EMI, CAGR and Depreciation for CUET Applied Maths

AnswerC · ₹6,000

1. Flat-rate interest is charged on the whole loan for the whole term: $1{,}20{,}000 \times 0.10 \times 2 = 24{,}000$.

2. EMI $= \frac{1{,}20{,}000 + 24{,}000}{24} = 6{,}000$ rupees.

AnswerB · ₹18,000

1. Flat-rate interest $= P \times r \times t$, on the full amount every year.

2. $50{,}000 \times 0.12 \times 3 = 18{,}000$ rupees.

AnswerC · ₹46,000

1. Interest $= 40{,}000 \times 0.075 \times 2 = 6{,}000$.

2. Repaid in all: $40{,}000 + 6{,}000 = 46{,}000$ rupees.

AnswerC · ₹3,100

1. Interest $= 90{,}000 \times 0.08 \times 3 = 21{,}600$ rupees.

2. EMI $= \frac{90{,}000 + 21{,}600}{36} = \frac{1{,}11{,}600}{36} = 3{,}100$ rupees.

AnswerA · ₹1,08,000

1. At a flat rate, interest is on the whole loan for every year: $P \times r \times t$.

2. $2{,}40{,}000 \times 0.09 \times 5 = 1{,}08{,}000$ rupees.

AnswerB · ₹1,20,000

1. Flat interest is charged on the full sum for the whole term.

2. $5{,}00{,}000 \times 0.08 \times 3 = 1{,}20{,}000$ rupees.

AnswerC · ₹17,222.22

1. She repays $5{,}00{,}000 + 1{,}20{,}000 = 6{,}20{,}000$ in all.

2. $\frac{6{,}20{,}000}{36} \approx 17{,}222.22$ rupees.

AnswerB · ₹2,160

1. The loan is $30{,}000 - 6{,}000 = 24{,}000$, and its interest for a year is $24{,}000 \times 0.08 = 1{,}920$.

2. EMI $= \frac{24{,}000 + 1{,}920}{12} = 2{,}160$ rupees.

AnswerA · Only I

1. I: that is what 'flat' means; repayments do not reduce the sum interest is charged on. True.

2. II: the reducing-balance method charges interest only on what is still owed, so at the same rate it costs less, not more. False.

AnswerB · 10%

1. Repaid in all: $24 \times 3{,}000 = 72{,}000$, so the interest is 12,000 over 2 years.

2. Rate $= \frac{12{,}000}{60{,}000 \times 2} = 10\%$ a year.

AnswerA · ₹2,300

1. The loan is $45{,}000 - 9{,}000 = 36{,}000$; interest for 1.5 years is $36{,}000 \times 0.1 \times 1.5 = 5{,}400$.

2. Instalment $= \frac{41{,}400}{18} = 2{,}300$ rupees.

AnswerD · The flat rate

1. At a flat rate, interest is charged on the whole loan for the whole term, even after much of it is repaid.

2. By reducing balance, interest is charged only on what is still owed, so it is less.

AnswerA · ₹16,606

1. $EMI = \frac{P \cdot i (1 + i)^{n}}{(1 + i)^{n} - 1} = \frac{5{,}00{,}000 \times 0.01 \times 1.4308}{0.4308}$.

2. $\frac{7{,}154}{0.4308} \approx 16{,}606$ rupees.

AnswerD · The bank's loan, by about ₹22,184

1. Bank: $36 \times 16{,}606 - 5{,}00{,}000 = 97{,}816$ of interest. Dealer: ₹1,20,000.

2. The bank is cheaper by $1{,}20{,}000 - 97{,}816 = 22{,}184$ rupees: a flat rate charges interest on money already repaid.

AnswerB · ₹2,000

1. In the first month the whole loan is still owed.

2. Interest $= 2{,}00{,}000 \times 0.01 = 2{,}000$ rupees.

AnswerC · ₹32,000

1. Total paid: $24 \times 8{,}000 = 1{,}92{,}000$.

2. Interest $= 1{,}92{,}000 - 1{,}60{,}000 = 32{,}000$ rupees.

AnswerD · $E = \frac{P\,i\,(1 + i)^{n}}{(1 + i)^{n} - 1}$

1. The instalments are an annuity whose present value is the loan: $E \times \frac{1 - (1 + i)^{-n}}{i} = P$.

2. So $E = \frac{P\,i}{1 - (1 + i)^{-n}} = \frac{P\,i\,(1 + i)^{n}}{(1 + i)^{n} - 1}$.

AnswerB · ₹7,886

1. The first month's interest is 1% of the whole loan: $1{,}00{,}000 \times 0.01 = 1{,}000$.

2. The rest of the EMI, $8{,}886 - 1{,}000 = 7{,}886$ rupees, repays the loan.

AnswerA · Only I

1. I: interest is charged only on what is still owed, which shrinks, so the interest part shrinks and the fixed EMI leaves more for the principal. True.

2. II: the EMI is fixed but its split changes, so the principal repaid grows each month. False.

AnswerD · ₹46,166

1. The first EMI pays 750 of interest and $4{,}584 - 750 = 3{,}834$ of the loan.

2. Still owed: $50{,}000 - 3{,}834 = 46{,}166$ rupees.

AnswerD · $i = 0.0075,\ n = 36$

1. Both must be per instalment period, a month: $i = \frac{0.09}{12} = 0.0075$.

2. There are $3 \times 12 = 36$ monthly instalments: $n = 36$.

AnswerA · Both A and R are true, and R is the correct explanation of A.

1. R describes the reducing-balance method. True.

2. A smaller balance means less interest each month, and since the EMI is fixed, more of it repays the loan: A is true because of R.

AnswerC · ₹3,834

1. Interest for the first month: $50{,}000 \times 0.015 = 750$.

2. The rest repays the loan: $4{,}584 - 750 = 3{,}834$ rupees.

AnswerC · ₹8,886

1. Monthly rate $i = 0.01$, $n = 12$: $E = \frac{1{,}00{,}000 \times 0.01 \times 1.1268}{0.1268}$.

2. $E = \frac{1{,}126.8}{0.1268} \approx 8{,}886$ rupees.

AnswerB · ₹28,810

1. $E = \frac{50{,}000 \times 0.1 \times 1.21}{1.21 - 1} = \frac{6{,}050}{0.21}$.

2. $E \approx 28{,}810$ rupees.

AnswerA · ₹10,356

1. $i = 0.01$, $n = 6$: $E = \frac{60{,}000 \times 0.01 \times 1.0615}{0.0615}$.

2. $E = \frac{636.9}{0.0615} \approx 10{,}356$ rupees.

AnswerB · ₹40,211

1. $E = \frac{P\,i\,(1 + i)^{n}}{(1 + i)^{n} - 1} = \frac{1{,}00{,}000 \times 0.1 \times 1.331}{0.331}$.

2. $E = \frac{13{,}310}{0.331} \approx 40{,}211$ rupees.

AnswerD · 20%

1. The return is the gain in price plus the income: $(230 - 200) + 10 = 40$ rupees.

2. Rate of return $= \frac{40}{200} \times 100 = 20\%$.

AnswerB · 20%

1. $\frac{E}{B} = 1.44$, and $\sqrt{1.44} = 1.2$.

2. CAGR $= 1.2 - 1 = 20\%$.

AnswerC · ₹88,200

1. At a constant 5% a year: $80{,}000 \times (1.05)^{2}$.

2. $= 80{,}000 \times 1.1025 = 88{,}200$ rupees.

AnswerC · 7%

1. The real return is roughly the nominal return minus inflation.

2. $12\% - 5\% = 7\%$.

AnswerD · 15%

1. Gain $= (440 - 400) + 20 = 60$ rupees.

2. Rate of return $= \frac{60}{400} = 15\%$.

AnswerD · 20%

1. Return $= \frac{60 - 50}{50} \times 100$.

2. 20%.

AnswerB · −10%

1. Return $= \frac{54 - 60}{60} \times 100$.

2. $-10\%$: the NAV fell.

AnswerD · ₹14,400

1. The NAV grew from 50 to 72, a factor of $\frac{72}{50} = 1.44$.

2. $10{,}000 \times 1.44 = 14{,}400$ rupees.

AnswerC · ₹300

1. ₹6 on each of 50 shares.

2. $50 \times 6 = 300$ rupees.

AnswerC · ₹600

1. Each share gains $252 - 240 = 12$ rupees.

2. $50 \times 12 = 600$ rupees.

AnswerA · 2.5%

1. Dividend yield is the dividend on a share as a percentage of its price.

2. $\frac{6}{240} \times 100 = 2.5\%$.

AnswerB · 6%

1. The nominal rate counts rupees; the real rate counts what those rupees can buy.

2. Real rate ≈ nominal rate − inflation $= 10\% - 4\% = 6\%$.

AnswerA · 10%

1. CAGR is the ratio of the end value to the start value raised to the power $\frac{1}{n}$, minus 1: $(1.4641)^{1/4} - 1$.

2. $1.1^{4} = 1.4641$, so CAGR $= 1.1 - 1 = 10\%$.

AnswerD · $\left(\frac{E}{B}\right)^{1/n} - 1$

1. CAGR is the steady yearly rate $r$ with $B(1 + r)^{n} = E$.

2. So $1 + r = \left(\frac{E}{B}\right)^{1/n}$ and $r = \left(\frac{E}{B}\right)^{1/n} - 1$.

AnswerA · 26%

1. CAGR $= 2^{1/3} - 1$.

2. $= 1.26 - 1 = 0.26$, about 26% a year.

AnswerB · −3%

1. Gain $= (470 - 500) + 15 = -15$ rupees: a loss.

2. Rate of return $= \frac{-15}{500} = -3\%$.

AnswerA · 14.47%

1. $\frac{E}{B} = \frac{75{,}000}{50{,}000} = 1.5$.

2. CAGR $= (1.5)^{1/3} - 1 = 0.1447$, about 14.47% a year.

AnswerA · 10%

1. $\frac{E}{B} = \frac{29{,}282}{20{,}000} = 1.4641 = (1.1)^{4}$.

2. CAGR $= 1.1 - 1 = 10\%$.

AnswerD · 7.5%

1. The investor paid $50 \times 240 = 12{,}000$ and gained $300 + 600 = 900$.

2. $\frac{900}{12{,}000} \times 100 = 7.5\%$.

AnswerA · 12.92%

1. $CAGR = \left(\frac{72}{50}\right)^{1/3} - 1 = (1.44)^{1/3} - 1$.

2. $1.1292 - 1 = 0.1292$, about 12.92% a year.

AnswerA · Only I

1. I: $\frac{20\% + (-20\%)}{2} = 0\%$. True.

2. II: sales end at $1.2 \times 0.8 = 0.96$ of the start, a fall, so the CAGR is negative, about $-2\%$. False.

AnswerD · Average 15%, CAGR about 9.54%

1. Average growth $= \frac{50\% + (-20\%)}{2} = 15\%$.

2. CAGR $= \left(\frac{120}{100}\right)^{1/2} - 1 = 1.0954 - 1 \approx 9.54\%$: the steady rate that actually gets from 100 to 120.

AnswerB · ₹18,000

1. Linear depreciation is (cost − residual value) ÷ useful life.

2. $\frac{2{,}00{,}000 - 20{,}000}{10} = 18{,}000$ rupees a year.

AnswerA · Only I

1. I: that is why the method is called linear, or straight-line: book value against time is a straight line. True.

2. II: a fixed amount is a growing share of a shrinking book value, so the percentage rises. A fixed percentage is a different method. False.

AnswerA · ₹45,000

1. Five years of depreciation: $5 \times 15{,}000 = 75{,}000$.

2. Book value $= 1{,}20{,}000 - 75{,}000 = 45{,}000$ rupees.

AnswerC · ₹88,000

1. The linear method takes the same amount off every year.

2. $4 \times 22{,}000 = 88{,}000$ rupees.

AnswerA · ₹50,000

1. Spread the loss of value, cost minus residual value, evenly over the useful life.

2. $\frac{3{,}50{,}000 - 50{,}000}{6} = \frac{3{,}00{,}000}{6} = 50{,}000$ rupees a year.

AnswerA · ₹70,000

1. Spread the loss in value, cost minus scrap value, evenly over the 8 years.

2. $\frac{6{,}00{,}000 - 40{,}000}{8} = 70{,}000$ rupees.

AnswerC · ₹3,90,000

1. Three years of depreciation: $3 \times 70{,}000 = 2{,}10{,}000$.

2. $6{,}00{,}000 - 2{,}10{,}000 = 3{,}90{,}000$ rupees.

AnswerC · ₹1,28,000

1. Each year the book value falls by the same ₹18,000: in 4 years, $4 \times 18{,}000 = 72{,}000$.

2. Book value $= 2{,}00{,}000 - 72{,}000 = 1{,}28{,}000$ rupees.

AnswerA · 9%

1. Rate of depreciation = yearly depreciation ÷ cost, as a percentage.

2. $\frac{18{,}000}{2{,}00{,}000} \times 100 = 9\%$.

AnswerB · 10 years

1. Useful life = (cost − residual value) ÷ yearly depreciation.

2. $\frac{50{,}000 - 5{,}000}{4{,}500} = \frac{45{,}000}{4{,}500} = 10$ years.