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Eicher Aptitude Placement Papers - Eicher Aptitude Interview Questions and Answers - Jul 22, 2022

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Eicher Aptitude Placement Papers - 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 thirteen) = 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 way of Amit, Raju and Ram = ( 50 + 55 +seventy five ) = Rs. 180

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

Therefore, (3x + 4x + 5x ) = a hundred and eighty

12x = one hundred eighty 

x = 15 

Therefore, amount paid thru,

Amit = Rs. Forty five

Raju = Rs. 60

Ram = Rs. Seventy five

But clearly as given within the query, Amit pays Rs. 50, Raju pays Rs. Fifty five and Ram will pay Rs. Eight@Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he desires to pay Rs. Five to Raju settle the quantity.

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 fifty six 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 normally undergo in mind, whilst ever x% loss, 

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

at the same time as ever x% earnings,

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

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

= 80% of 1200 

= (80/one hundred) * 1200

= 960.

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

Let the charge price = Rs one hundred

then, Marked fee = Rs a hundred thirty five

Required advantage = 8%, 

So Selling fee = Rs 108

Discount = one hundred thirty five - 108 = 27

Discount% = (27/one hundred thirty 5)*one hundred = 20%.

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

Assume precise 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 income = 5x + 5000 = 25,000 + 5000

Sham's gift income = Rs. 30,000.

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

Required biggest extensive variety need to be divisible through the L.C.M. Of 12, 15 and 18 

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

12 = 2 × 2 × 3

15 =five × three

18 = 2 × 3 × 3 

L.C.M. = 180 

Now divide 9999 with the aid of 100 eighty, we get the rest as ninety nine 

The required biggest quantity = (9999 – ninety nine) =9900 

Number 9900 is exactly divisible via one hundred eighty.

Q8. If Books Bought At Prices Ranging From Rs. Two hundred To Rs. 350 Are Sold At Prices Ranging From Rs. Three hundred 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 income = Rs. (3400 - 1600) = Rs. 1800.

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

Cost Price of  dozen oranges Rs. (five + four) = Rs. 9.

Sell fee of two dozen oranges = Rs. Eleven. 

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

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

Q10. A Wall Is four.Five 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 may be protected nice via the use of square sized wallpaper portions.

Different sized squares aren't allowed.

Length = four.5 m = 450 cm; 

Height = 3.5 m = 350 cm

Maximum rectangular length feasible me HCF of 350 and 450

We can see that 350 and 450 can be divided via manner of 50.

On dividing through manner of fifty, we get 7 and nine.

Since we cannot divide similarly,

             HCF = 50 = length of aspect of square

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

           Square place=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 four Days. The Ratio Of Daily Wages For Man To Women Is five : 4 And The Ratio For Man To Dau

Assume that the every day wages of guy, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.

Multiply (no. Of days) with (assumed day by day wage) of all and sundry to calculate the value of x. 

[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 100 and five

[15x + 8x + 12x] = one hundred and 5

35x = 100 and five

x = 3

Hence, man's every day wage = 5x = 5 x three = Rs. 15 

Wife's every day income = 4x = 4 x 3 = Rs. 12

Daughter's each day wage = 3x = 3 x 3 = Rs. Nine.

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

All 3 digit numbers divisible via three are :

102, one 0 5, 108, 111, ..., 999.

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

102 and distinction  'd' as three.

Let it incorporates n terms. Then,

102 + (n - 1) x3 = 999 

102 + 3n-3 = 999

3n = 900 or n = 3 hundred

Sum of AP = n/2 [2*a  + (n-1)*d]

Required sum = three hundred/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/third Of That Price, There Would Be A Loss Of 20%?

SP2 = 2/three SP1

CP = 100

SP2 = eighty

2/3 SP1 = 80

SP1 = 120

one hundred --- 20   => 20%.

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

Here finest variety which can divide me the HCF

Remainders are terrific so actually subtract remainders from numbers

17 - four = thirteen; forty two - three = 39; ninety 3 - 15 = seventy eight

Now allow's find out HCF of thirteen, 39 and 78

By direct observation we are able to see that each one numbers are divisible via thirteen.

? HCF = thirteen = required best variety.

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 item fees 'X' to A

Cost charge of B = 1.2X

Cost fee of C = zero.Seventy 5(1.2X) = 0.9X

Cost charge of D = 1.Four(zero.9X) = 1.26X = 252

Amount paid by using A for the object = Rs. Two hundred.

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

Numerators = 36, forty eight and seventy two.

Seventy two is largest quantity among them. Seventy two isn't divisible with the aid of 36 or forty eight

Start with desk of 72.

Seventy  x 2 = one hundred forty 4 = divisible with the aid of seventy two, 36 and 48

? LCM of numerators = a hundred and forty four

Denominators = 225, a hundred and fifty and sixty five

We can see that they can be divided with the resource of five.

On dividing through 5 we get forty five, 30 and 13

We can not divide similarly.

So, HCF = GCD = 5

LCM of fraction =a hundred and forty 4/five.

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 time period, a=35, amount of terms,

n= 12,d=2.

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

=6(70+23)

= 552 kms.

One hundred sixty@A Man Walking At The Rate Of five 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 five:6 Such That Their Lcm Is 48

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

Also K is the greatest not unusual issue (HCF) as five and six have no other not unusual element

? 5K x 6K = 480 x K

K = sixteen = 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. Forty four

Let C.P. = Rs. C. 

Then, 832 - C = C - 448

2C = 1280 => C = 640

Required S.P. = a hundred and fifty% of Rs. 640 = one hundred and fifty/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 = one hundred

then C.P. = 25

Profit = seventy five

Profit% = (75/25) * a hundred = three hundred%.

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 forty five lemons be Rs. X. 

Then, 80 : 40 = a hundred and twenty : x or x = 40×100 twenty/eighty= 60 

For Rs.60, lemons offered = forty five 

For Rs.24, lemons supplied  =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 top quantity.

Then E=  (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (three, 2), (3, 4), (4, 1), (four,three),(5, 2), (five, 6), (6, 1), (6, 5) 

n(E) = 15.

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




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