## Infosys Aptitude Interview Questions & Answers

1. 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 ?

200/800 ×100 = 25%

2. Question 2. A Man Invested Rs. 4940 In Rs. 10 Shares Quoted At Rs. 9.50. If The Rate Of Dividend Be 14%, His Annual Income Is

Market Value of a share = Rs.9.50

Investment = Rs.4940

Number of shares = 4940/9.50 = 520

Face Value of a share = Rs.10

dividend = 14%

dividend per share = 10*14/100  = Rs. 1.4

His annual income = 520 × 1.4 = Rs.728

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4. Question 3. 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

Amount after 2 years = Rs 720

Amount after 7 years = Rs 1020

Therefore, Interest for 5 years = Rs 300

Interest for 1 year = Rs 60

And Interest for 2 years = Rs 120

SO Principal = 720-120 = Rs 600

Also, 120 = (600*R*2)/100 = R = 10%

Amount after 2 years = Rs 720

Amount after 7 years = Rs 1020

Therefore, Interest for 5 years = Rs 300

Interest for 1 year = Rs 60

And Interest for 2 years = Rs 120

SO Principal = 720-120 = Rs 600

Also, 120 = (600*R*2)/100 = R = 10%

5. Question 4. A Train With 90 Km/h Crosses A Bridge In 36 Seconds. Another Train 100 Metres Shorter Crosses The Same Bridge At 45 Km/h. What Is The Time Taken By The Second Train To Cross The Bridge ?

Train A, Speed = 90kmph

=90*(5/18)m/s = 25m/s = 25m/s, t=36s

Let length, L = x+y = time*speed = 25*36 = 900m

=800m, Speed= 45*(5/18) = (25/2) m/s

t= (Distance/Speed) = (800/(25/2)) = (1600/25) = 64 seconds

6. Question 5. Gaurav’s Age After 15 Years Will Be 5 Times His Age 5 Years Back. What Is The Present Age Of Gaurav ?

Let Gaurav’s present age be x years. Then,

Gaurav’s age after 15 years = (x + 15) years.

Gourav’s age 5 years back = (x – 5) years.

Therefore x + 15 = 5 (x – 5) x + 15 = 5x – 25 4x = 40 x = 10.

Hence, Gaurav’s present age = 10 years.

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8. 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 = ?.

B – A = A – C

(B + C) = 2A

Now given that, (B + C) = 50

So, 50 = 2A and therefore A = 25.

Question is (B – A) = ?

Here we know the value(age) of A (25), but we don’t know the age of B.

Therefore, (B-A) cannot be determined.

9. Question 7. Ramesh Travels 760 Km To His Home, Partly By Train And Partly By Car He Takes 8 Hours, If He Travels 160 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 ?

Let speeds be x and y for train and car respectively.

Then 8 = (160/8) + (600/y) …..(1)

And 8(1/5) = (240/x) + ((760-240)/y) …..(2)

Solving for x and y, we get 100 and 80 km/hr.

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11. 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. 5. How Many Students Attended The Picnic?

By direction options,500/25=20 ,500/20=25

By mathematical method, the main steps are: xy = 500 …(1) and (x−5) (y+5) = 500 …(2),

From eqn. 2, x−y = 5 or y = x−5 Put in eqn 1, x(x−5) = 500 or x2-5x-500=0 ,

i.e. x = 25 and attended ones = x − 5 = 20

12. Question 9. (17)3.5 X (17)? = 178 Solve X ?

Let (17)3.5 x (17)x = 178.

Then, (17)3.5 + x = 178.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

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14. 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?

You can use A.P.,Tn =a+(n-1)d ,6000=a+2d…..(1) and 7000 = a + 6d …..(2)

Eqn (2) – Eqn (1) ⇒ 1000=4d,

i.e. d = 250 and a = 6000 − 500 = 5500

T10 =5500 + 9 × 250 =7750

15. Question 11. The Average Score Of Boys In An Examination In A School Is 71 And That Of The Girls Is 73. The Average Score Of The School Is 71.8. The Ratio Of The Number Of Boys To That Of The Girls That Appeared In The Examination Is

71.8 = (71x+73y)/(x+y)

71.8 (x+ y) = 71x + 73y

0.8x = 1.2y

x:y = 12:8 which is equals to 3:2

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17. Question 12. The Mean Monthly Salary Paid To 75 Workers In A Factory Is Rs. 5,680. The Mean Salary Of 25 Of Them Is Rs. 5,400 And That Of 30 Others Is Rs. 5,700. The Mean Salary Of The Remaining Workers Is

5680*75 = (5400*25+5700*30+x(75-25-30))/75

4,26,00 = 1,35,000 +1, 71,000 + 20x

X = 1,20,000/20, = 6,000

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19. 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 ?

B+ C’s 1 day’s work = ½ and B’s 1 day’s work = 1/20

Therefore, C’s 1 day’s work = (1/12) – (1/20) = 4/120 = 1/30

Monet will be distributed according to the ratio of work done i.e A: B: C: D

= 1/32 : 1/20 : 1/30 : 1/24 = 15 :24:16:20

Therefore, C’s Share = 16/(15+24+16+20) = Rs 16/3

20. 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 ?

CP/SP = 100/(100±x) , i.e. Total CP = 417 (500*100/200) + 568(500*100/88)≅ 985

Since CP

P% ≅ 15/985 X 100 ≅ 1.5 %

21. 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:

C.P. = Rs. 3000.

S.P. = Rs.3600 x 10 = Rs. 3000.

100 + (10 x 2)

Gain = 0%.

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23. 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 ?

Let 232 = x. Then, (232 + 1) = (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then,

(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 – x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

24. Question 17. How Many Of The Following Numbers Are Divisible By 132 ?264, 396, 462, 792, 968, 2178, 5184, 6336

132 = 4 x 3 x 11

So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.

264   11,3,4 (/)

396   11,3,4 (/)

462   11,3 (X)

792   11,3,4 (/)

968   11,4 (X)

2178   11,3 (X)

5184   3,4 (X)

6336   11,3,4 (/)

Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.

Required number of number = 4.

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26. 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 ?

Let the smaller number be x. Then larger number = (x + 1365).

x + 1365 = 6x + 15

5x = 1350

x = 270

Smaller number = 270

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28. Question 19. If The Number 517*324 Is Completely Divisible By 3, Then The Smallest Whole Number In The Place Of * Will Be:

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = (22 + x),

which must be divisible by 3.

x = 2.

29. Question 20. The Sum Of First 45 Natural Numbers Is:

Let Sn =(1 + 2 + 3 + … + 45). 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 46/2 = (45 x 23)/2

= 45 x (20 + 3)

= 45 x 20 + 45 x 3

= 900 + 135

= 1035.

Shortcut Method:

Sn = n(n + 1)/2=45(45 + 1)/2= 1035.

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