ISEE Upper Quant practice test

The legs are 9 and 12. The hypotenuse is $\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15$. The perimeter is $9+12+15=36$.

The total is $12+18+10+20=60$, and $60÷4=15$.

When $x=2$, $y=2(2)+1=5$. The y-intercept of $y=2x+1$ is 1, so $k=1$. Thus $5>1$.

Column A is negative, while $x^2$ is positive for any nonzero value of $x$. Column B is greater.

The composite numbers are $4,6,8,9,$ and 10. Including 1 gives 6 favorable outcomes among 10 equally likely sections, so the probability is ${\displaystyle \frac{6}{10}}={\displaystyle \frac{3}{5}}$.

Volume is $\pi r^{2}h$. For $P$, $\pi (3{)}^2(5)=45\pi$. For $Q$, $\pi (5{)}^2(3)=75\pi$. Column B is greater.

The possible values of $x$ are 3, 4, 5, and 6. Their distances from 5 are 2, 1, 0, and 1, so the set of possible absolute values is {0, 1, 2}.

$\sqrt{64+9}=\sqrt{73}$ and $\sqrt{64}+\sqrt{9}=8+3=11$. Because $8^2=64<73<121=11^2$, it follows that $8<\sqrt{73}<11$. Therefore $\sqrt{64}+\sqrt{9}$ is greater.

The original width is 216 divided by $18=12$ meters. After the path is added, the width is 14 meters. The new area is 18 times $14=252$ square meters.

Column A expands to $x^2+2x+1$. Its difference from Column B is $2x$, which is positive for every positive $x$, so Column A is greater.

Four notebooks total $4\times 3=\mathrm{\$}12$. Adding $7 gives $19 for five notebooks, so the mean is $19÷5=\mathrm{\$}3.80$. Keeping $3 ignores the new price; $4 would require a total of $20; $5 overweights the new notebook.

Let $x$ and $y$ be the token counts in Bin 1 and Bin 2. Then ${\displaystyle \frac{x}{3}}={\displaystyle \frac{y}{5}}$, so $x:y=3:5$. The 48 tokens make 8 equal parts, each containing 6 tokens. Bin 1 therefore contains $3\times 6=18$ tokens.

The base is 14 units and the perpendicular height is 7 units, so the area is ${\displaystyle \frac{1}{2}}\times 14\times 7=49$ square units.

The slope is $(-2-6)\mathrm{/}(5-1)=-2$. Using $(1,6)$ gives $6=-2(1)+b$, so $b=8$.

Perpendicular to slope $-2$ has slope ${\displaystyle \frac{1}{2}}$. With $y$-intercept 1, the equation is $y={\displaystyle \frac{1}{2}}x+1$.

Since $3k-5=7$, $k=4$. Substituting into the function gives $f(5)=4(5)-5=15$.

Each term is twice the previous term plus 1, so the next term is $2(47)+1=95$, less than 96.

The probability of landing on a highlighted section is ${\displaystyle \frac{5}{12}}$. The expected number is $24\times {\displaystyle \frac{5}{12}}=10$.

The midpoint of $AB$ has $x$-coordinate 1, so the symmetry line is $x=1$. Point $P$ is 3 units left of this line, so its reflection is 3 units to the right at $(4,1)$.

If $x=1$, the mean is ${\displaystyle \frac{3+7+11+1}{4}}=5.5$, which is less than 7. If $x=9$, the mean is ${\displaystyle \frac{3+7+11+9}{4}}=7.5$, which is greater than 7. The relationship cannot be determined.

Noah earns $30+40=70$ points out of $30+50=80$ possible points. Since $70÷80=0.875$, he earns $87.5\mathrm{\%}$.

$V=\pi r^{2}h$. Doubling $r$ multiplies $V$ by 4; doubling $h$ multiplies $V$ by 2. Column A is greater.

If $x=3$, then both quantities are 10. If $x=4$, Column A is 13 and Column B is 11. Since the relationship changes, it cannot be determined.

Equal rows require $t$ to divide 45 and $d$ to divide 35. The smallest allowed divisors greater than 1 are $t=3$ and $d=5$, so the smallest possible product is $3\times 5=15$.

Squaring gives $x+9=(x-3{)}^2$, or $x^2-7x=0$. The candidates are 0 and 7, but only 7 satisfies the original equation.

Quantity A is $25\pi -9\pi =16\pi$. Since $\pi >3$, $16\pi >48$. Quantity B is $8\cdot 6=48$, so Quantity A is greater.

The original sum is $7(26)=182$ and the remaining sum is $6(24)=144$, so the removed number is 38.

The parabola opens downward with vertex at $x=-b\mathrm{/}(2a)=8\mathrm{/}2=4$, which lies in $[-1,6]$. Then $y(4)=-16+32-7=9$. Endpoint checks: $y(-1)=-16$ and $y(6)=5$, so the maximum is 9.

The path leaves a planted rectangle measuring 12 by 7, with area 84. Removing the 2 by 3 fountain leaves $84-6=78$ square feet.

June sales are $80\times 1.75=140$. July sales are $140\times 0.25=35$. Since $35<80$, Column B is greater.

Without the restriction, there are $3\cdot 4\cdot 3=36$ sandwiches because the spread has three possibilities. Egg with pesto is forbidden for each of the 3 breads, so $36-3=33$ sandwiches are allowed.

A negative exponent takes the reciprocal, so $(-n{)}^{-3}=1\mathrm{/}(-n{)}^3=-1\mathrm{/}{n}^3$.

Every odd square leaves remainder 1 modulo 8, and 1 is less than 2.

Let $t={\displaystyle \frac{y}{x}}$. Since $x<y$, $t>1$, and ${\displaystyle \frac{x}{y}}={\displaystyle \frac{1}{t}}$. For any $t>1$, $t+{\displaystyle \frac{1}{t}}$ is greater than 2, so Quantity A is greater.

A $4\times 4\times 3$ box has surface area $2(16+12+12)=80$, less than 100, while a $48\times 1\times 1$ box has surface area $2(48+48+1)=194$, greater than 100. The relationship cannot be determined.

Reducing two times decreases the total, so the mean decreases. With 18 times the median is the average of the 9th and 10th ordered values. The two corrected times remain the two largest, so the middle pair is unchanged and the median stays 50 minutes.

Among the factor pairs of 96, the closest pair satisfying the difference condition is 6 and 16. The least possible perimeter is therefore $2(6+16)=44$ units.