Ramesh repays a loan of Rs.6500 by paying Rs.40 in first month and then increasing the payment by Rs.30 every month. In how many months will he be able to clear the loan?

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11 Jun 2013 Problem Ramesh repays a loan of Rs.6500 by paying Rs.40 in first month and then increasing the payment by Rs.30 every month. In how many months will he be able to clear the loan?: OPtion
1) 20
2) 30
3) 40
4) 50
5) 55
6) 60
7) 65
8) 70
9) 75
10) None of these; Correct Option: 1 Best Solution (22)

IFFAT IRAM 11 years AGO

let a be the first instalment, d be the increase in monthly payment, n be the no. of months.
the given problem is in A.P.
a=40 , d=30, Sum=6500
Sum=n[2a+(n-1)d]/2
6500=n[80+30(n-1)]/2
13000=n[30n+50]
1300=n(3n+5)
3n^2 + 5n - 1300 = 0
3n^2 + 65n - 60n - 1300 = 0
(3n+65)(n-20)=0
no. of months = 20

Like? Yes (5) | No (1)

Udhaya kumar 11 years AGO

40+70+...+610
20 times

Like? Yes (1) | No

serinth 11 years AGO

n/2(2*40 + (n-1)30)=6500
=> -3n^2+5n-1300 =0
=> n=20
months take to clear the loan = 20

Like? Yes (1) | No

HARMEET KAUR 11 years AGO

If x months are reqd , then
(x/2)*(2*40 + (x-1)*30)= 6500
solving it, we get
x=20

Like? Yes (2) | No (1)

Raman 11 years AGO

put n=20 in formula of sum of AP series
sum = (20/2)*[2*40 +19*30) = 10*(80+570)=10*650=6500 which is equal to given sum.

Like? Yes (2) | No (1)

Faiyyaz Khan 11 years AGO

compound entrest

Like? Yes (1) | No

Shivakrishna 11 years AGO

TOTAL SUM CAN BE CALCULATED BY TAKING SERIES IN A.P
6500=40+(40+30)+(40+30+30)+...........
Sn=n/2(2a+(n-1)d)
6500=n/2(2*40+(n-1)30) by solving the eqn we get n=20&n=-65/3...so valid n value will be n=20

Like? Yes (1) | No (1)

Balakrishnan M 11 years AGO

40+70+100+130+....(up to n months)=6500
Sn=(n/2)(2a+(n-1)d)
6500=(n/2)(2*40+(n-1)30)
13000=80n+30n^2-30n
3n^2+5n-1300=0
Solve this quadratic Equation,
n=20.
So 20 months will he be pay able to clear the loan.

Like? Yes (1) | No (1)

shailesh kumar mishra 11 years AGO

it is a series with a=40,d=30 and n to get so equation has to be solved and we get n=20

Like? Yes (1) | No (1)

Gotlooru Venkata Kartheek Kumar 11 years AGO

These terms all are in ap
40,70,100,130........
So
Sum =(n/2)(2a+(n-1)d)
So,
6500=(n/2)(2(40)+(n-1)30)
13000=n(80+30n-30)
13000=n(50+30n)
1300=n(5+3n)
3n*n+5n-1300=0

By substituting 20,we get answer

Like? Yes (1) | No (1)

C.Manohar 11 years AGO

Sn=n/2[2a+(n-1)d]
40+70+100+130+......A.P
a=40
d=30
find n=?

Like? Yes (1) | No (1)

Vinay Sipani 11 years AGO

40 + 70 + 100 + ... = 6500
=> (n/2)(80 + (n-1)30)= 6500
=> n (15n + 25) = 6500
=> n = 20

Like? Yes (1) | No (1)

Satyaranjan 11 years AGO

Here we get a AP series like 40+70+100+.... n terms=6500
So Sn=6500, here a=40 and d=30, so by solving with Sn=(n/2)[2a+(n-1)d]=6500 we get n=20

Like? Yes (1) | No (1)

lakshmi chaitanya 11 years AGO

by calculation we get 6500 fo 20 months

Like? Yes (2) | No (2)

Sneha Agarwal 11 years AGO

applying A.P.
3n^2 + 5n - 1300 = 0
solving we get n= 20, n= -65/3
n=20

Like? Yes (1) | No (1)

sanjay singh 11 years AGO

arrangingthe problem as n/2(2a + (n-1)d)=6500
a=40,d=30
we get n=20

Like? Yes (1) | No (1)

ROHITCHAURASIYA 11 years AGO

increse evry month 30 and got it

Like? Yes (1) | No (1)

Devendra Marghade 11 years AGO

If n=number of month to clear the loan, then
6500=(n/2)[2*40+(n-1)30]
13000=n(50+30n) or 3n^2+5n-1300=0
3n^2-60n+65n-1300=0
3n(n-20)+65(n-20)=0
So n=20

Like? Yes (1) | No (2)

Ankit Pandey 11 years AGO