Free PSAT Math Practice: 20 Questions with Answers

Let the number of women should be added to the town be \(x\), then:
\(\frac{(528+x)}{550} = 1.2 → 528 + x = 550 × 1.2 = 660 → x = 132\)

The perimeter of the rectangle is:
\(2x + 2y = 30 → x + y =15→ x = 15 – y\)
The area of the rectangle is:
\(x × y = 50 → (15 – y)(y) = 50→ y^2 – 15y + 50 = 0\)
Solve the quadratic equation by factoring method.
\((y – 5)(y – 10) = 0→y = 5\)
(Unacceptable, because y must be greater than 5) or \(y = 10\)
If \(\space y = 10 →x × y = 50→x × 10 = 50→x = 5\)

sin⁡\(β = \frac{opposite \space side}{hypotenuse}\)
To find the hypotenuse, we need to use the Pythagorean theorem.
\(a^2+b^2=c^2→c=\sqrt{(a^2+b^2 )}\)
sin\(⁡(β) =\ \frac{b}{c}= \frac{b}{\sqrt{(a^2+b^2 )}}\)

Combine like terms inside the absolute value:
\(\left|\frac{x}{2}-2x+10\right|=\left|-\frac{3x}{2}+10\right|\).
An absolute value less than 5 gives
\(-5<-\frac{3x}{2}+10<5\).
Subtract 10:
\(-15<-\frac{3x}{2}<-5\).
Multiply by \(-\frac{2}{3}\), reversing both inequality signs:
\(10>x>\frac{10}{3}\).
In increasing order, \(\frac{10}{3}<x<10\).

The intersection of two functions is the point with 2 for \(x\). Then:
\(f(2)=g(2) \space and \space g(2)=(2×(2))-3=4-3=1\)
\(Then, f(2)=1→a(2)^2+b(2)+c=1→4a+2b+c=1 (i)\)
The value of x in the vertex of the parabola is:
\(x=- \frac{b}{2a}→-2=- \frac{b}{2a}→b=4a \space (ii)\)
In the point (-2, 5), the value of the \(f(x)\) is 5.
\(f(-2)=5→a(-2)^2+b(-2)+c=5→4a-2b+c=5 (iii)\)
Using the first two equations:
\(4a+2b+c=1\)
\(4a-2b+c=5\)
Equation 1 minus equation 2 is:
\((i)-(iii) →4b=-4→b=-1 \space \space (iv)\)
Plug in the value of b in the second equation:
\(b=4a →a =\frac{b}{4}= – \frac{1}{4}\)
Plug in the values of a and b in the first equation. Then:
\(4 (-\frac{1}{4}) +2(-1)+c=1→-1-2+c=1→c=1+3→c=4\)
the product of a, b and c= \((-\frac{1}{4}) ×(-1)×4=1\)

Let \(L\) be the ladder’s length. The 30-foot ground distance is adjacent to the 60-degree angle, and the ladder is the hypotenuse.
\(\cos60^\circ=\frac{30}{L}=\frac12\), so \(L=60\) feet.

Let \(x\) be the length of an edge of a cube, then the volume of a cube is: \(V=x^3 \)
The surface area of a cube is:
\(SA=6x^2\)
The volume of cube A is \(1/2\) of its surface area. Then:
\(x^3 = \frac{6x^2}{2}→x^3=3x^2\)
divide both sides of the equation by \(x^2\). Then:
\(\frac{x^3}{x^2} = \frac{3x^2}{x^2} →x=3\)

Multiply both sides by \(x\).
\(x×(3x+6y)=x× (\frac{-3y^2+15}{x})→3x^2+6xy=-3y^2+15 \)
\(→3x^2+6xy+3y^2=15→3×(x^2+2xy+y^2 )=15→x^2+2xy+y^2= \frac{15}{3}\)
\(x^2+2xy+y^2=(x+y)^2, Then:\)
\(→(x+y)^2=5\)

Number of biology books: \(35\)
Total number of books; \(35+95+80=210\)
the ratio of the number of biology books to the total number of books is:
\(\frac{35}{210}= \frac{1}{6}\)

\(α=180^\circ-112^\circ=68^\circ\)
\( β=180^\circ-135^\circ=45^\circ\)
\( x+α+β=180^\circ→x=180^\circ-68^\circ-45^\circ=67^\circ\)

A. \(f(x)=x^2-5 \space if \ x=1→f(1)=(1)^2-5=1-5=-4≠5 \)
B. \(f(x)=x^2-1 \space if \ x=1→f(1)=(1)^2-1=1-1=0≠5 \)
C. \(f(x)=\sqrt{(x+2)} \space if \ x=1→f(1)=\sqrt{(1+2)}=\sqrt 3≠5\)
D. \(f(x)=\sqrt{x}+4 \space if \ x=1→f(1)=\sqrt 1+4=5\)

\(xp+2yq=26→xp=26-2yq \space (1)\)
\(xp+yq=17 \space (2)\)
\((1) \space in \space (2) →26-2yq+yq=17→26-yq=17→yq=26-17=9\)

Let \(x\) be all expenses, then
\(\frac{22}{100}x =$660 →x= \frac{100×$660}{22}=$3000\)
He spent for his rent: \(\frac{27}{100}×$3000 = $810\)

\(\frac{1}{8}= 0.125→C=5\)
\(\frac{1}{20}=0.05→D=5→C×D=5×5=25\)

Let P be the circumference of circle A, then;
\(2πr_{A}=18π→r_{A}=9\)
\(r_{A} = 3 r_{B}→r_{B}= \frac{9}{3} = 3\)
→ Area of circle B is;
\(πr^2_{B}=9π\)

\(f(x)=x^2+2x-5\)
\(f(3t^2 )=(3t^2 )^2+2(3t^2 )-5=9t^4+6t^2-5\)

\(y\) is the intersection of the three circles. Therefore, it must be odd (from circle A), negative (from circle B), and multiple of 5 (from circle C).
From the options, only\( -5\) is odd, negative, and multiple of 5.

let \(x\) be the total number of marbles in the box, then the number of red marbles is: \(x – 132\)
\(p = \frac{1}{4}= \frac{x – 132}{x}\)
Use cross multiplication to solve for \(x\).
\( x=4(x-132)→x=4x-528→3x=528→x=176 \)