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# irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons

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irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons left.hiw many blue balloons does irene have at first?

Guest Jul 6, 2015

#1
+26544
+10

Perhaps I've gone wrong somewhere or misinterpreted part of the question!

Edit:  Of course, I made a numerical mistake towards the end.  I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764.  Never-the-less, this still doesn't result in integer numbers of balloons!

.

Alan  Jul 6, 2015
Sort:

#1
+26544
+10

Perhaps I've gone wrong somewhere or misinterpreted part of the question!

Edit:  Of course, I made a numerical mistake towards the end.  I should have had 13b/12 = 764 (giving b = 705.231), not 13b/4 = 764.  Never-the-less, this still doesn't result in integer numbers of balloons!

.

Alan  Jul 6, 2015
#2
+19054
+5

irene had a total of 1686 red,blue and green balloons for sale.the ratio of red to blue was 2:3.after irene sold 3/4 blue and 1/2 red.she had 922 balloons left.hiw many blue balloons does irene have at first ?

$$\small{\text{ \begin{array}{lrcl} (1) & r + b + g &=& 1686 \\\\ (2) & \dfrac{r}{b} &=& \dfrac23 \qquad \text{ so } \qquad r = \dfrac23 \cdot b\\\\ \hline \\ (2) \text{ in } (1): \quad & \dfrac23 \cdot b + b+ g &=& 1686 \\\\ (I) & \dfrac53 \cdot b + g &=& 1686 \\\\ \hline \\ (3) & \dfrac14 \cdot b + \dfrac12 \cdot r +g &=& 922\\\\ (2) \text{ in } (3): \quad & \dfrac14 \cdot b +\dfrac12 \cdot \dfrac23 \cdot b + g &=& 922 \\\\ & \dfrac14 \cdot b +\dfrac13 \cdot b + g &=& 922 \\\\ & \dfrac3{12} \cdot b +\dfrac4{12} \cdot b + g &=& 922 \\\\ (II) & \dfrac7{12} \cdot b + g &=& 922 \\\\ \hline \\ (I)-(II): & \dfrac53 \cdot b - \dfrac7{12} \cdot b &=& 1686 - 922 \\\\ & \dfrac53 \cdot b - \dfrac7{12} \cdot b &=& 764 \\\\ & \dfrac{20}{12} \cdot b - \dfrac7{12} \cdot b &=& 764 \\\\ & \dfrac{13}{12} \cdot b &=& 764 \\\\ & b &=& \dfrac{ 764\cdot 12 } {13} \\\\ &\mathbf{ b }& \mathbf{=}& \mathbf{705} \end{array} }}$$

Irene does  have at first 705 blue balloons.

check: 176 ( blue ) + 235 ( red ) + 511 ( green ) = 922

705 ( blue )  + 470 ( red ) + 511 ( green ) = 1686

176 / 705 = 1 / 4

235 / 470 = 1 / 2

705 * ( 2 / 3 ) = 470

heureka  Jul 6, 2015
#3
+83935
+5

No, Alan...I worked it out, too....there's no "whole" number answer possible based on the available info........

Let x =  initial number red      ....the (3/2)x = initial number blue  ....  y=  initial  number green

Before the sale, we have:

x + (3/2)x   + y = 1686         →              (5/2)x + y = 1686     (1)

After the sale, we have:

(1/2)x + (1/4)(3/2)x + y = 922       →     (7/8)x + y =    922      (2)

Subtract (2) from (1)

(13/8)x = 764       x =  about 470   initial red

(3/2)x =  about 705  initial blue

Rats!!!.....heureka beat me to it  !!!!!

CPhill  Jul 6, 2015

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