Top 100+ Rattanindia Interview Questions And Answers
Question 1. The Ages Of Two Persons Differ By 1616 Years. 66 Years Ago, The Elder One Was 33 Times As Old As The Younger One. What Are Their Present Ages Of The Elder Person?
Answer :
Let gift age of the elder man or woman =x and
gift age of the younger character =x−sixteen
(x−6)=three(x−sixteen−6)
⇒x−6=3x−66
⇒2x=60
⇒x=60/2=30.
Question 2. Present Age Of A Father Is 33 Years More Than Three Times The Age Of His Son. Three Years Hence, Father's Age Will Be 1010 Years More Than Twice The Age Of The Son. What Is Father's Present Age?
Answer :
Let the present age the son =x
Then, gift age of the father =3x+3
Given that, 3 years as a result, father's age could be 1010 years more than two times the age of the son
⇒(3x+3+three)=2(x+3)+10
⇒x=10
Father's present age
=3x+3=three×10+3=33.
HR Management Interview Questions
Question three. Kamal Was forty four Times As Old As His Son 88 Years Ago. After 88 Years, Kamal Will Be Twice As Old As His Son. Find Out The Present Age Of Kamal?
Answer :
Let age of the son before 88 years =x
Then, age of Kamal before 88 years in the past =4x
After 88 years, Kamal might be two times as antique as his son
⇒4x+16=2(x+sixteen)
⇒x=eight
Present age of Kamal
=4x+8=4×8+8=forty.
Question four. The Average Age Of 36 Students In A Group Is 14 Years. When Teacher's Age Is Included To It, The Average Increases By One. Find Out The Teacher's Age In Years?
Answer :
common age of 36 students in a set is 14
Sum of the ages of 36 college students = 36 × 14
When instructor's age is protected to it, the average will increase through one
=> average = 15
Sum of the ages of 36 students and the instructor = 37 × 15
Hence instructors age
= 37 × 15 - 36 × 14
= 37 × 15 - 14(37 - 1)
= 37 × 15 - 37 × 14 + 14
= 37(15 - 14) + 14
= 37 + 14
= 51.
HR Management Tutorial
Question 5. The Average Of Five Numbers Id 27. If One Number Is Excluded, The Average Becomes 25. What Is The Excluded Number?
Answer :
Sum of five numbers = 5 × 27
Sum of 4 numbers after except for one wide variety = four × 25
Excluded wide variety
= five × 27 - four × 25
= 135 - 100 = 35.
Aptitude Interview Questions
Question 6. The Average Of Six Numbers Is X And The Average Of Three Of These Is Y. If The Average Of The Remaining Three Is Z, Then?
Answer :
Average of 6 numbers = x
=> Sum of 6 numbers = 6x
Average of the three numbers = y
=> Sum of those three numbers = 3y
Average of the ultimate 3 numbers = z
=> Sum of the last 3 numbers = 3z
Now we understand that 6x = 3y + 3z
=> 2x = y + z.
Question 7. The Average Age Of A Husband And His Wife Was 23 Years At The Time Of Their Marriage. After Five Years They Have A One Year Old Child. What Is The Average Age Of The Family ?
Answer :
Total age of husband and spouse (on the time in their marriage) = 2 × 23 = forty six
Total age of husband and wife after five years + Age of the 1 year old toddler
= forty six + 5 + five + 1 = 57
Average age of the own family = 57/three = 19.
HR Interview Questions
Question 8. If A Man Rows At The Rate Of five Kmph In Still Water And His Rate Against The Current Is 3 Kmph, Then The Man's Rate Along The Current Is?
Answer :
Let the charge along with the contemporary is xx km/hr
x+3/2=five
⇒x+3=10
⇒x=7 kmph.
Question 9. A Boatman Can Row ninety six Km Downstream In eight Hr. If The Speed Of The Current Is four Km/hr, Then Find In What Time Will Be Able To Cover eight Km Upstream?
Answer :
Speed downstream =ninety six/8 = 12 kmph
Speed of current = four km/hr
Speed of the boatman in nevertheless water = 12-four = 8 kmph
Speed upstream = 8-4 = 4 kmph
Time taken to cover eight km upstream =8/4 = 2 hours.
Abaxis Aptitude Interview Questions
Question 10. If The First Day Of A Year (other Than Leap Year) Was Friday, Then Which Was The Last Day Of That Year?
Answer :
Given that first day of a everyday yr turned into Friday
Odd days of the noted 12 months = 1 (Since it's far an regular yr)
Hence First day of the following year = (Friday + 1 Odd day) = Saturday
Therefore, remaining day of the cited yr = Friday.
Question eleven. If 1st October Is Sunday, Then 1st November Will Be?
Answer :
Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 peculiar days
(As we are able to lessen multiples of seven from atypical days if you want to not trade some thing)
Hence 1st November = (Sunday + three bizarre days) = Wednesday.
BHEL Aptitude Interview Questions
Question 12. 1.12.91 Is The First Sunday. Which Is The Fourth Tuesday Of December 91?
Answer :
Given that 1.12.Ninety one is the primary Sunday
Hence we can anticipate that three.12.Ninety one is the first Tuesday
If we upload 7 days to 3.12.91, we will get second Tuesday
If we upload 14 days to a few.12.Ninety one, we will get third Tuesday
If we upload 21 days to three.12.Ninety one, we will get fourth Tuesday
=> fourth Tuesday = (3.12.Ninety one + 21 days) = 24.12.91.
HR Management Interview Questions
Question 13. If The Seventh Day Of A Month Is Three Days Earlier Than Friday, What Day Will It Be On The Nineteenth Day Of The Month?
Answer :
Given that 7th day of a month is three days in advance than Friday
=> Seventh day is Tuesday
=> 14th is Tuesday
=> 19th is Sunday.
Question 14. A Rope Can Make 70 Rounds Of The Circumference Of A Cylinder Whose Radius Of The Base Is 14cm. How Many Times Can It Go Round A Cylinder Having Radius 20 Cm?
Answer :
Let the desired wide variety of rounds be x
More radius, less rounds(Indirect share)
Hence we will write as
(radius) 14 : 20 :: xx : 70
⇒14×70=20x
⇒14×7=2x
⇒x=7×7=forty nine.
Question 15. A Garrison Of 500 Persons Had Provisions For 27 Days. After three Days A Reinforcement Of three hundred Persons Arrived. For How Many More Days Will The Remaining Food Last Now?
Answer :
Given that citadel had provision for 500 men and women for 27 days
Hence, after three days, the remaining meals is enough for 500 individuals for twenty-four days
Remaining persons after 3 days = 500 + 300 = 800
Assume that once three days,the last food is sufficient for 800 individuals for xx days
More guys, Less days (Indirect Proportion)
(men) 500 : 800 :: xx : 24
⇒500×24=800x
⇒5×24=8x
⇒x=5×three=15.
Mu Sigma Aptitude Interview Questions
Question sixteen. A Hostel Had Provisions For 250 Men For forty Days. If 50 Men Left The Hostel, How Long Will The Food Last At The Same Rate?
Answer :
A hostel had provisions for 250 guys for forty days
If 50 men leaves the hostel, ultimate men = 250 - 50 = two hundred
We need to find out how lengthy the food will ultimate for these two hundred guys.
Let the desired variety of days = xx days
More men, Less days (Indirect Proportion)
(guys) 250 : 200 :: xx : forty
⇒250×40=200x
⇒5×forty=4x
⇒x=five×10=50.
Question 17. What Is The Least Multiple Of 7 Which Leaves A Remainder Of 4 When Divided By 6, nine, 15 And 18 ?
Answer :
LCM of 6, nine, 15 and 18 = 90
Required Number = (90k + four) which is a a couple of of seven
Put k = 1. We get variety as (90 × 1) + four = ninety four. But this isn't always a multiple of seven
Put ok = 2. We get variety as (ninety × 2) + 4 = 184. But this isn't a multiple of 7
Put okay = 3. We get quantity as (90 × three) + four = 274. But this isn't a more than one of 7
Put ok = 4. We get wide variety as (ninety × 4) + 4 = 364. This is a more than one of 7
Hence 364 is the solution.
IBM Aptitude Interview Questions
Question 18. Three Numbers Which Are Co-prime To Each Other Are Such That The Product Of The First Two Is 119 And That Of The Last Two Is 391. What Is The Sum Of The Three Numbers?
Answer :
Since the numbers are co-prime, their HCF = 1
Product of first two numbers = 119
Product of final two numbers = 391
The middle quantity is commonplace in each of those products. Hence, if we take HCF of 119 and 391, we get the common center variety.
HCF of 119 and 391 = 17
=> Middle Number = 17
First Number =119/17=7
Last Number =391/17=23
Sum of the 3 numbers = 7 + 17 + 23 = 47.
Aptitude Interview Questions
Question 19. Six Bells Start Ringing Together And Ring At Intervals Of 4, eight, 10, 12, 15 And 20 Seconds Respectively. How Many Times Will They Ring Together In 60 Minutes ?
Answer :
LCM of four, 8, 10, 12, 15 and 20 = 120
a hundred and twenty seconds = 2 mins
Hence all the six bells will ring together in each 2 mins
Hence, number of instances they may ring together in 60 minutes =1+60/2=31.
Question 20. 3 Litre Of Water Is Added To eleven Litre Of A Solution Containing forty two% Of Alcohol In The Water. The Percentage Of Alcohol In The New Mixture Is?
Answer :
We have a 11 litre answer containing 42% of alcohol within the water.
=> Quantity of alcohol within the solution =eleven×42/a hundred
Now three litre of water is introduced to the solution.
=> Total quantity of the new answer = eleven + 3 = 14
Percentage of alcohol within the new solution =11×forty two/one hundred/14×one hundred
=11×three/one hundred=33%.
Capgemini Aptitude Interview Questions
Question 21. ? + 3699 + 1985 - 2047 = 31111?
Answer :
Let x+3699+1985−2047=31111x+3699+1985−2047=31111
x=31111−3699−1985+2047=27474.
Question 22. What Is The Smallest 6 Digit Number Exactly Divisible By 111?
Answer :
Smallest 6 digit number = one hundred thousand
a hundred thousand÷111 = 900, the rest = a hundred.
Hence eleven have to be delivered to a hundred thousand to get the smallest 6 digit variety exactly divisible by way of 111
Therefore, smallest 6 digit quantity exactly divisible through 111= one hundred thousand + 11 = 100011
Question 23. A And B Starts A Business Investing Rs.85000 And Rs.15000 Respectively. Find Out The Ratio In Which The Profits Should Be Shared?
Answer :
Here A's and B's capitals are there for same time.Hence
A : B = 85000 : 15000
= 85 : 15
= 17 : three.
Cognizant Aptitude Interview Questions
Question 24. P, Q, R Enter Into A Partnership. P Initially Invests 25 Lakh And Adds Another 10 Lakh After One Year. Q Initially Invests 35 Lakh And Withdraws 10 Lakh After 2 Years. R's Investment Is Rs 30 Lakh. In What Ratio Should The Profit Be Divided At The End Of three Years?
Answer :
P : Q : R
=(25×1+35×2):(35×2+25×1)=(25×1+35×2):(35×2+25×1) :(30×three):(30×three)
=95:95:90=19:19:18
HR Interview Questions
Question 25. 30% Of The Men Are More Than 25 Years Old And 80% Of The Men Are Less Than Or Equal To 50 Years Old. 20% Of All Men Play Football. If 20% Of The Men Above The Age Of 50 Play Football, What Percentage Of The Football Players Are Less Than Or Equal To 50 Years?
Answer :
Let total wide variety of guys = one hundred
Then,
20 men play football.
80 guys are much less than or equal to 50 years vintage.
Remaining 20 men are above 50 years old.
Number of football players above 50 years antique
=20×20/100=4
Number of soccer gamers less than or equal to 50 years old
=20−four=sixteen
Required percentage
=sixteen/20×100=eighty%.
Question 26. An Event Manager Has Ten Patterns Of Chairs And Eight Patterns Of Tables. In How Many Ways Can He Make A Pair Of Table And Chair?
Answer :
He has 10 styles of chairs and eight styles of tables
A chair may be decided on in 10 ways.
A desk may be selected in 8 methods.
Hence one chair and one desk can be selected in 10×810×8 ways=80 methods.
Ikanos Aptitude Interview Questions
Question 27. 25 Buses Are Running Between Two Places P And Q. In How Many Ways Can A Person Go From P To Q And Return By A Different Bus?
Answer :
He can pass in any of the 25 buses (25 approaches).
Since he can't come returned in the same bus, he can return in 24 methods.
Total range of methods =25×24=six hundred.
Abaxis Aptitude Interview Questions
Question 28. In How Many Different Ways Can five Girls And five Boys Form A Circle Such That The Boys And The Girls Alternate?
Answer :
Around a circle, 5 boys can be arranged in 4! Approaches.
Given that the men and the women alternate.
Hence there are 5 locations for the ladies. Therefore the ladies may be organized in 5! Approaches.
Total variety of methods
=4!×5!=24×a hundred and twenty=2880.
Question 29. Two Pipes A And B Can Fill A Tank In 2 And 6 Minutes Respectively. If Both The Pipes Are Used Together, Then How Long Will It Take To Fill The Tank?
Answer :
Part crammed by way of first pipe in 11 minute =half of
Part crammed via 2nd pipe in 11 minute =1/6
Net component crammed by way of pipe A and B in eleven minute
=half+1/6=2/3
i.E, pipe A and B collectively can fill the tank in three/2 mins =1.5minutes.
INautix Aptitude Interview Questions
Question 30. One Ball Is Picked Up Randomly From A Bag Containing eight Yellow, 7 Blue And 6 Black Balls. What Is The Probability That It Is Neither Yellow Nor Black?
Answer :
Total quantity of balls, n(S) = 8 + 7 + 6 = 21
n(E) = Number of ways in which a ball can be selected that is neither yellow nor black
= 7 (? there are only 7 balls which might be neither yellow nor black)
P(E) = n(E)/n(S)=7/21=1/3.
Question 31. Two Dice Are Rolled Together. What Is The Probability Of Getting Two Numbers Whose Product Is Even?
Answer :
Total range of outcomes feasible when a die is rolled = 6 (? any individual face out of the 6 faces)
Hence, Total wide variety of outcomes feasible while cube are rolled, n(S) = 6 × 6 = 36
Let E = the occasion of getting two numbers whose product is even
= (1,2), (1,4), (1,6), (2,1), (2,2), (2,three), (2,four), (2,five), (2,6),
(3,2), (three,4), (3,6), (4,1), (4,2), (four,3), (four,four), (4,5), (4,6),
(five,2),(five,4), (five,6), (6,1), (6,2), (6,three), (6,4), (6,5), (6,6)
Hence, n(E) = 27
P(E) = n(E)/n(S)=27/36=three/four.
Question 32. Arun Bought A Computer With 15%15% Discount On The Labelled Price. He Sold The Computer For Rs. 2880rs. 2880 With 20p.C20% Profit On The Labelled Price. At What Price Did He Buy The Computer?
Answer :
selling fee =2880
labelled rate =2880×a hundred/120=2400
charge at which he offered the pc
=2400−2400×15/100=2040.
Question 33. If Selling Price Of An Article Is Rs. 250,rs. 250, Profit Percentage Is 25%.25%. Find The Ratio Of The Cost Price And The Selling Price?
Answer :
promoting rate =250
income =25%
price fee =250×100125=two hundred
Required ratio =200:250=four:5.
BHEL Aptitude Interview Questions
Question 34. A Can Run 220 Metres In forty one Seconds And B In forty four Seconds. By How Many Seconds Will B Win If He Has 30 Metres Start?
Answer :
B has a start of 30 metre
=> A has to run 220 metre and B has to run (220-30)=one hundred ninety metre
Given that A takes 41 seconds to cover this 220 metre
B takes 44 seconds to cowl 220 metre
=> B takes 44220 seconds to cover 1 metre
=> B takes forty four/220×a hundred ninety=420×190 = 38 seconds to cowl a hundred ninety metre
=> B takes forty four/220 seconds to cowl 1 metre
=> B takes forty four/220×one hundred ninety=four/20×one hundred ninety = 38 seconds to cowl a hundred ninety metre
i.E., A beats B through (41-38)= three seconds.
Question 35. In One Km Race A Beats B By 4 Seconds Or forty Metres. How Long Does B Take To Run The Kilometer?
Answer :
This manner, B takes 4 seconds to run 40 metres
=> B takes four/40=1/10 seconds to run 1 metre
=> B takes 1/10×one thousand=a hundred seconds to run one thousand metre.
Question 36. Find The Odd Man Out. 3576, 1784, 888, 440, 216, one zero five, 48?
Answer :
3576
(3576-8)/2 = 1784
(1784-8)/2 = 888
(888-eight)/2 = 440
(440-eight)/2 = 216
(216-8)/2 = 104
(104-8)/2 = 48
Hence, one hundred and five is incorrect. 104 have to have are available in place of a hundred and five.
Mu Sigma Aptitude Interview Questions
Question 37. Andrews Earns An Interest Of Rs. 1596 For The Third Year And Rs. 1400 For The Second Year On The Same Sum. Find The Rate Of Interest If It Is Lent At Compound Interest?
Answer :
Interest earned in 3rd year = Rs. 1596
Interest earned in 2d year = Rs. 1400
i.E, in 3rd year, Andrews receives additional hobby of (Rs. 1596 - Rs. 1400) = Rs.196
This way, Rs.196 is the interest acquired for Rs.1400 for 1 yr
R=a hundred×SI/PT=a hundred×196/1400×1=196/14=14%.
Question 38. In A Group Of Ducks And Cows, The Total Number Of Legs Is 2828 More Than Twice The Number Of Heads. Find The Total Number Of Cows?
Answer :
Let variety of ducks =d=d
Number of cows =c=c
Then, overall variety of legs =2d+4c=2d+4c
Total variety of heads =d+c=d+c
Given that general number of legs is 2828 extra than two times the wide variety of heads
⇒second+4c=28+2(d+c)⇒2c=28⇒c=14⇒2d+4c=28+2(d+c)⇒2c=28⇒c=14
i.E., total quantity of cows =14.
Question 39. A Room Has Equal Number Of Men And Women. Eight Women Left The Room, Leaving Twice As Many Men As Women In The Room. What Was The Total Number Of Men And Women Present In The Room Initially?
Answer :
Let preliminary quantity of men = preliminary wide variety of women =x
2(x−8)=x
⇒2x−sixteen=x
⇒x=16
Total range of women and men
=2x=2×16=32.
Question forty. John Buys one hundred Shares Of Par Value Rs. Five Each, Of A Company, Which Pays An Annual Dividend Of 12% At Such A Price That He Gets 10% On His Investment. Find The Market Value Of A Share?
Answer :
Face value of each share = Rs.5
Total dividend obtained by means of John = one hundred×five×12/100 = Rs.60
Let marketplace price of 100 shares = Rs.X
x×10/one hundred=60
x = six hundred
ie, Market fee of one hundred shares = Rs.Six hundred
Hence, Market value of each percentage = Rs.6
IBM Aptitude Interview Questions
Question forty one. To Produce An Annual Income Of Rs. 800 From A 8% Stock At 90, The Amount Of Stock Needed Is?
Answer :
Since face price isn't given, take it as Rs.100.
As it is an 8% inventory, earnings (dividend) according to stock = Rs.Eight
ie, For an profits of Rs.Eight,amount of stock wanted = Rs.A hundred
For an income of Rs.800, amount of inventory wanted = one hundred×800/8=ten thousand.
Question 42. A Man Sells 4000 Common Shares Of A Company X (each Of Par Value Rs. 10), Which Pays A Dividend Of 40% At Rs. 30 Per Share. He Invests The Sale Proceeds In Ordinary Shares Of Company Y (each Of Par Value Rs. 25) That Pays A Dividend Of 15%. If The Market Value Of Company Y Is Rs. 15, Find The Number Of Shares Of Company Y Purchased By The Man?
Answer :
market Value of Company X (his promoting rate) = Rs.30
Total stocks bought = 4000
Amount he receives = Rs.(4000 × 30)
He invests this quantity in everyday stocks of Company Y
Market Value of Company Y(His purchasing charge) = 15
Number of shares of agency Y which he purchases = 4000×30/15=8000.
Capgemini Aptitude Interview Questions
Question 43. A Company Declared A Semi-annual Dividend Of 12%. Find The Annual Dividend Of Sam Owing 2000 Shares Of The Company Having A Par Value Of Rs. 10 Each?
Answer :
semi-annual dividend = 10×12/100 = Rs.1.2
Total semi-annual dividend = 2000 × 1.2 = Rs.2400
Total annual dividend = 2 × Rs.2400 = Rs.4800.
Question forty four. Arun Invested Rs. 333000 In 5½ % Stocks At a hundred and ten .If Brokerage Is Rs.1, What Is His Annual Income From His Investment?
Answer :
Investment = Rs.333000
Since face value isn't always given, we can take it as Rs.A hundred
and dividend in step with percentage = Rs.11/2
Market Value = a hundred and ten + 1 = 111
Number of stocks purchased = 333000/111 = 3000
Total earnings = 3000×eleven/2 = Rs.16500.
Question 45. The Distance Between Two Cities A And B Is 330330 Km. A Train Starts From A At 88 A.M. And Travels Towards B At 6060 Km/hr. Another Train Starts From B At ninety nine A.M. And Travels Towards A At 7575 Km/hr. At What Time Will They Meet?
Answer :
Assume that they meet xx hours after 88 a.M.
Then, teach 1,1, starting from A, travels xx hours until the trains meet.
Distance travelled by means of educate eleven in xx hours =60x km=60x km
Train 2,2, beginning from B, travels (x−1)(x−1) hours till the trains meet.
Distance travelled by using train 22 in (x−1)(x−1) hours =seventy five(x−1) km=seventy five(x−1) km
Total distance travelled
= Distance travelled through train 11 + Distance travelled by means of educate 22
⇒330=60x+seventy five(x−1)
⇒12x+15(x−1)=sixty six
⇒12x+15x−15=sixty six
⇒27x=sixty six+15=81
⇒3x=nine
⇒x=three.
Question forty six. A Train Travelled At An Average Speed Of 100100 Km/hr, Stopping For 33 Minutes After Every 7575 Km. How Long Did It Take To Reach Its Destination 600600 Km From The Starting Point?
Answer :
Time had to travel 600 km =600/a hundred=6 hour
Now we need to find out the wide variety of stops within the 600600 km journey. Given that the educate stops after every 7575 km.
Six hundred/seventy five=eight
It way, the educate stops 7 times before 600 km and eleven time simply after 600 km. Hence we need to take simplest 7 stops into consideration for the 600 km journey.
Hence, total preventing time in the 600 km journey
=7×three=21minutes
Total time had to attain the vacation spot
=6 hours +21 minutes
=6 hours 21 minutes.
Question forty seven. P Takes Twice As Much Time As Q Or Thrice As Much Time As R To Finish A Piece Of Work. They Can Finish The Work In 2 Days If Work Together. How Much Time Will Q Take To Do The Work Alone?
Answer :
Let P takes x days to finish the work
Then Q takes x/2 days and R takes x/3 days to complete the paintings
Amount of labor P does in 1 day = 1/x
Amount of work Q does in 1 day = 2/x
Amount of labor R does in 1 day = three/x
Amount of labor P,Q and R do in 1 day = 1/x + 2/x + three/x = 1/x (1 + 2 + 3) = 6/x
6/x = 2
=> x = 12
=> Q takes 12/2 days = 6 days to complete the paintings.
Question 48. P Works Twice As Fast As Q. If Q Alone Can Complete A Work In 12 Days, P And Q Can Finish The Work In --- Days?
Answer :
Work completed by Q in 1 day = 1/12
Work carried out by using P in 1 day = 2 × (1/12) = 1/6
Work finished through P and Q in 1 day = 1/12 + 1/6 =1/4
=> P and Q can finish the work in 4 days.
Question forty nine. Two Stations P And Q Are 110110 Km Apart On A Straight Track. One Train Starts From P At 77a.M. And Travels Towards Q At 2020 Kmph. Another Train Starts From Q At 88 A.M. And Travels Towards P At A Speed Of 2525 Kmph. At What Time Will They Meet?
Answer :
Assume both trains meet xx hours after 77 a.M.
Distance blanketed by way of train starting from P in xx hours =20x km=20x km
Distance protected by using train starting from Q in (x−1)(x−1) hours =25(x−1) km=25(x−1) km
Total distance =110 km=one hundred ten km
⇒20x+25(x−1)=a hundred and ten
⇒45x=one hundred thirty five
⇒x=three
Hence, they meet 33 hours after 77 a.M.
I.E., they meet at 10 a.M.
Question 50. A Train Crosses A Post In 1515 Seconds And A Platform 100100 Metre Long In 2525 Seconds. Its Length Is?
Answer :
The teach can cowl a distance same to its duration in 1515 seconds.
The train can cover (a distance equal to its duration + 100100 metre) in 2525 seconds.
Therefore, it is able to be concluded that the teach can journey 100100 metre in (25−15)=10 seconds and consequently its speed is 100/10=10 m/s
Hence, length of the teach =10×15=150 metre.
