Loading....
Coupon Accepted Successfully!

 

Solved Examples

The following examples are taken from papers to show the kind of questions that are asked and the short cuts used for solving them.

Do these sums yourself before looking at the solutions given below. Also the student must ensure that he understands these sums, since

many examination questions are based on principles explained in the following examples.


Important note:
It is sums like this that students find difficult to do. Build an understanding of the techniques described below to be able

to attempt them. If you start thinking about squares or prime numbers in the exam, you will almost certainly lose time, so be sure to mug

them up!

Examples 1

Try the sums yourself before looking at the solutions given below.
 

Example 1 :

A certain number when divided by 175 leaves 140 as remainder. If the same number is divided by 25, the remainder will be:

10 2. 15 3. 5 4. 7
Solution

A certain number when divided by 175 leaves 140 as remainder…this means that the number is of the form 175k + 140. (We use the
property Dividend = Divisor ´ quotient + remainder).
Now if we divide by 25, we get (175k + 140)/25
= 175k/25 + 140/25.
We see that the first term is perfectly divisible by 25 so there is no remainder but the second term gives 15 as remainder.
Hence the answer would be 15.

Examples 2

Example 2 :

Let S be the set of integers X, and it is given that 100 < X < 200, X is odd and is divisible by 3 but not by 7. How many elements does S have in all?
16 2. 12 3. 11 4. 13

Solution

This sum was asked in CAT 2000 and is an example of a deceptive sum: it looks fairly simple but 95% of the students in the classroom get
it wrong. Here is how to do it.
First we have to look at odd numbers between 100 and 200 which are divisible by 3. These numbers are 105, 111, 117, …, 189, 195.
These are 16 terms.

Now delete those terms which are divisible by 7; there are three of them, 105, 147 and 189.
The correct answer would thus be 16 – 3 = 13 or (d). Did you get it right?

 

 

Examples 3

Example 3 :

The remainder when 784 is divided by 342 is:
0 2. 1 3. 49 4. 341

Solution

In this question, we have to know that 73 = 343 and that 84 = 3 × 28. So the term can be written as 784 = (73)28. So the sum becomes
(73)28/(73 – 1).

This is of the form: 73n/(73 – 1).
This means that the remainder must be 1, since 73/342 would give the remainder 1.
Notice that the student must know the squares and cubes of numbers by heart, otherwise such sums are quite difficult to do.
This sum was asked in CAT 1999.

Examples 4

Example 4 :

Let X, Y, Z be different integers, which are odd and positive. Then which of the following statements is not always true?
1. xyz2 is odd
2. (x – y )2z is even
3. (x + y – z ) 2 (x + y) is even
4. (x – y) (y + z ) (x + y – z ) is odd

Solution

1. xyz2 is odd
2. (x – y )2z is even
3. (x + y – z ) 2 (x + y) is even
4. (x – y) (y + z ) (x + y – z ) is odd

In such sums, we take any 3 odd and positive numbers and apply in the choices. We can, for example, take
x = 1, y= 3, and z = 5.
Choice (a): 152 is odd (true);
Choice (b): (–2)10 is even (true).
Choice (c): any number multiplied by an even number
(x + y) is even (true);
Choice (d): this must be even as well, since (y + z) is even.
Hence the correct answer would be (d).

The sum can also be done by remembering the qualities of odd and even numbers described in the chapter. We see that (odd + odd = even)
so any number multiplied by any sum of the numbers must be even, and we are easily able to spot (4) as the answer.

 

Examples 5

Example 5 :

Let S be the set of prime numbers greater than or equal to two and less than 100. If all such existing terms are then multiplied, then how many consecutive zeros would the final digit end with?

1 2. 4 3. 5 4. 10
Solution

A time saver. Note that there is only one even prime number. When it is multiplied by another prime (5) we will get just one zero.
No matter how many numbers there are, when prime numbers are multiplied, the resultant number can have only one zero.

 

Examples 6

Example 6 :

If |r – 6| = 11 and |2q – 12| = 8, what is the minimum possible value of q/r?
1. – 2/5  2. 2/17  3. 10/17
4. None of these

Solution

Another sum that more than 95% students get wrong in their first try.
Note that in cases of mod in an equation, we always get 2 values. In this case we will get 2 values of r and 2 values of q.
The values of r are: r = 17 or r = –5 and q = 10 or q = 2.
So there are 4 cases of the value of q/r. The least one is 10/(–5) = –2 and hence the answer would be (d). Did you get it right in the first go?

Examples 7

Example 7 :

There are four consecutive even numbers. The product of first and fourth is 216. Find the product of second and third numbers.

  168 2. 192 3. 224 4. 242
Solution

Since all the numbers are consecutive even numbers, we can say that the numbers are x, x + 2, x + 4 and x + 6.
Now look at the factors of 216 and their difference should be 6.
A quick break up into factors yields 12 and 18. So the required product is 14 × 16 = 224.
The sum can also be done by solving the quadratic
x(x + 6) = 216 but it is quicker if you can think of factors of 216.

 

Examples 8

Example 8 :

Consider a number system whose base is b. If in such a system 54 stands for 8 ´ 8 in decimal system, then 111 in that system will represent ?

  143 2. 157 3. 171 4. 145
Solution

This is a sum of change of base. We have to find out the base first from the first statement:
64 = 5 × b1 + 4 × b0, hence 5b + 4 = 64, so base b = 12.
Hence 111 = (1 × 120) + (1 × 121) + (1 ´ 122) = 157.

Examples 9

Example 9 :

Let a, b, c, be distinct digits. Consider a 2-digit number ‘ab’ and a 3-digit number ‘ccb’, both defined under the usual decimal number system.
If (ab)2 = ccb and ccb > 300, then the value of ‘b’ is 1 2. 0 3. 5 4. 6

Solution

This sum, asked in CAT 2001, illustrates he importance of learning the squares and square roots mentioned above. Try doing it without the
squares, and it will take you ages to do the sum. But once you know the squares, the question should take less than 30 seconds, because you will realise that that there is only
one number that fulfills the condition.

That number is 212 = 441 and b = 1.
Can you tell the answer in 10 seconds, if the condition is changed to ccb < 300? 

Examples 10

Example

Arun thinks his weight is more than 60 kg but less than 72 kg. His brother does not agree with him and thinks Arun’s weight is more than
65 kg but less than 70 kg. Their mother feels Arun’s weight cannot be more than 68 kg. If all of them are correct in their estimation and his
weight is an integer only, what is the average of the different probable values of Arun’s weight?
1. 69 kg 2. 67 kg 3. 68 kg 4. Data inadequate

Solution

In this sum, we list down the estimates of each member:
Arun’s estimate = 61 to 71;
His brother thinks he is less than 70, so the range reduces to 66 to 69.
By his mother’s estimate, the range becomes 66 to 68.
Average of 66, 67 and 68 = 67, hence answer is (b).

 

Examples 11

Example 11 :

In a long division sum, the dividend is 261498 and the successive remainders from the first to the last are 523, 360 and 123. Find the divisor and the quotient.


1394 2. 697 3. 297 4. 597
Solution

Clearly, the divisor must be greater than 523 since that is one of the remainders.

So the first number to be divided should be 2614 and the number must divide (2614 – 523) = 2091.
Now we look at the choices and see which of the choices divides 2091 and we find 697 × 3 = 2091, hence it must be the answer.

 

  

Examples 12

Example 12 :

A man has a certain number of mangoes. He sells half of what he had and one more to A, half of the remainder and one more to B, half of the remainder and one more to C, half of the remainder and one more to D. Then he has only 1 left. How many did he have at the beginning?

  1. 66 2. 56 3. 46 4. 36
Solution

This kind of sum has often been asked in exams.
In such sums, it is advisable to start from the back (try not to use x and form an equation, because you have four steps to solve). Before D,
he had (1+1) × 2 = 4 mangoes.
Before C he had (4+1) ×2 = 10 mangoes.
Before B he had (10+1) × 2 = 22 mangoes.
Before A he had (22+1) × 2 = 46.
We can also do this from the choices and work backwards.

Examples 13

Example 13 :

In a party there are 30 persons present, and every person shakes hand with all the other persons. What is the total number of handshakes that will take place?


1.29 2. 30 3. 31 4. 435
Solution

There are 30 persons. Imagine one person shaking hands with all the others and leaving. Then the second person shakes hand with the remaining and leaves, and so on.

 

 

Examples 14

Example 14 :

Find the greatest four-digit number which is a multiple of 12.

Solution

The greatest four-digit number is 9999. Divide it by 12 and get the remainder 3. This means this number is 3 more than a multiple of 12.
Subtract 3 from the number and get 9996 which is the greatest four-digit number which is a multiple of 12.

 

Examples 15

Example 15 :

Find the smallest six-digit number which is a multiple of 17.

Solution

The smallest six-digit number is 100000, when this is divided by 17 the remainder is 6.

As per the previous question if we subtract 6 from it, that number would definitely be a multiple of 17. But it would be a five-digit number
because we have already taken the smallest six-digit number.
This means some number should be added.
Now what smallest number should be added in the remainder 6 so that it becomes a multiple of 17?
The number is 17 – 6 = 11 is that smallest number. Therefore 100000 + 11 = 100011 is the answer.

After understanding the above sums, let us now practice our knowledge. Given below are two speed exercises which you should do in
about 30 seconds per question. The third exercise tests your knowledge of number systems. Look at the answers only after you have done the
exercises to see your mistakes and to improve further.





Test Your Skills Now!
Take a Quiz now
Reviewer Name