Download Grade 6 Math Circles Fall 2014 - Oct. 21/22 Solution Set: Algebra

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Faculty of Mathematics
Centre for Education in
Waterloo, Ontario
Mathematics and Computing
Grade 6 Math Circles
Fall 2014 - Oct. 21/22
Solution Set: Algebra - Solving Equations
1. (a) x + 4 = 10
x + 4 = 10 → x + 4 − 4 = 10 − 4 → x = 6
(b) x − 2 = 0
x−2=0→ x−2+2=0+2→ x=2
(c) 10 − y = 2
10 − y = 2 → 10 − y + y = 2 + y → 2 + y = 10 → 2 + y − 2 = 10 − 2 → y = 8
(d) 6 + 10 − 4 + z = 12
6 + 10 − 4 + z = 12 → 12 + z = 12 → 12 + z − 12 = 12 − 12 → z = 0
(e) x + 5 = 15, y + x = 12, 10 − 5 + x − y + 6 + z = 20
• x + 5 = 15 → x + 5 − 5 = 15 − 5 → x = 10
• y + x = 12 → y + 10 = 12 → y + 10 − 10 = 12 − 10 → y = 2
• 10 − 5 + x − y + 6 + z = 20 →
19 + z − 19 = 20 − 19 → z = 1
10 − 5 + 10 − 2 + 6 + z = 20 →
19 + z = 20 →
2
2. Tim had a dozen eggs in the fridge one night. When he went to make breakfast the next
morning, there were only half of a dozen eggs left. How many eggs mysteriously went missing?
Start your answer by representing this problem with an algebraic equation.
Let x be the number of eggs that went missing. Then the corresponding algebraic equation
for this word problem is:
12 − x = 6 → 12 − x + x = 6 + x → 6 + x = 12 → 6 + x − 6 = 12 − 6 → x = 6
Therefore, 6 eggs went missing.
3.
• What is a variable?
A variable is any random letter, a, b, x, y, etc., representing an unknown number in an
equation. We call it a variable because the number it can be varies.
• What is the reverse operation of subtraction; what is the reverse operation of multiplication?
The reverse operation of subtraction is addition; the reverse operation of multiplication
is division.
4. Collect the like terms on both sides of this equation. You do not have to solve this equation:
x + 2y + 3 + 8 + 3x − y = z + 5 − 1 + 2z + 5 − z
4x + y + 11 = 2z + 9 → 4x + y + 11 − 11 = 2z + 9 − 11 → 4x + y = 2z − 2 →
4x + y − 2z = 2z − 2 − 2z → 4x + y − 2z = −2
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5. There is an assortment of 10 flowers with roses, tulips, and daisies. There must be exactly 6
roses in the assortment, at least 2 tulips, and any amount of daisies. What are all the possible
arrangements?
Let x be the number of daisies. There are 3 possible assortments.
If there are 2 tulips: 6 + 2 + x = 10 → 8 + x = 10 → 8 + x − 8 = 10 − 8 → x = 2
If there are 2 tulips in the assortment, then there are 2 daisies as well. Solution is similar for
other 2 assortments.
6. Husky is 6 years older than Colly, and Colly is 3 years younger than Poodle. If Husky is 19
years old, how old is Poodle? Represent this question with an algebraic equation.
Let x be the age of Poodle. Since Colly is 3 years younger than Poodle, we know that the
age of Colly is then, x − 3. Since Husky is 6 years older than Colly, we know that the age
of Husky is then, (x − 3) + 6. But, we know that Husky is 19 years old, so we can make the
following algebraic equation:
(x − 3) + 6 = 19 → x − 3 + 6 = 19 → x + 3 = 19 → x + 3 − 3 = 19 − 3 → x = 16
Therefore, Poodle is 16 years old.
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7. Two consecutive numbers have a sum of 25. What are these two numbers? Represent this
problem with an algebraic equation.
Let x be the lower of the two numbers. We know that the next consecutive number is one
more than x, so we let (x + 1) be that number. Our equation is then:
x + (x + 1) = 25 → x + x + 1 = 25
2x + 1 = 25 → 2x + 1 − 1 = 25 − 1
2x = 24 →
2x
24
=
→ x = 12
2
2
Since x is the lower number and we know the other number is just one more than x, our two
numbers are 12 and 13. Let’s check our answer. “LS” means “left side of the equation”, and
“RS” means “right side of the equation”:
LS = x + (x + 1)
= 12 + 13
= 25
RS = 25
Therefore, since LS = RS, our answer is correct.
20
= 5, represent the right side as a fraction and then use cross multiplicax
tion to solve for x.
8. For the equation,
20
5
= → 5x = 20 →
x
1
5x
20
=
→ x=4
5
5
5
9. Solve the following equations:
(a) 5x = 25
5x = 25 →
5x
25
=
→ x=5
5
5
(b) 9y = 36
9y = 36 →
(c)
x
=4
6
x
=4→
6
(d)
16
8
= → 16 = 8x →
x
1
8x
16
=
→ x=2
8
8
3y
=9
5
3y
=9→
5
(f)
6∗x
= 6 ∗ 4 → x = 24
6
16
=8
x
16
=8→
x
(e)
9y
36
=
→ y=4
9
9
5 ∗ 3y
= 5 ∗ 9 → 3y = 45 →
5
z
1
=
6
3
z
1
= → 3z = 6 →
6
3
3z
6
= → z=2
3
3
3y
45
=
→ y = 15
3
3
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10. Mr. Johnson was printing out test papers for his class, but realized he accidentally printed 3
times the amount he actually needed. If he printed out 63 test papers, what is the size of Mr.
Johnson’s classroom (assuming each student gets one test paper).
Let x be the number of students in Mr. Johnson’s class. Then our equation is:
3x = 63
Solving for x we get
3x = 63 →
3x
63
=
→ x = 21
3
3
Therefore, there are 21 students in Mr. Johnson’s class.
11. Solve for each variable in this system of equations:
• z + 2x - y = 10
• 6y + x - 4 = 11
• 3x + 1 = 10
• Our third equation only has one variable, so we solve for x first:
3x + 1 = 10 → 3x + 1 − 1 = 10 − 1
9
3x
= → x=3
3x = 9 →
3
3
• We can then plug in our value for x into the second equation:
6y + x − 4 = 11 → 6y + 3 − 4 = 11
6y − 1 = 11 → 6y − 1 + 1 = 11 + 1
6y = 12 →
6y
12
=
→ y=2
6
6
• We then plug in our values for x and y into our first equation:
z + 2x − y = 10 → z + 2(3) − 2 = 10
z + 6 − 2 = 10 → z + 4 = 10
z + 4 − 4 = 10 − 4 → z = 6
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12. A classroom of 50 students is divided into two groups, with one of the groups having 8 more
students than the other. What is the size of each group?
Let x be the size of the smaller group. Since the other group has 8 more students, the size of
this group is x + 8. We know that there are 50 students in total, so our algebraic equation
looks like this:
x + (x + 8) = 50 → x + x + 8 = 50 → 2x + 8 = 50
2x
42
→ 2x + 8 − 8 = 50 − 8 → 2x = 42 →
=
→ x = 21
2
2
Therefore, the size of our smaller group is 21, and the size of the larger group is 29.
13. * Solve the following equations
(a)
x−1
=1
5
x−1
x−1
=1 → 5∗
=1∗5 → x−1=5 → x−1+1=5+1 → x=6
5
5
(b)
4x − 2
=7
2
4x − 2
4x − 2
=7 → 2∗
= 7 ∗ 2 → 4x − 2 = 14
2
2
→ 4x − 2 + 2 = 14 + 2 → 4x = 16 →
(c)
4x
16
=
→ x=4
4
4
x+2−1+x
=3
3
x+2−1+x
2x + 1
2x + 1
=3 →
=3 → 3∗
= 3 ∗ 3 → 2x + 1 = 9
3
3
3
→ 2x + 1 − 1 = 9 − 1 → 2x = 8 →
2x
8
=
→ x=4
2
2
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14. * The area of a pig pen is 1000m2 , with the width being 20m.
(a) What is the length of the pig pen in metres?
Let l be the length of the pig pen. We know the equation for area is A = l ∗ w, with A
being the area and w being the width. Since we are given the area and width of the pig
pen, we can solve for l with the following equation:
l ∗ 20 = 1000 →
l ∗ 20
1000
=
→ l = 50
20
20
Therefore, the length of the pig pen is 50m.
(b) If a pig is 2 metres in length and a total of 500 pigs can fit in the pen exactly, what is
the width of each pig?
Let w be the width of each pig. The area a pig will take up in the pig pen is w ∗ 2, since the
length of each pig is 2m. Since 500 pigs can fit in the pig pen exactly, and the area of the pig
pen is 1000m2 , we can use the following algebraic equation to solve for w:
500 ∗ 2 ∗ w = 1000 →
500 ∗ 2 ∗ w
1000
2∗w
2
=
→ 2∗w =2 →
=
→ w=1
500
500
2
2
Therefore, the width of each pig is 1m.
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15. * $50 is to be split among three friends. Friend B gets double the amount of friend A, and
friend C gets $10 more than friend B. How much money does each friend get?
Let x be the amount in dollars that friend A gets. Since friend B gets double the amount of
dollars of friend A, friend B will get $2x. Since friend C gets 10 more dollars than friend B,
friend C will get $(2x + 10). We then have the following algebraic equation:
x + 2x + (2x + 10) = 50
x + 2x + 2x + 10 = 50
x + 2x + 2x + 10 − 10 = 50 − 10
x + 2x + 2x = 40
5x = 40
5x
40
=
5
5
x=8
This tells us that friend A will get $8, friend B will get $16, and friend C will get $26. To
check our answer:
LS = x + 2x + (2x + 10)
= 8 + 2(8) + (2(8) + 10)
= 8 + 16 + 26
= 50
RS = 50
Therefore, since LS = RS, our answer is correct.
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16. * Two consecutive, odd numbers have a sum of 52. What are these two numbers. Represent
this problem with an algebraic equation (Hint: any odd number can be written as 2n + 1).
Let x be the lower of our two odd numbers. We know that the next consecutive odd number
is 2 more than x, so we let (x + 2) be that number. Our equation is then:
x + (x + 2) = 52
However, we need to ensure our x is odd, so we must use our hint which says that any odd
number can be written as 2n + 1. So we let x = 2n + 1. Our equation then becomes:
2n + 1 + (2n + 1 + 2) = 52 → 2n + 1 + 2n + 1 + 2 = 52
4n + 4 = 52 → 4n + 4 − 4 = 52 − 4
4n = 48 →
4n
48
=
→ n = 12
4
4
Now that we know n = 12, we plug this number into our equation for x (remember we let
x = 2n + 1):
x = 2n + 1 = 2(12) + 1 = 24 + 1 = 25
Since 25 is our lower number, our next odd number must then be 27. Let’s check our answer:
LS = x + (x + 2)
= 25 + 27
= 52
RS = 52
Therefore, since LS = RS, our answer is correct.
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17. * Solve the following equations
(a) x2 = 4
x2 = 4 →
√
x2 =
√
4 → x=2
(b) x3 = 8
x3 = 8 →
√
3
x3 =
√
3
8 → x=2
(c) (x + 1)(x − 3) = 0
In this case, if the inside of either bracket has a value of 0, this equation will work. So,
if x + 1 = 0 → x = −1
if x − 3 = 0 → x = 3
Therefore, x = −1 or x = 3. Let’s plug in either value of x to see if they both work
(note that (x + 1)(x - 3) is the same as (x + 1) * (x - 3) ).
For x = −1:
LS = (-1 + 1)(-1 - 3) = (0)(-4) = 0 = RS
For x = 3:
LS = (3 + 1)(3 - 3) = (4)(0) = 0 = RS
Therefore, since LS = RS for both of our values of x, our answer is correct.