IBM Aptitude interview Questions
International Business Machines Corporation(IBM) is an American multinational technology company headquartered in Armonk, New York, with operations in over 170 countries
IBM conducts 3-4 rounds to select fresher as Software Developer in their organization.
- Aptitude test
- Technical interview
- HR interview
In this Post we’ll see the Aptitude interview questions asked at IBM
1) If 20 men can construct a divider of 112 meters in length in 6 days, what length of a comparable divider can be worked by 25 men in 3 days?
- 69 meters.
- 58 meters.
- 70 meters.
- 76 meters
Answer: 3. 70 Meters.
Solution:
20 men in 6 days can build 112 meters
25 men in 30 days can build=112*(25/20)*(3/6)
= 70 meters
2) In a race of 600 meters, A can beat B by 60 meters and in a race of 500 meters; B can beat C by 50 meters. By what number of meters will A beat C in a race of 400 meters?
- 76 meters.
- 89 meters.
- 71 meters.
- 84 meters.
Answer: 1. 76 Meters
Solution:
Let’s assume A finishes the 600 m race in 60 sec, then
600/60 = 10 m/sec is his speed
B traveled (600-60 = 540 m in 60 sec, therefore
540/60 = 9 m/sec is B’s speed
“in a race of 500 metres, B can beat C by 50 metres.”
500/9 = 55.56 sec is B’s time to finish a 500 m race
C traveled 500-50 = 450 m in 55.56 sec, therefore
450/55.56 = 8.1 m/sec is C’s speed
By how many will A beat C in a race of 400 metres?
400/10 = 40 sec for A to run a 400 m race
C will travel 8.1*40 = 324 m in 40 sec therefore
C will be 400-324 = 76 m behind when A crosses the finish line
3) On the off chance that the accumulated dividends on a specific total of cash for a long time at 10% for each annum be Rs. 993, what might be the basic intrigue?
- Rs.840
- Rs.590
- Rs.695
- Rs.900
Answer: 4. Rs. 900
Solution:
Let P = Principal
A – Amount
We have a = P (1 + R/100)3 and CI = A – P
ATQ 993 = P (1 + R/100)3 – P
P = Rs 3000/ –
Presently SI @ 10% on Rs 3000/ – for 3 yrs = (3000 x 10 x 3)/100
= Rs 900/ –
4) What yearly portion will release an obligation of Rs. 4600 due in 4 years at 10% straightforward intrigue?
- 1000
- 1330
- 1600
- None of these
Answer: 1. 1000
Solution:
Give the yearly portion a chance to be Rs.100. The principal portion will be paid one year from now i.e. 3 years before it is in reality due. The second portion will be paid a long time from now i.e. 2 years before it is in reality due.
The third portion will be paid 1 year before it is in reality due.
The fourth portion will be paid on the day the sum is in reality due.
On the main portion, the intrigue will be paid for a long time, on the second for a long time, on the third for 1 year, on the fourth for multi-year. Altogether an enthusiasm for a long time will be paid (3 + 2 + 1 + 0) on Rs. 100 @ 10%. Intrigue = (100 × 6 × 10)/100 = Rs. 60 and the essential is Rs 100 × 4 = Rs 400. The aggregate credit that can be released is Rs. 400 + 60 = Rs. 460. Here the system of Chain Rule will be connected. I.e. for Rs. 460 the portion required is Rs. 100, for Rs. 4600 the portion required is 4600 × 100/460 = Rs. 1000.
5) A number whose fifth part expanded by 5 is equivalent to its fourth part lessened by 5, is
- 140.
- 180.
- 200.
- 270
Answer: 3. 200.
Solution:
X/5 + 5 = x/4 – 5
⇒ x/5 – x/4 = 10
X/20 = 10
⇒ x = 200
6) Two numbers are with the end goal that the proportion between them is 3:5, however in the event that each is expanded by 10, the proportion between them ends up 5:7. The numbers are
- 7, 5
- 7, 12
- 13, 29
- 15, 25
Answer: 4. 15, 25
Solution:
NO. Are in the ratio of 3:5
Let no be 3x and 5x
(3x+10): (5x+10) =5:7
X=5
NO are (15, 25)
7) A man pushes downstream 30 km and upstream 18 km, taking 5 hours each time. What is the speed of the stream (current)?
- 1.2 KM/HR
- 1.7 KM/HR
- 2.8 KM/HR
- 4.8 KM/HR
Answer: 1. 1.2 KM/HR
Solution:
Let x=speed of boat and y=speed of current
=30/ (x+y)=18/(x-y)=5 by solving y=1.2 km/hr
8) A train 125 meter long is running at 50 km/hr. In what time will it pass a man running at 5 km/hr in a similar bearing in which the train is going?
- 15 sec
- 10 sec
- 60 sec
- 55 sec
Answer: 2. 10 sec.
Solution:
Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s
Time=125/12.5=10sec
9) A is twice as fast as B is thrice as fast as C. The journey covered by C in 42 minutes, what will be covered by A is
- 21 MIN
- 64 MIN
- 17 MIN
- 40 MIN
Answer: 3. 17 MIN.
Solution:
B is thrice as fast as C
C covered in 42 minutes
B covered in 42/3=14 min
A is twice as fast as B
A covers in 14*(1/2) = 7 min
10) A can complete a work in 40 days and B in 28 days. In the event that A and B together take every necessary step, at that point roughly in how long will a similar function be finished?
- 17 days
- 14 days
- 16 days
- 29 days
Answer: 3. 16 days
Solution:
A’s 1day’s work = 1/40
B’s 1day’s work = 1/28
They can cooperate in = 1/40 + 1/28 = 16 days (estimate)
11) Teena is more youthful than Rani by 6 years. On the off chance that the proportion of their ages is 6:8, discover the time of Teena:
- 18 years
- 16 years
- 17 years
- 19 years
Answer: 1. 18 years
Solution:
On the off chance that Rani Age is x, at that point Teena age is x-6,
So (x-6)/x = 6/8
=> 8x-48 = 6x
=> 2x = 48
=> x = 24
So, Teena age is 24-6 = 18 years
12) A man purchases a book for Rs.29.50 and offers it for Rs 31.10. Discover his gain percent.
- 8.1%
- 5.4%
- 9.8%
- 2.4%
Answer: 2. 5.4%
Solution:
So we have C.P. = 29.50
S.P. = 31.10
Gain = 31.10 – 29.50 = Rs. 1.6
Gain %=( Gain/Cost*100)%
= (1.6/29.50*100)%=5.4%
13) Think about the arrangement: 464, 232, 240, 120, ____, 64. What number should fill the clear?
- 127
- 128
- 138
- 126
Ans. 2. 128
Step-by-step explanation:
Given
Think about the arrangement: 464, 232, 240, 120, ____, 64. What number should fill the clear?
- First consider the numbers given in the series
- 464, 232, 240, 120,…. , 64
- Consider the number 464. Divide by 2 we get 232
- Add 8 to the given number, so it will be 232 + 8 = 240
- Again divide the number 240 / 2 we get 120
- Add 8 to the number, so 120 + 8 = 128
- Again divide the number 128 / 2 we get 64
- So the process is first divide by 2 and add 8 to the number obtained.
- So the missing number is 128
14) Look at the series: A4, __, C16, D32, E64. What number should fill the blank?
- B16
- D4
- B8
- B10
Answer: 3
Solution:
The letters Increase by 1; the numbers are duplicated by 2.
15) A quick typist can type some issue in 2 hours and a moderate typist can type the same in 3 hours. In the event that both kinds consolidate, in what amount of time will they wrap up?
- 2.12 hr
- 1 .29 hr
- 1.12 hr
- 1.20 hr.
Answer: 3
Solution:
The quick typist’s work done in 1 hr = 1/2 The moderate typist’s work done in 1 hr = 1/3 If they work to join, work is done in 1 hr = 1/2+1/3 = 5/6 So, the work will be finished in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min
16) Two trains running in inverse ways cross a man remaining on the stage in 27 seconds and 17 seconds separately and they cross each other in 23 seconds. The ratio of their speed is:
- 2: 3
- 3: 2
- 3: 6
- None of these
Answer: B
17) What is the aggregate of all numbers somewhere in the range of 100 and 1000 which are distinct by 14?
- 353936
- 35392
- 35372
- 35322
Answer: 2
Solution:
The number nearest to 100 which is more noteworthy than 100 and divisible by 14 is 112, which is the principal term of the arrangement which must be summed. The number nearest to 1000 which is under 1000 and distinct by 14 is 994, which is the last term of the arrangement. 112 + 126 + …. + 994 = 14(8+9+ … + 71) = 35392
18) Gavaskar’s average in his initial 50 innings was 50. After the 51st innings, his average was 51. What number of runs did he score in his 51st inning? (assuming that he lost his wicket in his 51st innings)
- 101
- 103
- 98
- 100
Answer: 1. 101
Solution:
Add up to score after 50 innings = 50*50 = 2500 Total score after 51 innings = 51*51 = 2601. So, runs made in the 51st innings = 2601-2500 = 101 If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings.
19) I drove 60 km at 30 kmph and after that an extra 60 km at 50 km ph. Register my normal speed over my 120 km.
- 34 KM/HR
- 37 ½ KM/HR
- 42 KM/HR
- 35 ½ KM/HR
Answer: 2
Solution:
37 ½ km ph Solution: Time required for the initial 60 km = 120 min. The Time required for the second 60 km = 72 min. Add up to time required = 192 min Average speed = (60*120)/192 = 37 1/2
20) A can complete a specific work in a similar time in which B and C together can do it. In the event that A and B together could do it in 10 days and C alone in 50 days, at that point B alone could do it in.
- 14 DAYS
- 15 DAYS
- 20 DAYS
- 25 DAYS
Answer: 4
21) A boat can go at a speed of 13 km/hr in still water. On the off chance that the speed of the stream is 4 km/hr, discover the time taken by the vessel to go 68 km downstream.
- 2 HOURS
- 3 HOURS
- 4 HOURS
- 5 HOURS
Answer: 3
Solution:
Speed downstream = (13 + 4) km/hr = 17 km/hr. Time taken to movement 68 km downstream = (68/17) hrs = 4 hrs
22) A, B, C are the partner in a business. During a specific year. A got 33% of the benefit. B got one-fourth of the benefit and C got the rest of the Rs. 5000. What amount of measure of cash did A get?
- Rs. 1000
- Rs. 2000
- Rs. 4000
- Rs. 5000
Answer: 3
Solution:
Lets expect Total benefit x
x * (1-1/3-1/4) = 5000
=> x*(12-4-3)/12 = 5000
x = 5000*12/5 = Rs. 12000
so An’s offer = Rs. (1/3*12000) = Rs. 4000
23) A man possesses 2/3 of the statistical surveying bureau business and offers 3/4 of his offers for Rs. 75000. What is the value of Business?
- 12000
- 150000
- 160000
- 170000
Answer: 2
Solution:
3/4 of his offer = 75000
So his offer = 100000.
2/3 of business esteem = 100000
So add up to esteem = 150000
24) From its total company, A business organization burned through Rs.20, 000 for publicizing, half of the rest of commissions and had Rs.6000 cleared out. What was its aggregate salary?
- 32000
- 17000
- 39000
- 47000
Answer: 1
Solution:
Let add up to salary is X
X=20,000+(X-20,000/2) +6000
X-X/2=20,000-10,000+6000
X/2=16,000
X=32,000
25) Nirmal and Kapil began a business contributing Rs. 9000 and Rs. 12000 separately. Following a half year, Kapil pulled back portion of his speculation. In the event that following a year, the aggregate benefit was Rs. 4600, what was Kapil’s share initially?
- Rs 2300
- Rs 2800
- Rs 3500
- Rs 2200
Answer: 1 Rs. 2300
Solution:
Nirmal: Kapil = 9000*12:(12000*6+6000*6) = 1:1
Kapil share = Rs. [4600 *(1/2)) = Rs. 2300
26) Anirudh, Harish, and Sahil put a sum of Rs.1, 35,000 in the proportion 5:6:4 Anirudh contributed has the capital for 8 months. Harish contributed for a half year and Sahil contributed for 4 months. On the off chance that they acquire a benefit of Rs.75, 900, then what is the offer of Sahil in the Profit?
- Rs. 13200
- Rs. 15700
- Rs.14200
- Rs. 15800
Answer: 2. Rs.13200
Solution:
Anirudh contribute for 8 months, Harish contributed for 6 and
sahil for 4 months in the proportion of 5:6:4
so proportion = 5*8 : 6*6 : 4*4
=> 40:36:16
=> 10:9:4
So sahil’s profit= (4/23)*75900 = 13200
27) A begins riding his bicycle at 10 am with a speed of 20kmph and B likewise begins at 10 am with a speed of 40kmph from a similar point in a similar way. Returns south at 12 o’clock and B turns north at 11 am. What will be the Distance between A and B at 2 pm?
- 250 km
- 160 km
- 170 km
- 145.6 km
Answer: 2. 160 km
Solution:
At 12 O’clock, A cover 40km and on the opposite side B at 11 o clock cover 40km, again they went towards each other (which is really the separation between them), that is A needs to make a trip 2hr (From 12 to 2 at 20km/hr.) i.e. 2*20=40km and opposite side B needs to Travelled out 3hr (From 11 to 2 at 40km/hr.) i.e. 3*40=120Km.
At that point Then the total distance traveled by them is the Actual distance between them i.e. 40+120= 160Km (Ans.)
28) 60 liters of diesel is required to movement 600 km utilizing an 800 cc motor. In the event that the volume of diesel required to cover a separation changes specifically as the limit of the motor, at that point what number of liters of diesel is required to movement 800 km utilizing 1200 cc motor?
- 90 liters.
- 100 liters.
- 120 liters.
- None of these
Answer: 3 120 liters.
Solution:
Let keep 800 cc steady and compute the measure of diesel for 800 km
800*60/600=80 liters.
Presently, ascertain diesel required for new separation i.e. 800 km,
80*1200/800=120 liters.
29) If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance traveled by him is:
- 50 km
- 56 km
- 70 km
- 80 km
Answer: 1. 50 km
Solution:
Let the real separation voyaged be x km.
At that point, x/10 = (x + 20)/14.
=> 14x = 10x + 200
=> 4x = 200.
=> x = 50 km.
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