IBM Placement Papers And exam pattern

IBM Placement Papers And exam pattern

IBM Placement Papers And exam pattern:

Hello Everyone !

This is Zed Aay and If you are called for a off campus drive with IBM then this post is for you.

As usual expect huge crowd to appear for Written test. The selection process comprises of :

1. Written Test (Aptitude+ Number Series).
2. English Assessment Test. (Basically Business Writing).
3. Face to face Interview (Technical+ HR).

Points to remember about:

1. Aptitude question would be average from Simple Interest, Time and work, Speed time distance, Profit and loss. And few more easy topics. But the twist is you will be given fixed time of 2 minutes 15 seconds to attempt each question. After this allotted time the question window will disappear regardless of your attempt. So keep an eye on time.

Maximum Eliminations are done here. If you are through, 50% you get the job.

2. English assessment is not an ordinary English test. It is entirely a business letter type. For this practise paper given in other posts of IndiaBix. Another huge eliminations are done here. And yes good news is if you are through this round, 95%u get the job.

3. Face to face interview is a piece of cake. Believe me IBM HRs are the best HRs in the world. You will feel so comfortable talking to them.

So there is no point in being nervous. Just answer every confidently and try to be you. Don’t try to over emphasize your weaknesses and talk as much as possible. Try to be interactive inspite of giving one word answer.

This interview is meant to eliminate the dumbos only. So don’t worry. You Are in.

IBM Placement Paper Aptitude Test:

1. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
A. 4 days
B. 5 days
C. 6 days
D. 7 days
Explanation:
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2.
Solving these two equations, we get: x = 1/100 and y = 1/100.
(15 men + 20 boy)’s 1 day’s work = (15/100 + 20/200) = 1/4.
.’. 15 men and 20 boys can do the work in 4 days.

2. The ratio between the speeds of two trains is 7: 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A. 70 km/hr.
B. 75 km/hr.
C. 84 km/hr.
D. 87.5 km/hr.
Explanation:
Let the speed of two trains be 7x and 8x km/hr.
Then, 8x = (400/4) = 100.
=>x = 100 = 12.5
.’. Speed of first train = (7 x 12.5) km/hr. = 87.5 km/hr.

3. A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:
A. 4 days
B. 6 days
C. 8 days
D. 12 days
Explanation:
Suppose A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2.
=> 6/x = 1/2.
=> x = 12.
So, B takes (12/2) = 6 days to finish the work.

4. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
A. 17 kg
B. 20 kg
C. 26 kg
D. 31 kg
Explanation:
Let A, B, C represent their respective weights.
Then, we have:
A + B + C = (45 x 3) = 135 …. (i)
A + B = (40 x 2) = 80 …. (ii)
B + C = (43 x 2) = 86 …. (iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv)
Subtracting (i) from (iv), we get: B = 31.
=> B’s weight = 31 kg.

5. Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
A. 3.5
B. 4.5
C. 5.6
D. 6.5
Explanation:
Cost Price of 1 toy = Rs.375/12 = Rs. 31.25.
Selling Price of 1 toy = Rs. 33
So, Gain = Rs. (33 – 31.25) = Rs. 1.75
Profit % = (1.75/31.25 x 100) % = 28/5% = 5.6%.

6. A number consists of two digits. If the digit’s interchange places and the new number is added to the original number, then the resulting number will be divisible by:
A. 3
B. 5
C. 9
D. 11
Explanation:
Let the ten’s digit be x and unit’s digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
=> (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

7. A boat covers a certain distance downstream in 1 hour, while it comes back in 1 hours. If the speed of the stream be 3 kmph, what is the speed of the boat in still water?
A. 12 kmph
B. 13 kmph
C. 14 kmph
D. 15 kmph
E. None of these
Explanation:
Let the speed of the boat in still water be x kmph. Then,
Speed downstream = (x + 3) kmph,
Speed upstream = (x – 3) kmph.
.’. (x + 3) x 1 = (x – 3) x 3/2
=> 2x + 6 = 3x – 9
=> x = 15 kmph.

8. The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5: 8, then the second number is:
A. 20
B. 30
C. 48
D. 58
Explanation:
Let the three parts be A, B, C. Then,
A: B = 2: 3 and B: C = 5: 8
=> (5 x 3/5): (8 x 3/5) = 3: 24/5
=> A: B: C = 2: 3: 24/5 = 10: 15: 24
=> B = 98 x 15/49 = 30.

9. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
A. 34
B. 40
C. 68
D. 88
Explanation:
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

10. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A. 6 hours
B. 10 hours
C. 15 hours
D. 30 hours
Explanation:
Suppose, first pipe alone takes x hours to fill the tank.
Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.
.’. 1/x + 1/(x-5) = 1/(x-9).
=> (x – 5 + x)/x(x-5) = 1/(x-9).
=> (2x – 5) (x – 9) = x (x – 5).
=> x^2 – 18x + 45 = 0.
=> (x – 15) (x – 3) = 0.
=> x = 15. [neglecting x = 3]