Quantitative Aptitude - Arithmetic Ability Questions

80x900/100 + 256 x 4 x ? = 1331 - 99

720 + 1024 x ? = 1232

? = 1232 - 720/1024

? = 512/1024 = 0.5

? = 0.5 

Choose 2 juniors and 2 seniors.

14C2*23C2 = 23023

Let present Anirudh's age be 'A' and Bhavana's age be 'B'

Given seven years, their ratio is 7:9

=> A-7 : B-7 = 7 : 9

=> 9A - 7B = 14 .......(1)

And given,

Chandhu is 12 years older than A nad 12 years younger than B

=> C = A + 12....(2)

=> B = C + 12

=> B = A + 12 + 12 .....(From (2))

=> B = A + 24 ....(3)

Put (3) in (1)

9A - 7(A+24) = 14

9A - 7A - 168 = 14

2A = 182

=> A = 91 years

=> C = A + 12 = 91 +12 = 103 years.

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

Using pythagarous theorem,
distance travelled by first train = 36x5/18x30 = 300m
distance travelled by second train = 48x5/18x30 = 400m
so distance between them =√( 90000 + 160000) = √250000 = 500mts.

In these type of fractions, the decimal moves left the number of zeroes in the divisor times. Herein the given question 18/1000, there are 3 zeros in 1000 (divisor). Hence, the quotient is 0.018.

The quotient of 18/1000 = 0.018.

Let Shilpa's present age is p years

According to question,

p = 4(x+4) – 4(x–4)

p = 4x+16 – 4x+16

p = 32

Shilpa's present age = 32 years

Rajeev is four years older than Shilpa

=> Shilpa's present age = Rajeev's age 4 years before.

=> Rajeev's age 4 years before = 32 years.

Let the daughter age be 'x'
Therefore, Son age = 5x, Mother age = 25x, Person age = 50x.

Thus we get, x+5x+25x+50x = 81
81x = 81
x = 1

Daughter's age = 1, Son's = 5, Mother's = 25, Person's age = 50.