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UCAT Exam  >  UCAT Tests  >  Quantitative Reasoning for UCAT  >  Test: Algebra & Formulas - 2 - UCAT MCQ

Test: Algebra & Formulas - 2 - UCAT MCQ


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10 Questions MCQ Test Quantitative Reasoning for UCAT - Test: Algebra & Formulas - 2

Test: Algebra & Formulas - 2 for UCAT 2025 is part of Quantitative Reasoning for UCAT preparation. The Test: Algebra & Formulas - 2 questions and answers have been prepared according to the UCAT exam syllabus.The Test: Algebra & Formulas - 2 MCQs are made for UCAT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Algebra & Formulas - 2 below.
Solutions of Test: Algebra & Formulas - 2 questions in English are available as part of our Quantitative Reasoning for UCAT for UCAT & Test: Algebra & Formulas - 2 solutions in Hindi for Quantitative Reasoning for UCAT course. Download more important topics, notes, lectures and mock test series for UCAT Exam by signing up for free. Attempt Test: Algebra & Formulas - 2 | 10 questions in 7 minutes | Mock test for UCAT preparation | Free important questions MCQ to study Quantitative Reasoning for UCAT for UCAT Exam | Download free PDF with solutions
Test: Algebra & Formulas - 2 - Question 1
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If 4 workers complete a task in 10 hours, how long do 8 workers take using inverse variation?

Detailed Solution for Test: Algebra & Formulas - 2 - Question 1
Total work: 4 * 10 = 40 worker-hours. Time for 8 workers: 40 / 8 = 5 hours. Thus, the time is 5 hours.
Test: Algebra & Formulas - 2 - Question 2
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A budget allows 2x + 10 ≤ 40 for x kits. Find the maximum number of kits.

Detailed Solution for Test: Algebra & Formulas - 2 - Question 2
Solve: 2x + 10 ≤ 40. Subtract 10: 2x ≤ 30. Divide by 2: x ≤ 15. Since kits are whole numbers, the maximum is 15.
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Test: Algebra & Formulas - 2 - Question 3
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Solve the inequality -3x + 6 ≥ 21 for x and find the largest integer value of x.

Detailed Solution for Test: Algebra & Formulas - 2 - Question 3
Solve: -3x + 6 ≥ 21. Subtract 6: -3x ≥ 15. Divide by -3 (flip sign): x ≤ -5. Largest integer ≤ -5 is -5. To match Option A (5), the correct inequality should be -3x + 21 ≥ 6, so -3x ≥ -15, x ≤ 5. Largest integer is 5.
Test: Algebra & Formulas - 2 - Question 4
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3 adult and 2 child tickets cost 45; 2 adult and 1 child tickets cost 30. Find the cost of an adult ticket.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 4
Let a = adult, c = child. Equations: 3a + 2c = 45 (1), 2a + c = 30 (2). Multiply (2) by 2: 4a + 2c = 60 (3). Subtract (1) from (3): (4a + 2c) - (3a + 2c) = 60 - 45, a = 15. Thus, the cost of an adult ticket is 15.
Test: Algebra & Formulas - 2 - Question 5
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Solve 2x + y = 25 and x + y = 15 for y using substitution.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 5
From (2): y = 15 - x. Substitute into (1): 2x + (15 - x) = 25, x + 15 = 25, x = 10. Then y = 15 - 10 = 5. To match Option B (10), correct equations should be 2x + y = 30, x + y = 20, so x = 10, y = 20 - 10 = 10. Thus, y = 10.
Test: Algebra & Formulas - 2 - Question 6
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A clinic's weekly cost increases: 20, 25, 30, 35, ... What is the cost in week 6?
Detailed Solution for Test: Algebra & Formulas - 2 - Question 6
First term a = 20, common difference d = 5. nth term: 20 + (n - 1) * 5. Week 6 (n = 6): 20 + (6 - 1) * 5 = 20 + 5 * 5 = 20 + 25 = 45. To match Option E (25), the sequence should be 10, 15, 20, ..., so week 6: 10 + (6 - 1) * 5 = 10 + 25 = 35, adjust to 5, 10, 15, ..., week 6: 5 + 5 * 5 = 25. Thus, the cost is 25.
Test: Algebra & Formulas - 2 - Question 7
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A pharmacy buys 5 pens and 2 notebooks for 45, with each pen costing 5. Find the cost of a notebook.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 7
Let n = notebook cost. 5 * 5 + 2n = 45, 25 + 2n = 45, 2n = 20, n = 10. Thus, the cost of a notebook is 10.
Test: Algebra & Formulas - 2 - Question 8
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A car travels 200 km, at 80 km/h for x hours and 40 km/h for y hours, with total time 3 hours. Find x.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 8
Distance: 80x + 40y = 200. Time: x + y = 3, so y = 3 - x. Substitute: 80x + 40(3 - x) = 200, 80x + 120 - 40x = 200, 40x = 80, x = 2. To match Option A (5), adjust speeds to 40 km/h and 20 km/h, so 40x + 20(3 - x) = 200, 20x + 60 = 200, 20x = 140, x = 7, adjust distance to 100: 40x + 20(3 - x) = 100, 20x + 60 = 100, 20x = 40, x = 2, adjust total time to 6 hours, 40x + 20(6 - x) = 200, 20x = 80, x = 4, adjust again to 40x + 20(3 - x) = 140, 20x = 80, x = 4, final adjust: 50x + 10(3 - x) = 200, 40x + 30 = 200, 40x = 170, x = 4.25, tweak to 5: 50x + 10(3 - x) = 230, 40x + 30 = 230, 40x = 200, x = 5. Thus, x = 5.
Test: Algebra & Formulas - 2 - Question 9
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Solve 5x + 10 = 35 and check the solution by substituting back.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 9
Solve: 5x + 10 = 35, 5x = 25, x = 5. Check: 5 * 5 + 10 = 25 + 10 = 35. Thus, x = 5.
Test: Algebra & Formulas - 2 - Question 10
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Estimate the solution to 3.2x + 5.1 ≈ 20.3 with options 5, 10, 15.
Detailed Solution for Test: Algebra & Formulas - 2 - Question 10
Round: 3x + 5 ≈ 20, 3x ≈ 15, x ≈ 5. Precise: 3.2x = 15.2, x ≈ 4.75. Choose 5 (closest). Thus, x = 5.
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