Quantitative Aptitude - Arithmetic Ability Questions

The year 1985 is an ordinary year. So, it has 1 odd day.

So, the day on 19th June, 1985 will be 1 day after the day on 19th June, 1984.

But, 19th June,1984 is Monday

S0, 19th June, 1985 is Tuesday.

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

216004989600=0.043

Sony and her brother's age sum is 31
Let Sony's mother's age be 'x'.
Now the conditions given for both of their ages related to their mom is
1/5(x-15) + 3/5(x-10) = 31
x-15 + 3x-30 = 155
4x-45 = 155
4x = 200
x = 50
So, Mother's age = 50.

Cross verification : 15 years ago age = 35 => Sony age = 7
10 years ago age = 40 => bro age = 24

Sum = 24 + 7 = 31.

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14=4/7.

Given numbers are 1, 2 and 3

Factors of 1 = 1

Factors of 2 = 1, 2

Factors of 3 = 1, 2 and 3

Common factors = 1

Hence, the HCF of 1, 2 and 3 = 1.

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25 => x = 5
Hence, B's age = 2x = 10 years.

4 x 27 x 3125 = 22 × 33 × 55 ;

8 x 9 x 25 x 7 = 23 × 32 × 52 × 7

16 x 81 x 5 x 11 x 49 = 24 × 34 × 5 × 11 × 72

H.C.F = 22 × 32 × 5 = 180.