Questions on Word Problems: Problems with consecutive odd even integers answered by real tutors!

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Question 151899This question is from textbook Algebra 2
: 1.Jay's test scores were 78,80,95, and 81. He took his last test for the quarter on Friday. What must he get to maintain a B average (between 80 and 89)
2. Ernie's average after 3 tests is at least 72%. If two of his test scores were 55 & 85, what could he have scored on his 3rd test?
3.Marissa has a square patio. She hires Ben to make it bigger. Ben uses 1'x1' tiles and increases the patio length 6' and doubles the patio's width. if the perimeter o the new patio is between 30' and 45', what could the dimensions of the original patio be?
This question is from textbook Algebra 2
: 1.Jay's test scores were 78,80,95, and 81. He took his last test for the quarter on Friday. What must he get to maintain a B average (between 80 and 89)
2. Ernie's average after 3 tests is at least 72%. If two of his test scores were 55 & 85, what could he have scored on his 3rd test?
3.Marissa has a square patio. She hires Ben to make it bigger. Ben uses 1'x1' tiles and increases the patio length 6' and doubles the patio's width. if the perimeter o the new patio is between 30' and 45', what could the dimensions of the original patio be?

Answer by jojo14344(883) About Me  (Show Source):
You can put this solution on YOUR website!
1.Jay's test scores were 78,80,95, and 81. He took his last test for the quarter on Friday. What must he get to maintain a B average (between 80 and 89)
.
Since B average is bet. 80-89, we'll use the 80 average (minimum) for the eqn. Atleast if he gets higher than this is better right?
Test=T[1], We'll do averaging,
(T[1]+T[2]+T[3]+T[5])/5=80
(78+80+95+81+T[5])/5=80, adding & cross multiply:
334+T[5]=5*80
T[5]=400-334 ----> T[5]=66 ---------> FINAL SCORE to get ave at least 80 and maintain B average.
.
2. Ernie's average after 3 tests is at least 72%. If two of his test scores were 55 & 85, what could he have scored on his 3rd test?
.
Same thing, Averaging
(T[1]+T[2]+T[3])/3=72
(55+85+T[3])/3=72, adding & cross multiply,
140+T[3]=3*72
T[3]=216-140 ----> T[3]=76 ------> FINAL SCORE to get 72 AVERAGE.
.
3.If the square patio has X dimensions, the new dimensions now are:
Length=x+6 ----------> increased by 6'
Width=2x ------------> doubled
This becomes a rectangle now since it has 2 diff dimensions.
The loophole with this you can get different orig. dimensions because the New Perimeter ranges from 30'-45'. Why? Simply we 'll base our computation in solving for X for the New Perimeter.
Okay, we'll pick the New Perimeter as 42ft, then
P=2(L+W)
42=2(x+6+2x)
42=2(3x+6) -------------> 42=6x+12 ---------> 42-12=6x
cross(30)5/cross(6)=cross(6)x/cross(6)
x=5ft
SIZE of the orig square patio is 5ftx5ft based on the NEW PERIMETER of 42ft
Thank you,
Jojo