Interview Questions.

Question 1. If (x – three)2 + (y – 5)2 + (z – four)2 = zero, Then The Value Of X2/nine + Y2/25 + Z2/16 Is

Answer :

(x – three)2 + (y – 5)2 + (z – 4)2 = zero

⇒ x – three = 0 ⇒ x= three

Y – 5 = zero ⇒ y = five

Z – four = 0 ⇒ z = 4

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

Answer :

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

⇒ 8 + 2P = 12

⇒ 2P = 12 – 8 = four

⇒ P = 2

Aptitude Interview Questions
Question three. The Straight Line 2x + 3y = 12 Passes Through:

Answer :

The normal way to resolve those type of questions is to put x = 0 once and locate y coordinate. This could constitute the point wherein the road cuts the Y axis.

Similarly put y = 0 as soon as and locate x coordinate. This would represent the factor wherein the line cuts the X axis. Then be part of these points and you may get the graph of the road.

So while we positioned x = zero we get y = 4.

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

So when we be a part of those factors we see that we get a line in 1st quadrant, which whilst extended each facets would go to 4th and 2d quadrants.

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

Answer :

∠A + ∠B = 65°

∴ ∠C = 180° - 65° = a hundred and fifteen°

∠B + ∠C = 140°

∴ ∠B = a hundred and forty° - one hundred fifteen° = 25°

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

Answer :

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

Number of two‘s at tenths location (xy2z) xy will range from 10 – ninety nine (90 nos) and z from zero-nine (10 nos) so general 90*10 = 900

Number of two‘s at masses area(x2yz) x will range from 1-9(nine approaches) & yz from 00-99(100 approaches) so total nine*a hundred =900

Number of 2‘s at hundreds vicinity(2xyz) xyz will range from 000-999 so total = 1000

Therefore, overall range of two‘s = (900+900+900+1000) =3700.

Wipro Aptitude Interview Questions
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?

Answer :

Let the rate of old guy be: x m/min and that of younger man be: y m/min

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

Let younger guy catches old guy after ‘t’ mins.

So distance travelled via younger guy is ‘t’min = ty = 1.5tx

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

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

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

Alternate technique

Old man 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-way of the journey at 10:15 AM.

Question 7. What Is The Next Numbers For The Given Series? Eleven 23 forty seven eighty three 131 ?

Answer :

Given collection: eleven, 23, 47, eighty three, 131

1st range: 11

2d number: eleven+12*1=23

third range: 23+12*2=47

4th wide variety: forty seven+12*three=eighty three

fifth quantity: 83+12*five=131

sixth quantity: 131+five*12=191.

Value Labs Aptitude Interview Questions
Question 8. What Is The Chance That A Leap Year Selected At Random Contains fifty three Fridays ?

Answer :

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

So the opportunity 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?

Answer :

As the rectangular of sum of digits is eighteen greater than that of the wide variety, so the rectangular of the sum of digit must be greater than or identical to 28 (18+10 as 10 is the smallest 2-digit quantity) and ought to be less than or equal to 117 (18+99 as 99 is the biggest -digit variety)

So the feasible squares are:

36 and subsequently the possible number may be (36-18) =18 or (1+8)2 = 81 =! 36 and therefore now not possible.

49 and subsequently the viable range can be (49-18) =31 or (3+1)2 = sixteen =! 49 and for this reason no longer feasible.

Sixty four and subsequently the feasible range may be (64-18) =46 or (four+6)2 = a hundred = 81 =! Sixty four and therefore now not viable.

Eighty one and therefore the possible number can be (81-18) =sixty three or (6+three)2 = 81 = 81 and consequently viable.

100 and hence the possible number may be (one hundred-18) =eighty two or (8+2)2 = one hundred = one hundred and subsequently no longer possible.

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

Abaxis Aptitude Interview Questions
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.

Answer :

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

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

As wide 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.

Question eleven. B Moves By Taking three 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 forty Sec To Reach Top .Locate Speed Of Escalator.

Answer :

As B actions 3 steps ahead and then 1 step backward so in overall four seconds he movements handiest 2 steps ahead so in 116 seconds he movements fifty eight steps ahead now in next 2 seconds he actions 2 steps so in 118 seconds he movements overall 60 steps ahead.

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

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

So in forty second total flow=10*(2+4a)

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

therefore a=1step/sec. 

TCS Aptitude Interview Questions
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?

Answer :

As A and B completed a piece together in 5 days

Work executed with the aid of them in a day (A + B), 1/5

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

so, their paintings per day (2A + B/2) = 1/four

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

or 1-day paintings of A = 1/10

so A by myself can entire the paintings in 10 days.

Aptitude Interview Questions
Question thirteen. If Given Equation Is 137+276=435, How Much Is 731+672=.... Find The Result.

Answer :

In decimal number gadget; 137 + 276 = 413 however right here its 435 (> 413) so the bottom gadget have to be less than 10 and as the highest digit within the sum is 7 so the bottom ought to be more than 7.

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

So 7 + 6 = five + 8(1 deliver is forwarded) and subsequently the it is in octal quantity device.

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?

Answer :

Cost charge: Rs 1920

Profit = 20% = Rs 1920 x zero.20 = 384

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

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

Answer :

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

Rest  digits may be selected in 5c1 * 4c1 = 20 approaches.

UNDP Interview Questions
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.

Answer :

The opportunity of cube displaying an atypical nos = ½ and

the opportunity of coin displaying head = ½;

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

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

Answer :

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

Now for the 2nd ultimate digit the short reduce is

1021-tenths area digit*unit area digit of the strength= 2(1) = 2

further, for the second one no 3081 it's miles eight(1) = eight

so the closing  digits are 21+eighty one=102.

Therefore, closing 2 digits is: 02.

Sailpoint Interview Questions
Question 18. How Many Prime Numbers Between 1 And a hundred Are Factors Of 7150?

Answer :

Since, 7150 = 2×5^2×11×thirteen.

So, there are four distinct top numbers which can be under a hundred. 

Wipro Aptitude Interview Questions
Question 19. If Meeting O Is On Saturday, Then Meeting K Must Take Place On?

Answer :

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

Question 20. 3 15 _ 51 fifty three 159 161

Answer :

Observe the sequence:

five * three = 15

51 + 2 = 53; fifty three * three = 159; 159 + 2 = 161

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

Oracle Hyperion Interview Questions
Question 21. Fifty fifth Word Of Shuvank In Dictionary ?

Answer :

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

Nos of words beginning with A: 6! = 720

Nos of words starting with AH: five! = a hundred and twenty

Nos of words beginning with AHK: 4! = 24

Nos of words beginning with AHN: 4! = 24

Nos of phrases beginning with AHSK: three! = 6

Nos of words beginning with AHSN: 3! = 6

24+24+6 = 54, so the next word (fifty fifth) might be the primary word beginning form AHSN and may be AHSNUV.

Question 22. Mani Sells Vegetables And He Marks Up The Prices At five% Above His Cost Price. Also The Weighing Stones Used By Him Weigh Only ninety% Of The Correct Weight. Find His Effective Percentage Of Mark-up.

Answer :

Let the fee fee be a hundred per 1 kg

As he will sell 1 kg in one zero five but because of blunders in weighing stones he'll promote only 900 grams in a hundred and five however he has paid 900*(a hundred/1000) =ninety rs for 900 grams.

Therefore, internet profit= Rs (105-ninety) = Rs 15

% percent= (15/90) *100% =sixteen.Sixty seven%

Question 23. Car A Leaves City C At 5 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 nine Km Ahead Of Car A. Speed Of Car B Is 60 Kmph.

Answer :

Let after t time two motors will met.

So A will journey distance of 40t with 40kmph

B will journey the space of 60t with 60kmph

And also A is in advance 80 km (forty*2=80) from B

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

Also time taken via B to cover 9kms greater is nine/60 = 9mins

Additional distance is nine min

For extra time= (nine/20) *60=27 min

So correct answer = 4hrs 27 min

= 4 (27/60) hrs = 4.45 hrs

3i Infotech Interview Questions
Question 24. 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 5 Hours. If Both The Outlets Are Used At The Same Time, Approximately What Is The Number Of Hours Required To Fill The Pool?

Answer :

Assume tank capability is forty five Liters.

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

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

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

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

Value Labs Aptitude Interview Questions
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?

Answer :

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 feasible combos are (6,1) (7,2) (eight,3), (nine,4)

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

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

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

so,863 is the solution.

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

Answer :

Find the square root of 63520. It might be 252. _ _ so the closest best square is 252^2 = 63504

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

Question 27. Find The Missing Numbers In The Series: zero,2,five,?,17,28,?

Answer :

The difference between nos are: 2 , three , _ , _ ,eleven

The variations are top nos i.E 2, three, 5, 7, eleven so the next difference will be thirteen

Therefore, nos are: (five + 5) = 10 & (28 + thirteen) = forty one.

Abaxis Aptitude Interview Questions
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 five Kmph What Is The Speed Of The Boat In Still Water?

Answer :

Let the speed of boat in still water: x kmph

As distance is steady; (x+5) *30=(x-five) *forty five

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:

Answer :

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

So overall decrease deck remains: 2

Those eight who refuses to take a seat within the decrease deck will sit in upper deck

So total higher deck take a seat remains: five

These 7 human beings can sit in 5 upper deck and a pair of decrease deck in: 7c5 * 2c2 methods i.E. 21 ways.

Question 30. Two Merchants Sell An Article Each For Rs.A thousand.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?

Answer :

Selling Price of Article = Rs. One thousand

For 1st service provider, 10% income 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.Nine

For 2nd merchant, 10% earnings is on S.P i.E. Profit = zero.10 * Rs a thousand = Rs. A hundred

so the income of second merchant is better than the first merchant through Rs. (100 – ninety.Nine) = Rs. 9.1 (approx.).