SSLC (English Medium) कक्षा ८ - Tamil Nadu Board of Secondary Education Question Bank Solutions

Find the product of the terms

3x²y , −3xy³, x²y²

The missing terms in the product −3m³n × 9(__) = _________ m⁴n³

If the area of a square is 36x⁴y² then, its side is ____________

The compound interest on ₹ 5000 at 12% p.a for 2 years, compounded annually is ___________

The compound interest on ₹ 8000 at 10% p.a for 1 year, compounded half yearly is ____________

The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was _________

If the compound interest is calculated quarterly, the amount is found using the formula __________

The difference between the C.I and S.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is ___________

Depreciation value is calculated by the formula, `"P"(1 - "r"/100)^"n"`

If the present population of a city is P and it increases at the rate of r% p.a, then the population n years ago would be `"P"(1 + "r"/100)^"n"`

The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450

The time taken for ₹ 1000 to become ₹ 1331 at 20% p.a, compounded annually is 3 years

The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522

Find the compound interest on ₹ 3200 at 2.5% p.a for 2 years, compounded annually

Find the compound interest for `2 1/2` years on ₹ 4000 at 10% p.a, if the interest is compounded yearly

A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the principal

In how many years will ₹ 3375 become ₹ 4096 at `13 1/3` p.a if the interest is compounded half-yearly?

Find the C.I. on ₹ 15000 for 3 years if the rates of interest are 15%, 20% and 25% for the I, II and III years respectively

The number of conversion periods in a year, if the interest on a principal is compounded every two months is ___________

The time taken for ₹ 4400 to become ₹ 4851 at 10%, compounded half yearly is _______