+0

# 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

0
296
3

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
+26322
+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
+26322
+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
+18712
+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
+78577
+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

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details