**Directions:** Each of the questions given below is followed by series of options among which you have to choose the best one. You can also review your choices by clicking into answer section.

**1. What is the smallest prime number?**

A. 0

B. 1

C. 2

D. 3

**Answer**

**Ans:**Option C

** Explanation:**

Smallest prime number is 2.

0 and 1 are neither prime numbers nor composite numbers.

**2. (1000) ^{9} ÷ 10^{24} =?**

A.10000

B.1000

C.100

D.10

**Answer**

**Ans:**Option B

** Explanation:**

Given Exp. = (1000)^{9 }/ 10^{24} = (10^{3})^{9 }/ 10^{24} = (10)^{27} / 10^{24} = 10^{(27-24)} = 10^{3} = 1000

**3. What is the largest 5 digit number exactly divisible by 94?**

A. 99922

B. 99924

C. 99926

D. 99928

**Answer**

**Ans:**Option A

** Explanation:**

Largest 5 digit number = 99999

99999 ÷ 94 = 1063, remainder = 77

Hence largest 5 digit number exactly divisible by 94 = 99999 – 77 = 99922

**4. 287 x 287 + 269 x 269 – 2 x 287 x 269 =?**

A.534

B.446

C.354

D.324

**Answer**

**Ans:**Option D

** Explanation:**

Given Exp. = a^{2} + b^{2} – 2ab, where a = 287 and b = 269

= (a – b)^{2} = (287 – 269)^{2}

= (18^{2})

= 324

**5. If P and Q are odd numbers, then which of the following is even?**

A. P + Q

B. PQ

C. P + Q + 1

D. PQ + 2

**Answer**

**Ans:**Option A

** Explanation:**

The sum of two odd numbers is an even number

Hence P + Q is an even number

**6. The sum of all two digit numbers divisible by 5 is:**

A.1035

B.1245

C.1230

D.945

**Answer**

**Ans:**Option D

** Explanation:**

Required numbers are 10, 15, 20, 25… 95

This is an A.P. in which a = 10, d = 5 and l = 95.

t_{n} = 95

a + (n – 1)d = 95

10 + (n – 1) x 5 = 95

(n – 1) x 5 = 85

(n – 1) = 17

n = 18

Required Sum = n/2 (a + l) = 18/2 x (10 + 95) = (9 x 105) = 945.

**7. 108 + 109 + 110 + … + 202 =?**

A. 14615

B. 14625

C. 14715

D. 14725

**Answer**

**Ans:**Option D

** Explanation:**

Number of terms of an arithmetic progression

n= (l−A.d+1

where, n = number of terms, a= the first term , l = last term, d= common difference

Sum of first n terms in an arithmetic progression

Sn=n2[ 2a+(n−1)d ] =n2[ a+l ]where a = the first term, d= common difference, l=t_{n}=nth term = a+(n−1)d

a=108l=202d=109−108=1n=(l−A.d+1=(202−108)1+1=94+1=95Sn=n2[ a+l ]=952[ 108+202 ]=95×3102=95×155=14725

**8.** (**854 x 854 x 854 – 276 x 276 x 276) / (854 x 854 + 854 x 276 + 276 x 276) =?**

A.1130

C.565

D.1156

E. None of these

**Answer**

**Ans:**Option B

** Explanation:**

Given Exp. = (a^{3} – b^{3})/ (a^{2} + ab + b^{2})

= (a – B. = (854 – 276) = 578

**9. Which one of the following numbers is completely divisible by 99?**

A. 115909

B. 115919

C. 115939

D. 115929

**Answer**

**Ans:**Option D

** **If a number is divisible by two co-prime numbers, then the number is divisible by their product also.

If a number is divisible by more than two pairwise co-prime numbers, then the number is divisible by their product also.

If a number is divisible by another number, then it is also divisible by all the factors of that number.

We know that 99 = 9 × 11 where 9 and 11 are co-prime numbers. Also 9 and 11 are factors of 99. Hence if a number is divisible by 9 and 11, the number will be divisible by their product 99 also. If a number is not divisible by 9 or 11, it is not divisible by 99. So,

115929 is divisible by both 9 and 11 => 115929 is divisible by 99

115939 is not divisible by 9 and 11 => 115939 is not divisible by 99

115919 is not divisible by 9 and 11 => 115919 is not divisible by 99

115909 is not divisible by 9 and 11 => 115909 is not divisible by 99

Hence, 115929 is the answer

**10. If x and y are the two digits of the number 653xy such that this number is divisible by 80, then x + y =?**

A. 2 or 6

B.4

C.4 or 8

D.8

E.None of these

**Answer**

**Ans:**Option A

** Explanation:**

80 = 2 x 5 x 8

Since 653xy is divisible by 2 and 5 both, so y = 0.

Now, 653x is divisible by 8, so 13x should be divisible by 8.

This happens when x = 6.

x + y = (6 + 0) = 6.

**11. 12 + 22 + 32 + … + 82 =?**

A. 204

B. 200

C. 182

D. 214

**Answer**

**Ans:**Option A

** Explanation:**

(Reference: Power Series : Important formulas)

12+22+32+⋯+n2=∑n2=n(n+1)(2n+1)6

12+22+32+⋯+82=n(n+1)(2n+1)6=8(8+1)[(2×8)+1]6=8×9×176=4×9×173=4×3×17=204

**12. The sum of even numbers between 1 and 31 is:**

A.6

B.28

C.240

D.512

**Answer**

**Ans:**Option C

** Explanation:**

Let S_{n} = (2 + 4 + 6 + … + 30). This is an A.P. in which a = 2, d = 2 and l = 30

Let the number of terms be n. Then,

a + (n – 1)d = 30

2 + (n – 1) x 2 = 30

n = 15.

S_{n} = n/2 (a + l) = 15/2 x (2 + 30) = (15 x 16) = 240.

**13. What least number should be subtracted from 13601 such that the remainder is divisible by 87?**

A. 27

B. 28

C. 29

D. 30

**Answer**

**Ans:**Option C

** Explanation:**

13601 ÷ 87 = 156, remainder = 29

Hence 29 is the least number which can be subtracted from 13601 such that the remainder

is divisible by 87

**14. 3251 + 587 + 369 – ? = 3007**

A.1250

B.1300

C.1375

D.1200

E.None of these

**Answer**

**Ans:**Option D

** Explanation:**

3251 Let 4207 – x = 3007

+ 587 Then, x = 4207 – 3007 = 1200

+ 369

—-

4207

—-

**15. If (64) ^{2} – (36)^{2} = 10x, then x = ?**

A. 200

B. 220

C. 210

D. 280

**Answer**

**Ans:**Option D

** Explanation:**

a^{2}−b^{2}=(a−b)(a+b)

(64)^{2} – (36)^{2} = (64 – 36)(64 + 36) = 28 × 100

Given that (64)^{2} – (36)^{2} = 10x

28 × 100 = 10x

x = 280