# Bill Gate has a poultry farm. A hen lay some odd number of eggs in a straight line, placed at an interral of 10 m. These eggs have to be assembled aro

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Bill Gate has a poultry farm. A hen lay some odd number of eggs in a straight line, placed at an interral of 10 m. These eggs have to be assembled around the middle egg. Bill can carry only one egg at a time. He carried the job with one of the end eggs by carrying them in succession. In carrying all the eggs he covered a distance of 3km, find the number of eggs a hen laid.

1. 29 eggs
2. 23 eggs
3. 25 eggs
4. 27 eggs

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Correct Answer - Option 3 : 25 eggs

Given

A Hen lay some odd no. of eggs in a straight line at an interval of 10 m.

These eggs have to be assembled around the middle egg.

Bill can carry one egg at a time.

Formula Used

Sn = a + (n - 1)d.

Calculation

Let there be (2n + 1) eggs. Clearly one egg lies in the middle and n eggs on each side of it in a row. Let P be the mid- egg and let A and B be the end eggs on the the left and right of P respectively. Clearly, there are n intervals each of length 10m on both the sides of P. Now, suppose starts from A. He picks up the end egg on the left of mid-egg and goes to the mid-egg, drops it and goes to (n - 1)th egg on left, picks it up, goes to the mid egg and drops it. This process is repeated till he collects all eggs on the left of the the mid egg at the mid egg. So, distance covered in collecting eggs on the left of the mid eggs.

= 10 × n + 2[10 × (n - 1) + 10 × (n - 2) +………+ 10 × 2 + 10 × 1]

After collecting all eggs on left of the mid egg Bill goes to the egg B on the right side of the mid egg, picks it up, goes to the mid egg and drops it. Then he goes to (n-1) egg on the right and the process is repeated till he collects all eggs at the mid-egg.

Distance covered in collecting the eggs on the right side of the mid egg.

= 2[10 × n + 10 × (n - 1) + 10 × (n - 2) +……+ 10 × 2 + 10 × 1]

Total distance covered

= 10 × n + 2[10 × (n - 1) + 10 × (n - 2) + .....+ 10 × 2 + 10 × 1] + 2[10 × n + 10 × (n - 1) +........+ 10 × 2 + 10 × 1]

⇒ 4[10 × n + 10 × (n - 1) +........+ 10 × 2 + 10 × 1] - 10 × n

⇒ 40[1 + 2 + 3 +........+ n] - 10n = 40 {n/2(1 + n)} - 10n = 20n2 + 10n.

But, the total distance covered is 3 km = 3000m.

20n2 + 10n = 3000

⇒ 2n2 + n - 300 = 0

⇒ (n - 12) (2n + 25) = 0

⇒ n = 12

∴ The number of eggs = 2n + 1 = 25.