10 Most Common 8th Grade IAR Math Questions

C
\(4\%\) of the volume of the solution is alcohol. Let \(x\) be the volume of the solution.
Then: \(4\%\) of \(x = 24 ml ⇒ 0.04 x = 24 ⇒ x = 24 ÷ 0.04 = 600\)

B
Use the area of the rectangle formula \((s = a × b)\).
To find the area of the shaded region subtracts the smaller rectangle from the bigger rectangle.
\(S_{1} – S_{2} = (10 ft × 8ft) – (5ft × 8ft) ⇒ S_{1} – S_{2} = 40ft\)

B
Use the formula for Percent of Change
\(\frac{New \space Value-Old \space Value}{Old \space Value}× 100 \%\)
\(\frac{28-40}{40}× 100 \% = –30 \% \)(negative sign here means that the new price is less than old price).

C
Use simple interest formula:
\(I=prt\)
(I = interest, p = principal, r = rate, t = time)
\(I=(5000)(0.05)(4)=1000\)

8
Use the formula of rectangle prism volume.
\(V = (length) (width) (height) ⇒ 2000 = (25) (10) (height) ⇒
height = 2000 ÷ 250 = 8\)

B
Use this formula: Percent of Change
\(\frac{New \space Value-Old \space Value}{Old \space Value}× 100 %\)
\(\frac{16000-20000}{20000}× 100 % \)
\(= 20 \% \space and \space \frac{12800-16000}{16000}× 100 \% = 20 \%\)

B
To find the area of the shaded region subtract the smaller circle from the bigger circle.
\(S_{bigger}-S_{smaller} =π(r_{bigger})^2 -π(r_{smaller})^2⇒S_{bigger}-S_{smaller}=π(6)^2-π(4)^2\)
\(⇒ 36 π – 16π = 20 π\)

18
\(a = 6 ⇒\) area of the triangle is:
\(\frac{1}{2}(6×6)=\frac{36}{2}=18 \space cm^2\)

C
\($9×10=$90\)
Petrol use: \(10×2=20 \space liters\)
Petrol cost: \(20×$1=$20\)
Money earned: \($90-$20=$70\)

D
\(\frac{-x}{2}+ \frac{y}{4} = 1\)
\(\frac{-5y}{6}+ 2x = 4\)
Multiply the top equation by \(4\). Then,
\(-2x+ y = 4\)
\(\frac{-5y}{6}+ 2x = 4\)
Add two equations.
\(\frac{1}{6}y=8→y=48\), plug in the value of y into the first equation →\(x=22\)

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