Mixed Quantitative Aptitude Questions Set 84

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

1. 3,20,88,273,?,559
   551
   500
   554
   505
   517
   Option C
   (3+1)*5 = 20
   (20+2)*4 = 88
   (88+3)*3 = 273
   (273+4)*2 = 554
   (554+5)*1 = 559

    

2. 27,54,86,?,187,272
   128
   153
   122
   110
   111
   Option 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

    

3. 21,?,77,150,295,584
   42
   36
   50
   38
   40
   Option E
   21*2-2 = 40
   40*3-3 = 77
   77*4 – 4 = 150
   150*5 – 5 = 295
   295*6 – 6 = 584

    

4. 1,2,6,?,49,174
   20
   27
   30
   33
   35
   Option D
   1+1^3 = 2
   2 + 2^2 = 6
   6 + 3^3 = 33
   33+4^2 = 49
   49 + 5^3 = 174

    

5. 105,121,146,182,?,295
   231
   222
   252
   202
   220
   Option A
   —4^2—–5^2—-6^2—–7^2—-

    

6. 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 hours
   3 hours
   4 hours
   6 hours
   8 hours
   Option 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.

    

7. 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?
   1/3
   1/5
   1/8
   1/6
   1/4
   Option 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

    

8. 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 : 10
   2 : 6 : 9
   5 : 6 : 10
   7 : 6 : 8
   3 : 5 : 11
   Option 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

    

9. 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.22.52
   Rs.24.64
   Rs.18.35
   Rs.17.15
   Rs.20.25
   Option 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

    

10. A letter is takenout at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. Find the probability that they are the same letter.
    17/88
    19/90
    11/86
    15/87
    19/100
    Option 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