Interest - Study Mode
[#111] If x, y, z are three sums of money such that y is the simple interest on x, z is the simple interest on y for the same time and at the same rate of interest, then we have.
Correct Answer
(B) y 2 = xz
Explanation
Solution: Let time be T years and rate be R% p.a. $$eqalign{
& { ext{Then, }}y{ ext{ is the S}}{ ext{.I}}{ ext{. on x}} cr
& Rightarrow frac{{x{ ext{RT}}}}{{100}} = y......(i) cr
& { ext{And, }}z{ ext{ is the S}}{ ext{.I}}{ ext{. on y}} cr
& Rightarrow frac{{y{ ext{RT}}}}{{100}} = z cr
& Rightarrow y = frac{{100z}}{{RT}}......(ii) cr
& { ext{From (i) and (ii) we have:}} cr
& frac{{x{ ext{RT}}}}{{100}} = frac{{100z}}{{{ ext{RT}}}} cr
& Rightarrow frac{{x{{ ext{R}}^2}{{ ext{T}}^2}}}{{{{left( {100}
ight)}^2}}} = z cr
& Rightarrow frac{{{y^2}}}{x} = z cr
& Rightarrow {y^2}= xz cr }$$
[#112] Arun borrowed a sum of money from Jayant at the rate of 8% per annum simple interest for the first four years, 10% per annum for the next 6 years and 12% per annum for the period beyond 10 years. If he pays a total of Rs. 12160 as interest only at the end of 15 years, how much money did he borrow?
Correct Answer
(A) Rs. 8000
Explanation
Solution: $$eqalign{
& { ext{Let the sum be Rs}}{ ext{. }}x cr
& { ext{Then,}} cr} $$ $$ {frac{{x imes 8 imes 4}}{{100}}} + {frac{{x imes 10 imes 6}}{{100}}} ,+ $$ xa0 xa0 $$ {frac{{x imes 12 imes 5}}{{100}}} $$ xa0 $$ = 12160$$ $$eqalign{
& Rightarrow 32x + 60x + 60x = 1216000 cr
& Rightarrow 152x = 1216000 cr
& Rightarrow x = 8000 cr} $$
[#113] Kruti took a loan at simple interest rate of 6 p.c.p.a. in the first year and it increased by 1.5 p.c.p.a. every year. If she pays Rs. 8190 as interest at the end of 3 years, what was her loan amount ?
Correct Answer
Rs. 35400
Explanation
Solution: Let the loan amount be Rs. x $$eqalign{
& { ext{Then,}} cr
& Rightarrow frac{{6x}}{{100}} + frac{{7.5x}}{{100}} + frac{{9x}}{{100}} = 8190 cr
& Rightarrow 22.5x = 819000 cr
& Rightarrow x = 36400 cr} $$
[#114] A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a different rate becomes three times in 12 years. The lower rate of interest per annum is = ?
Correct Answer
(D) $$16frac{2}{3}$$ %
Explanation
Solution: Rate of interest $$ = frac{{100(x - 1)}}{t}\% $$ Where x is the no. of times the sum becomes of itself. Here the sum is getting 3 times. Therefore, x = 3 Where t is the time taken by sum to become x times of itself. Here, t = 12 years By the short trick approach, we get $$eqalign{
& = frac{{100(3 - 1)}}{{12}} cr
& = frac{{200}}{{12}} cr
& = frac{{50}}{3} = 16frac{2}{3}\% cr} $$ Alternative Method : Let the principal be P and in the 2nd scenario, SI = 2P $$eqalign{
& { ext{Rate}} = frac{{{ ext{SI}} imes 100}}{{{ ext{Principal}} imes { ext{Time}}}} cr
& = frac{{2{ ext{P}} imes 100}}{{{ ext{P}} imes 12}} cr
& = frac{{50}}{3} cr
& = 16frac{2}{3}\% cr} $$
[#115] If Rs. 12000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for $$4frac{1}{2}$$ years at 16% per annum, the greater part is = ?
Correct Answer
(A) Rs. 8000
Explanation
Solution: $$eqalign{
& { ext{Let the first part = Rs}}{ ext{. }}x cr
& herefore { ext{Hence second part}} cr
& { ext{ = Rs}}{ ext{. }}left( {12000 - x}
ight) cr
& { ext{According to the question,}} cr
& Rightarrow frac{{x imes 12 imes 3}}{{100}} = frac{{left( {12000 - x}
ight) imes 9 imes 16}}{{2 imes 100}} cr
& Rightarrow 36x = 72left( {12000 - x}
ight) cr
& Rightarrow x = 24000 - 2x cr
& Rightarrow 3x = 24000 cr
& Rightarrow x = { ext{Rs}}{ ext{. 8000}} cr
& {{ ext{1}}^{{ ext{st}}}}{ ext{ part = Rs}}{ ext{. 8000}} cr
& {{ ext{2}}^{{ ext{nd}}}}{ ext{ part}} cr
& { ext{ = Rs}}{ ext{. }}left( {12000 - 8000}
ight) cr
& = { ext{Rs}}{ ext{. 4000 }} cr
& { ext{Hence maximum part}} cr
& { ext{ = Rs}}{ ext{. 8000}} cr} $$ Alternate Note : In such type of questions to save your valuable time follow the given below method. Let two parts P 1 and P 2 respectively According to the question, $$eqalign{
& Rightarrow {{ ext{P}}_1} imes frac{{36}}{{100}} imes 1 = {{ ext{P}}_2} imes frac{9}{2} imes frac{{16}}{{100}} imes 1 cr
& Rightarrow {{ ext{P}}_1} imes 4 = 8{{ ext{P}}_2} cr
& Rightarrow frac{{{{ ext{P}}_1}}}{{{{ ext{P}}_2}}} = frac{2}{1} cr
& Rightarrow {{ ext{P}}_1}{ ext{:}}{{ ext{P}}_2} = 2:1 cr
& { ext{Hence greater part}} cr
& { ext{ = }}frac{{12000}}{{left( {2 + 1}
ight)}} imes 2 cr
& = { ext{Rs}}{ ext{. }}8000 cr} $$