POTW #9 Question:
POTW #8 Solution:
Thursday, November 1, 2018
POTW #9 - Happy Belated Halloween
I hope everyone had a nice Halloween yesterday. Last week's POTWs is one of my favourites. Please check the solution below. This week's POTW (#9) is similar to a Number Sense Assessment OF Learning (unit test) question cough cough...
POTW #9 Question:
POTW #8 Solution:
POTW #9 Question:
POTW #8 Solution:
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Grade 7/8 POTW
ReplyDeleteSimilar to the last question, I felt like we did this last year. Anyways, this question involves substitution (this method).
First, 15 girls left. This meant that there were g - 15 girls remaining. However, the question also stated that twice as many boys as girls remained. This meant that b/2 = g. If b/2 = g, it also means b = 2g. In this situation, I substituted b with 2g. Then, 3/4 of the boys left and 1/3 of the girls left. This meant that 1/4 of boys and 2/3 of girls remained. 1/4 of 2 is 1/2 and 2/3 of 1 is 2/3. The lcd (lowest common denominator) between the 2 is 6. The number of boys were 3/6g while the number of girls were 4/6g. They also said that there were now 14 more girls than boys meaning that g - b = 14, 4/6 - 3/6 = 1/6, or 14 = 1/6. This meant that 14 x 3 = 42 boys and 14 x 4 = 56 girls. In total, there were 42 + 56 = 98 total students remaining to perform. Note* the 15 girls left actually wasn't important at all for this method.
BONUS: Finding out original number of students. 42 x 4 = 168. 56 x 1.5 = 84. This matches with 168/2 = 84. 15 + 84 = 99. 99 + 168 = 267. There were originally 267 students.
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By the way, there are MANY different ways of solving the question. I just approached it in one of them.
Feel free to prove me wrong, but I really don't think I assigned this one last year. But...I am getting old so my memory isn't what it used to be haha
DeleteFiona thought the same thing too. It seems I DID post this Q during our algebra unit last year. My bad. Thanks for being on the ball!
DeleteLet x=Total # of Girls and y=Total # of Boys
ReplyDeleteand x1, y1 be # of boys and girls (respectively) after step 1 and so on
Let z=number of remaining students
x1=x-15
y1=(x1*2)=2x-30
y2=(¼)y1=x/2-15/2
x2=(⅔)x1=2x/3-10
x2=y2+14
2x/3-10=x/2-7.5+14
subtract x/2 from both sides
x/6-10=-7.5+14
add 10 to both sides
x/6=16.5
Times 6 to both sides
x=99
Now plug in x
99-15=84
(⅔)84=56
56-14=y2
42=y2
42+56=98
Let us represent the starting number of girls as x, the number of boys as y and the total as z.
ReplyDeleteWe know that originally, when the 15 girls left, there were twice as many boys.
y=2(x-15)
Expanding this expression grants us this answer:
y=2x-30
Because we need a simultaneous equation to subtract this from, we move onto the next step. Next, 3/4 of the boys and 1/3 of the girls leave, leaving 14 more girls than boys.
2/3(x-15)=1/4b+14
Multiplying outward,
2/3x-10=1/4y+14
Finding the LCM on 3 and 4 (which is 12), we multiply eh equation by that.
8x-120=3y+168
Next, we substitute one simultaneous equation into the other (I tried addition of simultaneous equations and IT DIDN'T MEET ALL THE CRITERIA)
8x-120=6x-90+168
2x-120=-90=168
2x=198
x=99
Substituting x into simultaneous equation y=2(x-15),
y=2*84
y=168
Total=267
Doing the calculations, after a lot of people left, there were only 98 performing.
Let x be the number of boys and y be girls at the start of the meeting.
ReplyDelete15 girls left, so (y-15) remain.
The number of boys is twice the number of remaining girls, so the expression for the remaining boys is 2(y-15) and after using distributive property, we can simplify this to 2y-30.
Next, 3/4 of the boys left leaving only 1/4x remaining. Same thing with the girls, 1/3 leave so only 2/3(y-15) remain. Now, the number of girls is 14 more than the remaining boys. The equation would be:
girls 2/3(y-15)=1/4x+14 boys
To rid of the fractions, we can multiply by a common denominator, 12 getting:
8(y-15)=3x+168
Distribute
8y-120=3(2y-30)+168
8y-120=6y-90+168
8y-30=6y+168
8y=6y+198
2y=198
y=99
Now that we know how many girls there are, we can find the number of boys by taking the expression 2(y-15) and using substitution.
x=2(99-15)
x=2(84)
x=168
Total: 99+168
=267
(insert therefore sign here)There are a total of 267 students, but only 98 performed in the show.
We can write this question in algebra to make it less confusing.
ReplyDeleteLet y equal girls and x equal boys
Using this let statement, we can write it out algebraically.
At the start of the meeting, 15 girls left and twice as many boys remained:
2(y-15) or 2y-30
After that, 3/4 of the boys left and 1/3 of the remaining girls left. The number of girls is now 14 more than boys.
1/4x | 2/3(y-15)
2/3(y-15) = 1/4x+14 => 8(y-15) = 3x+168
Finding the number of girls...
8y-120 = 3(2y-30)+168
8y-120 = 6y-90+168
8y-30 = 6y-168
8y-6y = 168 + 30
2y = 198
y = 198/2
y = 99
There are 99 girls, which means...
x = 2(99-15)
x = 2(84)
x = 168
Total amount: 267
To find how many students were still there...
(99-15)-(3/4(99-15)) = 21
168 - (3/4)*168 = 42
= 63.
Aliyah
ReplyDeleteLet Y=Girls, X=Boys
Y = 87 to start(guess and check)
- 15 = 72
- 1/3 = 24
X = 40 (guess and check)
- 3/4
= 10
24 + 10 = 34 students stayed for the show.
I know i'm correct because using RWA 34 students is acceptable for a play.
15 girls left and twice as many guys remained. 3/4 boys and 1/3 girls remained. this left 14 more girls then guys. how many kids remained?
ReplyDeleteLet G=Girls and B=Guys
15 g x2 =30b
30 divided by 4 = 7 b
7 b x 3 =21
21 + 14 = 35
therefor there would be 35 students left for the play.
Anya
ReplyDeletey=girls(after the 15 girls left)
2y=boys(after the 15 girls left)
2/3*y=total# of girls left
1/4*2y=total# of boys left
(1/2*y)
2/3*y-1/2*y=14
y(2/3-1/2)=14
y(4/6-3/6)=14
y*1/6=14
y/6=14
6*14=84=y
girls= 84*2/3
=28*2
=56
Boys=84*1/2
=42
42+56=98
98 total kids left for the play
Let the number of total students at the beginning of the meeting be S.
ReplyDeleteLet the number of total girls at the beginning of the meeting be G.
Let the number of total boys at the beginning of the meeting be B.
Now, 15 girls left, therefore there are G - 15 girls, which we can now call g.
Boys are twice of g, therefore B = 2 * g.
After a while, ¾ of the boys left, therefore the remaining number of boys is ¼ of B, which we can call B/4.
⅓ of the girls left as well, therefore ⅔ of g remain.
Now, there are 14 more girls than boys, so:
B / 4 + 14 = ⅔ * g
Now we know that B = 2 * g, therefore:
(2 * g) / 4 + 14 = ⅔ * g
This can be written as:
⅔ * g - ½ * g = 14
or
(4 * g - 3 * g )/ 6=14 or g / 6 =14
or
g = 14 * 6 = 84
Therefore:
B = 2 * g=168
B / 4 + 2 * g / 3 = 168 / 4 + 168 / 3 = 42 + 56 = 98
The number of students who remained to perform are 98.
POTW:
ReplyDeleteMy POTW was done on paper. My answer is that 98 students stayed to perform in the play.