 # MOST IMPORTANT ABB GROUP APTITUDE INTERVIEW QUESTIONS

Which is the MOST IMPORTANT ABB GROUP APTITUDE INTERVIEW QUESTIONS? Here are the list of Questions with solutions you can refer to prepare for you ABB Group interview.

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Question 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

Question 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

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

Question 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°

Question 5. A Person Starts Writing All 4 Digits’ Numbers. How Many Times Had He Written The Digit 2?
Number of 2‘s at unit’s place (from 100-1 to 999-2) = 900
Number of 2‘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.

Question 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’ 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.

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

Question 8. What Is The Chance That A Leap Year Selected At Random Contains 53 Fridays ?
A leap year has 366 days, therefore 52 weeks (i.e. 52 Friday’s) + 2 days.
So the probability of 53 Fridays = 2/7.

Question 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 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.

Question 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 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.

Question 11. B Moves By Taking 3 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 40 Sec To Reach Top .find Speed Of Escalator.
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.

Question 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 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.

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

Question 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 price: Rs 1920
Profit = 20% = Rs 1920 x 0.20 = 384
Therefore, Selling Price = Rs 1920 + 384 = 2304.

Question 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 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.

Question 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 probability of dice showing an odd nos = ½ and
the probability of coin showing head = ½;
so the overall probability is: ½ * ½ = ¼.

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

Question 18. How Many Prime Numbers Between 1 And 100 Are Factors Of 7150?
Since, 7150 = 2×5^2×11×13.
So, there are 4 distinct prime numbers that are below 100.

Question 19. If Meeting O Is On Saturday, Then Meeting K Must Take Place On?
IJKLMNO if O is Saturday then I will be Sunday and K will be Tuesday.

Question 20. 3 15 _ 51 53 159 161
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).

Question 21. 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 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.

Question 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 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%

Question 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 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
For additional time= (9/20) *60=27 min
So correct answer = 4hrs 27 min
= 4 (27/60) hrs = 4.45 hrs

Question 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 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.

Question 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 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

Question 26. The Least Number That Must Be Subtracted From 63520 To Make The Result A Perfect Square, Is:
Find the square root of 63520. It will be 252. _ _ so the nearest perfect square is 252^2 = 63504
So the nos to be subtracted is: (63520 – 63504) = 16.

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

Question 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 speed of boat in still water: x kmph
As distance is constant; (x+5) *30=(x-5) *45
or, 2x+10=3x-15
x = 25 kmph

Question 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 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.

Question 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?