d Question 25 & 26 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Question 25 & 26 Cricket fans know there are 6 balls in one over. So, in 10 overs there will be 10×6=60 balls !

Run rate of 3.2 means a score of 3.2 runs in one over – so,   So, it means out of target of 282 runs , 32 runs have readily been scored in first 10 overs. Now match is of 50 overs, so question is 282-32=250 runs will be scored with what run rate in next 40 overs !

Or simply we can say that next 40 overs should give 250 runs i-e;   So, a required run rate is of 6.25 runs   in next 40 overs ! Here,

Rate of interest =R= 5% pa

Term of Calculation of interest = T = 1/2

(One year is taken as one term and half yearly calculations  will be 1/2 terms two times in one year)

Principal or initial amount of investment =P= 1600 Rs

We know the formula of Compound Interest calculation; Put values from given data;      Or (rounding off) Now we also know ;  Where,

A = final amount including interest

I = Interest amount

P= Principal amount

So, we have; So amount for first half term is Rs 40/-

Now for next half term of the same year A will become principal amount;

So data for next half term can be arranged as;

P = 1640

R = 5% pa

T = 1/2

Again applying;       Or ( rounding off) So, real  interest on Rs 1600/- till end of second half becomes Rs 81

So total Internet = interest in first half + new interest till end of  second half

Interest = 40+81=121