Quantitative Aptitude - Arithmetic Ability Questions

Let the age of the daughter today be ‘x’ years.
Mother’s age today = 3x years
After 10 years, we have
3x + 10 = 2(x + 10)
x = 10 years.

The fruit content in both the fresh fruit and dry fruit is the same.

Let 'p' kgs of dry grapes are obtained from the 60 kgs of fresh grapes

=> $$\begin{gathered} 60 x 40100 = p x 60100 \\ => p = 40 kgs. \end{gathered}$$

Therefore, 40 kgs of Dry grapes are obtained from 60 kgs of Fresh grapes.

Ratio of initial investments = 1/2 : 1/3 : 1/4 = 6 : 4 : 3.

Let their initial investments be 6x, 2x and 3x respectively.

A : B : C = (6x * 2 + 3x * 10) : (4x * 12) : (3x * 12) = 42 : 48 : 36 = 7 : 8 : 6.

B's share = Rs. (378*8/21)  = Rs. 144.

4*A + 7/B + 2*C + 5/D + 6*E =47.2506
=> 4 * A + 7/B + 2 *C + 5/D + 6*E = 40 + 7 + 0.2 + 0.05 + 0.0006
Comparing the terms on both sides, we get :
4*A =40, 7/B = 7, 2*C = 0.2, 5/D = 0.05, 6*E = 0.0006
A = 10,
B = 1,
C = 0.1,
D = 100,
E = 0.0001.
5*A + 3*B + 6*C + D + 3*E =(5*10) + (3*1) + (6*0.1) +100 +(3*0.0001)
=50 + 3 + 0.6 + 100 + 0.003 = 153.6003.

Let 1 man's 1 day work = x and 1 woman's 1 day work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10
Solving these two equations, we get:
x = 11/400 and y = 1/400
10 woman's 1 day work = (1/400 x 10) = 1/40.

Hence, 10 women will complete the work in 40 days.

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

=12C1*1C152C2+13C1*3C152C2

=12*252*51+13*3*252*51= 24+7852*51 = 126