Quantitative Aptitude - Arithmetic Ability Questions

The total marks of all 5 subjects = 82 x 5 = 410

Total marks of 1st two subjects = 86.5 x 2 = 173

Total marks of last two subjects = 84 x 2 = 168

Now, marks in 3rd subject = 410 - (173 + 168) = 410 - 341 = 69.

x = 30% of z = 30z/100 = 3z/10

y = 40% of z = 40z/100 = 4z/10

According to the question,(x/y)*100 = p%

p% = 3z/100/4z/100×100= 75%

Given total number of students = 80

Total teachers = 5

Number of coupons each student got = 15 x 80/100 = 12

Number of coupons each teacher got = 25 x 80/100 = 20

Therefore, total number of shopping coupons = 80 x 12 + 5 x 20 = 960 + 100 = 1060.

The given series is 1000, 1012, 988, 1024, 978

1000

1000 + 12 = 1012

1012 - 24 = 988

988 + 36 = 1024

1024 - 48 = 976 (Not 978)

Let the required two numbers be a, b

From the given data,

2a + 3b = 36 .....(1)

3a + 2b = 39 .....(2)

Now, by solving 1 and 2 eqnts, we get

5a = 45

a = 9 ....(3)

Put (3) in (1), we get

2(9) + 3b = 36

=> 18 + 3b = 36

3b = 36 - 18

=> b = 18/3 = 6

Required, difference between the numbers = a + b = 9 + 6 = 15.

We know that,

I = PTR/100

ATQ,

15800 = 14800 x T x 6/100

T = 15800/148x7

T = 15800/1036

T = 15.25 yrs.

Total number of elementary events = 20C2 ways =190

There are 7 white balls out of which one white can be drawn in 7C1 ways and one ball from remaining 13 balls can be drawn in 13C1 ways.

So,favourable events = 7C1*13C1

Therfore,required probability = (7C1*13C1)/20C2=91/90