ISEE Upper Quant practice test

Two sides have the same length, so the triangle is isosceles.

Tuesday has 15 laps and Wednesday has 11 laps. Their difference is $15-11=4$ laps.

Substitute 6 for $n$ in $3n$ to get $3\times6=18$.

0.5 of 40 is 20, so the quantities are equal.

There are 5 favorable outcomes among 14 equally likely badges, so the probability is $\dfrac{5}{14}$.

There are four cubes on the bottom layer and one cube on top, for 5 unit cubes and a volume of 5 cubic centimeters.

The commutative property permits the factors to change order, so $6\times4=4\times6$.

Any nonzero number raised to the zero power equals 1.

Angle 1 and Angle 2 are corresponding angles, so they have equal measures.

Combining like terms gives $5a^2+4a+6$. The coefficient of $a$ is 4 and the constant term is 6.

The total is $12+18+10+20=60$, and $60\div4=15$.

The distance changes by $18,18,0,$ and 24 miles. The greatest change occurs from hour 3 to hour 4.

If the perimeter is 26 and the length is 8, then 2(8 + w) = 26, so w = 5. The area is 8 x 5 = 40, equal to Column B.

Rise over run from $(0,1)$ to $(4,3)$ is $\dfrac{3-1}{4-0}=\dfrac{2}{4}=\dfrac{1}{2}$.

Line $r$ crosses the $y$-axis at $(0,3)$ and line $s$ at $(0,-1)$. Since $3 > -1$, line $r$ has the greater $y$-intercept. The slopes differ, but the question asks only about intercepts.

Substitute the input carefully: $r(-5)=(-5)^2+2(-5)=25-10=15$, so the requested value is 15.

The first differences are $+3, +5, +7, +9$, so the next difference is $+11$. Then $26 + 11 = 37$.

The probability of heads is $1/2$, and the probability of rolling an even number is $3/6=1/2$. These independent conditions occur together with probability $(1/2)(1/2)=1/4$.

Point $C$ is 3 units left of $x=5$, so its image is 3 units to the right at $x=8$. A vertical reflection leaves its $y$-coordinate unchanged, giving $(8,5)$.

In order, the lengths are 8, 12, 14, 17, 21, and 25. The median is the mean of the two middle values, so $(14+17)\div2=15.5$.

Since $P$ is $25\%$ greater than $N$, the difference $P-N$ is $\dfrac{1}{4}N$. Because $N$ is positive, $\dfrac{1}{4}N$ is greater than $\dfrac{1}{5}N$, so Quantity A is greater.

The radius is half the diameter, so $r=8$. Therefore, the volume is $\dfrac{4}{3}\pi(8^3)=\dfrac{2048}{3}\pi$ cubic centimeters.

The middle integer is $72 \div 3=24$.

The greatest common factor of 24 and 36 is 12. The least prime number greater than 10 is 11. Therefore, Column A is greater.

$4.5\times10^{-3}=0.0045$, so the two quantities are equal.

The base is 14 units and the perpendicular height is 7 units, so the area is $\dfrac{1}{2}\times14\times7=49$ square units.

The two greatest counts are 22 and 18, and all five bars total 80. Therefore, $\dfrac{22+18}{80}=\dfrac{40}{80}=0.50$, or 50 percent.

The constants must multiply to $-20$ and add to $-1$. The numbers $-5$ and 4 meet both conditions, giving $(z-5)(z+4)$.

A regular 12-gon has interior angle $180(12-2)/12=150^\circ$.

After a $25\%$ decrease, $90 represents $75\%$ of the original price. The original price was therefore $\$90 \div 0.75 = \$120$.

After a red marble is removed, 9 marbles remain, of which 3 are blue, so the probability is $\dfrac{3}{9}=\dfrac{1}{3}$.

Since $r^0=1$, subtracting the denominator’s exponent gives $r^{-2-(-6)}=r^4$.

The bells ring together every $\operatorname{lcm}(12,18)=36$ minutes. From $9{:}00$ through noon is 180 minutes, so they ring together at $0,36,72{,}108{,}144,$ and 180 minutes: 6 times. Column A is greater.

Since $y-x=(2x-3)-x=x-3$, and $x<0$, we know $x-3<0$. Therefore $y<x<0$, so $y$ is farther from 0 than $x$ is. Thus $y^2>x^2$, and Column A is greater.

Each smaller sphere has $\dfrac{1}{64}$ of the original volume, so its radius is scaled by $\sqrt[3]{\dfrac{1}{64}}=\dfrac{1}{4}$. Its radius is therefore $12\times\dfrac{1}{4}=3$ centimeters.

Ordered list (11 values): 0,0,1,1,1,1,2,2,2,3,5. The median is the 6th value, 1. The mean is $(0+0+1+1+1+1+2+2+2+3+5)/11 = 18/11 \approx 1.64$, which is greater than 1. The mode is 1, which is not greater than the mean.

The two interior points do not determine the circle’s center. Centers with $x$-coordinates on either side of 2 are possible.