Full Length ISEE Upper Level Mathematics Practice Test

Solve for $x$.
$3 \ \leq \ 2 \ x \ - \ 3 \ < \ 13⇒$ (add $3$ all sides) $3 \ + \ 3 \ \leq \ 2 \ x \ - \ 3 \ + \ 3 \ < \ 13 \ + \ 3⇒$
$6 \ \leq \ 2 \ x \ < \ 16⇒$ ⇒ (divide all sides by $2$)
$3 \ ≤ \ x \ < \ 8$
$x$ is between $3$ and $8$.
Choice C represent this inequality.

A linear equation is a relationship between two variables, $x$ and $y$, and can be written in the form of $y= m \ x \ + \ 𝑏$.
A non-proportional linear relationship takes on the form $y= m \ x \ + \ 𝑏$, where $b ≠ 0$ and its graph is a line that does not cross through the origin.

The correct answer is $576$ cm
Write the proportion and solve for missing side.
$\frac{Smaller \ triangle \ height}{Smaller \ triangle \ base }= \frac{Bigger \ triangle \ height}{Bigger \ triangle \ base}⇒$
$\frac{100 \ cm}{162 \ cm} = \frac{100 \ + \ 250 \ cm}{x }⇒ x=576$ cm

The correct answer is $- \ 7, - \ \frac{1}{2}$
$x{1,2} = \frac{− \ b \ ± \ \sqrt{b^2 \ − \ 4 \ a \ c}}{ 2a}$
$a \ x^2 \ + \ b \ x + \ c = 0$
$2 \ x^2 \ + \ 15 \ x \ + \ 7 = 0 ⇒$ then:
$a = 2, b = 15$ and $c = 7$
$x = \frac{− \ 15 \ + \ \sqrt{15^2 \ − \ 4 \ . \ 2 \ . \ 7}}{ 2 \ . \ 2 }== \frac{− \ 15 \ + \ 13}{ 4 }= \frac{− \ 2}{4}= - \ \frac{1}{2}$
$x =\frac{ − \ 15 \ − \ \sqrt{15^2 \ − \ 4 \ . \ 2 \ . \ 7}}{ 2 \ . \ 2}=\frac{ − \ 15 \ − \ 13}{4}=\frac{ − \ 28}{4}= – \ 7$

The correct answer is $2$
$|1 \ – \ (12 \ ÷ \ | 1 \ - \ 5 |) | =|1 \ – \ (12 \ ÷ \ | - \ 4 |) | =|1 \ – \ (12 \ ÷ \ 4) | = |1 \ – \ 3 | = | – \ 2 | = 2$

The correct answer is $350$
The ratio of boy to girls is $7:4$.
Therefore, there are $7$ boys out of $4$ students.
To find the answer, first divide the total number of students by $11$, then multiply the result by $7$.
$550 \ ÷ \ 11 = 50 ⇒ 50 \ × \ 7 = 350$

Translated $5$ units down and $4$ units to the left means: $(x.y) ⇒ (x \ − \ 4, y \ − \ 5)$

The correct answer is $7.685 \ × \ 10^{–7}$
$0.0000007685 = \frac{7.685}{10000000 }⇒7.685 \ × \ 10^{–7}$

The correct answer is $x^2 \ - \ 6 \ x \ - \ 27$
Use FOIL (First, Out, In, Last) method.
$(x \ - \ 9) \ (x \ + \ 3) = x^2 \ + \ 3 \ x \ - \ 9 \ x \ - \ 27 =x^2 \ - \ 6 \ x \ - \ 27$

The correct answer is $\frac{7}{48}$
$\frac{2}{25} = 0.08$ and $25\% = 0.25$ therefore $x$ should be between $0.08$ and $0.25$
Choice B, $\frac{7}{48} = 0.145$ is between $0.08$ and $0.25$

The correct answr is $s \ \sqrt{15}$
Use Pythagorean theorem: $a^2 \ + \ b^2 = c^2→$
$s^2 \ + \ ℎ^2=(4 \ s)^2→$
$s^2 \ + \ ℎ^2=16 \ s^2$
Subtracting $s^2$ from both sides gives: $ℎ^2=15 \ s^2$
Square roots of both sides:
$ℎ=\sqrt{15 \ s^2}=\sqrt{s^2} \times \sqrt{15}  =s \ \sqrt{15}$

The area of the trapezoid is $40 \ + \ 4 \ \sqrt{ 10 }$
Area $=\frac{1}{2} \ ℎ \ (𝑏1 \ + \ 𝑏2)=\frac{1}{2} \ (x) \ (16 \ + \ 12)=168 →$
$14 \ x=168→x=12$
$y=\sqrt{4^2 \ + \ 12^2}=\sqrt{16 \ + \ 144}=\sqrt{160}=\sqrt{16 \times 10 } =4 \ \sqrt{ 10 }$
The perimeter of the trapezoid is:
$12 \ + \ 16 \ + \ 12 \ + \ 4 \ \sqrt{ 10 } =40 \ + \ 4 \ \sqrt{ 10 }$

The correct answer is $9,860$ sq. ft.
Area $=$ w $×$ h
Area $= 145 \ × \ 68 = 9,860$ sq. ft.

The correct answer is $6$
Emily $=$ Daniel
Emily $= 3$ Claire
Daniel $= 12 \ +$ Claire
Emily $=$ Daniel $→$ Emily $= 12 \ +$ Claire
Emily $= 3$ Claire $→ 3$ Claire $= 12 \ +$ Claire $→$
$3$ Claire $–$ Claire $= 12$
$2$ Claire $= 12$
Claire $= 6$

The correct answer is $465$
$93 \ ÷ \ \frac{1}{5} = \frac{\frac{93}{1}}{\frac{1}{5}} = 93 \ × \ 5=465$

The correct answer is $- \ 3$
$3 \ – \ 12 \ ÷ \ (5^2 \ ÷ \ 5) =$
$3 \ – \ 12 \ ÷ \ (25 \ ÷ \ 5) =$
$3 \ – \ 12 \ ÷ \ (3) = - \ 9 \ ÷ \ 3= - \ 3$

The correct answer is $2 \ \frac{1}{10}$ Miles
$\frac{1}{2}$ of the distance is $4 \ \frac{1}{5}$ miles.
Then: $\frac{1}{2} \ × \ 4 \ \frac{1}{5}=\frac{1}{2} \ × \ \frac{21}{5}=\frac{21}{10}$
Converting $\frac{21}{10}$ to a mixed number gives:
$\frac{21}{10}=2 \ \frac{1}{10}$

The correct answer is $0.25$
$\frac{5}{32}=0.25$

The correct answer is $\frac{7}{15}$
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 $=\frac{number \ of \ desired \ outcomes}{number \ of \ total \ outcomes}=\frac{x}{30 \ + \ 20 \ + \ 40 \ + \ x}=\frac{1}{10}$
Use cross multiplication and solve for $x$.
$10 \ x=90 \ + \ x→9 \ x=90→x=9$
Since, number of purple marbles can be $9$, then, choice be the probability of randomly selecting a purple marble from the bag.
Use same method for other choices.
B. $\frac{1}{4}$
$\frac{x}{20 \ + \ 30 \ + \ 40 \ + \ x}=\frac{1}{4}→4 \ x=90 \ + \ x→3 \ x=90→x=30$
C. $\frac{2}{5}$
$\frac{x}{20 \ + \ 30 \ + \ 40 \ + \ x}=\frac{2}{5}→5 \ x=180 \ + \ 2 \ x→3 \ x=180→x=60$
D. $\frac{7}{15}$
$\frac{x}{20 \ + \ 30 \ + \ 40 \ + \ x}=\frac{7}{15}→15 \ x=630 \ + \ 7 \ x→8 \ x=630→x=78.75$
Number of purple marbles cannot be a decimal.

The correct answer is $0$
$(3.2 \ + \ 3.6 \ + \ 3.8) \ x = x$
$10.6 \ x = x$
Then $x = 0$

The correct answer is $\frac{1}{4}$
For sum of $6: \ (1$ & $5)$ and $(5$ & $1), \ (2$ & $4)$ and $(4$ & $2), \ (3$ & $3)$, therefore we have $5$ options.
For sum of $9: \ (3$ & $6)$ and $(6$ & $3), \ (4$ & $5)$ and $(5$ & $4)$, we have $4$ options.
To get a sum of $6$ or $9$ for two dice: $5 \ + \ 4 = 9$
Since, we have $6 \ × \ 6 = 36$ total number of options, the probability of getting a sum of $6$ and $9$ is $9$ out of $36$ or $\frac{9}{36}=\frac{1}{4}$.

The correct answer is $29$ Miles
average $=\frac{𝑠𝑢𝑚}{total}=\frac{25 \ + \ 40 \ + \ 22}{3}=\frac{87}{3}=29$ Miles

The correct answer is $10 \ 𝜋$
The distance of A to B on the coordinate plane is:
$\sqrt{(x_{1} \ − \ x_{2})^2 \ + \ (y_{1} \ − \ y_{2} )^2}= \sqrt{(12 \ − \ 4)^2 \ + \ (8 \ − \ 2)^2}=\sqrt{8^2 \ + \ 6^2} =\sqrt{64 \ + \ 36}=\sqrt{100} = 10$
The diameter of the circle is $10$ and the radius of the circle is $5$.
Then: the circumference of the circle is:
$2 \ \pi \ 𝑟=2 \ \pi (5)=10 \ 𝜋$

The correct answer is $966.4$
$48.32 \ ÷ \ 0.05 =966.4$

The correct answer is $± \ 2$
$4 \ x^2 \ + \ 8 = 24$
$4 \ x^2 = 16$
$x^2 = 4$
$x =± \ 2$

The correct answer is $$ 113.28$
$25\%$ of $x = 28.32$
$\frac{25}{100 } \ x =28.32$
$x = \frac{100 \ × \ 28.32}{25} = 113.28$

The correct answer is $5$
$\frac{x \ + \ 10 \ + \ 12 \ + \ 15 \ + \ 8}{5} = 10 →$
$x \ + \ 45 = 50 →$
$x = 50 \ – \ 45 = 5$

The correct answer is $254.34$ in$^2$
Diameter $= 18$
then: Radius $= 9$
Area of a circle $= π \ r^2 ⇒$
A $= 3.14 \ (9)^2 = 254.34$ in$^2$

The correct answer is $55$
$\frac{240}{6 } \ < \ x \ < \ \frac{390}{6}$
$40 \ < \ x \ < \ 65$
Then:
Only choice C is correct

The correct answer is $\frac{ - \ 6 \ x \ - \ 20}{6 \ x^2 \ - \ 9}$
Simplify:
$\frac{\frac{1}{3} \ - \ \frac{x \ + \ 4}{2}}{\frac{x^2}{2} \ - \ \frac{3}{4}} = \frac{\frac{1}{3} \ - \ \frac{x \ + \ 4}{2}}{\frac{2 \ x^2 \ - \ 3}{4}} = \frac{4 \ (\frac{1}{3} \ - \ \frac{x \ + \ 4}{2})}{2 \ x^2 \ - \ 3}⇒$
Simplify: $\frac{1}{3} \ - \ \frac{x \ + \ 4}{2}= \frac{2 \ - \ 3 \ (x \ + \ 4)}{6}= \frac{2 \ - \ 3 \ x \ - \ 12}{6}= \frac{ - \ 3 \ x \ - \ 10}{6}$
Then:
$\frac{4 \ (\frac{- \ 3 \ x \ - \ 10}{6})}{2 \ x^2 \ - \ 3} =\frac{2 \ (\frac{- \ 3 \ x \ - \ 10}{3})}{2 \ x^2 \ - \ 3} =\frac{ \frac{- \ 6 \ x \ - \ 20}{3}}{2 \ x^2 \ - \ 3} =\frac{- \ 6 \ x \ - \ 20}{3\ (2 \ x^2 \ - \ 3)}=\frac{ - \ 6 \ x \ - \ 20}{6 \ x^2 \ - \ 9}$

The correct answer is $8$ square feet
Formula of triangle area $= \frac{1}{2}$ (base $×$ height)
Since the angles are $45-45-90$, then this is an isosceles triangle, meaning that the base and height of the triangle are equal.
Triangle area $= \frac{1}{2}$ (base $×$ height) $= \frac{1}{2} \  (4 × 4) = 8$

The correct answer is $40$
$\frac{5}{25}= \frac{x}{200} → x = \frac{5 \ × \ 200}{25} = 40$

The correct answer is $2 \ (4 \ H \ - \ 2) = 25$

The correct answer is $952,691$
The difference of the file added, and the file deleted is:
$552,129 \ – \ 439,566 = 112,563$
$840,128 \ + \ 112,563 = 952,691$

The correct answer is $3,840$
$x^{\frac{1}{2}}$ equals to the root of $x$. Then: $5 \ + \ x^{\frac{1}{2}}=20→4 \ + \ \sqrt{x}=20→\sqrt{x}=16→x=256$
$x=256$ and $15 \ × \ x$ equals: $15 \ × \ 256=3,840$

The correct answer is $1,728$ cm$^3$
If the length of the box is $36$, then the width of the box is one third of it, $12$, and the height of the box is $4$ (one third of the width).
The volume of the box is:
$𝑉 = 𝑙𝑤ℎ = (36) \ (12) \ (4) = 1,728$

The correct answer is $(7, − \ 9)$
When a point is reflected over $x$ axes, the $(y)$ coordinate of that point changes to $(− \ y)$ while its $x$ coordinate remains the same.
C $(7, 9)$ → C’ $(7, − \ 9)$

The correct answer is $3$ days $7$ hours $7$ minutes
$7$ days $21$ hours $21$ minutes $– \ 4$ days $14$ hours $14$ minutes $= 3$ days $7$ hours $7$ minutes

The correct answer is $3$ sq decreased
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.
$36 \ − \ 33=3$

The correct answer is $100$ cm$^2$
The area of the circle is $25 \ π$ cm$^2$, then, its diameter is $10$ cm.
Area of circle $=\pi \ r^2=12 \ π→𝑟^2=25→r=5$
Radius of the circle is $5$ and diameter is twice of it, $10$.
One side of the square equals to the diameter of the circle. Then:
Area of square $=$ side $×$ side $=10 \ × \ 10=100$

The correct answer is $- \ 2 \ x^2 \ y^3 \ + \ 5 \ x^3 \ y^5$
$6 \ x^2 \ y^3 \ + \ 2 \ x^3 \ y^5 \ – \ (8 \ x^2 \ y^3 \ – \ 3 \ x^3 \ y^5) =$
$6 \ x^2 \ y^3 \ + \ 2 \ x^3 \ y^5 \ – \ 8 \ x^2 \ y^3 \ + \ 3 \ x^3 \ y^5 =$
$- \ 2 \ x^2 \ y^3 \ + \ 5 \ x^3 \ y^5$

The correct answer is $131.95$
$45\%$ of $91 = 40.95$
$91 \ + \ 40.95= 131.95$

The correct answer is $70$
Number of squares equal to: $\frac{35 \ ×\ 18}{3 \ × \ 3}=35 \ × \ 2=70$

The correct answer is $4 \ 𝜋$
Let P be circumference of circle A, then; $2 \ \pi \ r_{A} = 12 \ 𝜋→r_{𝐴}=6$
$r_{𝐴}=3 \ r_{B}→r_{B}=\frac{6}{3}=2→$
Area of circle B is; $𝜋 \ r_{B}^2=4 \ 𝜋$

The correct answer is $\left\{(6, 1), (3, 1), (0, 5), (4, 5)\right\}$
A set of ordered pairs represents $y$ as a function of $x$ if: $x_{1}=x_{2}→y_{1}=y_{2}$
In choice A: $(3, - \ 2)$ and $(3, 7)$ are ordered pairs with same $x$ and different $y$, therefore $y$ isn’t a function of $x$.
In choice B: $(4, 2)$ and $(4, 7)$ are ordered pairs with same $x$ and different $y$, therefore $y$ isn’t a function of $x$.
In choice C: $(5, 7)$ and $(5, 18)$ are ordered pairs with same $x$ and different $y$, therefore $y$ isn’t a function of $x$.

The correct answer is $$500$
David’s weekly salary is $$350$ plus $10\%$ of $$1500$.
Then: $10\%$ of $1,500=0.1 \ × \ 1,500=150$
$350 \ + \ 150=500$

The correct answer is $40$
Write a proportion and solve. $\frac{4}{5}=\frac{x}{50}$
Use cross multiplication: $5 \ x=200→x=40$