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# A college clock tower (perpendicular to the ground) casts a shadow tha

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A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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20 Nov 2019, 23:16
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66% (01:30) correct 34% (01:20) wrong based on 44 sessions

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A college clock tower (perpendicular to the ground) casts a shadow that is 120 meters long. What is the distance from the top of the tower to the farthest tip of its shadow?

(1) The height of the tower is 1/3 greater than the length of its shadow.

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.

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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 01:05
1

tower and shadow forms the 2 sides of a right triangle. tower tip to shadow farthest point becomes hypotenuse.

statement 1 gives the lenght of other side (tower). Now we have 2 sides of right triangle. so we can find hypotenuse. Statement 1 is sufficient.

Statement 2 gives nothing but a similar right triangle. with tower height (h/2) and shadow length (60). This doesn'nt provide any additional info to find the hypotenuse. Statement 2 not sufficient.
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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 01:25
1
#1
The height of the tower is 1/3 greater than the length of its shadow.
h=160
so hypo = 200
30:40:60 ∆
sufficient
#2
If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.
we have
(x/2)^2+3600=a^2
two unkowns x & a
insufficient
IMO A

A college clock tower (perpendicular to the ground) casts a shadow that is 120 meters long. What is the distance from the top of the tower to the farthest tip of its shadow?

(1) The height of the tower is 1/3 greater than the length of its shadow.

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.
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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 01:31
1
(1) The height of the tower is 1/3 greater than the length of its shadow.............from this we will get the second side.....from which we can findthe third side.........SUFFICIENT

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long............This repeats the given information.............so INSUFFICIENT

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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 01:53
1
A college clock tower (perpendicular to the ground) casts a shadow that is 120 meters long. What is the distance from the top of the tower to the farthest tip of its shadow?

Let Tower height = x
Basically, we have to find the hypotenuse of the right angled triangle formed by the tower and its shadow. Since only shadow length is given, it is to be seen whether it has any form of relation with either tower height or hypotenuse.

(1) The height of the tower is 1/3 greater than the length of its shadow.
$$x = 120 + \frac{1}{3} * 120$$
Since, x can be found hypotenuse can be found.

SUFFICIENT.

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.
Let half of tower height = y
No relations among them is available. The statement simply sales down the original and gives nothing new to help find the answer. Hence

INSUFFICIENT.

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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 02:49
1
We are given the length of the shadow cast by the tower as 120m. We are to determine the distance,x, from the top of the tower to the end of the shadow. In other words, x is the hypotenuse of the right-angled triangle formed by the tower and the end of its shadow.

Based on the information given, if we know the angle of elevation,Z, from the end of the shadow cast by the tower to tip of the tower, we can easily determine x, as shown below:
cosZ=120/x, hence x=120/cosZ .........................(1)

Likewise, we can determine x if we know the height of the tower, h. With h known, x can be determined as follows:
x=(h^2 + 120^2)^0.5 .....................................(2)

1. The height of the tower is 1/3 greater than the length of its shadow.
Statement 1 sufficient since we can determine the height of the tower from the information above and substitute it into equation (2) in order to determine x, the distance from the furthest tip of the shadow to the top of the tower.
h=120+1/3 * 120 = 160m

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.
This is not sufficient because the given information does not provide any additional information that can help us to determine the height of the tower nor the angle of elevation to the top of tower from the furthest tip of the shadow.

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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 03:36
1
Quote:
A college clock tower (perpendicular to the ground) casts a shadow that is 120 meters long. What is the distance from the top of the tower to the farthest tip of its shadow?

(1) The height of the tower is 1/3 greater than the length of its shadow.

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long.

(1) The height of the tower is 1/3 greater than the length of its shadow. sufic

$$x^2=h^2+(120)^2…h=4/3(120)…x^2=(4*120/3)^2+120^2$$

(2) If the tower were to sink into the ground so that only half of its original length remained above ground, the tower would cast a shadow 60 meters long. insufic

Ans (A)
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Re: A college clock tower (perpendicular to the ground) casts a shadow tha  [#permalink]

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21 Nov 2019, 15:08
1
One of the legs in right—angled triangle is 120 meters
—>We need to find the hypotenuse of the triangle???

(Statement1):
The height = (1/3)*120+ 120 = 160

—> the hypotenuse^{2} 120^{2}+160^{2}
Hypotenuse= 200
Sufficient

(Statement2): we got two similar right-angled triangles.
—> but there is info about just one leg of both triangles. We need additional info to get this question done.
Insufficient

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Re: A college clock tower (perpendicular to the ground) casts a shadow tha   [#permalink] 21 Nov 2019, 15:08
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