How to find the solution to an equation

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ISEE Upper Level Quantitative Reasoning › How to find the solution to an equation

Questions 1 - 10

Define .

$(g \circ f) \left ( \frac{1}{2} \right ) = 6$

Which is the greater quantity?

(a)

(b)

(a) is the greater quantity

It is impossible to determine which is greater from the information given

(a) and (b) are equal

(b) is the greater quantity

Explanation

, so, by substitution,

$f \left ( \frac{1}{2} \right ) = 6 \cdot \frac{1}{2} = 3$ .

By way of the definition of a composition of functions,

$(g \circ f) \left ( \frac{1}{2} \right ) = g \left [ f \left ( \frac{1}{2} \right ) \right ] = g(3 )$ .

Since $(g \circ f) \left ( \frac{1}{2} \right ) =6$ , it follows that .

Also, by substitution,

$f(3 )= 6 \cdot 3 = 18$ .

Therefore, .

Three consecutive integers add up to 36. What is the greatest integer of the three?

Explanation

To solve this problem, you can translate the question into an equation. It should look like: . Since we don't know the first number, we name it as x. Then, we add one to each following integer, which gives us x+1 and x+2. Then, combine like terms to get . Solve for x and you get 11. However, the question is asking for the greatest integer of the set, so the answer is actually 13 (because it is the x+2 term).

Solve for :

$t= \frac{100 + 3x}{y}$

$t= \frac{100 -3x}{y}$

$t= \frac{100 -x}{3y}$

$t= \frac{100 -y}{3x}$

$t= \frac{100 +y}{3x}$

Explanation

$\frac{yt}{y} = \frac{100 + 3x}{y}$

$t= \frac{100 + 3x}{y}$

Which is the greater quantity?

(a)

(b)

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

(a) is greater.

Explanation

Each can be rewritten as a compound statement. Solve separately:

$45 - t =- 60 \textrm{ or }45 - t = 60$

Similarly:

$45 - u = -50 \textrm{ or } 45 - u = 50$

Therefore, it cannot be determined with certainty which of and is the greater.

One-third of the sum of a number and sixty is ninety-three. What is the number?

Explanation

If we let be the number, "the sum of a number and sixty" can be written as

"One-third of the sum of a number and sixty" can be written as

$\frac{1}{3} (x+60)$

Set this equal to ninety-three and solve:

$\frac{1}{3} (x+60) = 93$

$3 \cdot \frac{1}{3} (x+60) =3 \cdot 93$

Two lines have -intercept . Line A has -intercept ; Line B has -intercept . Which is the greater quantity?

(A) The slope of Line A

(B) The slope of Line B

(A) is greater

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Explanation

To get the slope of each line, use the slope formula

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1 }}$

For Line A, $x_{1} =-5, y_{1} =0, x_{2} = 0, y_{2} = 9$ . Substitute in the slope formula.

The slope is

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{9-0}{0-(-5)} = \frac{9}{5}$

For Line B, $x_{1} =-7, y_{1} =0, x_{2} = 0, y_{2} = 9$ . Substitute in the slope formula.

The slope is

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{9-0}{0-(-7)} = \frac{9}{7}$

Since

$\frac{9}{5} = \frac{63}{35}> \frac{45}{35} = \frac{9}{7}$ ,

Line A has the greater slope, and (A) is greater.

$\left \lfloor N\right \rfloor$ refers to the greatest integer less than or equal to .

and are integers.

Which is greater?

(a) $\left \lfloor a\right \rfloor + \left \lfloor b \right \rfloor$

(b) $\left \lfloor a+ b \right \rfloor$

(b) is greater.

It is impossible to tell from the information given.

(a) is greater.

(a) and (b) are equal.

Explanation

(a) Since is an integer, $\left \lfloor a \right \rfloor = \left \lfloor x + 0.5 \right \rfloor = x$ .

Since is an integer, $\left \lfloor b \right \rfloor = \left \lfloor y + 0.5 \right \rfloor = y$ .

$\left \lfloor a\right \rfloor + \left \lfloor b \right \rfloor = x + y$

(b) By closure, is an integer, so

$\left \lfloor a+b \right \rfloor = \left \lfloor x + 0.5 +y+0.5 \right \rfloor = \left \lfloor x +y+1 \right \rfloor = x +y+1$ .

This makes (b) greater.

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

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

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

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

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

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:

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

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

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

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

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

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 . These numbers are , so we do the following: