Show your work, or, how my math abilities started to decline
I
think it's problematic the way schools teach Algebra. Those meaningless
show-your-work approaches, without knowing what Algebra is truly about.
The overuse of calculators and the piecemeal way of teaching without
the unification of the math concepts are detrimental to our children's
ability to think critically and logically.
Of course
eventually, it would be beneficial to students if they show their work
with the much more challenging word problems (harder Mathcounts state
team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.
How
do you improve problem solving skills with tons of worksheets by going
through 50 to 100 problems all look very much the same? It's called busy
work.
Quote: "Insanity: doing the same thing over and over again and expecting different results."
Quotes from Richard Feynman, the famous late Nobel-laureate physicist. Feynman relates his cousin's unhappy experience with algebra:
My
cousin at that time—who was three years older—was in high school and
was having considerable difficulty with his algebra. I was allowed to
sit in the corner while the tutor tried to teach my cousin algebra. I
said to my cousin then, "What are you trying to do?" I hear him talking
about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and
I'm trying to figure out what x is," and I say, "You mean 4." He says,
"Yeah, but you did it by arithmetic. You have to do it by algebra."And
that's why my cousin was never able to do algebra, because he didn't
understand how he was supposed to do it. I learned
algebra, fortunately, by—not going to school—by knowing the whole idea
was to find out what x was and it didn't make any difference how you did
it. There's no such a thing as, you know, do it by arithmetic, or you do it by algebra. It was a false thing that they had invented in
school, so that the children who have to study algebra can all pass it.
They had invented a set of rules, which if you followed them without
thinking, could produce the answer. Subtract 7 from both sides. If you
have a multiplier, divide both sides by the multiplier. And so on. A
series of steps by which you could get the answer if you didn't
understand what you were trying to do.
So I was lucky. I always learnt things by myself.
Showing posts with label algebra. Show all posts
Showing posts with label algebra. Show all posts
Sunday, May 25, 2025
Sunday, April 23, 2017
Tricky Algebra Mathcounts National Questions: Counting Backwards
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?
#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent $1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?
#25: 1998 AMC-8 Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
Solution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this: x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars.
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally,
62 dollars at the beginning.
#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.
Amy Jan Toy
? ? 36
First round Amy gave Jan and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 72
Second round Jan gave Amy and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars Amy Jan 36
That means that at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.
Download this year's Mathcounts handbook here.
#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?
#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent $1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?
#25: 1998 AMC-8 Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
Solution I: Use Algebra:
#24: Let A contains x ounces and B contains y ounces and x > y (given).
After first pouring, A has (x - y) ounces left and B has 2y ounces (double the original amount)
After second pouring, A has ( 2x - 2y)(double the amount) ounces and B has (3y - x) ounces left.
After third pouring, A has (3x - 5y) ounces left and B has (6y - 2x) (double the amount)
3x - 5y = 64 times 2 for each terms 6x - 10y = 128 ----equation 3
6y - 2x = 64 times 3 for each terms 18y - 6x = 192 ---- equation 4
equation 3 + equation 4 and you have 8y = 320 and y = 40 ; Plug in any equation and you get x = 88
88 - 40 = 48 ounces
Solution II: Solving it backwards:
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.
Solution I: Use Algebra
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
Solution II: Solving it backwards:
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.
Solution I: Use Algebra
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
At the second store, he would spend 1 + (x-2)/4 and would have (x-2)/2 - 1 - (x-2)/4 or (x-6)/4 left
At the third store, he would spend 1 + (x-6)/8 and would have (x-14)/8 left
It looks like there's a pattern. At the fourth store, he would spend (x-30)/16
and at the 5th store he would spent (x-62)/32 = 0 so x - 62 = 0 and x = 62 dollarsSolution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this: x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars.
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally,
62 dollars at the beginning.
#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.
Amy Jan Toy
? ? 36
First round Amy gave Jan and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 72
Second round Jan gave Amy and Toy double the amount of what each of them has, so
Amy Jan Toy
? ? 144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars Amy Jan 36
That means that at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.
Wednesday, July 6, 2016
Mathcounts Prep : Algebra Manipulation
Note that:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}\)
\(\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}\)
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}\)
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
Applicable questions:
Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?
Solution:
\(\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}\)= \(\dfrac {a} {b}\)
Question 2: If \(x^{2}+y^{2}=153\) and x + y = 15, what is xy?
Solution:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(15^{2}-2xy=153\)\(\rightarrow xy=36\)
Question 3: If \(\left( x+y\right) ^{2}=1024\) , \(x^{2}+y^{2}\) = 530 and x > y , what is x - y?
Solution:
\(\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}\)
1024 - 2xy = 530\(\rightarrow 2xy=494\)
\(\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}=36\)
x - y = 6
Question 4: x + y = 3 and \(x^{2}+y^{2}=89\), what is \(x^{3}+y^{3}\)?
Solution:
\(\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}\)
9 - 2xy = 89 \(\rightarrow -2xy=80\) so xy = -40
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}\)
\(x ^{3}+y^{3}\)= 387
Question #5: If \(x+\dfrac {1} {x}=5\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Solution:
\(\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}\)
\(5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}\)
125 - 3*5 = \(x^{3}+\dfrac {1} {x ^{3}}\)
The answer is 110.
Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and \(x^{2}+y^{2}+z^{2}=19\), what is the arithmetic mean of the three product xy + yz + xz?
Solution:
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
\(7^{2}-2\left( xy+yz+xz\right) =19\)
xy + yz + xz = 15 so their mean is \(\dfrac {15} {3}=5\)
More practice problems (answer key below):
#1:If x + y = 5 and xy = 3, find the value of \(\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}\).
#2: If x + y = 3 and \(x^{2}+y^{2}=6\), what is the value of \(x^{3}+y^{3}\)?
#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.
#4:If \(x-\dfrac {6} {x}\) = 11, find the value of \(x^{3}-\dfrac {216} {x^{3}}\)?
#5: If \(x+\dfrac {3} {x} = 9\), find the value of \(x^{3}+\dfrac {27} {x^{3}}\)?
#6:If \(x+\dfrac {1} {x} = 8\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Answers:
#1 :\(\dfrac {19} {9}\)
#2: 13.5
#3: \(\dfrac {2} {5}\)
#4: 1529 [ \(11^{3}\)+ 3 x 6 x 11 =1529]
#5: 648 [\(9^{3}\)-3 x 3 x 9 = 648]
#6: 488 [ \(8^{3}\)– 3 x 8 = 488]
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}\)
\(\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}\)
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}\)
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
Applicable questions:
Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?
Solution:
\(\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}\)= \(\dfrac {a} {b}\)
Question 2: If \(x^{2}+y^{2}=153\) and x + y = 15, what is xy?
Solution:
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(15^{2}-2xy=153\)\(\rightarrow xy=36\)
Question 3: If \(\left( x+y\right) ^{2}=1024\) , \(x^{2}+y^{2}\) = 530 and x > y , what is x - y?
Solution:
\(\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}\)
1024 - 2xy = 530\(\rightarrow 2xy=494\)
\(\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}=36\)
x - y = 6
Question 4: x + y = 3 and \(x^{2}+y^{2}=89\), what is \(x^{3}+y^{3}\)?
Solution:
\(\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}\)
9 - 2xy = 89 \(\rightarrow -2xy=80\) so xy = -40
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}\)
\(x ^{3}+y^{3}\)= 387
Question #5: If \(x+\dfrac {1} {x}=5\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Solution:
\(\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}\)
\(5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}\)
125 - 3*5 = \(x^{3}+\dfrac {1} {x ^{3}}\)
The answer is 110.
Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and \(x^{2}+y^{2}+z^{2}=19\), what is the arithmetic mean of the three product xy + yz + xz?
Solution:
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
\(7^{2}-2\left( xy+yz+xz\right) =19\)
xy + yz + xz = 15 so their mean is \(\dfrac {15} {3}=5\)
More practice problems (answer key below):
#1:If x + y = 5 and xy = 3, find the value of \(\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}\).
#2: If x + y = 3 and \(x^{2}+y^{2}=6\), what is the value of \(x^{3}+y^{3}\)?
#3: The sum of two numbers is 2. The product of the same two numbers is 5.
Find the sum of the reciprocals of these two numbers, and express it in simplest form.
#4:If \(x-\dfrac {6} {x}\) = 11, find the value of \(x^{3}-\dfrac {216} {x^{3}}\)?
#5: If \(x+\dfrac {3} {x} = 9\), find the value of \(x^{3}+\dfrac {27} {x^{3}}\)?
#6:If \(x+\dfrac {1} {x} = 8\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?
Answers:
#1 :\(\dfrac {19} {9}\)
#2: 13.5
#3: \(\dfrac {2} {5}\)
#4: 1529 [ \(11^{3}\)+ 3 x 6 x 11 =1529]
#5: 648 [\(9^{3}\)-3 x 3 x 9 = 648]
#6: 488 [ \(8^{3}\)– 3 x 8 = 488]
Monday, September 9, 2013
Chicken, Rabbit Questions : Algebra Without Using Variables
The problems below are much easier so if you are preparing for the state, try the online timed test here.
Just write down random name to enter the test site is fine. At the end of the test, you'll see the answers for the ones you get wrong. Everyone is welcome to take the test. Thanks a lot and have fun problem solving.
#1: There are 20 horses and chickens at Old Macdonald's farm. Together there are 58 legs. How many horses and how many chickens?
#4: 14 octopus and 19 sea otters.
#5: 6 cows and 12 chickens.
Just write down random name to enter the test site is fine. At the end of the test, you'll see the answers for the ones you get wrong. Everyone is welcome to take the test. Thanks a lot and have fun problem solving.
#1: There are 20 horses and chickens at Old Macdonald's farm. Together there are 58 legs. How many horses and how many chickens?
Solutions I :
#1: Using algebra, you have
H + C = 20---equation 1 and
4H +2C = 58---equation 2
To get rid of one variable you can times equation 1 by 2 to get rid of Chicken or times 4 to get rid of horses.
Times 2 and you have 2 H + 2 C = 40 (every term needs to be multiplied by 2)---equation 3
4 H + 2C = 58---equation 2
Using equation 2 - equation 3 and yo get 2H = 18 so H = 9 and from there, solve for C = 11
Solution II:
Solution II:
Let there be C chickens and (20-C) horses. [Since the sum of the number of chickens and horses is 20. If one has C number, the other has (20 - C)
2C + 4 (20 - C) = 58; 2C + 80 - 4C = 58; -2C = -22; C = 11
and from there, you get H = 20 - 11 = 9
Solution III:
Without using algebra, you can make all the animals be chickens first. In that case, you'll have 20 x 2 = 40 legs. Since you have 58 legs, you need to get rid of some chickens and bring in more horses.
2C + 4 (20 - C) = 58; 2C + 80 - 4C = 58; -2C = -22; C = 11
and from there, you get H = 20 - 11 = 9
Solution III:
Without using algebra, you can make all the animals be chickens first. In that case, you'll have 20 x 2 = 40 legs. Since you have 58 legs, you need to get rid of some chickens and bring in more horses.
You gain 2 legs by every transaction (-2 + 4 = 2). (58 - 40) / 2 = 9 so 9 horses and 11 chickens.
Other similar questions to practice (answer key below):
#1: Rabbits and ducks -- 30 animals and 86 feet.
#1: Rabbits and ducks -- 30 animals and 86 feet.
#2: There are 24 three-leg stools and four-leg tables. Together there are 86 legs.
#3: There are 43 bicycles and tricycles and together there are 100 wheels.
#4:
There are 33 octopus (8 arms) and sea otters and together they have 188
arms/or for sea otters--legs. How many octopus and how many sea
otters?
#5: There are 18 animals in the barnyard, some are cows and some are chickens. There are total 48 legs. How many chickens and how many cows?
#5: There are 18 animals in the barnyard, some are cows and some are chickens. There are total 48 legs. How many chickens and how many cows?
Answers:
#1: 13 rabbits and 17 ducks.
#2: 10 three-leg stools and 14 four-leg tables.
#3: 29 bicycles and 14 tricycles. #4: 14 octopus and 19 sea otters.
#5: 6 cows and 12 chickens.
Tuesday, April 16, 2013
This Week's Work : Week 8 -- for Inquisitive Young Mathletes
Assignment 1:
Using Algebra and Number Sense as Shortcuts from Mathcounts Mini
Watch the video and work on the activity sheet below the video on the same link for more practices.
Also, review the following:
\(x^{2}-y^{2}=\left( x+y\right) \left( x-y\right)\) \(\left( x+y\right) ^{2}=x^{2}+2xy+y^{2}\) \(\left( x-y\right) ^{2}=x^{2}-2xy+y^{2}\) \(\left( x+y\right) ^{2}-2xy =x^{2}+ y^{2}\)
\(\left( x-y\right) ^{2}+ 2xy =x^{2}+y^{2}\)
\(\left( x+y\right) ^{2}-4xy =\left( x-y\right) ^{2}\)
Assignment 2:
Pascal's Triangle from Math is Fun
Pascal's Triangle and Its Patterns
Assignment 3:
It'll be sent through e-mail.
Happy problem solving !!
Using Algebra and Number Sense as Shortcuts from Mathcounts Mini
Watch the video and work on the activity sheet below the video on the same link for more practices.
Also, review the following:
\(x^{2}-y^{2}=\left( x+y\right) \left( x-y\right)\) \(\left( x+y\right) ^{2}=x^{2}+2xy+y^{2}\) \(\left( x-y\right) ^{2}=x^{2}-2xy+y^{2}\) \(\left( x+y\right) ^{2}-2xy =x^{2}+ y^{2}\)
\(\left( x-y\right) ^{2}+ 2xy =x^{2}+y^{2}\)
\(\left( x+y\right) ^{2}-4xy =\left( x-y\right) ^{2}\)
Assignment 2:
Pascal's Triangle from Math is Fun
Pascal's Triangle and Its Patterns
Assignment 3:
It'll be sent through e-mail.
Happy problem solving !!
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