31+ ABB Group Aptitude Interview Questions And Answers

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ABB Group Aptitude Interview Questions

Q 1. If (x – 3)2 + (y – 5)2 + (z – 4)2 = 0, Then The Value Of X2/9 + Y2/25 + Z2/16 Is

(x – 3)2 + (y – 5)2 + (z – 4)2 = 0

⇒ x – 3 = 0 ⇒ x= 3

Y – 5 = 0 ⇒ y = 5

Z – 4 = 0 ⇒ z = 4

Q 2. 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 – 8 = 4

⇒ P = 2

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

The ordinary technique to remedy these kind of Qs is to place x = Zero as soon as and discover y coordinate. This would signify the purpose the place the road cuts the Y axis.

Similarly put y = Zero as soon as and discover x coordinate. This would signify the purpose the place the road cuts the X axis. Then be a part of these factors and you’ll get the graph of the road.

So after we put x = Zero we get y = 4.

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

So after we be a part of these factors we see that we get a line in 1st quadrant, which when prolonged each side would go to 4th and 2nd quadrants.

Q 4. In Δabc, ∠a + ∠b = 65°, ∠b + ∠c = 140°, Then Find ∠b.

∠A + ∠B = 65°

∴ ∠C = 180° – 65° = 115°

∠B + ∠C = 140°

∴ ∠B = 140° – 115° = 25°

Q 5. A Person Starts Writing All 4 Digits’ Numbers. How Many Times Had He Written The Digit 2?

Number of two‘s at unit’s place (from 100-1 to 999-2) = 900

Number of two‘s at tenths place (xy2z) xy will vary from 10 – 99 (90 nos) and z from 0-9 (10 nos) so total 90*10 = 900

Number of 2‘s at hundreds place(x2yz) x will vary from 1-9(9 ways) & yz from 00-99(100 ways) so total 9*100 =900

Number of 2‘s at thousands place(2xyz) xyz will vary from 000-999 so total = 1000

Therefore, total number of 2‘s = (900+900+900+1000) =3700.

Q 6. 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 The Old Man, If The Old Man Starts At 10.00am And The Young Man Starts At 10.05am?

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’ minutes.

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

And distance travelled by outdated man in ‘t+5’min = (t+5) x = tx + 5x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So younger man catches the outdated man at 10:05 AM + 10min i.e 10:15 min

Alternate methodology

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 may meet in mid-way of the journey at 10:15 AM.

Q 7. What Is The Next Numbers For The Given Series? 11 23 47 83 131 ?

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

1st quantity: 11

2nd quantity: 11+12*1=23

third quantity: 23+12*2=47

4th quantity: 47+12*3=83

fifth quantity: 83+12*5=131

sixth quantity: 131+5*12=191.

Q 8. What Is The Chance That A Leap Year Selected At Random Contains 53 Fridays ?

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

So the chance of 53 Fridays = 2/7.

Q 9. A Two Digit Number Is 18 Less Than The Square Of The Sum Of Its Digits. How Many Such Numbers Are There?

As the sq. of sum of digits is 18 greater than that of the quantity, so the sq. of the sum of digit should be higher than or equal to 28 (18+10 as 10 is the smallest 2-digit quantity) and ought to be lower than or equal to 117 (18+99 as 99 is the biggest two-digit quantity)

So the potential squares are:

36 and therefore the potential quantity may be (36-18) =18 or (1+8)2 = 81 =! 36 and therefore not potential.

49 and therefore the potential quantity may be (49-18) =31 or (3+1)2 = 16 =! 49 and therefore not potential.

64 and therefore the potential quantity may be (64-18) =46 or (4+6)2 = 100 = 81 =! 64 and therefore not potential.

81 and therefore the potential quantity may be (81-18) =63 or (6+3)2 = 81 = 81 and therefore potential.

100 and therefore the potential quantity may be (100-18) =82 or (8+2)2 = 100 = 100 and therefore not potential.

So solely 2 potential values i.e. 63 and 82.

Q 10. 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 therefore radius(r) = 50/2 cm

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

As variety 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.

Q 11. B Moves By Taking 3 Steps Forward And 1 Step Backward (every Step In One Second ) He Walks Up A Stationary Escalator In 118 Sec. However On Moving Escalator He Takes 40 Sec To Reach Top .discover Speed Of Escalator.

As B strikes Three steps ahead after which 1 step backward so in complete Four seconds he strikes solely 2 steps ahead so in 116 seconds he strikes 58 steps ahead now in subsequent 2 seconds he strikes 2 steps so in 118 seconds he strikes complete 60 steps ahead.

So no. of steps required to succeed in the highest of the escalator is 60.

now let d escalator strikes a steps per second so in Four seconds B strikes 2 steps (3steps ahead and 1 step backward)in these Four sec. escalator strikes 4a step so in Four sec. B strikes a complete of two+4a step.

so in 40 second complete transfer=10*(2+4a)

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

therefore a=1step/sec.

Q 12. A And B Completed A Work Together In 5 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 Do The Work?

As A and B accomplished a piece collectively in 5 days

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

with twice the pace of A and half the pace of B , they completes the work in Four days,

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

by fixing each 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 full the work in 10 days.

Q 13. If Given Equation Is 137+276=435, How Much Is 731+672=…. Find The Result.

In decimal quantity system; 137 + 276 = 413 however right here its 435 (> 413) so the bottom system ought to be lower than 10 and because the highest digit within the sum is 7 so the bottom should be higher than 7.

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

So 7 + 6 = 5 + 8(1 carry is forwarded) and therefore the it’s in octal quantity system.

Therefore: 731 + 672 = 1623.

Q 14. A Dealer Buys A Product At Rs.1920. He Sells At A Discount Of 20% Still He Gets The Profit Of 20%. What Is The Selling Price?

Cost worth: Rs 1920

Profit = 20% = Rs 1920 x 0.20 = 384

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

Q 15. How Many 3-digit Numbers Can Be Formed From The Digits 2,3,5,6,7 And 9 Which Are Divisible By 5 And None Of The Digit Is Repeated.?

As the quantity is divisible by 5, the unit digit of 3-digit quantity should be 5.

Rest two digits may be chosen in 5c1 * 4c1 = 20 methods.

Q 16. 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 chance of cube exhibiting an odd nos = ½ and

the chance of coin exhibiting head = ½;

so the general chance is: ½ * ½ = ¼.

Q 17. Find Last Two Digit Of (1021^3921)+(3081^3921)?

When a nos ends with 1 its final digit can be 1.

Now for the 2nd final digit the quick lower is

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

equally, for the second no 3081 it’s 8(1) = 8

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

Therefore, final 2 digits is: 02.

Q 18. How Many Prime Numbers Between 1 And 100 Are Factors Of 7150?

Since, 7150 = 2×5^2×11×13.

So, there are Four distinct prime numbers which are under 100.

Q 19. If Meeting O Is On Saturday, Then Meeting Okay Must Take Place On?

IJKLMNO if O is Saturday then I can be Sunday and Okay can be Tuesday.

Q 20. 3 15 _ 51 53 159 161

Observe the sequence:

5 * 3 = 15

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

So _ can be 15 + 2 = 17 (additionally 51/3 = 17).

Q 21. 55th Word Of Shuvank In Dictionary ?

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

Nos of phrases beginning with A: 6! = 720

Nos of phrases beginning with AH: 5! = 120

Nos of phrases beginning with AHK: 4! = 24

Nos of phrases beginning with AHN: 4! = 24

Nos of phrases beginning with AHSK: 3! = 6

Nos of phrases beginning with AHSN: 3! = 6

24+24+6 = 54, so the following phrase (55th) would be the first phrase beginning type AHSN and can be AHSNUV.

Q 22. 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 price worth be 100 per 1 kg

As he’ll promote 1 kg in 105 however because of error in weighing stones he’ll promote solely 900 grams in 105 however he has paid 900*(100/1000) =90 rs for 900 grams.

Therefore, internet revenue= Rs (105-90) = Rs 15

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

Q 23. Car A Leaves City C At 5 Pm And Drives At A Speed Of 40 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. Speed Of Car B Is 60 Kmph.

Let after t time two vehicles will met.

So A will journey distance of 40t with 40kmph

B will journey the gap of 60t with 60kmph

And additionally A is forward 80 km (40*2=80) from B

=> 60t – 40t = 80 => t = 4hrs

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

Additional distance is 9 min

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

So appropriate reply = 4hrs 27 min

= 4 (27/60) hrs = 4.45 hrs

Q 24. The Water From One Outlet, Flowing At A Constant Rate, Can Fill The Swimming Pool In 9 Hours. The Water From Second Outlet, Flowing At A Constant Rate Can Fill Up The Same Pool In Approximately In 5 Hours. If Both The Outlets Are Used At The Same Time, Approximately What Is The Number Of Hours Required To Fill The Pool?

Assume tank capability is 45 Liters.

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

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

If each pipes are opened collectively, then mixed capability is 14 liters/hour.

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

Q 25. Sum Of Three-digit Number Is 17. Sum Of Squared Of Digits Of The Given Num Is 109. If We Subtract 495 From That Num We Will Get A Number Written In Square Order. Find The Num?

Let the nos be: abc

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

Also sum of sq. 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 potential mixtures are (6,1) (7,2) (8,3), (9,4)

For 1st mixture (6,1); b = (17 – 6 – 1) = 10 which isn’t potential

For 2nd mixture (7,2); b = (17 – 7 – 2) = Eight however a^2+b^2+c^2 =! 109 so not potential

For third mixture (8,3) ; b = (17 – 8 – 3) = 6 additionally a^2+b^2+c^2 = 109 so it’s potential

so,863 is the reply.

Q 26. The Least Number That Must Be Subtracted From 63520 To Make The Result A Perfect Square, Is:

Find the sq. root of 63520. It can be 252. _ _ so the closest excellent sq. is 252^2 = 63504

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

Q 27. Find The Missing Numbers In The Series: 0,2,5,?,17,28,?

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

The variations are prime nos i.e 2, 3, 5, 7, 11 so the following distinction can be 13

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

Q 28. 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 5 Kmph What Is The Speed Of The Boat In Still Water?

Let the pace of boat in nonetheless water: x kmph

As distance is fixed; (x+5) *30=(x-5) *45

or, 2x+10=3x-15

x = 25 kmph

Q 29. 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 Upper Deck And 8 Refuse To Sit In The Lower Deck Is:

Those 5 who refuses to sit down within the higher deck will sit in decrease deck

So complete decrease deck stays: 2

Those Eight who refuses to sit down within the decrease deck will sit in higher deck

So complete higher deck sit stays: 5

These 7 individuals can sit in 5 higher deck and a couple of decrease deck in: 7c5 * 2c2 methods i.e. 21 methods.

Q 30. 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 A Profit Of 10%, Who Made More Profit And By What Amount?

Selling Price of Article = Rs. 1000

For 1st service provider, 10% revenue 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 service provider, 10% revenue is on S.P i.e. Profit = 0.10 * Rs 1000 = Rs. 100

so the revenue of 2nd service provider is larger than the first service provider by Rs. (100 – 90.9) = Rs. 9.1 (approx.).


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