Ratio, Proportions -- Beginning Problem Soving (SAT level)

Questions:
#1: At a county fair, if adults to kids ratio is 2 to 3 and there are 250 people at the fair. How many adults and how many kids? 

Solution I :  Adults to Kids = 2 : 3 (given). Let there be 2x adults and 3x kids (only when two numbers have the same multiples can you simplify the two number to relatively prime.
 2x + 3x = 250;  5x = 250; x = 50 so there are 2x or 2 * 50 = 100 adults and 3x or 3 * 50 = 150 kids.

Solution II:  From the given information, you know that every time when there are 5 parts (2 + 3 = 5), there
will be 2 parts for A, or   of the total, and 3 parts for B, or of the total.
  * 250 = 100 adults and * 250 = 150 kids

#2: In a mixture of peanuts and cashews, the ratio by weight of peanuts to cashews is 5 to 2. How many pounds of cashews will there be in 4 pounds of this mixture? (an actual SAT question)

Solution:
You can use either the first or second method (see above two solutions) but the second solution is much faster. 
* 4 = pounds

#3: Continue with question #1: How many more kids than adults go to the fair? 

Solution I: 
Use the method on #1 and then find the difference.
150 - 100 = 50 more kids.
  
Solution II: 
Using the method for #1: solution II, you know    of the all the people go to the fair are adults ,
and of the total people go to the fair are kids.
(  -   ) * 250 = 50 more kids.

#3: If the girls to boys ratio at an elementary school is 3 to 4 and there are 123 girls, how many boys are there at that elementary school? 
Solution:
This one is easy, you set up the equation. Just make sure the numbers line up nicely.
Let there be "x" boys. = . Cross multiply to get x. Or since 123 is 41 * 3;
4 * 41 = 164 boys

#4 : If the girls to boys ratio at an elementary school is 2 to 5 and there are 78 more boys than girls, how many girls are there at the elementary school? 

Solution I: 
Again, you can use the algebra and let there be 2x girls and 5x boys. 
According to the given, 5x - 2x = 3x = 78 so x = 26
Plug in and you get there are 2*26 = 52 girls and 5*26 = 130 boys.

Solution II: 
If you keep expanding the ratio using the same multiples, 2 : 5 = 4 : 10 : 6 : 15...
Do you see that the difference of boys and girls are always multiples of 3 ( 5 - 2  = 3).
  = 26 so there are 2 * 26 or 52 girls

Review: Shoestring method, Heron's formula, Pythagorean Triples and Others related to Area of a triangle or polygon

Please give me feedback if there is any error or what concepts to be included for future practice tests. Thanks a lot in advance.

Online practices on finding the area of a triangle or irregular polygons.
Click here for the timed online test. Type in your first name and only write down "number" answers.

Before trying out the online timed test, here are some reminders.

a. If the irregular polygon include the origin, then using that as a starting point for shoestring method might save some calculation time.

b. There are Pythagorean Triples hidden in lots of questions that ask for the area of a triangle given three side lengths. Thus, if you find the side length that matches the triples or multiples of the triples, it might be much easier to use that instead of heron's formula.

c. For quadrilaterals, if the diagonals are perpendicular to each other, such as rhombus, kite or others, it's much easier to use D1* D2 / 2 to get the area.

d. Sometimes breaking the polygon into different smaller parts is the easiest way to find the answer.

e. Deliberate practices are the key to steady progress.

Have fun problem solving!!

Line Intercept, Slope, Area/Geometry Questions

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This question is similar to 2010 #3 Mathcouts Chapter Team problem. 

Solution I: Using the given slope, you can find the equation for AB ,
which is: y = -2 x + b, plug in (4, 7) and you have 7 = -2 x 4 + b so b = 15
The y intercept for AB is (0, 15)
Now you have four points, (0, 0), (0, 15), (4, 7) and (15, 0).
Using shoestring method to find area of any polygon, you have the area of polygon
AEDO = 82.5 square units.

Solution II:

After finding the y intercept for AB.

You break the area into three parts.

The side of the shaded region is 4 x 7 = 28.

The area of triangle ACE = 4 * (15-7) / 2 = 16

The area of triangle
EBD = (external height)7 * (15-4) / 2 = 38.5

So the total area is 82.5 square units.




Applicable problems: Answer key below.

#1:  The slope of AB is -3 and it intercepts with CD at point E, which is (3, 3). If the x intercepts of CD is (10,0), what is the area of AEDO?

#2: The slope of AB is 2 and it intercepts with CD at point E, which is (-2, 2 ). If the x intercepts of CD is (-7,0), what is the area of AEDO?

#3:  The slope of AB is -3 and it intercepts with CD at point E, which is ( 5, 8 ). If the x intercepts of CD is (14,0), what is the area of AEDO? -- Andrew's question.

#4: The slope of AB is \(\dfrac {-3} {2}\)and it intercepts with CD at point E, which is ( 2, 7 ). If the x intercepts of CD is (16,0), what is the area of AEDO? --Daniel's question.

#5: The slope of AB is -4/5 and it intercepts with CD at point E, which is (5, 2 ). If the x intercepts of CD is (12,0), what is the area of AEDO?












Answer key: 
#1: 33
#2: 13
#3: 113.5
#4: 66
#5: 27

Trianges That Share the Same Vertex/Similar Triangles

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Download this year's Mathcounts handbook here.

Besides similar triangles, "triangles that share the same vertex" also appears regularly on geometry questions. 

Look , for example, at the left image. 
It's easy to see AD being the height to base CE for Δ AEC.

However, it's much harder to see the same AD being the external height of  Δ ABC if BC is the base.





Question to ponder based on the above image:
#1: If  BC : CE = 2 to 5 and the area of Δ ABE is 98, what is the area of Δ ABC and Δ ACE?

Solution:
Since both triangles share the same height AD, if you use BC and CE  as the bases, the area ratio stays constant as 2 to 5.  Thus the area of Δ ABC = (2/7) * 98 = 28 and the area of
Δ AEC = (5/7) *98 = 70.

Knowing the above concept would help you solve the ostensibly hard trapezoid question.

Question #2: If ABCD is a trapezoid where AB is parallel to CD,  AB = 12 units and CD = 18 units.
If the area of  Δ AEB = 60 square units, what is the area of ΔCED, Δ AED and ΔBEC?






Solution:  Δ AEB and ΔCED are similar (Why? Make sure you understand this?) Thus if AB to CD = 12 to 18 or 2 to 3 ratio (given), the ratio of the area of the two similar triangles is 2² to 3² = 4 to 9 ratio.
The area of  Δ AEB = 60 square units (given), so the area of ΔCED is 9 * (60/4) = 135 square units. 
Δ AEB and Δ AED share the same vertex. The height is the same with base BE and ED.
Thus the area ratio is still 2 to 3. The area of Δ AEB = 60 square units (given),
 the area of Δ AED = 3 * (60/2) = 90 square units.

Using the same reasoning you get the area of ΔBEC = 90 square units. [Keep in mind that Δ AED and ΔBEC don't have to be congruent, but they do have the same area.]

Similar Triangles I

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

There are numerous questions on similar triangles at competition math.
Practice how to spot them and use them to solve tougher chapter, median hard state questions.

For example, There are three similar triangles in this image.

Triangle ABC is similar to triangle BDC and triangle ADB.

Using consistent symbols will help you set up the right proportion comparisons much easier.

BD² = AD x DC
BA² = AD x AC
BC² = CD x AC  

All because of the similarities.
That is also where those ratios work.











#1: What is the height to the hypotenuse of a 3-4-5 right triangle? 
Solution: 
 The area of a triangle is base x height over 2.
 Area of a 3-4-5 right triangle  = (3 x 4) / 2 = (5 x height to the hypotenuse) / 2
 Both sides times 2 : (3 x 4) = 5 x height to the hypotenuse
 The height to the hypotenuse = 12/5 

The same goes with 6-8-10 right triangle, the height to the hypotenuse is 4.8.
It's very straightforward so make sure you know how and why that works.

Hope this helps.