Abb Group Aptitude Placement Papers - Abb Group Aptitude Interview Questions and Answers updated on 16.Aug.2022

When a nos ends with 1 its last digit will be 1.

Now for the 2nd last digit the short cut is

1021-tenths place digit*unit place digit of the power= 2(1) = 2

similarly, for the second no 3081 it is 8(1) = 8

so the last two digits are 21+81=102.

Therefore, last 2 digits is: 02.

In decimal number system; 137 + 276 = 413 but here its 435 (> 413) so the base system should be less than 10 and as the highest digit in the sum is 7 so the base must be greater than 7.

Add the LSB; 7+6 = 5 (there must be a carry) 

So 7 + 6 = 5 + 8(1 carry is forwarded) and hence the it is in octal number system.

Therefore: 731 + 672 = 1623.

Let the cost price be 100 per 1 kg

As he will sell 1 kg in 105 but due to error in weighing stones he will sell only 900 grams in 105 but he has paid 900*(100/1000) =90 rs for 900 grams.

Therefore, net profit= Rs (105-90) = Rs 15

% percentage= (15/90) *100% =16.67%

Selling Price of Article = Rs. 1000

For 1st merchant, 10% profit is on C.P or C.P + Profit = S.P

Therefore 1.1 * C.P = Rs.1000 or C.P = Rs. 909.1 and Profit = Rs. 90.9

For 2nd merchant, 10% profit is on S.P i.e. Profit = 0.10 * Rs 1000 = Rs. 100

so the profit of 2nd merchant is higher than the 1st merchant by Rs. (100 – 90.9) = Rs. 9.1 (approx.).

Since, 7150 = 2×5^2×11×13.

So, there are 4 distinct prime numbers that are below 100. 

The usual way to solve these type of questions is to put x = 0 once and find y coordinate. This would represent the point where the line cuts the Y axis.

Similarly put y = 0 once and find x coordinate. This would represent the point where the line cuts the X axis. Then join these points and you will get the graph of the line.

So when we put x = 0 we get y = 4.

When we put y = 0 we get x = 6.

So when we join these points we see that we get a line in 1st quadrant, which when extended both sides would go to 4th and 2nd quadrants.

As B moves 3 steps forward and then 1 step backward so in total 4 seconds he moves only 2 steps forward so in 116 seconds he moves 58 steps forward now in next 2 seconds he moves 2 steps so in 118 seconds he moves total 60 steps forward.

So no. of steps required to reach the top of the escalator is 60.

now let d escalator moves a steps per second so in 4 seconds B moves 2 steps (3steps forward and 1 step backward)in these 4 sec. escalator moves 4a step so in 4 sec. B moves a total of 2+4a step.

so in 40 second total move=10*(2+4a)

so, 10*(2+4a) =60

hence a=1step/sec. 

A leap year has 366 days, therefore 52 weeks (i.e. 52 Friday’s) + 2 days.

So the probability of 53 Fridays = 2/7.

Observe the sequence:

5 * 3 = 15

51 + 2 = 53; 53 * 3 = 159; 159 + 2 = 161

So _ will be 15 + 2 = 17 (also 51/3 = 17).

Given series: 11, 23, 47, 83, 131

1st number: 11

2nd number: 11+12*1=23

3rd number: 23+12*2=47

4th number: 47+12*3=83

5th number: 83+12*5=131

6th number: 131+5*12=191.

As the square of sum of digits is 18 more than that of the number, so the square of the sum of digit must be greater than or equal to 28 (18+10 as 10 is the smallest 2-digit number) and should be less than or equal to 117 (18+99 as 99 is the largest two-digit number)

So the possible squares are:

36 and hence the possible number can be (36-18) =18 or (1+8)2 = 81 =! 36 and hence not possible.

49 and hence the possible number can be (49-18) =31 or (3+1)2 = 16 =! 49 and hence not possible.

64 and hence the possible number can be (64-18) =46 or (4+6)2 = 100 = 81 =! 64 and hence not possible.

81 and hence the possible number can be (81-18) =63 or (6+3)2 = 81 = 81 and hence possible.

100 and hence the possible number can be (100-18) =82 or (8+2)2 = 100 = 100 and hence not possible.

So only 2 possible values i.e. 63 and 82.

Those 5 who refuses to sit in the upper deck will sit in lower deck

So total lower deck remains: 2

Those 8 who refuses to sit in the lower deck will sit in upper deck

So total upper deck sit remains: 5

These 7 people can sit in 5 upper deck and 2 lower deck in: 7c5 * 2c2 ways i.e. 21 ways.

IJKLMNO if O is Saturday then I will be Sunday and K will be Tuesday.

Cost price: Rs 1920

Profit = 20% = Rs 1920 x 0.20 = 384

Therefore, Selling Price = Rs 1920 + 384 = 2304.

The difference between nos are: 2 , 3 , _ , _ ,11

The differences are prime nos i.e 2, 3, 5, 7, 11 so the next difference will be 13

Therefore, nos are: (5 + 5) = 10 & (28 + 13) = 41.

Diameter = 50 cm hence radius(r) = 50/2 cm

Therefore; Circumference of cycle = 2*22/7*r

As number of revolutions per minute = 420

Therefore; Speed = 2*(22/7) *[25/(100*1000)]*60*420 km/hr

= 396/10 km/hr

= 39.6 km/hr.

Let after t time two cars will met.

So A will travel distance of 40t with 40kmph

B will travel the distance of 60t with 60kmph

And also A is ahead 80 km (40*2=80) from B

=> 60t - 40t = 80 => t = 4hrs

Also time taken by B to cover 9kms more is 9/60 = 9mins

Additional distance is 9 min

For additional time= (9/20) *60=27 min

So correct wer = 4hrs 27 min

= 4 (27/60) hrs = 4.45 hrs

Assume tank capacity is 45 Liters.

Given that the first pipe fills the tank in 9 hours. So its capacity is 45 / 9 = 5 Liters/ Hour. 

Second pipe fills the tank in 5 hours. So its capacity is 45 / 5 = 9 Liters/Hour. 

If both pipes are opened together, then combined capacity is 14 liters/hour. 

To fill a tank of capacity 45 liters, both pipes takes 45 / 14 = 3.21 Hours.

Find the square root of 6352@It will be 25@_ _ so the nearest perfect square is 252^2 = 63504

So the nos to be subtracted is: (63520 - 63504) = 16.

S H U V a N K (A H K N S U V)

Nos of words starting with A: 6! = 720

Nos of words starting with AH: 5! = 120

Nos of words starting with AHK: 4! = 24

Nos of words starting with AHN: 4! = 24

Nos of words starting with AHSK: 3! = 6

Nos of words starting with AHSN: 3! = 6

24+24+6 = 54, so the next word (55th) will be the first word starting form AHSN and will be AHSNUV.

Let the speed of old man be: x m/min and that of young man be: y m/min

If the distance of the office be D meter, then A/c: D = 30x = 20y or y = 1.5x

Let young man catches old man after ‘t’ mins.

So distance travelled by young man is ‘t’min = ty = 1.5tx

And distance travelled by old man in ‘t+5’min = (t+5) x = tx + 5x

Therefore, A/c: 1.5tx = tx + 5x or 0.5tx = 5x or t = 10min

So young man catches the old man at 10:05 AM + 10min i.e 10:15 min

Alternate method

Old man takes 30 min i.e. he travels from 10:00 AM to 10:30 AM

Young man takes 20 min i.e. he travels from 10:05 AM to 10:25 AM

From symmetry; they will meet in mid-way of the journey at 10:15 AM.

As A and B completed a work together in 5 days

Work done by them in a day (A + B), 1/5

with twice the speed of A and half the speed of B , they completes the work in 4 days,

so, their work per day (2A + B/2) = 1/4

by solving both the eqns: 2(2A+B/2) – (A+B) = 3A = 2*1/4 – 1/5 = 3/10

or 1-day work of A = 1/10

so A alone can complete the work in 10 days.

The probability of dice showing an odd nos = ½ and

the probability of coin showing head = ½;

so the overall probability is: ½ * ½ = ¼.

As the number is divisible by 5, the unit digit of 3-digit number must be 5.

Rest two digits can be selected in 5c1 * 4c1 = 20 ways.

When x = 6, (4 * 6)/3 + 2P = 12

⇒ 8 + 2P = 12

⇒ 2P = 12 – 8 = 4

⇒ P = 2

Let the nos be: abc

As sum of the digit is @Therefore, a+b+c=17----(1)

Also sum of square of digits is 109 i.e a^2+b^2+c^2=109----(2)

Also, (100a+10b+c) – 495 = (100c+10b+a)

or, (100a – a) + (10b – 10b) + (c – 100c) = 495

or, 99 (a-c) =495 or (a - c) = 495

The possible combinations are (6,1) (7,2) (8,3), (9,4)

For 1st combination (6,1); b = (17 – 6 - 1) = 10 which is not possible

For 2nd combination (7,2); b = (17 – 7 - 2) = 8 but a^2+b^2+c^2 =! 109 so not possible

For 3rd combination (8,3) ; b = (17 – 8 - 3) = 6 also a^2+b^2+c^2 = 109 so it is possible

so,863 is the wer.