ISEE Upper Math practice test

The numbers are already in order. With five numbers, the median is the middle value, which is 10.

Combining like terms gives $10a+9a=19a$ and $12-11=1$.

Since $8=2^{3}$, the equation $2^{n}=8$ gives $n=3$.

The two angles must total $180^\circ$, so the other angle is $180^\circ-65^\circ=115^\circ$.

On a number line, the farther-right coordinate is greater. The labeled values are $P=-3$, $R=0$, $S=5$, and $Q=10$, so $Q$ is greatest.

The spinner has $2+3+1=6$ equal sections, and 3 are blue. The probability is $\dfrac{3}{6}=\dfrac{1}{2}$.

For equal base area and height, the pyramid has one-third the prism’s volume. Thus, $84\div3=28\text{ cm}^3$.

The graph has rule $y=-3x+16$. Substituting $x=2$ gives $y=-3(2)+16=10$.

For a one-unit move right, the vertical change is 3. Thus rise divided by run is $3/1=3$.

There are 6 plants that are both full sun and short out of 40 plants, so the probability is $\dfrac{6}{40}=\dfrac{3}{20}$.

Subtract 17 from both sides: $42-17=25$.

Apply $F=\dfrac{9}{5}C+32$: $\dfrac{9}{5}(40)+32=72+32=104$.

Segment $RS$ is horizontal, so a perpendicular walkway through $T$ must be vertical. The point $(1,4)$ has the same $x$-coordinate as $T$.

Adding 1 to both sides gives $2x<20$, and division by positive 2 gives $x<10$.

Range $= 48 - 12 = 36$. Interquartile range $= Q_3 - Q_1 = 32 - 18 = 14$.

There are 5 ratio parts, each worth $25\div5=5$. The red portion is $2(5)=10$.

Convert each value to a decimal. 0.7 is already 0.7. $\dfrac{2}{3}\approx0.667$. $65\%=0.65$. $\dfrac{5}{8}=0.625$. Since $0.7>0.667>0.65>0.625$, the largest value is 0.7.

Subtracting $3x$ and 4 from both sides gives $2x=14$, so $x=7$.

From grade 6 through 11, kilograms rise steadily (positive association). Grade 12’s 14 kg falls far below the trend, so it is an outlier.

A quadrilateral’s interior angles total $360^\circ$. The fourth angle is $360^\circ-(82^\circ+97^\circ+111^\circ)=70^\circ$.

There are $5+3+4=12$ marbles in all. The marbles that are not blue are the 5 red and 4 green marbles, for 9 favorable outcomes. The probability is $\dfrac{9}{12}=\dfrac{3}{4}$.

The shortest side was multiplied by 3, so every side is multiplied by 3. The hypotenuse is $17(3)=51$.

Distributing gives $6x-15=4x+9$. Subtracting $4x$ from both sides gives $2x-15=9$, so $2x=24$ and $x=12$.

$36 = 2^{2} \times 3^{2}$. The distinct prime factors are 2 and 3, and $2 + 3 = 5$.

When a vertical axis is truncated above zero, bar heights exaggerate gaps between close totals. The labeled counts are accurate, but the visual comparison is distorted by the broken scale.

Rewrite $9.6\times10^{4}$ as $0.96\times10^{5}$. Then $7.2+0.96=8.16$, so $8.16\times10^{5}$.

The missing leg is $\sqrt{15^2-9^2}=12$.

Independence allows multiplying: $\dfrac{3}{5}\times\dfrac{2}{3}=\dfrac{2}{5}$.

$3(2x+4) = 6x + 12$ and $-2(x-5) = -2x + 10$. Sum: $(6x - 2x) + (12 + 10) = 4x + 22$. Distractor $4x+2$ mishandles $-2 \cdot -5$ as $-10$; $8x+22$ adds the $x$-terms wrong; $8x+2$ combines both errors.

The reflection gives $(3,4)$, and moving down 2 gives $(3,2)$.

Convert 80 centimeters to 0.8 meter. The area is 2.5 x 0.8 = 2.0 square meters.

The weekday interquartile range is $7000-4000=3000$ steps. The weekend interquartile range is $10000-5000=5000$ steps. The difference is $5000-3000=2000$ steps.

The temperature is $89-4h$. Solving $89-4h>-83$ gives $-4h>-172$, so reversing the sign yields $h<43$.

The call uses $30\%$ of $80\%$, which is $0.30\times80\%=24\%$ of the full battery. The remaining charge is $80\%-24\%=56\%$.

Rise is 6 and run is 4, so slope is $6/4=3/2$.

The scale factor is $15/9=5/3$. The larger perimeter is $42\times 5/3=70$.

Equally likely outcomes: 8. Favorable: HHH, HHT, HTH, THH — 4 outcomes. Probability $\dfrac{4}{8}=\dfrac{1}{2}$.

Expanding gives $q^2-4q-32=q^2-64$. Canceling $q^2$ and solving $-4q-32=-64$ gives $q=8$.

New volume is $\pi(3r)^2\left(\dfrac{h}{3}\right) = \pi\cdot 9r^2\cdot\dfrac{h}{3} = 3\pi r^2h = 3 \times 180\pi = 540\pi$. Scaling by 9 without the height change, or by $\dfrac{1}{3}$ only, yields the distractors.

Set $x-3=2$ so $x=5$. Then $f(2)=4(5)+1=21$.

The only way to have no odd result is for both spins to be even. That probability is $\dfrac{2}{5}\times\dfrac{2}{5}=\dfrac{4}{25}$, so the probability of at least one odd result is $1-\dfrac{4}{25}=\dfrac{21}{25}$.

Numeric GCF is 6; letter GCF is $p$ only → $6p$.

Let the width be $w$ and the length be $2w+3$. Since the perimeter is 42, $2(w+2w+3)=42$, so $3w+3=21$ and $w=6$. The length is 15, so the area is $15\cdot6=90$.

Since $4^b=(2^2)^b=2^{2b}$, the equation becomes $2^{a+2b}=2^{10}$, so $a+2b=10$. With $a+b=7$, subtracting the equations gives $b=3$.

At an intersection, $2x-1=x^2-6x+11$. This gives $x^2-8x+12=0$, so $(x-2)(x-6)=0$. The $x$-coordinates are 2 and 6, and their sum is 8.

The expected change is $(-2)\times\dfrac{1}{4}+1\times\dfrac{1}{2}+5\times\dfrac{1}{4}=\dfrac{5}{4}$ points.