Free Full Length ISEE Upper Level Mathematics Practice Test

The correct answer is:
$E = 6 \ + \ A$
$A = S \ – \ 4$

The correct answer is $36^\circ$
The sum of supplement angles is $180$.
Let $x$ be that angle.
Therefore, $x \ + \ 4 \ x = 180$
$5 \ x = 180$, divide both sides by $5: \ x = 36$

The correct answer is $47$
Let $x$ be the integer. Then:
$2 \ x \ – \ 3 = 91$
Add 5 both sides: $2x = 94$
Divide both sides by $2: \ x = 47$

The correct answer is $$120$ profit
Plug in the value of $x=30$ into both equations. Then:
$𝐶(x)=x^2 \ - \ 4 \ x=(30)^2 \ - 4 \ (30)=900 \ - \ 120=780$
$𝑅(x)=30 \ x=30 \ × \ 30=900$
$900 \ − \ 780=120$

The correct answer is $3$
Subtract $\frac{1}{4 \ b}$ and $\frac{1}{𝑏^2}$ from both sides of the equation. Then:
$\frac{1}{4 \ b^2} \ + \ \frac{1}{4 \ b} = \frac{1}{b^2}→\frac{1}{4 \ b^2} \ − \ \frac{1}{b^2}=− \ \frac{1}{4 \ 𝑏}$
Multiply both numerator and denominator of the fraction $\frac{1}{b^2}$ by $4$. Then:
$\frac{1}{4 \ b^2} \ − \ \frac{4}{4 \ b^2}=− \ \frac{1}{4 \ b}$
Simplify the first side of the equation: $− \ \frac{3}{4 \ b^2 }=− \ \frac{1}{4 \ b}$
Use cross multiplication method: $12 \ 𝑏=4 \ 𝑏^2→12=4 \ 𝑏→𝑏=3$

The correct answer is $108 \ x^8 \ y^6$
Simplify.
$4 \ x^2 \ y^3 \ (3 \ x^2 \ y)^3=4 \ x^2 \ y^3 \ (27 \ x^6 \ y^3) = 108 \ x^8 \ y^6$

The correct answer is $– \ 22 \ ≤ \ x \ ≤ \ 10$
First, multiply both sides of inequality by $2$.
Then: $\frac{|6 \ + \ x|}{2} \ ≤ \ 8→|6 \ +\ x| \ ≤ \ 16$
Since $6 \ + \ x$ can be positive or negative, then: $6 \ + \ x \ ≤ \ 16$ or $6 \ + \ x \ ≥ \ − \ 16$
Then: $x \ ≤ \ 10$ or $x \ ≥ \ − \ 22$

The correct answer is $(2, 2)$
Plug in each pair of numbers in the equation.
The answer should be $12$.
A. $(2, 1): \ \ \ 2 \ (2) \ + \ 5 \ (1) = 9$ No!
B. $(– \ 1, 3): \ \ \ 2 \ (–1) \ + \ 5 \ (2) = 8$ No!
C. $(– \ 2, 2): \ \ \ 2 \ (– \ 2) \ + \ 5 \ (2) = 6$ No!
D. $(2, 2): \ \ \ 2 \ (2) \ + \ 5 \ (2) = 12$ Yes!

The correct answer is $10$
$x\% \ 13 = 1.3$
$\frac{x}{100} \ 13 = 1.3 →$
$x = \frac{1.3 \ × \ 100}{13} = 10$

The correct answer is $(6 \ × \ 10^2) \ + \ (7 \ × \ 10^1) \ + \ 4$
Let’s review the choices provided:
A. $(6 \ × \ 10^3) \ + \ (7 \ × \ 10^2) \ + \ (4 \ × \ 10) = 6,000 \ + \ 700 \ + \ 40 = 6,740$
B. $(6 \ × \ 10^2) \ + \ (7 \ × \ 10^1) \ – \ 4 = 600 \ + \ 70 \ – \ 4= 666$
C. $(6 \ × \ 10^2) \ + \ (7 \ × \ 10^1) \ + \ 4 = 600 \ + \ 70 \ + \ 4 = 674$
D. $(6 \ × \ 10^1) \ + \ (7 \ × \ 10^1) \ + \ 4 = 60 \ + \ 70 \ + \ 4 = 134$
Only choice C equals to $674$.

The correct answer is $4$ cm$^2$
First draw an isosceles triangle.
Remember that two sides of the triangle are equal.
Let put $a$ for the legs. Then:
$𝑎=4⇒$ area of the triangle is $=\frac{1}{2} \ (4 \ × \ 4)=\frac{16}{2}=4$ cm$^2$

The correct answer is $13$ cm
Use Pythagorean Theorem: $a^2 \ + \ b^2 = c^2$
$5^2 \ + \ 12^2 = c^2 ⇒$
$25 \ + \ 144 = c^2 ⇒$
$c^2 = 169 ⇒ c=13$

The correct answer is $37.68$
$C = 2 \ π \ r$
$C = 2 \ π \ × \ 6 = 12 \ π$
$π = 3.14 → C = 12 \ π = 37.68$

The correct answer is $$27.5$
$\frac{5}{200} = \frac{x}{1,100}$
$x = \frac{5 \ × \ 1,100}{200} =27.5$

The correct answer is $1610$ minutes
$23$ hours $= 82,800$ seconds
$1610$ minutes $= 96,600$ seconds
$1$ days $= 24$ hours $= 86,400$ seconds
$3,900$ seconds

The correct answer is $1610$ minutes
$23$ hours $= 82,800$ seconds
$1610$ minutes $= 96,600$ seconds
$1$ days $= 24$ hours $= 86,400$ seconds
$3,900$ seconds

The correct answer is $2,841$
$85.23 \ ÷ \ 0.03 =2,841$

The correct answer is $666$ Miles
To find total number of miles driven by Ed that week, you only need to subtract $46,687$ from $47,353$.
$47,353 \ − \ 46,687=666$

The correct answer is $69\%$
Area of a circle equals: $𝐴=\pi \ r^2$
The new diameter is $30\%$ larger than the original then the new radius is also $30\%$ larger than the original.
$30\%$ larger than $𝑟$ is $1.3 \ r$.
Then, the area of larger circle is: $𝐴=\pi \ r^2=\pi \ (1.3 \ 𝑟)^2=\pi \ (1.69 \ 𝑟^2)=1.69 \ 𝜋 \ r^2$
$1.69 \ \pi \ r^2$ is $69\%$ bigger than $\pi \ r^2$.

The correct answer is $32$ cm$^2$
First draw an isosceles triangle.
Remember that two sides of the triangle are equal.
Let put $𝑎$ for the legs. Then:
$𝑎=5⇒$ area of the triangle is $=\frac{1}{2} \ (8 \ × \ 8)=\frac{64}{2}=32$ cm$^2$

The correct answer is $96$
$x \ + \ y = 24$
Then: $4 \ x \ + \ 4 \ y = 24 \ × \ 4 = 96$

The correct answer is $104$
The area of the non-shaded region is equal to the area of the bigger rectangle subtracted by the area of smaller rectangle.
Area of the bigger rectangle $= 12 \ × \ 10 = 120$
Area of the smaller rectangle $= 8 \ × \ 2 = 16$
Area of the non-shaded region $= 120 \ – \ 16 = 104$

The correct answer is $\frac{16}{x^3}$
$\frac{x^3}{16} ⇒$ reciprocal is : $\frac{16}{x^3}$

The correct answer is $5.5$
$2 \ x^2 \ (y \ + \ 8) = 2 \ (0.5)^2 \ (3 \ + \ 8) = 2 \ (0.25) \ (11) = 5.5$

The correct answer is $$200$
Use interest rate formula:
Interet $=$ principal $×$ rate $×$ time $=2,500 \ × \ 0.04 \ × \ 2=200$

The correct answer is $2 \ (w^2 \ + \  x) = 25$

The correct answer is $5$ hr. $37$ min
$29$ hr. $21$ min
$23$ hr. $44$ min
_________________
$5$ hr. $37$ min

The correct answer is $- \ \frac{ 5}{4}$
$4 \ \frac{1}{2} - 5 \frac{3}{4}=$
$4 \ \frac{2}{4} \ - \ 5 \frac{3}{4} =$
$\frac{18}{4} \ - \ \frac{23}{4} =$
$\frac{- \ 5}{4} =- \ \frac{ 5}{4}$

The correct answer is $(x \ – \ 4)$
Factor each trinomial $x^2 \ – \ 2 \ x \ – 8$ and $x^2 \ – \ 6 \ x \ + \ 8$
$x^2 \ – \ 2 \ x \ – 8 ⇒ (x \ – \ 4) \ (x \ + \ 2)$
$x^2 \ – \ 6 \ x \ + \ 8 ⇒ (x \ – \ 2) \ (x \ – \ 4)$

The correct answer is $(6, 1)$
$\begin{cases}- \ 2 \ x \ - \ 3 \ y =\ - \ 15\\4 \ x\ + \ y= 25\end{cases}⇒$ Multiplication $(3)$ in first equation
$⇒\begin{cases}- \ 2 \ x \ - \ 3 \ y =\ - \ 15\\12 \ x\ + \ 3 y= 75\end{cases}$
Add two equations together $⇒ 10 \ x =60 ⇒ x=6$ then; $y=1$

The correct answer is Ellis spends about $40$ hours of his $60-$hour work week on the road.
Ellis travels $\frac{2}{3}$ of $60$ hours. $\frac{2}{3} \ × \ 60=40$
Ellis will be on the road for $40$ hours.

The correct answer is $0.444$... 
$\frac{12}{27} =0.444$... 

The correct answer is $154$
Perimeter of a rectangle $=2$ (width $+$ length) $=2 \ (49 \ + \ 28)=154$

The correct answer is $658$
The ratio of boys to girls is $5:9$.
Therefore, there are $5$ boys out of $14$ students.
To find the answer, first divide the number of boys by $5$, then multiply the result by $14$.
$235 \ ÷ \ 5 = 47 ⇒ 47 \ × \ 10 = 658$

The correct answer is $20$
$\frac{240}{12} = 20$

The correct answer is $144$ square feet
The area of a $12$ feet $x \ 12$ feet room is $144$ square feet.
$12 \ × \ 12 = 144$

The correct answer is $281.2$
$\frac{38}{100} \ × \ 740 =x$
$x=281.2$

The correct answer is $19$
Michelle $=$ Karen $– \ 6$
Michelle $=$ David $– \ 4$
Karen $+$ Michelle $+$ David $= 67$
Karen $+ \ 6 =$ Michelle $⇒$ Karen $=$ Michelle $– \ 6$
Karen $+$ Michelle $+$ David $= 67$
Now, replace the ages of Karen and David by Michelle. Then:
Michelle $+ \ 6 \ +$ Michelle $+$ Michelle $+ \ 4 =67$
$3$ Michelle $+ \ 10 = 67 ⇒ 3$ Michelle $= 67 \ – \ 10$
$3$ Michelle $= 57$
Michelle $= 19$

The correct answer is $2$ ft.
Write a proportion and solve for the missing number.
$\frac{24}{8} =\frac{ 6}{x}→ 24 \ x=6 \ × \ 8=48$
$24 \ x=48→x=\frac{48}{24}=2$

The correct answer is $12.6$ square feet
$A = b \ h$
$A = 3 \ × \ 4.2 = 12.6$

The correct answer is $\ \begin{bmatrix}2 & 0\\0 & - \ 2\\- \ 3 & 3 \end{bmatrix}$
$\begin{bmatrix}2 & 6 \\- \ 1 & - \ 2\\- \ 5 & - \ 4 \end{bmatrix} \ + \ \begin{bmatrix}0 & - \ 6 \\1 & 0\\2 & 7 \end{bmatrix} = \ \begin{bmatrix}2 & 0\\0 & - \ 2\\- \ 3 & 3 \end{bmatrix}$

The correct answer is $8$
$7∎13=\sqrt{7^2 \ + \ 15}=\sqrt{49 \ + \ 15} =\sqrt{64} = 8$

The correct answer is $(150) \ (0.85) \ (0.70)$
To find the discount, multiply the number by $(100\% \ –$ rate of discount).
Therefore, for the first discount we get: $(150) \ (100\% \ – \ 15\%) = (150) \ (0.85)$
For the next $15\%$ discount: $(150) \ (0.85) \ (0.70)$

The correct answer is $69.11$ kg
Average $=\frac{ sum \ of \ terms}{ number \ of \ terms}$
The sum of the weight of all girls is: $21 \ × \ 65 = 1,365$ kg
The sum of the weight of all boys is: $30 \ × \ 72 = 2,160$ kg
The sum of the weight of all students is: $1,365 \ + \ 2,160 = 3,525$ kg
Average $= \frac{3,525 }{51} = 69.11$

The correct answer is $480$ Liters
Let $x$ be the capacity of one tank. Then, $\frac{3}{4} \ x=320→x=320 \ × \ \frac{3}{4}=240$ Liters
The amount of water in two tanks is equal to: $2 \ × \ 240=480$ Liters

The correct answer is $- \ 2 \ x^2 \ - \ 2 \ x \ + \ 12$
Use FOIL (First, Out, In, Last)
$(2 \ x \ + \ 6) \ (- \ x \ + \ 2) = - \ 2 \ x^2 \ + 4 \ x \ - \ 6 \ x \ + \ 12 =- \ 2 \ x^2 \ - \ 2 \ x \ + \ 12$

The correct answer is $65 \ π$
To find the area of the shaded region subtract smaller circle from bigger circle.
$S_{bigger} \ – \ S_{smaller} = π \ (r_{bigger} )^2 \ – \ π \ (r_{smaller} )^2 ⇒ S_{bigger} \ – \ S_{smaller }= π \ (9)^2 \ – \ π \ (4)^2$
$⇒ 81 \ π \ – \ 16 \ π = 65 \ π$