# CSC CSE-PPT Numerical Reasoning: Fractions/Decimals/Percentages, Ratio & Proportion, Number Series, Age/Work/Motion Word Problems, Interest & Statistics

> Covers the Numerical Ability section of the CSC CSE-PPT (Professional and Sub-Professional): PEMDAS, Fractions, Decimals, Percentages, Ratio and Proportion, Number & Alphanumeric Series, Age Problems, Work & Tank Problems, Distance-Speed-Time Motion Problems, Mixture Problems, Simple & Compound Interest, Markup/Discount, and Basic Statistics/Probability.

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- **Level:** intermediate
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- **Total Cards:** 32

## Flashcards (32 Cards)

| # | Front (Term / Prompt) | Back (Definition / Answer) | Example | Multiple-Choice Options |
| :--- | :--- | :--- | :--- | :--- |
| 1 | In CSC Algebra (Coin / Denomination Problems), a cashier has 30 bills consisting of PHP 50 and PHP 100 bills totaling PHP 2,100. How many PHP 100 bills are there? | 12 bills of PHP 100 (and 18 bills of PHP 50). If all 30 bills were PHP 50, the total would be PHP 1,500. The excess PHP 600 (2,100 - 1,500) divided by the PHP 50 difference per bill (100 - 50) gives 600 / 50 = 12 bills of PHP 100. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 56 committees. Since order within a committee does not matter, use Combinations: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 56.; B) 12 bills of PHP 100 (and 18 bills of PHP 50). If all 30 bills were PHP 50, the total would be PHP 1,500. The excess PHP 600 (2,100 - 1,500) divided by the PHP 50 difference per bill (100 - 50) gives 600 / 50 = 12 bills of PHP 100. ✓; C) 90. Since Base N = Percentage / Rate = 45 / 0.15 = 300, 30% of 300 = 90 (or directly by proportion: since 30% is double 15%, the result is 2 × 45 = 90).; D) 12 hours. Net filling rate per hour = 1/4 - 1/6 = (3 - 2)/12 = 1/12 of the tank per hour, so total time = 12 hours. |
| 2 | In CSC Fractions, how do you quickly compare which fraction is larger among 5/8, 7/11, and 2/3 without long division? | Use Cross-Multiplication: compare 5/8 vs. 7/11 -> 5×11 = 55 vs. 8×7 = 56, so 7/11 > 5/8. Next compare 7/11 vs. 2/3 -> 7×3 = 21 vs. 11×2 = 22, so 2/3 > 7/11. Thus 2/3 is the largest. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Use Cross-Multiplication: compare 5/8 vs. 7/11 -> 5×11 = 55 vs. 8×7 = 56, so 7/11 > 5/8. Next compare 7/11 vs. 2/3 -> 7×3 = 21 vs. 11×2 = 22, so 2/3 > 7/11. Thus 2/3 is the largest. ✓; B) 90. Since Base N = Percentage / Rate = 45 / 0.15 = 300, 30% of 300 = 90 (or directly by proportion: since 30% is double 15%, the result is 2 × 45 = 90).; C) Ana is 12 years old and Maria is 36 years old. Let Ana = x and Maria = 3x. In 12 years: 3x + 12 = 2(x + 12) -> 3x + 12 = 2x + 24 -> x = 12, and 3x = 36.; D) 38. Let the four consecutive even integers be x, x+2, x+4, x+6. Then 4x + 12 = 140 -> 4x = 128 -> x = 32. The largest integer is x + 6 = 38. |
| 3 | In CSC Numerical Reasoning (PEMDAS/GEMDAS), what is the exact value of 48 ÷ 6 × (2 + 2) - 3²? | 23. Step 1 (Parentheses & Exponents): 48 ÷ 6 × 4 - 9. Step 2 (Multiplication and Division strictly from LEFT TO RIGHT): (48 ÷ 6) × 4 - 9 = 8 × 4 - 9 = 32 - 9 = 23. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 8:03:00 AM. Find the LCM of 12, 18, and 30: 12 = 2²·3, 18 = 2·3², 30 = 2·3·5 -> LCM = 2² × 3² × 5 = 180 seconds = 3 minutes.; B) 23. Step 1 (Parentheses & Exponents): 48 ÷ 6 × 4 - 9. Step 2 (Multiplication and Division strictly from LEFT TO RIGHT): (48 ÷ 6) × 4 - 9 = 8 × 4 - 9 = 32 - 9 = 23. ✓; C) 215. The pattern is n³ - 1 for n = 1, 2, 3, 4, 5, 6: 1³-1=0, 2³-1=7, 3³-1=26, 4³-1=63, 5³-1=124, and 6³-1 = 216 - 1 = 215.; D) Maturity value A = P(1 + r)^t = 10,000 × (1.10)² = 10,000 × 1.21 = PHP 12,100. Total compound interest earned = PHP 12,100 - PHP 10,000 = PHP 2,100. |
| 4 | In CSC Motion Word Problems (Catch-Up / Pursuit), a courier leaves Manila at 8:00 AM traveling at 60 km/h. A second courier leaves the same spot at 9:30 AM traveling along the same route at 90 km/h. At what time will the second courier overtake the first? | 12:30 PM (after 3 hours of pursuit). In the 1.5-hour head start (8:00 to 9:30 AM), the first courier travels 60 × 1.5 = 90 km. Relative catch-up speed = 90 - 60 = 30 km/h. Catch-up time = 90 km / 30 km/h = 3 hours after 9:30 AM = 12:30 PM. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 38. Let the four consecutive even integers be x, x+2, x+4, x+6. Then 4x + 12 = 140 -> 4x = 128 -> x = 32. The largest integer is x + 6 = 38.; B) 12:30 PM (after 3 hours of pursuit). In the 1.5-hour head start (8:00 to 9:30 AM), the first courier travels 60 × 1.5 = 90 km. Relative catch-up speed = 90 - 60 = 30 km/h. Catch-up time = 90 km / 30 km/h = 3 hours after 9:30 AM = 12:30 PM. ✓; C) 23. Step 1 (Parentheses & Exponents): 48 ÷ 6 × 4 - 9. Step 2 (Multiplication and Division strictly from LEFT TO RIGHT): (48 ÷ 6) × 4 - 9 = 8 × 4 - 9 = 32 - 9 = 23.; D) 4 days. In inverse proportion, Workers₁ × Days₁ = Workers₂ × Days₂ -> 6 × 10 = 15 × D₂ -> D₂ = 60 / 15 = 4 days. |
| 5 | In CSC Probability & Combinatorics, how many 3-person committees can be formed from a pool of 8 qualified officers? | 56 committees. Since order within a committee does not matter, use Combinations: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 56. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) By 3: sum of digits is divisible by 3.; B) 56 committees. Since order within a committee does not matter, use Combinations: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 56. ✓; C) 63. First-level differences are +4, +8, +12, +16. Second-level differences are constant (+4), so the next first-level difference is +20, giving 43 + 20 = 63.; D) 8 liters. Using the solute balance equation: 0.30(20) + 1.00(x) = 0.50(20 + x) -> 6 + x = 10 + 0.5x -> 0.5x = 4 -> x = 8 liters. |
| 6 | In CSC Geometry, if the length of a rectangle is increased by 20% and its width is increased by 10%, by what percentage does its area increase? | 32%. New Area = (1.20 L) × (1.10 W) = 1.32 LW, which is a 32% increase. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 13 employees. Since 10 speak neither, 60 - 10 = 50 employees speak at least one language. By the Inclusion-Exclusion Principle, \|I ∩ C\| = \|I\| + \|C\| - \|I ∪ C\| = 35 + 28 - 50 = 13.; B) PHP 40,000. A 20% discount means the sale price is 80% (0.80) of the original price: Original Price = 32,000 / 0.80 = PHP 40,000 (do NOT add 20% of 32,000).; C) 3.6 days. In 4 days alone, Alex completes 4 × (1/10) = 4/10 = 2/5 of the work, leaving 3/5 of the work. Their combined rate is 1/10 + 1/15 = 5/30 = 1/6 per day. Time for remaining work = (3/5) ÷ (1/6) =…; D) 32%. New Area = (1.20 L) × (1.10 W) = 1.32 LW, which is a 32% increase. ✓ |
| 7 | In CSC Number Series (Squares/Cubes ± 1), what is the next term in the sequence: 0, 7, 26, 63, 124, ___? | 215. The pattern is n³ - 1 for n = 1, 2, 3, 4, 5, 6: 1³-1=0, 2³-1=7, 3³-1=26, 4³-1=63, 5³-1=124, and 6³-1 = 216 - 1 = 215. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 3 hours. When moving toward each other (or in opposite directions), relative speed is the sum of their speeds: v_rel = 50 + 70 = 120 km/h. Time = Distance / v_rel = 360 km / 120 km/h = 3 hours.; B) 32%. New Area = (1.20 L) × (1.10 W) = 1.32 LW, which is a 32% increase.; C) For any two numbers, A × B = GCF(A, B) × LCM(A, B). For 12 (2²·3), 15 (3·5), and 20 (2²·5), take the highest power of each prime: LCM = 2² × 3 × 5 = 60.; D) 215. The pattern is n³ - 1 for n = 1, 2, 3, 4, 5, 6: 1³-1=0, 2³-1=7, 3³-1=26, 4³-1=63, 5³-1=124, and 6³-1 = 216 - 1 = 215. ✓ |
| 8 | In CSC Venn Diagram Word Problems, in an office of 60 employees, 35 speak Ilocano, 28 speak Cebuano, and 10 speak neither language. How many employees speak BOTH Ilocano and Cebuano? | 13 employees. Since 10 speak neither, 60 - 10 = 50 employees speak at least one language. By the Inclusion-Exclusion Principle, \|I ∩ C\| = \|I\| + \|C\| - \|I ∪ C\| = 35 + 28 - 50 = 13. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) For any two numbers, A × B = GCF(A, B) × LCM(A, B). For 12 (2²·3), 15 (3·5), and 20 (2²·5), take the highest power of each prime: LCM = 2² × 3 × 5 = 60.; B) 11. Odd positions double each time (2, 4, 8, 16), while even positions decrease by 3 each time (20, 17, 14, 11). The 8th term is an even position: 14 - 3 = 11.; C) PHP 50,000. Using I = P·r·t, solve for P = I / (r·t) = 9,000 / (0.06 × 3) = 9,000 / 0.18 = PHP 50,000.; D) 13 employees. Since 10 speak neither, 60 - 10 = 50 employees speak at least one language. By the Inclusion-Exclusion Principle, \|I ∩ C\| = \|I\| + \|C\| - \|I ∪ C\| = 35 + 28 - 50 = 13. ✓ |
| 9 | In CSC Number Theory, how are the Greatest Common Factor (GCF) and Least Common Multiple (LCM) of two numbers A and B related, and what is the LCM of 12, 15, and 20? | For any two numbers, A × B = GCF(A, B) × LCM(A, B). For 12 (2²·3), 15 (3·5), and 20 (2²·5), take the highest power of each prime: LCM = 2² × 3 × 5 = 60. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) PHP 32,000. Total ratio parts = 2 + 3 + 4 = 9 parts. Value of 1 part = PHP 72,000 / 9 = PHP 8,000. Largest share (4 parts) = 4 × PHP 8,000 = PHP 32,000.; B) For any two numbers, A × B = GCF(A, B) × LCM(A, B). For 12 (2²·3), 15 (3·5), and 20 (2²·5), take the highest power of each prime: LCM = 2² × 3 × 5 = 60. ✓; C) 3.6 days. In 4 days alone, Alex completes 4 × (1/10) = 4/10 = 2/5 of the work, leaving 3/5 of the work. Their combined rate is 1/10 + 1/15 = 5/30 = 1/6 per day. Time for remaining work = (3/5) ÷ (1/6) =…; D) 280 sq. meters. Since Perimeter = 2(L + W) = 68, L + W = 34. With L = W + 6, we get 2W + 6 = 34 -> W = 14 m and L = 20 m. Area = 20 × 14 = 280 m². |
| 10 | In CSC Percentage Problems (Base, Rate, Percentage), if 45 is 15% of a number N, what is 30% of N? | 90. Since Base N = Percentage / Rate = 45 / 0.15 = 300, 30% of 300 = 90 (or directly by proportion: since 30% is double 15%, the result is 2 × 45 = 90). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 63. First-level differences are +4, +8, +12, +16. Second-level differences are constant (+4), so the next first-level difference is +20, giving 43 + 20 = 63.; B) 90. Since Base N = Percentage / Rate = 45 / 0.15 = 300, 30% of 300 = 90 (or directly by proportion: since 30% is double 15%, the result is 2 × 45 = 90). ✓; C) 12:30 PM (after 3 hours of pursuit).; D) 86. Required total sum for 5 quizzes = 5 × 86 = 430. Current sum of 4 quizzes = 82 + 85 + 88 + 89 = 344. Required 5th score = 430 - 344 = 86. |
| 11 | In CSC Work Problems with Inlet and Outlet Pipes, Pipe A can fill a tank in 4 hours, while Drain Pipe B can empty the full tank in 6 hours. If both pipes are opened simultaneously when the tank is empty, how long will it take to fill the tank? | 12 hours. Net filling rate per hour = 1/4 - 1/6 = (3 - 2)/12 = 1/12 of the tank per hour, so total time = 12 hours. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 8 liters. Using the solute balance equation: 0.30(20) + 1.00(x) = 0.50(20 + x) -> 6 + x = 10 + 0.5x -> 0.5x = 4 -> x = 8 liters.; B) 86. Required total sum for 5 quizzes = 5 × 86 = 430. Current sum of 4 quizzes = 82 + 85 + 88 + 89 = 344. Required 5th score = 430 - 344 = 86.; C) PHP 40,000. A 20% discount means the sale price is 80% (0.80) of the original price: Original Price = 32,000 / 0.80 = PHP 40,000 (do NOT add 20% of 32,000).; D) 12 hours. Net filling rate per hour = 1/4 - 1/6 = (3 - 2)/12 = 1/12 of the tank per hour, so total time = 12 hours. ✓ |
| 12 | In CSC Age Word Problems, 5 years ago, Carlo was 4 times as old as Ben. If the sum of their ages today is 45, how old is Carlo today? | Carlo is 33 years old today. Five years ago, the sum of their ages was 45 - 2(5) = 35. Since their ratio 5 years ago was 4 : 1 (5 parts), Ben was 35/5 = 7 and Carlo was 28 five years ago. Today, Carlo is 28 + 5 = 33 (and Ben is 12). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 48 km/h (NOT 50 km/h). For equal distances, Average Speed is the harmonic mean: 2·v₁·v₂ / (v₁ + v₂) = 2(40)(60) / (40 + 60) = 4,800 / 100 = 48 km/h.; B) 12 meters, using the 5-12-13 Pythagorean triple (h = √(13² - 5²) = √(169 - 25) = √144 = 12 m). Common CSC Pythagorean triples to memorize: (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).; C) 90. Since Base N = Percentage / Rate = 45 / 0.15 = 300, 30% of 300 = 90 (or directly by proportion: since 30% is double 15%, the result is 2 × 45 = 90).; D) Carlo is 33 years old today. Five years ago, the sum of their ages was 45 - 2(5) = 35. Since their ratio 5 years ago was 4 : 1 (5 parts), Ben was 35/5 = 7 and Carlo was 28 five years ago. Today, Carlo is 28 + 5 = 33 (and Ben is 12). ✓ |
| 13 | In CSC Motion Problems (Average Speed for a Round Trip), a driver travels from City A to City B at 40 km/h and returns along the same route at 60 km/h. What is the average speed for the entire round trip? | 48 km/h (NOT 50 km/h). For equal distances, Average Speed is the harmonic mean: 2·v₁·v₂ / (v₁ + v₂) = 2(40)(60) / (40 + 60) = 4,800 / 100 = 48 km/h. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 12 bills of PHP 100 (and 18 bills of PHP 50).; B) 48 km/h (NOT 50 km/h). For equal distances, Average Speed is the harmonic mean: 2·v₁·v₂ / (v₁ + v₂) = 2(40)(60) / (40 + 60) = 4,800 / 100 = 48 km/h. ✓; C) Use Cross-Multiplication: compare 5/8 vs. 7/11 -> 5×11 = 55 vs. 8×7 = 56, so 7/11 > 5/8. Next compare 7/11 vs. 2/3 -> 7×3 = 21 vs. 11×2 = 22, so 2/3 > 7/11. Thus 2/3 is the largest.; D) 63. First-level differences are +4, +8, +12, +16. Second-level differences are constant (+4), so the next first-level difference is +20, giving 43 + 20 = 63. |
| 14 | In CSC Number Series (Second-Order Differences), what is the next number in the sequence: 3, 7, 15, 27, 43, ___? | 63. First-level differences are +4, +8, +12, +16. Second-level differences are constant (+4), so the next first-level difference is +20, giving 43 + 20 = 63. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 12 hours. Net filling rate per hour = 1/4 - 1/6 = (3 - 2)/12 = 1/12 of the tank per hour, so total time = 12 hours.; B) 38. Let the four consecutive even integers be x, x+2, x+4, x+6. Then 4x + 12 = 140 -> 4x = 128 -> x = 32. The largest integer is x + 6 = 38.; C) 1/8 = 0.125 (12.5%); 3/8 = 0.375 (37.5%); 5/8 = 0.625 (62.5%); 7/8 = 0.875 (87.5%); 1/6 = 0.1667 (16.67%); 5/6 = 0.8333 (83.33%).; D) 63. First-level differences are +4, +8, +12, +16. Second-level differences are constant (+4), so the next first-level difference is +20, giving 43 + 20 = 63. ✓ |
| 15 | In CSC Motion Word Problems (Opposite Directions / Meeting), two buses leave stations 360 km apart at the same time and travel toward each other at 50 km/h and 70 km/h. In how many hours will they meet? | 3 hours. When moving toward each other (or in opposite directions), relative speed is the sum of their speeds: v_rel = 50 + 70 = 120 km/h. Time = Distance / v_rel = 360 km / 120 km/h = 3 hours. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Maturity value A = P(1 + r)^t = 10,000 × (1.10)² = 10,000 × 1.21 = PHP 12,100. Total compound interest earned = PHP 12,100 - PHP 10,000 = PHP 2,100.; B) 13 employees. Since 10 speak neither, 60 - 10 = 50 employees speak at least one language. By the Inclusion-Exclusion Principle, \|I ∩ C\| = \|I\| + \|C\| - \|I ∪ C\| = 35 + 28 - 50 = 13.; C) PHP 32,000. Total ratio parts = 2 + 3 + 4 = 9 parts. Value of 1 part = PHP 72,000 / 9 = PHP 8,000. Largest share (4 parts) = 4 × PHP 8,000 = PHP 32,000.; D) 3 hours. When moving toward each other (or in opposite directions), relative speed is the sum of their speeds: v_rel = 50 + 70 = 120 km/h. Time = Distance / v_rel = 360 km / 120 km/h = 3 hours. ✓ |
| 16 | In CSC Ratio and Proportion (Inverse Proportion), if 6 clerks can encode a voter registry in 10 days, how many days will it take 15 clerks working at the same rate to complete the job? | 4 days. In inverse proportion, Workers₁ × Days₁ = Workers₂ × Days₂ -> 6 × 10 = 15 × D₂ -> D₂ = 60 / 15 = 4 days. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Carlo is 33 years old today.; B) Maturity value A = P(1 + r)^t = 10,000 × (1.10)² = 10,000 × 1.21 = PHP 12,100. Total compound interest earned = PHP 12,100 - PHP 10,000 = PHP 2,100.; C) 4 days. In inverse proportion, Workers₁ × Days₁ = Workers₂ × Days₂ -> 6 × 10 = 15 × D₂ -> D₂ = 60 / 15 = 4 days. ✓; D) Use Cross-Multiplication: compare 5/8 vs. 7/11 -> 5×11 = 55 vs. 8×7 = 56, so 7/11 > 5/8. Next compare 7/11 vs. 2/3 -> 7×3 = 21 vs. 11×2 = 22, so 2/3 > 7/11. Thus 2/3 is the largest. |
| 17 | In CSC Age Word Problems, Maria is currently 3 times as old as her daughter Ana. In 12 years, Maria will be twice as old as Ana. How old are Maria and Ana now? | Ana is 12 years old and Maria is 36 years old. Let Ana = x and Maria = 3x. In 12 years: 3x + 12 = 2(x + 12) -> 3x + 12 = 2x + 24 -> x = 12, and 3x = 36. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 12:30 PM (after 3 hours of pursuit).; B) 280 sq. meters. Since Perimeter = 2(L + W) = 68, L + W = 34. With L = W + 6, we get 2W + 6 = 34 -> W = 14 m and L = 20 m. Area = 20 × 14 = 280 m².; C) 8:03:00 AM. Find the LCM of 12, 18, and 30: 12 = 2²·3, 18 = 2·3², 30 = 2·3·5 -> LCM = 2² × 3² × 5 = 180 seconds = 3 minutes.; D) Ana is 12 years old and Maria is 36 years old. Let Ana = x and Maria = 3x. In 12 years: 3x + 12 = 2(x + 12) -> 3x + 12 = 2x + 24 -> x = 12, and 3x = 36. ✓ |
| 18 | In CSC Divisibility Rules, how do you test instantly if a large number is divisible by 3, 4, 8, 9, and 11? | By 3: sum of digits is divisible by 3. By 4: last 2 digits form a number divisible by 4. By 8: last 3 digits form a number divisible by 8. By 9: sum of digits is divisible by 9. By 11: alternating sum and difference of digits is 0 or a multiple of 11. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) By 3: sum of digits is divisible by 3. By 4: last 2 digits form a number divisible by 4. By 8: last 3 digits form a number divisible by 8. By 9: sum of digits is divisible by 9. By 11: alternating sum and difference of digits is 0 or a multiple of 11. ✓; B) 28%. Remaining price factor = (1 - 0.20) × (1 - 0.10) = 0.80 × 0.90 = 0.72 (72% of original), so total discount = 100% - 72% = 28% (or formula: d1 + d2 - (d1·d2)/100 = 20 + 10 - 2 = 28%).; C) 12 hours. Net filling rate per hour = 1/4 - 1/6 = (3 - 2)/12 = 1/12 of the tank per hour, so total time = 12 hours.; D) 32%. New Area = (1.20 L) × (1.10 W) = 1.32 LW, which is a 32% increase. |
| 19 | In CSC Geometry, the perimeter of a rectangular lot is 68 meters and its length is 6 meters more than its width. What is the area of the lot? | 280 sq. meters. Since Perimeter = 2(L + W) = 68, L + W = 34. With L = W + 6, we get 2W + 6 = 34 -> W = 14 m and L = 20 m. Area = 20 × 14 = 280 m². | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 23. Step 1 (Parentheses & Exponents): 48 ÷ 6 × 4 - 9. Step 2 (Multiplication and Division strictly from LEFT TO RIGHT): (48 ÷ 6) × 4 - 9 = 8 × 4 - 9 = 32 - 9 = 23.; B) 28%. Remaining price factor = (1 - 0.20) × (1 - 0.10) = 0.80 × 0.90 = 0.72 (72% of original), so total discount = 100% - 72% = 28% (or formula: d1 + d2 - (d1·d2)/100 = 20 + 10 - 2 = 28%).; C) 280 sq. meters. Since Perimeter = 2(L + W) = 68, L + W = 34. With L = W + 6, we get 2W + 6 = 34 -> W = 14 m and L = 20 m. Area = 20 × 14 = 280 m². ✓; D) 3 hours. When moving toward each other (or in opposite directions), relative speed is the sum of their speeds: v_rel = 50 + 70 = 120 km/h. Time = Distance / v_rel = 360 km / 120 km/h = 3 hours. |
| 20 | In CSC Number Series (Alternating / Interleaved Series), what is the next number in the sequence: 2, 20, 4, 17, 8, 14, 16, ___? | 11. Odd positions double each time (2, 4, 8, 16), while even positions decrease by 3 each time (20, 17, 14, 11). The 8th term is an even position: 14 - 3 = 11. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 11. Odd positions double each time (2, 4, 8, 16), while even positions decrease by 3 each time (20, 17, 14, 11). The 8th term is an even position: 14 - 3 = 11. ✓; B) 3.6 days. In 4 days alone, Alex completes 4 × (1/10) = 4/10 = 2/5 of the work, leaving 3/5 of the work. Their combined rate is 1/10 + 1/15 = 5/30 = 1/6 per day. Time for remaining work = (3/5) ÷ (1/6) =…; C) 12 bills of PHP 100 (and 18 bills of PHP 50).; D) By 3: sum of digits is divisible by 3. |
| 21 | In CSC Work Problems, Alex can finish an audit report in 10 days, and Bea can finish it in 15 days. Alex works alone for 4 days and then Bea joins him. How many MORE days will it take them working together to finish the remaining work? | 3.6 days. In 4 days alone, Alex completes 4 × (1/10) = 4/10 = 2/5 of the work, leaving 3/5 of the work. Their combined rate is 1/10 + 1/15 = 5/30 = 1/6 per day. Time for remaining work = (3/5) ÷ (1/6) = 18/5 = 3.6 days. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) PHP 50,000. Using I = P·r·t, solve for P = I / (r·t) = 9,000 / (0.06 × 3) = 9,000 / 0.18 = PHP 50,000.; B) 3.6 days. In 4 days alone, Alex completes 4 × (1/10) = 4/10 = 2/5 of the work, leaving 3/5 of the work. Their combined rate is 1/10 + 1/15 = 5/30 = 1/6 per day. Time for remaining work = (3/5) ÷ (1/6) = 18/5 = 3.6 days. ✓; C) 56 committees. Since order within a committee does not matter, use Combinations: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 56.; D) 28%. Remaining price factor = (1 - 0.20) × (1 - 0.10) = 0.80 × 0.90 = 0.72 (72% of original), so total discount = 100% - 72% = 28% (or formula: d1 + d2 - (d1·d2)/100 = 20 + 10 - 2 = 28%). |
| 22 | In CSC Simple Interest Problems, how much principal must be invested at 6% simple annual interest to earn PHP 9,000 in interest over 3 years? | PHP 50,000. Using I = P·r·t, solve for P = I / (r·t) = 9,000 / (0.06 × 3) = 9,000 / 0.18 = PHP 50,000. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 280 sq. meters. Since Perimeter = 2(L + W) = 68, L + W = 34. With L = W + 6, we get 2W + 6 = 34 -> W = 14 m and L = 20 m. Area = 20 × 14 = 280 m².; B) PHP 50,000. Using I = P·r·t, solve for P = I / (r·t) = 9,000 / (0.06 × 3) = 9,000 / 0.18 = PHP 50,000. ✓; C) By 3: sum of digits is divisible by 3.; D) 215. The pattern is n³ - 1 for n = 1, 2, 3, 4, 5, 6: 1³-1=0, 2³-1=7, 3³-1=26, 4³-1=63, 5³-1=124, and 6³-1 = 216 - 1 = 215. |
| 23 | In CSC Percentage Problems, after a 20% discount, a laptop was sold for PHP 32,000. What was the original marked price before the discount? | PHP 40,000. A 20% discount means the sale price is 80% (0.80) of the original price: Original Price = 32,000 / 0.80 = PHP 40,000 (do NOT add 20% of 32,000). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) PHP 40,000. A 20% discount means the sale price is 80% (0.80) of the original price: Original Price = 32,000 / 0.80 = PHP 40,000 (do NOT add 20% of 32,000). ✓; B) 11. Odd positions double each time (2, 4, 8, 16), while even positions decrease by 3 each time (20, 17, 14, 11). The 8th term is an even position: 14 - 3 = 11.; C) 48 km/h (NOT 50 km/h). For equal distances, Average Speed is the harmonic mean: 2·v₁·v₂ / (v₁ + v₂) = 2(40)(60) / (40 + 60) = 4,800 / 100 = 48 km/h.; D) 56 committees. Since order within a committee does not matter, use Combinations: C(8, 3) = (8 × 7 × 6) / (3 × 2 × 1) = 56. |
| 24 | In CSC Compound Interest Problems, if PHP 10,000 is invested at 10% annual interest compounded annually for 2 years, what is the final maturity value and total compound interest earned? | Maturity value A = P(1 + r)^t = 10,000 × (1.10)² = 10,000 × 1.21 = PHP 12,100. Total compound interest earned = PHP 12,100 - PHP 10,000 = PHP 2,100. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Maturity value A = P(1 + r)^t = 10,000 × (1.10)² = 10,000 × 1.21 = PHP 12,100. Total compound interest earned = PHP 12,100 - PHP 10,000 = PHP 2,100. ✓; B) 12 meters, using the 5-12-13 Pythagorean triple (h = √(13² - 5²) = √(169 - 25) = √144 = 12 m). Common CSC Pythagorean triples to memorize: (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25).; C) For any two numbers, A × B = GCF(A, B) × LCM(A, B). For 12 (2²·3), 15 (3·5), and 20 (2²·5), take the highest power of each prime: LCM = 2² × 3 × 5 = 60.; D) Ana is 12 years old and Maria is 36 years old. Let Ana = x and Maria = 3x. In 12 years: 3x + 12 = 2(x + 12) -> 3x + 12 = 2x + 24 -> x = 12, and 3x = 36. |
| 25 | In CSC Fractions & Decimals, what are the exact decimal and percentage equivalents of 1/8, 3/8, 5/8, 7/8, 1/6, and 5/6? | 1/8 = 0.125 (12.5%); 3/8 = 0.375 (37.5%); 5/8 = 0.625 (62.5%); 7/8 = 0.875 (87.5%); 1/6 = 0.1667 (16.67%); 5/6 = 0.8333 (83.33%). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 1/8 = 0.125 (12.5%); 3/8 = 0.375 (37.5%); 5/8 = 0.625 (62.5%); 7/8 = 0.875 (87.5%); 1/6 = 0.1667 (16.67%); 5/6 = 0.8333 (83.33%). ✓; B) PHP 40,000. A 20% discount means the sale price is 80% (0.80) of the original price: Original Price = 32,000 / 0.80 = PHP 40,000 (do NOT add 20% of 32,000).; C) Carlo is 33 years old today.; D) 12 bills of PHP 100 (and 18 bills of PHP 50). |
| 26 | In CSC Geometry (Pythagorean Triples), a ladder 13 meters long leans against a vertical wall with its foot 5 meters from the base of the wall. How high up the wall does the ladder reach? | 12 meters, using the 5-12-13 Pythagorean triple (h = √(13² - 5²) = √(169 - 25) = √144 = 12 m). Common CSC Pythagorean triples to memorize: (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Use Cross-Multiplication: compare 5/8 vs. 7/11 -> 5×11 = 55 vs. 8×7 = 56, so 7/11 > 5/8. Next compare 7/11 vs. 2/3 -> 7×3 = 21 vs. 11×2 = 22, so 2/3 > 7/11. Thus 2/3 is the largest.; B) PHP 32,000. Total ratio parts = 2 + 3 + 4 = 9 parts. Value of 1 part = PHP 72,000 / 9 = PHP 8,000. Largest share (4 parts) = 4 × PHP 8,000 = PHP 32,000.; C) 12 meters, using the 5-12-13 Pythagorean triple (h = √(13² - 5²) = √(169 - 25) = √144 = 12 m). Common CSC Pythagorean triples to memorize: (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25). ✓; D) 12:30 PM (after 3 hours of pursuit). |
| 27 | In CSC Algebra (Consecutive Integers), the sum of four consecutive even integers is 140. What is the largest of these four integers? | 38. Let the four consecutive even integers be x, x+2, x+4, x+6. Then 4x + 12 = 140 -> 4x = 128 -> x = 32. The largest integer is x + 6 = 38. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 38. Let the four consecutive even integers be x, x+2, x+4, x+6. Then 4x + 12 = 140 -> 4x = 128 -> x = 32. The largest integer is x + 6 = 38. ✓; B) 86. Required total sum for 5 quizzes = 5 × 86 = 430. Current sum of 4 quizzes = 82 + 85 + 88 + 89 = 344. Required 5th score = 430 - 344 = 86.; C) 8:03:00 AM. Find the LCM of 12, 18, and 30: 12 = 2²·3, 18 = 2·3², 30 = 2·3·5 -> LCM = 2² × 3² × 5 = 180 seconds = 3 minutes.; D) Carlo is 33 years old today. |
| 28 | In CSC Successive Discounts, what single equivalent discount rate is equal to successive discounts of 20% and 10%? | 28%. Remaining price factor = (1 - 0.20) × (1 - 0.10) = 0.80 × 0.90 = 0.72 (72% of original), so total discount = 100% - 72% = 28% (or formula: d1 + d2 - (d1·d2)/100 = 20 + 10 - 2 = 28%). | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 215. The pattern is n³ - 1 for n = 1, 2, 3, 4, 5, 6: 1³-1=0, 2³-1=7, 3³-1=26, 4³-1=63, 5³-1=124, and 6³-1 = 216 - 1 = 215.; B) 8 liters. Using the solute balance equation: 0.30(20) + 1.00(x) = 0.50(20 + x) -> 6 + x = 10 + 0.5x -> 0.5x = 4 -> x = 8 liters.; C) 13 employees. Since 10 speak neither, 60 - 10 = 50 employees speak at least one language. By the Inclusion-Exclusion Principle, \|I ∩ C\| = \|I\| + \|C\| - \|I ∪ C\| = 35 + 28 - 50 = 13.; D) 28%. Remaining price factor = (1 - 0.20) × (1 - 0.10) = 0.80 × 0.90 = 0.72 (72% of original), so total discount = 100% - 72% = 28% (or formula: d1 + d2 - (d1·d2)/100 = 20 + 10 - 2 = 28%). ✓ |
| 29 | In CSC Statistics (Weighted Mean), a student scored 82, 85, 88, and 89 on four quizzes. What score must they get on the 5th quiz to achieve an average of 86 across all 5 quizzes? | 86. Required total sum for 5 quizzes = 5 × 86 = 430. Current sum of 4 quizzes = 82 + 85 + 88 + 89 = 344. Required 5th score = 430 - 344 = 86. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 1/8 = 0.125 (12.5%); 3/8 = 0.375 (37.5%); 5/8 = 0.625 (62.5%); 7/8 = 0.875 (87.5%); 1/6 = 0.1667 (16.67%); 5/6 = 0.8333 (83.33%).; B) 86. Required total sum for 5 quizzes = 5 × 86 = 430. Current sum of 4 quizzes = 82 + 85 + 88 + 89 = 344. Required 5th score = 430 - 344 = 86. ✓; C) 4 days. In inverse proportion, Workers₁ × Days₁ = Workers₂ × Days₂ -> 6 × 10 = 15 × D₂ -> D₂ = 60 / 15 = 4 days.; D) 48 km/h (NOT 50 km/h). For equal distances, Average Speed is the harmonic mean: 2·v₁·v₂ / (v₁ + v₂) = 2(40)(60) / (40 + 60) = 4,800 / 100 = 48 km/h. |
| 30 | In CSC Word Problems with LCM, three traffic lights flash every 12 seconds, 18 seconds, and 30 seconds, respectively. If they flash together at 8:00:00 AM, at what time will they next flash together? | 8:03:00 AM. Find the LCM of 12, 18, and 30: 12 = 2²·3, 18 = 2·3², 30 = 2·3·5 -> LCM = 2² × 3² × 5 = 180 seconds = 3 minutes. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 4 days. In inverse proportion, Workers₁ × Days₁ = Workers₂ × Days₂ -> 6 × 10 = 15 × D₂ -> D₂ = 60 / 15 = 4 days.; B) 3 hours. When moving toward each other (or in opposite directions), relative speed is the sum of their speeds: v_rel = 50 + 70 = 120 km/h. Time = Distance / v_rel = 360 km / 120 km/h = 3 hours.; C) 8:03:00 AM. Find the LCM of 12, 18, and 30: 12 = 2²·3, 18 = 2·3², 30 = 2·3·5 -> LCM = 2² × 3² × 5 = 180 seconds = 3 minutes. ✓; D) 12 meters, using the 5-12-13 Pythagorean triple (h = √(13² - 5²) = √(169 - 25) = √144 = 12 m). Common CSC Pythagorean triples to memorize: (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25). |
| 31 | In CSC Ratio and Proportion (Partitive Proportion), a bonus of PHP 72,000 is divided among three employees in the ratio 2 : 3 : 4. How much does the employee with the largest share receive? | PHP 32,000. Total ratio parts = 2 + 3 + 4 = 9 parts. Value of 1 part = PHP 72,000 / 9 = PHP 8,000. Largest share (4 parts) = 4 × PHP 8,000 = PHP 32,000. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) Ana is 12 years old and Maria is 36 years old. Let Ana = x and Maria = 3x. In 12 years: 3x + 12 = 2(x + 12) -> 3x + 12 = 2x + 24 -> x = 12, and 3x = 36.; B) PHP 50,000. Using I = P·r·t, solve for P = I / (r·t) = 9,000 / (0.06 × 3) = 9,000 / 0.18 = PHP 50,000.; C) 23. Step 1 (Parentheses & Exponents): 48 ÷ 6 × 4 - 9. Step 2 (Multiplication and Division strictly from LEFT TO RIGHT): (48 ÷ 6) × 4 - 9 = 8 × 4 - 9 = 32 - 9 = 23.; D) PHP 32,000. Total ratio parts = 2 + 3 + 4 = 9 parts. Value of 1 part = PHP 72,000 / 9 = PHP 8,000. Largest share (4 parts) = 4 × PHP 8,000 = PHP 32,000. ✓ |
| 32 | In CSC Mixture Problems, how many liters of pure alcohol (100%) must be added to 20 liters of a 30% alcohol solution to produce a 50% alcohol solution? | 8 liters. Using the solute balance equation: 0.30(20) + 1.00(x) = 0.50(20 + x) -> 6 + x = 10 + 0.5x -> 0.5x = 4 -> x = 8 liters. | High-yield exam verification (CSC CSE-PPT Numerical Reasoning): distinguish the controlling statutory/doctrinal rule, timeframe, and exception from common test distractors. | A) 8 liters. Using the solute balance equation: 0.30(20) + 1.00(x) = 0.50(20 + x) -> 6 + x = 10 + 0.5x -> 0.5x = 4 -> x = 8 liters. ✓; B) 32%. New Area = (1.20 L) × (1.10 W) = 1.32 LW, which is a 32% increase.; C) 1/8 = 0.125 (12.5%); 3/8 = 0.375 (37.5%); 5/8 = 0.625 (62.5%); 7/8 = 0.875 (87.5%); 1/6 = 0.1667 (16.67%); 5/6 = 0.8333 (83.33%).; D) 11. Odd positions double each time (2, 4, 8, 16), while even positions decrease by 3 each time (20, 17, 14, 11). The 8th term is an even position: 14 - 3 = 11. |

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