Question

A man arranges to pay a debt of 90,000 in 18 monthly instalments of 5000 cash
and the interest @ 5% per annum in the 19th instalment. Calculate the simple interest
he pays in the 19th instalment.
​

Answer 1

Given :  A man arranges to pay a debt of 90,000 in 18 monthly installments of 5000 cash  and the interest @ 5% per annum in the 19th installment.  

To Find : Calculate the simple interest  he pays in the 19th installment.

Solution:

Interest for 90000 for 1 month

interest for  85000 for 1 month

and so on

interest for 5000  for  1  month

Interest = (90000 * 5  * 1/12)/100 + (85000 * 5  * 1/12)/100 +.................+ (5000 * 5  * 1/12)/100

= (5000*5 / 1200) ( 18 + 17 + ...............................+ 1)

= (125/6) ( 18 ) (19)/ 2

= 375 * 19/2

= 3562.5  

Rs 3562.5   interest has to be paid in 19th installment

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Answer 2

Answer:

$\huge \boxed{x = 51}$

$= 2 \times 1200$

$= 2400$

(3600)

(3600)= 2(1200)

(3600)= 2(1200)= 2400

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n =

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get180−39d−160+29d .....(2)

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get180−39d−160+29d .....(2)0=20−10d

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get180−39d−160+29d .....(2)0=20−10dFurther solving for d

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get180−39d−160+29d .....(2)0=20−10dFurther solving for d10d = 20

(3600)= 2(1200)= 2400Let us take the first installment as a and common difference as d.So, using the formula for the sum of n terms of an A.p,s n = 2n subtracting (1) from (2), we get180−39d−160+29d .....(2)0=20−10dFurther solving for d10d = 20d= 20/10

= 2/160−58

160−58

160−58

160−58 = 2/(102)

= Rs 51