Inautix Aptitude Placement Papers - Inautix Aptitude Interview Questions and Answers
Q1. 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?
Let age of the son earlier than 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 gift =3x+10=three×20+10=70
Required ratio =70:30=7:three.
Q2. Today Is Thursday. The Day After fifty nine Days Will Be?
Fifty nine days = 8 weeks three days = three abnormal days
Hence if today is Thursday, After 59 days, it is going to be = (Thursday + three abnormal days)
= Sunday
Q3. What Day Of The Week Was 1 January 1901?
1 Jan 1901 = (1900 years + 1st Jan 1901)
We recognize that quantity of unusual days in four hundred years = 0
Hence the number of unusual days in 1600 years = 0 (Since 1600 is an excellent a couple of of 400)
Number of ordinary days in the duration 1601-1900
= Number of unusual days in 300 years
= 5 x 3 = 15 = 1
(As we will reduce best multiples of seven from extraordinary days with out affecting some thing)
1st Jan 1901 = 1 unusual day
Total wide variety of strange days = (zero + 1 + 1) = 2
2 bizarre days = Tuesday
Hence 1 January 1901 is Tuesday.
Q4. There Are Two Divisions A And B Of A Class, Consisting Of 36 And forty four Students Respectively. If The Average Weight Of Divisions A Is forty Kg And That Of Division B Is 35 Kg. What Is The Average Weight Of
Total weight of students in department A = 36 × forty
Total weight of students in division B = forty four × 35
Total students = 36 + 44 = eighty
Average weight of the whole magnificence
=(36×forty)+(forty four×35)/eighty
=(9×40)+(eleven×35)/20
=(9×8)+(eleven×7)/four
=72+seventy seven/four
=149/4
=37.25.
Q5. The Ratio Between The Speeds Of Two Trains Is 7:87:
Speed of second train =four hundred/four=100 km/hr
Speed of first teach : Speed of 2nd train =7:8
Therefore, velocity of first teach =a hundred/8×7=87.Five km/hr.
Q6. 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?
Part crammed through first tap in 11 hour =1/three
Part emptied by means of 2nd faucet 11 hour =1/8
Net part crammed by way of each those taps in 11 hour
=1/3-1/eight=5/24
i.E, the cistern receives stuffed in 24/5 hours =4.Eight hours.
Q7. 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. Eleven Each, Find His Gain In The Traction?
Face cost of every share = Rs.20
Market fee of every proportion = Rs.25
Number of shares = 12500
Amount required to purchase the stocks = 12500 × 25 = 312500
Mohan further sells the shares at a premium of Rs. 11 every
ie, Mohan further sells the shares at Rs.(20+eleven) = Rs.31 per share
general quantity he gets by way of promoting all of the shares = 12500 × 31 = 387500
His advantage = 387500 - 312500 = Rs.75000.
Q8. Kamal Will Complete Work In 20 Days. If Suresh Is 25% More Efficient Than Kamal, He Can Complete The Work In --- Days?
Work finished by way of Kamal in 1 day = 1/20
Work carried out with the aid of Suresh in 1 day = (1/20) × (one hundred twenty five/a hundred) = five/80 = 1/sixteen
=> Suresh can whole the paintings in 16 days.
Q9. Find The Odd Man Out. 445, 221, 109, 46, 25, eleven, 4?
To reap subsequent variety, subtract three from the previous variety and divide the result through 2
445
(445-three)/2 = 221
(221-3)/2 = 109
(109-three)/2 = 53
(53-three)/2 = 25
(25-3)/2 = 11
(eleven-three)/2 = four
Clearly, 53 need to have come in place of 46.
Q10. 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 a hundred Points?
While A rankings ninety factors, B rankings (ninety-15)=75 points and C ratings (90-30)= 60 factors
i.E., when B rankings seventy five factors, C rankings 60 factors
=> When B scores 100 points, C scores 60/75×100 = 80 points
i.E., in a sport of 100 factors, B can supply C (a hundred-80)=20 factors.
Q11. 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, Wh
Suppose B's investment =x.
Then A's funding =3x
Suppose B's length of funding =y
then A's duration of funding =2y
A : B =3x×2y:xy=6:1
Total income ×1/7=4000
=> Total earnings =4000×7=28000.
Q12. 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 sixty four Metres. How Many Metres Start Can B Give C?
While A covers 1000 m, B covers (one thousand-40)=960 m and C covers (one thousand-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 deliver C a begin of (1000-975) = 25 m.
Q13. January 1, 2004 Was A Thursday, What Day Of The Week Lies On January 1 2005?
Given that January 1, 2004 changed into Thursday.
Odd days in 2004 = 2 (because 2004 is a leap 12 months)
(Also observe that we have taken the whole year 2004 because we want to find out the extraordinary 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.
Q14. A Cistern Is Filled By Pipe A In eight 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?
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 correctly fill 1/8-1/12=1/24 of the cistern in 1 hour
i.E, the cistern can be full in 24 hrs.
Q15. 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?
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.
Q16. 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?
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.
Q17. What Is The Percentage Increase In The Area Of A Rectangle, If Each Of Its Sides Is Increased By 20%?
Change in area
=(20+20+20×20/100)%=44%
i.E., area is increased by 44%.
Q18. 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?
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.
Q19. 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. Aft
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.
Q20. Find The Odd Man Out. 6, 13, 18, 25, 30, 37, 40?
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.
Q21. 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?
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.
Q22. 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?
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 collectively, they may complete the work in 24/7 days = three three/7 days.
Q23. The Ratio Between The Length And The Breadth Of A Rectangular Park Is 3:23:
Let duration =3xkm,
breadth =2xkm
Distance travelled by the man at the rate of 1212 km/hr in 88 mins =2(3x+2x)=10x
Therefore,
12×eight/60=10x
x=four/25 km=a hundred and sixty m
Area =3x×2x=6x2
=6×1602=153600 m2.
Q24. A Fort Had Provision Of Food For one hundred fifty Men For forty five Days. After 10 Days, 25 Men Left The Fort. Find Out The Number Of Days For Which The Remaining Food Will Last?
Given that citadel had provision of food for 150 men for 45 days
Hence, after 10 days, the remaining food is sufficient for one hundred fifty men for 35 days
Remaining men after 10 days = a hundred and fifty - 25 = a hundred twenty five
Assume that when 10 days,the final food is enough for one hundred twenty five men for xx days
More guys, Less days (Indirect Proportion)
150 : 125 :: x : 35
one hundred fifty×35=125x
6×35=5X
X=6×7=forty two.
Q25. If Log10 five + Log10 (5x+1) = Log10 (x+5) + 1, Then X Is Equal To?
Log105 + log10 (5x+1) = log10 (x+five) + 1
=> log105 + log10(5x+1) = log10(x+5) + log10 10
=> log10[5(5x+1)] = log10[10(x+5)]
=> 5(5x+1) = 10(x+5)
=> 5x+1 = 2(x+five)
=> 5x + 1 = 2x + 10
=> 3x = nine
=> x = 3.
Q26. 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?
Age of Sobha's father whilst Sobha was born =38=38
Age of Sobha's mother when Sobha became born =36-four=32=36-four=32
Required difference of age =38-32=6.
Q27. 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?
Face cost of every percentage = Rs.20
Market value of every percentage = Rs.25
Number of stocks = 12500
Amount required to purchase the shares = 12500 × 25 = 312500.
Q28. The True Discount On A Bill Of Rs. 2160 Is Rs. 36
F = Rs. 2160
TD = Rs. 360
PW = F - TD = 2160 - 360 = Rs. 1800
True Discount is the Simple Interest on the existing value for unexpired time
=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360
Banker's Discount is the Simple Interest at the face cost of the bill for unexpired time
= Simple Interest on Rs. 2160 for unexpired time
=360/1800×2160=1/5×2160=Rs. 432.
Q29. How Many Words With Or Without Meaning, Can Be Formed By Using All The Letters Of The Word, 'delhi' Using Each Letter Exactly Once?
The word 'DELHI' has five letters and these types of letters are distinctive.
Total wide variety of phrases (without or with which means) that can be formed the use of a majority of these 5 letters the usage of each letter precisely as soon as
= Number of arrangements of five letters taken all at a time
= 5P<sub>five</sub> =five!=5×4×three×2×1=120.
Q30. 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 three Km/hr, Find The Speed Of The Boat In Still Water?
Let the velocity of the water in still water = x
Given that velocity of the move = three kmph
Speed downstream=(x+three) kmph
Speed upstream=(x-three) kmph
He travels a sure distance downstream in four hour and are available again in 6 hour.
Ie, distance travelled downstream in four hour = distance travelled upstream in 6 hour
in view that distance = velocity × time, we have
(x+three)four=(x-three)6
?(x+3)2=(x-3)three
?2x+6=3x-9
?X=6+nine=15 kmph.
Q31. John Purchased A Machine For Rs. 80,000.Rs. Eighty,0
value fee =80000+5000+1000=86000
earnings =25%
promoting rate =86000+86000×1/four=107500.
Q32. 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?
Let the specified actual distance be xx km
More scale distance, More real distance(direct percentage)
Hence we will write as
(scale distance) zero.6 : eighty.Five :: 6.6 : xx
0.6x=eighty.Five×6.6
zero.1x=eighty.Five×1.1
x=80.Five×eleven=885.5.
Q33. John's Salary Was Decreased By 50% And Subsequently Increased By 50%. How Much Percent Does He Loss?
Let John's preliminary salary = Rs.A hundred
After decreasing by 50%, John's revenue = Rs. 50 (because it becomes half)
After eventually increasing with the aid of 50%, John's revenue
=50×100+50/one hundred=50×one hundred fifty/100= Rs.Seventy five
Loss = a hundred-seventy five=Rs.25
Loss percent =25/a hundred×one hundred=25%.
Q34. The Ratio Of Two Numbers Is four :
Let the numbers be 4k and 5k
HCF of four and five = 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 = one hundred twenty.
Q35. 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
water crammed via 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 using the inlet pipe in 88 hours
i.E., potential of the tank
= water stuffed by the inlet pipe in 88 hours
=eight×60×4=1920=eight×60×4=1920 litre.
Q36. A Batsman Makes A Score Of 87 Runs In The seventeenth Inning And Thus Increases His Averages By
Let the common after 17 innings = x
Total runs scored in 17 innings = 17x
Average after 16 innings = (x-three)
Total runs scored in sixteen innings = 16(x-3)
Total runs scored in sixteen innings + 87 = Total runs scored in 17 innings
=> 16(x-3) + 87 = 17x
=> 16x - forty eight + 87 = 17x
=> x = 39.
Q37. 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 Pers
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 absolutely 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 successfully fill (1/6-1/12)=1/12 tank
=> In 4 hours, Tank A and B can correctly fill 1/12×four=1/3×four=1/3 of the tank.
Time taken to fill the ultimate (1-1/3)=23(1-13)=2/three of the tank =(2/three)(1/6)= four hours.
