How to find the part from the whole - ISEE Middle Level Quantitative Reasoning

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$60:100\rightarrow\frac{60}{100}$

Reduce.

$\frac{60}{100}\rightarrow \frac{6}{10}\rightarrow \frac{3}{5}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:135\rightarrow \frac{Red}{135}$

Now, we can create a proportion using our two ratios.

$\frac{3}{5}=\frac{Red}{135}$

Cross multiply and solve for .

$5 \times Red=3\times135$

Simplify.

Divide both sides of the equation by .

$\frac{5Red}{5}=\frac{405}{5}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$40:100\rightarrow\frac{40}{100}$

Reduce.

$\frac{40}{100}\rightarrow \frac{4}{10}\rightarrow \frac{2}{5}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:400\rightarrow \frac{Red}{400}$

Now, we can create a proportion using our two ratios.

$\frac{2}{5}=\frac{Red}{400}$

Cross multiply and solve for .

$5 \times Red=2\times400$

Simplify.

Divide both sides of the equation by .

$\frac{5Red}{5}=\frac{800}{5}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$50:100\rightarrow\frac{50}{100}$

Reduce.

$\frac{50}{100}\rightarrow \frac{5}{10}\rightarrow \frac{1}{2}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:200\rightarrow \frac{Red}{200}$

Now, we can create a proportion using our two ratios.

$\frac{1}{2}=\frac{Red}{200}$

Cross multiply and solve for .

$2 \times Red=1\times200$

Simplify.

Divide both sides of the equation by .

$\frac{2Red}{2}=\frac{200}{2}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$30:100\rightarrow\frac{30}{100}$

Reduce.

$\frac{30}{100}\rightarrow \frac{3}{10}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:300\rightarrow \frac{Red}{300}$

Now, we can create a proportion using our two ratios.

$\frac{3}{10}=\frac{Red}{300}$

Cross multiply and solve for .

$10 \times Red=3\times300$

Simplify.

Divide both sides of the equation by .

$\frac{10Red}{10}=\frac{900}{10}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$4:100\rightarrow\frac{4}{100}$

Reduce.

$\frac{4}{100}\rightarrow \frac{2}{50}\rightarrow \frac{1}{25}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:40\rightarrow \frac{Red}{400}$

Now, we can create a proportion using our two ratios.

$\frac{1}{25}=\frac{Red}{400}$

Cross multiply and solve for .

$25 \times Red=1\times400$

Simplify.

Divide both sides of the equation by .

$\frac{25Red}{25}=\frac{400}{25}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$35:100\rightarrow\frac{35}{100}$

Reduce.

$\frac{35}{100}\rightarrow \frac{7}{20}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:200\rightarrow \frac{Red}{200}$

Now, we can create a proportion using our two ratios.

$\frac{7}{20}=\frac{Red}{200}$

Cross multiply and solve for .

$20 \times Red=7\times200$

Simplify.

Divide both sides of the equation by .

$\frac{20Red}{20}=\frac{1400}{20}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$80:100\rightarrow\frac{80}{100}$

Reduce.

$\frac{80}{100}\rightarrow \frac{8}{10}\rightarrow \frac{4}{5}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:800\rightarrow \frac{Red}{800}$

Now, we can create a proportion using our two ratios.

$\frac{4}{5}=\frac{Red}{800}$

Cross multiply and solve for .

$5 \times Red=4\times800$

Simplify.

Divide both sides of the equation by .

$\frac{5Red}{5}=\frac{3200}{5}$

Solve.

There are red cars in the parking lot.

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Question

Red is a very popular car color. A production company manufactures cars and parks them in a lot behind the plant. There are cars in the parking lot and of them are red. How many red cars are in the parking lot?

Answer

We can use ratios and proportions to solve this problem. Percentages can be written as ratios. The word “percent” means for every hundred. In the problem, we are told that of the cars are red. In other words, for every hundred cars of them are red. We can write the following ratio:

$8:100\rightarrow\frac{8}{100}$

Reduce.

$\frac{8}{100}\rightarrow \frac{4}{50}\rightarrow \frac{2}{25}$

We know that there are cars in the parking lot. We can write the following ratio by substituting the variable for the number of red cars:

$Red:3200\rightarrow \frac{Red}{3200}$

Now, we can create a proportion using our two ratios.

$\frac{2}{25}=\frac{Red}{3200}$

Cross multiply and solve for .

$25 \times Red=2\times3200$

Simplify.

Divide both sides of the equation by .

$\frac{25Red}{25}=\frac{6400}{25}$

Solve.

There are red cars in the parking lot.

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Question

You are given a bag with 5 red marbles, 3 green marbles and 7 blue marbles. What percent of the bag are red marbles?

Answer

When finding the percent, we must take the part and divide it by the whole.

$Percent=\frac{part}{whole}$

In this case, 5 is the part and we must find the sum of everything to make the whole as shown in the denominator below.

$Percent=\frac{5}{5+7+3}=\frac{5}{15}$

We can now reduce the fraction and convert to percent.

$Percent=\frac{5}{15}=\frac{1}{3}=33.3%$

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Question

and are positive integers; . Which is the greater quantity?

(a) of

(b) of

Answer

of is equal to

$4B \cdot \frac{A}{100} = \frac{4AB}{100}$

of is equal to

$4A \cdot \frac{B}{100} = \frac{4AB}{100}$

The two are equal regardless of the value or relation of and .

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Question

and are positive integers; . Which is the greater quantity?

(a) $\frac{1}{2} %$ of

(b) 0.5 % of

Answer

$0.5 = \frac{5}{10} = \frac{5 \div 5}{10 \div 5 } = \frac{1}{2}$ , so 0.5% of a number is the same as $\frac{1}{2} %$ of the number. Therefore, in each choice, we are taking the same percent of a number. Since , 0.5%, or $\frac{1}{2} %$ , of is less than 0.5% of .

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Question

Harvey bought a suit at a 25% employee discount at the store where he works. The suit originally cost $350.00. How much did he end up paying?

Answer

Finding 25% of a number is the same as multiplying it by 0.25; to get the discount, multiply 0.25 by the original purchase price of $350.00

$0.25 \cdot \$350 = \$87.50$

To get the price Harvey paid, subtract the discount from the original price:

$\$350 - $87.50 = $ 262.50$

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Question

is a positive integer. Which is the greater quantity?

(a) The remainder if is divided by 5

(b) The remainder if is divided by 4

Answer

The information is insufficient.

For example, if :

$4X = 4 \cdot 10 = 40$ .

$40 \div 5 = 8 \textrm{ R }0$

$5X = 5 \cdot 10 = 50$ .

$50 \div 4 = 12\textrm{ R }2$

This gives the division in (b) the greater remainder.

But if :

$4X = 4 \cdot 12 = 48$ .

$48 \div 5 = 9 \textrm{ R }3$

$5X = 5 \cdot 12 = 60$ .

$60 \div 4 = 15\textrm{ R }0$

This gives the division in (a) the greater remainder.

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Question

$A + B \equiv 5 \textrm{ mod } 9$

Which of the following is true if ?

Answer

Two expressions are equivalent in modulo 9 arithmetic if and only if, when each is divided by 9, the same remainder is yielded.

$A + B \equiv 5 \textrm{ mod } 9$ ,

so

$8 + B \equiv 5$

$8 + B - 8 \equiv 5 - 8$

$B \equiv 5 - 8$

$B \equiv -3$

, so

is the correct choice.

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Question

is a positive even integer. Which is the greater quantity?

(a) The remainder if is divided by 6

(b) The remainder if is divided by 3.

Answer

Since is an even integer, by definition, there is an integer such that

.

$3 X = 3 \cdot 2Y = 6Y$ ; therefore,

$3X \div 6 = 6Y \div 6 = Y$ ; the remainder is 0.

Also,

$6X \div 3 =2X$ ; the remainder is 0.

The two remainders are both equal to 0.

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Question

is a positive odd integer. Which is the greater quantity?

(a) The remainder if is divided by 8

(b) The remainder if is divided by 4

Answer

$8X \div 4 = 2X$ , with a remainder of 0.

If $4X \div 8$ were to yield a remainder of 0, then $4X \div 8 = \frac{1}{2} X$ must be a whole number; this can only happen if is even. Since is odd, it follows that $\frac{1}{2} X$ is not a whole number, and $4X \div 8$ must yield a nonzero remainder. (a) must be the greater quantity.

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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

120% of 80% of 100 130% of 70% of 105

Answer

Remember that "percent" means "per 100" and "of" means to multiply.

So 120% of 80% of 100 =

and 130% of 70% of 105 =

Column A is greater.

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Question

Which is the greater quantity?

(a) $\frac{1}{2} % \textrm{ of }1,000$

(b)

Answer

$\frac{1}{2} % \textrm{ of }1,000$ can be rewritten as $0.5 % \textrm{ of }1,000$ , and restated as $0.005 \times 1,000$ .

$0.005 \times 1,000 = 5 > 0.5$

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Question

is a positive number. Which of the following is the greater quantity?

(A) 50% of

(B) 20% of

Answer

50% of is equal to .

20% of is equal to $0.2 \cdot 3N = 0.6 N$

Since is positive and ,

,

and (B) is greater.

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Question

What is 60% of 60% of 25,000?

Answer

60% of 25,000 is equal to $0.60 \times 25,000 = 15,000$

60% of that is $0.60 \times 15,000 = 9,000$

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