Question

Ashok borrowed * 12,000 at some rate per
cent compound interest. After a year, he paid
back 4,000. If compound interest for the
second year be 920, find :
(i) the rate of interest charged
(ii) the amount of debt at the end of the
second year.​

Answer 1

Step-by-step explanation:

${\Large{\bf{\purple{Question :}}}}$

Ashok borrowed Rs. 12,000 at some rate per

cent compound interest. After a year, he paid

back Rs. 4,000. If compound interest for the

second year be Rs. 920, find :

(i) the rate of interest charged

(ii) the amount of debt at the end of the

second year.

${\Large{\bf{\purple{Solution :}}}}$

Given :

• Sum (P) = Rs. 12000
• Time (T) = 1 year

To find :

• The rate of interest charged
• The amount of debt at the end of the second year

Calculations :

1. Let's take x be the rate of interest charged

$✯ \: \: \: \: \: {\large{\bf{\underline{\red{For \: 1st \: year \: :}}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: \frac{12000 \: × \: x \: × \: 1}{100}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: 120x}}}$

$✯ \: \: \: \: \: {\large{\bf{\underline{\red{For \: 2nd \: year :}}}}}$

➠ After a year Ashok paid back Rs. 4000

Therefore,

➠ P = Rs. 12000 + Rs. 120x - Rs. 4000

➠ P = Rs. 8000 + 120x

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: \frac{(8000 \: + \: 120x) \: × \: x \: × \: 1}{100}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: 80x \: + \: 1.20x²}}}$

➠ The compound interest for the 2nd year is Rs. 920

$✯ \: \: \: \: \: \: {\large{\bf{\underline{\red{It \: means \: :}}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Rs. \: 80x \: + \: 1.20x² \: = \: Rs. \: 920}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: 80x \: + \: 1.20x² \: - \: 920 \: = \: 0}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: 3x² \: + \: 200x \: - \: 2300 \: = \: 0}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: 3x² \: + \: 230x \: - \: 30x \: - \: 2300 \: = \: 0}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: x(3x \: + \: 230) \: - \: 10(3x \: + \: 230) \: = \: 0}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: (3x \: + \: 230) (x \: - \: 10) \: = \: 0}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: x \: = \: 10}}}$

$\: \: \: \: \: \: \: {\large{\boxed{\mathrm{\pink{Rate \: of \: interest \: = 10\%}}}}}$

2. The amount of debt at the end of 2nd year :

$✯ \: \: \: \: \: {\large{\bf{\underline{\red{For \: 1st \: year \: :}}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: Rs. \: 120x \: = \: Rs. \: 1200}}}$

$✯ \: \: \: \: \: {\large{\bf{\underline{\red{For \: 2nd \: year \: :}}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Interest \: (I) \: = \: Rs. (80x \: + \: 1.20x²) \: = \: Rs. 920}}}$

The amount of debt at the end of 2nd year is equal to the addition of the sum of the 2nd year and interest for 2 years

${\bold{\sf{: \: ⟹ \: \: \: \: \: \: Debt \: = \: Rs. \: 8000 \: + \: Rs. \: 1200 \: + \: Rs. \: 920}}}$

$: \: ⟹ \: \: \: \: \: \: {\large{\mathrm{\fbox{\pink{ Debt = \: Rs. 10120}}}}}$

Answer 2

Answer:

Principal (i) = 12000 Rs

After 1 year he paid back Rs 4000

Let the rate of interest be r Hence,

the principle after 1 year

= 12000(1+r/100)-4000

Now it is given that

C.I for 2nd year = 920

Hence,

{12000(1+r/100)-4000}(1+r/100)

= {12000(1+r/100)-4000 +920} => (1+r/100) = {12000(1+r/100)-4000 +920}/{12000(1+r/100)-4000}

=> (1+r/100) = {8920+120r}/{8000+120r}

=> r/100 = {8920+120r}/{8000+120r} - 1

=> r/100 = 920/8000+120r

=> 120r^2+8000r-92000 = 0 => 3r^2+200r-2300=0 => 3r^2 -30r+230r-2300=0

=> 3r(r-10) + 230(r-10)=0

=> (3r+230)(r-10)=0

Since rate cannot be negative Hence r = 10

Therefore rate of interest =10%

2. Now initial principle = 12000 C.I for 1st year

= 12000(1+10/100)-12000 = 13200-12000 = 1200

C.I. for 2nd year = 920

Hence sum of principle and interest of both years

=12000+1200+920

=14120

But he paid back 4000 Hence amount of debt at the end of 2nd year = 14120 - 4000 = 10120 Rs.