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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 #14 : Squares

The sidelength of Square A is three-sevenths that of Square B. What is the ratio of the area of Square B to that of Square A?

Possible Answers:

$7\textrm{ to }3$

$49 \textrm{ to } 9$

$14\textrm{ to }3$

$9\textrm{ to }7$

None of the other answers give the correct ratio.

Correct answer:

$49 \textrm{ to } 9$

Explanation:

Since the ratio is the same regardless of the sidelengths, then for simplicity's sake, assume the sidelength of Square B is 7. The area of Square B is therefore the square of this, or 49.

Then the sidelength of Square A is three-sevenths of 7, or 3. Its area is the square of 3, or 9.

The ratio of the area of Square B to that of Square A is therefore 49 to 9.

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Example Question #1 : How To Find The Area Of A Square

Five squares have sidelengths one foot, two feet, three feet, four feet, and five feet.

Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Possible Answers:

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

(A) is greater

(B) is greater

(A) and (B) are equal

Correct answer:

(A) is greater

Explanation:

The areas of the squares are:

$1^{2} = 1$ square foot

$2^{2} = 4$ square feet

$3^{2} = 9$ square feet

$4^{2}= 16$ square feet

$5^{2}= 25$ square feet

Therefore, we are comparing the mean and the median of the set $\left \{ 1, 4, 9, 16, 25\right \}$ .

The mean of this set is the sum divided by 5:

$(1 + 4 + 9 + 16 + 25) \div 5 = 55 \div 5 = 11$

The median is the middle element after arrangement in ascending order, which is 9.

This makes (A), the mean, greater.

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Example Question #6 : How To Find The Area Of A Square

Four squares have sidelengths one meter, 120 centimeters, 140 centimeters, and 140 centimeters. Which is the greater quantity?

(A) The mean of their areas

(B) The median of their areas

Possible Answers:

(A) and (B) are equal

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

(A) is greater

(B) is greater

Correct answer:

(B) is greater

Explanation:

The areas of the squares are:

$100 \cdot 100 = 10,000$ square centimeters (one meter being 100 centimeters)

$120 \cdot 120 = 14,400$ square centimeters

$140 \cdot 140 = 19,600$ square centimeters

The mean of these four areas is their sum divided by four:

$(10,000+14,400+19,600+19,600) \div 4$

$= 63,600 \div 4 =15,900$ square centimeters.

The median is the mean of the two middle values, or

$( 14,400+19,600 ) \div 2 = 34,000 \div 2 = 17,000$ square centimeters.

The median, (B), is greater.

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Example Question #9 : How To Find The Area Of A Square

The perimeter of a square is 12x+16 . Give the area of the square in terms of .

Possible Answers:

$144x^{2} + 256$

$9x^{2}+24x + 16$

$144x^{2} +384x+ 256$

None of the other responses gives a correct answer.

$9x^{2} + 16$

Correct answer:

$9x^{2}+24x + 16$

Explanation:

The length of one side of a square is one fourth its perimeter. Since the perimeter of the square is 12x+16 , the length of one side is

$(12x+16) \div 4 = 3x+ 4$

The area of the square is the square of this sidelength, or

$( 3x+ 4)^{2}= (3x)^{2}+ 2 \cdot 3x \cdot 4 + 4 ^{2} = 9x^{2}+24x + 16$

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

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

Possible Answers:

$2 ^{4 x}$

$2 ^{x+1}$

$2 ^{x+2}$

$2 ^{2 x}$

$2 ^{x+4}$

Correct answer:

$2 ^{2 x}$

Explanation:

The area of a square is the square of its sidelength. Therefore, square $2^{x}$ :

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

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Example Question #15 : Squares

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

Possible Answers:

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

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

$2^{2x+1}$

$2^{2x}$

$2^{2x-1}$

Correct answer:

$2^{2x-1}$

Explanation:

A square being a rhombus, its area can be determined by taking half the product of the lengths of its (congruent) diagonals:

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

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Example Question #223 : Geometry

The lengths of the sides of ten squares form an arithmetic sequence. One side of the smallest square measures sixty centimeters; one side of the second-smallest square measures one meter.

Give the area of the largest square, rounded to the nearest square meter.

Possible Answers:

20 square meters

18 square meters

24 square meters

16 square meters

22 square meters

Correct answer:

18 square meters

Explanation:

Let $a_{n}$ be the lengths of the sides of the squares in meters. $a_{1} = 60 \div 100 = 0.6$ and $a_{2} = 1$ , so their common difference is

d =1 - 0.6 = 0.4

The arithmetic sequence formula is

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

The length of a side of the largest square - square 10 - can be found by substituting $a_{1} = 0.6, n= 10, d = 0.4$ :

$a_{10} =0.6+ (10-1) 0.4 = 0.6 + 9 (0.4) = 0.6 + 3.6 = 4.2$

The largest square has sides of length 4.2 meters, so its area is the square of this, or $A = 4.2^{2} = 17.64$ square meters.

Of the choices, 18 square meters is closest.

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Example Question #21 : Squares

The areas of six squares form an arithmetic sequence. The smallest square has perimeter 16; the second smallest square has perimeter 20. Give the area of the largest of the six squares.

Possible Answers:

100

Correct answer:

Explanation:

The two smallest squares have perimeters 16 and 20, so their sidelengths are one fourth of these, or, respectively, 4 and 5. Their areas are the squares of these, or, respectively, 16 and 25. Therefore, in the arithmetic sequence,

$A_{1} = 16$

$A_{2} = 25$

and the common difference is d = 25 - 16 = 9 .

The area of the th smallest square is

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

Setting $A_{1} = 16, n = 6, d=9$ , the area of the largest (or sixth-smallest) square is

$A_{n}= 16+ (6-1)9 = 16 + 5 \cdot 9 = 16 + 45 = 61$

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Example Question #21 : Quadrilaterals

Which is the greater quantity?

(a) The area of a square with sides of length meters

(b) The area of a square with perimeter 160 X centimeters

Possible Answers:

(b) is the greater quantity

(a) is the greater quantity

It cannot be determined which of (a) and (b) is greater

(a) and (b) are equal

Correct answer:

(a) is the greater quantity

Explanation:

A square with perimeter 160 X centimeters has sides of length one-fourth of this, or $160 X \div 4 = 40 X$ centimeters. Since one meter is equal to 100 centimeters, divide by 100 to get the equivalent in meters - this is

$40 X \div 100 = 0.4X$

meters.

The square in (b) has sidelength less than that of the square in (a), so its area is also less than that in (a).

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Example Question #23 : Squares

On the coordinate plane, Square A has as one side a segment with its endpoints at the origin and at the point with coordinates (A, B) . Square B has as one side a segment with its endpoints at the origin and at the point with coordinates (-B, -A) . and are both positive numbers and B > A . Which is the greater quantity?

(a) The area of Square A

(b) The area of Square B

Possible Answers:

(a) and (b) are equal

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

(b) is the greater quantity

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The length of a segment with endpoints (0,0) and (A, B) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = A$ , $y_{2} = B$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{(A-0)^{2}+(B-0)^{2}}$

$d = \sqrt{A ^{2}+B^{2}}$

The length of a segment with endpoints (0,0) and (-B, -A) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = -B$ , $y_{2} = -A$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{(-B-0)^{2}+(-A-0)^{2}}$

$d = \sqrt{(-B )^{2}+(-A )^{2}}$

$d = \sqrt{B^{2}+A ^{2}}$

$d = \sqrt{A ^{2}+B^{2}}$

The sides are of equal length, so the squares have equal area. Note that the fact that B > A is irrelevant to the question.

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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 #14 : Squares

Example Question #1 : How To Find The Area Of A Square

Example Question #6 : How To Find The Area Of A Square

Example Question #9 : How To Find The Area Of A Square

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

Example Question #15 : Squares

Example Question #223 : Geometry

Example Question #21 : Squares

Example Question #21 : Quadrilaterals

Example Question #23 : Squares

All ISEE Upper Level Quantitative Resources

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