1. Sunny has two bags (X & Y) that contain black and white balls.In the Bag ‘X’ there are 6 black and 8 white balls and in the Bag ‘Y’ there are 6 black and 6 white balls. One ball is drawn out from any of these two bags. What is the probability that the ball drawn is white ?
A. 13/28
B. 15/28
C. 17/28
D. 23/28
E. None of these
Sol. Total balls in bag X = 14, Total balls in bag Y = 12
Probability selectng white ball from bag X = 1/2(8c1/14c1) = 2/7
Probability selectng white ball from bag Y = 1/2(6c1/12c1) = 1/4
Total Probability = 2/7 + 1/4 =15/28.
2. 16 26 42 136 ?
A. 534
B. 524
C. 494
D. 484
E. None of these
Sol. The series is ×1+10, ×2-10, ×3+10, ×4-10
3. 5, 6, 16, 57, 244, 1245
A. 7500
B. 7506
C. 7502
D. 7508
E. None of these
Sol. 5*1 + 1^2 = 6
6*2 + 2^2 = 16
16*3 + 3^2 = 57
57*4 + 4^2 = 244
244*5 + 5^2 = 1245
1245*6 + 6^2 = 7506
4. 95, 60, 36, 21, 13, _
A. 8
B. 9
C. 10
D. 11
E. 12
Sol. 95 - 60 = 35
60 - 36 = 24
36 - 21 = 15
21 - 13 = 8
13 - 10 = 3
35 - 24 = 11
24 - 15 = 9
15 - 8 = 7
8 - 3 = 5
5. The product of the ages of suresh and ramesh is 240. If ramesh would have been twice his age then he will be 4 yrs elder to suresh. What is ramesh's age now ?
A. 16
B. 12
C. 24
D. 18
E. None of these
Sol. Now assume age of suresh to be 'x'
And assume age of ramesh to be 'y'
According to the question, xy = 240
Also given that, 2y = x + 4
Solving we get, y = 12 yrs.
6. If a car travels a distance of 300 km in 8 hours, partly at a speed of 40 kmh-1 and partly at 30 kmh-1, what is the distance it travelled at the speed of 40 kmh-1?
A. 200 km
B. 210 km
C. 220 km
D. 240 km
E. None of these
Sol. Lets assume at 40 km/hr the car covers 'x' km
Then at 30 km/hr it will cover '300-x' km
According tot he question, x/40 + (300-x)/30 = 8
Solving we get, x = 240 km.
7. 12, 19, 26, 93, 356, _
A. 1800
B. 1795
C. 1786
D. 1805
E. None of these
Sol. 12*1 + (1*7) = 19
19*2 - (2*6) = 26
26*3 + (3*5) = 93
93*4 - (4*4) = 356
356*5 + (5*3) = 1795
8. 25, 32, 53, 88, 137, _
A. 195
B. 200
C. 205
D. 210
E. None of these
Sol. 25 + (7*1) = 32
32 + (7*3) = 53
53 + (7*5) = 88
88 + (7*7) = 137
137 + (7*9) = 200
9. 3, 5, 13, 43, 177, _
A. 890
B. 889
C. 891
D. 892
E. None of these
Sol. 3*1 + 2 = 5
5*2 + 3 = 13
13*3 + 4 = 43
43*4 + 5 = 177
177*5 + 6 = 891
10. In a school of X students ,when Rahul was absent , the number of students could be divided into groups of 6 each If both Rahul and Varun were absent ,the remaining could be divided into groups of 6 each. What is the minimum possible value of X ?
A. 63
B. 37
C. 51
D. 42
E. None of these
Sol. If 1 Student is absent the no. of students should be in the form of 6a + 1
7, 13, 19, 25, 31, 37, 43 …
If 2 Students are absent the no. of students should be in the form of 7b + 2
9, 16, 23, 30, 37, 44 …
So take the common term as 37.
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