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Top 100+ Zensar Technologies Aptitude Interview Questions And Answers

Question 1. The Average Score Of A Cricketer For 13 Matches Is 42 Runs. On the off chance that His Average Score For The First 5 Matches Is 54, Then What Was His Average Score (in Runs) For Last 8 Matches?

Answer :

All out Score = Average * Number of matches

All out score of 13 matches = 13 × 42 = 546

All out score of initial 5 matches = 5 × 54 = 270

In this way, all out score of last 8 matches = 546 – 270 = 276

Normal = 276/8 = 34.5

Question 2. Basic Interest On A Sum Of Money For 4 Years At 7 P.c.p.a Is Rs. 3584/ - . What Would Be The Compound Interest (intensified Annually) On The Same Amount Of Money For 2 Years At 4p.c.p.a?

Answer :

Leave P alone the aggregate

Along these lines,

3584 = (P * 7 * 4)/100 => P = 12800/ -

CI = 12800(1 + (4/100))2 - 12800

=> 13844.48 - 12800 = 1044.48/ -

Inclination Interview Questions

Question 3. 6*136/8+132/628/16-26.25=?

Answer :

6*136/8+132/628/16-26.25

⇒ 234/13=18

Question 4. (1331)1/3/275 *52 = ?

Answer :

(1331)1/3/275 *(5)2

= (113)1/3/11*25 *(25)

Question 5. (43)2 + 841 = (?)2 + 1465?

Answer :

Utilizing BODMAS

2 = (43)2 + 841 – 1465

= 1849 + 841 – 1465 = 1225

Along these lines, = √1225 = 35

Wipro Aptitude Interview Questions

Question 6. (216)4 ÷ (36)4 × (6)5 = (6) ?

Answer :

(6) = (216)4 ÷ (36)4 × (6)5 =

2164/364 * 65 =

612/68 * 65=

69 Therefore, = 9

Question 7. (1097.63 + 2197.36 - 2607.24) ÷ 3.5 = ?

Answer :

(1097.63 + 2197.36 - 2607.24) ÷ 3.5 = 687.75 ÷ 3.5

= 196.5

ABB Group Aptitude Interview Questions

Question 8. A Certain Number Of Capsules Were Purchased For Rs. 216/ - . 15 More Capsules Could Have Been Purchased In The Same Amount If Each Capsule Was Cheaper By Rs. 10/ - . What Was The Number Of Capsules Purchased?

Answer :

Leave x alone the expense of each case and y be the no of containers

Thusly, x = 216/y and x - 10 = 216/(y+15)

Putting the estimation of x from eq. 1 into eq. 2, we get

216/y - 216/(y+15) = 10

y2 +15y - 324 = 0

Settling for y, we get y = 12

Question 9. What Will Come In Place Of Question Mark (?)in The Given Question?

Answer :

Here the distinction between each term is as 13+1, 33+1, 53+1, 73+1, etc

4 6 34 160 504 1234

+2 +28 +126 +344 +730

13 + 1 33 + 1 53 + 1 73 + 1 93 + 1

So = 160

VMware Aptitude Interview Questions

Question 10. An And B Are Two Numbers. 6 Times Square Of B Is 540 More Than The Square Of A. On the off chance that The Respective Ratio Between An And B Is 3:2. What Is The Value Of B?

Answer :

We are given that

6B2 = A2 + 540 and A : B = 3 : 2

Put the estimation of A from second condition into the principal condition, we get

6B2 = (1.5B)2 + 540

→ (6 – 2.25)B2 = 540 → 3.75B2 = 540 → B2 = 144 → B = 12

Question 11. Funnel A Can Fill A Tank In 20 Minutes And Pipe B In 30 Minutes Respectively. Funnel C Can Empty The Same In 40 Minutes. On the off chance that All The Three Pipes Are Opened Together, Find The Time Taken To Fill The Tank?

Answer :

Net part filled in 1 hour = ( 1/20 – 1/30 – 1/40) = 7/120

In this manner tank will be filled in = 120/7 = 17 1/7 minutes.

Worth Labs Aptitude Interview Questions

Question 12. Cost Of 5 Mangoes + Cost Of 4 Oranges = Cost Of 7 Mangoes + Cost Of 1 Orange. Discover The Ratio Of Cost Of Mango To Cost Of Orange ?

Answer :

⇒ 5 mangoes + 4 oranges = 7 mangoes + 1 orange

⇒4 orange – 1 orange = 7 mangoes – 5 mangoes

⇒ 3 orange = 2 mangoes

Thus Required Ratio is

⇒ Cost of mango: cost of orange = 3 : 2

Inclination Interview Questions

Question 13. There Are 6 Red Shoes and 4 Green Shoes. On the off chance that Two Of Red Shoes Are Drawn Randomly What Is The Probability Of Getting Red Shoes?

Answer :

All out number of shoes=6 + 4=10;

Leave S alone the example space.

At that point, n(s) = Number of methods of drawing 2 shoes out of

10 =10C2 =(10 x 9)/(2 x 1) =45

Let E = Event of drawing 2 shoes which are red.

In this way n(E) = Number of methods of drawing 2 red shoes out of

6 shoes.= 6C2 =(6 x 5)/(2 x 1) =15 P(E) = n(E)/n(S) = 1/3

Question 14. To 15 Liters Of Water Containing 20% Alcohol, We Add 5 Liters Of Pure Water. What Is % Alcohol?

Answer :

Case 1: Initial amount of Water = 15 liters Quantity of Alcohol = 20% = (20/100) * 15 = 3 liters

Case 2: Quantity of Water in the wake of including 5 liters of water = 15 + 5 = 20 liters There of the % of liquor in water = (3/20) * 100 = 15%

Question 15. A Certain Number Of Two Digits Is Three Times The Sum Of Its Digits. In the event that 45 Be Added To It, The Digits Are Reversed. The Number Is?

Answer :

Let x and y are the digits of the number with x in ten's place and y in one's place.

Thus, 10x+y is the estimation of the number.

Given 10x + y = 3 (x + y)

⇒7x – 2y = 0 … .(I)

And furthermore given that,

10x + y + 45 = 10y + x (? digits are reversed⇒ x in one's and y in ten's place).

⇒ 9x – 9y = - 45

⇒ x – y = - 5 … (ii)

Put estimation of y from eq. (I) to eq. (ii)

⇒ x – (7x/2) = - 5

⇒ x = 2

What's more, y = 7

So the number is 27

Abaxis Aptitude Interview Questions

Question 16. Water Flows Into A Tank 200 M × 150 M Through A Rectangular Pipe Of 1.5m × 1.25 M @ 20 Kmph . In What Time (in Minutes) Will The Water Rise By 2 Meters?

Answer :

We realize that, recipe:

Volume of any Cuboid = Length × Breadth × Height

As indicated by the given inquiry:

⇒ Volume of the water required in the tank = (200 × 150 × 2) m3 = 60000 m3 . . .

∴ Length of water section flown in1 min =(20 × 1000)/60 m = 1000/3 m

∴ Volume flown every moment = 1.5 × 1.25 × (1000/3) m3 = 625 m3 .

∴ Required time = (60000/625)min = 96min.

Consequently, the necessary answer is 96 minutes.

Question 17. A Boat Travels 20 Km Upstream In 4 Hours And 18 Km Downstream In 6 Hours. Discover The Speed Of The Boat In Still Water?

Answer :

Rate downstream= 20/4 = 5 kmph

Rate upstream= 18/6 = 3 kmph

Speed in still water = 1/2 (5 + 3) = 4 kmph.

Hurray Aptitude Interview Questions

Question 18. A Person Is 80 Years Old In 490 And Only 70 Years Old In 500 In Which Year Is He Born ?

Answer :

Since we are given that an individual is senior in 490 than in 500.

Henceforth, he more likely than not been conceived in BC.

Henceforth, we can say that he more likely than not been conceived in BC 490+80

= BC 570

Wipro Aptitude Interview Questions

Question 19. Grass In Lawn Grows Equally Thick And In A Uniform Rate. It Takes 40 Days For 40 Cows And 60 Days For 30 Cows To Eat The Whole Of The Grass. What number of Cows Are Needed To Eat The Grass In 96 Days?

Answer :

Let amount of grass at first = g

Leave r alone the rate at which grass develop in 1 day,

Leave c alone the amount of grass a dairy animals eat in 1 day

we can reason from '. It takes 40 days for 40 cows and 60 days for 30 bovines to eat the entire of the grass' that

⇒ g + 40r = 40 × 40c = 1600c - [1]

furthermore,

⇒ g + 60r = 60 × 30c = 1800c

⇒ g = 1800c - 60r - 2

Placing this estimation of g in eqn [1]

we have

⇒ 1800c – 60r + 40r = 1600c

⇒ 200c = 20r

⇒ C = 0.1 r

⇒ r = 10c

Leave m alone the quantity of days required by 20 dairy animals to eat the whole of the field = m

At that point

We have the eqn as

⇒ g + 96r = 96nc

⇒ 96nc = 1800c - 60r + 96r (as we have g=1800c-60r from eqn 2)

⇒ 96nc = 1800c + 36r

⇒ 96nc = 1800c+ 36 × 10c

⇒ 96nc = 1800c + 360c

⇒ 96nc = 2160c

Thus we have 96nc = 2160c

Isolating both the sides by 96c

We get

⇒ n = 22.5

Question 20. A Man Buys Spirit At Rs. 60 Per Liter, Adds Water To It And Then Sells It At Rs. 75 Per Liter. What Is The Ratio Of Spirit To Water If His Profit In The Deal Is 37.5%?

Answer :

We have SP = 75 for each liter

Benefit Per penny = 37.5%

Henceforth

⇒ CP = SP × 100/(100 + 37.5)

⇒ CP = 7500/137.5 = 54.5454

Henceforth, to make the CP from 60 to 54.54,

Thus,

Required proportion is

= 54.54 : 60 - 54.54

= 54.54 : 5.454

= 10:1

Yash Technologies Aptitude Interview Questions

Question 21. In What Ratio Must Rice At Rs.9.30 Per Kg Be Mixed With Rice At Rs. 10.80 Per Kg So That The Mixture Be Worth Rs.10 Per Kg ?

Answer :

We have to blend rice assortments costing Rs 9.3 and Rs 10.8 per kg to make the blend at Rs 10/kg

To do as such,

We have to include these assortments in the proportion

⇒10.8 - 10 : 10 - 9.3

= 0.8 : 0.7

= 8 : 7

Question 22. Three Types Of Tea The A, B, C Costs Rs. 95/kg, 100/kg And70/kg Respectively. What number of Kgs Of Each Should Be Blended To Produce 100 Kg Of Mixture Worth Rs.90/kg, Given That The Quantities Of B And C Are Equal?

Answer :

Let the amount of tea a = x

Let the amount of tea b = y

Let the amount of tea c = z

Given : y = z ...... (I)

x + y + z = 100 ....... (ii)

Cost of Mixture = (Sum of Cost of individual things in a blend/complete amount of blend)

90 = (95x + 100y + 70z)/(x + y + z)

90x + 90y + 90z = 95x + 100y + 70z

5x + 10y = 20z

5x = 20y - 10y = 10y............................... (from Eq. (I), y =z )

x = 2y .................... (iii)

From Equation (I), (ii) and (iii)

x + y + z = 100

2y + y + y = 100

4y = 100

y = 25

Question 23. A Man Buys 12 Liters Of Solution A Which Contains 20% Of The Liquid And The Rest Is Water. He Then Mixes It With 10 Liters Of Another Solution B With 30% Of Liquid. What Is The % Of Water In The New Mixture?

Answer :

% of Liquid in Solution A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Liters

Volume of Water in Solution A = (80/100) × 12 = 9.6 Liters

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Liters

Volume of Water in Solution B = (70/100) × 10 = 7 Liters

On the off chance that Solution An is blended in with Solution B

⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of

Question 24. A Fuel Dealer Mixes Two Brands Of Fuel Which Cost In The Ratio 2:3. A Solution Containing 30% Brand An And Remaining Brand B Yields Profit Of 10% When Sold At Rs 297. What Is Cost Of Brand B?

Answer :

Let amount of arrangement sold be 1 liter

At the point when the arrangement is sold for 297, the benefit percent = 10%

Consequently CP of arrangement = SP × 100/(100+P%)

⇒CP = 297 × 100/(100+10)

⇒CP = 270

Proportion of amounts of brand An and brand B is 3:7

Leave their costs alone 2c and 3c.

⇒ 0.3×2c + 0.7×3c = 270

⇒ 2.7c = 270

⇒ c = 100

Thus cost of Brand B = 3c = 300

ABB Group Aptitude Interview Questions

Question 25. On the off chance that 9 Men Working 7.5 Hours A Day Can Finish A Piece Of Work In 20 Days, Then How Many Days Will Be Taken By 12 Men, Working 6 Hours A Day To Finish The Work? It Is Being Given That 2 Men Of Latter Type Work As Much As 3 Men Of The Former Type?

Answer :

Work done by 9 men in a day = 1/20

∴Work done by 1 man in a day for 7.5 hours = (1/20) ÷ 9 = 1/180

∴ Work done by 1 man in an hour = (1/180) ÷ (7.5) = 1/1350

Given that 2 men of last kind are equivalent to 3 men of previous sort ⇒ 1 man of last kind = 1.5 men of previous sort.

∴ Work done by 1 man of last kind in an hour = 1.5 × (1/1350) = 1/900

∴ Work done by 1 man in 6 hours = 6 × (1/900) = 1/150

∴ Work done by 12 men in a day working 6 hours/day = 12 × (1/150) = 2/25

Days required to complete the work = 25/2 = 12.5 days.

Question 26. A Man's Basic Pay For A 40 Hours' Week Is Rs. 200. Additional minutes Is Paid At 25% Above The Basic Rate. In A Certain Week, He Worked Overtime And His Total Was Rs. 300. He Therefore, Worked For A Total Of (in Hours)?

Answer :

Fundamental compensation every hour of typical work = 200/40 = Rs.5

After some time is paid at 25% more = (125/100) × 5 =Rs.6.25 every hour

Man was paid Rs.200 for his normal work and Rs.100 for after some time (complete Rs.300).

Number of normal hours = 40

Number of extra time hours at 6.25 every hour = 100 ÷ 6.25 = 16 hours.

Complete number of hours worked = 40+16 = 56 hours.

Question 27. A Retailer Buys A Radio For Rs.225. His Overhead Expenses Are Rs.15 And He Sells The Radio For Rs.300. What Is The Profit Percent Of The Retailer?

Answer :

Selling cost = cost + benefit

Cost = 225 + 15 = 240

for example 240 + X * 240 * 1/100 = 300 240 + 24X/10 = 300

(taking LCM) 2400 + 24X = 3000

24 X = 3000 – 2400

24X = 600

X = 600/24

X = 25

VMware Aptitude Interview Questions

Question 28. A Truck Covers A Distance Of 640 Km In 10 Hrs. A Car Covers The Same Distance In 8 Hrs. What Is The Respective Ratio Between The Speed Of The Truck And The Car?

Answer :

Both truck and vehicle covers a separation = 640 km

Time taken by truck is 10 hrs.

∴ Speed of truck = 640/10 = 64 km/hr

Time taken via vehicle is 8 hrs.

∴ Speed of vehicle = 640/8 = 80 km/hr

∴ Ratio between speed of truck and vehicle =

Speed of truck : speed of vehicle = 64 : 80 = 4 : 5

Question 29. Two Trains From The Points An And B Moving In Opposite Direction, At The Point They Meet The Second Train Travels 120 Kms More Than The First. The Speeds Are 50kmph And 60kmph Respectively Find The Distance Between An And B?

Answer :

We realize that

Separation = time × speed

Assume first train ventured to every part of the separation x kms from A.

At that point second train will venture to every part of the separation (x + 120) kms from B.

Let they met each other after t time

∴ (x + 120)/60 = (x/50)

⇒ x = 600 kms

In any case, the separations between two point is (x + x + 120)

= 2x +120

= 2× 600 + 120

= 1320 km

Question 30. Consistently A Cyclist Meets A Train At A Particular Crossing. The Road Is Straight Before The Crossing And Both Are Traveling In The Same Direction. The Cyclist Travels With A Speed Of 10 Kmph. One Day The Cyclist Comes Early By 25 Min. What's more, Meets The Train 5km Before The Crossing. What Is The Speed Of The Train?

Answer :

Cyclist speed is 10kmph and he'll take 30 min to cover 5 kms.

On the off chance that Cyclist come 25 min early he would be 5 min before intersection.

So it implies train takes 5 min to arrive at intersection.

We have speed of train determined as

⇒ Speed= separation/time

⇒ Speed = 5km/5 min

⇒ Speed = 1 km/min

Or then again 60 km/hr

Question 31. The Average Speed Of A Bus From Koyambedu To Salem Is 57 Km Per Hour. The Bus Is Scheduled To Leave Koyambedu Bus Station At 10 Pm And Reach Salem At 4.35 Am On The Next Day. The Distance Between Salem And Koyambedu Bus Station Is 342 Km. In transit In Between Koyambedu And Salem A Halt Is Scheduled Compulsorily. Discover The Duration Of This Halt Scheduled?

Answer :

Given, separation among Salem and Koyambedu = 342 km.

Speed of transport = 57 kmph

Time = Distance/Speed

⇒ Time required by the transport to make a trip from Koyambedu to Salem = 342/57 hours = 6 hours.

Time really taken by the transport = 6 hours and 35 minutes.

Transport takes 35 minutes extra ⇒ The mandatory end time of transport = 35 minutes

Question 32. Discover The Equation Whose Roots Are 9 And 5 ?

Answer :

Finding the underlying foundations of given conditions x2 – 14x + 45 = 0 x2 – 9 x – 5 x + 45 = 0 (x – 5) (x – 9) = 0

Question 33. What Will Come In Place Of Question Mark (?) In The Following Question? 76% Of 1285 = 35% Of 1256 +?

Answer :

Given articulation is-

76% of 1285 = 35% of 1256 +?

⇒ (76/100) × 1285 = (35/100) × 1256 +?

⇒ 97660/100 = 43960/100 +?

⇒ ? = (97660/100) – (43960/100)

⇒ ? = 53700/100 = 537

Worth Labs Aptitude Interview Questions

Question 34. What Will Come In Place Of Question Mark In The Following Equation?

Answer :

(5863 - √2704) × 0.5 = ?

(5863 - √2704) × 0.5 = x

⇒ (5863 – 52) × 0.5 = x

⇒ x = 2905.5

Question 35. Let 13 And 273 Are The Hcf And Lcm Of Two Numbers Respectively, And If One Of Them Is Less Than 140 And Greater Than 60. At that point What Will Be That Number?

Answer :

Leave those 2 numbers alone p and q .

We are given that HCF (p, q) = 13 and

LCM (p, q) = 273.

We realize that result of numbers = HCF × LCM

Henceforth, p × q = 13 × 273

⇒ p × q = 13 × 13 × 3 × 7

P can't be 13 × 3 or 13 × 13 or 7 × 3 or 13 or 3 or 7 as these numbers don't divide 60 and 140.

Thus we can say that p = 13 × 7 = 91

Question 36. The Least Number Which When Divided By 4, 6, 8, 12 And 16 Leaves A Remainder Of 2 In Each Case Is?

Answer :

We should locate a number which is totally distinguishable by 4, 6, 8, 12 and 16.

To discover such number, we have to discover the LCM of every one of these numbers

LCM of (4,6) , 8 , 12 , 16

= 12 , 8 , 12 , 16

= LCM of (12 , 8) , LCM of (12 , 16)

= 24 , 48

= LCM of (24 , 48)

= 48

Thus 48 + 2 ie 50 will leave a rest of 2 whenever isolated by any of the accompanying numbers : 4, 6, 8, 12 and 16

Abaxis Aptitude Interview Questions

Question 37. Locate The Number Of Sides Of A Regular Convex Polygon Whose Interior Angle Is 40 Degrees?

Answer :

We have inside edge as 40 degrees.

Consequently Let n be the quantity of sides of that polygon,

At that point equation is

⇒180 × (n-2) = 40 × n

⇒180n – 360 = 40n

⇒140n = 360

Since, no number estimation of n is conceivable, henceforth such a polygon can not exist

Question 38. There Is A 5 Digit No. Entirety Of 3 Pairs Of Digits Is Eleven Each. Last Digit Is 3 Times The First One. third Digit Is 3 Less Than The Second. fourth Digit Is 4 More Than The Second One. Locate The Number?

Answer :

We are given a 5 digit number

Leave first digit alone 'X'

at that point fifth digit is '3X'

leave second digit alone 'Y'

at that point third digit is 'Y - 3'

what's more, fourth digit is 'Y + 4'

at that point the no is '(X)(Y)(Y - 3)(Y + 4)(3X)'

from the above we can say 3X <= 9

so X<=3 and any of the digit in the number is <= 9

and furthermore given that 3 sets aggregate is 11...

Likewise Y–3 ≥ 0 and Y+4 ≤ 9