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ISEE Upper Level Quantitative : Equations

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

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All ISEE Upper Level Quantitative Resources

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Example Questions

Example Question #1 : Equations

A company makes toy boats. Their monthly fixed costs are $1500. The variable costs are $50 per boat. They sell boats for $75 a piece. How many boats must be sold each month to break even?

Possible Answers:

100

Correct answer:

Explanation:

The break-even point is where the costs equal the revenues

Fixed Costs + Variable Costs = Revenues

1500 + 50x = 75x

Solving for x results in x = 60 boats sold each month to break even.

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Example Question #2 : Equations

Sally sells cars for a living. She has a monthly salary of $1,000 and a commission of $500 for each car sold. How much money would she make if she sold seven cars in a month?

Possible Answers:

$5,500

$4,500

$4,000

$5,000

Correct answer:

$4,500

Explanation:

The commission she gets for selling seven cars is $500 * 7 = $3,500 and added to the salary of $1,000 yields $4,500 for the month.

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Example Question #3 : Equations

Solve the following system of equations: x – y = 5 and 2x + y = 4.

What is the sum of x and y?

Possible Answers:

Correct answer:

Explanation:

Add the two equations to get 3x = 9, so x = 3. Substitute the value of x into one of the equations to find the value of y; therefore x = 3 and y = –2, so their sum is 1.

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Example Question #4 : Equations

If x = 1/3 and y = 1/2, find the value of 2x + 3y.

Possible Answers:

13/6

6/5

Correct answer:

13/6

Explanation:

Substitute the values of x and y into the given expression:

2(1/3) + 3(1/2)

= 2/3 + 3/2

= 4/6 + 9/6

= 13/6

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Example Question #2 : Equations

For what value(s) of is the expression $\frac{x^{2}-1}{x^{2}+9}$ undefined?

Possible Answers:

The expression is undefined for x= 1 and x= -1

The expression is undefined for x = -3 and x=3

The expression is undefined for x= 1

The expression is undefined for x = -3

The expression is defined for all real values of

Correct answer:

The expression is defined for all real values of

Explanation:

The expression is undefined for exactly those values of which yield a denominator of 0 - that is, for which

$x^{2} +9 =0$

However, for all real ,

$x^{2} \geq 0$ ,

and, subsequently,

$x^{2} +9 \geq 9 >0$

meaning the denominator is always positive. Therefore, the expression is defined for all real values of .

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Example Question #3 : Equations

Albert has thirteen bills in his wallet, each one a five-dollar bill or a ten-dollar bill. What is the fewest number of ten-dollar bills that he can have and have more than $100.

Possible Answers:

Correct answer:

Explanation:

Let be the number of ten-dollar bills Albert has; then he has 13-x five-dollar bills.

He then has 10x + 5(13-x) dollars in his wallet, which must be greater than $100. Set up and solve an inequality:

10x + 5(13-x) > 100

10x + 65-5x > 100

5x > 35

x>7

Therefore, the lowest whole number of ten-dollar bills that Albert can have is eight.

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Example Question #4 : Equations

Solve for :

$x (x+ 5) = \left ( x+2\right )^{2}$

Possible Answers:

x = 9

x = 5

x = 4

$x = 4 \textrm{ or }x=9$

$x = 4 \textrm{ or }x=5$

Correct answer:

x = 4

Explanation:

Expand both products, the left using distribution, the right using the binomial square pattern:

$x (x+ 5) = \left ( x+2\right )^{2}$

$x \cdot x+x \cdot 5= x^{2} + 2 \cdot 2 \cdot x +2^{2}\right )$

$x^{2}+5 x = x^{2} + 4x +4$

$x^{2} +5 x-x^{2} = x^{2} + 4x +4 -x^{2}$

5 x= 4x +4

Note that the quadratic terms can be eliminated, yielding a linear equation.

5 x-4x = 4x +4 -4x

x = 4

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Example Question #8 : Equations

Solve for :

4x + 2(x-7) = 2 (3x -5)

Possible Answers:

x =-4

The equation has no solution.

x = 0

x = 4

x = 6

Correct answer:

The equation has no solution.

Explanation:

4x + 2(x-7) = 2 (3x -5)

$4x + 2 \cdot x-2 \cdot7 = 2 \cdot 3x -2 \cdot5$

4x + 2 x-14 = 6x -10

$\left (4 + 2 \right )x-14 = 6x -10$

6x-14 = 6x -10

6x-6x-14 = 6x -6x -10

-14 = -10

This identically false statement alerts us to the fact that the original equation has no solution.

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Example Question #9 : Equations

(2x-7)(x + 4) = 27

Possible Answers:

$x = -5\frac{1}{2} \textrm{ or } x = 5$

$x =17 \textrm{ or } x =23$

$x =-3\frac{1}{2} \textrm{ or } x =4$

$x =-4 \textrm{ or } x =3\frac{1}{2}$

$x = 5\textrm{ or } x = 5\frac{1}{2}$

Correct answer:

$x = -5\frac{1}{2} \textrm{ or } x = 5$

Explanation:

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

(2x-7)(x + 4) = 27

$2x \cdot x+ 2x \cdot 4 -7\cdot x -7\cdot 4 = 27$

$2x^{2}+ 8x -7 x -28 = 27$

$2x^{2}+ x -28 = 27$

$2x^{2}+ x -28-27 = 27-27$

$2x^{2}+ x- 55 = 0$

Use the method to split the middle term into two terms; we want the coefficients to have a sum of 1 and a product of 2 (-55) = -110 . These numbers are -10,11 , so we do the following:

$2x^{2}-10x+ 11x- 55 = 0$

$2x\left (x- 5 \right) + 11 \left ( x- 5 \right) = 0$

$\left (2x+ 11 \right) \left ( x- 5 \right) = 0$

Set each expression equal to 0 and solve:

2x + 11 = 0

2x = -11

$x = -5\frac{1}{2}$

x-5= 0

x = 5

The solution set is $\left \{ -5 \frac{1}{2}, 5\right \}$ .

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Example Question #10 : Equations

Solve for :

$x\left (3x+7 \right ) =20$

Possible Answers:

$x=4 \frac{1}{3} \textrm{ or } x = 20$

$x=-4 \textrm{ or }x = 1\frac{2}{3}$

$x=-1\frac{2}{3} \textrm{ or }x = 4$

$x=-1\frac{1}{3} \textrm{ or }x = 5$

$x=-5 \textrm{ or }x = 1\frac{1}{3}$

Correct answer:

$x=-4 \textrm{ or }x = 1\frac{2}{3}$

Explanation:

First, rewrite the quadratic equation in standard form by distributing the through the product on the left, then collecting all of the terms on the left side:

$x\left (3x+7 \right ) =20$

$x\left \cdot \; 3x+x\left \cdot\; 7 =20$

$3x ^{2}+7x =20$

$3x ^{2}+7x -20=20 -20$

$3x ^{2}+7x -20=0$

Use the method to factor the quadratic expression $3x ^{2}+7x -20$ ; we are looking to split the linear term by finding two integers whose sum is 7 and whose product is 3 (-20) = -60 . These integers are -5,12 , so:

$\left (3x ^{2}-5x \right ) + \left ( 12x -20 \right ) =0$

$x \left (3x-5 \right ) +4 \left ( 3x -5 \right ) =0$

$\left (x +4 \right )\left (3x-5 \right )=0$

Set each expression equal to 0 and solve:

x+4 = 0

x=-4

3x-5=0

3x=5

$x = \frac{5}{3} = 1\frac{2}{3}$

The solution set is $\left \{ -4, 1\frac{2}{3}\right \}$ .

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Equations

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #1 : Equations

Example Question #2 : Equations

Example Question #3 : Equations

Example Question #4 : Equations

Example Question #2 : Equations

Example Question #3 : Equations

Example Question #4 : Equations

Example Question #8 : Equations

Example Question #9 : Equations

Example Question #10 : Equations

All ISEE Upper Level Quantitative Resources

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