**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 The Age Of His Mother. What Is The Present Age Of The Mother?****Answer :**Let present age of the mother =5x

Then, present age of the person =2x

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

5x+8=4x+16

X=8

present age of the mother =5x=40.

**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.

**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=6x

^{2}=6×1602=153600 m

^{2}.**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.

**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

=> 16(x-3) + 87 = 17x

=> 16x – 48 + 87 = 17x

=> x = 39.

**Question 9. 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 value for unexpired time

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

Banker’s Discount is the Simple Interest on the face value of the bill for unexpired time

= Simple Interest on Rs. 2160 for unexpired time

=360/1800×2160=1/5×2160=Rs. 432.

**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 together can effectively fill (1/6-1/12)=1/12 tank

=> In 4 hours, Tank A and B can effectively fill 1/12×4=1/3×4=1/3 of the tank.

Time taken to fill the remaining (1-1/3)=23(1-13)=2/3 of the tank =(2/3)(1/6)= 4 hours.

**Question 11. 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 effectively fill 1/8-1/12=1/24 of the cistern in 1 hour

i.e, the cistern will be full in 24 hrs.

**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+1/40=1/8 of the tank in 1 hr

i.e., Pipe A and B together can fill the tank in 8 hours.

**Question 13. A Boat Covers A Certain Distance Downstream In 4 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 speed of the water in still water = x

Given that speed of the stream = 3 kmph

Speed downstream=(x+3) kmph

Speed upstream=(x-3) kmph

He travels a certain distance downstream in 4 hour and come back in 6 hour.

ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour

since distance = speed × time, we have

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

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

?2x+6=3x-9

?x=6+9=15 kmph.

**Question 14. What Day Of The Week Was 1 January 1901?****Answer :**1 Jan 1901 = (1900 years + 1st Jan 1901)

We know that number of odd days in 400 years = 0

Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)

Number of odd days in the period 1601-1900

= Number of odd days in 300 years

= 5 x 3 = 15 = 1

(As we can reduce perfect multiples of 7 from odd days without affecting anything)

1st Jan 1901 = 1 odd day

Total number of odd days = (0 + 1 + 1) = 2

2 odd 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 odd days

Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days)

= Sunday

**Question 16. January 1, 2004 Was A Thursday, What Day Of The Week Lies On January 1 2005?****Answer :**Given that January 1, 2004 was Thursday.

Odd days in 2004 = 2 (because 2004 is a leap year)

(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)

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

**Question 17. A Fort Had Provision Of Food For 150 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 150 men for 45 days

Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days

Remaining men after 10 days = 150 – 25 = 125

Assume that after 10 days,the remaining food is sufficient for 125 men for xx days

More men, Less days (Indirect Proportion)

150 : 125 :: x : 35

150×35=125x

6×35=5X

X=6×7=42.

**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 required actual distance be xx km

More scale distance, More actual distance(direct proportion)

Hence we can write as

(scale distance) 0.6 : 80.5 :: 6.6 : xx

0.6x=80.5×6.6

0.1x=80.5×1.1

x=80.5×11=885.5.

**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 at the starting point after 2772 seconds.

i.e., after 46 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 greatest number which divides 7654, 8506 and 9997 and leaves same remainder

= HCF of 2343, 1491, 852

= 213

Now we need to find out the common remainder.

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

7654 ÷ 213 = 35, remainder = 199.

**Question 21. The Ratio Of Two Numbers Is 4 : 5. If The Hcf Of These Numbers Is 6, What Is Their Lcm?****Answer :**Let the numbers be 4k and 5k

HCF of 4 and 5 = 1

Hence HCF of 4k and 5k = k

Given that HCF of 4k and 5k = 6

=> k = 6

Hence the numbers are (4 × 6) and (5 × 6)

= 24 and 30

LCM of 24 and 30 = 120.

**Question 22. If Log10 5 + Log10 (5x+1) = Log10 (x+5) + 1, Then X Is Equal To?****Answer :**log

_{10}5 + log_{10}(5x+1) = log_{10}(x+5) + 1=> log

_{10}5 + log_{10}(5x+1) = log_{10}(x+5) + log_{10}10=> log

_{10}[5(5x+1)] = log_{10}[10(x+5)]=> 5(5x+1) = 10(x+5)

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

=> 5x + 1 = 2x + 10

=> 3x = 9

=> x = 3.

**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 investment =x.

Then A’s investment =3x

Suppose B’s period of investment =y

then A’s period of investment =2y

A : B =3x×2y:xy=6:1

Total profit ×1/7=4000

=> Total profit =4000×7=28000.

**Question 24. John’s Salary Was Decreased By 50% And Subsequently Increased By 50%. How Much Percent Does He Loss?****Answer :**Let John’s initial salary = Rs.100

After decreasing by 50%, John’s salary = Rs. 50 (because it will become half)

After subsequently increasing by 50%, John’s salary

=50×100+50/100=50×150/100= Rs.75

Loss = 100-75=Rs.25

Loss percent =25/100×100=25%.

**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 all these letters are different.

Total number of words (with or without meaning) that can be formed using all these 5 letters using each letter exactly once

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

= 5P

_{5}=5!=5×4×3×2×1=120.**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 4 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 filled by the inlet pipe in 24hours

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

Therefore, water emptied by the leak in 66 hours

= water filled by the inlet pipe in 88 hours

i.e., capacity of the tank

= water filled by the inlet pipe in 88 hours

=8×60×4=1920=8×60×4=1920 litre.

**Question 27. A Cistern Can Be Filled By A Tap In 3 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 filled by first tap in 11 hour =1/3

Part emptied by second tap 11 hour =1/8

Net part filled by both these taps in 11 hour

=1/3-1/8=5/24

i.e, the cistern gets filled in 24/5 hours =4.8 hours.

**Question 28. John Purchased A Machine For Rs. 80,000.rs. 80,000. 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 :**cost price =80000+5000+1000=86000

profit =25%

selling price =86000+86000×1/4=107500.

**Question 29. A, B And C Are The Three Contestants In One Km Race. If A Can Give B A Start Of 40 Metres And A Can Give C A Start Of 64 Metres. How Many Metres Start Can B Give C?****Answer :**While A covers 1000 m, B covers (1000-40)=960 m and C covers (1000-64)=936 m

i.e., when B covers 960 m, C covers 936 m

When B covers 1000 m, C covers 936/960×1000 = 975 m

i.e., B can give C a start of (1000-975) = 25 m.

**Question 30. In A Game Of 90 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 scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points

i.e., when B scores 75 points, C scores 60 points

=> When B scores 100 points, C scores 60/75×100 = 80 points

i.e., in a game of 100 points, B can give C (100-80)=20 points.

**Question 31. Find The Odd Man Out. 6, 13, 18, 25, 30, 37, 40?****Answer :**The difference between two successive terms from the beginning are 7, 5, 7, 5, 7, 5

Hence, in place of 40, right number is 37+5=42.

**Question 32. Find The Odd Man Out. 445, 221, 109, 46, 25, 11, 4?****Answer :**To obtain next number, subtract 3 from the previous number and divide the result by 2

445

(445-3)/2 = 221

(221-3)/2 = 109

(109-3)/2 = 53

(53-3)/2 = 25

(25-3)/2 = 11

(11-3)/2 = 4

Clearly, 53 should have come in place 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 value of each share = Rs.20

Market value of each share = Rs.25

Number of shares = 12500

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

**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 value of each share = Rs.20

Market value of each share = Rs.25

Number of shares = 12500

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

Mohan further sells the shares at a premium of Rs. 11 each

ie, Mohan further sells the shares at Rs.(20+11) = Rs.31 per share

total amount he gets by selling all the shares = 12500 × 31 = 387500

His gain = 387500 – 312500 = Rs.75000.

**Question 35. The Ratio Between The Speeds Of Two Trains Is 7:87:8. If The Second Train Runs 400400 Km In 44hours, What Is The Speed Of The First Train?****Answer :**Speed of second train =400/4=100 km/hr

Speed of first train : Speed of second train =7:8

Therefore, speed of first train =100/8×7=87.5 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 done by P in 1 day = 1/24

Work done by Q in 1 day = 1/6

Work done by R in 1 day = 1/12

Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24

=> Working together, they will complete the work in 24/7 days = 3 3/7 days.

**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 done by Kamal in 1 day = 1/20

Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16

=> Suresh can complete the work in 16 days.

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