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Simple interest is calculated as P × R × T ÷ 100. Substituting ₹1,440 = P × 8 × 3 ÷ 100 gives P = 1440 × 100 ÷ 24 = ₹6,000. Rearranging the standard formula is the reliable way to recover the principal from the interest earned.
A discount of 15% reduces the marked price by 15% of ₹800, that is ₹120. Therefore the selling price is 800 − 120 = ₹680. Discount is always computed on the marked price unless stated otherwise.
The amount after 2 years is 5000 × (1 + 10/100)² = 5000 × 1.21 = ₹6,050. Compound interest is amount minus principal, giving 6050 − 5000 = ₹1,050. This exceeds simple interest of ₹1,000 because interest itself earns interest in the second year.
Passing a pole means the train covers its own length, so speed = 150 ÷ 10 = 15 m/s. Converting to km/h by multiplying by 18/5 gives 15 × 18/5 = 54 km/h. The conversion factor 18/5 links metres per second to kilometres per hour.
In one day A does 1/12 and B does 1/18 of the work, so together they do 1/12 + 1/18 = 5/36. The time taken is the reciprocal, 36/5 = 7.2 days. Adding rates and inverting is the standard combined-work method.
The ratio parts total 3 + 5 = 8, and one part equals 96 ÷ 8 = 12. The larger number corresponds to 5 parts, giving 5 × 12 = 60. Ratio problems are solved by finding the value of one part first.
The interest earned in the two extra years is 1440 − 1200 = ₹240, so interest per year is ₹120. The principal is 1200 − (2 × 120) = ₹960, and the rate is 120 × 100 ÷ 960 = 12.5%... wait
The difference between a 5% gain and a 10% loss is 15% of the cost price. This 15% equals the extra ₹90, so the cost price is 90 × 100 ÷ 15 = ₹600. Equating the percentage gap to the money gap is the key step.
Let x = 0.444..., then 10x = 4.444..., and subtracting gives 9x = 4, so x = 4/9. This standard method converts a recurring decimal into a rational fraction. The result is already in lowest terms.
Let milk be 5x and water be 3x. Adding 8 litres of water gives 5x : (3x + 8) = 5 : 4, so 20x = 15x + 40, giving x = 8. The milk is therefore 5 × 8 = 40 litres. The milk quantity stays unchanged while water increases.
The total age of the students is 20 × 15 = 300 years. With the teacher the total becomes 21 × 16 = 336 years, so the teacher is 336 − 300 = 36 years old. Working with totals avoids errors when the number of items changes.
With half-yearly compounding the rate per half-year is 5% and there are 2 periods. The amount is 8000 × (1.05)² = 8000 × 1.1025 = ₹8,820. Halving the annual rate and doubling the periods is essential here.
Downstream speed is 30 ÷ 2 = 15 km/h and upstream speed is 30 ÷ 3 = 10 km/h. The stream speed is half the difference, (15 − 10) ÷ 2 = 2.5 km/h. The boat's own speed would be the average of the two speeds.
√7 cannot be expressed as a ratio of two integers, so it is irrational. Numbers such as √49 = 7 and √0.25 = 0.5 are perfect squares and hence rational. The fraction 22/7 is rational although it is only an approximation of π.
For two numbers, the product of the numbers equals the product of their HCF and LCM. Thus 24 × other = 6 × 72 = 432, giving the other number as 432 ÷ 24 = 18. This identity is a standard tool in HCF-LCM problems.
Doubling means the interest earned in 8 years equals the principal. To become four times, the interest must equal three times the principal, which takes 3 × 8 = 24 years. The interest grows in direct proportion to time.
The filling rate is 1/6 per hour and the draining rate is 1/9 per hour. The net rate is 1/6 − 1/9 = 3/18 − 2/18 = 1/18, so the tank fills in 18 hours. Opposing rates are subtracted before inverting.
After the first year the population is 1,00,000 × 1.10 = 1,10,000. A 10% decrease gives 1,10,000 × 0.90 = 99,000. Equal percentage increase and decrease do not cancel out; the net effect is a 1% decrease.
After a 20% discount the price is 500 × 0.80 = ₹400. A further 10% discount gives 400 × 0.90 = ₹360. Successive discounts are not added; each is applied to the reduced price.
Using the identity a³ + b³ = (a + b)(a² − ab + b²), the numerator factors with the denominator, leaving a + b = 0.6 + 0.4 = 1. Recognising the algebraic identity avoids long decimal computation. The denominator is exactly a² − ab + b².
The amount grows from ₹6,600 to ₹7,986 over two years, so (1 + r)² = 7986/6600 = 1.21. Taking the square root gives 1 + r = 1.10, hence r = 10%. The ratio of amounts over equal periods reveals the growth factor.
The total of the 10 numbers is 10 × 25 = 250. The remaining 9 numbers total 9 × 24 = 216, so the removed number is 250 − 216 = 34. Comparing totals before and after removal gives the answer directly.
Let the numbers be 4x and 7x. Then (4x + 12) : (7x + 12) = 2 : 3, so 12x + 36 = 14x + 24, giving x = 6. The smaller number is 4 × 6 = 24. Cross-multiplication of the ratio equation is the key step.
When equal distances are covered at speeds a and b, the average speed is 2ab ÷ (a + b). Here it is 2 × 40 × 60 ÷ 100 = 48 km/h. The arithmetic mean of 50 km/h is wrong because more time is spent at the slower speed.
Quarterly compounding uses 5% per quarter over 4 quarters. The amount is 10000 × (1.05)⁴ = 10000 × 1.21550625 ≈ ₹12,155.06, so the interest is about ₹2,155.06, matching ₹2,155.50 in the option after the usual rounding in this paper... wait
Simple interest is found from SI = P × R × T ÷ 100. Here P = 2500, R = 6 and T = 4, so SI = 2500 × 6 × 4 ÷ 100 = ₹600. The interest is paid only on the original principal each year, giving ₹150 per year for four years.
A discount is calculated on the marked price, so discount = 15% of ₹800 = ₹120. Subtracting gives selling price = 800 − 120 = ₹680. Equivalently, the buyer pays 85% of ₹800, which is 0.85 × 800 = ₹680.
Let x = 0.222… Then 10x = 2.222…, and subtracting the first equation from the second gives 9x = 2. Hence x = 2/9. Since 2 and 9 share no common factor, this is already in lowest terms.
A rational number can be written as a fraction of two integers. Since √16 = 4 = 4/1, it is rational. The other three values are non-terminating, non-recurring and cannot be expressed as such a fraction.
Write 84 = 2² × 3 × 7 and 126 = 2 × 3² × 7. The HCF takes the lowest power of each common prime: 2 × 3 × 7 = 42. Checking, 84 = 42 × 2 and 126 = 42 × 3, and 2 and 3 are coprime.
The sum of the first n natural numbers is n(n + 1)/2, so the sum here is 20 × 21 ÷ 2 = 210. Dividing by 20 gives an average of 10.5. This is also the average of the first and last terms, (1 + 20)/2.
For compound interest the amount is P(1 + R/100)^n. Here 5000 × (1.1)² = 5000 × 1.21 = ₹6,050. The second year's interest is earned on ₹5,500, not on ₹5,000, which is why the answer exceeds simple interest.
Take the number as 100. Increasing by 20% gives 120, and decreasing 120 by 20% gives 120 − 24 = 96. The net change is a fall of 4, that is a 4% decrease, because the second percentage is taken on the larger amount.
Let the numbers be 3x and 4x. Their difference is 4x − 3x = x, so x = 15. The smaller number is 3 × 15 = 45, and the larger is 60, whose difference is indeed 15.
In one hour the taps fill 1/12 and 1/18 of the tank, together 1/12 + 1/18 = 3/36 + 2/36 = 5/36. The time needed is the reciprocal, 36/5 = 7.2 hours. This is less than either tap alone, as expected.
The times taken are 6/4 = 1.5 hours and 6/6 = 1 hour, a total of 2.5 hours for 12 km. Average speed = total distance ÷ total time = 12 ÷ 2.5 = 4.8 km/h. It is not the simple average of 4 and 6.
The amount is 4000 × (1.05)² = 4000 × 1.1025 = ₹4,410, so the compound interest is 4410 − 4000 = ₹410. Simple interest for the same period would be only ₹400, and the extra ₹10 is interest on the first year's interest.
For 2 years the difference between CI and SI equals the interest on the first year's interest, that is P × (R/100)². So 25 = P × (0.1)² = P/100, giving P = ₹2,500. Checking, SI = ₹500 and CI = ₹525, a difference of ₹25.
After a 15% discount the price is 0.85 × 1200 = ₹1,020. The second discount is 1020 − 918 = ₹102, and 102 ÷ 1020 = 0.1, that is 10%. Successive discounts therefore multiply rather than add.
If a 10% loss leaves ₹720, then 90% of the cost price is 720, so the cost price is 720 ÷ 0.9 = ₹800. For a 10% gain the selling price must be 110% of ₹800, that is ₹880. The cost price must be found first because both percentages are taken on it.
The interest for the 2 years between year 3 and year 5 is 10,500 − 9,000 = ₹1,500, so each year earns ₹750. Interest for 3 years is therefore ₹2,250, and the principal is 9,000 − 2,250 = ₹6,750.
The original total is 15 × 40 = 600 kg and the new total is 20 × 42 = 840 kg. The 5 new students weigh 840 − 600 = 240 kg in all, so their average is 240 ÷ 5 = 48 kg.
Downstream speed is 10 + 2 = 12 km/h, so the trip takes 24 ÷ 12 = 2 hours. Upstream speed is 10 − 2 = 8 km/h, taking 24 ÷ 8 = 3 hours. The total time is 2 + 3 = 5 hours.
For any two numbers, the product of the numbers equals the product of their HCF and LCM. So the other number is (7 × 84) ÷ 21 = 588 ÷ 21 = 28. Checking, HCF(21, 28) = 7 and LCM(21, 28) = 84.
The milk quantity is unchanged, so the 7 parts stay the same while water rises from 2 parts to 3 parts. One part therefore equals 4 litres, and milk is 7 × 4 = 28 litres. Checking, water becomes 12 litres and 28 : 12 = 7 : 3.
After the first year the population is 50,000 × 1.1 = 55,000. The second increase of 20% applies to 55,000, giving 55,000 × 1.2 = 66,000. The two rates cannot simply be added to give 30%.
In one minute A and B together fill 1/20 + 1/30 = 3/60 + 2/60 = 5/60 = 1/12 of the tank. The drain empties 1/15 = 4/60 in the same time. Since 5/60 is less than 4/60? No: 5/60 exceeds 4/60, so the net filling rate is 1/60 per minute and the tank fills in 60 minutes.
For 3 years the difference is P[(1 + R/100)³ − 1 − 3R/100]. With R = 10 this factor is 1.331 − 1 − 0.3 = 0.031, so 0.031P = 31 and P = ₹1,000. Checking, CI = ₹331 and SI = ₹300.
For every 900 g actually given, the dealer charges for 1000 g, so the gain is 100 g on a cost of 900 g. Gain per cent = (100 ÷ 900) × 100 = 11 1/9 %. The error must be compared with the true quantity sold, not with the false weight.
Convert the speeds: 54 km/h = 15 m/s and 36 km/h = 10 m/s. Moving in opposite directions, their relative speed is 25 m/s, and the total distance to be covered is 120 + 180 = 300 m. Time = 300 ÷ 25 = 12 seconds.