10 Most Common TASC Math Questions
B. 132 cm
C. 156 cm
D. 175 cm
9-In five successive hours, a car traveled 40 km,45 km,50 km,35 km, and 55 km. In the next five hours, it traveled with an average speed of 65 km per hour. Find the total distance the car traveled in 10 hours.
A. 425 km
B. 450 km
C. 550 km
D. 600 km
10- How long does a 420–miles trip take moving at 65 miles per hour (mph)?
A. 4 hours
B. 4 hours and 24 minutes
C. 6 hours and 24 minutes
D. 8 hours and 30 minutes
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Answers:
1- D
Write the equation and solve for B: 0.60 A=0.20 B, divide both sides by 0.20, then:
\(\frac{0.60}{0.20 } A=B\), therefore: B=3A, and B is 3 times of A or it’s \(300\%\) of A.
2- B
Use this formula: Percent of Change \(\frac{New \ Value-Old \ Value}{Old \ Value}×100\% \)
\(\frac{15000-20000}{20000}×100\%=-25\%\) and \(\frac{11200-15000}{15000}×100\%=-25\% \)
3- D
If the length of the box is 45, then the width of the box is one-third of it, 15, and the height of the box is 5 (one-third of the width). The volume of the box is:
V=lwh =(45)(15)(5)=3375
4- D
To find the number of possible outfit combinations, multiply the number of options for each factor: \(6×3×5=90\)
5- C
Use simple interest formula: I=prt (I=interest,p=principal,r=rate,t=time)
I=(8,000)(0.045)(5)=1,800
6- B
\(6^4\)=6×6×6×6=1,296
7- C
\(x=\frac{15}{20}=0.75=75\% \)
8- C
The perimeter of the trapezoid is 56. Therefore, the missing side (height) is =56-18-12-14=12. Area of a trapezoid: \(A=\frac{1}{2 } h (b_{1}+b_{2})=\frac{1}{2}(12)(12+14)=156\)
9- C
Add the first 5 numbers. 40+45+50+35+55=225
To find the distance traveled in the next 5 hours, multiply the average by the number of hours.
Distance=Average×Rate = 65×5 =325
Add both numbers. 325+225=550
10- C
Use distance formula: Distance =Rate×time ⇒ 420=65×T, divide both sides by \(65. 420÷65=T ⇒ T=6.4\) hours. Change hours to minutes for the decimal part. 0.4 hours =0.4×60 =24 minutes.
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