# 300+ TOP Eicher Aptitude Interview Questions and Answers

Q1. Ajay Bought 15 Kg Of Dal At The Rate Of Rs 14.50 Per Kg And 10 Kg At The Rate Of Rs 13 Per Kg. He Mixed The Two And Sold The Mixture At The Rate Of Rs 15 Per Kg. What Was His Total Gain In This Tracti

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.

Q2. Amit, Raju And Ram Agree To Pay Their Total Electricity Bill In The Proportion 3 : 4 :

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.

Q3. A Trader Mixes 26 Kg Of Rice At Rs. 20 Per Kg With 30 Kg Of Rice Of Other Variety At Rs. 36 Per Kg And Sells The Mixture At Rs. 30 Per Kg. His Profit Percent Is?

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%.

Q4. A Man Buys An Item At Rs. 1200 And Sells It At The Loss Of 20 Percent. Then What Is The Selling Price Of That Item?

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.

Q5. A Shopkeeper Fixes The Marked Price Of An Item 35% Above Its Cost Price. The Percentage Of Discount Allowed To Gain 8% Is?

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%.

Q6. Salaries Of Ram And Sham Are In The Ratio Of 4 :

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.

Q7. Find The Largest Number Of 4-digits Divisible By 12, 15 And 18?

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.

Q8. If Books Bought At Prices Ranging From Rs. 200 To Rs. 350 Are Sold At Prices Ranging From Rs. 300 To Rs. 425, What Is The Greatest Possible Profit That Might Be Made In Selling Eight Books ?

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

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

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

Q9. A Man Buys Oranges At Rs 5 A Dozen And An Equal Number At Rs 4 A Dozen. He Sells Them At Rs 5.50 A Dozen And Makes A Profit Of Rs 5

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.

Q10. A Wall Is 4.5 Meters Long And 3.5 Meters High. Find The Number Of Maximum Sized Wallpaper Squares, If The Wall Has To Be Covered With Only The Square Wall Paper Pieces Of Same Size?

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.

Q11. A Man, His Wife And Daughter Worked In A Graden. The Man Worked For 3 Days, His Wife For 2 Days And Daughter For 4 Days. The Ratio Of Daily Wages For Man To Women Is 5 : 4 And The Ratio For Man To Dau

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.

Q12. The Sum Of All 3 Digit Numbers Divisible By 3 Is?

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.

Q13. What Profit Percent Is Made By Selling An Article At A Certain Price, If By Selling At 2/3rd Of That Price, There Would Be A Loss Of 20%?

SP2 = 2/3 SP1

CP = 100

SP2 = 80

2/3 SP1 = 80

SP1 = 120

100 — 20   => 20%.

Q14. Find The Largest Number To Divide All The Three Numbers Leaving The Remainders 4, 3, And 15 Respectively At The End?

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.

Q15. ‘a’ Sold An Article To ‘b’ At A Profit Of 20%. ‘b’ Sold The Same Article To ‘c’ At A Loss Of 25% And ‘c’ Sold The Same Article To ‘d’ At A Profit Of 40%. If ‘d’ Paid Rs 252 For The Article, Then Find

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.

Q16. Find The Lcm Of Following Three Fractions:36/,48/,72/?

Numerators = 36, 48 and 72.

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

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.

Q17. The Speed Of A Car Increases By 2 Kms After Every One Hour. If The Distance Travelling In The First One Hour Was 35 Kms. What Was The Total Distance Travelled In 12 Hours?

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?

Q18. The Two Given Numbers A And B Are In The Ratio 5:6 Such That Their Lcm Is 48

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.

Q19. The Profit Earned By Selling An Article For Rs. 832 Is Equal To The Loss Incurred When The Same Article Is Sold For Rs. 44

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.

Q20. If The Cost Price Is 25% Of Selling Price. Then What Is The Profit Percent?

Let the S.P = 100

then C.P. = 25

Profit = 75

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

Q21. By Selling 45 Lemons For Rs 40, A Man Loses 20%. How Many Should He Sell For Rs 24 To Gain 20% In The Traction ?

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.

Q22. Two Dice Are Tossed.the Probability That The Total Score Is A Prime Number Is?

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.