Quantitative Aptitude - Arithmetic Ability Questions

   Sum = [(B.D.xT.D.)/ (B.D.-T.D.)]

              = [(B.D.xT.D.)/B.G.]

T.D./B.G.  = Sum/ B.D.

   =1650/165

               =10

Thus, if B.G. is Re 1, T.D. = Rs. 10.

If B.D.is Rs. ll, T.D.=Rs. 10.

If B.D. is Rs. 165, T.D. = Rs. [(10/11)xl65]

            =Rs.150

 And, B.G. = Rs. (165 - 150) = Rs, 15.

19th April, 1986 = (1985 years + Period from 1.1.1986 to 19.4.1986)

Odd days in 1600 years = 0

Odd days in 300 years = 1

85 years has 21 leap years + 64 ordinary years.

Number of odd days in 85 years =( 21 x 2 + 64) = 1 odd day.

Jan. Feb. March April

(31 + 28 + 31 + 19) = 109 days

109 days = 15 weeks + 4 days= 4 odd day.

Total number of odd days = (1 + 0 + 1 + 4) = 6.

Given day is Saturday.

kl>k+1l+1 for (k,l) > 0 and  k > l 

Let     k = x+1    and   l = x

Therefore, x+1x>(x+1)+1(x)+1

 (x + 1) > x

Therefore, log(x+1)log(x)>log(x+2)log(x+1)

⇒logxx+1 >logx+1x+2

Area =    = 96

Sum of Area =  

Let the speed of the faster train be 'S' kmph

Then speed of the slower train will be '(S-5)' kmph

Time taken by faster train = 350/S hrs

Time taken by slower train = 350/(S-5) hrs

 350s-5 - 350s = 2 hrs 20 min = 213 = 73

=> S = 30 km/hr.

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.
Let k be the quantity of third variety of oil in the final mixture.
Let the ratio of initial quantity of oils be 'y'
From given details,
4x + 5 = 5y ..... (1)
5x + 10 = 7y .....(2)
8x + k = 9y ......(3)

By solving (1) & (2), we get
x = 5 & y = 5

From (3) => k = 5

Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red = 7C1 + 3C115C1 = 7 + 315 = 1015 = 23