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Mathematics 22 Online
OpenStudy (anonymous):

A stamp collector bought 160 stamps for $25.00. The purchase included 5¢ stamps, 15¢ stamps, and 25¢ stamps. The number of 15¢ stamps is three times the number of 5¢ stamps. How many of each type of stamp was purchased?

OpenStudy (anonymous):

I found 25 cents and 15 cents, but i cant seem to find 5 cents

OpenStudy (irishboy123):

how did you solve this for 25c and 15c?

OpenStudy (anonymous):

160-4x = 160 -(4*30) = 40 for 25 cents 160 - 3x = 160 - (3*30) = 90 for 15 cents

OpenStudy (welshfella):

how did you get 4x?

OpenStudy (anonymous):

let 5cents stamps to be x and so for 15cents and 25 cents as y ,z respectively

OpenStudy (anonymous):

i addded 3x + x

OpenStudy (anonymous):

now 15 cts stamps are thrice the time of 5 cts => y=3x;

OpenStudy (anonymous):

now again total no. stamps is 160 hence x + y + z =160 also total amount is 25 $ so 5x+15y+25z=25

OpenStudy (anonymous):

solve all three equetions for solutions

OpenStudy (anonymous):

okaay

OpenStudy (anonymous):

would i plug in 25 for x y and z?

OpenStudy (anonymous):

no x for 5 , y for 15 and 25 for z

OpenStudy (anonymous):

from 1st eq. y=3x puting in 2nd and 3rd we have x+3x+z=160 => 4x + z = 160 ............1 and 5x + 15*3x + 25 z = 2500 ..........2 (since in doller not in cents) => 50x + 25z =2500 or 2x + z = 100 .........2

OpenStudy (anonymous):

wouldnt it be .5 .15 .25?

OpenStudy (anonymous):

now from new equation 4x +z =160 and 2x + z =100 2(100-z) + z = 160 200 -2z +z =160 => z=40 put in 2x +z = 100 => x=(100-40 ) / 2 x = 30 and y = 3*30 as 3x y=90 hence solved as 30 stamp for 5 c 40 stamps for 25 c and 90 stamps for 15 c

OpenStudy (anonymous):

yes .5 .25 .15 for 25 or 5 25 15 for 2500 both are same

OpenStudy (anonymous):

oh okay, thank you :)

OpenStudy (anonymous):

it s okay ;)

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