Quantitative Aptitude - Arithmetic Ability Questions

100C10 + 90C20 + 70C30 + 40C40 = 100!10!x20!x30!x40!

Converting each of the given fractions into decimal form, we get

 5/9 = 0.55, 7/11 = 0.63,  8/15 = 0.533, 11/17 = 0.647

Clearly, 0.647>0.63>0.55>0.533

so , 11/17 > 7/11 > 5/9 > 8/15

After 10 days : 150 men had food for 35 days.

Suppose 125 men had food for x days.

Now, Less men, More days (Indirect Proportion)

∴ 125 : 150 :: 35 : x    ⇔   125 x x = 150 x 35

 ⇒x=150×35125=42

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

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.

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.

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.

 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