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Example 1

Example

Solution

Apply BODMAS
Solve the bracket first: 23/10 + 13/5 = 49/10
Solving of: 3/7 of 49/10 = 21/10
Solving division: 1/(21/10) = 10/21 and 1/5 × 5/7 = 1/7
Hence we have: 10/21 + 1/7 – 2/7 = 7/21 or 1/3.


Example 2

Example

Consider a 99-digit number created by writing side by side the first four natural numbers as follows: 1234567890111213......5354. The
above number when divided by 8 will leave a remainder of:
1. 6 2. 4 3. 2 4. 0

Solution

Check out the rule for divisibility of 8 (given above), hence we have to check for the last 3 digits. Since 354 on division by 8 leaves a
remainder of 2, the answer is 2; tick (3).

 

Example 3

Example

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
5. 32

Solution

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. Before C he had (4+1) ×× 2 = 10. Before B he had (10+1) ×× 2 = 22 and before A he had (22+1) ×× 2 = 46. We can also
do this from the choices.

 

Example 4

Example

Solution

The numerator can be written in the form (a – b)2 or (0.47 – 0.35)2.
Hence (0.12)2/0.12 = 0.12.

 

 

 

Example 5

Example

Solution

Use the formulas given to see that the term can be written as (a3 – b3)/(a2 + ab + b2).
Since a3 – b3 = (a – b) (a² + b² + ab), we can cancel out the terms and are left with (a – b).
Hence the answer is (0.7541 – 0.2459) = 0.5082.


Example 6

Example

Let M = 21! + 23! + 25! + 27! + 29! + … 53!. Which of the following is definitely false about M?
1. The last two digits of M will be 00.
2. M is divisible by 50.
3. (M ÷ 10) will be a prime number.
4. M is divisible by 55.
5. none of these

Solution

Each multiples of 5 will result in one 0 at the end. It can be seen that 21! will have 4 zeroes at the end of it and therefore factorial of all the
bigger number will have at least 4 zeroes at the end. Thus the 1st and the 3rd options are true. As each of these numbers is a multiple of 11 also,
the total will also be a multiple of 11, 4th option is also right. 3rd option is definitely false, if four zeroes are at the end it can be verified that even
if it is divided by 10, still three zeroes will be at the end and is not prime. Hence answer is (3).

 

Example 7

Example

What will be the remainder when 12492 is divided by 1727?
1. 12 2. 144 3. 1 4. 0 5. 14

Solution

12492 can also be written as (123)164 = 1728164, which when divided by 1727 gives a remainder of 1. Thus, the remainder will be 1 for all
powers.





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