Question

Competitors in a beanbag race run on a straight track. Beanbags are placed at various distances from the starting line: $20$ m, $24$ m, $28$ m and so on. The final beanbag is placed $40$ m from the starting line.Start $0, 20, 24, 28$ meters, $40 .$ Competitors run to the first beanbag, pick it up and then return it to the starting line. They repeat this until all the beanbags have been returned to the starting line.

$a .$ Find the number of beanbags that each competitor must collect.

$b .$ Find the total distance that each competitor runs.

Accepted Answer

Starting point $= 0$ m

Distance of first beanbag from the starting point $= 20$ m

Distance of second beanbag from the starting point $= 24$ m

Distance of third beanbag from the starting point $= 28$ m

Distance of last beanbag from the starting point $= 40$ m

Thus the given series is in A.P.

Let the total number of beanbags $= n$

$∴ a .$ Number of beanbags is given as

$40 = 20 + (n - 1) 4$

$\Rightarrow 10 = 5 + n - 1$

$\Rightarrow 10 = n + 4$

$\Rightarrow n = 6$

Thus, number of beanbags that each competitor must collect $= 6$

$b .$ If $d$ is the distance of a beanbag from the starting point

$∴$ The distance covered by the competitor in bringing back the beanbag to the starting point $= 2 b$

$∴$ Distance covered by each competitor in bringing back all the beanbags back to the starting point is given as

$d = 2 \times \frac{6}{2} \times (2 \times 20 + (6 - 1) 4)$

$= 6 \times (40 + 5 \times 4)$

$= 6 \times (40 + 20)$

$= 6 \times 60$

$\Rightarrow d = 360$ m

$1$	$3$
$+ 2$	$+ 4$
$+ 7$	$+ 5$
$+ 9$	$+ 8$
Total $= 19$	Total $= 20$

Important Questions on Arithmetic Progressions

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