# Ibm Aptitude Placement Papers - Ibm Aptitude Interview Questions and Answers updated on 06.Dec.2023

3/4 of his offer = 75000

So his offer = 100000.

2/3 of business esteem = 100000

So add up to esteem = 150000

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.

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

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.

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

Let the real separation voyaged be x km.

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

=> 14x = 10x + 200

=> 4x = 200.

=> x = 50 km.

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.

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

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

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

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

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.

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%

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

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

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

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

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

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

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

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/ -

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

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

Time=125/12.5=10sec

20 men in 6 days can build 112 meters

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

= 70 meters

X/5 + 5 = x/4 - 5

⇒ x/5 - x/4 = 10

X/20 = 10

⇒ x = 200

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)

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

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