Elgi Aptitude Placement Papers - Elgi Aptitude Interview Questions and Answers
Q1. The G.C.D Of 1.08, 0.36 And zero.Nine Is?
Given numbers are 1.08 , zero.36 and 0.90
H.C.F of 108, 36 and 90 is 18 [ ? G.C.D is nothing but H.C.F]
Therefore, H.C.F of given numbers = 0.18.
Q2. How Many Numbers Between 20 And 451 Are Divisible By nine?
The required numbers are 27, 36, 45……450.
This is an A.P. With a = 27 and d = 9
Let it has n phrases.
Then Tn = 450 = 27 + (n-1) x9
450 = 27+ 9n - nine
9n = 432
n = forty eight.
Q3. What Is The Difference Between The Compound Interests On Rs. 5000 For 1 half Years At 4% Per Annum Compounded Yearly And Half-every year?
C.I. When hobby
= Rs. 5304.
C.I. When interest is
compounded half of-every year=rs.5000(1+2/100)^three
= Rs. 5306.04
Difference = Rs. (5306.04 - 5304) = Rs. 2.04.
Q4. The Difference Between Simple And Compound Interest Compounded Annually On A Certain Sum Of Money For 2 Years At 4% Per Annum Is Re.
Q5. In How Many Ways Can The Letters Of The Word 'capital' Be Arranged In Such A Way That All The Vowels Always Come Together?
CAPITAL = 7
Vowels = three (A, I, A)
Consonants = (C, P, T, L)
5 letters which may be organized in 5P5=five!
Vowels A,I = three!/2!
No.Of arrangements = 5! X three!/2!=360
Q6. If An Integer N Is Divisible By three, 5 And 12, What Is The Next Larger Integer Divisible By All These Numbers?
If n is divisible via three, five and 12 it need to a multiple of the lcm of 3, five and 12 that's 60.
N = 60 ok
n + 60 is also divisible by using 60 on account that
n + 60 = 60 k + 60 = 60(okay + 1).
Q7. When The Integer N Is Divided By eight, The Remainder Is
When n is divided by 8, the remainder is 3 may be written as
n = 8 okay + 3
multiply all phrases by way of 6
6 n = 6(eight okay + three) = eight(6k) + 18
Write 18 as 16 + 2 on the grounds that sixteen = 8 * 2.
= 8(6k) + sixteen + 2
Factor eight out.
= 8(6k + 2) + 2
The above suggests that if 6n is divided with the aid of 8, the the rest is two.
Q8. Akash Leaves Mumbai At 6 Am And Reaches Bangalore At 10 Am . Prakash Leaves Bangalore At 8 Am And Reaches Mumbai At 11:30 Am. At What Time Do They Cross Each Other?
Time taken by means of Akash = four h
Time taken by way of Prakash = three.Five h
For your convenience take the made from times taken by both as a distance.
Then the distance = 14km
Since, Akash covers half of of the space in 2 hours(i.E at 8 am)
Now, the rest 1/2 (i.E 7 km) might be coverd with the aid of both prakash and akash
Time taken by them = 7/7.5 = fifty six min
Thus , they'll cross every other at 8 : 56am.
Q9. The Compound Interest On A Certain Sum For 2 Years At 10% Per Annum Is Rs. Fifty two
Let the sum be Rs. P.
= Rs. 500.
Q10. On A Sum Of Money, The Simple Interest For 2 Years Is Rs. 660,even as The Compound Interest Is Rs.696.30,the Rate Of Interest Being The Same In Both The Cases. The Rate Of Interest Is?
Difference in C.I and S.I for two years
S.I for one years = Rs330.
S.I on Rs.330 for 1 yr =Rs. 36.30
Q11. Which Of These Numbers Is Not Divisible By three?
One can also wer this query the usage of a calculator and take a look at for divisibility by means of 3.
However we also can check for divisibilty by means of adding the digits and if the end result is
divisible by3 then the variety is divisible through 3.
3 + three + nine = 15 , divisible through three.
3 + four + 2 = 9 , divisible by way of 3.
Five + five + 2 = 12 , divisible by means of three.
1 + 1 + 1 + 1 = 4 , not divisible with the aid of 3.
Q12. A Sum Of Money Invested At Compound Interest Amounts To Rs. 800 In 3 Years And To Rs. 840 In four Years. The Rate Of Interest Per Annum Is?
S.I. On Rs.800 for 1 yr
=Rs[840 - 800]
Q13. The Sum Of Two Number Is 15 And Sum Of Their Square Is 1
Let the wide variety be x and (15 - x)
Then, x2 + (15 - x)2= 113
x2- 15x + 56 = zero
(x-7) (x-8) = zero.
Q14. Two Boys Starting From The Same Place Walk At A Rate Of 5kmph And 5.5kmph Respectively. What Time Will They Take To Be 8.5km Apart, If They Walk In The Same Direction?
In this kind of questions we want to get the relative speed among them,
The relative speed of the lads = five.5kmph – 5kmph
= zero.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / zero.5 kmph = 17 hrs.
Q15. The Compound Interest On Rs.30000 At 7% Per Annum Is Rs.434
Amount = Rs.(30000+4347) = Rs.34347
allow the time be n years
Then,30000(1+7/100)^n = 34347
(107/100)^n = 34347/30000 = 11449/10000 = (107/a hundred)^2
n = 2years.
Q16. Jagan Went To Another Town Covering 240 Km By Car Moving At 60 Kmph. Then He Covered 400km By Train Moving At one hundred Kmph And Then Rest 2 hundred Km He Covered By A Bus Moving At 50 Kmph. The Average Speed Dur
By car 240 km at 60 kmph
Time taken = 240/60 = four hr.
By teach 240 km at 60 kmph
Time taken = 400/a hundred = 4 hr.
By bus 240 km at 60 kmph
Time taken = 200/50 = four hr.
So overall time = four + 4 + four = 12 hr.
And total pace = 240+four hundred+2 hundred = 840 km
Average speed of the entire adventure = 840/12 = 70 kmph.
Q17. What Should Be The Value Of * In 985*865, If Number Is Divisible By 9?
Nine + eight + 5 + * + 8 + 6 + 5 = 9x
forty one + * = 9x
Nearest value of 9x should be forty five
forty one + * = 45
* = four.
Q18. 2 Men And 7 Boys Can Do A Piece Of Work In 14 Days; three Men And eight Boys Can Do The Same In eleven Days. Then, eight Men And 6 Boys Can Do Three Times The Amount Of This Work In?
(2 x 14) men +(7 x 14) boys = (3 x eleven) guys + (eight x eleven) boys
=>5 guys= 10 boys => 1man= 2 boys
Therefore, (2 guys+ 7 boys) = (2 x 2 +7) boys = eleven boys
( 8 men + 6 boys) = (eight x 2 +6) boys = 22 boys.
Let the required wide variety of days be x.
More boys , Less days (Indirect share)
More paintings , More days (Direct percentage)
Boys22:11Work1 : three?? 14:x
Therefore, (22 * 1 * x) = (eleven * 3 * 14)
=> x = 21
Hence, the desired wide variety of days = 21.
Q19. A Certain Number Of Men Can Finish A Piece Of Work In 100 Days. If There Were 10 Men Less, It Would Take 10 Days More For The Work To Be Finished. How Many Men Were There Originally?
Originally allow there be x men.
Less men, More days(Indirect Proportion)
Therefore, (x-10) : x :: 100 :a hundred and ten
=> (x - 10) * one hundred ten = x * a hundred=> x= 110.
Q20. A Man Whose Speed Is 4.5 Kmph In Still Water Rows To A Certain Upstream Point And Back To The Starting Point In A River Which Flows At 1.5 Kmph, Find His Average Speed For The Total Journey ?
Speed of Man = four.Five kmph
Speed of movement = 1.Five kmph
Speed in DownStream = 6 kmph
Speed in UpStream = 3 kmph
Average Speed = (2 x 6 x 3)/nine = 4 kmph.
Q21. Find Compound Interest On Rs. 7500 At 4% Per Annum For 2 Years, Compounded Annually?
Amount = Rs [7500*(1+(4/100)2] = Rs (7500 * (26/25) * (26/25) ) = Rs. 8112.
Therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.
Q22. In A Group Of 6 Boys And four Girls, Four Children Are To Be Selected. In How Many Different Ways Can They Be Selected Such That At Least One Boy Should Be There ?
We might also have (1 boy and three girls)or(2boys and a pair of girls)or(3 boys and 1 female)or(4 boys).
Required wide variety of approaches = (6C1×4C3) + (6C2×4C2) + (6C3×4C1)+(6C4)
Q23. A And B Runs Around A Circular Track. A Beats B By One Round Or 10 Minutes. In This Race, They Had Completed four Rounds. If The Race Was Only Of One Round, Find The A's Time Over The Course?
B runs around the tune in 10 min.
I.E ,Speed of B = 10 min in line with spherical
Therefore, A beats B by using 1 round
Time taken by using A to finish four rounds
= Time taken with the aid of B to complete three rounds
= 30 min
Therefore, A's velocity = 30/four min in keeping with round = 7.5 min in keeping with spherical
Hence, if the race is handiest of 1 round A's time over the route = 7 min 30 sec.
Q24. A Man Covered A Certain Distance At Some Speed. Had He Moved three Kmph Faster, He Would Have Taken 40 Minutes Less. If He Had Moved 2 Kmph Slower, He Would Have Taken forty Minutes More. The Distance (in Km
Let distance = x km and regular fee = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = forty/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by using (ii), we get:
x = forty km.
Q25. In A Palace, Three Different Types Of Coins Are There Namely Gold, Silver And Bronze. The Number Of Gold, Silver And Bronze Coins Is 18000, 9600 And 3600 Respectively. Find The Minimum Number Of Rooms
Gold cash = 18000 , Silver coins = 9600 , Bronze cash = 3600
Find a number which exactly divide a lot of these numbers
That is HCF of 18000, 9600& 3600
All the cost has 00 at stop so the aspect will also have 00.
HCF for a hundred and eighty, ninety six & 36.
180 = three x three x five x 2 x 2
96 = 2 x 2 x 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x three
Common elements are 2x2×3=12
Therefore, Actual HCF is 1200
Gold Coins 18000/1200 can be in 15 rooms
Silver Coins 9600/1200 can be in 8 rooms
Bronze Coins 3600/1200 will be in 3 rooms
Total rooms may be (15+eight+3) = 26 rooms.
Q26. Six Bells Commence Tolling Together And Toll At Intervals Of 2, 4, 6, 8 10 And 12 Seconds Respectively. In 30 Minutes, How Many Times Do They Toll Together?
L.C.M. Of 2, 4, 6, eight, 10, 12 is one hundred twenty.
So, the bells will toll together after every 120 seconds(2 minutes).
In half-hour,they'll collectively (30/2)+1=16 instances.
Q27. A Sum Of Money Amounts To Rs.6690 After three Years And To Rs.10,half After 6 Years On Compound Interest.Find The Sum?
Let the sum be Rs.P.Then
P(1+R/100)^three=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/a hundred)^3=10025/6690=3/2.
Substituting this fee in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.
Q28. A Train Covers A Distance In 50 Minutes, If It Runs At A Speed Of 48kmph On An Average. Find The Speed At Which The Train Must Run To Reduce The Time Of Journey To forty Minutes?
We are having time and pace given,
so first we are able to calculate the space.
Then we are able to get new velocity for given time and distance.
Lets resolve it.
Time = 50/60 hr = five/6 hr
Speed = 48 mph
Distance = S*T = forty eight * 5/6 = 40 km
New time may be forty minutes so,
Time = forty/60 hr = 2/3 hr
Now we understand,
Speed = Distance/Time
New velocity = 40*3/2 kmph = 60kmph.
Q29. A Person Takes 20 Minutes More To Cover A Certain Distance By Decreasing His Speed By 20%. What Is The Time Taken To Cover The Distance At His Original Speed ?
Let the gap and original speed be 'd' km and 'k' kmph respectively.
D/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/three
=> (5d - 4d)/4k = 1/3 => d = four/three ok
Time taken to cover the distance at unique speed
= d/okay = four/three hours = 1 hour 20 minutes.
Q30. A Man Reaches His Office 20 Min Late, If He Walks From His Home At three Km Per Hour And Reaches 30 Min Early If He Walks 4 Km Per Hour. How Far Is His Office From His House?
Let distance = x km.
Time taken at 3 kmph : dist/speed = x/3 = 20 min overdue.
Time taken at four kmph : x/four = 30 min in advance
distinction among time taken : 30-(-20) = 50 mins = 50/60 hours.
X/three- x/four = 50/60
x/12 = five/6
x = 10 km.
Q31. Find Compound Interest On Rs. 8000 At 15% Per Annum For 2 Years 4 Months, Compounded Annually?
Time = 2 years 4 months = 2(four/12) years = 2(1/three) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. Eleven@.
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.
Q32. A Man In A Train Notices That He Can Count 41 Telephone Posts In One Minute. If They Are Known To Be 50 Metres Apart, Then At What Speed Is The Train Travelling?
Number of gaps among forty one poles = forty
So overall distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute teach is shifting 2 km/minute.
Speed in hour = 2*60 = a hundred and twenty km/hour.
Q33. The Number X Is Exactly Divisible By 5 And The Remainder Obtained On Dividing The Number Y By 5 Is
Let x/5 = p and permit y whilst divided with the aid of 5 offers q as quotient and 1 as the rest.
Then, y = 5q + 1
Now x = 5p and y = 5q + 1
x + y = 5p + 5q + 1 = five(p + q) + 1.
Q34. What Is The Smallest Positive 2-digit Whole Number Divisible By three And Such That The Sum Of Its Digits Is nine?
Let xy be the complete variety with x and y the two digits that make up the variety.
The quantity is divisible by means of 3 may be written as follows
10 x + y = three k
The sum of x and y is same to nine.
X + y = nine
Solve the above equation for y
y = nine - x Substitute y = nine - x in the equation 10 x + y = 3 ok to reap.
10 x + nine - x = 3 ok
Solve for x
x = (k - three) / three
x is a high-quality integer smaller than 10
Let ok = 1, 2, 3, ... And choose the first value that offers x as an integer.
K = 6 offers x = 1
Find y using the equation y = nine - x = 8
The quantity we are looking for is eighteen.
It is divisible by using three and the sum of its digits is equal to 9 and it is the smallest and effective complete wide variety with such residences.
Q35. The Difference Of Two Numbers Is
Since their HCFs are 7, numbers are divisible by means of 7 and are of the form 7x and 7y
Difference = 14
=> 7x - 7y = 14
=> x - y = 2
product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = sixty three
Now, we've got
x * y = 63 , x - y = 2
=> x = nine , y = 7
The numbers are 7x and 7y
=> sixty three and 49.
Q36. The Least Number Which When Divided By 5, 6 , 7 And 8 Leaves A Remainder three, But When Divided By 9 Leaves No Remainder, Is?
L.C.M. Of 5, 6, 7, 8 = 840.
Required range is of the form 840k + three
Least cost of okay for which (840k + 3) is divisible by using nine is okay = 2.
Required range = (840 x 2 + three) = 1683.
Q37. An Employee May Claim Rs. 7.00 For Each Km When He Travels By Taxi And Rs. 6.00 For Each Km If He Drives His Own Car. If In One Week He Claimed Rs. 595 For Traveling ninety Km. How Many Kms Did He Travel
Let x and y be the respective km's travelled by guy thru taxi and by means of his personal automobile.
Given x + y = ninety => x = 90 - y
But in keeping with the question,
7x + 6y = 595
7(90-y) + 6y = 595
=> 630 - 7y + 6y = 595
=> y = 630 - 595 = 35
=> x = ninety - 35 = fifty five
Therefore, the distance travelled through taxi is fifty five kms.
Q38. Which Of The Following Has The Most Number Of Divisors?
Ninety nine = 1 x 3 x three x eleven
one zero one = 1 x one hundred and one
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x thirteen
So, divisors of ninety nine are 1, 3, nine, 11, 33, .99
Divisors of 101 are 1 and one zero one
Divisors of 176 are 1, 2, four, eight, 11, sixteen, 22, forty four, 88 and 176
Divisors of 182 are 1, 2, 7, thirteen, 14, 26, ninety one and 182.
Hence, 176 has the maximum wide variety of divisors.
Q39. A Man Walks At A Speed Of 2 Km/hr And Runs At A Speed Of 6 Km/hr. How Much Time Will The Man Require To Cover A Distance Of 20 half Km, If He Completes Half Of The Distance, I.E., (10 1/four) Km On Foot A
We recognize that
Time = Distance/velocity
Required time = (10 1/4)/2 + (10 1/four)/6
= forty one/8 + forty one/6
= 287/24 = eleven.9 hours.
Q40. If 20 Men Can Build A Wall fifty six Meters Long In 6 Days , What Length Of A Similar Wall Can Be Built By 35 Men In three Days?
Let the specified period be x meters
More men, More period built (Direct percentage)
Less days, Less period built (Direct Proportion)
Days6 : 3?? 56 :x
=> (20 x 6 x X)=(35 x 3 x 56)
=> x = 49
Hence, the desired duration is forty nine m.
Q41. Ramu Rides His Bike At An Average Speed Of forty five Km/hr And Reaches His Desitination In Four Hours. Somu Covers The Same Distance In Six Hours. If Ramu Covered His Journey At An Average Speed Which Was 9
Distance travelled by way of Ramu = forty five x four = one hundred eighty km
Somu travelled the equal distance in 6 hours.
His pace = a hundred and eighty/6 = 30 km/hr
Hence within the conditional case, Ramu's velocity = forty five - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.
Therefore tour time of Ramu and Somu might be five hours and four.Five hours respectively.
Hence distinction in the time taken = 0.5 hours = 30 minutes.
Q42. The Least Perfect Square, Which Is Divisible By Each Of 15, 20 And 36 Is?
LCM of 15, 20 and 36 is one hundred eighty
Now 180 = three x three x 2 x 2 x 5
To make it ideal square, it should
be improved from 5.
So required no. = 32 x 22 x fifty two = 900.
Q43. L.C.M Of Two Prime Numbers X And Y (x>y) Is sixteen
H. C. F of top numbers is 1.
Product of numbers = 1 x 161 = 161.
Let the numbers be a and b . Then , ab= 161.
Now, co-primes with product 161 are (1, 161) and (7, 23).
Since x and y are high numbers and x >y ,
we've got x=23 and y=7.
Therefore, 3y-x = (3 x 7)-23 = -2.
Q44. A Person X Started At three Hours Earlier At 40km/h From A Place P, Then Another Person Y Followed Him At 60km/h. Started His Journey At three O'clock, Afternoon. What Is The Diference In Time When X Was30 K
Time ( whilst X changed into 30 km beforehand of Y) = (a hundred and twenty-30)/20 =4.5h
Time ( while Y became 30 km ahead of X) = (a hundred and twenty+30)/20 = 7.Five h
Thus, required distinction in time = 3h.
Q45. The Compound Interest On A Sum Of Money For 2 Years Is Rs.832 And The Simple Interest On The Same Sum For The Same Period Is Rs.800 .The Difference Between The Compound Interest And Simple Interest Fo
distinction in C.I and S.I in 2years =Rs.32
S.I for 1year =Rs.Four hundred
S.I for Rs.400 for 12 months =Rs.32
distinction among in C.I and S.I for 3rd year
=S.I on Rs.832= Rs.(832*8*1)/a hundred=Rs.66.56
Q46. If A Positive Integer N Is Divided By five, The Remainder Is
n divided via five yields a the rest equal to a few is written as follows
n = 5 ok + 3 , wherein ok is an integer.
Add 2 to each facets of the above equation to obtain
n + 2 = five k + five = five(okay + 1)
The above shows that n + 2 divided with the aid of 5 yields a remainder identical to zero.
Q47. Find The Compound Interest On Rs. 16,000 At 20% Per Annum For 9 Months, Compounded Quarterly?
Principal = Rs. 16000; Time = 9 months =3 quarters;
Rate = 20% in keeping with annum = 5% in line with quarter.
Amount = Rs. [16000 x (1+(5/100))3] = Rs. 18522.
CJ. = Rs. (18522 - 16000) = Rs. 2522.
Q48. The Largest four Digit Number Exactly Divisible By 5, 6 And 7 Is?
The required range should be divisible by L.C.M. Of five,6 and 7.
L.C.M. Of 5, 6 and seven = five x 6 x 7 = 210
Let us divide 9999 by using 210.
210) 9999 (forty seven
Required wide variety = 9999 – 129 = 9870.
Q49. The H.C.F And L.C.M Of Two Numbers Are 11 And 385 Respectively. If One Number Lies Between seventy five And one hundred twenty five , Then That Number Is?
Product of numbers = 11 x 385 = 4235
Let the numbers be 11a and 11b . Then , 11a x 11b = 4235 => ab = 35
Now, co-primes with product 35 are (1,35) and (5,7)
So, the numbers are ( eleven x 1, 11 x 35) and (11 x five, 11 x 7)
Since one number lies 75 and one hundred twenty five, the appropriate pair is (55,77)
Hence , required number = seventy seven.
Q50. If N Is An Integer, When (2n + 2)^2 Is Divided By four The Remainder Is?
We first extend (2n + 2)2
(2n + 2)^2 = 4n^2 + 8n + four
Factor 4 out.
= 4(n^2 + 2n + 1)
(2n + 2)2 is divisible by four and the the rest is identical to 0.