The batting average for 40 innings of a crick | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Total score of the player for $40$ innings $= 50 	imes 40 = 2000$ runs.Let the highest score be $H$ and the lowest score be $L$. Given,$H - L = 1

Vedclass · Accepted Answer

$174$

Vedclass · Answer

(D) Total score of the player for $40$ innings $= 50 	imes 40 = 2000$ runs.Let the highest score be $H$ and the lowest score be $L$. Given,$H - L = 172$.Total score of the remaining $38$ innings $= 38 	imes 48 = 1824$ runs.The sum of the highest and lowest scores is $H + L = 2000 - 1824 = 176$.Now,we have a system of two linear equations:$1) H + L = 176$$2) H - L = 172$Adding these two equations:$(H + L) + (H - L) = 176 + 172$$2H = 348$$H = 174$.Therefore,the highest score of the player is $174$.

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. The highest score of the player is

Similar Questions

$A$ cricketer,whose bowling average was $12.4$ runs/wicket,takes $5$ wickets for $22$ runs in a match,thereby decreasing his average by $0.4$. The number of wickets taken by him before this match was:

If the mean of $a, b, c$ is $M$ and $ab + bc + ca = 0$,then the mean of $a^2, b^2, c^2$ is

If the mean of $5$ observations $x, x+2, x+4, x+6$ and $x+8$ is $11,$ then the mean of the last three observations is

The average of five consecutive positive integers is $n$. If the next two integers are also included,the average of all these integers will

The mean of $50$ observations was $36$. It was found later that an observation $48$ was wrongly taken as $23$. The corrected new mean is

Vedclass Test Series

Exam Paper Generator

Online Exam Module