4GMAT Diagnostic Test Q8 - Problem Solving : Simple and Compound Interest
- 1. GMAT QUANTITATIVE REASONING SIMPLE & COMPOUND INTEREST PROBLEM SOLVING Diagnostic Test
- 2. Question Robin invested $1000 in a 12% simple interest savings deposit for 3 years. He also invested an equal amount in a 10% compound interest savings deposit for 3 years. At the end of 3 years, how much more interest did he get from the simple interest deposit? A. $31 B. $60 C. $39 D. $29 E. $390
- 3. Part 1 Simple & Compound Interest Formulae
- 5. Formulae recap Simple Interest Pnr 100 Simple Interest, SI =
- 6. Formulae recap Simple Interest Pnr 100 Simple Interest, SI = P – Principal; n – number of years; r – rate of interest % p.a.
- 7. Formulae recap Simple Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a.
- 8. Formulae recap Simple Interest Compound Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a.
- 9. Formulae recap Simple Interest Compound Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a. Amount A = P 1+ r 100 n
- 10. Formulae recap Simple Interest Compound Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a. Amount A = P 1+ r 100 n P – Principal; n – number of years; r – rate of interest % p.a.
- 11. Formulae recap Simple Interest Compound Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a. Amount A = P 1+ r 100 n P – Principal; n – number of years; r – rate of interest % p.a. Compound Interest CI = A - P
- 12. Formulae recap Simple Interest Compound Interest Pnr 100 Simple Interest, SI = Amount accrued A = P + SI P – Principal; n – number of years; r – rate of interest % p.a. Amount A = P 1+ r 100 n P – Principal; n – number of years; r – rate of interest % p.a. Compound Interest CI = A - P Note In simple interest, the formula given computes the simple interest. In compound interest, the formula given computes the amount accrued.
- 13. Part 2 Compute simple and compound interest
- 14. At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest
- 15. $1000 in a 12% simple interest savings deposit for 3 years At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest
- 16. $1000 in a 12% simple interest savings deposit for 3 years Pnr 100 Simple Interest, SI = At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest
- 17. $1000 in a 12% simple interest savings deposit for 3 years Pnr 100 Simple Interest, SI = At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest Simple Interest, SI =
- 18. $1000 in a 12% simple interest savings deposit for 3 years Pnr 100 Simple Interest, SI = At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest Simple Interest, SI = 1000×3×12 100
- 19. $1000 in a 12% simple interest savings deposit for 3 years Pnr 100 Simple Interest, SI = At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest Simple Interest, SI = 1000×3×12 100 = $360
- 20. $1000 in a 12% simple interest savings deposit for 3 years Pnr 100 Simple Interest, SI = $360 Simple Interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 1: Compute simple interest Simple Interest, SI = 1000×3×12 100 = $360
- 21. Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit?
- 22. $1000 in a 10% compound interest savings deposit for 3 years Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit?
- 23. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit?
- 24. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A =
- 25. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3
- 26. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3
- 27. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3 = 1000 110 100 3
- 28. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3 = 1000 110 100 3 = $1331
- 29. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3 = 1000 110 100 3 = $1331 Compound Interest CI = A - P
- 30. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3 = 1000 110 100 3 = $1331 Compound Interest CI = A - P = 1331 - 1000 = $331
- 31. $1000 in a 10% compound interest savings deposit for 3 years Amount A = P 1+ r 100 n $331 Compound Interest Step 2: Compute compound interest At the end of 3 years, how much more interest did he get from the simple interest deposit? Amount A = 1000 1+ 10 100 3 = 1000 100+10 100 3 = 1000 110 100 3 = $1331 Compound Interest CI = A - P = 1331 - 1000 = $331
- 32. At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 3: Compute the difference
- 33. At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 3: Compute the difference 01 Simple interest = $360
- 34. At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 3: Compute the difference 01 Simple interest = $360 02 Compound interest = $331
- 35. Difference = 360 – 331 = $29 At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 3: Compute the difference 01 Simple interest = $360 02 Compound interest = $331
- 36. Correct Answer choice D. Difference = 360 – 331 = $29 At the end of 3 years, how much more interest did he get from the simple interest deposit? Step 3: Compute the difference 01 Simple interest = $360 02 Compound interest = $331
- 37. Alternative Method to compute CI Do not use formula
- 38. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1
- 39. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 Principal for 1st year = $1000 Rate of interest is10%
- 40. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100
- 41. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100
- 42. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1
- 43. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1
- 44. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10%
- 45. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110
- 46. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210
- 47. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2
- 48. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2
- 49. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2 Principal for 3nd year = $1210 Rate of interest is10%
- 50. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2 Principal for 3nd year = $1210 Rate of interest is10% Interest for year 3 = 10% of 1210 = $121
- 51. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2 Principal for 3nd year = $1210 Rate of interest is10% Interest for year 3 = 10% of 1210 = $121 Amount at the end of year 2 = 1210 + 121 = $1331
- 52. Computing iteratively $1000 @ 10% p.a. compound interest for 3 years I: Amount end of year 1 II: Amount end of year 2 III: Amount end of year 3 Principal for 1st year = $1000 Rate of interest is10% Interest for year 1 = 10% of 1000 = $100 Amount at the end of year 1 = 1000 + 100 = $1100 Principal for year 2 = Amount end of year 1 Principal for 2nd year = $1100 Rate of interest is10% Interest for year 2 = 10% of 1100 = $110 Amount at the end of year 2 = 1100 + 110 = $1210 Principal for year 3 = Amount end of year 2 Principal for 3nd year = $1210 Rate of interest is10% Interest for year 3 = 10% of 1210 = $121 Amount at the end of year 2 = 1210 + 121 = $1331 CI = 1331 – 1000 = $331
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