Answer: Option D. -> y2 = zx
Let the time is 't' years and the rate of interest is R%
According to the question,
$$\eqalign{
& {\bf{Case(I):}} \cr
& y = \frac{{x \times {\text{R}} \times {\text{t}}}}{{100}}.....(i) \cr
& {\bf{Case(II):}} \cr
& z = \frac{{y \times {\text{R}} \times {\text{t}}}}{{100}}.....(ii) \cr} $$
By dividing equation (i) by equation (ii)
$$\eqalign{
& \frac{y}{z} = \frac{{x \times {\text{R}} \times {\text{t}}}}{{y \times {\text{R}} \times {\text{t}}}} \cr
& \Rightarrow {y^2} = zx \cr} $$

Answer: Option C. -> Rs. 8000
Let the money borrowed by Nitin = Rs. P
According to the question,
$$\eqalign{
& \frac{{{\text{P}} \times 6 \times 3}}{{100}} + \frac{{{\text{P}} \times 9 \times 5}}{{100}} + \frac{{{\text{P}} \times 13 \times 3}}{{100}} = {\text{Rs}}{\text{. }}8160 \cr
& \Rightarrow \frac{{{\text{18P}}}}{{100}} + \frac{{{\text{45P}}}}{{100}} + \frac{{{\text{39P}}}}{{100}} = {\text{Rs}}{\text{. 1860}} \cr
& \Rightarrow \frac{{{\text{102P}}}}{{100}} = {\text{Rs}}{\text{. 8160}} \cr
& \Rightarrow {\text{P = Rs}}{\text{.}}\frac{{8160 \times 100}}{{102}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,{\text{ = Rs}}{\text{. 8000}} \cr} $$
Alternate
Note : In such type of questions to save your valuable time follow the given below method.
Let principal = Rs. 100
Total Interest
$$\frac{{100 \times 6 \times 3}}{{100}} + \frac{{100 \times 9 \times 5}}{{100}} + $$      $$\frac{{100 \times 13 \times 3}}{{100}}$$
$$\eqalign{
& = 18 + 45 + 39 \cr
& = 102{\text{ units}} \cr
& {\text{According to the question,}} \cr
& {\text{102 units = Rs}}{\text{. 8160}} \cr
& {\text{1 unit = Rs}}{\text{. }}\frac{{8160}}{{102}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = Rs}}{\text{. 80}} \cr
& {\text{100 units = Rs}}{\text{. 8000}} \cr
& {\text{Hence sum = Rs}}{\text{. 8000}} \cr} $$
Alternate
Total rate of interest in 11 years
$$\eqalign{
& \left( {6 \times 3} \right)\% + \left( {5 \times 9} \right)\% + \left( {3 \times 13} \right)\% {\text{ }} \cr
& \Rightarrow 102\% = 8160 \cr
& \Rightarrow 100\% = 8000 \cr
& \therefore {\text{Sum = Rs}}{\text{. 8000}} \cr} $$

Answer: Option D. -> $$7\frac{1}{2}$$ years
$$\eqalign{
& {\text{Let}}\,{\text{sum}} = x. \cr
& {\text{Then,}} \cr
& {\text{S}}{\text{.I}}{\text{.}} = \frac{{\text{9}}}{{{\text{16}}}}x \cr
& {\text{Let rate}} = {\text{R}}\% \,\text{and} \cr
& \text{time} = R\,\text{years} \cr
& \therefore \left( {\frac{{x \times {\text{R}} \times {\text{R}}}}{{100}}} \right) = \frac{{9x}}{{16}} \cr
& \Rightarrow {{\text{R}}^2} = \frac{{900}}{{16}} \cr
& {\text{R}} = \frac{{30}}{4} = 7\frac{1}{2} \cr
& {\text{Hence,}} \cr
& {\text{time}} = 7\frac{1}{2}\text{years} \cr} $$

Answer: Option A. -> Rs. 992
$$\eqalign{
& {\text{Increased interest in 3 years}} \cr
& {\text{ = 3}} \times {\text{3}} \cr
& {\text{ = 9% }} \cr
& {\text{Hence increased amount}} \cr
& {\text{ = }}\frac{{800 \times 9}}{{100}} \cr
& = {\text{Rs}}{\text{. }}72 \cr
& {\text{Total amount}} \cr
& {\text{ = }}\left( {920 + 72} \right) \cr
& = {\text{Rs}}{\text{. }}992 \cr} $$

Answer: Option B. -> Rs. 1500
Let the sum lent to C be Rs. x.
Then,
$$\eqalign{
& \left( {\frac{{2500 \times 7 \times 4}}{{100}}} \right) + \left( {\frac{{x \times 7 \times 4}}{{100}}} \right) = 1120 \cr
& \Rightarrow \frac{{7x}}{{25}} = \left( {1120 - 700} \right) \cr
& \Rightarrow x = \left( {\frac{{420 \times 25}}{7}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 1500. \cr} $$

Answer: Option D. -> Rs. 1, 70, 000
$$\eqalign{
& {\text{Let the sum be Rs}}{\text{. }}x{\text{ and}} \cr
& {\text{original rate be R}}\% \cr
& {\text{Then,}} \cr
& \Rightarrow \frac{{x \times \left( {{\text{R}} + 1} \right) \times 3}}{{100}} - \frac{{x \times {\text{R}} \times 3}}{{100}} = 5100 \cr
& \Rightarrow 3{\text{R}}x + 3x - 3{\text{R}}x = 510000 \cr
& \Rightarrow 3x = 510000 \cr
& \Rightarrow x = 170000. \cr
& {\text{Hence,}} \cr
& {\text{Sum}} = {\text{Rs}}.170000 \cr} $$

Answer: Option A. -> Rs. 200
Let the annual installment be Rs. x.
Then,
$$ \Rightarrow \left[ {x + \left( {\frac{{x \times 3 \times 4}}{{100}}} \right)} \right] + $$     $$\left[ {x + \left( {\frac{{x \times 2 \times 4}}{{100}}} \right)} \right] + $$     $$\left[ {x + \left( {\frac{{x \times 1 \times 4}}{{100}}} \right)} \right] + $$     $$x = 848$$
$$\eqalign{
& \Leftrightarrow \frac{{28x}}{{25}} + \frac{{27x}}{{25}} + \frac{{26x}}{{25}} + x = 848 \cr
& \Leftrightarrow 106x = 848 \times 25 \cr
& \Leftrightarrow 106x = 21200 \cr
& \Leftrightarrow x = 200 \cr} $$
Short Cut Method : The annual payment that will discharge a debt of Rs. A due in t years at the rate of interest r % p.a. is.
$$\eqalign{
& \frac{{100{\text{A}}}}{{100t + \frac{{rt\left( {t - 1} \right)}}{2}}} \cr
& \therefore {\text{Annual installment}} \cr
& = {\text{Rs}}{\text{.}}\left[ {\frac{{100 \times 848}}{{100 \times 4 + \frac{{4 \times 4 \times 3}}{2}}}} \right] \cr
& = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 848}}{{424}}} \right) \cr
& = {\text{Rs}}{\text{. }}200 \cr} $$

Answer: Option C. -> Rs. 17800
Total price of TV = Rs. 16000
Initial payment = Rs. 4000
Remaining amount = Rs. 12000
Simple interest in 15 months for Rs. 12000
$$\eqalign{
& \Rightarrow {\text{S}}{\text{.I}}{\text{. = }}\frac{{{\text{P}} \times {\text{R}} \times {\text{T}}}}{{100}} \cr
& \Rightarrow {\text{S}}{\text{.I}}{\text{. = }}\frac{{12000 \times 12 \times 15}}{{100 \times 12}} \cr
& \Rightarrow {\text{S}}{\text{.I}}{\text{. = Rs}}{\text{. 1800}} \cr} $$
⇒ With S.I. total amount to be paid for principal amount Rs. 12000
= Rs. (12000 + 1800)
= Rs. 13800
= Therefore, total amount he pays for the TV is
= 4000 + 13800
= Rs. 17800

Answer: Option A. -> 6%
$$\eqalign{
& \frac{{\text{P}}}{{{\text{S}}{\text{.I}}{\text{.}}}} = \frac{{10}}{3} \cr
& {\text{Let Principal = 10}} \cr
& {\text{S}}{\text{.I}}{\text{. for 5 years = 3}} \cr
& {\text{S}}{\text{.I}}{\text{. for 1 year = 0}}{\text{.6}} \cr
& {\text{Rate = }}\frac{{{\text{S}}{\text{.I}}{\text{.}}}}{{{\text{Principal}}}} \times 100 \cr
& {\text{Rate = }}\frac{{0.6}}{{10}} \times 100 \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = 6\% \cr} $$

Answer: Option D. -> Rs. Nil
10% of Rs. 3 Lacs = 30000
6% of Rs. 3 Lacs = 18000
1 month interest income = 2000
∴ 1 year interest income = 2000 × 12 = 24000
Profit of Bank = 24000 - 18000 = 6000
Profit of Post Office = 30000 - 24000 = 6000
∴ Ratio of profit = 6000 : 6000 = 1 : 1
So, amount deposited = Rs. 150000 each
And difference = 0