Quantitative Aptitude - Arithmetic Ability Questions

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

By trial and error method, when we substitute 

x = 0

Both LHS and RHS are equal.

The weight will be 250g plus (1.55 - 0.65)/0.10 units of 100g

250 + 900 = 1150

This is the maximum weight that can be sent at that price. But, weights exceeding

250 + 800 will also get charged this amount (that is what the ‘part thereof’ implies).

Hence a package weighing 1145 will be charged $1145

P ( neither A nor B) = PA and B  

=  PA∩B=  PA∪B= 1-PA∪B  

= 1-61100=39100

250 numbers between 101 and 350 i.e. n(S)=250

n(E)=100th digits of 2 = 299−199 = 100

P(E)= n(E)/n(S) = 100/250 = 0.40

Let the present ages of Arun and Harish be 4x and 3x years respectively.

Then, 4x + 6 = 26 => x = 5

Harish's age = 3x = 15 years.

Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.