1. In a certain series, each number except the first and second is obtained by adding the previous two numbers. If the first no is 2 and sixth no is 26, then the seventh number is:

a) 52

b) 46

c) 40

d) 42

Explanation: 1

^{st}no. = 2

2

^{nd}no. = x (let)

3

^{rd}no. = x + 2

4

^{th}no. = x + x + 2 = 2x + 2

5

^{th}no. = x + 2 + 2x + 2 = 3x + 4

6

^{th}no.,

3x + 4 + 2x + 2 = 26

5x = 20

x = 4

7

^{th}no. = 26 + 3x + 4 = 26 + 3 × 4 + 4 = 42

2. If (-1)^{n} + (-1)^{4n} = 0, then n is

a) Any positive integer

b) Any negative integer

c) Any odd natural number

d) Any even natural number

Explanation: Putting n = 1, 2, 3, 4, 5 . . . .

n = 1 in the given equation

(-1)

^{1}+ (-1)

^{4 × 1}= 0

-1 + 1 = 0

n = 2,

(-1)

^{2}+ (-1)

^{4 × 2}= 0

1 + 1 ≠ 0

n = 3,

-1 + 1 = 0

We will see that the given condition will be satisfy only for any odd natural number.

3. Sunil entered a shopping center and spent one half of the money that he had. When he finished his purchase he found that he had as many paise as he had rupees and half as many rupees as he had paise when he went. How much money did he have when he entered?

a) Rs. 75.50

b) Rs. 98.98

c) Rs. 99.98

d) None of these

Explanation: Here we have to verify the given two conditions for the given options.

Condition :1

As given, after finished his purchase he had the rupees as much as the paise when he went.

Condition :2

After finishing purchase, he had Half as many rupees as he had paise when he went.

Solve it through option checking method.

Case :1

Now suppose the answer is option c

Then he spent Rs. $$\frac{{99.98}}{2}$$ = Rs. 49.99

Condition :1

Here we have 99 paise so the rupees should be 99 when he went. He initially had 99 Rs. Hence condition 1 is satisfied.

Condition :2

He had 98 paise to begin with. So the rupees he spent should be $$\frac{{98}}{2}$$ = Rs. 49 as per condition 2.

It turns out that the rupees actually he spent was 49.

Therefore, condition 2 is satisfied as well.

4. 11 * X = $$\frac{{693}}{9}$$

a) 3

b) 5

c) 7

d) 9

Explanation:

$$\eqalign{ & X = \frac{{ {693} }}{{ {9 \times 11} }} \cr & X = 7 \cr} $$

5. What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

a) 840

b) 841

c) 820

d) 814

Explanation: Length of the room

= 15 m 17 cm

= 15 × 100 + 17

= 1517 cm.

Breadth of the room

= 9 m 2 cm

= 902 cm.

The HCF of the 1517 and 902 will be size of square tiles.

HCF of 1517 and 902 = 41 cm.

Area of the room

= length × breadth

= 1517 × 902 cm

^{2}

Area of tiles = 41 × 41 cm

^{2}

Number of tiles required

= $$\frac{{1517 \times 902}}{{41 \times 41}}$$

= 814 tiles

6. Three numbers which are co-prime to each other are such that the product of the first two is 119 and that of the last two is 391. What is the sum of the three numbers?

a) 47

b) 43

c) 53

d) 51

Explanation: Since the numbers are co-prime, their HCF = 1

Product of first two numbers = 119

Product of last two numbers = 391

The middle number is common in both of these products.

Hence if we take HCF of 119 and 391, we get the common middle number.

HCF of 119 and 391 = 17

So, Middle Number = 17

First Number = $$\frac{{119}}{{17}}$$ = 7

Last Number = $$\frac{{391}}{{17}}$$ = 23

Sum of the three numbers

= 7 + 17 + 23

= 47

7. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

a) 36 minutes 32 seconds

b) 36 minutes 22 seconds

c) 36 minutes 12 seconds

d) 46 minutes 12 seconds

Explanation: LCM of 252, 308 and 198 = 2772

They all will be again at the starting point after 2772 seconds or 46 minutes 12 seconds.

8. What is the least number when divided by 8, 12, 20 and 30 leaves a remainder 4 and when divided by 21 leaves a remainder 7 ?

a) 124

b) 360

c) 364

d) Cannot be determined

Explanation: Lcm of 8, 12,20,30 is 120. As they leave a remainder 4

So it should be 120 + 4. But when being divided by 21, remainder is 7

So the number will be exactly divisible by (21 + 7) = 28.

The number will be in the form of (120k + 4) which is divisible by 28

Where k is any positive integer. Now, if we divide 120 by 28, the remainder is 8

We need a remainder of 24, because (24 + 4) = 28

Now (8 × 3) = 24

k = 3

The number would be (120 × 3) + 4 = 364

9. Six bells start ringing together and ring at intervals of 4, 8, 10, 12, 15 and 20 seconds respectively. how many times will they ring together in 60 minutes ?

a) 31

b) 15

c) 16

d) 30

Explanation: LCM of 4, 8, 10, 12, 15 and 20 = 120

120 seconds = 2 minutes

All the six bells will ring together in every 2 minutes

Number of times they will ring together in 60 minutes

= $$1 + \frac{{60}}{2}$$

= 31

10. What is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder?

a) 1108

b) 1683

c) 2007

d) 3363

Explanation: LCM of 5, 6, 7 and 8 = 840

Hence the number can be written in the form (840k + 3) which is divisible by 9

If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9

If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9

1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder