Question 1. A Man, His Wife And Daughter Worked In A Graden. The Man Worked For three Days, His Wife For 2 Days And Daughter For four Days. The Ratio Of Daily Wages For Man To Women Is 5 : 4 And The Ratio For Man To Daughter Is five : 3. If Their Total Earnings Is Mounted To Rs. One hundred and five, Then Find The Daily Wage Of The Daughter?

Answer :

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

Multiply (no. Of days) with (assumed daily wage) of all and sundry to calculate the cost of x.

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

[15x + 8x + 12x] = one zero five

35x = a hundred and five

x = 3

Hence, guy's every day salary = 5x = five x three = Rs. 15

Wife's day by day salary = 4x = four x three = Rs. 12

Daughter's day by day wage = 3x = 3 x three = Rs. Nine.

Question 2. Amit, Raju And Ram Agree To Pay Their Total Electricity Bill In The Proportion three : four : five. Amit Pays First Day's Bill Of Rs. 50, Raju Pays Second Day's Bill Of Rs. Fifty five And Ram Pays Third Day's Bill Of Rs. Seventy five. How Much Amount Should Amit Pay To Settle The Accounts?

Answer :

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

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

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

12x = 180

x = 15

Therefore, quantity paid via,

Amit = Rs. 45

Raju = Rs. 60

Ram = Rs. Seventy five

But definitely as given inside the query, Amit can pay Rs. 50, Raju can pay Rs. 55 and Ram pays Rs. 80. Hence, Amit will pay Rs. 5 less than the real quantity to be paid. Hence he wishes to pay Rs. 5 to Raju settle the amount.

Aptitude Interview Questions

Question three. Salaries Of Ram And Sham Are In The Ratio Of four : 5. If The Salary Of Each Is Increased By Rs. 5000, Then The New Ratio Becomes 50 : 60. What Is Sham's Present Salary?

Answer :

Assume authentic 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 gift earnings = 5x + 5000 = 25,000 + 5000

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

Question 4. Two Dice Are Tossed.The Probability That The Total Score Is A Prime Number Is?

Answer :

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

Let E = Event that the sum is a high wide variety.

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

n(E) = 15.

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

Question 5. Find The Lcm Of Following Three Fractions:36/,48/,72/?

Answer :

Numerators = 36, 48 and 72.

72 is largest number amongst them. 72 isn't divisible with the aid of 36 or forty eight

Start with desk of 72.

Seventy two x 2 = 144 = divisible by means of 72, 36 and forty eight

? LCM of numerators = a hundred and forty four

Denominators = 225, a hundred and fifty and 65

We can see that they may be divided by using five.

On dividing by means of five we get forty five, 30 and 13

We can't divide similarly.

So, HCF = GCD = five

LCM of fraction =one hundred forty four/five.

Sales Interview Questions

Question 6. Find The Largest Number To Divide All The Three Numbers Leaving The Remainders four, three, And 15 Respectively At The End?

Answer :

Here finest range that can divide manner the HCF

Remainders are exclusive so certainly subtract remainders from numbers

17 - four = thirteen; forty two - 3 = 39; 93 - 15 = 78

Now allow's discover HCF of 13, 39 and 78

By direct remark we can see that every one numbers are divisible by means of thirteen.

? HCF = 13 = required finest wide variety.

Question 7. The Two Given Numbers A And B Are In The Ratio 5:6 Such That Their Lcm Is 480. Find Their Hcf?

Answer :

Let K be not unusual component. So 2 numbers are 5K and 6K

Also K is the finest commonplace thing (HCF) as five and 6 haven't any other common component

? 5K x 6K = 480 x K

K = 16 = HCF.

Sales and advertising Interview Questions

Question 8. 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?

Answer :

Wall can be blanketed most effective by means of using rectangular sized wallpaper portions.

Different sized squares are not allowed.

Length = four.5 m = 450 cm;

Height = 3.Five m = 350 cm

Maximum rectangular length viable way HCF of 350 and 450

We can see that 350 and 450 can be divided with the aid of 50.

On dividing by using 50, we get 7 and 9.

Since we can't divide similarly,

HCF = 50 = size of side of rectangular

Number of squares =Wall place/=450 x 350/= sixty three

Square place=50 x 50.

Question 9. Find The Largest Number Of four-digits Divisible By 12, 15 And 18?

Answer :

Required largest range have to be divisible by means of the L.C.M. Of 12, 15 and 18

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

12 = 2 × 2 × three

15 =5 × three

18 = 2 × 3 × three

L.C.M. = one hundred eighty

Now divide 9999 by using one hundred eighty, we get remainder as ninety nine

The required biggest wide variety = (9999 – ninety nine) =9900

Number 9900 is exactly divisible with the aid of 180.

Tele Sales Interview Questions

Question 10. 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?

Answer :

C.P. Of fifty six 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) % = five%.

Question eleven. By Selling forty five Lemons For Rs 40, A Man Loses 20%. How Many Should He Sell For Rs 24 To Gain 20% In The Transaction ?

Answer :

Let S.P. Of 45 lemons be Rs. X.

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

For Rs.60, lemons bought = 45

For Rs.24, lemons bought =4560×24= 18.

Manufacturing Industries Interview Questions

Question 12. 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 ?

Answer :

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

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

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

Aptitude Interview Questions

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

Answer :

Let the S.P = a hundred

then C.P. = 25

Profit = seventy five

Profit% = (75/25) * one hundred = 300%.

Question 14. 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 50. How Many Oranges Does He Buy?

Answer :

Cost Price of two dozen oranges Rs. (five + 4) = Rs. 9.

Sell price of 2 dozen oranges = Rs. 11.

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

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

Question 15. The Sum Of All 3 Digit Numbers Divisible By three Is?

Answer :

All three digit numbers divisible by way of three are :

102, one hundred and five, 108, 111, ..., 999.

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

102 and distinction 'd' as 3.

Let it consists of 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 = three hundred/2[2*102 + 299*3] = 165150.

Manufacturing Industrial Engineer Interview Questions

Question sixteen. 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?

Answer :

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

This is an A.P with first time period, a=35, variety of terms,

n= 12,d=2.

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

=6(70+23)

= 552 kms.

Question 17. A Man Walking At The Rate Of 5 Km/hr Crosses A Bridge In 15 Minutes. The Length Of The Bridge (in Metres) Is?

Answer :

speed= (5x5/18)m/sec=25/18 m/sec.

Distance blanketed in 15 minutes= (25/18 x 15 x 60)m= 1250 m.

ABB Group Aptitude Interview Questions

Question 18. A Man On Tour Travels First a hundred and sixty Km At sixty four Km/hr And The Next a hundred and sixty Km At eighty Km/hr. The Average Speed For The First 320 Km Of The Tour Is?

Answer :

Total time taken = (160/64 + a hundred and sixty/eight)hrs

= 9/2 hrs.

Average pace = (320 x 2/nine) km.Hr

= 71.Eleven km/hr.

Sales Interview Questions

Question 19. If The Product And H.C.F. Of Two Numbers Are 4107 And 37 Respectively, Then Find The Greater Number?

Answer :

4107 is the rectangular of 37.

So let numbers be 37x and 37y.

37x × 37y = 4107

xy = three

3 is the product of (1 and 3)

x = 1 and y = three

37x = 37 × 1 =37

37y = 37 × 3 = 111

Greater quantity = 111.

Question 20. The Traffic Lights At Three Different Road Crossings Change After Every 40 Sec, seventy two Sec And 108 Sec Respectively. If They All Change Simultaneously At five : 20 : 00 Hours, Then Find The Time At Which They Will Change Simultaneously?

Answer :

Traffic lights at three special avenue crossings alternate after every forty sec, seventy two sec and 108 sec respectively.

Therefore, discover the L.C.M. Of 40, seventy two and 108.

L.C.M. Of 40, 72 and 108 = 1080

The visitors lights will trade again after 1080 seconds = 18 min

The subsequent simultaneous change takes place at five : 38 : 00 hrs.

Yahoo Aptitude Interview Questions

Question 21. John, Smith And Kate Start At Same Time, Same Point And In Same Direction To Run Around A Circular Ground. John Completes A Round In 250 Seconds, Smith In 300 Seconds And Kate In one hundred fifty Seconds. Find After What Time Will They Meet Again At The Starting Point?

Answer :

L.C.M. Of 250, 300 and one hundred fifty = 1500 sec

Dividing 1500 by 60 we get 25, which imply 25 minutes.

John, Smith and Kate meet after 25 minutes.

Question 22. The Product Of Two Numbers Is 2028 And Their H.C.F. Is 13. The Number Of Such Pairs Is?

Answer :

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (three, four).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the desired numbers are (thirteen x 1, thirteen x 12) and (13 x 3, thirteen x 4).

Clearly, there are 2 such pairs.

Question 23. Three Number Are In The Ratio Of three : four : 5 And Their L.C.M. Is 2400. Their H.C.F. Is?

Answer :

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = forty.

The numbers are (3 x 40), (four x 40) and (five x forty).

Hence, required H.C.F. = forty.

Tech Mahindra Aptitude Interview Questions

Question 24. The Sum Of Two Numbers Is 528 And Their H.C.F Is 33. The Number Of Pairs Of Numbers Satisfying The Above Condition Is?

Answer :

Let the desired numbers be 33a and 33b.

Then 33a +33b= 528 => a+b = sixteen.

Now, co-primes with sum 16 are (1,15) , (3,thirteen) , (5,11) and (7,9).

Therefore, Required numbers are ( 33 x 1, 33 x 15), (33 x three, 33 x thirteen), (33 x 5, 33 x eleven), (33 x 7, 33 x 9)

The number of such pairs is four.

Sales and advertising and marketing Interview Questions

Question 25. The L.C.M Of Two Numbers Is 495 And Their H.C.F Is five. If The Sum Of The Numbers Is a hundred, Then Their Difference Is?

Answer :

Let the numbers be x and (a hundred-x).

Then,x(100-x)=5*495

=> x2-100x+2475=zero

=> (x-fifty five) (x-forty five) = zero

=> x = fifty five or x = forty five

The numbers are forty five and fifty five

Required distinction = (fifty five-45) = 10.

Question 26. A Rectangular Courtyard 3.78 Meters Long 5.25 Meters Wide Is To Be Paved Exactly With Square Tiles, All Of The Same Size. What Is The Largest Size Of The Tile Which Could Be Used For The Purpose?

Answer :

three.78 meters =378 cm = 2 × three × 3 × 3 × 7

five.25 meters=525 cm = 5 × five × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest length of rectangular tiles that may be paved exactly with square tiles is 21 cm.

Capgemini Aptitude Interview Questions

Question 27. What Is The Probability Of Getting fifty three Mondays In A Leap Year?

Answer :

1 12 months = 12 months . A jump 12 months has three hundred and sixty six days

A yr has fifty two weeks. Hence there may be fifty two Sundays for certain.

52 weeks = fifty two x 7 = 364days

366 – 364 = 2 days

In a soar yr there may be 52 Sundays and 2 days can be left.

These 2 days may be:

1. Sunday, Monday

2. Monday, Tuesday

three. Tuesday, Wednesday

4. Wednesday, Thursday

five. Thursday, Friday

6. Friday, Saturday

7. Saturday, Sunday

Of these overall 7 outcomes, the beneficial outcomes are 2.

Hence the possibility of having 53 days = 2/7.

Tele Sales Interview Questions

Question 28. A Man Bought An Article And Sold It At A Gain Of 5 %. If He Had Bought It At five% Less And Sold It For Re 1 Less, He Would Have Made A Profit Of 10%. The C.P. Of The Article Was?

Answer :

Let original Cost fee is x

Its Selling price = (a hundred and five/100) * x = 21x/20

New Cost rate = (95/a hundred) * x = 19x/20

New Selling fee = (a hundred and ten/one hundred )* (19x/20 )= 209x/2 hundred

[(21x/20) - (209x/200)] = 1

=> x = two hundred.

Question 29. Every Year Before The Festive Season,a Shopkeeper Increases The Price Of The Product By 35% And Then Introduce Two Successive Discount Of 10% And 15% Respectively.What Is Percentage Loss And Percentage Gain?

Answer :

Let cp= one hundred,

35 % growth in sp=a hundred thirty five

10 % discount in one hundred thirty five((a hundred thirty five*10)/a hundred)=13.5

so 1st sp=(one hundred thirty five-thirteen.5)=121.Five, once more 15 % bargain in 1st sp((121.5*15)/one hundred)=18.225

2nd sp=(121.Five-18.225)=103.275,

so in the end cp=one hundred,sp=103.275 ,benefit by three.27%.

HCL Aptitude Interview Questions

Question 30. In A Scheme, A Pack Of Three Soaps With Mrp Rs.45 Is Available For Rs.42. If It Still Gives A Profit Of 5% To The Shopkeeper, Then The Cost Price Of The Pack Is ?

Answer :

Given M.P=forty five,S.P=forty two, Profit = 0.05

Let C.P=x , Then

Profit = (42-x)/x = 0.05

=> x = forty.

Question 31. A Shopkeeper Sells One-1/3 Of His Goods At A Profit Of 10%, Another One-1/3 At A Profit Of 20%, And The Rest At A Loss Of 6%.What Is His Overall Profit Percentage?

Answer :

Let the shopkeeper purchase 300g for Rs.Three hundred. Now he sells 100g for Rs.One hundred ten, another 100g for Rs120, and the relaxation 100g for Rs94. Sir

Therefore, the entire amount he gets = Rs.A hundred and ten + Rs.A hundred and twenty + Rs.94 = 324.

Therefore, the shopkeeper spends Rs.300 and gets again Rs.324.

Therefore, his profit percentage = 24/300x100 % = 8%.

Question 32. By Mixing Two Qualities Of Pulses In The Ratio 2: three And Selling The Mixture At The Rate Of Rs 22 Per Kilogram, A Shopkeeper Makes A Profit Of 10 %. If The Cost Of The Smaller Quantity Be Rs 14 Per Kg, The Cost Per Kg Of The Larger Quantity Is?

Answer :

Cost Price of five kg = Rs.(14*2 + x*three) = (28 + 3x).

Sell charge of 5 kg = Rs. (22x5) = Rs. One hundred ten.

[110 - (28 + 3x)/(28 + 3x) ]* 100 =10

[82-3x/28 + 3x]= 1 / 10

820 - 30x = 28 +3x

33x = 792

x = 24.

Deloitte Aptitude Interview Questions

Question 33. 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%?

Answer :

SP2 = 2/3 SP1

CP = one hundred

SP2 = eighty

2/three SP1 = eighty

SP1 = a hundred and twenty

a hundred --- 20 => 20%.

Manufacturing Industries Interview Questions

Question 34. 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 Transaction?

Answer :

Cost rate of 25 kg = Rs. (15 x 14.50 + 10 x thirteen) = Rs. 347.50.

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

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

Question 35. The Profit Earned By Selling An Article For Rs. 832 Is Equal To The Loss Incurred When The Same Article Is Sold For Rs. 448. What Should Be The Sale Price For Making 50% Profit ?

Answer :

Let C.P. = Rs. C.

Then, 832 - C = C - 448

2C = 1280 => C = 640

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

Question 36. 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?

Answer :

Here always recollect, while ever x% loss,

it way S.P. = (a hundred - x)% of C.P

when ever x% profit,

it method S.P. = (one hundred + x)% of C.P

So right here could be (one hundred - x)% of C.P.

= 80% of 1200

= (eighty/100) * 1200

= 960.

Manufacturing Industrial Engineer Interview Questions

Question 37. '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 How Much Did 'a' Pay For It?

Answer :

Let the object fees 'X' to A

Cost price of B = 1.2X

Cost fee of C = 0.Seventy five(1.2X) = 0.9X

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

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

Question 38. A Shopkeeper Fixes The Marked Price Of An Item 35% Above Its Cost Price. The Percentage Of Discount Allowed To Gain 8% Is?

Answer :

Let the cost rate = Rs one hundred

then, Marked price = Rs a hundred thirty five

Required advantage = eight%,

So Selling charge = Rs 108

Discount = 135 - 108 = 27

Discount% = (27/135)*a hundred = 20%.