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The length of a rope, to which a cow is tied, is increased from 19 m

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The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 14 Mar 2016, 08:58
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The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.

A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m

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Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 14 Mar 2016, 11:28
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Length of rope will act as the radius of circular region which the cow can graze
Additional ground which the cow will be able to graze =\(\pi\) * ( 30^2 - 19^2)
= 22/7 (900 - 361)
=22/7 * 539
=22 * 77
= 1694 sq m

Answer B
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The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 14 Mar 2016, 11:49
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Bunuel wrote:
The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.

A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m


Since cow can only eat around the area in circular motion, considering it is tied in field, the extra area will be of shape of donut which is available due to extra length.
\(\frac{22}{7} * (30^2 - 19^2)\) => \(\frac{22}{7}*(49)(11)\) => 1694
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Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 17 May 2016, 04:37
Hi, Guyes I have found this problem from a website but I couldn't find the explanation of this problem. Please help me if anybody is there.
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Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 17 May 2016, 09:27
Bunuel wrote:
The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.

A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m


Attachment:
Circle.PNG
Circle.PNG [ 13.42 KiB | Viewed 4694 times ]


Area of the shaded region is

\(\frac{22}{7}(R^2 - r^2)\) R = 30 and r = 19

So, \(\frac{22}{7}(30^2 -19^2)\)

Or, \(\frac{22}{7}(30 - 19)(30 + 19)\)

Or, \(\frac{22}{7}*11*49\) = 1694

Hence IMHO (B) 1694 :-D
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Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 17 May 2016, 09:46
Bunuel wrote:
The length of a rope, to which a cow is tied, is increased from 19 m to 30 m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. Use \(\pi = \frac{22}{7}\) in your calculations.

A. 1594 sq m
B. 1694 sq m
C. 1696 sq m
D. 1756 sq.m
E. 1896 sq.m


Hi,

All though the exact method has been given above, ONE easy way to exploit the number properties..

The area will be a multiple of 11 and there is no fraction to cancel out this 11..
so our answer should be a MULTIPLE of 11...

ONLY B is a multiple of 11...

Property of a number to be div by 11:-
the difference in the SUM of alternate digits, that is, SUM of ODD digits- SUM of EVEN digits should be a multiple of 11 including 0...

1694... 4+6-(1+9)=0..
B
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Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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New post 23 Nov 2019, 12:26
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Re: The length of a rope, to which a cow is tied, is increased from 19 m   [#permalink] 23 Nov 2019, 12:26
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