A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 meters between two consecutive saplings. Mithilesh, the country's greatest forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner, he covers a total distance of 1.32 kms. How many saplings does he plant in the process if he ends at the starting point?

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Solution

The correct option is A 11
To plant the $1^{s t}$ sapling, Mithilesh will cover 20 m; to plant the $2^{n d}$ sapling he will cover 40 m and so on.
Let n be number of saplings planted.
Then, $20 + 40 + 60 \dots + 20 \times n = 1320$
$20 \times (1 + 2 + 3 \dots + n) = 1320$
We get, n = 11

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The correct option is A 11To plant the 1st sapling, Mithilesh will cover 20 m; to plant the 2nd sapling he will cover 40 m and so on. Let n be number of saplings planted. Then, 20+40+60⋯+20×n=1320 20×(1+2+3⋯+n)=1320 We get, n = 11

The correct option is A 11
To plant the $1^{s t}$ sapling, Mithilesh will cover 20 m; to plant the $2^{n d}$ sapling he will cover 40 m and so on.
Let n be number of saplings planted.
Then, $20 + 40 + 60 \dots + 20 \times n = 1320$
$20 \times (1 + 2 + 3 \dots + n) = 1320$
We get, n = 11