Infosys Aptitude Placement Papers - Infosys Aptitude Interview Questions and Answers updated on 27.Jul.2024

Q1. 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:

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.

Q2. A Man Invested Rs. 4940 In Rs. 10 Shares Quoted At Rs. 9.5

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

Q3. (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

Q4. 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

Q5. The Average Score Of Boys In An Examination In A School Is 71 And That Of The Girls Is 7

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

Q6. Some Students Planned A Picnic. The Budget For Food Was Rs. 5

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

Q7. 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 Traction ?

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

Since CP<SP .Therefore, Profit = 100-985 = 15

P% ≅ 15/985 X 100 ≅ 1.5 %

Q8. 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.

Q9. 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%.

Q10. 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.

Q11. 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%

Q12. 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.

Q13. 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%

Q14. 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

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

Q15. The Difference Of Two Numbers Is 136

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

x + 1365 = 6x + 15

5x = 1350

x = 270

Smaller number = 270

Q16. 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 Th

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.

Q17. 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

Q18. The Mean Monthly Salary Paid To 75 Workers In A Factory Is Rs. 5,68

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

Q19. 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.

Q20. 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.