Essar Aptitude Placement Papers - Essar Aptitude Interview Questions and Answers
Q1. Find The Next Term Of The Following Series. 1, 1, three, 9, eleven, 121,?
1*1=1, 1+2=three, three*three=nine, nine+2=11, eleven*eleven=121, 121+2=123
Q2. Two Cards Are Drawn Together From A Pack Of fifty two Cards. The Probability That One Is A Spade And One Is A Heart, Is ?
Let S be the pattern area.
Then, n(S) = 52C2 = 52 * fifty one/ (2*1)= 1326.
Let E = event of having 1 spade and 1 heart. N(E) = variety of approaches of choosing 1 spade out of thirteen and 1 heart out of thirteen = (13C1 x 13C1)
= (thirteen x 13) = sixteen@P(E) =n (E)/n(S) = 169/ 1326= thirteen/102
Q3. Two Number Are In The Ratio 3 :
Let the numbers be 3x and 5x.
Then ,(3x-nine)/(5x-9)=12/thirteen 23(3x – nine) = 12(5x – nine) 9x = ninety nine x = 11.
The smaller wide variety = (three x eleven) = 33.
Q4. The Salaries A, B, C Are In The Ratio 2 : three :
Let A = 2k, B = 3k and C = 5k.
A’s new income = a hundred and fifteen/ a hundred of 2k =[115/ 100 x 2k]= 23k/ 10
B’s new revenue = a hundred and ten/ 100 of 3k = [110/ 100x 3k] = 33k/ 10
C’s new salary = one hundred twenty/ a hundred of 5k = [120/ 100 x 5k] = 6k
New ratio = 23k : 33k : 6k = 23 : 33 : 60
Q5. After Decreasing 24% In The Cost Price Of An Article,its Costs Rs.Nine
CP* (76/100) = 912 => CP = 912 * one hundred/seventy six
CP= 12 * one hundred
=> CP = 1200
price price of article = Rs. 1200
Q6. 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 ?
Clearly, there are fifty two playing cards, out of which there are 12 face cards. P (getting a face card)=12/fifty two=3/thirteen
Q7. 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 F
Original fraction = (x - 4)/x
In case II,
eight(x - four - 2) = x + 1
⇒ 8x - forty eight = x + 1
⇒ 7x = 49 ⇒ x = 7
= (7 - 4)/7 = three/7
Q8. Ratio Of Ashok's Age To Pradeep's Age Is 4 :
Given A/p= four/three Also A = 26 after 6 years,
so his present age = 20years,
Substituting we get P = 15 years.
Q9. Two Trains Each one hundred 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 ?
Velocity of the faster teach = 2x m/sec.
Relative velocity of train = (x + 2x) m/sec = 3x m/sec.
Total distance = (a hundred + one hundred)m = 200m
3x = 200/eight
=> 24x = 2 hundred => x = 25/three
So speed of the faster educate = 2 * 25/3 m/sec
= 50/three m/sec
= 50/three * 18/5 = 60 km/hr.
Q10. A Bag Contains four White, five Red And 6 Blue Balls. Three Balls Are Drawn At Random From The Bag. The Probability That All Of Them Are Red, Is?
Let S be the pattern area. Then, n(S)= wide variety of approaches of drawing 3 balls out of 15 = 15C3 = (15 x 14 x thirteen)/ (three x 2 x 1) = 455.
Let E = event of having all the three red balls. N(E) = 5C3 = 5C2 = (five x four)/ (2 x 1)= 10.
P(E) = n(E) / n(S)= 10/ 455 = 2/ ninety one
Q11. 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?
Here, the first divisor (289) is a multiple of second divisor (17)
∴ Required the rest = Remainder obtained on dividing 18 with the aid of 17 = 1
Q12. 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?
A & B in the future paintings = 1/8
A by myself sooner or later work = 1/12
B alone one day work = (1/eight - 1/12) = ( three/24 - 2/24)
=> B one day work = 1/24
so B can entire the paintings in 24 days.
Q13. 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 ?
X sooner or later work = 1/20
y at some point work = 1/12
paintings carried out via x in four days = 4 * 1/20 = 1/5
left paintings = (1-1/5) = 4/five
x and y one day paintings = (1/20 + 1/12) = eight/60 = 2/15
=> time required to do 2/15 a part of paintings with the aid of x and y = 1 day
so for entire work = 1/(2/15) = 15/2
so for four/5 part of work x and y will take =( 4/five*15/2 ) = 6 days.
=> How lengthy did the work closing = four day + 6 day = 10 days.
Q14. 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 seventy five Kmph. When Will Be Overtake The Thief
As Theft is determined at three:00pm but Thief stole the automobile at 2:30.
This me thief covered far on this 30 min hole.
Distance travelled via thief in 30 min = 60 * half = 30 km
Owner Discovered Car at 3:00pm
Now relative pace = (75-60)km/hr = 15km/hr
Time had to tour 30km with the aid of 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
Q15. X And Y Entered Into Partnership With Rs. 700 And Rs. Six hundred Respectively. After three Months X Withdrew 2/7 Of His Stock But After three Months, He Puts Back three/5 Of What He Had Withdrawn. The Profit At The End
X’s income : Y’s profit
= seven-hundred × three + 500 × 3 + 620 × 6 : 600 × 12
= 2,a hundred + 1,500 + three,720 : 7,two hundred
= 7,320 : 7,200
= sixty one : 60
X’s share within the earnings = 61/(60+61) × 726 = 366
Q16. The Incomes Of Chanda And Kim Are In The Ratio 9 : four And Their Expenditures Are In The Ratio 7 :
Let the earning of Chanda and Kim be 9x and expenses 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.
Q17. The Product Of Two Numbers Is 9375 And The Quotient, When The Larger One Is Divided By The Smaller, Is
Let the numbers be x and y.
Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = sixty two@y = 2@x = 15y = (15 x 25) = 375.
Sum of the numbers = x + y = 375 + 25 = 400.
Q18. 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/three Of The Work In 13 Days. Who Will Complete The Work First ?
X can do 1/four of labor in = 10 days
so x can do complete paintings in = (10 x 4) = forty days.
Y can do (40% or 40/a hundred)of work in = forty days
so Whole paintings can be executed via Y = (40x100/forty)= a hundred days.
Z can do 1/3 of work in = thirteen days
Whole paintings will be performed by using Z in (13 x 3) = 39 days.
So compare x , y ,z work evaluate = y > x > z
so Z can entire the work first.
Q19. A Man Sitting In A Train Travelling At The Rate Of 50 Km/hr Observes That It Takes nine Sec For A Goods Train Travelling In The Opposite Direction To Pass Him. If The Goods Train Is 187.Five M Long, Find It
Let required speed be x.
So,187.5/ (x+50)*five/18 =nine
Q20. 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 T
(A & B)'s 1 day paintings = 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 paintings = 1/sixteen
Work performed by using A, B and C in 10 days = (10*1/sixteen) = 5/eight
so left work = (1?Five/eight)=three/8
A's 1 day's work (1/sixteen?1/24)=1/48
=> 1/forty eight part of paintings is done with the aid of A = 1 day.
So, three/8 part of paintings can be accomplished by A = (48?Three/8) = 6*three = 18 days.
Q21. The Fourth Proportional To 5, eight, 15 Is?
Let the fourth proportional to 5, 8, 15 be x.
Then, 5 : eight : 15 : x 5x = (8 x 15) x=(eight*15)/5=24
Q22. 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 ?
First discover the 1 day work of each (A & B)
A 1 day's paintings = 1/10
B 1 day's paintings = 1/15
So (A + B) 1 day's work = (1/10+1/15)
= (3/30+2/30) = five/30 = 1/6
So Both (A & B) together can end work in 6 days
Q23. Ravi's Salary Was Reduced By 25%.Percent Increase To Be Effected To Bring The Salary To The Original Level Is
Let's expect Ravi salary = a hundred
It get reduced by 25% => Salary = seventy five
75(1 + P/a hundred) = one hundred
1+ P/100 = four/three
P = one hundred/three = 33 1/three%.
You can use without delay formulation i.E
[(R*100)/(100-R)]% Where 'R' is decresed %
so positioned 25 at area of 'R'
=> [(25 * 100)/(100 - 25)]% => [(25 *100)/75]%
=>a hundred/three% = 33 1/3%
Q24. In A Bag, There Are Coins Of 25 P, 10 P And five P In The Ratio Of 1 : 2 :
Let the wide variety of 25 p, 10 p and five p coins be x, 2x, 3x respectively.
Then, sum of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/a hundred 60x/a hundred=30
x=(30*a hundred)/60=50 Hence, the variety of five p coins = (three x 50) = 150.
Q25. Solve 2 3 10 39 178 885 ?
The logic is ×1+1, ×2+four, ×three+9, ×4+sixteen, ×5+25,….
So following the common sense we get 178 is inaccurate rather it need to be 172.
Q26. If (4x - three)/x + (4y - three)/y + (4z - three)/z = zero, Then The Value Of 1/x + 1/y + 1/z Is?
(4x - 3)/x + (4y - 3)/y + (4z - 3)/z = zero
=> 4x/x - three/x + 4y/y - 3/y + 4z/z - three/z = zero
=> three/x + 3/y + 3/z = four + 4 + 4 = 12
=> 1/x + 1/y + 1/z = 12/3 = 4
Q27. Fresh Fruit Contains 68% Water And Dry Fruit Contains 20% Water. How Much Dry Fruit Can Be Obtained From one hundred Kg?
@Given sparkling fruit has sixty eight% water,
=> Remaining 32% might be fruit content material.
@Given Dry fruit has 20% water
=> Remaining 80% is fruit content material.
Here expect weight of dry fruit = x kg.
"fruit content in both the fresh fruit and dry fruit is the same"
Fruit % in clean-fruit = fruit% in dry-fruit
so (32/a hundred) * a hundred = (80/a hundred )* x
=> x = forty kg.
Q28. 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 ?
Find someday 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 paintings = 1/forty ----(3)
2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40)
=> (A+B+C) = (6+four+three)/2*a hundred and twenty
=> (A+B+C) = thirteen/240 -----------(4)
By eq. (2) and (4)
A + 1/30 = thirteen/240
=> A = 13/240 - 1/30 = (13-eight)/240 = 1/forty eight
then A's 1 day work = 1/forty eight
so A alon can end the job = 48 days