Puzzles & Brain Teasers (www. SAT math.com) Oct.28 The Hare and the Tortoise Haretown and Tortoiseville are 96 miles apart. A hare travels at 9 miles per hour from Haretown to Tortoiseville, while a tortoise travels at 3 miles per hour from Tortoiseville to Haretown. If both set out at the same time, how many miles will the hare have to travel before meeting the tortoise en route? Answer 72 Solution: The hare and the tortoise are together covering the distance at 12 miles per hour (i.e., on adding their speeds). So, they will cover the distance of 96 miles in 8 hours. Thus, in 8 hours, they will meet and the hare will have traveled 72 miles. Alternative Solution through Equations: Note that : Distance = Speed × Time Let t be the time before the hare and the tortoise meet.
Transcript
Puzzles & Brain Teasers (www. SAT math.com) Oct.28
The Hare and the Tortoise
Haretown and Tortoiseville are 96 miles apart. A hare travels at 9 miles per hour from Haretown to Tortoiseville, while a tortoise travels at 3 miles per hour from Tortoiseville to Haretown.
If both set out at the same time, how many miles will the hare have to travel before meeting the tortoise en route?
Answer 72
Solution:
The hare and the tortoise are together covering the distance at 12 miles per hour (i.e., on adding their speeds). So, they will cover the distance of 96 miles in 8 hours. Thus, in 8 hours, they will meet and the hare will have traveled 72 miles.
Alternative Solution through Equations:
Note that : Distance = Speed × Time
Let t be the time before the hare and the tortoise meet. In t hours, the hare will travel 9 t miles. In t hours, the tortoise will travel 3 t miles.
Now, 9 t + 3 t = 96 So, t = 96 ⁄ 12 = 8 hours.
Thus, distance traveled by hare before meeting = 9 × 8 = 72 miles.
PROBLEMS ON VARIATION(//www.kwiznet.com/p/takeQuiz.php) Oct. 29,09
Q 1: A farmer has enough grain to feed 50 cattle for 10 days. He sells 10 cattle. For how many days will the grain last now?
55 days
50 days
40 days
Q 2: In an army camp, there is food for 8 weeks for 1200 people. After 3 weeks, 300 more soldiers joined the camp. For how many more weeks will the food last?
another 4 weeks
another 3 weeks
another 5 weeks
Q 3: The groceries in a home of 4 members are enough for 30 days. If a guest comes and stays with them, how many days will the groceries last?
24 days
26 days
28 days
Q 4: It is known that current (A) in an electric circuit is inversely proportional to the resistance (R) in the circuit. When the resistance is 3 ohms, the current is 2 amperes. Find the resistance if the current is 5 amperes; and find the current when the r
1.2 ohms and 1.2 amperes respectively
2 ohms and 1 amperes respectively
1 ohms and 1 amperes respectively
Q 5: If y varies inversely as x; and x=6 when y=-3, find y when x=-9.
6
2
3
Q 6: 24 people can construct a house in 15 days. But the owner would like to finish the work in 12 days. How many more workers should he employ?
8
7
6
Q 7: Some people working at the rate of 6 hours a day can complete the work in 19 1/2 days. As they have received another contract, they want to finish this work early. Now they start working 6 1/2 hours a day. In how many days will they finish this work?
18 days
16 days
20 days
Q 8: 35 carts are needed to transport the entire grain in a warehouse at the rate of 8 bags a truck. But the truck drivers objected saying it was too heavy and carried only 7 bags a truck. How many more trucks are needed to transport the grain?
4
9
5
Direct Proportionality: synchronized, one-way relationship between items a) as the number of items in a backpack increases, the weight of the backpack increases
b) as the population of a city decreases, the amount of tax money collected decreases
Inverse Proportionality: synchronized, opposite relationship between items a) as the number of workers increases, the time needed to complete a project decreases b) as the amount of available food decreases, the number of hungry people increases
SAMPLE PRACTICE QUESTION FOR PROPORTIONS: SAT, PSAT, ACT, GRE, GMAT, SSAT
1. If s is inversely proportional to y2 and s = 1/4 when y = 1/2, then what is the non-negative value of y when s = 4?
Solution A) First, find the change in s. s increased from 1/4 to 4, a 16-fold increase in s. s is inversely proportional to y2, so y2 must decrease by a factor of 16.
B) Next, find y2. original y = 1/2, so original y2 = 1/4 original y2 = 1/4 --> new y2 = (1/4) ÷ 16 = 1/64
C) Finally, solve for new y. new y2 = 1/64 new y = √1/64 = 1/8
PRACTICE QUESTIONS FOR PROPORTIONS: SAT, PSAT, ACT, GRE, GMAT, SSAT
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1. If m is directly proportional to x and m = 2 when x = 12, then what is the value of x when m = 2-1?
2. If x and y are inversely proportional and x = 6 when y = 1/4, what is the value of x when y = 1/12?
3. If s and p2 are inversely proportional and s = 9 when p2 = 6, what is the value of p when s = 6?
Solving proportions is simply a matter of stating the ratios as fractions, setting the two fractions equal to each other, cross-multiplying, and solving the resulting equation. You'll probably start out by just solving proportions, like this:
Find the unknown value in the proportion: 2 : x = 3 : 9.
2 : x = 3 : 9
First, I convert the colon-based odds-notation ratios to fractional form:
.
Then I solve the proportion:
9(2) = x(3) 18 = 3x 6 = x
Find the unknown value in the proportion: (2x + 1) : 2 = (x + 2) : 5
(2x + 1) : 2 = (x + 2) : 5
First, I convert the colon-based odds-notation ratios to fractional form:
Once you've solved a few proportions, you'll likely then move into word problems where you'll first have to invent the proportion, extracting it from the word problem, before solving it.
If twelve inches correspond to 30.48 centimeters, how many centimeters are there in thirty inches?
I will set up my ratios with "inches" on top, and will use "c" to stand for the number of centimeters for which they've asked me.
12c = (30)(30.48) 12c = 914.4 c = 76.2
Thirty inches corresponds to 76.2 cm.
I could have used any letter I liked for my variable. I chose to use "c" because this will help me to remember what the variable is representing. An "x" would only tell me that I'm looking for "some unknown value"; a "c" can remind me that I'm looking for "centimeters". Warning: Don't fall into the trap of feeling like you "have" to use "x" for everything. You can use whatever you find most helpful.
A metal bar ten feet long weighs 128 pounds. What is the weight of a similar bar that is two feet four inches long?
First, I'll need to convert the "two feet four inches" into a feet-only measurement. Since four inches is four-twelfths, or one-third, of a foot, then:
2 feet + 4 inches = 2 feet + 1/3 feet = 7/3 feet
I will set up my ratios with the length values on top, set up my proportion, and then solve for the required weight value:
Since this is a "real world" word problem, I should probably round or decimalize my exact fractional solution to get a practical "real world" sort of number.
The bar will weigh 448/15 , or about 29.87, pounds.
The tax on a property with an assessed value of $70 000 is $1 100. What is the assessed value of a property if the tax is $1 400?
I will set up my ratios with the assessed valuation on top, and I will use "v" to stand for the value that I need to find. Then:
One piece of pipe 21 meters long is to be cut into two pieces, with the lengths of the pieces being in a 2 : 5 ratio. What are the lengths of the pieces?
I'll label the length of the short piece as "x". Then the long piece, being the total piece less what was cut off for the short piece, must have a length of 21 – x.
Referring back to my set-up for my equation, I see that I defined "x" to stand for the length of the shorter piece. Then the length of the longer piece is given by:
21 – x = 21 – 6 = 15
Now that I've found both required values, I can give my answer:
The two pieces have lengths of 6 meters and 15 meters.
In the last exercise above, if I had not defined what I was using "x" to stand for, I could easily have overlooked the fact that "x = 6" was not the answer the exercise was wanting. Try always to clearly define and label your variables. Also, be sure to go back and re-check the word problem for what it actually wants. This exercise did not ask me to find "the value of a variable" or "the length of the shorter piece". By re-checking the original exercise, I was able to provide an appropriate response, being the lengths of the two pieces, including the correct units ("meters").
A proportion is a special form of an algebra equation. It is used to compare two ratios or make equivalent fractions. A ratio is a comparison between two values. Such as the following:
1 apple: 3 oranges
This ratio compares apples to oranges. It means for every apple there are 3 oranges.
A proportion will help you solve problems like the one below.
Jane has a box of apples and oranges in the ratio of 2:3. If she has six apples, how many oranges does she have?
Before we begin to set up proportions for a word problem, we will concentrate on solving proportions. Remember, a proportion is a comparison between two ratios. The proportion shown below compares two ratios which are in the fraction form.
The four parts of the proportion are separated into two groups, the means and the extremes, based on their arrangement in the proportion. Reading from left-to-right and top-to-bottom, the extremes are the very first number, and the very last number. This can be remembered because they are at the extreme beginning and the extreme end. Reading from left-to-right and top-to-bottom, the means are the second and third numbers. Remembering that "mean" is a type of average may help you remember that the means of a proportion are "in the middle" when reading left-to-right, top-to-bottom. Both the means and the extremes are illustrated below.
Ratio and Proportion (http://www.mathleague.com/help/ratio/ratio.htm) Oct. 29, 09
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Ratio
A ratio is a comparison of two numbers. We generally separate the two numbers in the ratio with a colon (:). Suppose we want to write the ratio of 8 and 12.We can write this as 8:12 or as a fraction 8/12, and we say the ratio is eight to twelve.
Examples:
Jeannine has a bag with 3 videocassettes, 4 marbles, 7 books, and 1 orange.
1) What is the ratio of books to marbles?Expressed as a fraction, with the numerator equal to the first quantity and the
denominator equal to the second, the answer would be 7/4.Two other ways of writing the ratio are 7 to 4, and 7:4.
2) What is the ratio of videocassettes to the total number of items in the bag?There are 3 videocassettes, and 3 + 4 + 7 + 1 = 15 items total. The answer can be expressed as 3/15, 3 to 15, or 3:15.
Comparing Ratios
To compare ratios, write them as fractions. The ratios are equal if they are equal when written as fractions.
Example:
Are the ratios 3 to 4 and 6:8 equal? The ratios are equal if 3/4 = 6/8.These are equal if their cross products are equal; that is, if 3 × 8 = 4 × 6. Since both of these products equal 24, the answer is yes, the ratios are equal.
Remember to be careful! Order matters!A ratio of 1:7 is not the same as a ratio of 7:1.
Examples:
Are the ratios 7:1 and 4:81 equal? No! 7/1 > 1, but 4/81 < 1, so the ratios can't be equal.
Are 7:14 and 36:72 equal?Notice that 7/14 and 36/72 are both equal to 1/2, so the two ratios are equal.
Proportion
A proportion is an equation with a ratio on each side. It is a statement that two ratios are equal. 3/4 = 6/8 is an example of a proportion.
When one of the four numbers in a proportion is unknown, cross products may be used to find the unknown number. This is called solving the proportion. Question marks or letters are frequently used in place of the unknown number.
Example:
Solve for n: 1/2 = n/4.Using cross products we see that 2 × n = 1 × 4 =4, so 2 × n = 4. Dividing both sides by 2, n = 4 ÷ 2 so that n = 2.
Rate
A rate is a ratio that expresses how long it takes to do something, such as traveling a certain distance. To walk 3 kilometers in one hour is to walk at the rate of 3 km/h. The fraction expressing a rate has units of distance in the numerator and units of time in the denominator.Problems involving rates typically involve setting two ratios equal to each other and solving for an unknown quantity, that is, solving a proportion.
Example:
Juan runs 4 km in 30 minutes. At that rate, how far could he run in 45 minutes?Give the unknown quantity the name n. In this case, n is the number of km Juan could run in 45 minutes at the given rate. We know that running 4 km in 30 minutes is the same as running n km in 45 minutes; that is, the rates are the same. So we have the proportion 4km/30min = n km/45min, or 4/30 = n/45. Finding the cross products and setting them equal, we get 30 × n = 4 × 45, or 30 × n = 180. Dividing both sides by 30, we find that n = 180 ÷ 30 = 6 and the answer is 6 km.
Converting rates
We compare rates just as we compare ratios, by cross multiplying. When comparing rates, always check to see which units of measurement are being used. For instance, 3 kilometers per hour is very different from 3 meters per hour! 3 kilometers/hour = 3 kilometers/hour × 1000 meters/1 kilometer = 3000 meters/hourbecause 1 kilometer equals 1000 meters; we "cancel" the kilometers in converting to the units of meters.
Important:
One of the most useful tips in solving any math or science problem is to always write out the units when multiplying, dividing, or converting from one unit to another.
Example:
If Juan runs 4 km in 30 minutes, how many hours will it take him to run 1 km?Be careful not to confuse the units of measurement. While Juan's rate of speed is given in terms of minutes, the question is posed in terms of hours. Only one of these units may be used in setting up a proportion. To convert to hours, multiply 4 km/30 minutes × 60 minutes/1 hour = 8 km/1 hourNow, let n be the number of hours it takes Juan to run 1 km. Then running 8 km in 1 hour is the same as running 1 km in n hours. Solving the proportion,8 km/1 hour = 1 km/n hours, we have 8 × n = 1, so n = 1/8.
Average Rate of Speed
The average rate of speed for a trip is the total distance traveled divided by the total time of the trip.
Example:
A dog walks 8 km at 4 km per hour, then chases a rabbit for 2 km at 20 km per hour. What is the dog's average rate of speed for the distance he traveled?The total distance traveled is 8 + 2 = 10 km.Now we must figure the total time he was traveling.For the first part of the trip, he walked for 8 ÷ 4 = 2 hours. He chased the rabbit for 2 ÷ 20 = 0.1 hour. The total time for the trip is 2 + 0.1 = 2.1 hours.The average rate of speed for his trip is 10/2.1 = 100/21 kilometers per hour.
SOLVING VARIATION PROBLEMS (http://hubpages.com/hub/SOLVING-VARIATION--PROBLEMS) Oct 29,0963rate or flag this page
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SOLVING VARIATION PROBLEMS
DIRECT VARIATION
It is a special function which can be expressed as the equation y = kx where k is a constant. The equation y = kx is read "y varies directly as x" or "y is proportional to x. The constant k is called the the constant of variation or constant of proportionality.
Illustration Number One :
The circumference ( C ) of a circle varies directly as the diameter ( d ). The direct variation is written as C = ∏ d . The constant of variation is ∏ .
Illustration Number Two :
A teacher makes $10 per hour. The total wage of the teacher is directly proportional to the number of hours (h ) worked. The equation of variation is W = 10h. The constant of proportionality is 10.
A direct variation equation can also be written in the form y = kx^n, where n is a positive number. For example the equation y = kx^2 is read " y varies directly as the square of x..
Illustration Number Three :
The area ( A ) of a circle varies directly as the square of a radius ( r ) of the circle.
The direct variation equation is A = ∏ r ^2. The constant of variation is ∏.
Sample Problem Number One :
Given that V varies directly as r and that V = 50 when r = 5 , find the constant of variation and the equation of variation.
First, write the basic direct variation equation: V = kr
Then, replace V and r by the given values : 50 = k * 5
Solve for k : (1/5) 50 = 5k (1/5 ==è k = 10
Write the direct variation equation by substituting the values of k into the basic direct variation equation: V = 10r
Sample Problem Number Two :
The tension ( T ) in a spring varies directly as the distance ( x ) it is stretched.
If T = 20 lbs. when x = 5 inches. Find T when x = 10 inches.
Write the basic direct variation equation : T = kx
Replace T and x by the given value then solve for k :
20 = 5 * k =è (1/5) 20 = 5k ( 1/5) ===è k = 4
Write the direct variation equation by substituting the value of k into the basic direct
Variation equation : T = 4x
To find T when x = 10 inches. Substitute 10 for x in the equation to solve for T.
T = 4 (10) = 40 lbs.
INVERSE VARIATION
It is a function which can be expressed as the equation y = k/x where k is a constant. The equation y = k/x is read " y varies inversely as x" or "y is inversely proportional to x ." In general, an inverse variation equation can be written y = k/x^n where n is a positive number .
Illustration Number Four :
The equation y = k/x^2 is read "y varies inversely as the square of x."
Given that P varies inversely as the square of x and that P = 10 when x = 2,
Find the variation constant and the equation of variation :.
Set the inverse variation equation : P = k/x^2
Substitute the given values to corresponding variables in the equation and solve for k :
10 = k/2^2 ==è ( 4 ) 10 = k/4 (4) =è k = 40
The constant of variation is 40. The inverse variation equation is P = 40/x^2
Sample Problem Number Three :
The length ( L ) of a rectangle with fixed area is inversely proportional to the width.
If L = 10 W = 4, find the length when w = 7.
Write inverse variation equation : L = k/W
Substitute the given values to the equation and solve for k :
10 = k/4 =è k = 40
L = 40/7 or L = 5 and 5/7
JOINT VARIATION
It is a variation wherein a variable varies directly as the product of two or more
other variables. A joint variation can be expressed as the equation Z = kXY, where K is a constant . The equation is read as "Z varies jointly as X and Y.
Illustration Number Five :
The area ( A ) of a triangle varies jointly as the base and the height. The joint variation equation is written as A = ½ bh. The constant of variation is ½.
COMBINED VARIATION
It is a variation wherein two or more types of variation occurs at the same time.
For example in Physics, the volume ( V ) of a gas varies directly as the temperature ( T ) and inversely as the pressure ( P ). The combind variation equation is written as
V = k T/ P
Sample Problem Number Four :
The pressure P of a gas varies directly as the temperature T and inversely as the volume ( V ). When T = 50 degrees and V = 200 in ^3 P = 30lb/in. Find the pressure of a gas when T = 70 degrees and V = 300 in^3.
Write first the basic combined variation equation : P = kT/V
Replace the variables by the given values then solve for k :
