10 Most Common ISEE Upper-Level Math Questions

B. s \(\sqrt{7}\)

C. 5s

D. 5s\(^2\)

4-

There are also purple marbles in the bag. Which of the following can NOT be the probability of randomly selecting a purple marble from the bag?

A. \(\frac{1}{10}\)

B. \(\frac{1}{4}\)

C. \(\frac{2}{5}\)

D. \(\frac{7}{15}\)

5- A square measures 6 inches on one side. By how much will the area be increased if its length is increased by 5 inches and its width decreased by 3 inches.

A. 1 sq decreased

B. 3 sq decreased

C. 6 sq decreased

D. 9 sq decreased

6- If a box contains red and blue balls in the ratio of \(2: 3\) red to blue, how many red balls are there if 90 blue balls are in the box?

A. 40

B. 60

C. 80

D. 30

7- How many \(3 × 3\) squares can fit inside a rectangle with a height of 54 and width of 12?

A. 72

B. 52

C. 62

D. 42

8- David makes a weekly salary of $220 plus \(8\%\) commission on his sales. What will his income be for a week in which he makes sales totaling $1100?

A. $328

B. $318

C. $308

D. $298

9- \(4x^2y^3 + 5x^3y^5 – (5x^2y^3 – 2x^3y^5) = \)

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A. \(–x^2y^3\)

B. \(6x^2y^3 – x^3y^5\)

C. \(7x^2y^3\)

D. \(7x^3y^5 – x^2y^3\)

10- If the area of trapezoid is 126 cm, what is the perimeter of the trapezoid?

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Answers:

2- D
\(\frac{1}{3}\) of the distance is \(5\frac{1}{4}\) miles. Then:
\(\frac{1}{3} × 5 \frac{1}{4} = \frac{1}{3} × \frac{21}{4}= \frac{21}{12}\)
Converting \(\frac{21}{12}\) to a mixed number gives:
\(\frac{21}{12}= 1\frac{9}{12}=1\frac{3}{4}\)

3- A
Use Pythagorean theorem:
\(a^2+b^2=c^2→s^2+h^2=(5s)^2→s^2+h^2=25s^2\)
Subtracting (s^2) from both sides gives: \( h^2=24s^2\)
Square roots of both sides: \(h=\sqrt{24s^2}=\sqrt{4×6×s^2 } =\sqrt 4 × \sqrt6 × \sqrt{s^2 }=2 × s × \sqrt6 = 2s\sqrt6\)

4- D
Let \(x\) be the number of purple marbles. Let’s review the choices provided:
A. \(\frac{1}{10}\), if the probability of choosing a purple marble is one out of ten, then:
\(Probability=\frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes}=\frac{x}{20+30+40+x}=\frac{1}{10}\)
Use cross multiplication and solve for x. \(10x=90+x→9x=90→x=9\)
Since the number of purple marbles can be 9, then, the choice is the probability of randomly selecting a purple marble from the bag.
Use the same method for other choices.
B. \(\frac{1}{4}\)
\(\frac{x}{20+30+40+x}=\frac{1}{4}→4x=90+x→3x=90→x=30\)
C. \(\frac{2}{5}\)
\(\frac{x}{20+30+40+x}=\frac{2}{5}→5x=180+2x→3x=180→x=60\)
D. \(\frac{7}{15}\)
\(\frac{x}{20+30+40+x}=\frac{7}{15}→15x=630+7x→8x=630→x=78.75\)
The number of purple marbles cannot be a decimal.

5- B
The area of the square is 36 square inches.
\(Area of square=side×side=6×6=36\)
The length of the square is increased by 5 inches and its width decreased by 3 inches. Then, its area equals:
Area of \(rectangle=width×length=11×3=33\)
The area of the square will be decreased by 3 square inches.
\(36-33=3\)

6- B
Write a proportion and solve.
\(\frac{2}{3}=\frac{x}{90}\)
Use cross multiplication:
\(3x=180→x=60\)

7- A
Number of squares equal to: \(\frac{54×12}{3×3} = 18 × 4 = 72\)

8- C
David’s weekly salary is $220 plus \(8\%\) of $1,100. Then:
\(8\% \space of \space 1,100=0.08 × 1,100 = 88\)
\(220+88=308\)

9- D
\(4x^2y^3 + 5x^3y^5 – (5x^2y^3 – 2x^3y^5) = 4x^2y^3 + 5x^3y^5 – 5x^2y^3 + 2x^3y^5 = – x^2y^3 + 7 x^3y^5\)

10- 13
The area of the trapezoid is:
\(Area=\frac{1}{2}h(b_1+b_2 )=\frac{1}{2}(x)(13+8)=126\)
\(→10.5x=126→x=12\)
\(y=\sqrt{5^2 +12^2 } = \sqrt{25+144} = \sqrt{169} = 13\)

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