**Directions(1-5):** Find the missing term ‘?’ of the following series.

- 8,13,26,51,92,?
133153140132150Option B

+(1^2+2^2)

+(2^2+3^2)

+(3^2+4^2)

+(4^2+5^2)

+(5^2+6^2)

? = 153 - 630,210,84,42,?,28,56
1520281613Option C

/3

/2.5

/2

/1.5

/1

/0.5

? = 28 - 463,453,433,403,?,313
333363350344362Option B

-10.-20,-30,-40,-50

? = 363 - 24,16,28,74,?,1187.5
222255241263272Option D

*0.5+4

*1.5+4

*2.5+4

*3.5+4

*4.5+4

? = 263 - 23,40,61,86,?,148
951008811590Option D

+17,+21,+25,+29,+33

? = 115 - At what rate of simple interest must Rs.15900 be deposited for 4 years to receive the same interest that is obtained by investing Rs.15000 at 12% per annum compound interest compounded for 2 years ?
86754Option B

Compound interest earned = 15000*{(1+0.12)^2 – 1} = Rs.3816

Let rate be r%.

Simple interest earned = 15900*4*r/100 = 3816

=> r = 6 - If the numerator of a fraction is increased by 40% and denominator in increased by 25% then the resultant fraction will become 7/5. What is the original fraction?
3/511/97/24/35/4Option E

Let the original fraction be x/y. (x*1.4)/(y*1.25) = 7/5

=> x/y = 5/4 - Find the area of the circular field having radius equal to the breadth of a room with area of four walls 840 m^2 and height 10 m if it is given that ratio of length and breadth of the room is 4:3 resp.(Take pi = 3)
900 m^2818 m^2919 m^2972 m^2700 m^2Option D

2(l+b)*h = 840

=> 2(4x+3x)*10 = 840

=> x = 6

Breadth = 3*6 = 18 m

Area of the circular field = pi*(18)^2 = 972 m^2 - Present ages of A and B are in the ratio of 2:3 resp. Present ages of C and D are in the ratio of 8:7 resp. Present average age of A ,B and C is 46 years. Present average age of B,C and D is 48 years. Find the difference between the present ages of B and C.
4 years2 years5 years8 years6 yearsOption E

Let the present ages of A,B,C and D are 2x,3x,8y and 7y resp.

Now, 2x+3x+8y = 3*46 5x+8y = 138 —-(1)

Also, 3x+8y+7y = 3*48 —–(2)

On solving (1) and (2), we get

x = 18

y = 6

Difference between present ages of B and C = 54 â€“ 48 = 6 years - A cricket team of 11 players is to be selected from a group of 8 batsmen, 5 bowlers and 3 wicketkeepers. In how many ways a team a having at least 4 bowlers and exactly 1 wicketkeeper is selected?
500555588610585Option C

Required number of ways = 3C1*5C4*8C6 + 3C1*5C5*8C5

= 420+168 = 588