Practice Quantitative Aptitude questions on Installments, covering equal installments, present value, compound interest, simple interest, repayment schedules, and calculating installment amounts for loans and purchases. Takes about 20–25 minutes and is useful for placement tests, TCS NQT, SSC, banking and MBA entrance exams.
Question 1
A shopkeeper borrowed Rs. 12 lakh at a simple interest rate of 5% per annum for 4 years. If he repays the whole amount in equal monthly instalments, find the monthly instalment.
Explanation
Total interest = 5% × 4 = 20% of the principal.
Total amount = 120% of Rs. 1,200,000 = Rs. 1,440,000.
Number of months = 4 × 12 = 48.
Monthly instalment = Rs. 1,440,00048 = Rs. 30,000.
Total amount = 120% of Rs. 1,200,000 = Rs. 1,440,000.
Number of months = 4 × 12 = 48.
Monthly instalment = Rs. 1,440,00048 = Rs. 30,000.
Shortcut Method
Interest = 5 × 4 = 20%
Amount = 120% of Rs. 1,200,000 = Rs. 1,440,000
∴ Rs. 1,440,000 ÷ 48 = Rs. 30,000 ✓
Amount = 120% of Rs. 1,200,000 = Rs. 1,440,000
∴ Rs. 1,440,000 ÷ 48 = Rs. 30,000 ✓
Question 2
A shopkeeper borrowed Rs. 9 lakh at a simple interest rate of 8% per annum for 3 years. If he repays the whole amount in equal monthly instalments, find the monthly instalment.
Explanation
Total interest = 8% × 3 = 24% of the principal.
Total amount = 124% of Rs. 900,000 = Rs. 1,116,000.
Number of months = 3 × 12 = 36.
Monthly instalment = Rs. 1,116,00036 = Rs. 31,000.
Total amount = 124% of Rs. 900,000 = Rs. 1,116,000.
Number of months = 3 × 12 = 36.
Monthly instalment = Rs. 1,116,00036 = Rs. 31,000.
Shortcut Method
Interest = 8 × 3 = 24%
Amount = 124% of Rs. 900,000 = Rs. 1,116,000
∴ Rs. 1,116,000 ÷ 36 = Rs. 31,000 ✓
Amount = 124% of Rs. 900,000 = Rs. 1,116,000
∴ Rs. 1,116,000 ÷ 36 = Rs. 31,000 ✓
Question 3
A sum of money is borrowed and repaid in two equal annual instalments of Rs. 7,200 each. If the rate of compound interest is 20% per annum, compounded annually, what was the sum borrowed?
Explanation
Present value of the 1st instalment = Rs. 7,200 ÷ (1 + 20100) = Rs. 7,200 × 100120 = Rs. 6,000.
Present value of the 2nd instalment = Rs. 7,200 ÷ 1 + 201002 = Rs. 5,000.
Sum borrowed = Rs. 6,000 + Rs. 5,000 = Rs. 11,000.
Present value of the 2nd instalment = Rs. 7,200 ÷ 1 + 201002 = Rs. 5,000.
Sum borrowed = Rs. 6,000 + Rs. 5,000 = Rs. 11,000.
Shortcut Method
1st: 7,200 × 100120 = 6,000
2nd: 6,000 × 100120 = 5,000
∴ Sum = Rs. 11,000 ✓
2nd: 6,000 × 100120 = 5,000
∴ Sum = Rs. 11,000 ✓
Question 4
Meera cleared her loan with two equal yearly payments of Rs. 11,025 each. If the rate of compound interest is 5% per annum, compounded annually, what was the sum borrowed?
Explanation
Present value of the 1st instalment = Rs. 11,025 ÷ (1 + 5100) = Rs. 11,025 × 100105 = Rs. 10,500.
Present value of the 2nd instalment = Rs. 11,025 ÷ 1 + 51002 = Rs. 10,000.
Sum borrowed = Rs. 10,500 + Rs. 10,000 = Rs. 20,500.
Present value of the 2nd instalment = Rs. 11,025 ÷ 1 + 51002 = Rs. 10,000.
Sum borrowed = Rs. 10,500 + Rs. 10,000 = Rs. 20,500.
Shortcut Method
1st: 11,025 × 100105 = 10,500
2nd: 10,500 × 100105 = 10,000
∴ Sum = Rs. 20,500 ✓
2nd: 10,500 × 100105 = 10,000
∴ Sum = Rs. 20,500 ✓
Question 5
A trader took a loan from a bank for stock purchase, to be repaid in two equal annual instalments of Rs. 2,420 each. If the rate of interest was 10% per annum, compounded yearly, the total interest paid was:
Explanation
Present value of the 1st instalment = Rs. 2,420 ÷ (1 + 10100) = Rs. 2,420 × 100110 = Rs. 2,200.
Present value of the 2nd instalment = Rs. 2,420 ÷ 1 + 101002 = Rs. 2,000.
Sum borrowed = Rs. 2,200 + Rs. 2,000 = Rs. 4,200.
Total paid = 2 × Rs. 2,420 = Rs. 4,840.
Total interest = Rs. 4,840 − Rs. 4,200 = Rs. 640.
Present value of the 2nd instalment = Rs. 2,420 ÷ 1 + 101002 = Rs. 2,000.
Sum borrowed = Rs. 2,200 + Rs. 2,000 = Rs. 4,200.
Total paid = 2 × Rs. 2,420 = Rs. 4,840.
Total interest = Rs. 4,840 − Rs. 4,200 = Rs. 640.
Shortcut Method
P = Rs. 2,200 + Rs. 2,000 = Rs. 4,200
paid = Rs. 4,840
∴ Interest = Rs. 640 ✓
paid = Rs. 4,840
∴ Interest = Rs. 640 ✓
Question 6
A company repays a bank loan through two equal yearly payments of Rs. 3,600 each. If the rate of interest was 20% per annum, compounded yearly, the total interest paid was:
Explanation
Present value of the 1st instalment = Rs. 3,600 ÷ (1 + 20100) = Rs. 3,600 × 100120 = Rs. 3,000.
Present value of the 2nd instalment = Rs. 3,600 ÷ 1 + 201002 = Rs. 2,500.
Sum borrowed = Rs. 3,000 + Rs. 2,500 = Rs. 5,500.
Total paid = 2 × Rs. 3,600 = Rs. 7,200.
Total interest = Rs. 7,200 − Rs. 5,500 = Rs. 1,700.
Present value of the 2nd instalment = Rs. 3,600 ÷ 1 + 201002 = Rs. 2,500.
Sum borrowed = Rs. 3,000 + Rs. 2,500 = Rs. 5,500.
Total paid = 2 × Rs. 3,600 = Rs. 7,200.
Total interest = Rs. 7,200 − Rs. 5,500 = Rs. 1,700.
Shortcut Method
P = Rs. 3,000 + Rs. 2,500 = Rs. 5,500
paid = Rs. 7,200
∴ Interest = Rs. 1,700 ✓
paid = Rs. 7,200
∴ Interest = Rs. 1,700 ✓
Question 7
What annual instalment will discharge a debt of Rs. 5,520 due in 4 years at 10% simple interest per annum?
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 10 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 10 × 6100 x = 460100 x.
460100 x = Rs. 5,520, so x = Rs. 5,520 × 100460 = Rs. 1,200.
Total of all instalments with interest = 4 x + x × 10 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 10 × 6100 x = 460100 x.
460100 x = Rs. 5,520, so x = Rs. 5,520 × 100460 = Rs. 1,200.
Shortcut Method
Units: instalment = 100
Total = 4×100 + 10 × 6 = 460 units
∴ Rs. 5,520 ÷ 460 × 100 = Rs. 1,200 ✓
Total = 4×100 + 10 × 6 = 460 units
∴ Rs. 5,520 ÷ 460 × 100 = Rs. 1,200 ✓
Question 8
A debt of Rs. 6,360 due in 3 years is to be cleared by equal annual instalments. If simple interest is charged at 6% per annum, find each instalment.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 6 × (2 + 1 + 0) ÷ 100 = 3 x + 6 × 3100 x = 318100 x.
318100 x = Rs. 6,360, so x = Rs. 6,360 × 100318 = Rs. 2,000.
Total of all instalments with interest = 3 x + x × 6 × (2 + 1 + 0) ÷ 100 = 3 x + 6 × 3100 x = 318100 x.
318100 x = Rs. 6,360, so x = Rs. 6,360 × 100318 = Rs. 2,000.
Shortcut Method
Units: instalment = 100
Total = 3×100 + 6 × 3 = 318 units
∴ Rs. 6,360 ÷ 318 × 100 = Rs. 2,000 ✓
Total = 3×100 + 6 × 3 = 318 units
∴ Rs. 6,360 ÷ 318 × 100 = Rs. 2,000 ✓
Question 9
A sum of Rs. 2,520 is payable after 3 years. If the borrower wants to pay it in 3 equal annual instalments and the rate of simple interest is 5% per annum, what is each instalment?
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 2,520, so x = Rs. 2,520 × 100315 = Rs. 800.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 2,520, so x = Rs. 2,520 × 100315 = Rs. 800.
Shortcut Method
Units: instalment = 100
Total = 3×100 + 5 × 3 = 315 units
∴ Rs. 2,520 ÷ 315 × 100 = Rs. 800 ✓
Total = 3×100 + 5 × 3 = 315 units
∴ Rs. 2,520 ÷ 315 × 100 = Rs. 800 ✓
Question 10
A dealer owes Rs. 11,200 due 4 years hence. He clears the debt in 4 equal yearly instalments, simple interest being 8% per annum. Find the instalment.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 8 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 8 × 6100 x = 448100 x.
448100 x = Rs. 11,200, so x = Rs. 11,200 × 100448 = Rs. 2,500.
Total of all instalments with interest = 4 x + x × 8 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 8 × 6100 x = 448100 x.
448100 x = Rs. 11,200, so x = Rs. 11,200 × 100448 = Rs. 2,500.
Shortcut Method
Units: instalment = 100
Total = 4×100 + 8 × 6 = 448 units
∴ Rs. 11,200 ÷ 448 × 100 = Rs. 2,500 ✓
Total = 4×100 + 8 × 6 = 448 units
∴ Rs. 11,200 ÷ 448 × 100 = Rs. 2,500 ✓
Question 11
Find the equal annual instalment that discharges a debt of Rs. 5,400 due in 5 years at 4% simple interest per annum.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 5 x + x × 4 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 4 × 10100 x = 540100 x.
540100 x = Rs. 5,400, so x = Rs. 5,400 × 100540 = Rs. 1,000.
Total of all instalments with interest = 5 x + x × 4 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 4 × 10100 x = 540100 x.
540100 x = Rs. 5,400, so x = Rs. 5,400 × 100540 = Rs. 1,000.
Shortcut Method
Units: instalment = 100
Total = 5×100 + 4 × 10 = 540 units
∴ Rs. 5,400 ÷ 540 × 100 = Rs. 1,000 ✓
Total = 5×100 + 4 × 10 = 540 units
∴ Rs. 5,400 ÷ 540 × 100 = Rs. 1,000 ✓
Question 12
What yearly instalment will clear a debt of Rs. 6,450 due in 4 years at 5% simple interest per annum?
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 5 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 5 × 6100 x = 430100 x.
430100 x = Rs. 6,450, so x = Rs. 6,450 × 100430 = Rs. 1,500.
Total of all instalments with interest = 4 x + x × 5 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 5 × 6100 x = 430100 x.
430100 x = Rs. 6,450, so x = Rs. 6,450 × 100430 = Rs. 1,500.
Shortcut Method
Units: instalment = 100
Total = 4×100 + 5 × 6 = 430 units
∴ Rs. 6,450 ÷ 430 × 100 = Rs. 1,500 ✓
Total = 4×100 + 5 × 6 = 430 units
∴ Rs. 6,450 ÷ 430 × 100 = Rs. 1,500 ✓
Question 13
A farmer must clear a debt of Rs. 9,900 due in 3 years by three equal annual instalments. If the rate of simple interest is 10% per annum, each instalment is:
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 10 × (2 + 1 + 0) ÷ 100 = 3 x + 10 × 3100 x = 330100 x.
330100 x = Rs. 9,900, so x = Rs. 9,900 × 100330 = Rs. 3,000.
Total of all instalments with interest = 3 x + x × 10 × (2 + 1 + 0) ÷ 100 = 3 x + 10 × 3100 x = 330100 x.
330100 x = Rs. 9,900, so x = Rs. 9,900 × 100330 = Rs. 3,000.
Shortcut Method
Units: instalment = 100
Total = 3×100 + 10 × 3 = 330 units
∴ Rs. 9,900 ÷ 330 × 100 = Rs. 3,000 ✓
Total = 3×100 + 10 × 3 = 330 units
∴ Rs. 9,900 ÷ 330 × 100 = Rs. 3,000 ✓
Question 14
Annual instalments clear a debt of Rs. 10,080 due in 5 years at 6% simple interest per annum. Find each instalment.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 5 x + x × 6 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 6 × 10100 x = 560100 x.
560100 x = Rs. 10,080, so x = Rs. 10,080 × 100560 = Rs. 1,800.
Total of all instalments with interest = 5 x + x × 6 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 6 × 10100 x = 560100 x.
560100 x = Rs. 10,080, so x = Rs. 10,080 × 100560 = Rs. 1,800.
Shortcut Method
Units: instalment = 100
Total = 5×100 + 6 × 10 = 560 units
∴ Rs. 10,080 ÷ 560 × 100 = Rs. 1,800 ✓
Total = 5×100 + 6 × 10 = 560 units
∴ Rs. 10,080 ÷ 560 × 100 = Rs. 1,800 ✓
Question 15
What equal annual instalment will settle a debt of Rs. 7,488 due in 3 years, if simple interest is 4% per annum?
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 4 × (2 + 1 + 0) ÷ 100 = 3 x + 4 × 3100 x = 312100 x.
312100 x = Rs. 7,488, so x = Rs. 7,488 × 100312 = Rs. 2,400.
Total of all instalments with interest = 3 x + x × 4 × (2 + 1 + 0) ÷ 100 = 3 x + 4 × 3100 x = 312100 x.
312100 x = Rs. 7,488, so x = Rs. 7,488 × 100312 = Rs. 2,400.
Shortcut Method
Units: instalment = 100
Total = 3×100 + 4 × 3 = 312 units
∴ Rs. 7,488 ÷ 312 × 100 = Rs. 2,400 ✓
Total = 3×100 + 4 × 3 = 312 units
∴ Rs. 7,488 ÷ 312 × 100 = Rs. 2,400 ✓
Question 16
A merchant owes Rs. 6,976 after 4 years. He plans to pay it in 4 equal annual instalments at 6% simple interest per annum. What is each instalment?
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 6 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 6 × 6100 x = 436100 x.
436100 x = Rs. 6,976, so x = Rs. 6,976 × 100436 = Rs. 1,600.
Total of all instalments with interest = 4 x + x × 6 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 6 × 6100 x = 436100 x.
436100 x = Rs. 6,976, so x = Rs. 6,976 × 100436 = Rs. 1,600.
Shortcut Method
Units: instalment = 100
Total = 4×100 + 6 × 6 = 436 units
∴ Rs. 6,976 ÷ 436 × 100 = Rs. 1,600 ✓
Total = 4×100 + 6 × 6 = 436 units
∴ Rs. 6,976 ÷ 436 × 100 = Rs. 1,600 ✓
Question 17
Find the annual instalment needed to clear a debt of Rs. 5,220 due in 5 years at 8% simple interest per annum.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 5 x + x × 8 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 8 × 10100 x = 580100 x.
580100 x = Rs. 5,220, so x = Rs. 5,220 × 100580 = Rs. 900.
Total of all instalments with interest = 5 x + x × 8 × (4 + 3 + 2 + 1 + 0) ÷ 100 = 5 x + 8 × 10100 x = 580100 x.
580100 x = Rs. 5,220, so x = Rs. 5,220 × 100580 = Rs. 900.
Shortcut Method
Units: instalment = 100
Total = 5×100 + 8 × 10 = 580 units
∴ Rs. 5,220 ÷ 580 × 100 = Rs. 900 ✓
Total = 5×100 + 8 × 10 = 580 units
∴ Rs. 5,220 ÷ 580 × 100 = Rs. 900 ✓
Question 18
A debt of Rs. 6,930 due in 3 years is to be cleared in equal yearly instalments at 5% simple interest per annum. Find the instalment.
Explanation
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 6,930, so x = Rs. 6,930 × 100315 = Rs. 2,200.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 6,930, so x = Rs. 6,930 × 100315 = Rs. 2,200.
Shortcut Method
Units: instalment = 100
Total = 3×100 + 5 × 3 = 315 units
∴ Rs. 6,930 ÷ 315 × 100 = Rs. 2,200 ✓
Total = 3×100 + 5 × 3 = 315 units
∴ Rs. 6,930 ÷ 315 × 100 = Rs. 2,200 ✓
Question 19
A sum Rs. P was borrowed and paid back in two equal yearly instalments, each of Rs. 2,704. If interest was 4% per annum, compounded annually, then the value of P is:
Explanation
Present value of the 1st instalment = Rs. 2,704 ÷ (1 + 4100) = Rs. 2,704 × 100104 = Rs. 2,600.
Present value of the 2nd instalment = Rs. 2,704 ÷ 1 + 41002 = Rs. 2,500.
P = Rs. 2,600 + Rs. 2,500 = Rs. 5,100.
Present value of the 2nd instalment = Rs. 2,704 ÷ 1 + 41002 = Rs. 2,500.
P = Rs. 2,600 + Rs. 2,500 = Rs. 5,100.
Shortcut Method
PV: Rs. 2,600 + Rs. 2,500
∴ P = Rs. 5,100 ✓
∴ P = Rs. 5,100 ✓
Question 20
Rs. P was taken as a loan and cleared by two identical year-end payments of Rs. 3,600. If interest was 25% per annum, compounded annually, then the value of P is:
Explanation
Present value of the 1st instalment = Rs. 3,600 ÷ (1 + 25100) = Rs. 3,600 × 100125 = Rs. 2,880.
Present value of the 2nd instalment = Rs. 3,600 ÷ 1 + 251002 = Rs. 2,304.
P = Rs. 2,880 + Rs. 2,304 = Rs. 5,184.
Present value of the 2nd instalment = Rs. 3,600 ÷ 1 + 251002 = Rs. 2,304.
P = Rs. 2,880 + Rs. 2,304 = Rs. 5,184.
Shortcut Method
PV: Rs. 2,880 + Rs. 2,304
∴ P = Rs. 5,184 ✓
∴ P = Rs. 5,184 ✓
Question 21
A person borrowed Rs. 6,300 at 5% annual simple interest, repayable in 3 equal annual instalments. What will be the annual instalment?
Explanation
Amount due = P + simple interest = Rs. 6,300 + Rs. 945 = Rs. 7,245.
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 7,245, so x = Rs. 7,245 × 100315 = Rs. 2,300.
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 3 x + x × 5 × (2 + 1 + 0) ÷ 100 = 3 x + 5 × 3100 x = 315100 x.
315100 x = Rs. 7,245, so x = Rs. 7,245 × 100315 = Rs. 2,300.
Shortcut Method
Amount = Rs. 7,245
Units: 3 × 100 + 5 × 3 = 315
∴ Rs. 7,245 ÷ 315 × 100 = Rs. 2,300 ✓
Units: 3 × 100 + 5 × 3 = 315
∴ Rs. 7,245 ÷ 315 × 100 = Rs. 2,300 ✓
Question 22
A farmer borrowed Rs. 4,300 at 5% annual simple interest, repayable in 4 equal annual instalments. What will be the annual instalment?
Explanation
Amount due = P + simple interest = Rs. 4,300 + Rs. 860 = Rs. 5,160.
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 5 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 5 × 6100 x = 430100 x.
430100 x = Rs. 5,160, so x = Rs. 5,160 × 100430 = Rs. 1,200.
Let the annual instalment be Rs. x, paid at the end of each year. The instalment paid at the end of year k earns simple interest for the remaining years.
Total of all instalments with interest = 4 x + x × 5 × (3 + 2 + 1 + 0) ÷ 100 = 4 x + 5 × 6100 x = 430100 x.
430100 x = Rs. 5,160, so x = Rs. 5,160 × 100430 = Rs. 1,200.
Shortcut Method
Amount = Rs. 5,160
Units: 4 × 100 + 5 × 6 = 430
∴ Rs. 5,160 ÷ 430 × 100 = Rs. 1,200 ✓
Units: 4 × 100 + 5 × 6 = 430
∴ Rs. 5,160 ÷ 430 × 100 = Rs. 1,200 ✓
Question 23
A refrigerator is sold for Rs. 46,000 cash, or for Rs. 21,000 cash down followed by 6 monthly instalments of Rs. 4,400 each (the first paid one month after purchase). What is the rate of interest per annum under the instalment plan?
Explanation
Balance after the down payment = Rs. 46,000 − Rs. 21,000 = Rs. 25,000.
Let the rate be R% per annum. Amount of the balance for 6 months = Rs. 25,000 + Rs. 25,000 × 612 × R100 = Rs. 25,000 + Rs. 150,0001200 R.
Instalments: the one paid at the end of month j earns interest for (6 − j) months. The interest months total 5 + 4 + 3 + 2 + 1 + 0 = 15.
Total = 6 × Rs. 4,400 + Rs. 4,400 × R × 151200 = Rs. 26,400 + Rs. 66,0001200 R.
Equating: Rs. 25,000 + Rs. 150,0001200 R = Rs. 26,400 + Rs. 66,0001200 R, which gives R = 20%.
Let the rate be R% per annum. Amount of the balance for 6 months = Rs. 25,000 + Rs. 25,000 × 612 × R100 = Rs. 25,000 + Rs. 150,0001200 R.
Instalments: the one paid at the end of month j earns interest for (6 − j) months. The interest months total 5 + 4 + 3 + 2 + 1 + 0 = 15.
Total = 6 × Rs. 4,400 + Rs. 4,400 × R × 151200 = Rs. 26,400 + Rs. 66,0001200 R.
Equating: Rs. 25,000 + Rs. 150,0001200 R = Rs. 26,400 + Rs. 66,0001200 R, which gives R = 20%.
Shortcut Method
Balance = Rs. 25,000
R = 1200(6×Rs. 4,400 − Rs. 25,000)Rs. 150,000 − Rs. 4,400×15
∴ R = 20% ✓
R = 1200(6×Rs. 4,400 − Rs. 25,000)Rs. 150,000 − Rs. 4,400×15
∴ R = 20% ✓
Question 24
A refrigerator is sold for Rs. 39,000 cash, or for Rs. 21,000 cash down followed by 6 monthly instalments of Rs. 3,200 each (the first paid one month after purchase). What is the rate of interest per annum under the instalment plan?
Explanation
Balance after the down payment = Rs. 39,000 − Rs. 21,000 = Rs. 18,000.
Let the rate be R% per annum. Amount of the balance for 6 months = Rs. 18,000 + Rs. 18,000 × 612 × R100 = Rs. 18,000 + Rs. 108,0001200 R.
Instalments: the one paid at the end of month j earns interest for (6 − j) months. The interest months total 5 + 4 + 3 + 2 + 1 + 0 = 15.
Total = 6 × Rs. 3,200 + Rs. 3,200 × R × 151200 = Rs. 19,200 + Rs. 48,0001200 R.
Equating: Rs. 18,000 + Rs. 108,0001200 R = Rs. 19,200 + Rs. 48,0001200 R, which gives R = 24%.
Let the rate be R% per annum. Amount of the balance for 6 months = Rs. 18,000 + Rs. 18,000 × 612 × R100 = Rs. 18,000 + Rs. 108,0001200 R.
Instalments: the one paid at the end of month j earns interest for (6 − j) months. The interest months total 5 + 4 + 3 + 2 + 1 + 0 = 15.
Total = 6 × Rs. 3,200 + Rs. 3,200 × R × 151200 = Rs. 19,200 + Rs. 48,0001200 R.
Equating: Rs. 18,000 + Rs. 108,0001200 R = Rs. 19,200 + Rs. 48,0001200 R, which gives R = 24%.
Shortcut Method
Balance = Rs. 18,000
R = 1200(6×Rs. 3,200 − Rs. 18,000)Rs. 108,000 − Rs. 3,200×15
∴ R = 24% ✓
R = 1200(6×Rs. 3,200 − Rs. 18,000)Rs. 108,000 − Rs. 3,200×15
∴ R = 24% ✓
Question 25
A scooter priced at Rs. 480,000 is bought by paying some amount as down payment. On the balance, simple interest of 10% is charged as a lump sum, and the total is paid in 10 equal annual instalments of Rs. 22,000 each. How much is the down payment?
Explanation
Total paid in instalments = 10 × Rs. 22,000 = Rs. 220,000.
This is the balance plus 10% interest, so balance × 110100 = Rs. 220,000.
Balance = Rs. 220,000 × 100110 = Rs. 200,000.
Down payment = Rs. 480,000 − Rs. 200,000 = Rs. 280,000.
This is the balance plus 10% interest, so balance × 110100 = Rs. 220,000.
Balance = Rs. 220,000 × 100110 = Rs. 200,000.
Down payment = Rs. 480,000 − Rs. 200,000 = Rs. 280,000.
Shortcut Method
Total = 10 × Rs. 22,000 = Rs. 220,000
Balance = Rs. 220,000 ÷ 1.10 = Rs. 200,000
∴ Down = Rs. 480,000 − Rs. 200,000 = Rs. 280,000 ✓
Balance = Rs. 220,000 ÷ 1.10 = Rs. 200,000
∴ Down = Rs. 480,000 − Rs. 200,000 = Rs. 280,000 ✓
