What You'll Learn This Hour
- 1 Apply the Rate × Time = Distance triangle to any motion or work problem in under 60 seconds
- 2 Set up combined-rate equations for work problems involving two or more workers/pipes
- 3 Solve mixture and concentration problems using the weighted average approach
- 4 Distinguish ratio part-to-part vs part-to-whole and handle direct/inverse variation confidently
Core Concepts
1. Rate × Time = Distance (or Work)
This single formula handles motion problems AND work problems. For work, replace "Distance" with "Jobs Done."
Rate
D ÷ T
miles/hr, jobs/hr
Time
D ÷ R
hours, days
Distance / Work
R × T
miles, jobs
Tip: Always check units. If speed is in km/hr and time is in minutes, convert before computing.
2. Work Problems — Combined Rates
When two workers (A and B) work together, add their individual rates. Never add their times.
If A completes job in a hours and B in b hours:
Combined rate = 1/a + 1/b
Time together = 1 ÷ (1/a + 1/b) = ab/(a+b)
For a draining pipe, subtract its rate from the filling rate.
3. Mixture & Concentration Problems
The key principle: total amount of substance = concentration × volume, and this amount is conserved when you mix.
C₁ × V₁ + C₂ × V₂ = C_mix × (V₁ + V₂)
# C = concentration (decimal), V = volume
You can also use alligation: the ratio of volumes is inversely proportional to the distance from each concentration to the target.
4. Ratio and Proportion
A ratio a:b means for every a units of the first quantity, there are b units of the second. Total parts = a + b.
Part-to-Part
Boys : Girls = 3 : 2
Boys = 3k, Girls = 2k for some k
Part-to-Whole
Boys are 3/5 of total students
Total parts = 3 + 2 = 5
5. Direct and Inverse Variation
Direct Variation
y = kx → y/x = k
As x increases, y increases proportionally. Example: cost = price × quantity.
Inverse Variation
y = k/x → xy = k
As x increases, y decreases. Example: speed × time = constant distance.
Visual Frameworks
The R×T=D Triangle
Cover the unknown variable to see the formula for it
Mixture Diagram (Alligation)
0.30·V₁ + 0.50·V₂ = 0.40·(V₁+V₂) → V₁/V₂ = 1:1
Worked Examples
A train travels from City A to City B at 60 mph. On the return trip, it travels at 90 mph. What is the train's average speed for the entire round trip?
Step-by-Step Solution:
Step 1 — Name the distance. Let the one-way distance = d miles. (A specific value is not needed; it will cancel.)
Step 2 — Find each leg's time.
Time A→B = d/60. Time B→A = d/90.
Step 3 — Average speed = Total Distance ÷ Total Time.
Total distance = 2d.
Total time = d/60 + d/90 = 3d/180 + 2d/180 = 5d/180 = d/36.
Step 4 — Compute.
Avg speed = 2d ÷ (d/36) = 2d × 36/d = 72 mph.
KEY INSIGHT: Average speed is NEVER the arithmetic mean of two speeds when you travel equal distances. Use the harmonic mean formula: 2ab/(a+b) = 2(60)(90)/(60+90) = 10800/150 = 72. ✓
Pipe A can fill a tank in 4 hours. Pipe B can fill the same tank in 6 hours. Pipe C can drain a full tank in 12 hours. If all three pipes are open simultaneously, how long does it take to fill the tank?
Step-by-Step Solution:
Step 1 — Assign rates.
Rate A = 1/4 tank/hr (filling). Rate B = 1/6 tank/hr (filling). Rate C = −1/12 tank/hr (draining).
Step 2 — Combined net rate.
Net = 1/4 + 1/6 − 1/12.
Common denominator = 12: 3/12 + 2/12 − 1/12 = 4/12 = 1/3 tank/hr.
Step 3 — Time = 1 ÷ Net Rate.
Time = 1 ÷ (1/3) = 3 hours.
KEY INSIGHT: Draining pipes subtract from the combined rate. Always convert each worker/pipe to a rate (fraction of job per unit time), combine, then take the reciprocal for time.
In a company, the ratio of managers to engineers to analysts is 2 : 5 : 3. If there are 120 employees total, how many more engineers are there than analysts?
Step-by-Step Solution:
Step 1 — Identify total parts.
Total parts = 2 + 5 + 3 = 10 parts.
Step 2 — Value of one part.
120 employees ÷ 10 parts = 12 employees per part.
Step 3 — Count each group.
Managers = 2 × 12 = 24. Engineers = 5 × 12 = 60. Analysts = 3 × 12 = 36.
Step 4 — Answer the question.
Engineers − Analysts = 60 − 36 = 24 more engineers.
KEY INSIGHT: Always sum the ratio parts first to find the "1 part" value. Then multiply each ratio term by that value. This avoids algebra entirely on most ratio problems.
GMAT Traps to Avoid
Trap 1: Adding Times, Not Rates
If A takes 3 hrs and B takes 5 hrs together, the time is NOT (3+5)=8 hrs. Add the RATES: 1/3+1/5 = 8/15, so together = 15/8 = 1.875 hrs.
Trap 2: Forgetting Drain Pipes Subtract
When a pipe drains while others fill, its contribution is NEGATIVE. Net rate = sum of fill rates minus sum of drain rates. A common error is adding all rates as positive.
Trap 3: Part vs Whole in Ratios
If boys:girls = 3:2, boys are 3/5 of the TOTAL, not 3/2. Always compute the total parts (denominator) before finding a fraction of the whole.
Trap 4: Average Speed ≠ Arithmetic Mean
For equal distances, use the harmonic mean: 2ab/(a+b). The arithmetic mean (a+b)/2 is only correct when time is equal on each leg, NOT distance.
Trap 5: Confusing Direct and Inverse Variation
"Inversely proportional" means y = k/x (product is constant). GMAT problems sometimes describe an inverse relationship in plain English — e.g., "more workers, fewer days" — without using the word "inverse." Always set up xy = k and verify the relationship makes sense before solving.
Practice Questions
12 GMAT-style questions. Attempt each before revealing the answer.
Q1. Train X leaves Station A at 8 AM traveling at 60 mph toward Station B. Train Y leaves Station B at 9 AM traveling at 80 mph toward Station A. The stations are 280 miles apart. At what time do the trains meet?
Show Answer ▾
Answer: (A) 10:00 AM
By 9 AM, Train X has traveled 60 × 1 = 60 miles. Remaining gap = 280 − 60 = 220 miles. After 9 AM, both trains approach each other at combined speed 60 + 80 = 140 mph. Time to close = 220/140 = 11/7 ≈ 1.57 hrs... Wait — let's recheck. 220 ÷ 140 = 1.571 hrs ≈ 1 hr 34 min → they meet around 10:34. But checking answer A (10:00 AM): at 10 AM, X has traveled 2 hrs × 60 = 120 mi; Y has traveled 1 hr × 80 = 80 mi; total = 200 ≠ 280. The correct meet time is 10:34 AM, closest to C (10:30 AM). Correct answer: C — 10:30 AM (approximate; 220/140 = 1h 34min after 9 AM = 10:34 AM, rounding to closest choice).
Q2. Worker A can complete a project in 10 days. Worker B can complete the same project in 15 days. How many days will it take if they work together?
Show Answer ▾
Answer: (C) 6 days
Rate A = 1/10 job/day. Rate B = 1/15 job/day. Combined = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 job/day. Time = 1 ÷ (1/6) = 6 days. Formula: ab/(a+b) = (10)(15)/(10+15) = 150/25 = 6. ✓ Trap avoided: 10+15=25 is wrong; never add times.
Q3. A chemist mixes a 20% salt solution with a 50% salt solution to get 60 liters of a 30% salt solution. How many liters of the 20% solution are used?
Show Answer ▾
Answer: (C) 40 liters
Let x = liters of 20% solution. Then (60−x) = liters of 50% solution. Equation: 0.20x + 0.50(60−x) = 0.30(60). Expand: 0.20x + 30 − 0.50x = 18. −0.30x = −12. x = 40 liters. Check: 0.20(40) + 0.50(20) = 8 + 10 = 18 = 0.30(60). ✓
Q4. The ratio of red to blue to green marbles is 3 : 4 : 5. If there are 48 blue marbles, how many total marbles are there?
Show Answer ▾
Answer: (B) 144
Blue part = 4. Blue marbles = 48. So 1 part = 48/4 = 12. Total parts = 3+4+5 = 12. Total marbles = 12 × 12 = 144. Trap: if you used 48 as the total parts multiplier directly, you get 576 — always find the value of 1 part first.
Q5. y varies inversely as x. When x = 4, y = 9. What is y when x = 12?
Show Answer ▾
Answer: (C) 3
Inverse variation: xy = k. So k = 4 × 9 = 36. When x = 12: y = 36/12 = 3. Trap: direct variation would give y = 27 (answer D) — that is wrong here. Inversely means the product stays constant, not the ratio.
Q6. A car travels 120 miles at 40 mph and then 120 miles at 60 mph. What is the car's average speed for the entire trip?
Show Answer ▾
Answer: (B) 48 mph
Total distance = 240 miles. Time leg 1 = 120/40 = 3 hrs. Time leg 2 = 120/60 = 2 hrs. Total time = 5 hrs. Avg speed = 240/5 = 48 mph. Using the harmonic mean formula: 2(40)(60)/(40+60) = 4800/100 = 48. ✓ Common trap: (40+60)/2 = 50 is wrong when distances are equal.
Q7. Machine P produces 100 widgets per hour. Machine Q produces 150 widgets per hour. If both machines work together for 8 hours, how many widgets are produced?
Show Answer ▾
Answer: (E) 2,000
Combined rate = 100 + 150 = 250 widgets/hour. Total in 8 hours = 250 × 8 = 2,000. This is a straightforward rate addition. No harmonic mean needed since we want total output, not a single rate derived from completing 1 job.
Q8. A 40-liter solution is 25% alcohol. How much pure alcohol must be added to make it 40% alcohol?
Show Answer ▾
Answer: (C) 10 liters
Initial alcohol = 0.25 × 40 = 10 liters. Let x = liters of pure alcohol added. New equation: (10 + x)/(40 + x) = 0.40. Solve: 10 + x = 0.40(40 + x) = 16 + 0.4x. 0.6x = 6. x = 10 liters. Check: (10+10)/(40+10) = 20/50 = 0.40. ✓
Q9. A pump can fill a pool in 6 hours. A drain can empty the pool in 9 hours. If the pool is half full and both are running, how long until the pool is completely full?
Show Answer ▾
Answer: (D) 9 hours
Net rate = 1/6 − 1/9 = 3/18 − 2/18 = 1/18 pool/hr. Pool is half full, so needs another 1/2 pool. Time = (1/2) ÷ (1/18) = (1/2) × 18 = 9 hours. Key: the starting point (half full) changes how much work remains, but not the rate.
Q10. The ratio of cats to dogs in a shelter is 5 : 3. If 10 more cats are added, the ratio becomes 2 : 1. How many dogs are in the shelter?
Show Answer ▾
Answer: (D) 30
Let cats = 5k, dogs = 3k. After adding 10 cats: (5k+10)/(3k) = 2/1. Cross multiply: 5k+10 = 6k. k = 10. Dogs = 3k = 30. Check: original cats = 50, dogs = 30, ratio 50:30 = 5:3 ✓. New cats = 60, ratio 60:30 = 2:1 ✓.
Q11. It takes 8 workers 12 days to build a wall. If the work rate per worker is constant, how many workers are needed to build the same wall in 6 days?
Show Answer ▾
Answer: (D) 16
Total work = workers × days = 8 × 12 = 96 worker-days. To finish in 6 days: workers needed = 96/6 = 16. This is an inverse variation: workers × days = constant (total work). As days decrease, workers must increase proportionally.
Q12. Two trains start simultaneously from opposite ends of a 400-mile track. Train A travels at 70 mph and Train B at 30 mph. How far from Train A's starting point do they meet?
Show Answer ▾
Answer: (C) 280 miles
Combined speed = 70 + 30 = 100 mph. Time to meet = 400/100 = 4 hours. Distance traveled by Train A = 70 × 4 = 280 miles from A's starting point. Train B covers 30 × 4 = 120 miles. Check: 280 + 120 = 400. ✓ The meeting point is always proportional to each train's speed relative to the combined speed.
Quick Reference Card
# Word Problems & Ratios — Formula Sheet
## Rate-Time-Distance / Work
Rate × Time = Distance (or Jobs)
Avg speed (equal dist) = 2ab / (a+b) ← harmonic mean
Avg speed (equal time) = (a+b) / 2 ← arithmetic mean
## Combined Work
Together = ab / (a+b) [two workers, a and b hours solo]
Net rate = Σ(fill rates) − Σ(drain rates)
Time = 1 / Net_rate
## Mixtures
C₁·V₁ + C₂·V₂ = C_mix·(V₁+V₂)
Alligation ratio: V₁/V₂ = (C₂ − C_mix) / (C_mix − C₁)
Pure substance added: treat as 100% concentration
## Ratios
a:b → Each unit = Total / (a+b)
Part-to-whole fraction = part / (sum of all parts)
Always find "1 part" value before scaling
## Variation
Direct: y = kx → y/x = constant
Inverse: y = k/x → xy = constant
Workers × Days = Total work (inverse variation)
## GMAT Time Savers
# Pick d=LCM when distance is unspecified
# Total work = (workers) × (days) = constant
# Trains approaching: time = gap / combined_speed
# Trains same direction: time = gap / speed_difference
Hour 3 of 24 — Keep going! You're building real momentum.