# Elico Aptitude Placement Papers - Elico Aptitude Interview Questions and Answers updated on 03.Feb.2023

310 x 999 = 310 x (1000-1)

= 310 x 1000 -310 x 1

= 310000 – 310

= 309690.

125% ---- 3400

=> 100% ---- ?

=> ? = 3400x100/125 = 2720

=> Cost price of the article = Rs. 2720

Profit when article sold at Rs. 3265 = 3265 - 2720 = 545

Hence, Profit% = Gain x 100/cost price

=> P% = 545 x 100/2720

=> P% = 20%.

Let the third number = N

Then the first number = 3N

And second number = 3N/2

According to the question [N + 3N + 3N/2] / 3=154.

Let C.P.= Rs. 100.

Then, Profit = Rs.220,

S.P. = Rs.320.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs.320.

Profit = Rs. (320 - 125) = Rs. 195

Required percentage = 195/320×100== 60.9 =~ 61%.

Let the cost price be x and selling price be y.

Loss = x – y

When the cost price doubles, the loss gets tripled.

So it becomes like this, 2x – y = 3(x-y)

=> x = 2y

Loss % = (loss/ C.P) x 100 = [(2y-y)/2y] x 100= 50 %.

The given number is 1, 3, 5………….99

This is an A.P with a = 1, d = 2

Let it contain n term 1 + (n-1)2 = 99

==> n = 50

Then required sum = n/2 (first term + last term)

==> 50/2(1 + 99) = 2500.

Sum = Rs.(50*100)/2*5=Rs.500

Amount=Rs.[500*(1+5/100)2]

= Rs. 551.25

C.I = Rs.(551.25-500)= Rs.51.25.

C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.

S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.

S.P. per kg = 819/50 = Rs. 16.38 = 16.30.

Let the distance between home and school is 'x'.

Let actual time to reach be 't'.

Thus, x/4 = t + 4 ---- (1)

and x/6 = t - 30 -----(2)

Solving equation (1) and (2)

=> x = 98 mts.

The formula to calculate simple interest is :

SI=(P x T x R)/100

In the given question Principal(P)=Rs. 14000, Time(T)=(3/12)yr. and Rate of interest(R)=10%

So, SI=(14000 x 1/4 x 10)/100

SI= Rs. 350.

Let Cost Price(C.P) = P

gain% = {(S.P-C.P)/C.P} x 100

25 = {(1540-P)/P} x 100

25/100 = (1540-P)/P

=> P = 4(1540)-4P

=> 5P = 4(1540)

=> P = 1232

So, Cost Price = Rs. 1232.

90 = 10 x 9

Clearly, 653xy is divisible by 10, so y = 0

Now, 653x0 is divisible by 9.

So, (6 + 5 + 3 + x + 0) = (14 + x) is divisible by @So, x = 4.

Hence, (x + y) = (4 + 0) = 4.

We know 60 min = 1 hr

Total northward Laxmi's distance = 20kmph x 1hr = 20 km

Total southward Prasanna's distance = 30kmph x 1hr = 30 km

Total distance between Prasanna and Laxmi is = 20 + 30 = 50 km.

Say total cost price of tea is x.

Then total profit at a rate of 15% is = (15x/100)

According to question,

15x/100 = 60

so x = 400

C.p of the tea is Rs. 400.

so total selling price will be = (400+60) = Rs.460

so the quantity of the tea will be = (460/5.75) = 80kg.

Dividend = 12 * 35 =420

Now dividend = 420 and divisor = 21

Therefore correct quotient = 420/21 = 20.

Let x men can do the in 12 days and the required number of days be z

More men, Less days     [Indirect Proportion]

Less work, Less days     [Direct Proportion  ]

men2x:xwork1:12} :: 12 : z

(2x×1×z)=(x×12×12)

z=3.

Let Rs. K invested in each scheme

Two years C.I on 20% = 20 + 20 + 20x20/100 = 44%

Two years C.I on 15% = 15 + 15 + 15x15/100 = 32.25%

Now,

(P x 44/100) - (P x 32.25/100) = 528.75

=> 11.75 P = 52875

=> P = Rs. 4500

Hence, total invested money = P + P = 4500 + 4500 = Rs. 9000.

Divisor = (5 x 46) = 230

10 x Quotient = 230=230= 23/10

Dividend = (Divisor x Quotient) + Remainder

= (230 x 23) + 46

= 5290 + 46

= 5336.

Let the required number of days be x.

Less persons, More days (Indirect Proportion)

More working hours per day, Less days (Indirect Proportion)

Persons30:39Working hours/day6:5}?? 12:x

=> 30 * 6 * x = 39 * 5 * 12

=> x= 13.

As we know that Speed = Distance / Time

for average speed = Total Distance / Total Time Taken

Thus, Total Distance = 325 + 470 = 795 km

Thus, Total Speed = 7.5 hrs

Average Speed = 795/7.5 => 106 kmph.

Let x = 6q + 3.

Then, x2 = (6q + 3)2

= 36q2 + 36q + 9

= 6(6q2 + 6q + 1) + 3

Thus, when x2 is divided by 6, then remainder = 3.

Less Men, me more Days {Indirect Proportion}

Let the number of days be x then,

27 : 36 :: 18 : x

[Please pay attention, we have written 27 : 36 rather than 36 : 27, in indirect proportion, if you get it then chain rule is clear to you :)]

=>27x = 36 * 18

=> x = 24

So 24 days will be required to get work done by 27 men.

Average speed = Total distance / Total time.

Total distance traveled by Joel = Distance covered in the first 3 hours + Distance covered in the next 5 hours.

Distance covered in the first 3 hours = 3 x 60 = 180 miles

Distance covered in the next 5 hours = 5 x 24 = 120 miles

Therefore, total distance traveled = 180 + 120 = 300 miles.

Total time taken = 3 + 5 = 8 hours.

Average speed = 300/8 = 37.5 mph.

we know that 1 mile = 1.6 kms

=> 37.5 miles = 37.5 x 1.6 = 60 kms

Average speed in Kmph = 60 kmph.

Since the car runs at 7/11 th of its own speed, the time it take is 11/7th of its usual speed.

let the usual time taken be t hours

then we can write, 11t / 7 = 22

or t  = 22 * 7 / 11 = 14 hours

Time saved  = 22 - 14 = 8hours.

Let d = divisor and q = quotient

First remove the last remainder:

d x q + 413 = 380606

d x q = 380193.

Let the number be P and Q

According to the question

(P + Q) = 2.5 (P - Q)

P + Q = 2.5P - 25Q

3.5Q = 1.5P

P / Q = 7/3 .....(i)

Now PQ = 84.

Speeds of two trains = 30 kmph and 58 kmph

=> Relative speed = 58 - 30 = 28 kmph = 28 x 5/18 m/s = 70/9 m/s

Given a man takes time to cross length of faster train = 18 sec

Now, required Length of faster train = speed x time = 70/9  x  18 = 140 mts.

The only part of this type of calculation that needs particular

care is that concerning the interest rate. The formula assumes that

r is a proportion, and so, in this case:

r = 0.08

In addition, we have P = 5,000 and n = 5, so:

V = P(1 + r)5 = 5,000 x (1 + 0.08)5 = 5,000 x 1.469328 = 7,346.64

Thus the value of the investment will be 7,346.64.

Let the C.P be 'x'

Then, the selling price S.P = x - 16x/100

= 84x/100 = 21x/25

Now, if the S.P is 60 more, then the profit is 14%

=> 21x/25 + 60 = x + 14x/100

=> 114x/100 - 21x/25 = 60

=> (57 - 42)x/50 = 60

=> 15x/50 = 60

x = 3000/15 = 200

Therefore, the Cost price C.P = x = Rs. 200.

Let 'd' be the distance between A and B

K -time = d/10 + d/9 = 19d/90 hours

L -time = 2d/12 = d/6 hours

We know that, 10 min = 1/6 hours

Thus, time difference between K and L is given as 10 minutes.

=> 19d/90 - d/6 = 1/6

=> (19d-15d)/90 = 1/6

=> 4d/90 = 1/6

Thus,

d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.

Time taken to cover 600 km = 600/100 = 6 hrs.

Number of stoppages = 600/75 - 1 = 7

Total time of stoppages = 4 x 7 = 28 min

Hence, total time taken = 6 hrs 28 min.

Let first number = P

And second number = Q

According to the question

P - Q = 10 ....(i)

And 3P - 4Q = 18...(ii)

on multiply eq (i) by 3 and then subtract eq (ii) from it

3P - 3Q = 54

3P - 4Q = 18

Q = 36

On putting the value of Q in the eq (i)

P = 18 + Q = 18 +36

P = 54.

Amount

=Rs.[8000x(1+5/100)²]

= Rs.[8000 x 21/20x21/20]

= Rs.8820.

Initial speed = 80km/hr

Total distance = 80 * 10 = 800km

new speed = 800/4 =200km/hr

Increase in speed = 200 - 80 = 120km/hr.

let 't' be the time after which they met since L starts.

Given K is 50% faster than L

50 t + 1.5*50(t-1) = 300

50 t +75 t = 300 + 75

t = 375 / 125 = 3 hrs past the time that L starts

So they meet at (9 + 3)hrs = 12:00 noon.

2056 x 987 = 2056 x (1000 - 13)

= 2056 x 1000 - 2056 x 13

= 2056000 - 26728

= 2029272.

net height climbed in 2 min = 20m - 4m = 16m

In 10 min the net height climbed = 16 * 10/2 = 80m

Remaining height = 96m - 80m = 16m

In the 11th min , the monkey will be ascending up.

Time taken to ascend the last 16min

16/20 = 4/5min

Total time taken = 10 + 4/5 = 54/5 min.

least cost price      = 200*8 = 1600

greatest sold price = 425 * 8 = 3400

profit required = 3400- 1600 = 1800.

Average speed=total distance/total time

Let distance to store be  K

then, total time =(K/20)+(K/30)=K/12

and, total time =(2K)

so average speed= 2K / (K/12) = 24kmph.

Let the C.P of one item is Rs. P

and that of other is Rs. (7500 - P)

According to the data given

C.P = S.P

=> Px(116/100) + (7500-P)x(86/100) = 7500

=> 30P = 105000

=> P = 3500

Required difference between selling prices

= Rs. [(3500/100) x 116] - [(4000/100) x 86]

= 4060-3440

= Rs. 620.

Length of train A = 48 x 9 x 5/18 = 120 mts

Length of train B = 48 x 24 x 5/18 - 120

=> 320 - 120 = 200 mts.

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000*(1+7/100)^n=34347

n= 2 years.

Let 'x' be the cost price.

Now Marked price = x + 30x/100 = 13x/10

10% discount = 10/100 x 13x/10 = 13x/100

Selling price = 13x/10 - 13x/100 = 117x/100

Given gain = 340

Here gain = 117x/100 - x = 17x/100 = 340 => x = Rs. 2000.

CP of 1000gm = Rs. 10

SP of 800gm = Rs. 12

SP of 1000gm =12x1000/800 = Rs. 15

Now take 1000gm as reference to calculate profit.

Profit=SP-CP=15-10=Rs. 5

Profit % = 5x100/10 = 50%.

Cost price of the article is given by

= 400x100/(20+20)

= Rs.1000.

Let x be the number and y be the quotient. Then,

x = 357 x y + 39

= (17 x 21 x y) + (17 x 2) + 5

= 17 x (21y + 2) + 5)

Required remainder = 5.

We know Compound Interest = C.I. = P1+r100t - 1

Here P = 2680, r = 8 and t = 2

C.I. = 26801 + 81002-1= 268027252-12= 26802725+12725-1= 2680 5225×225

= (2680 x 52 x 2)/625

= 445.95

Compound Interest = Rs. 445.95.

Let marked price = Rs. 100.

Then, C.P. = RS. 54,

S.P. = Rs. 85

Gain % = 31/64 x 100 = 48.4%.

Let the distance travelled by Tilak in first case or second case = d kms

Now, from the given data,

d/20 = d/22 + 36 min

=> d/20 = d/22 + 3/5 hrs

=> d = 132 km.

Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.

When 3 passengers income was 3x

expense= Rs.30

profit =20% of 30 = Rs.6

That me his earning is Rs.3@so that per passenger fare must be Rs.12.

When 4 passengers

earning = 12x4=Rs.48.

expense =Rs.24.

profit = Rs.24 = 100%.