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
Saturday, November 3, 2012
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.
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