Equations - ISEE Middle Level Quantitative Reasoning

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Question

Billy is at the store purchasing flowers for his mother, his grandmother, and his friend. He finds roses on sale by the half dozen (6), tulips selling by the dozen (12), and daisies selling groups of 18.

Billy wants to have the same number of flowers in each bouquet, so that he is able to give everyone the same number of each flower. How many bundles of roses, tulips, and daisies will he have to buy so he has the same amount of each? (Please answer by roses, tulips, then daisies.)

Answer

This is a least common multiple problem because we want to have the same number of each flower; meaning if I have 10 roses in each bouquet I should have 10 tulips and 10 daisies as well. In order to solve this problem we should break down the story problem. Let's look at the numbers we are having to work with: 6, 12, 18.

To do least common multiple we must look at the prime factors of each number and we can list them out. A factor is simply a number multiplied by a number to give us a product. A prime number is a number that contains only two factors, one of them being 1 and the other its own number.

So lets list the prime factors of 6, 12, and 18

$6:2\times 3$ (2 and 3 are both prime numbers, and factors of 6)

$12:2\times 2\times 3$ (2 x 2 = 4. 4x3=12 We have to say 2 x 2 because 4 is not a prime number).

$18:2\times 3\times 3$

Now we have the prime factors listed out for each of our numbers. Next is a fun trick. We must choose which number contains the most of each prime factor. In this case which number contains the most 2's? (12; because 12 has two 2 prime factors). Which number contains the most 3's? (18; because 18 has two 3 prime factors).

Our next step is to multiply the most of our prime factors so in this case: $2\times 2\times 3\times 3=36$

36 is our least common multiple. So now what do you think we can do with this number? Well the 36 means that is the lowest number of flowers we need of each type in order to have an equal amount of each for the boquets.

Knowing this, if we need 36 roses, and we are able to buy 6 roses per bundle. We need 6 bundles of roses, because 36 divided by 6 is 6. If we get 12 tulips by the bundle, we take 36 divided by 12 to give us 3 bundles of tulips needed. Lastly we can buy 18 daisies per bundle, 36 divided by 18 gives us 2 bundles needed giving us our answers 6, 3, 2 (bundles of roses, tulips, and daisies).

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Question

Solve for :

Answer

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

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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).

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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.

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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.

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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.

$\small 36+9=45$

$\small 45-14=31$

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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.

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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, .

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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.

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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.

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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.

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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,

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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.

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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.

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Question

Which is greater?

(A)

(B)

Answer

, and (B) is greater.

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Question

What is equal to?

Answer

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

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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.

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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.

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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.

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Question

Evaluate .

Answer

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