Equations - ISEE Middle Level Quantitative Reasoning

Card 0 of 20

Question

Solve for :

Answer

$5t \div 5 = 45\div 5$

Compare your answer with the correct one above

Question

Choose the best answer from the four choices given.

If $3z^{3}-2y=17$ ,

then what must $12z^{3}-8y$ equal?

Answer

Notice that $12z^{3}-8y$ is greater than $3z^{3}-2y$ by a factor of 4.

$4(3z^{3})=12z^{3}$ and

Thus, $12z^{3}-8y$ must equal 4 times the other side of the given equation, as well (17).

Compare your answer with the correct one above

Question

Choose the best answer from the four choices given.

Richard reaches into his pocket and determines that he has $3.85 in quarters and nickels. When he pulls out his fistful of change, though, he is dismayed to realize that what he thought were quarters are really pennies and what he thought to be nickels are really dimes. If he has 12 pennies, how much is his change worth?

Answer

If Richard has 12 pennies, that means that he thought he had 12 quarters ($3.00) and 85 cents worth of nickels (17 nickels). Since what he thought were nickels are really dimes, it means that he has $1.70 worth of dimes to go along with his 12 cents in pennies.

Compare your answer with the correct one above

Question

Using the information given in each question, compare the quantity in Column A to the quantity in Column B.

Column A Column B

the slope of the y-intercept

this equation of this equation

Answer

Divide each of the elements of the equation by 5 to get it into form.

$\frac{5y}{5}-\frac{9x}{5}=\frac{10}{5}$

$y-\frac{9}{5}x=2$

$y=\frac{9}{5}x+2$

Thus, the slope is 1.8 and the y-intercept is 2, so B is the correct answer.

Compare your answer with the correct one above

Question

Solve the following equation if $\small \small x=3$ :

$\small 4x^{2}+3x-14$

Answer

In order to solve this problem, we must substitute in for $\small x$ .

$\small 4(3)^{2}+3(3)-14$

Based on the order of operations we know that we must first deal with the exponents, so we look at $\small 3^{2}$ , which is $\small 9$ .

$\small 4(9)+3(3)-14$

Next we multiply.

$\small 4(9)=36$

$\small 3(3)=9$

$\small 36+9-14$

Finally we add subtract.

$\small 36+9=45$

$\small 45-14=31$

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a)

(b)

Answer

Rewrite in standard form:

$x \cdot x+x \cdot 1 = 30$

$x ^{2}+x = 30$

$x ^{2}+x- 30 = 30 - 30$

$x ^{2}+x- 30 = 0$

Factor the quadratic expression as , replacing the question marks with two integers whose product is and whose sum is 1. These integers are , so the equation becomes:

Set each linear binomial to 0 and solve:

$x-5= 0 \Rightarrow x= 5$

$x+ 6 = 0 \Rightarrow x = -6$

Therefore, it is not clear whether is greater than or less than 0.

Compare your answer with the correct one above

Question

$\left ( t- 10\right ) (2t-3) = -35$

Which is the greater quantity?

(a)

(b)

Answer

Rewrite in standard form:

$\left ( t- 10\right ) (2t-3) = -35$

$t \cdot 2t -t \cdot 3 -10 \cdot 2t + 10 \cdot 3 = -35$

$2t^{2} -3t -20 t + 30 = -35$

$2t^{2} -23t + 30 = -35$

$2t^{2} -23t + 30+ 35 = -35 + 35$

$2t^{2} -23t + 65 =0$

Use the -method to factor the quadratic expression, splitting the middle term using integers whose product is $2 \times65=130$ and whose sum is . These integers are , so the equation becomes:

$2t^{2} -10t - 13t + 65 =0$

Group and solve:

$\left ( 2t^{2} -10t \right )-\left ( 13t - 65 \right ) =0$

$2t \left ( t-5\right )-13\left ( t-5 \right ) =0$

$\left ( 2t -13 \right )\left ( t-5\right ) =0$

Set each linear binomial to 0 and solve:

$2t \div 2 = 13 \div 2$

$t = 6 \frac{1}{2}$

or

In either case, .

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a)

(b)

Answer

Rewrite in standard form:

$y \cdot y +y \cdot 7 = 30$

$y ^{2} +7y = 30$

$y ^{2} +7y- 30 = 30 - 30$

$y ^{2} +7y- 30 = 0$

Factor the quadratic expression as , replacing the question marks with two integers whose product is and whose sum is 7. These integers are , so rewrite:

Set each linear binomial to 0 and solve:

$y- 3 \Rightarrow y= 3$

$y+10= 0 \Rightarrow y = -10$

Both solutions are less than 5.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(a)

(b)

Answer

Rewrite in standard form:

$x \cdot 3x-x \cdot7 = 40$

$3x^{2}-7x = 40$

$3x^{2}-7x- 40 = 40 - 40$

$3x^{2}-7x- 40 = 0$

Use the -method to factor the quadratic expression, splitting the middle term using integers whose product is and whose sum is . These two numbers are , so the equation becomes:

$3x^{2}-15x+8x- 40 = 0$

$\left ( 3x^{2}-15x \right ) +\left ( 8x- 40 \right ) = 0$

$3x \left ( x-5 \right ) + 8 \left ( x-5 \right ) = 0$

$\left (3x+ 8 \right ) \left ( x-5 \right ) = 0$

Set each linear binomial to 0 and solve:

$3x\div 3= -8 \div 3$

$x = -2 \frac{2}{3}$

or

Therefore, it is not clear whether is greater than or less than 3.

Compare your answer with the correct one above

Question

$y^{2} + 50 = 15y$

Which is the greater quantity?

(a)

(b)

Answer

Rewrite in standard form:

$y^{2} + 50 = 15y$

$y^{2} + 50 -15y= 15y -15y$

$y^{2} -15y+ 50= 0$

Factor the quadratic expression as , replacing the question marks with two integers whose product is 50 and whose sum is . These integers are , so rewrite the equation as:

Set each linear binomial to 0 and solve:

$y -5= 0\Rightarrow y = 5$

$y -10= 0\Rightarrow y = 10$

It is unclear whether is equal to or greater than 5.

Compare your answer with the correct one above

Question

$\left (x-5 \right )^{2} = 36$

Which is the greater quantity?

(a)

(b)

Answer

Use the square root method to solve this equation:

$\left (x-5 \right )^{2} = 36$

$x-5 =- 6 \textrm{ or }x-5 = 6$

Solve each equation separately:

or

In either case,

Compare your answer with the correct one above

Question

$\frac{x}{3}-9=11$

Column A Column B
x y

Answer

First, off you need to find the quantities of x and y. For the first equation, add 9 to both sides so that $\frac{x}{3}=20.$ Multiply both sides by 3 so that Then, solve the second equation. First, subtract 5 from both sides so that . Divide both sides by 5 so that Therefore, the quantity in Column A is greater.

Compare your answer with the correct one above

Question

$\frac{x}{2}+7=19$

Column A Column B

x y

Answer

First, you must solve for x and y. To solve for x, first subtract 7 from both sides, so that you get $\frac{x}{2}=12$ . Multiply both sides by 2, giving you 24 for x. To solve for y, first add 14 to both sides, which gives you Divide both sides by 3 to give you 7. Therefore, Column A is greater.

Compare your answer with the correct one above

Question

Which is greater?

(A)

(B)

Answer

, and (B) is greater.

Compare your answer with the correct one above

Question

What is equal to?

Answer

$2s +2 t = 2 (s+t) =2 \cdot 21 = 42$

Compare your answer with the correct one above

Question

How many elements of the set $\left { 1, 2, 3, 4, 5 \right }$ can be substituted for to make the inequality a true statement?

Answer

$4y \div 4 < 15 \div 4$

$y < 3 \frac{3}{4}$

Three elements of the set—1, 2, and 3—fit this criterion.

Compare your answer with the correct one above

Question

Which is the greater quantity?

(A)

(B)

Answer

Note that we are not given that and are whole numbers.

Setting to , , or fulfills the condition that and , but in these cases, , , , respectively. It cannot be determined which is greater.

Compare your answer with the correct one above

Question

If , then how many integers can be substituted for to make the equation $x^{2} - y^{2} = 0$ a true statement?

Answer

If , the equation can be restated and solved as follows:

$x^{2} - y^{2} = 0$

$x^{2} - 12^{2} = 0$

$x^{2} -144 = 0$

$x^{2} -144 + 144 = 0 + 144$

$x^{2} = 144$

Both and make this true, so both make the original statement true. "Two" is the correct choice.

Compare your answer with the correct one above

Question

Evaluate .

Answer

Compare your answer with the correct one above

Question

Evaluate .

Answer

Compare your answer with the correct one above

Tap the card to reveal the answer