 7. Hence the simplified denominator in each case becomes 5*7 = 35. As the denominators are same, we simply proceed further for calculating the numerators

4*7-4*5 and 4*5-2*7

=28-20 and 20-14

=8 and 6

As the first value is greater than the second one, answer option is “A”

Q3)      Column A        Column B

   (-7)⁸                  (-8)⁷

Solution:

We need not calculate the absolute values here. Observe that first value yields positive result as the corresponding exponent is even (ie.,8) and second value yields negative result as the corresponding exponent is odd (ie.,7). As positive number is always greater than negative number, first value is greater than the second. Hence answer option is “A”.

Q4)      Column A        Column B

1 –14/27          8/9 – 1/81

Solution:

First Value: 1 –14/27= (27-14)/27= 13/27

Second Value: observe the denominators 9 and 81, so make a common denominator 81 by multiplying numerator and denominator of 8/9 with 9 ie

(8*9)/(9*9) – 1/81=(72-1)/81= 71/81

Now the two given vales are: 13/27 and 71/81.

Observe that the denominators of first and second value are 27 and 81 respectively. Make the first denominator as 81 by multiplying with 3 on top and bottom

First value: (13*3)/(27*3)= 39/81

Now the two given vales are: 39/81 and 71/81.

As the second value is greater than the first one, answer is option “B”

Q5)      Column A        Column B

    0.6               √(0.6)

Solution:

As the case here is related to absolute fractions, this takes converse of the logic applied to normal numbers.

The rule for normal numbers is: x > √x and for absolute fractions: x < √x

Ie., 0.6 <  √(0.6) hence, answer option is “B”

Q6)         Column A              Column B

  √31 + √25            √(31+25)

Solution:

No need to try for getting absolute or approximate values of the square roots here.

Squaring both the values will give a quicker solution.

Square of first value: 31+25+2√(31*25)

Square of second value: 31+25

As the first value has 2√(31*25) in addition to the second value, first one is greater. Answer option is “A”