How to Prepare for Quantitative Aptitude for CAT (2014)

First Things First

Chapter 4. SQUARES AND CUBES OF NUMBERS

SQUARES AND SQUARE ROOTS

When any number is multiplied by itself, it is called as the square of the number.

Thus, 3 × 3 = 3² = 9

Squares have a very important role to play in mathematics. In the context of preparing for CAT and other similar aptitude exams, it might be a good idea to be able to recollect the squares of 2 digit numbers.

Let us now go through the following table carefully:

Table 4.1

So, how does one get these numbers onto one’s finger tips? Does one memorize these values or is there a simpler way?

Yes indeed! There is a very convenient process when it comes to memorising the squares of the first 100 numbers.

First of all, you are expected to memorise the squares of the first 30 numbers. In my experience, I have normally seen that most students already know this. The problem arises with numbers after 30. You do not need to worry about that. Just follow the following processes and you will know all squares upto 100.

Trick 1: For squares from 51 to 80 – (Note: This method depends on your memory of the first thirty squares.)

The process is best explained through an example.

Suppose, you have to get an answer for the value of 67². Look at 67 as (50 + 17). The 4 digit answer will have two parts as follows:

The last two digits will be the same as the last two digits of the square of the number 17. (The value 17 is derived by looking at the difference of 67 with respect to 50.)

Since, 17² = 289, you can say that the last two digits of 67² will be 89. (i.e. the last 2 digits of 289.) Also, you will need to carry over the ‘2’ in the hundreds place of 289 to the first part of the number.

The first two digits of the answer will be got by adding 17 (which is got from 67 – 50) and adding the carry over (2 in this case) to the number 25. (Standard number to be used in all cases.) Hence, the first two digits of the answer will be given by 25 + 17 + 2 = 44.

Hence, the answer is 67²= 4489.

Similarly, suppose you have to find 76².

Step 1: 76 = 50 + 26.

Step 2: 26² is 676. Hence, the last 2 digits of the answer will be 76 and we will carry over 6.

Step 3: The first two digits of the answer will be 25 + 26 + 6 = 57.

Hence, the answer is 5776.

This technique will take care of squares from 51 to 80 (if you remember the squares from 1 to 30). You are advised to use this process and see the answers for yourself.

SQUARES FOR NUMBERS FROM 31 TO 50

Such numbers can be treated in the form (50 – x) and the above process modified to get the values of squares from 31 to 50. Again, to explain we will use an example. Suppose you have to find the square of 41.

Step 1: Look at 41 as (50 – 9).

Again, similar to what we did above, realise that the answer has two parts—the first two and the last two digits.

Step 2: The last two digits are got by the last two digits in the value of (– 9)² = 81. Hence, 81 will represent the last two digits of 41².

Step 3: The first two digits are derived by 25 – 9 = 16 (where 25 is a standard number to be used in all cases and – 9 comes from the fact that (50 – 9) = 41).

Hence, the answer is 1681.

Note: In case there had been a carry over from the last two digits it would have been added to 16 to get the answer.

For example, in finding the value of 36² we look at 36 = (50 – 14)

Now, (–14)² = 196. Hence, the last 2 digits of the answer will be 96. The number ‘1’ in the hundreds place will have to be carried over to the first 2 digits of the answer.

The, first two digits will be 25 – 14 + 1 = 12

Hence, 36² = 1296.

With this process, you are equipped to find the squares of numbers from 31 to 50.

FINDING SQUARES OF NUMBERS BETWEEN 81 TO 100

Suppose you have to find the value of 82². The following process will give you the answers.

Step 1: Look at 82 as (100 – 18). The answer will have 4 digits whose values will be got by focusing on getting the value of the last two digits and that of the first two digits.

Step 2: The value of the last two digits will be equal to the last two digits of ( – 18)².

Since, ( – 18)² = 324, the last two digits of 82² will be 24. The ‘3’ in the hundreds place of (– 18)² will be carried over to the other part of the answer (i.e. the first two digits).

Step 3: The first two digits will be got by: 82 + (– 18) + 3 Where 82 is the original number; (– 18) is the number obtained by looking at 82 as (100 – x); and 3 is the carry over from (– 18)².

Similarly, 87² will give you the following thought process:

87 = 100 – 13 Æ (– 13)² = 169. Hence, 69 are the last two digits of the answer Æ Carry over 1. Consequently, 87 + (–13) + 1 = 75 will be the first 2 digits of the answer.

Hence, 87² = 7569.

With these three processes you will be able to derive the square of any number up to 100.

Properties of squares:

1.When a perfect square is written as a product of its prime factors each prime factor will appear an even number of times.

2.The difference between the squares of two consecutive natural numbers is always equal to the sum of the natural numbers. Thus, 41² – 40² = (40 + 41) = 81.

This property is very useful when used in the opposite direction—i.e. Given that the difference between the squares of two consecutive integers is 81, you should immediately realise that the numbers should be 40 and 41.

3.The square of a number ending in 1, 5 or 6 also ends in 1, 5 or 6 respectively.

4.The square of any number ending in 5: The last two digits will always be 25. The digits before that in the answer will be got by multiplying the digits leading up to the digit 5 in the number by 1 more than itself.

Illustration:

85² = ___25.

The missing digits in the above answer will be got by 8 × (8 + 1) = 8 × 9 = 72. Hence, the square of 85 is given by 7225.

Similarly, 135² = ___25. The missing digits are 13 × 14 = 182. Hence, 135² = 18225.

5.The value of a perfect square has to end in 1, 4, 5, 6, 9 or an even number of zeros. In other words, a perfect square cannot end in 2, 3, 7, or 8 or an odd number of zeros.

6.If the units digit of the square of a number is 1, then the number should end in 1 or 9.

7.If the units digit of the square of a number is 4, then the units digit of the number is 2 or 8.

8.If the units digit of the square of a number is 9, then the units digit of the number is 3 or 7.

9.If the units of the square of a number is 6, then the unit’s digit of the number is 4 or 6.

10.The sum of the squares of the first ‘n’ natural numbers is given by

[(n) (n + 1) (2n + 1)]/6.

11.The square of a number is always non-negative.

12.Normally, by squaring any number we increase the value of the number. The only integers for which this is not true are 0 and 1. (In these cases squaring the number has no effect on the value of the number).

Further, for values between 0 to 1, squaring the number reduces the value of the number. For example 0.5² < 0.5.

Say, you have to Find the Square Root of a Given Number

Say 7056

Step 1: Write down the number 7056 as a product of its

Prime factors.7056 = 2 × 2 × 2 × 2 × 21 × 21

= 2⁴ × 21²

Step 2: The required square root is obtained by halving the values of the powers.

Hence, = 2² × 21¹

CUBES AND CUBE ROOTS

When a number is multiplied with itself two times, we get the cube of the number.

Thus, x × x × x = x³

Method to find out the cubes of 2 digit numbers: The answer has to consist of 4 parts, each of which has to be calculated separately.

The first part of the answer will be given by the cube of the ten’s digit.

Suppose you have to find the cube of 28.

The first step is to find the cube of 2 and write it down.

2³ = 8.

The next three parts of the number will be derived as follows.

Derive the values 32, 128 and 512.

(by creating a G. P. of 4 terms with the first term in this case as 8, and a common ratio got by calculating the ratio of the unit’s digit of the number with its tens digit. In this case the ratio is 8/2 = 4.)

Now, write the 4 terms in a straight line as below. Also, to the middle two terms add double the value.

Properties of Cubes

1.When a perfect cube is written in its standard form the values of the powers on each prime factor will be a multiple of 3.

2.In order to find the cube root of a number, first write it in its standard form and then divide all powers by 3.

Thus, the cube root of 3⁶ × 5⁹ × 17³ × 2⁶ is

given by 3² × 5³ × 17 × 2²

3.The cubes of all numbers (integers and decimals) greater than 1 are greater than the number itself.

4.0³ = 0, 1³ = 1 and – 1³ = –1. These are the only three instances where the cube of the number is equal to the number itself.

5.The value of the cubes of a number between 0 and 1 is lower than the number itself. Thus, 0.5³ < 0.5² < 0.5.

6.The cube of a number between 0 and –1 is greater than the number itself. (–0.2)³ > –0.2.

7.The cube of any number less than –1, is always lower than the number. Thus, (–1.5)³ < (–1.5).