Jason is Superior to Everyone Else

Definitions

100

System of Equations

2 or more equations that are related.

200

Elimination

Adding or Subtracting 2 equations to remove a variable.

300

Substitution

Substituting the 2 equations into each other in order to remove a variable

400

Ratio

The relationship between parts of one thing to parts of another thing.

500

Rate

The amount of something per amount something else

Substitution

100

x + y = 6

2x + 4y = 18

You could probably solve another way and we wouldn't know. But good sportsmanship? :)

x = 3

y = 3

200

Solve the following system using the substitution method.

2y=12x+8

3x+4y=70

Divide the first equation by 2 in order to solve for y. y=6x+4. 

Substitute 6x+4 into y for the second equation. 

3x+4(6x+4)=120. 3x+24x+16=120. 27x+16=70.

27x=54. Divide by 27 on both sides. x=2. 

Plug 2 into x for y=6x+4. 6(2)+4=16, so y=16. The solution is (2,16).

300

Which of the following must be the value of x such that 2x/3+2y=6 and y+16/y=4?

(A) 0 (B) -2 (C) 3 (D) -3

Multiply the first equation by 3 to get rid of the fraction. 2x+6y=18.

Multiply the second equation by y to get rid of the y in the denominator. y^2+16=4y. To make the right side of the equation equal to 0, subtract 3y on both sides. y^2-4y+16=0.

Factor the equation. (y-4)^2=0. y must equal 4.

Substitute y=4 into the first equation. 2x+6(4)=18.

2x+24=18. Subtract 24 on both sides. 2x=-6. Divide by 2 on both sides. x=-3. 

400

Which of the following values of a would result in no solutions for the system 

9x-14y=-3, 2x-ay=-6 

(A) -9/14 (B) -28/9 (C) 9/14 (D) 28/9

A linear system of equations with no solutions would result in two distinct lines with the same slope.

If you put both equations into slope-intercept form, you get y=9/14x+3/14 and y=2/ax+6/a. If the equations need the same slope, 9/14=2/a. Multiply both sides by a. 9a/14=2. 2÷9/14=2*14/9. a=28/9.

500

Let (a,b) be some point in the coordinate plane such that y≤ -15x+3000 and y≤5x. What is the maximum possible value of b?

Substitute 5x for y. 5x≤ -15x+3000. 

Add 15x to both sides. 20x≤3000. 

Divide by 20 on both sides. x≤3000.

y≤5x, so y≤5(150).

y≤750, so b is at most 750.

Elimination

100

Solve the following system using the elimination method.

3x+2y=11

-x+y=3

Multiply the second equation by 2 to eliminate y. This leaves -2x+2y=6. 

Subtract -2x+2y=6 from the first equation 3x+2y=11. This leaves 5x=5. Divide by 5 on both sides, making x=1. 

Substitute x=1 into either equation. -1+y=3. y=4. The solution is (1,4).

200

Solve the following system using the elimination method. 

4x+3y=24

12x+5y=100

Multiply the first equation by 3 to eliminate x, resulting in 12x+9y=72.

Subtract the second equation from the first, leaving 4y=-28. Divide by 4 on both sides to get y=-7.

Plug -7 into y for either equation. 4x+3(-7)=24.   4x-21=24. Add 21 to both sides. 4x=45. Divide by 4 on both sides. x=45/4. The solution is (45/4,-7).

300

Solve the following system using the elimination method. 

10x+3y=500

60x+18y=3000

If you divide the second equation by 6, you are left with the first equation. Therefore, there are infinitely many solutions as long as x=-3y/10+50 and y=-10x/3+500/3.

400

Solve for x using the elimination method. 

3csc(3𝜋/2)+8x=45

6csc(3𝜋/2)+21x=120

Assume that csc(3𝜋/2) is some variable y. This leaves us with 3y+8x=45 and 6y+21x=120. 

Multiply the first equation by 2 to eliminate y. This gives us 6y+16x=90. 

Subtract 6y+16x=90 from 6y+21x=120, giving us 5x=30. Divide by 5 on both sides. x=6.

500

A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at each rate?

Let x be the amount invested into the fund paying 9% interest and y be the amount invested into the fund paying 11% interest. 

Given this, x+y=12000 and 0.09x+0.11y=1180.

Multiply the first equation by 0.09 to eliminate x. This gives us 0.09x+0.09y=1080. Subtract this from the second equation, giving us 0.02y=100. Divide both sides by 0.02. y=5000.

Substitute 5000 into y for the first equation.   12000-5000=7000. Thus, $5,000 was invested at 9% interest, and $7,000 was invested at 11% interest.

Word Problems

100

William, Xing Mei, Yuki, and Zack run a race.  In how many different ways can they finish?

(A)  4  (B)  16  (C)  24  (D)  32  (E)  64

There are 4! ways to finish the race.

4!  =  4 ⋅ 3 ⋅ 2 ⋅ 1

  =  24

200

Phone Company A charges 50 + 3x dollars for an international phone plan, where x is the number of minutes spent talking.  Phone Company B charges 60 + 2x dollars for an international phone plan, where x is the number of minutes spent talking.  What is the price at which both companies charges the same amount?

(A)  $10  (B)  $20  (C)  $30  (D)  $80  (E)  $110

Charge of company A  =  50 + 3x

Charge of company B  =  60 + 2x

50 + 3x  =  60 + 2x

3x - 2x  =  60 - 50

x  =  10 (Number of minutes)

Charge of company A  =  50 + 3(10)

=  50 + 30

=  $80

Hence the price is $80.

300

Lenny's average score after 3 tests is 88. What score on the 4th test would bring Lenny's average upto exactly 90?

(A)  92  (B)  93  (C)  94  (D)  95  (E)  96

Sum of 3 test scores  =  3 (88)  =  264

Sum of 4 test scores  =  4(90)  =  360

4th test score  =  360 - 264

  =  96

400

William, Xing Mei, Yuki, and Zack run a race.  In how many different ways can they finish?

(A)  4  (B)  16  (C)  24  (D)  32  (E)  64

There are 4! ways to finish the race.

4!  =  4 ⋅ 3 ⋅ 2 ⋅ 1

  =  24

500

Darius ran 1/6 as many times around the track as Ezekiel.  Darius ran around the track 2 ⅔ times.  How many times did Ezekiel run around the track?

(A)  8  (B)  10  (C)  12  (D)  14  (E)  16

Let "x" be the number of times that Ezekiel around the track, then 

Darius ran around the track  =  (1/6)x  =  x/6

Now, Darius ran around the track  =  2 ⅔ times

  =  8/3

We need to show 8/3 in terms of 1/6.

For that, let us multiply the numerator and denominator by 2.

So, we get 

  =  (8/3) ⋅ (2/2)

  =  16/6

  =  (1/6) ⋅ 16

Hence Ezekiel run around the track 16 times.

Work and Ratios

100

A recipe calls for 4 parts sugar per 11 parts flour.

How many cups of sugar are needed for 132 cups of flour?

Set up a ratio with sugar in the numerator and flour in the denominator. 4/11=x/132.

Cross multiply. 4*132=11x. 4*132=528. 11x=528. Divide by 11 on both sides. x=48.

48 cups of sugar are needed for the recipe.

200

Chad and George are painting a room. Chad can finish the job alone in 8 hours, while George can finish the job alone in 10 hours. How long, in hours, would it take for both to finish the job at the same time without breaks? 

(A) 9/40 (B) 4 (C) 40/9 (D) 37/8

Chad can finish 1/8 of the job in an hour. 

George can finish 1/10 of the job in an hour. 

Together, their rate would be 1/8+1/10 of the job per hour. 1/8+1/10=9/40. This means that they can complete 9/40 of the job per hour. 

Divide 1 job by 9/40 of the job per hour to find the number of hours. The job would take 40/9 hours.

300

In the synthesis reaction for NH₃, 1 mole of N₂ and 3 moles of H₂ are needed for every 2 moles of NH₃. If the reaction goes to completion with 1.5 moles of N₂, how many moles of H₂ were used in the reaction?

The amount of moles of NH₃ is irrelevant. Given that 1 mole of N₂ is needed for every 3 moles of H₂, the amount of H₂ is 3 times the amount of H₂. Therefore, 4.5 moles of H₂ were used because 1.5 times 3 is 4.5.

400

Karl and Dave are construction workers. Dave can finish a particular job in 10 hours. Together, they can finish that same job in 4 hours. How long, in hours, would it take Karl to finish the job alone?

(A) 20/3 (B) 6 (C) 2/9 (D) 3/20

If Dave can finish the job in 10 hours, he can finish 1/10 of the job per hour. 

If both workers can finish the job in 4 hours, they would finish 1/4 of the job in an hour. 

To find Karl's rate, we can subtract Dave's rate from the collective rate. 1/4-1/10=3/20.

Karl can finish 3/20 of the job in one hour. One job divided by 3/20 of the job per hour gets us 20/3 hours.

500

In a box of blocks, there are 3 red blocks for every 2 green blocks. There are 5 blue blocks for every 2 red blocks and 7 yellow blocks for every green block. If there are 56 yellow blocks in the box, how many blue blocks are in the box?

Red:green - 3:2, Blue:red - 5:2, Yellow:green - 7:1

Yellow:green - 14:2, Red:yellow - 3:14, 

Blue:red - 15:6, Red:yellow - 6:28

Yellow:blue - 28:15 = 56:30

There are 30 blue blocks.