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Test: Problem Solving- 2 - CUET Commerce MCQ


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10 Questions MCQ Test General Test Preparation for CUET - Test: Problem Solving- 2

Test: Problem Solving- 2 for CUET Commerce 2024 is part of General Test Preparation for CUET preparation. The Test: Problem Solving- 2 questions and answers have been prepared according to the CUET Commerce exam syllabus.The Test: Problem Solving- 2 MCQs are made for CUET Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Problem Solving- 2 below.
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Test: Problem Solving- 2 - Question 1

A racecar driver has completed 12 1/2 laps of a 50 lap race. What fractional part of the race remains?

Detailed Solution for Test: Problem Solving- 2 - Question 1

The driver has 50 - 12 1/2 or 37 1/2 laps to go out of the . Therefore, (37 1/2)/50 is the fractional part of the race remaining. This is equivalent to 3/4, which is answer C.
a. (1/4) This is the fraction of the race that the driver has completed.
b. (1/5) This comes from mistakenly dividing by , the latter number coming from adding both lap amounts. The race is only .
c. Correct
d. (4/5) This comes from taking the fraction in response B and subtracting it from 1.
e. (75/2) This comes from , but that’s how many laps are left - the driver is only in one race, not !

Test: Problem Solving- 2 - Question 2

If M is the set of positive multiples of 2 less than 150 and N is the set of positive multiples of 9 less than 150, how many members are there in M n N?

Detailed Solution for Test: Problem Solving- 2 - Question 2

The only elements in both sets are multiples of 2 x 9 or 18. The multiples of 18 less than 150 are 18, 36, 54, 72, 90, 108, 126 and 144. These are the 8 elements that are in the intersection of both sets. Thus, B is the correct response.
a. (0) There are elements that are in both sets.
b. (8) Correct
c. (9) There are 8 elements in both sets, not 9.
d. (18) This is the number that the elements are multiples of.
e. (74) This is the number of multiples of 2 less than 150

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Test: Problem Solving- 2 - Question 3

At Bruno’s Video World, the regular price for a DVD is d dollars. How many DVDs can be purchased for x dollars when the DVDs are on sale at 20% off the regular price?

Detailed Solution for Test: Problem Solving- 2 - Question 3

Since you have x dollars to spend, you must divide x by the sale price of a CD, which is (100%-20%)d or 80%d . 80%d can be written as a fraction in lowest terms as 4/5d . Therefore, you are dividing x/(4/5d) = 5x/4d. This is answer E.
a. This is the percent of the original price times x.
b. This is the reciprocal of the sale fraction times x.
c. This is the fraction of the original price, which represents the sale price.
d. This comes from incorrectly handling the fraction in the denominator.
e. Correct

Test: Problem Solving- 2 - Question 4

If x ≠ 2y, then 

Detailed Solution for Test: Problem Solving- 2 - Question 4

Both fractions each = -1, so adding -1 twice gives an answer of -2. This is choice E.
a. Both fractions are not equal to (x-2y) , so we do not have like terms.
b. The other parts of the fractions are not eliminated, so this is not possible.
c. Each fraction equals -1, so this result is not possible.
d. Each fraction equals -1, so adding twice gives .
e. Correct

Test: Problem Solving- 2 - Question 5

If Dave drove one-third of the distance of his trip on the first day, and 60 miles on the second day, he figured out that he still had 1/2 of the trip to drive. What was the total length, in miles, of his trip?

Detailed Solution for Test: Problem Solving- 2 - Question 5

We can model this situation by the equation 1/3x + 60 = 1/2x . Solving for x, we get x=360 , which is answer A.
a. Correct
b. This is the number of miles he drove before he figured out how much of his trip was left.
c. This amount is how much he drove the first day.
d. This is how much he drove the second day.
e. This amount is half of the distance he covered by the end of the second day.

Test: Problem Solving- 2 - Question 6

If x2 - y2 = 48, the 2/3(x+y)(x-y) = 

Detailed Solution for Test: Problem Solving- 2 - Question 6

The difference of two squares (x2-y2) equals 48, and the factors are (x+y)(x-y), so we can just replace both factors in the second expression by 48 and multiply by 2/3 to get 32 . This is choice D.
a. This involves canceling the 3 into the 48 , but forgetting to multiply by 2 .
b. This comes from multiplying 3/2 by 48.
c. This is 48 x 2 , but it was not divided by 3.
d. Correct
e. This is 4/3 times 48.

Test: Problem Solving- 2 - Question 7

Eddie is 7 years older than Brian. If Brian is x years old, then how old was Eddie 11 years ago?

Detailed Solution for Test: Problem Solving- 2 - Question 7

We can model this by getting Eddie’s age now and then figuring out how old he was 11 years ago. Right now, he is x+7 . Eleven years ago, Eddie was x + 7 -11 = x - 4 years old. This is answer B.
a. The correct calculation should be x+ 7 (to get Eddie’s age now) - 11 , which is x - 4 .
b. Correct
c. This is an incorrect calculation of Eddie’s present age.
d. This response does not relate their ages properly.
e. This also does not express their age relationship properly.

Test: Problem Solving- 2 - Question 8

Working alone at their respective constant rates, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

Test: Problem Solving- 2 - Question 9

According to the graph below, the greatest change in the profit of the Sports Shack occurred between which two consecutive months?

Detailed Solution for Test: Problem Solving- 2 - Question 9

A January and February. The profit in January was $20,000. The profit in February was-$20,000 . This represents a change in profit of -$40,000, which is the greatest that the graph shows.

Test: Problem Solving- 2 - Question 10

If a, b, and c are not equal to zero, what is the difference between the maximum and minimum value of S? S

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