How Old Are They?

Mary’s and Bob’s ages combined are twice Jane’s age. Mary is 8 years older than Bob. Jane’s age plus Bob’s age is 20 years. How old are they?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let b,j, and m represent the ages of Bob, Jane, and Mary respectively.
b+m=2j\qquad (1)
m-b=8\qquad\; (2)
b+j=20\qquad\; (3)
Add Eq. (1) and Eq. (2)
b+m=2j
m-b=8
——————
2m=2j+8
m=j+4\qquad (4)
Add Eq. (2) and Eq. (3)
m-b=8
b+j=20
——————
m+j=28\quad\;\; (5)
Substitute the value of m in Eq. (4)  into Eq. (5)
(j+4)+j=28
2j=24
j=12
Jane’s age equals 12.
Substitute the value of Jane’s age into Eq. (4)
m=12+4=16
Mary’s age equals 16.
Substitute the value of Jane’s age into Eq. (3)
b+12=20
b=8
Bob’s age equals 8.

Answer: Bob is 8, Jane 12, Mary 16

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Arc of a Circle

\overline{AB} is a diameter of a circle of radius 1 unit. \overline{CD} is a chord perpendicular to \overline{AB} that cuts \overline{AB} at E. If the arc CAD is 2/3 of the circumference of the circle, what is the length of the segment \overline{AE}?
Source: NCTM Mathematics Teacher, September 2006

Solution
image
Circumference = 2\pi(r)=2\pi(1)=2\pi. Since measure of major arc CAD=2/3(\mathrm{circumference}), measure of minor arc CBD=1/3(\mathrm{circumference})=2\pi/3. Central angle COD which tends arc CBD therefore measures 2\pi/3 radians or 120^\circ. Triangle COD is isosceles, hence altitude OE is also the angle bisector of central angle COD. Triangle OEC is a 30^\circ\!\textrm{-}60^\circ\!\textrm{-}90^\circ triangle with hypotenuse OC equal 1. Thus, side \overline{OE} has length equal 1/2.
AE=AO+OE=1+1/2=3/2 units

Answer: 3/2 units

Alternative solution
Since chord \overline{AB} is a diameter and chord \overline{CD} is perpendicular to \overline{AB}, arc AD is a mirror image of arc AC in the \overleftrightarrow{AB} line of reflection. Points A,C, and D divide the circumference into three equal arcs each measuring 120^\circ. Hence, ACD is an equilateral triangle where the circumcenter O is also the centroid of the triangle.
By the Concurrency of Medians of a Triangle theorem
AO=(2/3)AE
AE=(3/2)AO=(3/2)(1)=3/2

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Multiplicative Inverse

If \alpha=2+i, then in the form of a+bi, what does \alpha^{-1} equal?
Source: NCTM Mathematics Teacher, September 2006

Solution
By definition of multiplication inverse
\alpha\alpha^{-1}=1
\alpha^{-1}=\dfrac{1}{2+i}
=\dfrac{1}{2+i}\times\dfrac{2-i}{2-i}
=\dfrac{2-i}{2^2-i^2}
=\dfrac{2-i}{4+1}
=\dfrac{2-i}{5}
=\dfrac{2}{5}-\dfrac{1}{5}i

Answer: \dfrac{2}{5}-\dfrac{1}{5}i

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Second Number Larger Than The First

Suppose a six-sided die has sides numbered 1 through 6. If a person throws the die two times, what is the probability that the second number will be larger than the first?
Source: NCTM Mathematics Teacher, September 2006

Solution
The desirable outcomes are
1:\left \{2,3,4,5,6\right \} where 1 is the first number and 2,3,4,5,6 are the second number
2:\left \{3,4,5,6\right \}
3:\left \{4,5,6\right \}
4:\left \{5,6\right \}
5:\left \{6\right \}
6:\left \{\textrm{not possible}\right\}
Total number of desirable outcomes
5+4+3+2+1=15
Number of possible outcomes
6\times 6=36
Probability that the second number is larger than the first = 15/36

Answer: 15/36

Alternative solution
First, the die is thrown and the first number \left \{1,2,3,4,5,6\right \} comes up each with a probability equal 1/6. Then, the die is thrown again and we want the second number larger than the first.
1:\left \{2,3,4,5,6\right \} probability = 5/6
2:\left \{3,4,5,6\right \} probability = 4/6
3:\left \{4,5,6\right \} probability = 3/6
4:\left \{5,6\right \} probability =2/6
5:\left \{6\right \} probability = 1/6
6:\left \{\textrm{not possible}\right \} probability = 0/6
Probability of rolling the die two times and the second number is larger than the first
\dfrac{1}{6}\times \dfrac{5}{6}+\dfrac{1}{6}\times \dfrac{4}{6}+\dfrac{1}{6}\times \dfrac{3}{6}+\dfrac{1}{6}\times \dfrac{2}{6}+\dfrac{1}{6}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{0}{6}=\dfrac{1}{6}\left (\dfrac{5+4+3+2+1+0}{6}\right )
=\dfrac{1}{6}\times \dfrac{15}{6}
=15/36

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Values of Integers

If m and n are integers such that 2m-n=3, what are the possible values of m-2n?
(a) -3 only
(b) 0 only
(c) only multiples of 3
(d) any integer
(e) none of these
Source: NCTM Mathematics Teacher, September 2006

Solution
Let x be an integer such that x=m-2n. We have the following two equations
2m-n=3\qquad\qquad (1)
m-2n=x\qquad\qquad (2)
Multiply Eq. (1) by -2 and add to Eq. (2)
-4m+2n=-6
m-2n=x
————————
-3m=x-6
Divide both sides by -3
m=-x/3+2
For m to be an integer, x must be a multiple of 3.

Answer: (c) only multiples of 3

Alternative solution
m-2n=m-2n+(3)-3
=m-2n+(2m-n)-3
=3m-3n-3
=3(m-n-1)

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Length of Side of Square

If the numerical value of the area of a square plus two times the numerical value of its perimeter is equal to 20, what is the length of one side of the square?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let a represent the length of one side of a square. The following equation is true
a^2+2(4a)=20
a^2+8a-20=0
(a-2)(a+10)=0
a=2 or a=-10 (extraneous solution)
The length of one side of the square equals 2 units.

Answer: 2 units

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Lemonade Stand

John has a lemonade stand. He sells a small lemonade for 50 cents and a large lemonade for \$1. A small serving contains 1 cup of lemonade; a large contains 1.5 cups. At the end of the day, John has made \$9 and sold 15.5 cups of lemonade. How many small and large lemonades has he sold?
Source: NCTM Mathematics Teacher, September 2006

Solution
Let x represent the number of small lemonades and y the number of large lemonades sold. We have the following system of equations
x(.50)+y(1)=9\qquad\qquad\; (1)
x(1)+y(1.5)=15.5\qquad\quad\, (2)
Multiply Eqs. (1) and (2) by 2
x+2y=18\qquad\quad\:\: (3)
2x+3y=31\qquad\quad (4)
Multiply Eq. (3) by -2 and add to Eq. (4)
-2x-4y=-36
2x+3y=31
————————-
-y=-5
y=5
Substitute the value of y=5 into Eq. (3)
x+2(5)=18
x=8
John has sold 8 small and 5 large lemonades.

Answer: 8 small and 5 large

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Red and Yellow Plums

One store sold red plums at 4 for \$1 and yellow plums at 3 for \$1. A second store sold red plums at 4 for \$1 and yellow plums at 6 for \$1. You bought m red plums and n yellow plums from each store, spending a total of \$10. How many plums in all did you buy?
Source: NCTM Mathematics Teacher, September 2006

Solution
In dollars the following equation depicts how you spent \$10 at the two stores
m(1/4)+n(1/3)+m(1/4)+n(1/6)=10
Collect like terms and simplify
m(1/2)+n(1/2)=10
m+n=20
At each store you purchase 20 plums; in all you bought 40 plums.

Answer: 40

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Value of Sum

What is the value of the following sum?
2006[(-1)^1+(-1)^2+(-1)^3+(-1)^4+\cdots+(-1)^{2006}]
Source: NCTM Mathematics Teacher, September 2006

Solution
The square bracket in the sum expression contains 1003 pairs of consecutive powers (-1)^n+(-1)^{n+1} which pairwise sum to zero
(-1)^1+(-1)^2=-1+1=0
(-1)^3+(1-)^4=-1+1=0
(-1)^5+(-1)^6=-1+1=0
\cdots
(-1)^{2005}+(-1)^{2006}=-1+1=0
2006[(-1)^1+(-1)^2+(-1)^3+\cdots+(-1)^{2006}]=2006[0+0+0+\cdots+0]=0

Answer: 0

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Biggest Pizza Slice

An 8-inch pizza is cut into 3 equal slices. A 10-inch pizza is cut into 4 equal slices. A 12-inch pizza is cut into 6 equal slices. A 14-inch pizza is cut into 8 equal slices. From which pizza would you take a slice if you want as much pizza as possible?
Source: NCTM Mathematics Teacher, September 2006

Solution
Areas of pizza slices
8-inch: \dfrac{\pi\times 4^2}{3}=\dfrac{\pi\times 16}{3}=\dfrac{\pi\times 128}{24}
10-inch: \dfrac{\pi\times 5^2}{4}=\dfrac{\pi\times 25}{4}=\dfrac{\pi\times 150}{24}
12-inch: \dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 6^2}{6}=\dfrac{\pi\times 144}{24}
14-inch: \dfrac{\pi\times 7^2}{8}=\dfrac{\pi\times 49}{8}=\dfrac{\pi\times 147}{24}
Since \dfrac{150}{24} is the largest fraction, you want to take a slice from the 10-inch pizza.

Answer: 10-inch pizza

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