# Interview Questions.

Abb Group Aptitude Placement Papers - Abb Group Aptitude Interview Questions and Answers

## Abb Group Aptitude Placement Papers - Abb Group Aptitude Interview Questions and Answers

Q1. Find Last Two Digit Of (1021^3921)+(3081^3921)?

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

Now for the 2d closing digit the quick cut is

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

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

so the remaining  digits are 21+81=102.

Therefore, closing 2 digits is: 02.

Q2. If Given Equation Is 137+276=435, How Much Is 731+672=.... Find The Result.

In decimal wide variety device; 137 + 276 = 413 but right here its 435 (> 413) so the base system should be less than 10 and as the best digit inside the sum is 7 so the base should be extra than 7.

Add the LSB; 7+6 = five (there ought to be a deliver)

So 7 + 6 = five + eight(1 convey is forwarded) and therefore the it's miles in octal quantity device.

Therefore: 731 + 672 = 1623.

Q3. Mani Sells Vegetables And He Marks Up The Prices At 5% Above His Cost Price. Also The Weighing Stones Used By Him Weigh Only 90% Of The Correct Weight. Find His Effective Percentage Of Mark-up.

Let the fee fee be one hundred in step with 1 kg

As he will sell 1 kg in a hundred and five however due to error in weighing stones he'll sell simplest 900 grams in one hundred and five however he has paid 900*(100/a thousand) =ninety rs for 900 grams.

Therefore, internet earnings= Rs (a hundred and five-ninety) = Rs 15

% percent= (15/ninety) *one hundred% =16.Sixty seven%

Q4. Two Merchants Sell An Article Each For Rs.1000.One Of Them Computes Profit As A % Of Cost Price, While The Second Calculates It Incorrectly As A % Of Selling Price. If Both Of Them Claim To Have Made

Selling Price of Article = Rs. One thousand

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

Therefore 1.1 * C.P = Rs.A thousand or C.P = Rs. 909.1 and Profit = Rs. Ninety.Nine

For 2nd merchant, 10% earnings is on S.P i.E. Profit = 0.10 * Rs one thousand = Rs. 100

so the income of 2d merchant is better than the 1st service provider through Rs. (one hundred – ninety.9) = Rs. 9.1 (approx.).

Q5. How Many Prime Numbers Between 1 And 100 Are Factors Of 7150?

Since, 7150 = 2×5^2×11×13.

So, there are four awesome prime numbers that are under 100.

Q6. The Straight Line 2x + 3y = 12 Passes Through:

The normal way to solve those kind of questions is to place x = zero once and find y coordinate. This might represent the factor wherein the road cuts the Y axis.

Similarly placed y = 0 once and find x coordinate. This could constitute the point wherein the road cuts the X axis. Then be part of those factors and you may get the graph of the road.

So while we put x = zero we get y = four.

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

So while we join these factors we see that we get a line in 1st quadrant, which whilst extended each sides would visit 4th and second quadrants.

Q7. B Moves By Taking three Steps Forward And 1 Step Backward (each Step In One Second ) He Walks Up A Stationary Escalator In 118 Sec. However On Moving Escalator He Takes forty Sec To Reach Top .Discover Speed

As B moves three steps ahead and then 1 step backward so in general four seconds he actions handiest 2 steps forward so in 116 seconds he actions 58 steps forward now in subsequent 2 seconds he actions 2 steps so in 118 seconds he actions overall 60 steps ahead.

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

Now let d escalator movements a steps in line with 2nd so in 4 seconds B actions 2 steps (3steps ahead and 1 step backward)in those four sec. Escalator moves 4a step so in four sec. B movements a complete of 2+4a step.

So in forty 2d general flow=10*(2+4a)

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

therefore a=1step/sec.

Q8. What Is The Chance That A Leap Year Selected At Random Contains 53 Fridays ?

A jump yr has three hundred and sixty six days, therefore 52 weeks (i.E. 52 Friday’s) + 2 days.

So the opportunity of fifty three Fridays = 2/7.

Q9. Three 15 _ fifty one 53 159 161

Observe the series:

5 * three = 15

fifty one + 2 = fifty three; fifty three * 3 = 159; 159 + 2 = 161

So _ may be 15 + 2 = 17 (additionally fifty one/three = 17).

Q10. A Motor Boat Covers A Certain Distance Downstream In 30 Minutes, While It Comes Back In 45 Minutes. If The Speed Of The Stream Is five Kmph What Is The Speed Of The Boat In Still Water?

Let the speed of boat in still water: x kmph

As distance is consistent; (x+five) *30=(x-five) *45

or, 2x+10=3x-15

x = 25 kmph

Q11. What Is The Next Numbers For The Given Series? Eleven 23 forty seven 83 131 ?

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

1st variety: 11

second quantity: eleven+12*1=23

3rd wide variety: 23+12*2=47

4th quantity: 47+12*three=eighty three

5th quantity: eighty three+12*five=131

sixth number: 131+five*12=191.

Q12. A Two Digit Number Is 18 Less Than The Square Of The Sum Of Its Digits. How Many Such Numbers Are There?

As the rectangular of sum of digits is eighteen greater than that of the wide variety, so the square of the sum of digit should be extra than or equal to 28 (18+10 as 10 is the smallest 2-digit wide variety) and have to be less than or same to 117 (18+99 as 99 is the most important -digit wide variety)

So the feasible squares are:

36 and therefore the possible wide variety can be (36-18) =18 or (1+eight)2 = eighty one =! 36 and for this reason no longer possible.

49 and hence the feasible wide variety may be (forty nine-18) =31 or (three+1)2 = sixteen =! Forty nine and for this reason no longer viable.

Sixty four and therefore the feasible range may be (sixty four-18) =forty six or (4+6)2 = one hundred = 81 =! 64 and as a result no longer feasible.

81 and as a result the possible variety may be (81-18) =sixty three or (6+three)2 = eighty one = eighty one and subsequently possible.

100 and therefore the feasible wide variety can be (100-18) =82 or (8+2)2 = 100 = 100 and consequently not possible.

So best 2 viable values i.E. Sixty three and eighty two.

Q13. 20 Passengers Are To Travelled By A Doubled Decked Bus Which Can Accommodate 13 In The Upper Deck And 7 In The Lower Deck. The Number Of Ways That They Can Be Distributed If 5 Refuse To Sit In The Upp

Those five who refuses to take a seat in the top deck will take a seat in lower deck

So overall lower deck stays: 2

Those 8 who refuses to sit within the lower deck will take a seat in upper deck

So overall higher deck sit stays: 5

These 7 people can take a seat in five higher deck and 2 decrease deck in: 7c5 * 2c2 ways i.E. 21 ways.

Q14. If Meeting O Is On Saturday, Then Meeting K Must Take Place On?

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

Q15. A Dealer Buys A Product At Rs.192

Cost price: Rs 1920

Profit = 20% = Rs 1920 x 0.20 = 384

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

Q16. Find The Missing Numbers In The Series: zero,2,5,?,17,28,?

The difference among nos are: 2 , three , _ , _ ,11

The variations are high nos i.E 2, three, 5, 7, 11 so the following difference will be thirteen

Therefore, nos are: (five + five) = 10 & (28 + 13) = forty one.

Q17. A Boy Is Cycling Such That The Wheel Of The Cycle Are Making 420 Revolutions Per Minute. If The Diameter Of The Wheel Is 50 Cm, Find The Speed Of The Boy.

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

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

As quantity of revolutions according to minute = 420

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

= 396/10 km/hr

= 39.6 km/hr.

Q18. Car A Leaves City C At five Pm And Drives At A Speed Of forty Kmph. 2 Hours Later Another Car B Leaves City C And Drives In The Same Direction As Car A. In How Much Time Will Car B Be 9 Km Ahead Of Car A. S

Let after t time two cars will met.

So A will tour distance of 40t with 40kmph

B will tour the space of 60t with 60kmph

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

Also time taken by using B to cowl 9kms extra is nine/60 = 9mins

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

So accurate wer = 4hrs 27 min

= 4 (27/60) hrs = 4.45 hrs

Q19. The Water From One Outlet, Flowing At A Constant Rate, Can Fill The Swimming Pool In nine Hours. The Water From Second Outlet, Flowing At A Constant Rate Can Fill Up The Same Pool In Approximately In five H

Assume tank capability is 45 Liters.

Given that the primary pipe fills the tank in nine hours. So its capacity is forty five / 9 = five Liters/ Hour.

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

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

To fill a tank of potential forty five liters, each pipes takes 45 / 14 = three.21 Hours.

Q20. The Least Number That Must Be Subtracted From 63520 To Make The Result A Perfect Square, Is:

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

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

Q21. 55th Word Of Shuvank In Dictionary ?

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

Nos of words starting with A: 6! = 720

Nos of phrases beginning with AH: 5! = 120

Nos of words starting with AHK: 4! = 24

Nos of words starting with AHN: four! = 24

Nos of words beginning with AHSK: 3! = 6

Nos of phrases starting with AHSN: 3! = 6

24+24+6 = fifty four, so the following word (fifty fifth) might be the primary word starting shape AHSN and might be AHSNUV.

Q22. 2 Workers, One Old And One Young, Live Together And Work At The Same Office. The Old Man Takes 30 Mins Whereas The Young Man Takes Only 20 Mins To Reach The Office. When Will The Young Man Catch Up Th

Let the velocity of vintage man be: x m/min and that of younger guy be: y m/min

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

Let younger man catches vintage man after ‘t’ minutes.

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

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

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

So younger man catches the antique man at 10:05 AM + 10min i.E 10:15 min

Alternate method

Old guy takes 30 min i.E. He travels from 10:00 AM to ten:30 AM

Young guy takes 20 min i.E. He travels from 10:05 AM to 10:25 AM

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

Q23. A And B Completed A Work Together In five Days. Had A Worked At Twice The Speed And B At Half The Speed, It Would Have Taken Them Four Days To Complete The Job. How Much Time Would It Take For A Alone To

As A and B finished a work together in five days

Work done with the aid of them in an afternoon (A + B), 1/five

with two times the speed of A and 1/2 the speed of B , they completes the work in four days,

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

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

or 1-day work of A = 1/10

so A alone can complete the paintings in 10 days.

Q24. A Die Is Rolled And A Coin Is Tossed. Find The Probability That The Die Shows An Odd Number And The Coin Shows A Head.

The probability of dice showing an peculiar nos = ½ and

the opportunity of coin displaying head = ½;

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

Q25. How Many three-digit Numbers Can Be Formed From The Digits 2,three,five,6,7 And nine Which Are Divisible By five And None Of The Digit Is Repeated.?

As the quantity is divisible through 5, the unit digit of three-digit range ought to be 5.

Rest two digits may be decided on in 5c1 * 4c1 = 20 ways.

Q26. If 4x/3 + 2p = 12 For What Value Of P, X = 6?

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

⇒ 8 + 2P = 12

⇒ 2P = 12 – eight = 4

⇒ P = 2

Q27. Sum Of Three-digit Number Is

Let the nos be: abc

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

Also sum of rectangular 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, ninety nine (a-c) =495 or (a - c) = 495

The viable combos are (6,1) (7,2) (eight,3), (9,four)

For 1st aggregate (6,1); b = (17 – 6 - 1) = 10 which isn't always viable

For second combination (7,2); b = (17 – 7 - 2) = 8 but a^2+b^2+c^2 =! 109 so no longer possible

For third mixture (8,three) ; b = (17 – eight - 3) = 6 also a^2+b^2+c^2 = 109 so it's far viable

so,863 is the wer.