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Initially, the markers and pens in a drawer were in the ratio of 5 : 7

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Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 06 Mar 2020, 03:43
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 06 Mar 2020, 03:47
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Bunuel wrote:
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?

(A) 29
(B) 43
(C) 49
(D) 50
(E) 51


Markers = 5x
Pens = 7x

Pen remaining = 7x - 6

Markers, 5x = 35 i.e. x = 7

i.e. Pens = 7x = 49

ie.. Pen remaining = 49-6 = 43

Answer: Option B
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 06 Mar 2020, 04:11
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Bunuel wrote:
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?

(A) 29
(B) 43
(C) 49
(D) 50
(E) 51


Solution:


    • The ratio of markers to pens = \(5: 7\)
    • Let \(5x\), and \(7x\) be the number of markers and pens in the drawer.
      o Number of pens after removal of 6 pens = \(7x – 6\)
    • Number of markers = \(35\)
      o \(5x =35\)
      o \(x =7\).
    • Number of pens left = \(7*7 – 6 = 43\).
Hence, the correct answer is Option B.
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 12 Mar 2020, 00:47
Given,
Markers is 5x and Pens is 7x with multiplier x.
It's also given that 5x is 35.
Hence multiplier x=7
And Pen remaining after being removed = 7x - 6= 7x7-6 = 49-6= 43.

Answer Is Option B
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 14 Mar 2020, 08:41
Bunuel wrote:
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?

(A) 29
(B) 43
(C) 49
(D) 50
(E) 51


Marker : Pen
5x :7x

Since we know that there were 35 markers in the drawer, x = 7 ( 5*7 = 35)
Therefore, 7(7) = 49 (Total number of Pens)
After 6 pens were removed, the balance is 43.

Answer is B.
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7  [#permalink]

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New post 15 Mar 2020, 03:59
Bunuel wrote:
Initially, the markers and pens in a drawer were in the ratio of 5 : 7. Then, 6 pens were removed. If there are 35 markers in the drawer, how many pens are left?

(A) 29
(B) 43
(C) 49
(D) 50
(E) 51


Since the ratio of markers to pens was 5 : 7, the number of markers can be represented by 5x and the number of pens can be represented by 7x where x is some positive integer. We can create the equation for the number of markers:

5x = 35

x = 7

Originally, there were 35 markers and 7 x 7 = 49 pens. Since 6 pens were removed, 43 pens were left in the drawer.

Answer: B
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Re: Initially, the markers and pens in a drawer were in the ratio of 5 : 7   [#permalink] 15 Mar 2020, 03:59

Initially, the markers and pens in a drawer were in the ratio of 5 : 7

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