# 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