10 Most Common PSAT Math Questions

C
First, find the factors of numerator and denominator of the expression. Then simplify.
\(\frac{5x^2+75x-80}{x^2-1}=\frac{5(x^2+15x-16)}{(x-1)(x+1)}\)
\(\frac{5(x+16)(x-1)}{ (x-1)(x+1)}=\frac{5(x+16)}{ (x+1)} \)
\(=\frac{(5x+80)}{ (x+1)} \)

A
If \(r=55→ \frac{x^2+6x-55}{x-5}\) = \(\frac{(x+11)(x-5)}{x-5}\) \(= x + 11\)
For all other options, the numerator expression is not divisible by \((x-5)\).

D
Plug in the values of \(x\) and \(y\) in the equation of the parabola. Then:
\(12=a(2)^2+5(2)+10→12=4a+10+10→12=4a+20\)
\(→4a=12-20=-8→a=\frac{-8}{4}=-2→a^2=(-2)^2=4\)

50
\(\frac{y}{4} = x – \frac{2}{5}x + 10 \)
Multiply both sides of the equation by \(5\). Then:
\(5×\frac{y}{4} = 5× (x – \frac{2}{5}x + 10)\)
\(→y=5x-2x+50→y=3x+50\)
Now, subtract \(3x\) from both sides of the equation. Then:
\(y – 3x = 50\)

4
First, factorize the numerator and simplify.
\(\frac{(x-3)(x+3)}{x+3}+2x+8=17\)
\(→x-3+2x+8=15→3x+5=17\)
Subtract \(5\) from both sides of the equation. Then:
\(→3x=17-5→x=\frac{12}{3} x=4\)

25
\(\frac{100x^2 y^4}{4x^2 y^4}\)
Remove \(x^2 y^4\) from both numerator and denominator.
\(\frac{100x^2 y^4}{4x^2 y^4}=\frac{100}{4}=\frac{50}{2}=25\)

1
Remember that, the reflection of the point \((x,y)\) over the line \(y=x\) is the point \((y,x)\). Then:
The reflected point of \(A(2,-1)\), is \((-1,2)\)
The reflected point of \(B(6,3)\) is point \((3,6)\)
Therefore, the slope of the reflected line is:
\(\frac{y_{2} – y_{1}}{x_{2} – x_{1}}=\frac{6-2}{3-(-1)}=\frac{4}{4} \space or\space 1\)

C

The amount of petrol consumed after \(x\) hours is: \(6 × x = 6x\)
Petrol remaining after \(x\) hours driving: \(80 – 6x\)

D
Let \(x\) be the integer. Then:
\(2x – 5 = 83\)
Add 5 both sides: \(2x = 88\)
Divide both sides by \(2:\) \(x = 44\)

B
\(mean) = \frac{sum \space of \space terms}{number \space of \space terms}=\frac{9+12+15+16+19+16+14.5}{7}=14.5\)