The length of a rope, by which a cow is tied...

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The length of a rope, by which a cow is tied...

by AAPL » Sat Jan 13, 2018 1:16 pm
The length of a rope, by which a cow is tied, is decreased from 22m to 14m. What is the decrease in the area that can be grazed by the cow?

$$A.\ 288\pi$$
$$B.\ 144\pi$$
$$C.\ 200\pi$$
$$D.\ 120\pi$$
$$E.\ 72\pi$$

The OA is A.

I don't have clear this PS question,

I know that the grazed's area will be the form of a circle, right? Then I need to determine the 2 areas, first one with a radius of 22m and the second with a radius of 14m. Finally I need to substract first area minus second area and get the result.

I appreciate if any expert explain it for me. Thank you so much.

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by DavidG@VeritasPrep » Sun Jan 14, 2018 12:09 pm
AAPL wrote:The length of a rope, by which a cow is tied, is decreased from 22m to 14m. What is the decrease in the area that can be grazed by the cow?

$$A.\ 288\pi$$
$$B.\ 144\pi$$
$$C.\ 200\pi$$
$$D.\ 120\pi$$
$$E.\ 72\pi$$

The OA is A.

I don't have clear this PS question,

I know that the grazed's area will be the form of a circle, right? Then I need to determine the 2 areas, first one with a radius of 22m and the second with a radius of 14m. Finally I need to substract first area minus second area and get the result.

I appreciate if any expert explain it for me. Thank you so much.
Yep. So you could actually calculate the values.
22^2 * Pi = 484 * Pi
14^2 * Pi = 196 * Pi
484* Pi - 196* Pi = 288* Pi

But it's enough to see that 22^2 will have a units digit of 4 and 14^2 will have a units digit of 6. Thus, we can see that difference must have a units digit of 8. Only A works.
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by DrMaths » Mon Jan 15, 2018 4:42 am
The area will be proportional (by pi) to 22^2 - 14^2 = (2 x 11)^2 - (2 x 7)^2 = 4 (11^2 - 7^2) = 4(121 - 49) = 4 x 72 = 288.
So answer = 288 x pi

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by EconomistGMATTutor » Tue Jan 16, 2018 11:33 am
The length of a rope, by which a cow is tied, is decreased from 22m to 14m. What is the decrease in the area that can be grazed by the cow?

$$A.\ 288\pi$$
$$B.\ 144\pi$$
$$C.\ 200\pi$$
$$D.\ 120\pi$$
$$E.\ 72\pi$$

The OA is A.

I don't have clear this PS question,

I know that the grazed's area will be the form of a circle, right? Then I need to determine the 2 areas, first one with a radius of 22m and the second with a radius of 14m. Finally I need to substract first area minus second area and get the result.

I appreciate if any expert explain it for me. Thank you so much.
Hi AAPL,
Let's take a look at your question.

The rope was originally 22m long, therefore we will find the grazed area using 22m. The cow can graze in a circular area and the rope will act as the radius of the circle, therefore,
$$Area_1=\pi r_1^2$$
$$Area_1=\pi\left(22\right)^2$$
$$Area_1=484\pi$$

Now, find the grazed area for the rope length 14m.
$$Area_2=\pi r_2^2$$
$$Area_2=\pi\left(14\right)^2$$
$$Area_2=196\pi$$

The decrease in area can be calculated as:
$$=Area_1-Area_2$$
$$=484\pi-196\pi=288\pi$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.
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by Scott@TargetTestPrep » Thu Jan 18, 2018 1:38 pm
AAPL wrote:The length of a rope, by which a cow is tied, is decreased from 22m to 14m. What is the decrease in the area that can be grazed by the cow?

$$A.\ 288\pi$$
$$B.\ 144\pi$$
$$C.\ 200\pi$$
$$D.\ 120\pi$$
$$E.\ 72\pi$$
When a cow is tied by a rope, its grazing area is a circle with the length of the rope as the radius of the circle. Thus, the initial radius is 22m, and the initial area is π x 22^2 = 484π.

Once the rope is decreased to 14m, the new area is π x 14^2 = 196π.

Thus, the decrease in area is 484Ï€ - 196Ï€ = 288Ï€.

Answer: A

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