A monkey standing on the ground wants to climb to the top of a vertical pole $13$ m tall. He climbs $5$ m in $1$ s and then slips downwards $3$ m in the next second. He again climbs $5$ m in $1$ s and slips by $3$ m in the next second as so on. How much time will the monkey take to reach the top of the pole?

7 s e c

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9 s e c

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11 s e c

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13 s e c

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Solution

The correct option is B $9 s e c$
The monkey takes 1 second to go up $5 m$ and then, he comes down by $3 m$ in next one second.
So, every $2 s e c$ he moves up by $2 m$ .
To move up $8 m$ , he will take $8 s e c$ .
From $8 m$ point, he will take one second to move another $5 m$ ,
i.e., height of $8 + 5 =$ $13 m$ . So, he has reached the top.
Hence, he reaches the top in $8 s + 1 s = 9 s$ .

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A monkey standing on the ground wants to climb to the top of a vertical pole 13 m tall. He climbs 5 m in 1 s and then slips downwards 3 m in the next second. He again climbs 5 m in 1 s and slips by 3 m in the next second as so on. How much time will the monkey take to reach the top of the pole?