Essar Aptitude Interview Questions & Answers

  1. 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 – 3)/x + (4y – 3)/y + (4z – 3)/z = 0

    => 4x/x – 3/x + 4y/y – 3/y + 4z/z – 3/z = 0

    => 3/x + 3/y + 3/z = 4 + 4 + 4 = 12

    => 1/x + 1/y + 1/z = 12/3 = 4

  2. 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 multiple of second divisor (17)

    ∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1

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  4. Question 3. An Equation Of The Form Ax + By + C = 0 Where A ≠ 0, B ≠ 0, C = 0 Represents A Straight Line Which Passes Through?

    Answer :

    Ax+by+c = 0

    When c = 0

    ax+by = 0

    by = -ax ⇒ y = – ax/b

    when x = 0, y = 0 i.e., this line passes through the origin (0,0).

  5. Question 4. The Numerator Of A Fraction Is 4 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 – 4)/x

    In case II,

    8(x – 4 – 2) = x + 1

    ⇒ 8x – 48 = x + 1

    ⇒ 7x = 49 ⇒ x = 7

    ∴Original fraction

    = (7 – 4)/7 = 3/7

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  7. Question 5. The Fourth Proportional To 5, 8, 15 Is?

    Answer :

    Let the fourth proportional to 5, 8, 15 be x.

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

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  9. 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 fresh fruit has 68% water,

    => Remaining 32% will be fruit content.

    2. Given Dry fruit has 20% water

    => Remaining 80% is fruit content.

    Here assume weight of dry fruit = x kg.

    “fruit content in both the fresh fruit and dry fruit is the same”

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

    so (32/100) * 100 = (80/100 )* x

    => x = 40 kg.

  10. Question 7. The Salaries A, B, C Are In The Ratio 2 : 3 : 5. 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 salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10

    B’s new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10

    C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k 

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

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  12. Question 8. A Thief Steals A Car At 2.30 P.m And Drives It At 60 Kmph. The Theft Is Discovered At 3 P.m And The Owner Sets Off In Another Car At 75 Kmph. When Will Be Overtake The Thief

    Answer :

    As Theft is discovered at 3:00pm but Thief stole the car at 2:30.

    This means thief covered some distance in this 30 min gap.

    Distance travelled by thief in 30 min = 60 * 1/2 = 30 km

    Owner Discovered Car at 3:00pm

    Now relative speed = (75-60)km/hr = 15km/hr 

    Time needed to travel 30km by the speed of 15km/hr.

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

    So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief

  13. Question 9. One Card Is Drawn At Random From A Pack Of 52 Cards. What Is The Probability That The Card Drawn Is A Face Card ?

    Answer :

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

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  15. 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 sample space.

    Then, n(S) = 52C2 = 52 * 51/ (2*1)= 1326.

    Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) 

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

  16. Question 11. A Bag Contains 4 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)= number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1) = 455.

    Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10.

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

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  18. Question 12. Two Trains Each 100 M Long, Moving In Opposite Directions, Cross Each Other In 8 Seconds. If One Is Moving Twice As Fast The Other, Then The Speed Of The Faster Train Is ?

    Answer :

    speed of the faster train = 2x m/sec.

    Relative speed of train = (x + 2x) m/sec = 3x m/sec.

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

    3x = 200/8 

    => 24x = 200 => x = 25/3

    So speed of the faster train = 2 * 25/3 m/sec

    = 50/3 m/sec 

    = 50/3 * 18/5 = 60 km/hr.

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  20. Question 13. Two Number Are In The Ratio 3 : 5. 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-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11.

    The smaller number = (3 x 11) = 33.

  21. 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 5 P Coins Are There ?

    Answer :

    Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

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

    x=(30*100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.

  22. 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 * 100/76

    CP= 12 * 100 

    => CP = 1200

    cost price of article = Rs. 1200

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  24. Question 16. A And B Can Finish A Piece Of Work In 20 Days .b And C In 30 Days And C And A In 40 Days. In How Many Days Will A Alone Finish The Job ?

    Answer :

    Find one day work for all three

    (A+B)’s 1 day work = 1/20 —-(1)

    (B+C)’s 1 day work = 1/30 —-(2)

    and (C+A)’s 1 day work = 1/40 —-(3)

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

    => (A+B+C) = (6+4+3)/2*120 

    => (A+B+C) = 13/240 ———–(4)

    By eq. (2) and (4)

    A + 1/30 = 13/240 

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

    then A’s 1 day work = 1/48

    so A alon can finish the job = 48 days

  25. 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)

    = (3/30+2/30) = 5/30 = 1/6

    So Both (A & B) together can finish work in 6 days

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  27. 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 assume Ravi salary = 100

    It get reduced by 25% => Salary = 75

    75(1 + P/100) = 100 

    1+ P/100 = 4/3

    P = 100/3 = 33 1/3%.

    Method: 2

    You can use directly formula i.e 

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

    so put 25 at place of ‘R’

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

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

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  29. 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 = 400.

  30. Question 20. Find The Next Term Of The Following Series. 1, 1, 3, 9, 11, 121,?

    Answer :

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

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  32. Question 21. A’ And ‘b’ Complete A Work Togather In 8 Days.if ‘a’ Alone Can Do It In 12 Days.then How Many Day ‘b’ Will Take To Complete The Work?

    Answer :

    A & B one day work = 1/8 

    A alone one day work = 1/12

    B alone one day work = (1/8 – 1/12) = ( 3/24 – 2/24) 

    => B one day work = 1/24

    so B can complete the work in 24 days.

  33. 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 187.5 M Long, Find Its Speed.

    Answer :

    Let required speed be x.

    So,187.5/{ (x+50)*5/18} =9

  34. Question 23. X Can Do 1/4 Of A Work In 10 Days, Y Can Do 40% Of The Work In 40 Days And Z Can Do 1/3 Of The Work In 13 Days. Who Will Complete The Work First ?

    Answer :

    x can do 1/4 of work in = 10 days

    so x can do whole work in = (10 x 4) = 40 days.

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

    so Whole work can be done by Y = (40×100/40)= 100 days.

    Z can do 1/3 of work in = 13 days

    Whole work will be done by Z in (13 x 3) = 39 days.

    so compare x , y ,z work compare = y > x > z 

    so Z can complete the work first.

  35. 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 one day work = 1/20

    y one day work = 1/12

    work done by x in 4 days = 4 * 1/20 = 1/5

    left work = (1-1/5) = 4/5

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

    => time required to do 2/15 part of work by x and y = 1 day 

    so for whole work = 1/(2/15) = 15/2

    so for 4/5 part of work x and y will take =( 4/5*15/2 ) = 6 days. 

    => How long did the work last = 4 day + 6 day = 10 days.

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  37. Question 25. X And Y Entered Into Partnership With Rs. 700 And Rs. 600 Respectively. After 3 Months X Withdrew 2/7 Of His Stock But After 3 Months, He Puts Back 3/5 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 profit : Y’s profit

    = 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12

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

    = 7,320 : 7,200

    = 61 : 60

    X’s share in the profit = 61/(60+61) × 726 = 366

  38. 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 work = 1/30

    (B & C)’s 1 day work = 1/24

    (C & A)’s 1 day work = 1/20

    so 2 (A + B + C)’s 1 day’s work = (1/30+1/24+1/20) = 15/120 = 1/8

    => (A + B + C)’s 1 day’s work = 1/16

    Work done by A, B and C in 10 days = (10*1/16) = 5/8

    so left work = (1?5/8)=3/8

    A’s 1 day’s work (1/16?1/24)=1/48

    => 1/48 part of work is done by A = 1 day.

    So, 3/8 part of work will be done by A = (48?3/8) = 6*3 = 18 days.

  39. Question 27. Solve 2 3 10 39 178 885 ?

    Answer :

    The logic is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,….

    So following the logic we get 178 is wrong instead it should be 172.

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  41. Question 28. The Incomes Of Chanda And Kim Are In The Ratio 9 : 4 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 × 10000 = Rs. 70,000.

  42. Question 29. Ratio Of Ashok’s Age To Pradeep’s Age Is 4 : 3. Ashok Will Be 26 Years Old After 6 Years. How Old Is Pradeep Now ?

    Answer :

    Given A/p= 4/3 Also A = 26 after 6 years,

    so his present age = 20years,

    Substituting we get P = 15 years.