+0  
 
0
52
3
avatar

Kate had three boxes, A, B and C, containing a total of 1512 necklaces.

The number of necklaces in Box A to the total number of necklaces was 2:7.

Kate sold 190 necklaces from Box B and sold 1/4 of the necklaces in Box C.

The number of necklaces left in Box B to the number of necklaces left in Box C was 2:1.

How many necklaces were there in Box C at first?

 Oct 26, 2021
 #1
avatar
0

smileyHello I can't help you but I want to tell you that you can do it I believe in you have a good daysmiley

You can do it

 Oct 26, 2021
edited by Guest  Oct 26, 2021
 #2
avatar
0

Not funny

Guest Oct 26, 2021
 #3
avatar
0

A + B + C ==1512 necklaces.

 

A / 1512 ==2 / 7, solve for A

 

A ==432 Necklacesin box A

 

1512   -  432 ==1,080 Necklaces in box B + C

 

[B - 190] ==Necklaces left in box B

 

[C  - 1/4C] ==3/4C - Necklaces left in box C

 

B  +  C  ==1080..........................(1)

 

[B - 190] / [3/4C] ==2/1............(2), solve for B, C

 

B==724  and  C ==356 - number of necklaces in box C at first.

 Oct 26, 2021

7 Online Users