Quantitative Aptitude - Arithmetic Ability Questions

Apply Allegation Method and first calculate the ratio in which they have to be mixed.

= 8 : 12 = 2 : 3

Thus, the two varieties of oil should be mixed in the ratio 2 : 3. So, if 240 liters of the second variety are taken, then 160 liters of the first variety should be taken.

length of the diagonal = 52 x (15/60) =13m 

sum of length and breadth = 68 x (15/60) = 17m

l2+b2=13 or l+b =17

area =lb = 1/2[(2lb)] = 1/2[(l+b)² - (l²+b²)] = 1/2[17² -169]

=1/2 x 120 = 60 sq.m

Let A's age 10 years ago = x years.
Then, B's age 10 years ago = 2x years.
(x + 10) / (2x+ l0) = 3/4
=> x = 5.
So, the total of their present ages =(x + 10 + 2x + 10)
= (3x + 20) = 35 years.

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

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.

Prime factors of 

252 = 2 x 2 x 3 x 3 x 7; 

308 = 2 x 2 x 7 x 11; 

198 = 2 x 3 x 3 x 11

Therefore LCM of 252, 308 and 198 is 2 x 2 x 3 x 3 x 7 x 11 = 4 x 9 x 77 = 2772.

2172+1832

= 200+172+200-172

It is in the form of a+b2+a-b2=2a2+b2

= 2(40000 + 289) = 80578.