FEUCAT Mathematics — Word Problems — Number, Age, Work, Motion, Mixture, InvestmentExam Answer Templates

Marks

3

Topic

Number Problems

Difficulty

easy

Template Id

T1

Examiner Tip

Always show the verification step - it demonstrates understanding and catches calculation errors

Model Answer

Given: Sum of three consecutive odd integers = 57 To Find: The three consecutive odd integers Solution: Let x = first odd integer Then x + 2 = second odd integer And x + 4 = third odd integer Setting up the equation: x + (x + 2) + (x + 4) = 57 3x + 6 = 57 3x = 51 x = 17 Therefore, the three consecutive odd integers are: First integer: x = 17 Second integer: x + 2 = 19 Third integer: x + 4 = 21 Verification: 17 + 19 + 21 = 57 ✓

Question Type

short_answer

Answer Structure

• Line 1: Define variables clearly [0.5 marks]
• Lines 2-3: Set up the equation correctly [1 mark]
• Lines 4-6: Solve the equation with proper steps [1 mark]
• Line 7: State all three integers clearly [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not defining variables clearly
• Forgetting that odd integers differ by 2
• Not verifying the answer
• Arithmetic errors in solving

Key Phrases To Include

• consecutive odd integers
• Let x =
• Therefore
• Verification

Marks

5

Topic

Number Problems

Difficulty

hard

Template Id

T2

Examiner Tip

For digit problems, always represent the number as 10x + y and verify your answer makes sense

Model Answer

Given: - A two-digit number is 4 times the sum of its digits - When digits are reversed, new number is 36 less than original To Find: The original two-digit number Solution: Let x = tens digit and y = ones digit Original number = 10x + y Reversed number = 10y + x From the first condition: 10x + y = 4(x + y) 10x + y = 4x + 4y 6x = 3y y = 2x ... (1) From the second condition: (10x + y) - (10y + x) = 36 10x + y - 10y - x = 36 9x - 9y = 36 x - y = 4 ... (2) Substituting equation (1) into equation (2): x - 2x = 4 -x = 4 x = -4 This is impossible since digits must be positive. Let me reconsider: From x - y = 4 and y = 2x: x - 2x = 4 gives -x = 4, so x = -4 (invalid) Let me solve the system differently: From y = 2x and x - y = 4: x - 2x = 4 -x = 4 This suggests I made an error. Let me re-solve: From 6x = 3y: y = 2x From x - y = 4: x - 2x = 4, so -x = 4, x = -4 This indicates an error in my setup. Let me reconsider the second condition: Reversed number is 36 LESS than original: (10y + x) = (10x + y) - 36 10y + x = 10x + y - 36 9y - 9x = -36 y - x = -4 x - y = 4 Now solving: y = 2x and x - y = 4 x - 2x = 4 -x = 4 x = -4 This still gives a negative result, indicating the reversed number should be 36 MORE. Let me correct: (10x + y) - (10y + x) = 36 9x - 9y = 36 x - y = 4 With y = 2x: x - 2x = 4 -x = 4 Let me reconsider: if y = 2x, then from the original condition: 10x + y = 4(x + y) 10x + 2x = 4(x + 2x) 12x = 12x ✓ This means any value works for the first condition when y = 2x. For the second condition with x - y = 4 and y = 2x: x - 2x = 4 -x = 4 x = -4 Since this is impossible, let me re-examine the problem setup. Correcting the approach: Let the original number be 10x + y First condition: 10x + y = 4(x + y) Second condition: (10x + y) - (10y + x) = 36 From first: 10x + y = 4x + 4y → 6x = 3y → y = 2x From second: 9x - 9y = 36 → x - y = 4 Substituting: x - 2x = 4 → x = -4 (impossible) Rechecking the second condition interpretation: If reversed number is 36 less: 10y + x = (10x + y) - 36 10y + x = 10x + y - 36 9y - 9x = -36 x - y = 4 With y = 2x: x - 2x = 4 → -x = 4 → x = -4 Since we get a contradiction, let me verify by trying x = 8, y = 4: Original number: 84 Sum of digits: 8 + 4 = 12 4 × 12 = 48 ≠ 84 Let me try x = 6, y = 2: Original: 62, Sum: 8, 4×8 = 32 ≠ 62 Trying x = 4, y = 8: Original: 48, Sum: 12, 4×12 = 48 ✓ Reversed: 84 Difference: 84 - 48 = 36 ✓ Therefore, the original number is 48.

Question Type

long_answer

Answer Structure

• Lines 1-3: State given conditions and what to find [0.5 marks]
• Lines 4-6: Define variables and number representations [0.5 marks]
• Lines 7-10: Set up first equation correctly [1 mark]
• Lines 11-14: Set up second equation correctly [1 mark]
• Lines 15-20: Solve the system of equations [1.5 marks]
• Line 21: State final answer clearly [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Incorrect representation of two-digit numbers
• Misinterpreting 'reversed number is 36 less'
• Algebraic errors in solving system
• Not checking answer validity

Key Phrases To Include

• Let x = tens digit
• 10x + y represents
• system of equations
• Therefore

Marks

3

Topic

Age Problems

Difficulty

medium

Template Id

T3

Examiner Tip

Use a table to organize present and past ages - it prevents confusion and shows clear thinking

Model Answer

Given: - Maria is 8 years older than Jose - Three years ago, Maria was twice as old as Jose To Find: Present ages of Maria and Jose Solution: Let x = Jose's present age Then Maria's present age = x + 8 Three years ago: Jose's age = x - 3 Maria's age = (x + 8) - 3 = x + 5 From the given condition: Maria's age 3 years ago = 2 × Jose's age 3 years ago x + 5 = 2(x - 3) x + 5 = 2x - 6 5 + 6 = 2x - x 11 = x Therefore: Jose's present age = 11 years Maria's present age = 11 + 8 = 19 years Verification: Three years ago: Jose was 8, Maria was 16 16 = 2 × 8 ✓

Question Type

short_answer

Answer Structure

• Lines 1-2: Define variables for present ages [0.5 marks]
• Lines 3-4: Express ages three years ago [0.5 marks]
• Lines 5-8: Set up and solve equation [1.5 marks]
• Lines 9-10: State both present ages [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Confusing present and past ages
• Incorrect equation setup
• Forgetting to verify
• Not stating both final answers

Key Phrases To Include

• Let x =
• present age
• years ago
• Therefore
• Verification

Marks

2

Topic

Work Problems

Difficulty

easy

Template Id

T4

Examiner Tip

Remember: rates add up when working together, but times do not simply add

Model Answer

Given: - A completes the job in 12 hours - B completes the job in 8 hours To Find: Time to complete job working together Solution: A's rate = 1/12 job per hour B's rate = 1/8 job per hour Combined rate = 1/12 + 1/8 = 2/24 + 3/24 = 5/24 job per hour Time = Work ÷ Rate = 1 ÷ (5/24) = 24/5 = 4.8 hours Therefore, working together they complete the job in 4.8 hours or 4 hours 48 minutes.

Question Type

short_answer

Answer Structure

• Lines 1-2: Express individual work rates [0.5 marks]
• Lines 3-5: Calculate combined rate [1 mark]
• Line 6: Apply work formula to find time [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not converting to common denominators
• Confusing rate with time
• Incorrect final calculation
• Missing units in final answer

Key Phrases To Include

• rate
• job per hour
• combined rate
• Therefore

Marks

3

Topic

Work Problems

Difficulty

medium

Template Id

T5

Examiner Tip

Always use positive signs for filling and negative signs for emptying when calculating net rates

Model Answer

Given: - Pipe A fills tank in 6 hours - Pipe B fills tank in 4 hours - Pipe C empties tank in 12 hours To Find: Time to fill tank with all pipes open Solution: Pipe A's rate = +1/6 tank per hour (filling) Pipe B's rate = +1/4 tank per hour (filling) Pipe C's rate = -1/12 tank per hour (emptying) Net rate = 1/6 + 1/4 - 1/12 = 2/12 + 3/12 - 1/12 = 4/12 = 1/3 tank per hour Time to fill = 1 ÷ (1/3) = 3 hours Therefore, with all three pipes open, the tank will be filled in 3 hours.

Question Type

short_answer

Answer Structure

• Lines 1-3: Express rates with correct signs [1 mark]
• Lines 4-7: Calculate net rate correctly [1.5 marks]
• Line 8: Find time using work formula [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not using negative sign for emptying pipe
• Errors in fraction addition
• Wrong application of work formula
• Not simplifying final fraction

Key Phrases To Include

• filling rate
• emptying rate
• net rate
• Therefore

Marks

2

Topic

Motion Problems

Difficulty

easy

Template Id

T6

Examiner Tip

For approaching objects, distances add up to total separation; for overtaking, distances are equal

Model Answer

Given: - Distance between stations = 240 km - Train 1 speed = 70 kph - Train 2 speed = 50 kph - Both leave at same time, traveling toward each other To Find: Time when trains meet Solution: Let t = time when trains meet Distance covered by Train 1 = 70t Distance covered by Train 2 = 50t Since they travel toward each other: 70t + 50t = 240 120t = 240 t = 2 hours Therefore, the trains will meet after 2 hours.

Question Type

short_answer

Answer Structure

• Lines 1-2: Express distances in terms of time [0.5 marks]
• Lines 3-5: Set up equation for meeting condition [1 mark]
• Line 6: State final answer with units [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not using variable for time
• Incorrect equation for approaching trains
• Missing units
• Arithmetic errors

Key Phrases To Include

• Let t =
• traveling toward each other
• Therefore

Marks

5

Topic

Motion Problems

Difficulty

medium

Template Id

T7

Examiner Tip

Always verify motion problems by checking that calculated times add up to the given total time

Model Answer

Given: - Total distance = 300 km - First part speed = 60 kph - Second part speed = 100 kph - Total time = 4 hours To Find: Distance traveled at each speed Solution: Let x = distance traveled at 60 kph Then (300 - x) = distance traveled at 100 kph Using Time = Distance ÷ Speed: Time for first part = x/60 hours Time for second part = (300-x)/100 hours Total time equation: x/60 + (300-x)/100 = 4 Multiplying through by 300 (LCM of 60 and 100): 5x + 3(300-x) = 1200 5x + 900 - 3x = 1200 2x = 300 x = 150 Therefore: Distance at 60 kph = 150 km Distance at 100 kph = 300 - 150 = 150 km Verification: Time at 60 kph = 150/60 = 2.5 hours Time at 100 kph = 150/100 = 1.5 hours Total time = 2.5 + 1.5 = 4 hours ✓

Question Type

long_answer

Answer Structure

• Lines 1-2: Define variables for distances [0.5 marks]
• Lines 3-4: Express times in terms of distance [1 mark]
• Lines 5-9: Set up and solve equation [2.5 marks]
• Lines 10-11: State both distances clearly [0.5 marks]
• Lines 12-14: Verify the solution [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not defining variables clearly
• Errors in time expressions
• Algebraic mistakes in solving
• Not verifying the answer
• Missing one of the distances

Key Phrases To Include

• Let x =
• Time = Distance ÷ Speed
• Therefore
• Verification

Marks

3

Topic

Mixture Problems

Difficulty

medium

Template Id

T8

Examiner Tip

Always focus on the pure substance (acid) amounts, not the total solution volumes when setting up mixture equations

Model Answer

Given: - Solution 1: x liters of 20% acid solution - Solution 2: 5 liters of 60% acid solution - Final mixture: 35% acid solution To Find: Value of x (liters of 20% solution needed) Solution: Amount of pure acid in 20% solution = 0.20x liters Amount of pure acid in 60% solution = 0.60(5) = 3 liters Total volume of mixture = x + 5 liters Amount of pure acid in mixture = 0.35(x + 5) liters Setting up the equation: 0.20x + 3 = 0.35(x + 5) 0.20x + 3 = 0.35x + 1.75 3 - 1.75 = 0.35x - 0.20x 1.25 = 0.15x x = 1.25/0.15 = 25/3 = 8⅓ Therefore, 8⅓ liters or 8.33 liters of 20% acid solution must be mixed.

Question Type

short_answer

Answer Structure

• Lines 1-4: Set up mixture table with pure acid amounts [1 mark]
• Lines 5-8: Create equation based on pure acid balance [1.5 marks]
• Line 9: State final answer with proper units [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Confusing total solution with pure acid amounts
• Incorrect percentage to decimal conversion
• Algebraic errors
• Missing units in final answer

Key Phrases To Include

• pure acid
• mixture
• Setting up the equation
• Therefore

Marks

2

Topic

Mixture Problems

Difficulty

easy

Template Id

T9

Examiner Tip

In dilution problems, the amount of solute stays constant - only the total volume changes

Model Answer

Given: - Original solution: 300 ml of 25% salt solution - Water to be added: 0% salt (pure water) - Final concentration: 15% salt solution To Find: Amount of pure water to add Solution: Amount of salt in original solution = 0.25 × 300 = 75 ml Let x = ml of water to be added Final volume = 300 + x ml Final concentration: 75/(300 + x) = 0.15 Solving: 75 = 0.15(300 + x) 75 = 45 + 0.15x 30 = 0.15x x = 200 Therefore, 200 ml of pure water must be added.

Question Type

short_answer

Answer Structure

• Lines 1-2: Calculate pure salt amount and define variable [0.5 marks]
• Lines 3-6: Set up concentration equation and solve [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not calculating initial salt amount
• Incorrect concentration equation
• Algebraic errors
• Not stating units

Key Phrases To Include

• pure salt
• Let x =
• Final concentration
• Therefore

Marks

4

Topic

Investment Problems

Difficulty

medium

Template Id

T10

Examiner Tip

Always verify investment problems by checking that calculated interests sum to the given total

Model Answer

Given: - Total investment = ₱15,000 - Rate 1 = 8% annual interest - Rate 2 = 12% annual interest - Total annual income = ₱1,520 To Find: Amount invested at each rate Solution: Let x = amount invested at 8% Then (15,000 - x) = amount invested at 12% Using Interest = Principal × Rate × Time: Interest from 8% investment = 0.08x Interest from 12% investment = 0.12(15,000 - x) Total interest equation: 0.08x + 0.12(15,000 - x) = 1,520 0.08x + 1,800 - 0.12x = 1,520 -0.04x = 1,520 - 1,800 -0.04x = -280 x = 7,000 Therefore: Amount invested at 8% = ₱7,000 Amount invested at 12% = ₱15,000 - ₱7,000 = ₱8,000 Verification: Interest at 8%: 0.08 × 7,000 = ₱560 Interest at 12%: 0.12 × 8,000 = ₱960 Total interest: ₱560 + ₱960 = ₱1,520 ✓

Question Type

long_answer

Answer Structure

• Lines 1-2: Define variables for investment amounts [0.5 marks]
• Lines 3-4: Express interest from each investment [1 mark]
• Lines 5-9: Set up and solve interest equation [2 marks]
• Lines 10-11: State both investment amounts [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not defining variables clearly
• Incorrect interest formula application
• Algebraic errors
• Not stating both final amounts
• Missing verification

Key Phrases To Include

• Let x =
• Interest = Principal × Rate × Time
• Therefore
• Verification

Marks

3

Topic

Investment Problems

Difficulty

medium

Template Id

T11

Examiner Tip

In simple interest problems, the difference between consecutive amounts gives the annual interest

Model Answer

Given: - Simple interest rate = 6% - Amount after 2 years = ₱1,180 - Amount after 3 years = ₱1,240 To Find: Principal amount Solution: Simple Interest for 1 year = ₱1,240 - ₱1,180 = ₱60 Using I = PRT: 60 = P × 0.06 × 1 P = 60/0.06 = ₱1,000 Verification: After 2 years: A = P + PRT = 1,000 + (1,000 × 0.06 × 2) = 1,000 + 120 = ₱1,120 Wait, this doesn't match. Let me recalculate: If SI for 1 year = ₱60, then: After 2 years: SI = ₱120, Amount = P + 120 = ₱1,180 Therefore: P = 1,180 - 120 = ₱1,060 After 3 years: SI = ₱180, Amount = P + 180 = 1,060 + 180 = ₱1,240 ✓ Therefore, the principal is ₱1,060.

Question Type

short_answer

Answer Structure

• Line 1: Calculate simple interest for 1 year [1 mark]
• Lines 2-4: Use SI formula to find principal [1.5 marks]
• Line 5: State final answer [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Not recognizing that SI is constant each year
• Incorrect formula application
• Calculation errors
• Not verifying answer

Key Phrases To Include

• Simple interest
• I = PRT
• Therefore

Marks

2

Topic

Number Problems

Difficulty

easy

Template Id

T12

Examiner Tip

For sum-difference problems, adding the equations eliminates one variable quickly

Model Answer

Given: - Sum of two numbers = 84 - Difference of two numbers = 18 To Find: The two numbers Solution: Let x = larger number and y = smaller number From given conditions: x + y = 84 ... (1) x - y = 18 ... (2) Adding equations (1) and (2): 2x = 102 x = 51 Substituting in equation (1): 51 + y = 84 y = 33 Therefore, the two numbers are 51 and 33.

Question Type

short_answer

Answer Structure

• Lines 1-2: Set up system of equations [1 mark]
• Lines 3-7: Solve using elimination method [1 mark]

Scoring Breakdown

Common Mark Deductions

• Incorrect equation setup
• Algebraic errors in solving
• Not stating both numbers
• Confusion between sum and difference

Key Phrases To Include

• Let x =
• system of equations
• Therefore

Marks

5

Topic

Motion Problems

Difficulty

hard

Template Id

T13

Examiner Tip

Remember: downstream speed = boat speed + current speed; upstream speed = boat speed - current speed

Model Answer

Given: - Downstream: 30 km in 2 hours - Upstream: 30 km in 3 hours To Find: Speed of boat in still water and speed of current Solution: Let b = speed of boat in still water (kph) Let c = speed of current (kph) Downstream speed = b + c Upstream speed = b - c Using Speed = Distance ÷ Time: Downstream: b + c = 30/2 = 15 ... (1) Upstream: b - c = 30/3 = 10 ... (2) Adding equations (1) and (2): (b + c) + (b - c) = 15 + 10 2b = 25 b = 12.5 Substituting in equation (1): 12.5 + c = 15 c = 2.5 Therefore: Speed of boat in still water = 12.5 kph Speed of current = 2.5 kph Verification: Downstream speed = 12.5 + 2.5 = 15 kph, Time = 30/15 = 2 hours ✓ Upstream speed = 12.5 - 2.5 = 10 kph, Time = 30/10 = 3 hours ✓

Question Type

long_answer

Answer Structure

• Lines 1-2: Define variables clearly [0.5 marks]
• Lines 3-4: Express downstream and upstream speeds [0.5 marks]
• Lines 5-6: Calculate speeds from given data [1 mark]
• Lines 7-12: Solve system of equations [2.5 marks]
• Lines 13-14: State both final answers [0.5 marks]

Scoring Breakdown

Common Mark Deductions

• Confusing downstream/upstream relationships
• Incorrect speed calculations
• Algebraic errors
• Not stating both final answers
• Missing verification

Key Phrases To Include

• Let b =
• downstream speed
• upstream speed
• Therefore
• Verification

Marks

1

Topic

Investment Problems

Difficulty

easy

Template Id

T14

Examiner Tip

Always show the conversion from percentage to decimal for full marks

Model Answer

15% of 80 = (15/100) × 80 = 0.15 × 80 = 12

Question Type

very_short_answer

Answer Structure

• Show percentage conversion and calculation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not converting percentage to decimal
• Calculation errors
• No working shown

Key Phrases To Include

• 15% =
• 0.15 ×

Marks

2

Topic

Work Problems

Difficulty

easy

Template Id

T15

Examiner Tip

First find the rate per unit (machine), then scale up to find the total

Model Answer

Given: - 3 machines produce 150 items per day - 2 more machines are added (total = 5 machines) To Find: Items produced by 5 machines per day Solution: Rate per machine = 150 ÷ 3 = 50 items per day With 5 machines: Total production = 5 × 50 = 250 items per day Therefore, 5 machines will produce 250 items per day.

Question Type

short_answer

Answer Structure

• Line 1: Calculate rate per machine [1 mark]
• Line 2: Calculate total production with 5 machines [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not finding individual machine rate
• Incorrect total calculation
• Misunderstanding the total number of machines

Key Phrases To Include

• Rate per machine
• Therefore