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Re: The length of a rope, to which a cow is tied, is increased from 19 m [#permalink]
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]
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 13868 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]
Expert Reply
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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Re: The length of a rope, to which a cow is tied, is increased from 19 m [#permalink]
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