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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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 : Quadrilaterals

The area of a square is $36t ^{2} + 84t + 49$ .

Give the perimeter of the square.

Possible Answers:

|42t|

|6t+7|

|36t+49|

|26t|

|24t+28|

Correct answer:

|24t+28|

Explanation:

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

$36t ^{2} + 84t + 49$

$=( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}$

$= (6t+7)^{2}$

Therefore, the square root of the area is

$\sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|$ ,

which is the length of one side.

The perimeter of the square is four times this length, or

4 |6t+7| = |24t+28| .

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Example Question #1 : Squares

The perimeters of six squares form an arithmetic sequence. The second-smallest square has sides that are two inches longer than those of the smallest square.

Which, if either, is the greater quantity?

(a) The perimeter of the third-smallest square

(b) The length of one side of the largest square

Possible Answers:

(a) and (b) are equal

(a) is greater

(b) is greater

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

Correct answer:

(a) is greater

Explanation:

Let the length of one side of the first square be $s_{1}$ . Then the length of one side of the second-smallest square is $s_{2}= s_{1}+ 2$ , and the perimeters of the squares are

$P_{1}= 4s_{1}$

and

$P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8$

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the th-smallest square is

$P_{n} = P_{1} + (n-1)d$

Since d = 8 , this becomes

$P_{n} = P_{1} + 8 (n-1)$

The perimeter of the third-smallest square is

$P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16$

The perimeter of the largest, or sixth-smallest, square is

$P_{6} = P_{1} + 8 (6-1) = P_{1}+ 8 \cdot 5= P_{1} + 40$

The length of one side of this square is one fourth of this, or

$s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10$

Therefore, we are comparing $P_{1} + 16$ and $\frac{1}{4} P_{1} + 10$ .

Since a perimeter must be positive,

$\frac{1}{4} P_{1} < P_{1}$ ;

also, 10 < 16 .

Therefore, regardless of the value of $P_{1}$ ,

$P_{1} + 16 > \frac{1}{4} P_{1} + 10$ ,

and

$P_{3}> s_{6}$ ,

making (a) the greater quantity.

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Example Question #2 : Squares

The sidelength of a square is $2^{x}$ . Give its perimeter in terms of .

Possible Answers:

$2 ^{x+4}$

$2 ^{x+2}$

$2 ^{x+1}$

$2 ^{2 x}$

$2 ^{4 x}$

Correct answer:

$2 ^{x+2}$

Explanation:

The perimeter of a square is four times the length of a side, which here is $2 ^{x }$ :

$P = 4 \cdot 2^{x} = 2 ^{2} \cdot 2^{x} =2 ^{x+2}$

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Example Question #2 : Squares

A diagonal of a square has length $2^{x}$ . Give its perimeter.

Possible Answers:

$2 ^{x+\frac{3}{2}}$

$2 ^{x+\frac{1}{2}}$

$2^{2x}$

$2^{2x-1}$

$2^{3x}$

Correct answer:

$2 ^{x+\frac{3}{2}}$

Explanation:

The length of a side of a square can be determined by dividing the length of a diagonal by $\sqrt{2}$ - that is, $2 ^{\frac{1}{2}}$ . A diagonal has length $2^{x}$ , so the sidelength is

$2^{x} \div 2 ^{\frac{1}{2}}= 2 ^{x-\frac{1}{2}}$

Multiply this by four to get the perimeter:

$4 \cdot 2 ^{x-\frac{1}{2}} = 2^{2} \cdot 2 ^{x-\frac{1}{2}} = 2 ^{x-\frac{1}{2}+2} = 2 ^{x+\frac{3}{2}}$

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Example Question #1 : Squares

The perimeters of six squares form an arithmetic sequence. The smallest square has area 9; the second smallest square has area 25. Give the perimeter of the largest of the six squares.

Possible Answers:

None of the other responses gives the correct answer.

Correct answer:

Explanation:

The two smallest squares have areas 9 and 25, so their sidelengths are the square roots of these, or, respectively, 3 and 5. Their perimeters are the sidelengths multiplied by four, or, respectively, 12 and 20. Therefore, in the arithmetic sequence,

$P_{1} = 12$

$P_{2} = 20$

and the common difference is d = 20- 12 = 8 .

The perimeter of the th smallest square is

$P_{n}=P_{1}+ (n-1)d$

Setting $P_{1} = 12, n = 6, d=8$ , the perimeter of the largest (or sixth-smallest) square is

$P_{6}=P_{1}+ (6-1)8 = 12+ 5 \cdot 8 = 12 + 40 = 52$ .

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Example Question #1 : Squares

The perimeter of a square is one yard. Which is the greater quantity?

(a) The area of the square

(b) $\frac{1}{2}$ square foot

Possible Answers:

(a) and (b) are equal.

(b) is greater.

(a) is greater.

It is impossible to tell form the information given.

Correct answer:

(a) is greater.

Explanation:

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\frac{3}{4}$ feet. The area of the square is $\frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\frac{9}{16} > \frac{1}{2}$ , making (a) greater.

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Example Question #211 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Square 1 is inscribed inside a circle. The circle is inscribed inside Square 2.

Which is the greater quantity?

(a) Twice the area of Square 1

(b) The area of Square 2

Possible Answers:

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

(a) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

Let be the sidelength of Square 1. Then the length of a diagonal of this square - which is $\sqrt{2}$ times this sidelength, or $s \sqrt{2}$ , by the $45^{\circ }-45^{\circ }-90^{\circ }$ Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2.

Therefore, Square 1 has area $A = s^{2}$ , and Square 2 has area $A = \left ( s \sqrt{2 } \right )^{2} = 2s^{2}$ , twice that of Square 1.

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Example Question #212 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Which is the greater quantity?

(A) The area of a square with sidelength one foot

(B) The area of a rectangle with length nine inches and height fourteen inches

Possible Answers:

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

(A) is greater

(A) and (B) are equal

(B) is greater

Correct answer:

(A) is greater

Explanation:

The area of a square is the square of its sidelength, which here is 12 inches:

$A = 12^{2} = 144$ square inches.

The area of a rectangle is its length multiplied by its height, which, respectively, are 9 inches and 14 inches:

$A = 9 \cdot 14 = 126$ square inches.

The square has the greater area.

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Example Question #213 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

A square lawn has sidelength twenty yards. Give its area in square feet.

Possible Answers:

$3,200\textrm{ ft}^{2}$

$4,000\textrm{ ft}^{2}$

$4,200\textrm{ ft}^{2}$

$3,600\textrm{ ft}^{2}$

$2,800\textrm{ ft}^{2}$

Correct answer:

$3,600\textrm{ ft}^{2}$

Explanation:

20 yards converts to $20 \times3 = 60$ feet. The area of a square is the square of its sidelength, so the area in square feet is $60^{2} = 60 \times 60 = 3,600$ square feet.

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Example Question #5 : How To Find The Area Of A Square

Rectangle A and Square B both have perimeter 2 meters. Rectangle A has width 25 centimeters. The area of Rectangle A is what percent of the area of Square B?

Possible Answers:

$66 \frac{2}{3} \%$

$87 \frac{1}{2} \%$

$80 \%$

$83 \frac{1}{3} \%$

$75 \%$

Correct answer:

$75 \%$

Explanation:

The perimeter of a rectangle can be given by the formula

P = 2(L+W)

Rectangle A has perimeter 2 meters, which is equal to 200 centimeters, and width 25 centimeters, so the length is:

2(L+W) = P

2(L+25) = 200

2L + 50= 200

2L = 150

L = 75

The dimensions of Rectangle A are 75 centimeters and 25 centimeters, so its area is

$75 \times 25 = 1,875$ square centimeters.

The sidelength of a square is one-fourth its perimeter, which here is

$\frac{200}{4} = 50$ centimeters; its area is therefore

$50 ^{2} = 2,500$ square centimeters.

The area of Rectangle A is therefore

$\frac{1,875}{2,500} \times 100 = 75 \%$

that of Square B.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : ISEE Upper Level (grades 9-12) Quantitative Reasoning

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #1 : Quadrilaterals

Example Question #1 : Squares

Example Question #2 : Squares

Example Question #2 : Squares

Example Question #1 : Squares

Example Question #1 : Squares

Example Question #211 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #212 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #213 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #5 : How To Find The Area Of A Square

All ISEE Upper Level Quantitative Resources

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