Five questions on ratio and proportion, mixtures, time and work, and time, speed and distance.
Question 1 · Type in the answer (+3 / 0)
A container holds 80 litres of pure milk. A certain quantity is drawn off and replaced with water. This operation is carried out three times in all, with the same quantity each time. The container then holds 33.75 litres of milk. How many litres were drawn off each time? Enter the answer in litres.
Show the solutionSolution
Answer: 20
If x litres are replaced each time, the milk left after n operations is 80 × (1 − x/80)ⁿ.
So 1 − x/80 = 3/4, giving x/80 = 1/4 and x = 20 litres.
Check: 80 → 60 → 45 → 33.75 litres of milk.
Question 2 · Multiple choice (+3 / −1)
A and B together can finish a job in 12 days, B and C in 15 days, and C and A in 20 days. All three start together; A leaves after 2 days, and B leaves 3 days before the job is finished. In how many days is the job finished?
Show the solutionSolution
Answer: B. 16 1/4 days
Adding the pair rates: 2(A + B + C) = 1/12 + 1/15 + 1/20 = (5 + 4 + 3)/60 = 1/5, so A + B + C = 1/10 per day.
Individual rates: A = 1/10 − 1/15 = 1/30, B = 1/10 − 1/20 = 1/20, C = 1/10 − 1/12 = 1/60.
Let the job take T days. A works 2 days, B works (T − 3) days and C works T days: 2/30 + (T − 3)/20 + T/60 = 1.
Multiply by 60: 4 + 3(T − 3) + T = 60, so 4T − 5 = 60 and T = 65/4 = 16 1/4 days.
Question 3 · Type in the answer (+3 / 0)
Two cars start at the same moment from towns P and Q, 300 km apart, and drive towards each other at constant speeds. After they meet, the car from P takes 1 hour 20 minutes to reach Q and the car from Q takes 3 hours to reach P. What is the speed of the faster car? Enter the answer in km/h.
Show the solutionSolution
Answer: 90
Let the speeds be a (from P) and b (from Q), and let them meet after t hours.
After the meeting, the car from P covers the stretch the other car covered in t hours: a × 4/3 = b × t. Similarly, b × 3 = a × t.
Dividing: (a/b)² = 3 ÷ 4/3 = 9/4, so a : b = 3 : 2, and t = √(4/3 × 3) = 2 hours.
In 2 hours they cover 300 km together, so a + b = 150 km/h. With a : b = 3 : 2, a = 90 km/h and b = 60 km/h.
Shortcut: meeting time = √(t₁ × t₂) and the speed ratio = √(t₂ ÷ t₁).
Question 4 · Multiple choice (+3 / −1)
Ira, Jai and Kabir start a business with capitals in the ratio 3 : 4 : 5. After 4 months Ira doubles the capital invested, and after 8 months Kabir withdraws half of the capital invested. At the end of the year the profit is ₹3,16,000. How much more than Jai does Ira receive?
Show the solutionSolution
Answer: C. ₹24,000
Profit is shared in the ratio of capital × months.
Ratio 60 : 48 : 50, total 158. One part = 3,16,000 ÷ 158 = ₹2,000.
Ira gets 60 × 2,000 = ₹1,20,000 and Jai gets 48 × 2,000 = ₹96,000.
Difference = ₹24,000.
Question 5 · Multiple choice (+3 / −1)
In a 1,200 m race, A beats B by 120 m, or equivalently by 15 seconds. In an 800 m race, B beats C by 50 m. All three run at constant speeds. By how many seconds would A beat C in a 1,200 m race?
Show the solutionSolution
Answer: C. 25 seconds
B covers the 120 m gap in 15 seconds, so B's speed = 120 ÷ 15 = 8 m/s.
B runs 1,200 m in 1,200 ÷ 8 = 150 s, so A runs it in 150 − 15 = 135 s.
When B runs 800 m, C runs 750 m, so C's speed = 8 × 750/800 = 7.5 m/s.
C runs 1,200 m in 1,200 ÷ 7.5 = 160 s.
A beats C by 160 − 135 = 25 seconds.
Full worked solution
Repeated replacement: quantity left = initial × (1 − x/V)ⁿ, where x is drawn from volume V each time.
Time and work: add the pair rates and halve to get the combined rate; subtract to get each person's rate; then sum (rate × days worked) = 1.
Two bodies moving towards each other who take t₁ and t₂ after meeting: speed ratio = √(t₂ ÷ t₁), meeting time = √(t₁t₂).
Partnership: profit is shared in the ratio of capital × time, adding up each change in capital.
Races: 'beats by d metres or t seconds' means the loser covers d metres in t seconds, which fixes the loser's speed.