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Math Expert V
Joined: 02 Sep 2009
Posts: 59622
The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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Difficulty:   15% (low)

Question Stats: 84% (02:02) correct 16% (02:26) wrong based on 45 sessions

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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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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

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Joined: 16 Apr 2015
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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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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
Intern  Joined: 17 Apr 2016
Posts: 19
Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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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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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 [ 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 _________________
Thanks and Regards

Abhishek....

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Math Expert V
Joined: 02 Aug 2009
Posts: 8291
Re: The length of a rope, to which a cow is tied, is increased from 19 m  [#permalink]

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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

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] 23 Nov 2019, 12:26
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