Question 1. A Sphere And A Cube Have The Same Surface Area. Find The Ratio Of Their Volumes?

Answer :

SA of a sphere: 4πr²

SA of a dice: 6x²

4πr² = 6x²

r²/x² = 6 / 4π

(r/x)² = 3 / 2π

r/x = (three / 2π)^zero.Five

Volume of a sphere: 4/three πr³

Volume of a cube: x³

Find the ratio which means (four/three πr³)/x³

= (4π/three)(r³/x³)

= (4π/3)(r/x)³

= (4π/three)(3 / 2π)^1.Five

= (2²π/three)[3^1.5 / (2^1.5)(π^1.5)]

= √2√three / √π

= √(6/π).

Question 2. Four Horses Are Tethered At four Corners Of A Square Plot Of Side 63 Meters So That They Just Cannot Reach One Another. The Area Left Ungrazed Is?

Answer :

region of rectangular = sixty three*sixty three=3969 m^2

location inside the rectangular this is grazed =4*place of quadrants of 4 circles

= four*(1/four)*(22/7)*(63/2)*(63/2) =3118.5

vicinity left ungrazed = 3969-3118.5=850.Five.

Aptitude Interview Questions

Question 3. How Many No Between 1000 To 9900 In Which Four Digits Of Number Are Different?

Answer :

Number of 4 digit numbers having all one of a kind digits = nine*9*eight*7 = 4536

smallest wide variety : 1234, biggest quantity : 9876, all are between one thousand and 9900.

Question 4. A Train 360 M Long Is Running At A Speed Of forty five Km/hr. In What Time Will It Pass A Bridge one hundred forty M Long?

Answer :

Formula for converting from km/hr to m/s: X km/hr = X *( five/18) m/s.

So Speed = 45 * five/18 m/sec = 25/2 m/sec.

Total distance to be covered = (360 + one hundred forty) m = 500 m.

Required time = 500 * 2/25 sec = 40 sec.

Question five. The G.C.D. Of 1.08, zero.36 And 0.Nine Is?

Answer :

Given numbers are 1.08 , zero.36 and 0.Ninety

H.C.F of 108, 36 and 90 is 18 ( G.C.D is nothing however H.C.F)

So H.C.F of given numbers = 0.18.

HR Interview Questions

Question 6. The Slant Height Of A Right Circular Cone Is 10 M And Its Height Is 8 M. Find The Area Of Its Curved Surface?

Answer :

l = 10m

h = 8m

so r = √(l^2 - h^2) = √(10^2 - 8^2) = 6m

So Curved surface region = (π * r * l) = ( π * 6 * 10) m2 = 60π m2.

Question 7. How Many Times In A Day, Are The Hands Of A Clock In Straight Line But Opposite In Direction?

Answer :

The palms of a clock factor in opposite directions (within the identical directly line) eleven times in each 12 hours. (Because among five and 7 they factor in opposite guidelines at 6 o'clock only).

So, in an afternoon, the fingers factor within the contrary guidelines 22 instances.

Wipro Aptitude Interview Questions

Question eight. In Covering A Distance Of 30 Km, Abhay Takes 2 Hours More Than Sameer. If Abhay Doubles His Speed, Then He Would Take 1 Hour Less Than Sameer. Abhay's Speed Is?

Answer :

Let Abhay's velocity be x km/hr.

So 30/x - 30/2x = 3

=> 6x = 30

x = five km/hr.

Question nine. The Sum Of The Present Ages Of A Father And His Son Is 60 Years. Six Years Ago, Father's Age Was Five Times The Age Of The Son. After 6 Years, Son's Age Will Be?

Answer :

Let the present a while of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 6 = 5(x - 6)

=> fifty four - x = 5x - 30

=> 6x = eighty four

=> x = 14.

Son's age after 6 years = (x+ 6) = two decades.

Abaxis Aptitude Interview Questions

Question 10. Sum Of Three Even Consecutive Numbers Is forty eight, And Then Least Number Is?

Answer :

Let the numbers be 2n, 2n+2 and 2n+4

2n + (2n+2) + (2n+4) = 48

6n = forty eight-6 = forty two, n = 7

Hence the numbers are -- > 14, sixteen and 18

The least range is 14.

Question 11. Walking At 3/4 Of His Usual Speed ,a Man Is Late By half of Hr. The Usual Time Is?

Answer :

Usual pace = S

Usual time = T

Distance = D

New Speed is ¾ S

New time is four/3 T

four/three T ? T = 5/2

T=15/2 = 7 ½.

Yahoo Aptitude Interview Questions

Question 12. If 7 Spiders Make 7 Webs In 7 Days, Then 1 Spider Will Make 1 Web In How Many Days?

Answer :

Let the specified quantity days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

Spiders 1 : 7

> :: 7 : x

Webs 7 : 1

1 x 7 x x = 7 x 1 x 7

=> x = 7.

Aptitude Interview Questions

Question thirteen. A Student Multiplied A Number By three/5 Instead Of 5/three. What Is The Percentage Error In The Calculation?

Answer :

Let the variety be x.

Then, error = x*five/3 - x*three/five = x * sixteen/15

So Error% = ( x*sixteen/15 * three/5x * 100)% = sixty four%.

Question 14. If A's Height Is forty% Less Than That Of B, How Much Percent B's Height Is More Than That Of A?

Answer :

Excess of B's top over A's = [(40/(100 - 40)] x one hundred%

= sixty six.Sixty six%

Question 15. From A Group Of 7 Men And 6 Women, Five Persons Are To Be Selected To Form A Committee So That At Least three Men Are There On The Committee. In How Many Ways Can It Be Done?

Answer :

We may have (three guys and 2 girls) or (4 guys and 1 girl) or (five guys best).

Required quantity of methods = (7C3 * 6C2)+(7C4 * 6C1)+(7C5)

=>[(7*6*5)/(3*2*1) * (6*5)/(2*1)]+(7C3 * 6C1) + (7C2)

=> 525 + [(7*6*5)/(3*2*1)* 6]+[(7*6)/(2*1)]

= (525 + 210 + 21)

= 756.

Oracle Aptitude Interview Questions

Question sixteen. In How Many Different Ways Can The Letters Of The Word 'mathematics' Be Arranged So That The Vowels Always Come Together?

Answer :

In the phrase 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we've got MTHMTCS (AEAI).

Now, we should arrange eight letters, out of which M happens twice, T occurs twice and the relaxation are distinct.

So,Number of methods of arranging those letters = (eight!/2!*2!)=10080.

Now, AEAI has 4 letters wherein A occurs 2 instances and the rest are one of a kind.

Number of methods of arranging these letters = (four!/2)=12.

=> Required wide variety of phrases = (10080 * 12) = 120960.

Question 17. What Will Be The Day Of The Week fifteenth August, 2010?

Answer :

fifteenth August, 2010 = (2009 years + Period 1.1.2010 to 15.Eight.2010)

Odd days in 1600 years = 0

Odd days in four hundred years = 0

nine years = (2 bounce years + 7 ordinary years) = (2 * 2 + 7 * 1) = eleven odd days four bizarre days.

Jan. Feb. March April May June July Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

227 days = (32 weeks + 3 days) 3 strange days.

Total variety of strange days = (zero + 0 + 4 + three) = 7 zero odd days.

Given day is Sunday.

TCS Aptitude Interview Questions

Question 18. A Thief Is Noticed By A Policeman From A Distance Of 200 M. The Thief Starts Running And The Policeman Chases Him. The Thief And The Policeman Run At The Rate Of 10 Km And eleven Km Per Hour Respectively. What Is The Distance Between Them After 6 Minutes ?

Answer :

locate the relative velocity of the thief and policeman = (eleven-10)km/hr = 1 km/hr

Distance included in 6 mins = (1/60)*6=1/10km => 100meters

So distance among them after 6 mins = (2 hundred-100)=a hundred meters.

HR Interview Questions

Question 19. An Industrial Loom Weaves zero.128 Metres Of Cloth Every Second. Approximately, How Many Seconds Will It Take For The Loom To Weave 25 Metre Of Cloth ?

Answer :

Lets anticipate the time required to weave 25 meters = x sec.

Rule: More fabric approach More time (Direct Proportion)

=> 0.128:1::25:x

=>x=[(25*1)/0.128]

=>x=195.31

So time wished ~195 seconds.

Question 20. There Are Three Numbers, These Are Co-high To Each Other Are Such That The Product Of The First Two Is 551 And That Of The Last Two Is 1073. What Will Be The Sum Of Three Numbers?

Answer :

Given that numbers are co primes,

and products have the middle quantity in commonplace.

=> Middle number = H.C.F. Of 551 and 1073 = 29

so first wide variety is = 551/29 = 19

=> Third range = 1073/29 = 37

Therefore, sum of these numbers is = (19 + 29 + 37) = 85.

Infosys Aptitude Interview Questions

Question 21. When A Student Weighing forty five Kgs Left A Class, The Average Weight Of The Remaining 59 Students Increased By 200g. What Is The Average Weight Of The Remaining fifty nine Students?

Answer :

Lets assume common weight of the 59 students = x kgs.

=> Total weight of the 59 = fifty nine*x

Given when the weight of this scholar who left is added, the whole weight of the magnificence = (59x + forty five)

The average weight decreases by 0.2 kgs, whilst those college students covered.

=> (59x+45)/60=(x-zero.2)

=> (59x+45)= 60(x-zero.2)

=> 59x + 45 = (60A - 12)

=> forty five + 12 = 60x - 59x

=> x = fifty seven.

Question 22. Nirmal And Kapil Started A Business Investing Rs. 9000 And Rs. 12000 Respectively. After 6 Months, Kapil Withdrew Half Of His Investment. If After A Year, The Total Profit Was Rs. 4600, What Was Kapil's Share Initially ?

Answer :

Nirmal:Kapil = 9000*12:(12000*6+6000*6) = 1:1

Kapils percentage = Rs. [4600 *(half of)) = Rs. 2300.

Question 23. A Is Able To Do A Piece Of Work In 15 Days And B Can Do The Same Work In 20 Days. If They Can Work Together For four Days, What Is The Fraction Of Work Left?

Answer :

Total work done by A + B in 1 day = 1/15 + 1/(20 ) = 7/60

Work carried out in 4 days = 7/60 × 4 = 7/15

Therefore, fraction of labor left = 1 - 7/15 = 8/15.

IBM Aptitude Interview Questions

Question 24. Population Of A Village Increased By 5% From 2007 To 2008 And By 25% From 2005 To 2009. If The Population Of The Village Was 480 In 2007, What Was Its Population In 2009?

Answer :

Population in 2007 = 480

In 2008 = 1.05 × 480 = 504

In 2009 = 1.25 × 504 = 630.

Wipro Aptitude Interview Questions

Question 25. 24% Of a hundred and fifty × three/4= ?

Answer :

24% of one hundred fifty × three/4= x = 36 × four/3 = forty eight.

Three/17 of 20% of 510 + 7 = x2

3/17 of 20% of 510 + 7 = x2

three/17 × 20/one hundred × 510 + 7 = 18 + 7 = 25 => x = five.

Question 26. In The Month Of March, Hiten Spent 45% Of His Monthly Salary On Paying Bill And Rent. Out Of The Remaining Salary, He Invested 60% In Ppf And The Remaining He Deposited In Bank. He Deposited Rs. 15,four hundred In Bank. If In April, He Got An Increment Of 10%, What Was His Salary In April?

Answer :

Let the preliminary salary of Hiten be ‘S’. Then:

(0.Fifty five × zero.Four) S = 15400 => S = 70000

After increment, Hiten’s revenue = 1.1 × 70000 = 77000.

Capgemini Aptitude Interview Questions

Question 27. A Person Covers A Certain Distance By Travelling At A Uniform Speed Of 120 Km/h For 90 Minutes. At What Speed Will He Have To Travel In Order To Cover The Same Distance In 1 Hour 20 Minutes? (in Km/h)

Answer :

Distance = one hundred twenty × ninety/60 = a hundred and eighty km

Speed required to cover one hundred eighty km in 1 hr 20 minutes

( four/three hrs) = a hundred and eighty × ¾ = 135 km/hr.

Abaxis Aptitude Interview Questions

Question 28. In Jar A, one hundred twenty Litres Milk Was Mixed With 24 Litre Water. 12 Litre Of This Mixture Was Taken Out And 3 Litre Water Was Added. If 27 Litre Of Newly Formed Mixture Is Taken Out, What Will Be The Resultant Quantity Of Water In The Jar? (in Litre)

Answer :

Ratio of milk : water in Jar A = 120 : 24 = five : 1

12 lts of this combination is taken out => milk = 5/6 × 12 = 10 lts and water = 2 lts taken out

three lts of water brought = 24 – 2 + 3 = 25 lts => new ratio of milk : water

= one hundred ten : 25 = 22 : 5

Now 27 lts of this combination is again taken out =>

water taken out = 5/27 × 27 = five lts

water left = 25 – five = 20 lts.

Question 29. A Boat, Whose Speed In 15 Km/hr In Still Water Goes 30 Km Downstream And Comes Back In A Total Of 4 Hours 30 Minutes. What Is The Speed Of The Stream? (in Km/hr)?

Answer :

Speed of boat = 15 ok/h

Let the velocity of stream be ‘S’

Given : 30/(15+S) + 30/(15-S) = nine/2

Solving we get, S = five km/h.

Cognizant Aptitude Interview Questions

Question 30. Six Years Ago, The Ratio Of The Ages Of Kunal And Sagar Was 6 : 5. Four Years Hence, The Ratio Of Their Ages Will Be 11:10. What Is Sagar's Present Age?

Answer :

Let the a long time of Kunal and Sagar be K and S respectively.

Given : (K-6)/(S-6) = 6/5 => 5K – 6S = -6 ……(i)

And: (K+4)/(S+4) = 11/10 => 10K – 11 S = four …..(ii)

Solving (i) & (ii) we get: S = sixteen years.