Eicher Aptitude Placement Papers - Eicher Aptitude Interview Questions and Answers updated on 24.May.2024

Cost price of 25 kg = Rs. (15 x 14.50 + 10 x 13) = Rs. 347.50. 

Sell price of 25 kg = Rs. (25 x 15) = Rs. 375. 

profit = Rs. (375 — 347.50) = Rs. 27.50.

Toatal bill paid by Amit, Raju and Ram = ( 50 + 55 +75 ) = Rs. 180

Let amount paid by Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively.

Therefore, (3x + 4x + 5x ) = 180

12x = 180 

x = 15 

Therefore, amount paid by,

Amit = Rs. 45

Raju = Rs. 60

Ram = Rs. 75

But actually as given in the question, Amit pays Rs. 50, Raju pays Rs. 55 and Ram pays Rs. 8@Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he needs to pay Rs. 5 to Raju settle the amount.

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

Gain =(80/1600*100) % = 5%.

Here always remember, when ever x% loss, 

it me S.P. = (100 - x)% of C.P

when ever x% profit,

it me S.P. = (100 + x)% of C.P

So here will be (100 - x)% of C.P.

= 80% of 1200 

= (80/100) * 1200

= 960.

Let the cost price = Rs 100

then, Marked price = Rs 135

Required gain = 8%, 

So Selling price = Rs 108

Discount = 135 - 108 = 27

Discount% = (27/135)*100 = 20%.

Assume original salaries of Ram and Sham as 4x and 5x respectively.

Therefore,

(4x + 5000)/= 50

(5x + 5000) 60

60 (4x + 5000) = 50 (5x + 5000)

10 x = 50,000 

5x = 25, 000

Sham's present salary = 5x + 5000 = 25,000 + 5000

Sham's present salary = Rs. 30,000.

Required largest number must be divisible by the L.C.M. of 12, 15 and 18 

L.C.M. of 12, 15 and 18

12 = 2 × 2 × 3

15 =5 × 3

18 = 2 × 3 × 3 

L.C.M. = 180 

Now divide 9999 by 180, we get remainder as 99 

The required largest number = (9999 – 99) =9900 

Number 9900 is exactly divisible by 180.

Least Cost Price = Rs. (200 * 8) = Rs. 1600.

Greatest Selling Price = Rs. (425 * 8) = Rs. 3400.

Required profit = Rs. (3400 - 1600) = Rs. 1800.

Cost Price of 2 dozen oranges Rs. (5 + 4) = Rs. 9.

Sell price of 2 dozen oranges = Rs. 11. 

If profit is Rs 2, oranges bought = 2 dozen. 

If profit is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens.

Wall can be covered only by using square sized wallpaper pieces.

Different sized squares are not allowed.

Length = 4.5 m = 450 cm; 

Height = 3.5 m = 350 cm

Maximum square size possible me HCF of 350 and 450

We can see that 350 and 450 can be divided by 50.

On dividing by 50, we get 7 and 9.

Since we cannot divide further,

             HCF = 50 = size of side of square

Number of squares =Wall area/=450 x 350/= 63

           Square area=50 x 50.

Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.

Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x. 

[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105

[15x + 8x + 12x] = 105

35x = 105

x = 3

Hence, man's daily wage = 5x = 5 x 3 = Rs. 15 

Wife's daily wage = 4x = 4 x 3 = Rs. 12

Daughter's daily wage = 3x = 3 x 3 = Rs. 9.

All 3 digit numbers divisible by 3 are :

102, 105, 108, 111, ..., 999.

This is an A.P. with first element 'a' as 

102 and difference  'd' as 3.

Let it contains n terms. Then,

102 + (n - 1) x3 = 999 

102 + 3n-3 = 999

3n = 900 or n = 300

Sum of AP = n/2 [2*a  + (n-1)*d]

Required sum = 300/2[2*102 + 299*3] = 165150. 

Here greatest number that can divide me the HCF

Remainders are different so simply subtract remainders from numbers

17 - 4 = 13; 42 - 3 = 39; 93 - 15 = 78

Now let's find HCF of 13, 39 and 78

By direct observation we can see that all numbers are divisible by 13.

? HCF = 13 = required greatest number.

Let the article costs 'X' to A

Cost price of B = 1.2X

Cost price of C = 0.75(1.2X) = 0.9X

Cost price of D = 1.4(0.9X) = 1.26X = 252

Amount paid by A for the article = Rs. 200.

Numerators = 36, 48 and 72.

72 is largest number among them. 72 is not divisible by 36 or 48

Start with table of 72.

72 x 2 = 144 = divisible by 72, 36 and 48

? LCM of numerators = 144

Denominators = 225, 150 and 65

We can see that they can be divided by 5.

On dividing by 5 we get 45, 30 and 13

We cannot divide further.

So, HCF = GCD = 5

LCM of fraction =144/5.

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms)

This is an A.P with first term, a=35, number of terms,

n= 12,d=2.

Required distance = 12/2[2 x 35+{12-1) x 2]

=6(70+23)

= 552 kms.

160@A Man Walking At The Rate Of 5 Km/hr Crosses A Bridge In 15 Minutes. The Length Of The Bridge (in Metres) Is?

Let K be common factor. So 2 numbers are 5K and 6K

Also K is the greatest common factor (HCF) as 5 and 6 have no other common factor

? 5K x 6K = 480 x K

K = 16 = HCF.

Let C.P. = Rs. C. 

Then, 832 - C = C - 448

2C = 1280 => C = 640

Required S.P. = 150% of Rs. 640 = 150/100 x 640 = Rs. 960.

Let the S.P = 100

then C.P. = 25

Profit = 75

Profit% = (75/25) * 100 = 300%.

Let S.P. of 45 lemons be Rs. x. 

Then, 80 : 40 = 120 : x or x = 40×120/80= 60 

For Rs.60, lemons sold = 45 

For Rs.24, lemons sold  =4560×24= 18.

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.