Inautix Aptitude Placement Papers - Inautix Aptitude Interview Questions and Answers updated on 14.Jun.2024

Let age of the son before 1010 years =x=x and

age of the father before 1010 years =3x=3x

(3x+20)=2(x+20)

Age of the son at present =x+10=20+10=30

Age of the father at present =3x+10=3×20+10=70

Required ratio =70:30=7:3.

59 days = 8 weeks 3 days = 3 odd days

Hence if today is Thursday, After 59 days, it will be = (Thursday + 3 odd days)

= Sunday

1 Jan 1901 = (1900 years + 1st Jan 1901)

We know that number of odd days in 400 years = 0

Hence the number of odd days in 1600 years = 0 (Since 1600 is a perfect multiple of 400)

Number of odd days in the period 1601-1900

= Number of odd days in 300 years

= 5 x 3 = 15 = 1

(As we can reduce perfect multiples of 7 from odd days without affecting anything)

1st Jan 1901 = 1 odd day

Total number of odd days = (0 + 1 + 1) = 2

2 odd days = Tuesday

Hence 1 January 1901 is Tuesday.

Total weight of students in division A = 36 × 40

Total weight of students in division B = 44 × 35

Total students = 36 + 44 = 80

Average weight of the whole class

=(36×40)+(44×35)/80

=(9×40)+(11×35)/20

=(9×8)+(11×7)/4

=72+77/4

=149/4

=37.25.

Speed of second train =400/4=100 km/hr

Speed of first train : Speed of second train =7:8

Therefore, speed of first train =100/8×7=87.5 km/hr.

Part filled by first tap in 11 hour =1/3

Part emptied by second tap 11 hour =1/8

Net part filled by both these taps in 11 hour

=1/3-1/8=5/24

i.e, the cistern gets filled in 24/5 hours =4.8 hours.

Face value of each share = Rs.20

Market value of each share = Rs.25

Number of shares = 12500

Amount required to purchase the shares = 12500 × 25 = 312500

Mohan further sells the shares at a premium of Rs. 11 each

ie, Mohan further sells the shares at Rs.(20+11) = Rs.31 per share

total amount he gets by selling all the shares = 12500 × 31 = 387500

His gain = 387500 - 312500 = Rs.75000.

Work done by Kamal in 1 day = 1/20

Work done by Suresh in 1 day = (1/20) × (125/100) = 5/80 = 1/16

=> Suresh can complete the work in 16 days.

To obtain next number, subtract 3 from the previous number and divide the result by 2

445

(445-3)/2 = 221

(221-3)/2 = 109

(109-3)/2 = 53

(53-3)/2 = 25

(25-3)/2 = 11

(11-3)/2 = 4

Clearly, 53 should have come in place of 46.

While A scores 90 points, B scores (90-15)=75 points and C scores (90-30)= 60 points

i.e., when B scores 75 points, C scores 60 points

=> When B scores 100 points, C scores 60/75×100 = 80 points

i.e., in a game of 100 points, B can give C (100-80)=20 points.

Suppose B's investment =x.

Then A's investment =3x

Suppose B's period of investment =y

then A's period of investment =2y

A : B =3x×2y:xy=6:1

Total profit ×1/7=4000

=> Total profit =4000×7=28000.

While A covers 1000 m, B covers (1000-40)=960 m and C covers (1000-64)=936 m

i.e., when B covers 960 m, C covers 936 m

When B covers 1000 m, C covers 936/960×1000 = 975 m

i.e., B can give C a start of (1000-975) = 25 m.

Given that January 1, 2004 was Thursday.

Odd days in 2004 = 2 (because 2004 is a leap year)

(Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004, that is the whole year 2004)

Hence January 1, 2005 = (Thursday + 2 odd days) = Saturday.

Pipe A can fill 1/8 of the cistern in 1 hour.

Pipe B can empty 1/12 of the cistern in 1 hour

Both Pipe A and B together can effectively fill 1/8-1/12=1/24 of the cistern in 1 hour

i.e, the cistern will be full in 24 hrs.

Age of C << Age of A << Age of B

Given that sum of the ages of B and C is 5050 years.

Now we need to find out (B's age - A's age). But this cannot be determined with the given data.

Pipe A can fill 1/10 of the tank in 1 hr

Pipe B can fill 1/40 of the tank in 1 hr

Pipe A and B together can fill 1/10+1/40=1/8 of the tank in 1 hr

i.e., Pipe A and B together can fill the tank in 8 hours.

Change in area

=(20+20+20×20/100)%=44%

i.e., area is increased by 44%.

Let present age of the mother =5x

Then, present age of the person =2x

5x+8=2(2x+8)

5x+8=4x+16

X=8

present age of the mother =5x=40.

LCM of 252, 308 and 198 = 2772

Hence they all will be again at the starting point after 2772 seconds.

i.e., after 46 minutes 12 seconds.

The difference between two successive terms from the beginning are 7, 5, 7, 5, 7, 5

Hence, in place of 40, right number is 37+5=42.

9997 - 7654 = 2343

9997 - 8506 = 1491

8506 - 7654 = 852

Hence, the greatest number which divides 7654, 8506 and 9997 and leaves same remainder

= HCF of 2343, 1491, 852

= 213

Now we need to find out the common remainder.

Take any of the given numbers from 7654, 8506 and 9997, say 7654

7654 ÷ 213 = 35, remainder = 199.

Work done by P in 1 day = 1/24

Work done by Q in 1 day = 1/6

Work done by R in 1 day = 1/12

Work done by P,Q and R in 1 day = 1/24 + 1/6 + 1/12 = 7/24

=> Working together, they will complete the work in 24/7 days = 3 3/7 days.

Let length =3xkm,

Distance travelled by the man at the speed of 1212 km/hr in 88 minutes =2(3x+2x)=10x

Therefore,

12×8/60=10x

x=4/25 km=160 m

Area =3x×2x=6x2

=6×1602=153600 m2.

Given that fort had provision of food for 150 men for 45 days

Hence, after 10 days, the remaining food is sufficient for 150 men for 35 days

Remaining men after 10 days = 150 - 25 = 125

Assume that after 10 days,the remaining food is sufficient for 125 men for xx days

More men, Less days (Indirect Proportion)

150 : 125 :: x : 35

150×35=125x

6×35=5X

X=6×7=42.

log105 + log10 (5x+1) = log10 (x+5) + 1

=> log105 + log10(5x+1) = log10(x+5) + log10 10

=> log10[5(5x+1)] = log10[10(x+5)]

=> 5(5x+1) = 10(x+5)

=> 5x+1 = 2(x+5)

=> 5x + 1 = 2x + 10

=> 3x = 9

=> x = 3.

Age of Sobha's father when Sobha was born =38=38

Age of Sobha's mother when Sobha was born =36-4=32=36-4=32

Required difference of age =38-32=6.

Face value of each share = Rs.20

Market value of each share = Rs.25

Number of shares = 12500

Amount required to purchase the shares = 12500 × 25 = 312500.

F = Rs. 2160

TD = Rs. 360

PW = F - TD = 2160 - 360 = Rs. 1800

True Discount is the Simple Interest on the present value for unexpired time

=>Simple Interest on Rs. 1800 for unexpired time = Rs. 360

Banker's Discount is the Simple Interest on the face value of the bill for unexpired time

= Simple Interest on Rs. 2160 for unexpired time

=360/1800×2160=1/5×2160=Rs. 432.

The word 'DELHI' has 5 letters and all these letters are different.

Total number of words (with or without meaning) that can be formed using all these 5 letters using each letter exactly once

= Number of arrangements of 5 letters taken all at a time

= 5P<sub>5</sub> =5!=5×4×3×2×1=120.

Let the speed of the water in still water = x

Given that speed of the stream = 3 kmph

Speed downstream=(x+3) kmph

Speed upstream=(x-3) kmph

He travels a certain distance downstream in 4 hour and come back in 6 hour.

ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour

since distance = speed × time, we have

(x+3)4=(x-3)6

?(x+3)2=(x-3)3

?2x+6=3x-9

?x=6+9=15 kmph.

cost price =80000+5000+1000=86000

profit =25%

selling price =86000+86000×1/4=107500.

Let the required actual distance be xx km

More scale distance, More actual distance(direct proportion)

Hence we can write as

(scale distance) 0.6 : 80.5 :: 6.6 : xx

0.6x=80.5×6.6

0.1x=80.5×1.1

x=80.5×11=885.5.

Let John's initial salary = Rs.100

After decreasing by 50%, John's salary = Rs. 50 (because it will become half)

After subsequently increasing by 50%, John's salary

=50×100+50/100=50×150/100= Rs.75

Loss = 100-75=Rs.25

Loss percent =25/100×100=25%.

Let the numbers be 4k and 5k

HCF of 4 and 5 = 1

Hence HCF of 4k and 5k = k

Given that HCF of 4k and 5k = 6

=> k = 6

Hence the numbers are (4 × 6) and (5 × 6)

= 24 and 30

LCM of 24 and 30 = 120.

water filled by the inlet pipe in 24hours

= water emptied by the leak in 24-6=1824-6=18 hours.

Therefore, water emptied by the leak in 66 hours

= water filled by the inlet pipe in 88 hours

i.e., capacity of the tank

= water filled by the inlet pipe in 88 hours

=8×60×4=1920=8×60×4=1920 litre.

Let the average after 17 innings = x

Total runs scored in 17 innings = 17x

Average after 16 innings = (x-3)

Total runs scored in 16 innings = 16(x-3)

Total runs scored in 16 innings + 87 = Total runs scored in 17 innings

=> 16(x-3) + 87 = 17x

=> 16x - 48 + 87 = 17x

=> x = 39.

Tap A can fill the tank completely in 6 hours

=> In 1 hour, Tap A can fill 1/6 of the tank

Tap B can empty the tank completely in 12 hours

=> In 1 hour, Tap B can empty 1/12 of the tank

i.e., In one hour, Tank A and B together can effectively fill (1/6-1/12)=1/12 tank

=> In 4 hours, Tank A and B can effectively fill 1/12×4=1/3×4=1/3 of the tank.

Time taken to fill the remaining (1-1/3)=23(1-13)=2/3 of the tank =(2/3)(1/6)= 4 hours.