# 300+ TOP Ibm Aptitude Interview Questions and Answers

Q1. A Man Possesses 2/3 Of The Statistical Surveying Bureau Business And Offers 3/4 Of His Offers For Rs. 750

3/4 of his offer = 75000

So his offer = 100000.

2/3 of business esteem = 100000

So add up to esteem = 150000

Q2. 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 T

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.

Q3. 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 A

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

Q4. What Annual Installment Will Discharge A Debt Of Rs. 4600 Due In 4 Years At 10% Simple Interest ?

Let the annual instalment be Rs. @The first instalment will be paid one year from now i.e. 3 years before it is actually due. The second instalment will be paid two years from now i.e. 2 years before it is actually due.

The third instalment will be paid 1 year before it is actually due.

The fourth instalment will be paid on the day the amount is actually due.

On the first instalment the interest will be paid for 3 years, on the second for 2 years, on the third for 1 year, on the fourth for 0 year. In total an interest for 6 years will be paid (3 + 2 + 1 + 0) on Rs. 100 @ 10%. Interest = (100 × 6 × 10)/100 = Rs. 60 and the principal is Rs 100 × 4 = Rs 4@The total loan that can be discharged is Rs. 400 + 60 = Rs. 46@Here the technique of Chain Rule will be applied. I.e. for Rs. 460 the instalment required is Rs. 100, for Rs. 4600 the instalment required is 4600 × 100/460 = Rs. 1000.

Q5. 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

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

Q6. 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:

Let the real separation voyaged be x km.

At that point, x/10 = (x + 20)/14.

=> 14x = 10x + 200

=> 4x = 200.

=> x = 50 km.

Q7. Gavaskar’s Average In His Initial 50 Innings Was 5

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.

Q8. A Man Pushes Downstream 30 Km And Upstream 18 Km, Taking 5 Hours Each Time. What Is The Speed Of The Stream (current) ?

Let x=speed of boat and y=speed of current

=30/ (x+y)=18/(x-y)=5 by solving y=1.2 km/hr

Q9. 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

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

Q10. 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. 50

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

Q11. In A Race Of 600 Metres, A Can Beat B By 60 Metres And In A Race Of 500 Metres, B Can Beat C By 50 Metres. By How Many Metres Will A Beat C In A Race Of 400 Metres ?

A runs B runs C runs

600 metres race 600m 540 m

500 metres race 500 m 450m

Combing ratio A runs B runs C runs

300metres – 2700meters – 2430metres

Unitary A runs B runs C runs

Method 400mtres – 360 metres – 324 metres

∴ A beats C by 400-324 = 76 metres.

Q12. A Man Purchases A Book For Rs.29.50 And Offers It For Rs 31.

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%

Q13. 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

Nirmal: Kapil = 9000*12:(12000*6+6000*6) = 1:1

Kapil share = Rs. [4600 *(1/2)) = Rs. 2300

Q14. 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.

Speed downstream = (13 + 4) km/hr = 17 km/hr.

Time taken to movement 68 km downstream = (68/17) hrs = 4 hrs

Q15. Look At The Series: A4, __, C16, D32, E6

The letters Increase by 1; the numbers are duplicated by 2.

Q16. 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.

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

Q17. 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:

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

Q18. 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?

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/ –

Q19. 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 ?

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

Q20. 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 ?

Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s

Time=125/12.5=10sec

Q21. 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 ?

20 men in 6 days can build 112 meters

25 men in 30 days can build=112*(25/20)*(3/6)

= 70 meters

Q22. A Number Whose Fifth Part Expanded By 5 Is Equivalent To Its Fourth Part Lessened By 5, Is

X/5 + 5 = x/4 – 5

⇒ x/5 – x/4 = 10

X/20 = 10

⇒ x = 200

Q23. 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 ?

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)

Q24. What Is The Aggregate Of All Numbers Somewhere In The Range Of 100 And 1000 Which Are Distinct By 14 ?

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

Q25. 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 ?

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