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

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 ✓

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

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 ✓

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

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.

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

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.

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

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.

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

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 ✓

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

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.

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

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.

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

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 ✓

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

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.

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

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 ✓

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

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

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 ×

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.

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

Do

• Show key calculation step
• Include units in final answer
• Write answer clearly

Don't

• Skip showing any working
• Give answer without calculation
• Use unclear notation

Do

• Define variables if needed
• Show equation setup
• Display algebraic steps
• State final answer clearly

Don't

• Rush through steps
• Skip variable definition
• Forget units
• Omit final statement

Do

• State given information
• Define all variables clearly
• Show complete algebraic working
• Verify answer
• Use proper mathematical notation

Don't

• Skip any major step
• Make calculation errors
• Forget to verify
• Use poor organization

Do

• Use proper headings (Given, To Find, Solution)
• Create tables for complex problems
• Show every algebraic step
• Include complete verification
• State assumptions if any

Don't

• Rush the solution
• Skip intermediate steps
• Forget verification
• Use unclear variable names
• Make organizational mistakes