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Of the following, which best approximates:

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V
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Of the following, which best approximates:  [#permalink]

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New post 11 Nov 2018, 09:07
3
9
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

52% (02:32) correct 48% (02:18) wrong based on 301 sessions

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Of the following, which best approximates:

\(\frac{(0.1667)(0.8333)(0.3333)}{(0.2222)(0.6667)(0.1250)}\) ?

(A) 2.00
(8) 2.40
(C) 2.43
(D) 2.50
(E) 3.43

Project PS Butler : Question #11


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Re: Of the following, which best approximates:  [#permalink]

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New post 11 Nov 2018, 09:12
4
4
0.16667 = 1/6
0.3333 = 1/3
0.2222 = 2/9
0.6667 = 2/3
0.125 = 1/8
0.8333 = 5/6

Substituting this,

9*3*8*5/6*3*6*2*2 = 5/2 = 2.5

D is the answer.
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Of the following, which best approximates:  [#permalink]

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New post 11 Nov 2018, 09:15
2
3
We can see that,

0.16667 = 1/6 , 0.3333 = 1/3 , 0.2222 = 2/9, 0.6667 = 2/3 , 0.125 = 1/8 , 0.8333 = 5/6

Side note: If one does not remember, quick way to find these out would be as follows: Let x = 0.22222, 10*x = 2.22222, so 10*x - x = 2
Hence 9x = 2, or x = 2/9

If we substitute back we get 2.5 as the answer.

Hope that helps.
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Re: Of the following, which best approximates:  [#permalink]

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New post 11 Nov 2018, 10:35
HKD1710 wrote:
Of the following, which best approximates:

\(\frac{(0.1667)(0.8333)(0.3333)}{(0.2222)(0.6667)(0.1250)}\) ?

(A) 2.00
(8) 2.40
(C) 2.43
(D) 2.50
(E) 3.43

Project PS Butler : Question #11


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rounded 0.16 to 0.2 --> 1/5

80/100 = 4/5

30/100 = 3/10

= 1/5*4/5*3/10 = 12/ 250


0.2222 = 2/10 = 1/5

0.6667 = 7/10

0.125 = 1/8

1/5*7/10*1/8 = 7/400


12/250 *400/7 = 96/35 = 2.7


Closest to 2.7 is Option D 2.5
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Re: Of the following, which best approximates:  [#permalink]

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New post 03 Feb 2019, 09:27
3
2
HKD1710 wrote:
Of the following, which best approximates:

\(\frac{(0.1667)(0.8333)(0.3333)}{(0.2222)(0.6667)(0.1250)}\) ?

(A) 2.00
(8) 2.40
(C) 2.43
(D) 2.50
(E) 3.43



I ball parked the given values to 0.16 * 0.8 * 0.3 / ( 0.2 * 0.6 * 0.12)

When divided will give 8/3

Nearest to D
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Re: Of the following, which best approximates:   [#permalink] 03 Feb 2019, 09:27
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