A man borrows a loan of Rs 20,000 from a merchant at the rate of interest 1% per month for two month

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)
Step 2: New Principal
Now let X be the first installment.
After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)2-X (1+R/100)] = X
P [(1+R/100)2] = X+X (1+R/100)
Divide both sides by (1+r/100)2
So we left with
P= X/(1+r/100)2 + X/(1+r/100)
Now putting the value of P = 20000 and r = 1
we get the value of X is 10150.2487 = 10150.25
which is option 8
that's it.

Let,the each installment sum is x
Then,from the question we can write that,
2000(1+1/100)^2=x(1+1/100)+x
or,20402=(1.01)*x+x
or,(2.01)*x=20402
or,x=10150.24876
So, option 8)10150.25 is the right answer

say equal installment is X
Amount after 1 year= 20000*101/100=20200
start of 2nd yr principal is (20200-X)
so
(20200-X)*101/100=X
X= 20200*101/201
X= 10150.24

Interest on 20000 for 1st month = 200
If he pays 10150.25 as 1st installment then for 2nd month the principle becomes (10000+49.75 = 10049.75)
Interest on 10049.75 for 2nd month = 100.4975
So amount for 2nd month = 10150.2475 (almost 10150.25)
So correct answer is 10150.25

let 20000 = x+y
1st month pay = x + (20000*0.01) = x + 200
2nd month pay = y * 1.01 = 1.01y
x + 200 = 1.01 y .................eq.1
x + y = 200 .................eq.2
solve the eq. x = 9950.25
his installment = 9950.25 + 200 = 10150.25

Installment will be x
One month interest and amount is 20200(20000+200000/100)
two equal interest
every month 20200-x+(20200-x) 1% = x
2020000-100x+20200-x=100x
201 x =2040200
x=10150.24875

using EMI formula his installment will be 10150.25
so option(8)

here p=20000
i=1%=.01
n=2 month
usin EMI Formula;
M = [Pi(1 + i)^n] / [(1 + i)^n -1] =[(20000*.01)(1+.01)^2]/[(1+.01^2)-1]=10150.249
m=10150.25

let each installment is x
x/(1+1/100) + x/(1+1/100)^2 = 20000
x*(100/101) + x*(10000/10201) = 20000
x=10150.25

let the man gives the installment of Rs x to the merchant then
x*(1.01)+x=20000*(1.01)*(1.01)
(2.01)*x=20402
x=10150.2488

Ans is Rs.10150.25

Suppose 1st installment is x then another installment would be (20200-x)1.01 with interest of one month. Now x= 20402-1.01x now splve it

Answer is option 8)

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)
Step 2: New Principal
Now let X be the first installment.
After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)2-X (1+R/100)] = X
P [(1+R/100)2] = X+X (1+R/100)
Divide both sides by (1+r/100)2
So we left with
P= X/(1+r/100)2 + X/(1+r/100)
Now putting the value of P = 20000 and r = 1
we get the value of X is 10150.2487 = 10150.25

formula for installments:

A=P{r(1 + r)^n}/{(1 + r)^n - 1}

p: 20000
r: 1%=.01
n=2


Ans is 10150.25

ans 10150.248

There is an formula for equal installment calculation:
loan amount=equal install((100/100+r)+(100/100+r)^2+(100/100+r)^3+..................+(100/100+r)^n)

loan amount=principal=20000;
equal installment=to be find=x
1,2,....,n=no. of period=2 months
r=rate of interest

20000=x*((100/100+1)+(100/100+1)^2);
20000=x*(20100/101*101);
so, x=10150.24876
which is equivalent to 10150.25

20000 = x/(1.01) + x/(1.01)^2

gives, x = 10150.25

its 10150.25 Ans.

Let P be theprincipal for the first month so amount at the end of first month is P(1+r/100)^n
Let x be the installment paid after one month so principal for 2 month is P(1+r/100)-x=P1
Amount for 2 month is P1(1+r/100)
(P(1+r/100)-x)(1+r/100)
Now as the installment are equal so
(P(1+r/100)-x)(1+r/100)=x
Putting P=20000 and r=1 we get
x=10150.25

acc. to question
(p(1+r/100)-x)(1+r/100)=x
p(1+r/100)^2 = x(1+1+r/100)=x(2+r/100)
hence
x=20000(1.01)(1.01)/(2.01)
x=10150.25

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)
Step 2: New Principal
Now let X be the first installment.
After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)2-X (1+R/100)] = X
P [(1+R/100)2] = X+X (1+R/100)
Divide both sides by (1+r/100)2
So we left with
P= X/(1+r/100)2 + X/(1+r/100)
Now putting the value of P = 20000 and r = 1
we get the value of X is 10150.2487 = 10150.25
Read more at http://www.m4maths.com/placement-puzzles.php?ISSOLVED=&page=2&LPP=10&SOURCE=&MYPUZZLE=&UID=&TOPIC=&SUB_TOPIC=#1IwxrSHdKdrvFiB3.99

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)
Step 2: New Principal
Now let X be the first installment.
After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)2-X (1+R/100)] = X
P [(1+R/100)2] = X+X (1+R/100)
Divide both sides by (1+r/100)2
So we left with
P= X/(1+r/100)2 + X/(1+r/100)
Now putting the value of P = 20000 and r = 1
we get the value of X is 10150.2487 = 10150.25

Assume P to be the principal and r the rate of interest.

Step 1: P(1+r/100)=P1 (Amount for one year)
Step 2: New Principal
Now let X be the first installment.
After giving the first installment, the principal value will change and the new principal will be = P1 – X
P2 = P1 –X (1+R/100)
Step 3: Amount and Interest for the second year
Now the interest charged will be charged on this amount.
Amount at the end of second year: [P(1+r/100)-X][1+r/100] Step 4: Since the installments are equal, this new amount has to equal X.
Hence,
[P(1+r/100)-X][1+r/100]=X
On solving, we have
P [(1+R/100)2-X (1+R/100)] = X
P [(1+R/100)2] = X+X (1+R/100)
Divide both sides by (1+r/100)2
So we left with
P= X/(1+r/100)2 + X/(1+r/100)
Now putting the value of P = 20000 and r = 1
we get the value of X is 10150.2487 = 10150.25

one installment is 10150.25

1% of 20000 is 200
in first month he submitted 10000 with intrest 200 rupee
in second month he paid intrest on 20200-10150=10050
1% of 10050 is =100.50
then equal installment is equal is 10150.25.
emi=p*{r(1+r)^n/(1+r)^n-1}

installment = (P*r*(1+r)^n)/((1+r)^n -1)
and given
P=? 20000
r=12%
n=2 month
on putting value
installment= ?10150.25