Banking (Recurring Deposit Accounts) — Important Questions

Answer: (a) $\text{₹}12000$.

$2$ years $=24$ months, so total deposit $=500\times24=\text{₹}12000.$

Answer: (a) $78$.

$\dfrac{n(n+1)}{2}=\dfrac{12\times13}{2}=\dfrac{156}{2}=78.$

Answer: (a) $\text{₹}208$.

$I=400\times\dfrac{12\times13}{2\times12}\times\dfrac{8}{100}=400\times6.5\times0.08=\text{₹}208.$

Answer: (a) $\text{₹}12650$.

Total deposit $=1000\times12=\text{₹}12000$; $I=1000\times\dfrac{12\times13}{24}\times\dfrac{10}{100}=1000\times6.5\times0.10=\text{₹}650.$

$\text{MV}=12000+650=\text{₹}12650.$

Answer: (c) A is true — $\text{MV}=Pn+I$. R is false: each instalment earns interest for a different number of months, so interest uses the equivalent-months factor $\dfrac{n(n+1)}{2}$, not the whole sum for the whole period.

$I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=600\times\dfrac{20\times21}{24}\times\dfrac{9}{100}.$

$=600\times17.5\times0.09=\text{₹}945.$

Total deposit $=250\times24=\text{₹}6000.$

$I=250\times\dfrac{24\times25}{24}\times\dfrac{8}{100}=250\times25\times0.08=\text{₹}500.$

$\text{MV}=6000+500=\text{₹}6500.$

Here $P=800$, $n=36$, $r=7$.

(i) $I=800\times\dfrac{36\times37}{2\times12}\times\dfrac{7}{100}=800\times55.5\times0.07=\text{₹}3108.$

(ii) Total deposit $=800\times36=\text{₹}28800.$

$\text{MV}=28800+3108=\text{₹}31908.$

Here $P=700$, $n=24$.

$I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=700\times\dfrac{24\times25}{24}\times\dfrac{r}{100}=700\times25\times\dfrac{r}{100}=175r.$

So $175r=1050\Rightarrow r=\dfrac{1050}{175}=6.$

Rate $=6\%$ per annum.

$I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=150\times\dfrac{n(n+1)}{24}\times\dfrac{8}{100}.$

$=\dfrac{150\times8}{24\times100}\,n(n+1)=0.5\,n(n+1).$

Set $0.5\,n(n+1)=300\Rightarrow n(n+1)=600.$ Since $24\times25=600$, $n=24$ months.

Total deposit $=150\times24=\text{₹}3600$, so $\text{MV}=3600+300=\text{₹}3900.$

Here $P=2500$, $n=48$, $r=6$.

(i) Total deposit $=2500\times48=\text{₹}120000.$

(ii) $I=2500\times\dfrac{48\times49}{2\times12}\times\dfrac{6}{100}=2500\times98\times0.06=\text{₹}14700.$

(iii) $\text{MV}=120000+14700=\text{₹}134700.$

Let the monthly deposit be $\text{₹}P$, with $n=24$, $r=10$.

Interest $I=P\times\dfrac{24\times25}{2\times12}\times\dfrac{10}{100}=P\times25\times0.10=2.5P.$

Total deposit $=24P$, so $\text{MV}=24P+2.5P=26.5P.$

$26.5P=13250\Rightarrow P=\dfrac{13250}{26.5}=500.$

Monthly deposit $=\text{₹}500.$

Here $P=1200$, $n=30$, $r=8$.

(i) Total deposit $=1200\times30=\text{₹}36000.$

(ii) $\dfrac{n(n+1)}{2}=\dfrac{30\times31}{2}=465$ months.

(iii) $I=1200\times\dfrac{465}{12}\times\dfrac{8}{100}=1200\times38.75\times0.08=\text{₹}3720.$

(iv) $\text{MV}=36000+3720=\text{₹}39720.$