1. A two digit number ab is added to another number ba, which is obtained by reversing the digits then we get three digit number. Thus (a + b) equals to:

a) at least 18

b) 2ab

c) 2(a + b)

d) (a + b) ≥ 10

Explanation: When two 2 digit numbers are added and the resultant value is a three digit number. It means there must be a carry over (i.e. The sum of unit digits be greater than 9. Similarly, the sum of the tens digit is also greater than 9)

Required numbers = 64 + 46 = 110

So, Option 'd' is correct.

2. A gardener plants his garden with 5550 trees and arranged them so that there is one plant more per row as there are rows then number of trees in a row is:

a) 56

b) 74

c) 76

d) 75

Explanation: Let there be n rows, then number of trees in each row = (n + 1)

Total number of trees,

n × (n +1) = 5550

Now, at this moment this problem can be solved in two ways. First by finding the roots of quadratic equation. Second, by using the values from options.

74 × 75 = 5550

i.e. (n + 1) = 75

3. The sum of two numbers is 18. The greatest product of these two number can be:

a) 17

b) 80

c) 81

d) Can't determined

Explanation: a + b = 18

Maximum of (a × b) will be only when a = b

Thus, a = b = 9

Maximum of (a × b) = 9 × 9 = 81.

4. The unit digit of (316)^{34n} + 1 is :

a) 4

b) 5

c) 1

d) 7

Explanation: The unit digit of (316)

^{34n}, depends on the power of 6.

See the pattern, 6

^{2}= 36

6

^{3}= 216

6

^{4}= 1296

Any power of 6 will give unit digit 6.

The unit digit of (316)

^{34n}always 6.

So, unit digit of (316)

^{34n}+ 1 will be 7

5. A number when divided by 14 leaves reminder of 8, but when the same number is divided by 7, it will leave the remainder :

a) 3

b) 2

c) 1

d) 4

Explanation: When the number is divided by 14 it gives a remainder of 8,

The number = 14N + 8 (14N is divisible by 14)

When same number is divided by 7 it will give remainder 1

6. The HCF and LCM of 2^{4}, 8^{2}, 16^{2}, 20^{3} are :

a) 2^{3}, 32000

b) 2^{4}, 32000

c) 2^{4}, 25600

d) 2^{2}, 3200

Explanation: HCF of 2

^{4}, 8

^{2}, 16

^{2}, 20

^{3}= 2

^{4}

LCM of 2

^{4}, 8

^{2}, 16

^{2}, 20

^{3}= 2

^{4}= 2

^{8}× 125 = 32000

7. The four digit smallest positive number which when divided by 4, 5, 6 or 7, it leaves always the remainder as 3:

a) 1000

b) 1257

c) 1263

d) 1683

Explanation: The least possible number = (LCM of 4, 5, 6 and 7) + 3

= 420 + 3

= 423

The next higher number is,

(420m + 3), now we put a least value of m such that

(420m + 3) ≥ 1000

At m = 3,

value = 420 × 3 + 3 = 1263

8. A string of length 221 metre is cut into two parts such that one part is $$\frac{9}{4}$$ th as long as the rest of the string, then the difference between the larger piece and the shorter piece is

a) 58 m

b) 53 m

c) 85 m

d) 76 m

Explanation: Let one part of string is x.

x + $$\frac{{9{\text{x}}}}{4}$$ = 221 m

x = 68 meter

and, $$\frac{{9{\text{x}}}}{4}$$ = 153 m

Difference between two parts = 153 - 68 = 85 m

9. The sum of 100 terms of the series 1 - 3 + 5 - 7 + 9 - 11 .......... is:

a) 100

b) -200

c) 200

d) -100

Explanation: 1 - 3 + 5 - 7 + 9 - 11 .......... + 197 - 199

= (- 2) + (-2) + (-2) + .......... + (-2) (50 times)

= 50 × (-2) = -100

10. The remainder of $$\frac{{{6^{36}}}}{{215}}:$$

a) 0

b) 1

c) 2

d) 3

Explanation:

$$\eqalign{ & \frac{{ {{6^{36}}} }}{{215}},\,{\text{can}}\,{\text{be}}\,{\text{written}}\,{\text{as}} \cr & \frac{{{{\left( {{6^3}} \right)}^{12}}}}{{215}} \cr & \frac{{{{216}^{12}}}}{{215}},\,[216\,{\text{on}}\,{\text{divided}}\,{\text{by}}\,215,\,{\text{gives}}\,{\text{remainder}}\,1] \cr & \frac{{{1^{12}}}}{{215}} \cr & {\text{The}}\,{\text{remainder}}\,{\text{will}}\,{\text{be}}\,1 \cr} $$