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MK · Practice test 7 of 12

Mathematics Knowledge Practice Test 7

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  1. Question 1 easy

    Maria is 8 years older than Chloe. Four years ago, Maria was twice as old as Chloe was. How old is Chloe now?

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    Correct answer: A — 12

    Let c be Chloe's age now. Maria's age now is $$c + 8$$. Four years ago, their ages were $$(c - 4)$$ and $$(c + 8 - 4) = c + 4$$. According to the problem, $$c + 4 = 2(c - 4)$$. Expanding gives $$c + 4 = 2c - 8$$, which leads to $$c = 12$$

  2. Question 2 easy

    The sum of two numbers is 135. One of the numbers is twice the value of the other. What are the values of these two numbers?

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    Correct answer: C — 45 and 90

    Let the smaller number be x, so the larger number is $$2x$$. Their sum is $$x + 2x = 3x = 135$$. Solving for x gives $$x = 45$$. The second number is $$2(45) = 90$$. Thus, the numbers are 45 and 90.

  3. Question 3 easy

    Liam rented a small moving van for $40 per day, plus $0.60 per mile. It took him two days to move, and the total van cost was $320. How many miles did Liam drive?

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    Correct answer: B — 400

    The rental cost for 2 days at $40/day is 2 × $40 = $80. Subtract this from the total cost: $320 - $80 = $240 spent purely on mileage. Since mileage costs $0.60 per mile, divide $240 by $0.60 to find the total miles: $240 ÷ $0.60 = 400 miles.

  4. Question 4 medium

    Captain Vance and Sergeant Major Lee both donated money to a charity fund. Captain Vance gave three times as much as Sergeant Major Lee did. Between the two of them, they donated $360. How much money did Captain Vance give?

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    Correct answer: D — $270

    Let Lee's donation be x, making Vance's donation $$3x$$. Their combined donation is $$x + 3x = 4x = 360$$. Solving for x gives $$x = 90$$ (Lee's donation). Captain Vance's donation is 3 times this amount: 3 × $90 = $270.

  5. Question 5 medium

    Four friends decide to take a kayaking lesson. Each person has to pay for the cost of their own lesson, plus a $20 rental fee for the equipment. If the total group cost is $200, how much does the lesson cost each person?

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    Correct answer: B — $30

    Let L be the cost of the lesson for each person. Each person pays $$(L + 20)$$. For 4 people, the total equation is $$4(L + 20) = 200$$. Divide both sides by 4: $$L + 20 = 50$$. Subtract 20 from both sides: $$L = 30$$.

  6. Question 6 medium

    Leo spent $54 on a guitar. This amount is $6 less than three times the cost of his guitar picks. What was the price of the guitar picks?

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    Correct answer: C — $20

    Let p be the price of the guitar picks. According to the problem, $$3p - 6 = 54$$. Add 6 to both sides: $$3p = 60$$. Divide by 3: $$p = 20$$.

  7. Question 7 medium

    A pet adoption center has 44 animals consisting of cats and dogs. The number of dogs is four more than three times the number of cats. How many cats are at the adoption center?

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    Correct answer: B — 10

    Let c be the number of cats. The number of dogs is $$3c + 4$$. The total number of animals is cats + dogs = 44, so $$c + (3c + 4) = 44$$. Combine like terms: $$4c + 4 = 44$$. Subtract 4: $$4c = 40$$. Divide by 4: $$c = 10$$ cats.

  8. Question 8 hard

    Find three consecutive integers whose sum is 72.

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    Correct answer: B — 23, 24, and 25

    Let the three consecutive integers be x, $$x + 1$$, and $$x + 2$$. Their sum is $$x + (x + 1) + (x + 2) = 3x + 3 = 72$$. Subtract 3 from both sides: $$3x = 69$$. Divide by 3: $$x = 23$$. Thus, the three integers are 23, 24, and 25.

  9. Question 9 hard

    $120 is to be divided among Anna, Ben, and Clara. Ben's share is twice Anna's share, and Clara's share is $10 more than Ben's. What is the amount Clara receives?

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    Correct answer: C — $54

    Let Anna's share be a. Ben's share is $$2a$$, and Clara's share is $$2a + 10$$. The total sum is $$a + 2a + (2a + 10) = 120$$. Combine like terms: $$5a + 10 = 120$$. Subtract 10: $$5a = 110$$. Divide by 5: $$a = 22$$ (Anna's share). Clara receives 2(22) + 10 = 44 + 10 = $54.

  10. Question 10 hard

    A logistics warehouse stores 130 crates consisting of heavy machinery and fragile electronics. The number of fragile electronics crates is 10 more than three times the number of heavy machinery crates. Each heavy machinery crate weighs 20 kg, and each fragile electronics crate weighs 15 kg. If the total weight of all 130 crates combined is 2,100 kg, how many heavy machinery crates are in the warehouse?

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    Correct answer: B — 30

    Step 1: Set up a system of linear equations. Let m be the number of heavy machinery crates and e be the number of fragile electronics crates. We know that $$m + e = 130$$, and $$e = 3m + 10$$.

    Step 2: Substitute the expression for e into the first equation: $$m + (3m + 10) = 130$$.

    Step 3: Combine like terms and solve for m: $$4m + 10 = 130 \Rightarrow 4m = 120 \Rightarrow m = 30$$ heavy machinery crates. (Checking: $$e = 3(30) + 10 = 100$$ crates. Total weight $$= 30(20) + 100(15) = 600 + 1500 = 2{,}100$$ kg).