Top 100+ Infosys Aptitude Interview Questions And Answers
Question 1. A Dishonest Dealer Professes To Sell His Goods At The Cost Price But Uses A Weight Of 800gm Instead Of 1kg. Find His Real Gain Percent ?
Answer :
two hundred/800 ×a hundred = 25%
Question 2. A Man Invested Rs. 4940 In Rs. 10 Shares Quoted At Rs. Nine.50. If The Rate Of Dividend Be 14%, His Annual Income Is
Answer :
Market Value of a percentage = Rs.9.50
Investment = Rs.4940
Number of shares = 4940/nine.50 = 520
Face Value of a proportion = Rs.10
dividend = 14%
dividend per share = 10*14/one hundred = Rs. 1.Four
His annual profits = 520 × 1.Four = Rs.728
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Question three. A Sum Of Money Lent Out At Simple Interest Amounts To Rs. 720 After 2 Years And To Rs. 1,020 After A Further Period Of 5 Years. The Sum And The Rate % Are
Answer :
Amount after 2 years = Rs 720
Amount after 7 years = Rs 1020
Therefore, Interest for 5 years = Rs three hundred
Interest for 1 12 months = Rs 60
And Interest for two years = Rs a hundred and twenty
SO Principal = 720-one hundred twenty = Rs six hundred
Also, 120 = (six hundred*R*2)/a hundred = R = 10%
Amount after 2 years = Rs 720
Amount after 7 years = Rs 1020
Therefore, Interest for five years = Rs 300
Interest for 1 12 months = Rs 60
And Interest for two years = Rs one hundred twenty
SO Principal = 720-one hundred twenty = Rs 600
Also, one hundred twenty = (600*R*2)/a hundred = R = 10%
Question 4. A Train With ninety Km/h Crosses A Bridge In 36 Seconds. Another Train one hundred Metres Shorter Crosses The Same Bridge At forty five Km/h. What Is The Time Taken By The Second Train To Cross The Bridge ?
Answer :
Train A, Speed = 90kmph
=ninety*(five/18)m/s = 25m/s = 25m/s, t=36s
Let duration, L = x+y = time*pace = 25*36 = 900m
=800m, Speed= 45*(5/18) = (25/2) m/s
t= (Distance/Speed) = (800/(25/2)) = (1600/25) = sixty four seconds
Question 5. Gaurav's Age After 15 Years Will Be five Times His Age 5 Years Back. What Is The Present Age Of Gaurav ?
Answer :
Let Gaurav's gift age be x years. Then,
Gaurav's age after 15 years = (x + 15) years.
Gourav's age five years back = (x - five) years.
Therefore x + 15 = 5 (x - five) x + 15 = 5x - 25 4x = 40 x = 10.
Hence, Gaurav's gift age = 10 years.
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Question 6. A Is As Much Younger Than B. B As He Is Older Than C. If The Sum Of The Ages Is 50 Years. What Is The Difference Between The Ages Of B And A's? Solution: Given That: 1. The Difference Of Age B/w B And A = The Difference Of Age B/w A And C. 2. Sum Of Age Of B And C Is 50 I.E. (b + C) = 50. Question: B - A = ?.
Answer :
B - A = A - C
(B + C) = 2A
Now given that, (B + C) = 50
So, 50 = 2A and consequently A = 25.
Question is (B - A) = ?
Here we know the fee(age) of A (25), but we don't know the age of B.
Therefore, (B-A) can not be determined.
Question 7. Ramesh Travels 760 Km To His Home, Partly By Train And Partly By Car He Takes 8 Hours, If He Travels one hundred sixty Km By Train And The Rest By Car. He Takes 12 Minutes More, If He Travels 240 Km By Train And The Rest By Car. What Are The Speeds Of The Train And Of The Car ?
Answer :
Let speeds be x and y for teach and car respectively.
Then 8 = (160/eight) + (600/y) .....(1)
And eight(1/five) = (240/x) + ((760-240)/y) .....(2)
Solving for x and y, we get 100 and 80 km/hr.
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Question 8. Some Students Planned A Picnic. The Budget For Food Was Rs. 500. But, 5 Of Them Failed To Go And Thus The Cost Of Food For Each Member Increased By Rs. Five. How Many Students Attended The Picnic?
Answer :
By course alternatives,500/25=20 ,500/20=25
By mathematical method, the primary steps are: xy = 500 …(1) and (x−5) (y+5) = 500 …(2),
From eqn. 2, x−y = five or y = x−five Put in eqn 1, x(x−5) = 500 or x2-5x-500=0 ,
i.E. X = 25 and attended ones = x − 5 = 20
Question nine. (17)three.Five X (17)? = 178 Solve X ?
Answer :
Let (17)three.Five x (17)x = 178.
Then, (17)3.Five + x = 178.
3.Five + x = eight
x = (8 - 3.5)
x = 4.Five
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Question 10. After Being Set Up, A Company Manufactured 6000 Scooters In The Third Year And 7000 Scooters In The Seventh Year. Assuming That The Production Increases Uniformly By A Fixed Number Every Year, What Is The Production In The Tenth Year?
Answer :
You can use A.P.,Tn =a+(n-1)d ,6000=a+2nd.....(1) and 7000 = a + 6d .....(2)
Eqn (2) – Eqn (1) ⇒ a thousand=4d,
i.E. D = 250 and a = 6000 − 500 = 5500
T10 =5500 + 9 × 250 =7750
Question eleven. The Average Score Of Boys In An Examination In A School Is seventy one And That Of The Girls Is seventy three. The Average Score Of The School Is seventy one.8. The Ratio Of The Number Of Boys To That Of The Girls That Appeared In The Examination Is
Answer :
seventy one.Eight = (71x+73y)/(x+y)
71.8 (x+ y) = 71x + 73y
zero.8x = 1.2y
x:y = 12:eight that is equals to 3:2
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Question 12. The Mean Monthly Salary Paid To 75 Workers In A Factory Is Rs. Five,680. The Mean Salary Of 25 Of Them Is Rs. Five,four hundred And That Of 30 Others Is Rs. Five,seven hundred. The Mean Salary Of The Remaining Workers Is
Answer :
5680*seventy five = (5400*25+5700*30+x(seventy five-25-30))/75
four,26,00 = 1,35,000 +1, 71,000 + 20x
X = 1,20,000/20, = 6,000
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Question 13. A Sum Of Rs. 25 Was Paid For A Work Which A Can Do In 32 Days, B In 20 Days, B And C In 12 Days And D In 24 Days. How Much Did C Receive If All The Four Work Together ?
Answer :
B+ C's 1 day's work = ½ and B's 1 day's work = 1/20
Therefore, C's 1 day's paintings = (1/12) – (1/20) = four/120 = 1/30
Monet might be distributed in keeping with the ratio of work carried out i.E A: B: C: D
= 1/32 : 1/20 : 1/30 : 1/24 = 15 :24:16:20
Therefore, C's Share = sixteen/(15+24+16+20) = Rs sixteen/3
Question 14. A Man Sold Two Steel Chairs For Rs. 500 Each. On One, He Gains 20% And On Other, He Loses 12%. How Much Does He Gain Or Lose In The Whole Transaction ?
Answer :
CP/SP = a hundred/(a hundred±x) , i.E. Total CP = 417 (500*100/2 hundred) + 568(500*one hundred/88)≅ 985
Since CP<SP .Therefore, Profit = a hundred-985 = 15
P% ≅ 15/985 X 100 ≅ 1.Five %
Question 15. A Man Purchased A Cow For Rs. 3000 And Sold It The Same Day For Rs. 3600, Allowing The Buyer A Credit Of 2 Years. If The Rate Of Interest Be 10% Per Annum, Then The Man Has A Gain Of:
Answer :
C.P. = Rs. 3000.
S.P. = Rs.3600 x 10 = Rs. 3000.
100 + (10 x 2)
Gain = zero%.
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Question 16. It Is Being Given That (232 + 1) Is Completely Divisible By A Whole Number. Which Of The Following Numbers Is Completely Divisible By This Number ?
Answer :
Let 232 = x. Then, (232 + 1) = (x + 1).
Let (x + 1) be completely divisible via the natural quantity N. Then,
(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by means of N, because (x + 1) is divisible by N.
Question 17. How Many Of The Following Numbers Are Divisible By 132 ?264, 396, 462, 792, 968, 2178, 5184, 6336
Answer :
132 = 4 x 3 x 11
So, if the number divisible by way of all the three range four, 3 and eleven, then the wide variety is divisible with the aid of 132 also.
264 eleven,three,four (/)
396 eleven,3,four (/)
462 eleven,3 (X)
792 eleven,three,four (/)
968 11,four (X)
2178 eleven,3 (X)
5184 3,four (X)
6336 eleven,3,4 (/)
Therefore the following numbers are divisible by using 132 : 264, 396, 792 and 6336.
Required wide variety of quantity = 4.
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Question 18. The Difference Of Two Numbers Is 1365. On Dividing The Larger Number By The Smaller, We Get 6 As Quotient And The 15 As Remainder. What Is The Smaller Number ?
Answer :
Let the smaller variety be x. Then large number = (x + 1365).
X + 1365 = 6x + 15
5x = 1350
x = 270
Smaller quantity = 270
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Question 19. If The Number 517*324 Is Completely Divisible By 3, Then The Smallest Whole Number In The Place Of * Will Be:
Answer :
Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x),
which have to be divisible by 3.
X = 2.
Question 20. The Sum Of First 45 Natural Numbers Is:
Answer :
Let Sn =(1 + 2 + three + ... + forty five). This is an A.P. In which a =1, d =1, n = 45.
Sn = n[2a + (n - 1)d] = 45x [2 x 1 + (45 - 1) x 1]/2 =45x forty six/2 = (45 x 23)/2
= forty five x (20 + three)
= forty five x 20 + 45 x 3
= 900 + one hundred thirty five
= 1035.
Shortcut Method:
Sn = n(n + 1)/2=45(forty five + 1)/2= 1035.
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