What is the key to solving the following problem:
Given that the question marks below are replaced with the digits 1, 2, 3, 4, 5, and 6, what is the greatest possible positive difference that can be obtained? (Each digit is used once.)
? ? ?
- ? ? ?
————
Make the first number as large as possible by using the big digits (4,5, and 6) and the second number as small as possible by using the small digits (1,2, and 3).
Given that the question marks below are replaced with the digits 1, 2, 3, 4, 5, and 6, what is the greatest possible positive difference that can be obtained? (Each digit is used once.)
? ? ?
- ? ? ?
654 – 123 = 531
Key to solving the following problem: Which of the fractions 59/24, 79/32, 99/40, 149/66 is the smallest?
All the numerators end in 9; we can add 1 and get simpler fractions.
Which of the fractions 59/24, 79/32, 99/40, 149/66 is the smallest?
- (59+1)/24 = 60/24
(79+1)/32 = 80/32
(99+1)/ 40 = 100/40
…
2. Note all are equal to 5/2.
- Rewrite each fraction in the form:
59/24 = 5/2 - 1/24
…
149/60 = 5/2 - 1/60 - We have the biggest impact when we subtract the fraction with the smallest denominator, so 59/24 is farther from 5/2 than the other three fractions. Therefore, it is the smallest of the four fractions.
Key to solving the problem:
Four friends shared a bottle of water. The first friend drank 1/4 of the water. The second friend drank 1/3 of the remaining water. The third friend drank 1/2 of what was left. The fourth friend drank the last six ounces. How many ounces of water were originally in the bottle?
Work backwards or introducing a variable x.
Four friends shared a bottle of water. The first friend drank 1/4 of the water. The second friend drank 1/3 of the remaining water. The third friend drank 1/2 of what was left. The fourth friend drank the last six ounces. How many ounces of water were originally in the bottle?
1, The third friend must have started with 12 ounces of water, since he drank half of that and had 6 ounces left.
2, So now, we have that the second friend ended up with 12 ounces.
She ended up with 12 ounces of water, having consumed 1/3 of what she started with, meaning she had 2/3 remaining.
- Given that 12 ounces represent 2/3 of the starting amount, she must have started with 18 ounces of water.
- Now, we have that the first friend ended up with 18 ounces. So how much water did she start with?
She ended up with 18 ounces after drinking 1/4 of what she started with, meaning she had 3/4 remaining. Therefore, she must have started with 24 ounces.
Four friends shared a bottle of water. The first friend drank 1/4 of the water. The second friend drank 1/3 of the remaining water. The third friend drank 1/2 of what was left. The fourth friend drank the last six ounces. How many ounces of water were originally in the bottle?
If the first friend drinks 1/4 of the available water, how much water is in the bottle when it is passed to friend two? 3/4 of x
- This friend drinks 1/3 of the water she is given, leaving how much for friend three?
1/3 of the water is drunk, leaving 2/3 of the water for the next friend. That is (2/3)⋅(3/4)⋅x, or (1/2)⋅x. - Friend 3 is given a bottle with x/2 ounces of water and drinks 1/2 of this water which is x/4
- Good! Drinking half of the water leaves half of the water in question, or
(1/2)⋅(1/2)x=x/4 - x/4 = 6 ==> x = 24
Key to solving the following problem using algebra. My average on the first six history tests is 82. What do I have to score on the next test to raise my average to 85?
Use as few variables as possible.
6⋅82+x=85⋅7. Then we subtract 6⋅82 from both sides.
Shorten your calculation by writing 857 = 7(82+3)
x=85⋅7−6⋅82=103
Non-algebraic method of solving:
- Our new average is supposed to be 85.
- The first six tests are each three under the new desired average. So their total is 18 under the average.
- ==> Last test must be 18 over the average of 85, or else they won’t all average out to 85.
- Last test must be 85 + 18 = 103.
If Richard can paint 2 rooms in 3 days, how many rooms can he paint in 12 days?
8
If 5 technicians can produce 64 surfboards in 4 days, how many surfboards can 8 technicians produce in 10 days?
- one technician make in one day?
Achilleas 2015-06-13 13:48:55
Here, we have only 1/5 the work rate, so we can make 16/5 = 3.2 surfboards a day. (scale down!)
We have that one technician can make 3.2 surfboards in one day.
Given that one technician can make 3.2 surfboards in one day, how many surfboards can one technician make in 10 days?
Right, we have increased our time tenfold, so one technician can make 3.2 * 10 = 32 surfboards in 10 days. (scale up!)
Finally, how many can 8 technicians make in 10 days?
Achilleas 2015-06-13 13:50:44
We have increased our work rate, so 8 technicians can make 32 * 8 = 256 surfboards in 10 days. (scale up!)
Name some problem solving tactics.
- Trusting your instincts,
- Looking at a problem in different ways
- Working backwards,
- Using as few variables as possible
- Generalizing a known technique.
- Knowledge of mathematical facts, formulas, and procedures.
- Factorization
- Guess and check.
9.
Expand (x+10)(3x−9).
(x+10)(3x−9)=x(3x−9)+10(3x−9)
Using the dist. prop. once
=3x^2−9x+30x−90
=3x^2+21x−90.Using the dist. prop. two more times!Combining like terms
What is the grid (table) method for expanding: (x+10)(3x−9).
x 10
———————-
3x | 3x^2 30x
-9 | -9x -90
Name some problems in which factorization is helpful?
- Breaking integers down into their prime factors
2. Breaking quadratic expressions down into their binomial factors,
Express as a common fraction in simplest form:
2 + 4 +6 + .. + 50)/(3 + 6 + 9 ..+75
- Factor the numerator and the denominator.
(2 + 4 +6 + .. + 50)/(3 + 6 + 9 ..+75)
= (2 * (1+2 … 25))/(3*(1+2..25))
= 2/3
5!*6!/(6! + 5! + 5!)
the denominator factors as: 6!+5!+5!=6⋅5!+5!+5!=(6+1+1)5!=8⋅5!.
5!*6!/(6! + 5! + 5!) = 90
What is 4!
24
5!
120
6!
720
7!
5040
8!
40320
2009^2 − 2007^2
(2009+2007)(2009−2007)
= 8032
What is the least possible positive difference between two positive integers whose squares differ by 400?
x^2−y^2=400, and we want x−y to be as small as possible.
x−y must be an integer.
x-y cannot be 1 because it would be an odd number.
x+y has to be an odd number also
if we let x=101 and y=99 then we have x−y=2 and (x+y)(x−y)=2⋅200=400
