**Directions(1-5): **Find the missing term of the series.

- 3,20,88,273,?,559
551500554505517Option C

(3+1)*5 = 20

(20+2)*4 = 88

(88+3)*3 = 273

(273+4)*2 = 554

(554+5)*1 = 559 - 27,54,86,?,187,272
128153122110111Option A

—27—–32—–42—–59—–85—–

—5—–10—–17—–26–

5 == 4^2+1

10 == 3^2+1

17== 4^2+1

26== 5^2+1

? == 128 - 21,?,77,150,295,584
4236503840Option E

21*2-2 = 40

40*3-3 = 77

77*4 – 4 = 150

150*5 – 5 = 295

295*6 – 6 = 584 - 1,2,6,?,49,174
2027303335Option D

1+1^3 = 2

2 + 2^2 = 6

6 + 3^3 = 33

33+4^2 = 49

49 + 5^3 = 174 - 105,121,146,182,?,295
231222252202220Option A

—4^2—–5^2—-6^2—–7^2—- - If machine X can produce 1,000 bolts in 8 hours and machine Y can produce 1,000 bolts in 24 hours. In how many hours can machines X and Y, working together at these constant rates, produce 1,000 bolts?
5 hours3 hours4 hours6 hours8 hoursOption D

1/8 + 1/24 = 1/h => 4/24 = 1/6.

Working together, machines X and Y can produce 1,000 bolts in 6 hours. - A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?
1/31/51/81/61/4Option D

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

Now,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 – 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 – 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remaining part of work = (144-120)/144 = 1/6 - A, B, C started a business with their investments in the ratio 1:3 :5. After 4 months, A invested the same amount as before and B as well as C withdrew half of their investments. Find the ratio of their profits at the end of the year .
3 : 6 : 102 : 6 : 95 : 6 : 107 : 6 : 83 : 5 : 11Option C

Let their initial investments be x, 3x and 5x respectively.

Then, A:B:C = (x*4+2x*8) : (3x*4+(3x/2)*8) : (5x*4+(5x/2)*8)

= 20x : 24x : 40x = 5 : 6 : 10 - The difference between compound interest and simple interest on a sum for two years at 8% per annum, where the interest is compounded annually is Rs.16. if the interest were compounded half yearly , what is the difference between in two interests .
Rs.22.52Rs.24.64Rs.18.35Rs.17.15Rs.20.25Option B

For 1st year S.I = C.I.

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is Rs.200

i.e S.I on the principal for 1 year is Rs.200

Principle = Rs.(100*200)/(8*1) = Rs.2500

Amount for 2 years, compounded half-yearly Rs.[2500*(1+4/100)^4] = Rs.2924.4

C.I = Rs.424.64

Also, S.I = Rs.(2500*8*2/100) = Rs.400

S.I. = Rs.2500*8*2/100 = Rs.400

Required Difference = C.I – S.I = Rs. (424.64 – 400) = Rs.24.64 - A letter is takenout at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. Find the probability that they are the same letter.
17/8819/9011/8615/8719/100Option B

ASSISTANT == AAINSSSTT

STATISTICS == ACIISSSTTT

Here, N and C are not common and same letters can be A, I, S, T.

Therefore Probability of choosing A = 2C1/9C1×1C1/10C1 = 1/45

Probability of choosing I = 1/9C1×2C1/10C1 = 1/45

Probability of choosing S = 3C1/9C1×3C1/10C1 = 1/10

Probability of choosing T = 2C1/9C1×3C1/10C1 = 1/15

Hence, Required probability = 1/45+1/45+1/10+1/15 = 19/90