d Word Problems 1 & 2 (Average innings and scores, Average temperature) | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Word Problems 1 & 2 (Average innings and scores, Average temperature)

Problem 1: The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. Find his highest and lowest score.

Solution:

 

Let highest score be ‘y’ and lowest score be ‘z’ runs, Then by given condition highest score is more than lowest score by 172 runs i-e their difference is 172;

\Rightarrow y-z=172----(1)

Now total score for 40 innings can be calculated by multiplying 40 with given average score;

\Rightarrow Score for 40 Innings= 40 \times 50=2000 Runs

And score for innings after excluding highest and lowest score innings can also be calculated by multiplying remaining 38 innings with given average scores for these innings;

\Rightarrow Score for 38 Innings= 38 \times 48=1824 Runs

But if seen logically , total score of 40 innings should be equal to scores of 38 + 2 innings;

i-e Score for 40 Innings=Score Of 38 Innings + Score Of 2 Excluded Innings

\Rightarrow Score Of 2 Excluded Innings=Score for 40 Innings-Score Of 38 Innings

\Rightarrow Score Of 2 Excluded Innings=2000-1824=176

Or

\Rightarrow y+z=176----(2)

Adding eq (1) and eq(2);

\Rightarrow y-z=172----(1)

\Rightarrow y+z=176----(2)

__________________________________

\Rightarrow 2y=348

\Rightarrow y=174

Put this value of y in eq (1) or (2), we get;

\Rightarrow z=2

So, the highest score Inning was 174 runs and lowest score inning was 2 runs



(math test for jobs)

 

Problem 2: The average temperature for 1st, 2nd and 3rd June was 40 degrees. The average temperature for 2nd, 3rd and 4th June was 41 degree, That for 4th June 42 degree. What was the temperature on 1st June?

a) 30 deg             b) 39 deg              c) 40 deg                  d) 35 deg

Solution:

Let temperature on 1st, 2nd and 3rd June be Jt1, Jt2 and Jt3 respectively;

So average temperature

\frac{Jt1+Jt2+Jt3}{3}= 40 (Given)

\Rightarrow Jt1+Jt2+Jt3= 3\times40

\Rightarrow Jt1+Jt2+Jt3= 120 -----(1)

Now let temperature on 4th June be Jt4, then average temperature of 2nd, 3rd and 4th June is;

\Rightarrow \frac{Jt2+Jt3+Jt4}{3}=41(Given)

\Rightarrow Jt2+Jt3+Jt4=3 \times41

But temperature on 4th June = Jt4=42C (given) — put in above;

\Rightarrow Jt2+Jt3+42=123

\Rightarrow Jt2+Jt3=123-42

\Rightarrow Jt2+Jt3=81

put this value in eq (1) above;

\Rightarrow Jt1+81=120

\Rightarrow Jt1=120-81

\Rightarrow Jt1=39

So, temperature on 1st June was 39C!

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