Interview Questions.

Question 1. If (4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0, Then The Value Of 1/x + 1/y + 1/z Is?

Answer :

(4x - three)/x + (4y - three)/y + (4z - three)/z = 0

=> 4x/x - three/x + 4y/y - three/y + 4z/z - 3/z = zero

=> three/x + three/y + three/z = four + four + four = 12

=> 1/x + 1/y + 1/z = 12/3 = four

Question 2. A Number X When Divided By 289 Leaves 18 As A Remainder. The Same Number When Divided By 17 Leaves Y As A Remainder. The Value Of Y Is?

Answer :

Here, the first divisor (289) is a a couple of of second divisor (17)

∴ Required remainder = Remainder received on dividing 18 by way of 17 = 1

Aptitude Interview Questions
Question three. An Equation Of The Form Ax + By + C = 0 Where A ≠ zero, B ≠ zero, C = 0 Represents A Straight Line Which Passes Through?

Answer :

Ax+by way of+c = zero

When c = 0

ax+by way of = zero

through = -ax ⇒ y = - ax/b

while x = 0, y = zero i.E., this line passes thru the foundation (zero,zero).

Question four. The Numerator Of A Fraction Is four Less Than Its Denominator. If The Numerator Is Decreased By 2 And The Denominator Is Increased By 1, Then The Denominator Becomes Eight Times The Numerator. Find The Fraction?

Answer :

Original fraction = (x - four)/x

In case II,

8(x - 4 - 2) = x + 1

⇒ 8x - 48 = x + 1

⇒ 7x = forty nine ⇒ x = 7

∴Original fraction

= (7 - four)/7 = three/7

Renewable Energy Tutorial
Question five. The Fourth Proportional To 5, 8, 15 Is?

Answer :

Let the fourth proportional to five, 8, 15 be x.

Then, five : 8 : 15 : x 5x = (8 x 15) x=(eight*15)/5=24

Steam Boilers Interview Questions
Question 6. Fresh Fruit Contains 68% Water And Dry Fruit Contains 20% Water. How Much Dry Fruit Can Be Obtained From 100 Kg?

Answer :

1. Given clean fruit has sixty eight% water,

=> Remaining 32% can be fruit content.

2. Given Dry fruit has 20% water

=> Remaining eighty% is fruit content material.

Here expect weight of dry fruit = x kg.

"fruit content in each the sparkling fruit and dry fruit is the equal"

Fruit % in fresh-fruit = fruit% in dry-fruit

so (32/a hundred) * a hundred = (eighty/one hundred )* x

=> x = 40 kg.

Question 7. The Salaries A, B, C Are In The Ratio 2 : 3 : five. If The Increments Of 15%, 10% And 20% Are Allowed Respectively In Their Salaries, Then What Will Be New Ratio Of Their Salaries ?

Answer :

Let A = 2k, B = 3k and C = 5k.

A’s new profits = a hundred and fifteen/ a hundred of 2k =[115/ 100 x 2k]= 23k/ 10

B’s new income = a hundred and ten/ a hundred of 3k = [110/ 100x 3k] = 33k/ 10

C’s new income = 120/ a hundred of 5k = [120/ 100 x 5k] = 6k 

New ratio = 23k : 33k : 6k = 23 : 33 : 60

Metallurgy Interview Questions
Question eight. A Thief Steals A Car At 2.30 P.M And Drives It At 60 Kmph. The Theft Is Discovered At three P.M And The Owner Sets Off In Another Car At 75 Kmph. When Will Be Overtake The Thief

Answer :

As Theft is observed at 3:00pm however Thief stole the car at 2:30.

This manner thief protected far in this 30 min gap.

Distance travelled with the aid of thief in 30 min = 60 * half of = 30 km

Owner Discovered Car at 3:00pm

Now relative velocity = (seventy five-60)km/hr = 15km/hr 

Time had to journey 30km by the velocity of 15km/hr.

Time at which proprietor meets thief = 30/15 = 2 hrs

So After 2 hrs (i.E,at 5:00 pm) the proprietor will trap/overtake the thief

Question 9. One Card Is Drawn At Random From A Pack Of fifty two Cards. What Is The Probability That The Card Drawn Is A Face Card ?

Answer :

Clearly, there are fifty two playing cards, out of which there are 12 face playing cards. P (getting a face card)=12/52=3/13

Metasploit Interview Questions
Question 10. Two Cards Are Drawn Together From A Pack Of 52 Cards. The Probability That One Is A Spade And One Is A Heart, Is ?

Answer :

Let S be the pattern space.

Then, n(S) = 52C2 = 52 * fifty one/ (2*1)= 1326.

Let E = occasion of having 1 spade and 1 coronary heart. N(E) = variety of methods of selecting 1 spade out of 13 and 1 coronary heart out of 13 = (13C1 x 13C1) 

= (13 x 13) = 169. P(E) =n (E)/n(S) = 169/ 1326= thirteen/102

Question eleven. A Bag Contains four White, 5 Red And 6 Blue Balls. Three Balls Are Drawn At Random From The Bag. The Probability That All Of Them Are Red, Is?

Answer :

Let S be the sample space. Then, n(S)= range of approaches of drawing three balls out of 15 = 15C3 = (15 x 14 x thirteen)/ (three x 2 x 1) = 455.

Let E = event of having all of the three red balls. N(E) = 5C3 = 5C2 = (five x four)/ (2 x 1)= 10.

P(E) = n(E) / n(S)= 10/ 455 = 2/ 91

Renewable Energy Interview Questions
Question 12. Two Trains Each 100 M Long, Moving In Opposite Directions, Cross Each Other In eight Seconds. If One Is Moving Twice As Fast The Other, Then The Speed Of The Faster Train Is ?

Answer :

velocity of the quicker train = 2x m/sec.

Relative pace of educate = (x + 2x) m/sec = 3x m/sec.

Total distance = (100 + one hundred)m = 200m

3x = 2 hundred/8 

=> 24x = 2 hundred => x = 25/three

So speed of the faster train = 2 * 25/three m/sec

= 50/3 m/sec 

= 50/three * 18/5 = 60 km/hr.

Aptitude Interview Questions
Question thirteen. Two Number Are In The Ratio three : five. If 9 Is Subtracted From Each, The New Numbers Are In The Ratio 12 : 23. The Smaller Number Is?

Answer :

Let the numbers be 3x and 5x.

Then ,(3x-nine)/(5x-nine)=12/13 23(3x – nine) = 12(5x – nine) 9x = 99 x = 11.

The smaller number = (three x 11) = 33.

Question 14. In A Bag, There Are Coins Of 25 P, 10 P And 5 P In The Ratio Of 1 : 2 : 3. If There Is Rs. 30 In All, How Many five P Coins Are There ?

Answer :

Let the quantity of 25 p, 10 p and five p cash be x, 2x, 3x respectively.

Then, sum of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/a hundred 60x/100=30 

x=(30*100)/60=50 Hence, the range of five p cash = (3 x 50) = a hundred and fifty.

Question 15. After Decreasing 24% In The Cost Price Of An Article,its Costs Rs.912. Find The Actual Cost Of An Article?

Answer :

CP* (76/100) = 912 => CP = 912 * a hundred/seventy six

CP= 12 * a hundred 

=> CP = 1200

cost charge of article = Rs. 1200

HR Interview Questions
Question sixteen. A And B Can Finish A Piece Of Work In 20 Days .B And C In 30 Days And C And A In forty Days. In How Many Days Will A Alone Finish The Job ?

Answer :

Find one day work for all 3

(A+B)'s 1 day work = 1/20 ----(1)

(B+C)'s 1 day work = 1/30 ----(2)

and (C+A)'s 1 day paintings = 1/40 ----(three)

2(A+B+C) 1 day paintings = (1/20 + 1/30 + 1/40)

=> (A+B+C) = (6+four+3)/2*a hundred and twenty 

=> (A+B+C) = thirteen/240 -----------(4)

By eq. (2) and (four)

A + 1/30 = 13/240 

=> A = thirteen/240 - 1/30 = (13-8)/240 = 1/48

then A's 1 day paintings = 1/forty eight

so A alon can end the job = 48 days

Question 17. A Can Do A Work In 10 Days And B Can Do The Same Work In 15 Days. So How Many Days They Will Take To Finish The Same Work ?

Answer :

First find the 1 day work of both (A & B)

A 1 day's work = 1/10

and 

B 1 day's work = 1/15

So (A + B) 1 day's work = (1/10+1/15)

= (three/30+2/30) = 5/30 = 1/6

So Both (A & B) together can end paintings in 6 days

Sheet Metal Interview Questions
Question 18. Ravi's Salary Was Reduced By 25%.Percentage Increase To Be Effected To Bring The Salary To The Original Level Is

Answer :

Explanation:

Method: 1

Let's count on Ravi earnings = one hundred

It get reduced by using 25% => Salary = 75

75(1 + P/a hundred) = one hundred 

1+ P/a hundred = four/three

P = one hundred/3 = 33 1/3%.

Method: 2

You can use immediately method i.E 

[(R*100)/(100-R)]% Where 'R' is decresed %

so put 25 at location of 'R'

=> [(25 * 100)/(100 - 25)]% => [(25 *100)/75]%

=>100/three% = 33 1/3%

Steam Boilers Interview Questions
Question 19. The Product Of Two Numbers Is 9375 And The Quotient, When The Larger One Is Divided By The Smaller, Is 15. The Sum Of The Numbers Is?

Answer :

Let the numbers be x and y.

Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 625. Y = 25. X = 15y = (15 x 25) = 375.

Sum of the numbers = x + y = 375 + 25 = four hundred.

Question 20. Find The Next Term Of The Following Series. 1, 1, 3, nine, eleven, 121,?

Answer :

1*1=1, 1+2=3, 3*3=9, 9+2=eleven, 11*11=121, 121+2=123

Ikanos Aptitude Interview Questions
Question 21. A' And 'b' Complete A Work Togather In eight Days.If 'a' Alone Can Do It In 12 Days.Then How Many Day 'b' Will Take To Complete The Work?

Answer :

A & B in the future work = 1/eight 

A on my own in the future paintings = 1/12

B by myself someday work = (1/8 - 1/12) = ( three/24 - 2/24) 

=> B someday work = 1/24

so B can entire the paintings in 24 days.

Question 22. A Man Sitting In A Train Travelling At The Rate Of 50 Km/hr Observes That It Takes 9 Sec For A Goods Train Travelling In The Opposite Direction To Pass Him. If The Goods Train Is 187.Five M Long, Find Its Speed.

Answer :

Let required velocity be x.

So,187.5/ (x+50)*5/18 =nine

Question 23. X Can Do 1/4 Of A Work In 10 Days, Y Can Do forty% Of The Work In 40 Days And Z Can Do 1/3 Of The Work In thirteen Days. Who Will Complete The Work First ?

Answer :

x can do 1/four of work in = 10 days

so x can do entire paintings in = (10 x four) = 40 days.

Y can do (40% or 40/100)of labor in = 40 days

so Whole work may be carried out by way of Y = (40x100/40)= one hundred days.

Z can do 1/three of work in = 13 days

Whole paintings can be carried out by means of Z in (13 x 3) = 39 days.

So evaluate x , y ,z work evaluate = y > x > z 

so Z can whole the paintings first.

Question 24. X And Y Can Do A Piece Of Work In 20 Days And 12 Days Respectively. X Started The Work Alone And Then After 4 Days Y Joined Him Till The Completion Of The Work. How Long Did The Work Last ?

Answer :

X someday paintings = 1/20

y at some point work = 1/12

work completed by using x in four days = 4 * 1/20 = 1/5

left paintings = (1-1/five) = four/five

 x and y someday work = (1/20 + 1/12) = 8/60 = 2/15

=> time required to do 2/15 a part of paintings with the aid of x and y = 1 day 

so for complete paintings = 1/(2/15) = 15/2

so for four/5 part of paintings x and y will take =( four/5*15/2 ) = 6 days. 

=> How long did the work final = four day + 6 day = 10 days.

Metallurgy Interview Questions
Question 25. X And Y Entered Into Partnership With Rs. 700 And Rs. 600 Respectively. After three Months X Withdrew 2/7 Of His Stock But After three Months, He Puts Back 3/five Of What He Had Withdrawn. The Profit At The End Of The Year Is Rs. 726. How Much Of This Should X Receive ?

Answer :

X’s earnings : Y’s earnings

= seven hundred × three + 500 × 3 + 620 × 6 : 600 × 12

= 2,100 + 1,500 + 3,720 : 7,200

= 7,320 : 7,two hundred

= sixty one : 60

X’s share inside the income = sixty one/(60+61) × 726 = 366

Question 26. A And B Can Do A Piece Of Work In 30 Days, While B And C Can Do The Same Work In 24 Days And C And A In 20 Days. They All Work Together For 10 Days When B And C Leave. How Many Days More Will A Take To Finish The Work ?

Answer :

(A & B)'s 1 day paintings = 1/30

(B & C)'s 1 day paintings = 1/24

(C & A)'s 1 day paintings = 1/20

so 2 (A + B + C)'s 1 day's paintings = (1/30+1/24+1/20) = 15/one hundred twenty = 1/eight

=> (A + B + C)'s 1 day's work = 1/sixteen

Work carried out by using A, B and C in 10 days = (10*1/16) = 5/eight

so left paintings = (1?5/eight)=three/8

A's 1 day's work (1/16?1/24)=1/48

=> 1/forty eight part of work is achieved via A = 1 day.

So, 3/8 part of paintings can be done by using A = (48?3/eight) = 6*3 = 18 days.

Question 27. Solve 2 3 10 39 178 885 ?

Answer :

The common sense is ×1+1, ×2+four, ×three+nine, ×4+sixteen, ×five+25,….

So following the common sense we get 178 is wrong rather it have to be 172.

Metasploit Interview Questions
Question 28. The Incomes Of Chanda And Kim Are In The Ratio 9 : four And Their Expenditures Are In The Ratio 7 : 3. If Each Saves Rs. 2,000, Then Chanda's Expenditure Is

Answer :

Let the incomes of Chanda and Kim be 9x and expenditures be 7y and 3y respectively. 

Since = Income – Expenditure,

we get 9x – 7y = 2000 and 4x – 3y = 2000. 

Solving, we get, x = 8000 and y = 10000. 

So Chanda’s expenditure = 7y = 7 × ten thousand = Rs. 70,000.

Question 29. Ratio Of Ashok's Age To Pradeep's Age Is 4 : three. Ashok Will Be 26 Years Old After 6 Years. How Old Is Pradeep Now ?

Answer :

Given A/p= four/three Also A = 26 after 6 years,

so his present age = 20years,

Substituting we get P = 15 years.