Lessons 9 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 10, Parts 1 and 2, 1st edition. EXTREMUM PRESS, Tashkent, 2017
9
Problem solving: percentages and interest
Textbook, Part 1: pp. 53–57
GoalSolve percentage problems: equal monthly payments, number of periods in compound interest, finding the initial amount, and depreciation value.
New words
monthly payment · oylik to‘lovdepreciation · amortizatsiyadepreciation rate · amortizatsiya normasinumber of periods · davrlar soni
Explanation
If a loan is repaid in equal monthly payments, first find the interest I, then divide the total C + I by the number of months: monthly payment = (C + I) / (12n). In compound interest, shorter periods raise the number of periods kn while the rate per period drops to r/k; for example 2.5 years with monthly compounding gives kn = 12 · 2.5 = 30. If the final balance A is known, the initial amount is C = A / (1 + r/100)ⁿ. Property such as equipment, furniture or computers wears out and loses value: with depreciation rate r, the value after n years is A = C · (1 − r/100)ⁿ, i.e. each year the value drops by r% of its current value. So the yearly loss gets smaller each year, and the total loss over n years is less than n times the first year’s loss. Banks also use tables for quick calculation: the monthly payment for each 1000 units of loan is read from the table and scaled to the loan.
Worked examples
12 000 000 so‘m for 2 years at 8% simple interest per year, repaid in equal monthly payments: I = 12 000 000 · 8 · 2 / 100 = 1 920 000; total 13 920 000; 24 months, monthly payment 13 920 000 : 24 = 580 000 so‘m.
A piece of equipment worth 8 000 000 so‘m has a yearly depreciation rate of 10%. After 2 years its value is A = 8 000 000 · 0.9² = 6 480 000 so‘m; after 3 years 6 480 000 · 0.9 = 5 832 000 so‘m. The losses were 800 000, 720 000, 648 000 — the loss shrinks each year.
Class activity
Groups get the task “Loan or savings?”: with an imagined furniture price, rate and term, each group computes two ways (a loan with monthly payments vs saving in advance) and presents which is cheaper.
Practice
1
Write the formula for the value after n years of depreciation.
A = C · (1 − r/100)ⁿ, where r is the yearly depreciation rate (%).
2
6 000 000 so‘m is borrowed for 2 years at 10% simple interest and repaid in 24 equal monthly payments. How many so‘m is each payment?
300000
3
A tool worth 2 000 000 so‘m has a yearly depreciation rate of 20%. What is its value in so‘m after 2 years?
1280000
4
Why is the 2-year loss not simply twice the first year’s 20% loss? Explain with 2 000 000 so‘m.
Year 1 loses 400 000 so‘m, leaving 1 600 000. Year 2 loses 20% of 1 600 000 = 320 000. Total 720 000, not double (800 000), because the second year’s percent is taken from the reduced value.