Question 1. A Person's Present Age Is Two-fifth Of The Age Of His Mother. After 88 Years, He Will Be One-half of Of The Age Of His Mother. What Is The Present Age Of The Mother?

Answer :

Let gift age of the mom =5x

Then, present age of the character =2x

5x+eight=2(2x+8)

5x+8=4x+16

X=eight

gift age of the mom =5x=forty.

Question 2. A Is As Much Younger Than B And He Is Older Than C. If The Sum Of The Ages Of B And C Is 5050 Years, What Is Definitely The Difference Between B And A's Age?

Answer :

Age of C << Age of A << Age of B

Given that sum of the ages of B and C is 5050 years.

Now we need to find out (B's age - A's age). But this cannot be determined with the given data.

Aptitude Interview Questions

Question 3. Sobha's Father Was 3838 Years Of Age When She Was Born While Her Mother Was 3636 Years Old When Her Brother Four Years Younger To Her Was Born. What Is The Difference Between The Ages Of Her Parents?

Answer :

Age of Sobha's father when Sobha was born =38=38

Age of Sobha's mother when Sobha was born =36-4=32=36-4=32

Required difference of age =38-32=6.

Question 4. The Age Of Father 1010 Years Ago Was Thrice The Age Of His Son. Ten Years Hence, Father's Age Will Be Twice That Of His Son. What Is The Ratio Of Their Present Ages?

Answer :

Let age of the son before 1010 years =x=x and

age of the father before 1010 years =3x=3x

(3x+20)=2(x+20)

Age of the son at present =x+10=20+10=30

Age of the father at present =3x+10=3×20+10=70

Required ratio =70:30=7:3.

Question 5. The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3:23:2. If A Man Cycling Along The Boundary Of The Park At The Speed Of 1212 Km/hr Completes One Round In 88minutes, Then What Is The Area Of The Park (in Sq. M)?

Answer :

Let length =3xkm,

breadth =2xkm

Distance travelled by the man at the speed of 1212 km/hr in 88 minutes =2(3x+2x)=10x

Therefore,

12×8/60=10x

x=4/25 km=160 m

Area =3x×2x=6x2

=6×1602=153600 m2.

HR Interview Questions

Question 6. What Is The Percentage Increase In The Area Of A Rectangle, If Each Of Its Sides Is Increased By 20%?

Answer :

Change in area

=(20+20+20×20/100)%=44%

i.E., area is increased by 44%.

Question 7. There Are Two Divisions A And B Of A Class, Consisting Of 36 And 44 Students Respectively. If The Average Weight Of Divisions A Is 40 Kg And That Of Division B Is 35 Kg. What Is The Average Weight Of The Whole Class?

Answer :

Total weight of students in division A = 36 × 40

Total weight of students in division B = 44 × 35

Total students = 36 + 44 = 80

Average weight of the whole class

=(36×40)+(44×35)/80

=(9×40)+(11×35)/20

=(9×8)+(11×7)/4

=72+77/4

=149/4

=37.25.

ABB Group Aptitude Interview Questions

Question 8. A Batsman Makes A Score Of 87 Runs In The 17th Inning And Thus Increases His Averages By 3. What Is His Average After 17th Inning?

Answer :

Let the average after 17 innings = x

Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)

Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings

=> sixteen(x-3) + 87 = 17x

=> 16x - forty eight + 87 = 17x

=> x = 39.

Question nine. The True Discount On A Bill Of Rs. 2160 Is Rs. 360. What Is The Banker's Discount?

Answer :

F = Rs. 2160

TD = Rs. 360

PW = F - TD = 2160 - 360 = Rs. 1800

True Discount is the Simple Interest on the present price for unexpired time

=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360

Banker's Discount is the Simple Interest at the face fee of the bill for unexpired time

= Simple Interest on Rs. 2160 for unexpired time

=360/1800×2160=1/five×2160=Rs. 432.

Abaxis Aptitude Interview Questions

Question 10. Tap 'a' Can Fill The Tank Completely In 6 Hrs While Tap 'b' Can Empty It By 12 Hrs. By Mistake, The Person Forgot To Close The Tap 'b', As A Result, Both The Taps, Remained Open. After 4 Hrs, The Person Realized The Mistake And Immediately Closed The Tap 'b'. In How Much Time Now Onwards, Would The Tank Be Full?

Answer :

Tap A can fill the tank completely in 6 hours

=> In 1 hour, Tap A can fill 1/6 of the tank

Tap B can empty the tank completely in 12 hours

=> In 1 hour, Tap B can empty 1/12 of the tank

i.E., In one hour, Tank A and B collectively can correctly fill (1/6-1/12)=1/12 tank

=> In 4 hours, Tank A and B can correctly fill 1/12×four=1/three×4=1/three of the tank.

Time taken to fill the remaining (1-1/3)=23(1-13)=2/3 of the tank =(2/3)(1/6)= four hours.

Question eleven. A Cistern Is Filled By Pipe A In 8 Hrs And The Full Cistern Can Be Leaked Out By An Exhaust Pipe B In 12 Hrs. If Both The Pipes Are Opened In What Time The Cistern Is Full?

Answer :

Pipe A can fill 1/8 of the cistern in 1 hour.

Pipe B can empty 1/12 of the cistern in 1 hour

Both Pipe A and B together can efficiently fill 1/8-1/12=1/24 of the cistern in 1 hour

i.E, the cistern can be complete in 24 hrs.

Oracle Aptitude Interview Questions

Question 12. Two Pipes A And B Can Fill A Tank In 10 Hrs And 40 Hrs Respectively. If Both The Pipes Are Opened Simultaneously, How Much Time Will Be Taken To Fill The Tank?

Answer :

Pipe A can fill 1/10 of the tank in 1 hr

Pipe B can fill 1/40 of the tank in 1 hr

Pipe A and B together can fill 1/10+1/forty=1/8 of the tank in 1 hr

i.E., Pipe A and B together can fill the tank in eight hours.

Aptitude Interview Questions

Question thirteen. A Boat Covers A Certain Distance Downstream In four Hours But Takes 6 Hours To Return Upstream To The Starting Point. If The Speed Of The Stream Be 3 Km/hr, Find The Speed Of The Boat In Still Water?

Answer :

Let the velocity of the water in nonetheless water = x

Given that speed of the move = three kmph

Speed downstream=(x+3) kmph

Speed upstream=(x-three) kmph

He travels a certain distance downstream in four hour and are available lower back in 6 hour.

Ie, distance travelled downstream in four hour = distance travelled upstream in 6 hour

for the reason that distance = velocity × time, we've

(x+three)4=(x-3)6

?(x+three)2=(x-3)three

?2x+6=3x-nine

?X=6+nine=15 kmph.

Question 14. What Day Of The Week Was 1 January 1901?

Answer :

1 Jan 1901 = (1900 years + 1st Jan 1901)

We understand that quantity of strange days in four hundred years = 0

Hence the variety of peculiar days in 1600 years = zero (Since 1600 is an excellent more than one of four hundred)

Number of strange days within the period 1601-1900

= Number of atypical days in 300 years

= five x 3 = 15 = 1

(As we can lessen ideal multiples of 7 from strange days with out affecting some thing)

1st Jan 1901 = 1 ordinary day

Total range of extraordinary days = (0 + 1 + 1) = 2

2 abnormal days = Tuesday

Hence 1 January 1901 is Tuesday.

Question 15. Today Is Thursday. The Day After 59 Days Will Be?

Answer :

59 days = 8 weeks 3 days = 3 abnormal days

Hence if these days is Thursday, After fifty nine days, it will be = (Thursday + three unusual days)

= Sunday

Infosys Aptitude Interview Questions

Question sixteen. January 1, 2004 Was A Thursday, What Day Of The Week Lies On January 1 2005?

Answer :

Given that January 1, 2004 become Thursday.

Odd days in 2004 = 2 (due to the fact 2004 is a soar year)

(Also observe that we've got taken the whole year 2004 due to the fact we want to find out the atypical days from 01-Jan-2004 to 31-Dec-2004, this is the whole 12 months 2004)

Hence January 1, 2005 = (Thursday + 2 unusual days) = Saturday.

Question 17. A Fort Had Provision Of Food For a hundred and fifty Men For 45 Days. After 10 Days, 25 Men Left The Fort. Find Out The Number Of Days For Which The Remaining Food Will Last?

Answer :

Given that fort had provision of food for one hundred fifty guys for forty five days

Hence, after 10 days, the remaining food is sufficient for a hundred and fifty men for 35 days

Remaining men after 10 days = one hundred fifty - 25 = one hundred twenty five

Assume that once 10 days,the ultimate food is sufficient for a hundred twenty five men for xx days

More guys, Less days (Indirect Proportion)

150 : a hundred twenty five :: x : 35

a hundred and fifty×35=125x

6×35=5X

X=6×7=42.

IBM Aptitude Interview Questions

Question 18. On A Scale Of A Map 0.6 Cm Represents 6.6km. If The Distance Between Two Points On The Map Is 80.5 Cm ,that Is The The Actual Distance Between These Points?

Answer :

Let the desired real distance be xx km

More scale distance, More actual distance(direct percentage)

Hence we are able to write as

(scale distance) zero.6 : 80.5 :: 6.6 : xx

0.6x=eighty.Five×6.6

0.1x=eighty.5×1.1

x=80.5×eleven=885.Five.

HR Interview Questions

Question 19. A, B And C Start At The Same Time In The Same Direction To Run Around A Circular Stadium. A Completes A Round In 252 Seconds, B In 308 Seconds And C In 198 Seconds, All Starting At The Same Point. After What Time Will They Again At The Starting Point ?

Answer :

LCM of 252, 308 and 198 = 2772

Hence they all will be again on the starting point after 2772 seconds.

I.E., after forty six minutes 12 seconds.

Question 20. A Boy Divided The Numbers 7654, 8506 And 9997 By A Certain Largest Number And He Gets Same Remainder In Each Case. What Is The Common Remainder?

Answer :

9997 - 7654 = 2343

9997 - 8506 = 1491

8506 - 7654 = 852

Hence, the best quantity which divides 7654, 8506 and 9997 and leaves equal remainder

= HCF of 2343, 1491, 852

= 213

Now we need to find out the not unusual remainder.

Take any of the given numbers from 7654, 8506 and 9997, say 7654

7654 ÷ 213 = 35, remainder = 199.

Capgemini Aptitude Interview Questions

Question 21. The Ratio Of Two Numbers Is four : five. If The Hcf Of These Numbers Is 6, What Is Their Lcm?

Answer :

Let the numbers be 4k and 5k

HCF of four and 5 = 1

Hence HCF of 4k and 5k = okay

Given that HCF of 4k and 5k = 6

=> ok = 6

Hence the numbers are (four × 6) and (5 × 6)

= 24 and 30

LCM of 24 and 30 = 120.

Question 22. If Log10 five + Log10 (5x+1) = Log10 (x+5) + 1, Then X Is Equal To?

Answer :

log105 + log10 (5x+1) = log10 (x+5) + 1

=> log105 + log10(5x+1) = log10(x+five) + log10 10

=> log10[5(5x+1)] = log10[10(x+5)]

=> five(5x+1) = 10(x+5)

=> 5x+1 = 2(x+five)

=> 5x + 1 = 2x + 10

=> 3x = nine

=> x = three.

Question 23. A And B Started A Partnership Business. A's Investment Was Thrice The Investment Of B And The Period Of His Investment Was Two Times The Period Of Investments Of B. If B Received Rs 4000 As Profit, What Is Their Total Profit?

Answer :

Suppose B's funding =x.

Then A's funding =3x

Suppose B's length of funding =y

then A's period of investment =2y

A : B =3x×2y:xy=6:1

Total profit ×1/7=4000

=> Total income =4000×7=28000.

Impetus Aptitude Interview Questions

Question 24. John's Salary Was Decreased By 50% And Subsequently Increased By 50%. How Much Percent Does He Loss?

Answer :

Let John's preliminary profits = Rs.100

After lowering by using 50%, John's salary = Rs. 50 (as it becomes 1/2)

After in the end growing by means of 50%, John's revenue

=50×a hundred+50/a hundred=50×one hundred fifty/a hundred= Rs.Seventy five

Loss = one hundred-seventy five=Rs.25

Loss percent =25/100×a hundred=25%.

ABB Group Aptitude Interview Questions

Question 25. How Many Words With Or Without Meaning, Can Be Formed By Using All The Letters Of The Word, 'delhi' Using Each Letter Exactly Once?

Answer :

The word 'DELHI' has 5 letters and a lot of these letters are different.

Total variety of phrases (without or with which means) that can be fashioned the usage of a majority of these 5 letters the usage of each letter precisely once

= Number of preparations of 5 letters taken all at a time

= 5P<sub>5</sub> =five!=5×4×3×2×1=one hundred twenty.

Question 26. A Leak In The Bottom Of A Tank Can Empty The Full Tank In 6 Hours. An Inlet Pipe Fills Water At The Rate Of four Liters A Minute. When The Tank Is Full, The Inlet Is Opened And Due To The Leak, The Tank Is Empty In 24 Hours. How Many Liters Does The Tank Hold?

Answer :

water stuffed with the aid of the inlet pipe in 24hours

= water emptied through the leak in 24-6=1824-6=18 hours.

Therefore, water emptied by using the leak in sixty six hours

= water filled by way of the inlet pipe in 88 hours

i.E., capacity of the tank

= water stuffed by means of the inlet pipe in 88 hours

=eight×60×4=1920=8×60×4=1920 litre.

Question 27. A Cistern Can Be Filled By A Tap In three Hours While It Can Be Emptied By Another Tap In 8 Hours. If Both The Taps Are Opened Simultaneously, Then After How Much Time Will The Cistern Get Filled?

Answer :

Part crammed via first faucet in 11 hour =1/three

Part emptied by means of 2d tap 11 hour =1/eight

Net part filled by both those faucets in eleven hour

=1/three-1/8=five/24

i.E, the cistern receives filled in 24/5 hours =four.Eight hours.

Abaxis Aptitude Interview Questions

Question 28. John Purchased A Machine For Rs. Eighty,000.Rs. Eighty,000. After Spending Rs. 5000rs. 5000 On Repair And Rs. 1000rs. 1000 On Transport He Sold It With 25%25% Profit. What Price Did He Sell The Machine?

Answer :

fee rate =80000+5000+one thousand=86000

income =25%

selling fee =86000+86000×1/four=107500.

Question 29. A, B And C Are The Three Contestants In One Km Race. If A Can Give B A Start Of forty Metres And A Can Give C A Start Of 64 Metres. How Many Metres Start Can B Give C?

Answer :

While A covers one thousand m, B covers (a thousand-40)=960 m and C covers (a thousand-64)=936 m

i.E., whilst B covers 960 m, C covers 936 m

When B covers one thousand m, C covers 936/960×a thousand = 975 m

i.E., B can provide C a begin of (1000-975) = 25 m.

Question 30. In A Game Of ninety Points A Can Give B 15 Points And C 30 Points. How Many Points Can B Give C In A Game Of 100 Points?

Answer :

While A ratings ninety factors, B ratings (90-15)=75 factors and C rankings (90-30)= 60 factors

i.E., when B scores seventy five factors, C rankings 60 factors

=> When B rankings 100 points, C rankings 60/75×one hundred = eighty points

i.E., in a game of one hundred factors, B can deliver C (100-eighty)=20 points.

Question 31. Find The Odd Man Out. 6, thirteen, 18, 25, 30, 37, forty?

Answer :

The distinction among successive phrases from the beginning are 7, five, 7, five, 7, 5

Hence, in area of 40, proper number is 37+5=42.

Question 32. Find The Odd Man Out. 445, 221, 109, forty six, 25, 11, four?

Answer :

To achieve subsequent wide variety, subtract 3 from the preceding number and divide the end result through 2

445

(445-3)/2 = 221

(221-three)/2 = 109

(109-three)/2 = 53

(fifty three-three)/2 = 25

(25-3)/2 = 11

(11-3)/2 = 4

Clearly, fifty three ought to have are available in region of 46.

Question 33. 12500 Shares, Of Par Value Rs. 20 Each, Are Purchased From Ram By Mohan At A Price Of Rs. 25 Each. Find The Amount Required To Purchase The Shares?

Answer :

Face price of every share = Rs.20

Market cost of every proportion = Rs.25

Number of shares = 12500

Amount required to purchase the shares = 12500 × 25 = 312500.

Oracle Aptitude Interview Questions

Question 34. 12500 Shares, Of Par Value Rs. 20 Each, Are Purchased From Ram By Mohan At A Price Of Rs. 25 Each. If Mohan Further Sells The Shares At A Premium Of Rs. 11 Each, Find His Gain In The Transaction?

Answer :

Face fee of each proportion = Rs.20

Market price of each proportion = Rs.25

Number of shares = 12500

Amount required to purchase the stocks = 12500 × 25 = 312500

Mohan further sells the stocks at a premium of Rs. Eleven every

ie, Mohan similarly sells the shares at Rs.(20+eleven) = Rs.31 in keeping with percentage

general quantity he receives with the aid of promoting all of the stocks = 12500 × 31 = 387500

His benefit = 387500 - 312500 = Rs.75000.

Question 35. The Ratio Between The Speeds Of Two Trains Is 7:87:eight. If The Second Train Runs 400400 Km In 44hours, What Is The Speed Of The First Train?

Answer :

Speed of 2nd educate =400/4=one hundred km/hr

Speed of first train : Speed of 2d train =7:8

Therefore, velocity of first teach =one hundred/8×7=87.Five km/hr.

Question 36. P, Q And R Can Complete A Work In 24, 6 And 12 Days Respectively. The Work Will Be Completed In --- Days If All Of Them Are Working Together?

Answer :

Work achieved by P in 1 day = 1/24

Work performed by using Q in 1 day = 1/6

Work completed by way of R in 1 day = 1/12

Work executed with the aid of P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24

=> Working together, they may complete the paintings in 24/7 days = three three/7 days.

Infosys Aptitude Interview Questions

Question 37. Kamal Will Complete Work In 20 Days. If Suresh Is 25% More Efficient Than Kamal, He Can Complete The Work In --- Days?

Answer :

Work achieved by way of Kamal in 1 day = 1/20

Work performed with the aid of Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16

=> Suresh can entire the paintings in 16 days.