Free Quantitative Reasoning MCQs with Answers

131. The quotient of ,a number and five, equals nine less than one half of the number. What is the number?

• A. -20
• B. -10
• C. 20
• D. 30

Explanation: Let the number be represented as x. The equation formed is x/5 + 9 = x/2. Simplifying further, we get x = 30. Therefore, the correct answer is 30. Option A and B are incorrect as they assume negative values for the number which don't satisfy the equation. Option C is incorrect as substituting 20 in the equation does not satisfy both sides of the equation.

132. If point (7,b ) is in Quadrant I and point ( a, -3) is in Quadrant III, in which Quadrant is the point (a,b)?

• A. Quadrant I
• B. Quadrant II
• C. Quadrant III
• D. Quadrant IV

Explanation: To determine the quadrant in which the point (a, b) lies, analyze the given conditions. The point (7, b) in Quadrant I means x is positive, y (b) is positive. The point (a, -3) in Quadrant III means x (a) is negative, y is negative. For (a, b), x (a) is negative and y (b) is positive, placing it in Quadrant II. Other options are incorrect as they do not align with the sign conditions derived from the given points.

133. What is the average rate of change for the relation shown in the table?

• A. 1/3
• B. 1/2
• C. 2/3
• D. 5/6

Explanation: To find the average rate of change, we use the formula (y2 - y1) / (x2 - x1), which represents the change in y divided by the change in x. From the table, select two points, say (x1, y1) = (-3, -1) and (x2, y2) = (6, 5). The change in y is 5 - (-1) = 6, and the change in x is 6 - (-3) = 9. Therefore, the average rate of change is 6/9, which simplifies to 2/3. Options A, B, and D represent incorrect calculations of these differences.

134. In 2005, 120 students at Lincoln High School had smartphones. By 2010, 345 students in the same school had smartphones. Which of the following best describes the annual rate of change in the number of smartphones students had from 2005 to 2010 at Lincoln High School?

• A. The average increase in the number of smartphones per year is 40
• B. The average increase in the number of smart phones per year is 45
• C. The average increase in the number of smartphones per year is 50
• D. The average increase in the number of smartphones per year is 55

Explanation: To determine the annual rate of change, calculate the difference in the number of students with smartphones from 2005 to 2010, which is 345 - 120 = 225. Then, divide this difference by the number of years (5) to find the average annual increase. Thus, 225 divided by 5 equals 45. Therefore, the average increase in the number of smartphones per year is 45, making Option B the correct choice. Options A, C, and D are incorrect because they result in average increases that do not match the calculated result.

135. Which of the following equations represents a line that passes through point (7,6) and is parallel to the x- axis?

• A. x=6
• B. x=7
• C. y=7
• D. y=6

Explanation: A line parallel to the x-axis must have a constant y-value, as it does not rise or fall. The correct equation for the line passing through (7,6) is y=6. Option A and B are vertical lines parallel to the y-axis. Option C is a horizontal line but does not pass through (7,6).

136. y=2x+4x-y=-1Which ordered pair (x,y) satisfies the system of equations shown above?

• A. ( -2, -3)
• B. ( -3, -2)
• C. ( -1, 2)
• D. ( -2, 0)

Explanation: To solve the system of equations, substitute the values of x and y from the ordered pair into both equations. Option B, (-3, -2), satisfies both equations when substituted. The first equation becomes -2 = 2(-3) + 4, which simplifies to -2 = -6 + 4, or -2 = -2. The second equation becomes -3 - (-2) = -1, which simplifies to -1 = -1. Therefore, option B is correct. Options A, C, and D are incorrect as they do not satisfy both equations when substituted.

137. Which of the following expressions is equal to −1 for some values of x?

• A. |1-x|+6
• B. |1-x|+4
• C. |1-x|+2
• D. |1-x|-2

Explanation: Option A) 1 - x = -6 1 + 6 = x x= 7 Option B) - 4=1 - x X = 5 Option C) 1 - x = -2 1 + 2 = x Option D) 1 - x =2 1 - 2 = x = -1 The correct answer is Option D. When |1-x| = 2, the expression |1-x|-2 can be equal to -1 for x = -1. Options A, B, and C are incorrect because they do not account for the absolute value's behavior and the conditions required for the expression to equal -1.

138. At the beginning of a trip, the tank of Chloe's car was filled with 12 gallons of gas. When she travels constantly on the highway at 60 miles per hour, the car consumes 1 gallon of gas per 35 miles. If she traveled 5 hours and 15 minutes on the highway with a constant speed of 60 miles per hour, how many gallons of gas are left in the tank?

• A. 3
• B. 4
• C. 5
• D. 6

Explanation: For this to be solved we have to find the gallons of gas used If car travels 60 miles in one hour Car uses 1 gallon for 35 miles let's find the gallons it uses for 60 miles or in an hour 1 /35 x /60 ( cross multiplying) 60 *1 =35*x 12 /7=x Now multiply it with the hours of the journey Time of the journey : 5 hours and 15 minutes 15 minutes in hours :15/60 =¼Gallons used in the journey:5 ¼ * 12 /7 21/4 *12/7 252/28 =9 gallons Now add 9 to the options whichever adds up to 12 is the answer Option A) 3+9=12 Correct Option B) 9+4=13 Option C) 9+5=14 Option D) 6+9=15 None adds upto 12 except A so A is the answer.

139. In 2005 a house was purchased for $280,000 and in 2013 it was sold at $334,000. Assuming that the value of the house increased at a constant annual rate what will be the price of the house in the year 2018?

• A. $354,250
• B. $361,000
• C. $367,750
• D. $374,500

Explanation: To solve this problem, first determine the annual increase in the house's value between 2005 and 2013. The house was bought for $280,000 in 2005 and sold for $334,000 in 2013, which is an increase of $54,000 over 8 years. This gives an annual increase of $54,000 / 8 = $6,750.From 2013 to 2018, there are 5 years. Thus, the increase in value over these 5 years is $6,750 x 5 = $33,750.Adding this increase to the 2013 value gives the 2018 value: $334,000 + $33,750 = $367,750. Therefore, Option C is correct.Options A, B, and D are incorrect because they do not match the calculated increase based on the constant annual rate.

140. From 1990 to 2000, the population of city A rose from 12,000 to 28,000 and the population of city B rose from 18,000 to 24,000. If the population of the two cities increased at a constant rate, in what year was the population of both cities the same?

• A. 1996
• B. 1997
• C. 1994
• D. 1999

Explanation: To find the year when the populations of the two cities are equal, first calculate the annual increase in population for each city.For City A, the population increased from 12,000 to 28,000 over 10 years, which is an increase of 16,000. Therefore, the annual increase is 16,000 / 10 = 1,600 people per year.For City B, the population increased from 18,000 to 24,000 over 10 years, which is an increase of 6,000. Therefore, the annual increase is 6,000 / 10 = 600 people per year.The population of City A after x years from 1990 is given by 1,600x + 12,000, and for City B, it is 600x + 18,000. Set these two expressions equal:1,600x + 12,000 = 600x + 18,000Solving for x gives 1,000x = 6,000, hence x = 6.Adding 6 years to 1990 gives 1996 as the year when the populations are equal. Therefore, the correct answer is Option A, 1996. The other options are incorrect because they do not satisfy the condition where both populations are equal.