Mental Math Practices: All questions can be solved in your head reasonably fast (less than 5 minutes) if you understand the concepts well.
Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.
1. What is 12.5% of 128?
2. After 20% discount, you paid $144? What is the original price?
3. To make 20% profit, you charge your customer 144 dollars. What is the wholesale price?
4. You lost 20% in the stock market. What percentage increase will you break even?
5. A : B = 2 : 5 and A is 75 more than B, what is B?
6. A : B = 2 : 3 and A + B = 14, what is A and what is B?
7. A : B = 3 : 5 and B : C = 2 to 9, what is A : B?
8. Two triangles are similar and the length of their base ratio is 1: 2. The smaller triangle has a base that equals 5 and an area that is 110. What is the area of the larger triangle?
9. One side of the square increases 10% and the other side decreases 20 percent. What is the percentage change? Increase or decrease? By how much?
10. There are 60 students in a gym and 30% are boys. After a few more boys enter the gym, now 40% are boys. How many boys enter the room?
Friday, November 16, 2012
Saturday, November 3, 2012
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
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
Sunday, August 5, 2012
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.Questions: (Solutions below)
#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.
Wednesday, July 4, 2012
Sunday, June 24, 2012
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?
#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.
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.
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.
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:
Solution I:
Both m and c need to be positive so the only x that works is when x = 5.
Solution II:
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.
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