Beginning Algebra: II

Learn the basics here.

Here we are going to review and work on some other harder problems.

Question #1:   Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

Solution: 
Let there be x tickets sold for singles and y tickets sold for couples. According to the given:
20 x + 35 y = 2280 --- equation 1
x  + 2 y = 128 --- equation 2

equation 2 times 20 - equation 1 and you have 20 x + 40 y - (20 x + 35 y) = 20 x 128 - 2280 = 280
5y = 280 so y = 56 and plug in to equation 2 and you get x = 16

There will be 16 tickets sold for singles and 56 tickets sold for couples.
Always check your answer to see if that is right.

Question #2: There are 20 questions on an algebra test, you got 5 points for each question you answer correctly and - 2 points for each questions you answer wrong. 
If a student gets 79 for his score and he answers all the questions, how many questions does he answer right? 

Solution I: 
If the student gets all the questions right, he'll get 5 x 20 = 100 points; however, he only gets 79 points. 
For each questions he gets wrong, he not only doesn't have the 5 points but on top of that, he gets 2 points deducted, thus it's minus 7 total. 

(100 - 79) / 7 = 3 so the student gets 3 wrong and 17 questions right. 
Check if that's correct: 17 x 5 - 2 x 3 = 79

Solution II:  
Let the student get x questions right, and that means he gets (20 - x) questions wrong.
5x - 2 (20 -x) = 79;    5x - 40 + 2x = 79;   7 x = 119 so x = 17

Question #3: My piggy bank contains only nickels, dimes, and quarters, and contains 20 coins worth a total of $3.30. If the total value of the quarters is five times the total value of all the other coins, how many dimes are in the piggy bank?

Solution:  
According to the given: 
N + D + Q = 20  --- equation 1
The total value of the quarters is 5 times the total value of all the other coins, which mean that if the total value of N + D = x, the total value of Q is 5 x
x + 5x = 330 (Make sure to use the same unit) so x = 55 (cents)
5x = 5 x 55 = 275 and 275/ 25 = 11 so there are 11 quarters

Plug in equation 1 and you have N + D = 20 - 11 = 9
Since 5N + 10D = x = 55, D and N could be
D        N
5         1
4         3
3         5
2         7... etc   2 + 7 = 9 so there are 2 dimes (7 nickels and 11 quarters)


Other applicable problems: (answer key below)
#1: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number of nickels, then how many of each does he have?

#2:  Kim has $1.65 in nickels, dimes and quarters. She has 10 coins all together and the number of quarters is equal to the number of nickels and dimes combined. Haw many of each coin does she have?

#3:  A collection of 24 nickels, dimes, and quarters is worth $3.20. There are seven more nickels than dimes. How many of each are there in the collection?

#4: Continue with #3, if there are seven more dimes than nickels, how many of each are there in the collection? 











Answer key: 
#1: 6 quarters, 20 nickels and 15 dimes  
#2: 2 nickels, 3 dimes and 5 quarters 
#3: 9 quarters, 4 dimes and 11 nickels
#4: 7 quarters, 12 dimes and 5 nickels

Geometry : Dimensional Target type questions

Check out Mathcounts, the best middle school competition math up to the national level.

Questions: (Solutions below)

#1: A circle is circumscribed around a square and another circle is inscribed in the same square. Find the ratio of the area of the smaller circle to the area of the larger circle. Express your answer as a common fraction.

#2: A square pyramid has a base edge of 32 inches and an altitude of 1 foot. A square pyramid whose altitude is one-fourth of the original altitude is cut from the vertex. The volume of the remaining frustrum is what fractional part of the volume of the original pyramid?

#3: This question is similar to one SAT question that most of my high school students have problem with. 

  























Solutions:
The two questions above are very similar. They are both related to dimension change, and appear very often on competition math. Just remember that for any two figures, the ratio of their areas is simply (ratio of base * ratio of height), and the ratio of their volumes is (ratio of base area * ratio of height). In cases where the ratio of all the sides are the same - that is, when you're dealing with similar figures - the ratio of the areas is just (ratio of side)²2, whereas the ratio of the volumes is just (ratio of side)³.

#1: The ratio of the two radii is 1 to √ 2   (45-45-90 degree right triangle), so the area ratio is 
(1 over √ 2 )² = 1/2.

#2 : Since the two pyramids are similar, and you know that the height of the new smaller pyramid is 1/4 of the old height, the volume must be (1/4)³ = 1/64 of the original pyramid.
So the volume of the frustum is (1-1/64) = 63/64 of the original shape.

#3: Since the radius is a linear relationship. R1 to r2 = 1 : 2 (given), the area ratio is 1 : 4 (square both ratio)The larger measure is double the smaller one, so once you know the area of ABC, you can get the area of DEF by multiplying the area of ABC by 4 x 2 = 8 so 5 x 8 = 40 square units.

Similar Triangles II

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Weird but Delicious Math Questions

Check out Mathcounts here--the best competition math for middle school mathletes.

Problem: (Solution below)
#1: 1993 Mathcounts National Team Round #4 :The teacher whispers positive integer A to Anna, B to Brett, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says: "I know that Brett and Chris have different numbers." Then Brett says: "I already knew that all three of our numbers were different." Finally, Chris announces: "Now I know all three of our numbers." What is the product ABC?

#2: 2000 AMC10 #22: One morning each member of Angela’s family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

Solution: 

#1: For Anna to know that Brett and Chris have different numbers, she must have an odd number
because 14 - odd = odd and you can only get odd sum of two numbers if they are different, one odd
one even.

For Brett to say that he already know all three have different numbers, he not only must have an odd number, but also his number has to be larger or equal to 7 and is not the same as what A have.

Otherwise, A-B-C could be 1-1-12; 3-3-8; or 5-5-4.
It would exceed 14 if you have 7-7-__. (All numbers are positive) so Brett"s and Anna's numbers must
be different.

If Brett has 7, then the numbers could be A-B-C = 1-7-6 ; 3-7-4 or 5-7-2.

If Brett has 9, then the numbers could be A-B-C = 1-9-4; or 3-9-2.

If Brett has 11, then then numbers could be A-B-C = 1-11-2.

From the above possibilities you know Chris has to have 6 for him to be sure he knows all the numbers. 

So A-B-C = 1-7-6 and the product of ABC = 1 x 7 x 6 = 42

#2: Let there be m cups of mild, c cups of coffee. and x people in Angela's family.
According to the given, you can set up the following equation: 
  m + c = = 8. Two ways to solve this equation.

Solution I:

m + c =   Both sides times 12 x

Both m and c need to be positive so the only x that works is when x = 5.

Solution II: 

m + c  = 8, Both sides times 12 and you have 

3m + 2c = 96; m + c is a multiple of 8.

30      3

28      6

26      9

.

.

.

2       45;  from  (30 + 3 ) = 33 to (2 +45) = 47 only 40 is a multiple of 8

and 40 divided by 8 = 5 so 5 is the answer.

 

Mental Math Tricks I

Multiples of 11:

11 x 14 = 1 ___ 4 , the middle number is (1 + 4) so the answer is 154

Let's try a few calculations mentally.

11 x 33 = 3 __ 3, the middle number is ( 3 + 3 ) so the answer is 363

11 x 52 = 5 __ 2, the middle number is ( 5+2 ) so the answer is 572

11 x 27 = 2 __ 7 , the middle number is ( 2 + 7 ) so the answer is 297

What about if the sum of the middle number is > 9? Then you carry over the "1" to the digit on its left
as you are doing the addition.

Examples:

11 x 67 = 6 __ 7, the middle number is (6 + 7) = 13 so the answer is 737

11 x 89 = 8 __ 9, the middle number is (8 + 9) = 17 so the answer is 979

11 x 47 = 4 __ 7, the middle number is (4 + 7) = 11 so the answer is 517

Harder trick:

11 x 223 = ?  2 _ _3, Write the two digits on the far left and right. Now add two numbers together,
starting with the unit digit. When the sum is  larger than 9,carry over to the next digit to the left.
2 _5(3+2) 3;  2 4 (2 + 2) 53 so the answer is 2453.

11 x 40532 = 4_ _ _ _ 2; 4_ _ _ 5 (2 +3) 2;  4_ _ 8(5+3)52;  4_ 5(0+5)852;  44(4 +0)5852,
so the answer is 445852  (This is fun!!)

Other interesting pattern:

11 x 11 =  121
111 x 111 = 12321
1111 x 1111 = 1234321
etc… till  111111111 x 111111111 = 12345678987654321

Divisibility rules for 11: 

The difference  of the sum of alternative digits is a multiple of 11, including “0”                                               (0 x 11 = 0, a multiple of 11.)

Example:

61985   (6 + 9 + 5) – (1 + 8) = 11   The number is divisible by 11.

7469     (7 + 6) – (4 + 9) = 0   The number is divisible by 11.

Try these mentally:(Answers below.)

#1: 11 x 23 =

#2: 11 x 72 =

#3: 11 x 97 =

#4: 11 x 55 =

#5: 11 x 76 =

#6: 11 x 60 =

#7: Sum of the first multiples of 11 smaller than 150.

#8:11 x 3421 =

#9: 11 x 452360 =

#10:  11 x 204673=

#11: 11111 x 11111=

#12: 1111111 x 1111111=

#13: What is the sum of the digits of 11111111 x 11111111?

#14: What is n if 45732n is divisible by 11?

#15: How many solutions for distinct numbers A and B if 4A8B is divisible by 11?

#16: How many solutions for distinct numbers A and B if A7B2 is divisible by 11?  What is their sum?

#17: How many solutions for distinct numbers A and B if A3B41 is divisible by 11?  

Answer key:
#1: 253
#2: 792
#3: 1067
#4: 605
#5: 836
#6: 660
#7 1001 The easiest way to do this is to see that this is an arithmetic sequence, starting with 11, 22, 33...
143 (11 x 13). There are 13 terms and the median is 77 so 13 x 77 or 13 x 7 x 11 = 1001
To do Mathcounts well, you need to know 7 x 11 x 13 =1001 by heart.
#8: 37631
#9: 4975960
#10: 2251403
#11:  123454321
#12:  1234567654321
#13:  The number is 123456787654321 so the sum of the digit is (4 x 7) x 2 + 8 = 64
#14:  n = 5
#15:  A + B = 12 (9, 3); (8, 4); (7, 5); skip (6, 6) because A and B are distinct (5, 7); (4, 8), (3, 9)
#16:  90, 81, 72, 63, 54, 45, 36, 27, 18;  A can’t be “0” so there are 9 pairs and the numbers are equally spaced, an arithmetic sequence. Thus the sum is median times how many numbers.                                      54 x 9 = 486
#17: A + B + 1 = 7 so A + B = 6 ; There are (6, 0);(5, 1); (4, 2); (2, 4); (1, 5) 
A + B + 1 = 7 + 11 = 18; A + B = 17; There are (9, 8) and (8, 9) so total 7 pairs