SSC GD Ultra Hard Maths: 4 Solved Questions with Full Solutions

Quick answer key: 1) A — 40 seconds; 2) D — ₹10,000; 3) C — 30 litres; 4) D — 18 days. This SSC GD ultra-hard maths set covers train speed, compound interest, mixture replacement, and time-and-work rates with staggered departures. Try each problem before reading its full solution.

SSC GD Ultra Hard Maths: Four-Question Challenge

Ready to test your calculation and reasoning? Try to solve all four questions without looking at the explanations first. Write down what each number represents, keep units consistent, and check whether your final answer matches the options. The solutions below explain both the method and the reason it works.

Question 1: Train, platform and increased speed

A train crosses a pole in 18 seconds and a platform 270 metres long in 36 seconds. If the train's speed is increased by 20%, how much time will it take to cross a 450-metre platform?

1. A. 40 seconds
2. B. 45 seconds
3. C. 30 seconds
4. D. 36 seconds

Step-by-step solution

Step 1: Understand what “crosses a pole” means. A pole has no meaningful length for this calculation, so the train travels exactly its own length while crossing it. If its speed is v metres per second, its length is therefore 18v metres.

Step 2: Use the platform crossing to find the original speed. To completely pass a 270-metre platform, the train must travel its own length plus the platform length. In 36 seconds the distance is 36v, so:

36v = 18v + 270

Subtract 18v from both sides: 18v = 270, so the original speed is v = 15 metres per second. The train's length is 18 × 15 = 270 metres.

Step 3: Increase the speed by 20%. A 20% increase means multiplying by 1.20: 15 × 1.20 = 18 metres per second. Do not add 20 metres per second; the increase is a percentage of the original speed.

Step 4: Find the full distance across the new platform. The train must cover its 270-metre length plus the 450-metre platform: 270 + 450 = 720 metres.

Step 5: Calculate the time. Time = distance ÷ speed = 720 ÷ 18 = 40 seconds.

Correct answer: A. 40 seconds.

A quick check

At 18 metres per second, 40 seconds covers 18 × 40 = 720 metres exactly. That is the train's 270 metres plus the platform's 450 metres, so the result is consistent.

Question 2: Compound interest and the original principal

A sum of money becomes ₹13,310 in 3 years and ₹14,641 in 4 years at compound interest, compounded annually. Find the original principal.

1. A. ₹12,000
2. B. ₹11,000
3. C. ₹9,000
4. D. ₹10,000

Step-by-step solution

Step 1: Compare the amounts in consecutive years. From the end of year 3 to the end of year 4, one year of interest is added. The amount grows from ₹13,310 to ₹14,641. So the one-year growth factor is:

₹14,641 ÷ ₹13,310 = 1.10

A growth factor of 1.10 means the annual rate is 10%, because 1.10 = 1 + 10/100. You do not need to guess an interest rate from the options.

Step 2: Move backwards three years. The 3-year amount equals principal × 1.1 × 1.1 × 1.1. Thus:

Principal = ₹13,310 ÷ (1.1)³

(1.1)³ = 1.331, and ₹13,310 ÷ 1.331 = ₹10,000.

Correct answer: D. ₹10,000.

Why divide by the growth factor?

Compound interest multiplies the principal by the same growth factor once per year. To undo multiplication over three years, divide by that factor three times. Dividing by 1.10 once would only undo one year's interest, not all three years.

A quick check

Starting with ₹10,000 at 10% annually gives ₹11,000 after year 1, ₹12,100 after year 2, and ₹13,310 after year 3. One more 10% increase gives ₹14,641 after year 4, matching both amounts in the question.

Question 3: Mixture replacement

A vessel contains 80 litres of a milk-water mixture in the ratio 7:3. How many litres of the mixture must be removed and replaced with water so that the final ratio of milk to water becomes 7:9?

1. A. 25 litres
2. B. 20 litres
3. C. 30 litres
4. D. 35 litres

Step-by-step solution

Step 1: Find the starting amounts. The ratio has 7 + 3 = 10 equal parts. Each part is 80 ÷ 10 = 8 litres. So the vessel starts with 56 litres of milk and 24 litres of water.

Step 2: Use the fact that the final volume stays at 80 litres. Removing some mixture and adding back the same amount of water leaves the total at 80 litres. The final ratio 7:9 has 16 parts in all. Therefore the final milk amount must be 80 × 7/16 = 35 litres.

Step 3: Work out how much mixture was removed. The removed mixture has the same 7:3 ratio as the original. If x litres are removed, the fraction of the original mixture left is (80 − x)/80. The milk left is:

56 × (80 − x)/80 = 35

Divide both sides by 56: (80 − x)/80 = 35/56 = 5/8. So 80 − x = 50, which gives x = 30 litres.

Correct answer: C. 30 litres.

Check the final ratio directly

Removing 30 litres leaves 50 litres of the original mixture. That remaining mixture contains 50 × 7/10 = 35 litres of milk and 50 × 3/10 = 15 litres of water. Add the 30 litres of water: there are now 35 litres of milk and 45 litres of water. The ratio 35:45 simplifies to 7:9, exactly as required.

Question 4: Time and work with staggered departures

A, B and C can complete a piece of work individually in 20 days, 30 days and 60 days respectively. They start working together. A leaves after 4 days, and B leaves 3 days before the work is completed. In how many days is the entire work completed?

1. A. 10 days
2. B. 15 days
3. C. 12 days
4. D. 18 days

Step-by-step solution

Step 1: Express the work rates in convenient units. The least common multiple of 20, 30 and 60 is 60. Treat the whole job as 60 units. A completes 60 ÷ 20 = 3 units per day, B completes 60 ÷ 30 = 2 units per day, and C completes 60 ÷ 60 = 1 unit per day. Together their rate is 3 + 2 + 1 = 6 units per day.

Step 2: Calculate the work completed in the first 4 days. All three work together for 4 days, so they complete 4 × 6 = 24 units.

Step 3: Represent the middle phase. Let the total time be T days. A has left after day 4. B leaves 3 days before completion, so B and C work together from day 4 until day T − 3. That middle phase lasts T − 7 days, at a combined rate of 2 + 1 = 3 units per day. In the final 3 days, C works alone and completes 3 × 1 = 3 units.

Step 4: Add the work from all three phases. The whole job is 60 units, so:

24 + 3(T − 7) + 3 = 60

Simplify: 24 + 3T − 21 + 3 = 60, so 3T + 6 = 60. Therefore 3T = 54 and T = 18 days.

Correct answer: D. 18 days.

Check the phases

In 4 days together, they complete 24 units. The middle phase lasts 18 − 7 = 11 days, so B and C complete 11 × 3 = 33 units. C works alone for the final 3 days and completes 3 units. Total work = 24 + 33 + 3 = 60 units, exactly the whole job.

All answers at a glance

• Train and platform: A — 40 seconds.
• Compound interest: D — ₹10,000.
• Milk-water replacement: C — 30 litres.
• Staggered time and work: D — 18 days.

Common mistakes to avoid

• Train question: For a platform, use train length plus platform length. For a pole, use only train length. Convert kilometres per hour to metres per second if units require it; here the derived speed is already in metres per second.
• Compound interest: The ratio of consecutive year-end amounts gives one year's multiplier. It is not the total four-year interest rate.
• Mixture replacement: The removed liquid contains both milk and water. Do not assume that removing 30 litres removes 30 litres of milk.
• Staggered time and work: Count each phase separately; after A leaves, B and C work together, then C finishes alone.
• Every question: Substitute the answer back into the original conditions. A short check often catches a rushed calculation.

How to practise these SSC GD maths questions

First attempt each problem with a timer, but do not sacrifice clear working for speed. After checking the answer, write down the specific idea that solved it: distance equals speed times time, consecutive compound amounts reveal the annual multiplier, a well-mixed replacement removes each ingredient in the same proportion, and staggered work requires a separate rate for each phase. Revisit only the step you missed, then solve a fresh question of the same type. This turns a correct answer into a repeatable exam method.

For SSC GD quantitative aptitude, practise arithmetic, percentages, ratio and proportion, interest, time and distance, and mixtures in short focused sets. Review wrong answers by cause—concept, setup, calculation, or reading—so you know what to improve next.