# Infosys Aptitude Placement Papers - Infosys Aptitude Interview Questions and Answers updated on 06.Dec.2023

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.

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

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

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

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

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

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 %

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.

C.P. = Rs. 3000.

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

100 + (10 x 2)

Gain = 0%.

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

which must be divisible by 3.

x = 2.

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.

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%

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

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

x + 1365 = 6x + 15

5x = 1350

x = 270

Smaller number = 270

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.

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

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

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.

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.