A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs. 72 more. Then sum is:

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30 Jul 2014 Simple Interest A sum was put at simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched Rs. 72 more. Then sum is:: OPtion
1) Rs. 2000
2) Rs. 1600
3) Rs. 1500
4) Rs. 2200
5) Rs. 1200
6) Rs. 1800
7) Rs. 1000
8) Rs. 1400
9) Rs. 2600
10)None of these; Correct Option: 5 Best Solution (144)

Sundar Kumar 10 years AGO

At 3% more rate, the increase in SI for 2 years = Rs.72 (given)
So, at 3% more rate, the increase in SI for 1 year = 72/2= Rs.36/-
i.e. Rs.36 is 3% of the invested sum
So, 1% of the invested sum = 36/3
Therefore, the invested sum = 36 × 100/3 = Rs.1200.
Answer: Option 5) 1200

Like? Yes (56) | No (8)

Sandip Dutta 10 years AGO

let sum is x.
now (x*2*3)/100=72
so x=1200

Like? Yes (38) | No (10)

GOKULARAMANAN 10 years AGO

s.i=pnr/100;
n=2;
r=3;
s.i=72;
so,p=1200;

Like? Yes (17) | No (9)

natuva aditya 10 years AGO

SI=PTR/100------(1)
SI=72, T=2 YEARS, R=3%, P=?
substituting the terms in equation(1),
we get the sum P is Rs.1200

Like? Yes (10) | No (2)

Swarnabha Dey 10 years AGO

Ans = 5) 1200
let the sum of money = p
rate of interest = r%
we know interest (i)= (p*r*t )/100
know,
{p(r+3)*2}/100 -2pr/100 = 72
=> p(r+3)/100 - pr/100 = 72
=> (pr+3p-pr)/100 = 36
=> 3p= 3600
=> p =1200
so the principle is Rs. 1200 :)

Like? Yes (11) | No (4)

Siddhant Modi 10 years AGO

actual interest=(x*y*2)/100
new interest=x*(y+3)*2/100
Acc. to problem
new interest -actual interest=72
on solving
6x=7200
=> x=1200

Like? Yes (6) | No (1)

Avinash Kumar 10 years AGO

let sum be y rs. rate be r %
now. for normal condition.;
I= (prt)/100 = (pr*2)/100...........1
if interest rate is increased 3%;
I'=(p(r+3)*2)/100...........2 {this is 72 Rs. more than I]
now equating equation 1 and 2;
I+72= I'
=> (pr*2)/100 +72 = (p(r+3)*2)/100
=> 7200 = 6p
=> p = 1200 Rs :)

Like? Yes (5) | No (1)

Mahesh Mulani 10 years AGO

Let x be the actual rate
Than i2-i1 = 2/100 (p (x+3-x))
72= 2/100 (3p)
P= 1200

Like? Yes (3) | No

anushree dey 10 years AGO

sum=rs.x;
rate = r% then,
(x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

Like? Yes (3) | No (1)

Hanish 10 years AGO

P*(R+3)*2/100-P*R*2/100=72
=> 2PR+6P-2PR=7200
=> P=1200
ThereFore ans is opt 5)

Like? Yes (2) | No

Abhishek 10 years AGO

Let rate be x
P*x*2/100 +72=P*(x+3)*2/100
2Px/100+72=(2Px+6P)/100
P=1200

Like? Yes (2) | No

Nidhi Ruthia 10 years AGO

by using si formula, si=(p*r*t)/100
i calculate p as sum, because other values are already given

Like? Yes (2) | No

shilpareddy 10 years AGO

p*2*r/100 +72=p*2*(r+3)/100
by solving abov equ we get p as 1200 which is the sum

Like? Yes (2) | No

Udhaya kumar 10 years AGO

At 3% more rate, the increase in SI for 2 years = Rs.72 (given)
So, at 5% more rate, the increase in SI for 1 year = 72/2 = Rs.36/-
i.e. Rs.36 is 3% of the invested sum
So, 1% of the invested sum = 36/3
Therefore, the invested sum = 36 × 100/3 = Rs.1200.

Like? Yes (4) | No (2)

rangababu gudimetla 10 years AGO

x(3%)=36 for 1 year..by this x=1200

Like? Yes (3) | No (1)

kedarisetti aparna 10 years AGO

s.i=ptr/100
so,s.i=(p*2*r)/100
but
s.i+72=(p*2*(r+3))/100
on equating and substituting
p=1200

Like? Yes (2) | No

sompon temtheerakij 10 years AGO

1 year interest = 72/2 = Rs. 36
the sum = 36*100/3 = Rs. 1200

Like? Yes (2) | No (1)

Vivek Jain 10 years AGO

=>I=S*R*2/100 ...(i)
=>I+72=S*(R+2)*2/100 ...(ii)
(i)/(ii), we get
=>I/(I+72) = R/(R+3)
=>I/R=24 ...(iii)
from (i), we have
=>S=50*I/R
=>S=50*24 from (iii)
Answer 5) Sum =Rs. 1200

Like? Yes (3) | No (2)

sai 10 years AGO

I=ptr/100
I=pr*2/100
pr=50I
now rate increased by 3% so r =r+3
interest increased by 72 so,
2*p*(r+3)/100=72+I
I=1200

Like? Yes (2) | No (1)

Hanuman Prasad 10 years AGO

p*(x)*(r+3) p*t*r
--------- - -------- =72
100 100

2pr-2pr+6p=7200
p=1200

Like? Yes (1) | No

Ragaven 10 years AGO

N-numbeq of years
P-intrest
P+(R+3)X2/100=PxRx2/100=72
By solving this we get rs1200

Like? Yes (1) | No

dushyant singh pawar 10 years AGO

Let amount be A , sum be P , rate be R and Interest be S.

now according to the conditions given in the question

A=P+((P*R*2)/100)

A+72=P+((P*(R+3)*2)/100)

or

P+(P*R/50) +72=P+(P(R+3))/50

or

P*R+3600=P*R+3P

P=1200 [OPTION 5]

Like? Yes (1) | No

SAYAN CHATTERJEE 10 years AGO

let, p=principal
r=rate of interest
t=time
p*2/100(r+3-r)=72
so, p=1200

Like? Yes (1) | No

Praful chikhle 10 years AGO

There is a direct formula,
Sum= (More Interest * 100)/(Time More Rate)
Therefore
Sum= (72*100)/(2*3)= Rs 1200

Option (5)

Like? Yes (1) | No

sangeetha 10 years AGO

((p*(y+3)*2)/100)-((p*y*2)/100)=72
(p*2*3)/100=72
by solving p=1200

Like? Yes (1) | No

DIVANSHU 10 years AGO

S.I=P*R*T/100
72=P*3*2/100
P=1200

Like? Yes (1) | No

mandeep goel 10 years AGO

p*r*t/100=72
p*3*2/100=72
p=1200

Like? Yes (1) | No

vadthyavath 10 years AGO

I=PTR/100
I+72=PT(R+3)/100
divide above 2 eqns
I+72/I=R+3/R
1+72/I=1+3/R
I/R=72/3=24

I=PTR/100
24R=PTR/100
P=24*100/T=24*100/2=1200

Like? Yes (1) | No

prabhat kumar 10 years AGO

p*2*r/100=si --------(1)

p*(r+3)*2/100=si+72 ------(2)

by solving 1 & 2
p=1200

Like? Yes (1) | No

BITU KUMAR NIRALA 10 years AGO

let pricipal p,rate r
then according to question
(2*x*(r+3))/100 -(2*x*r)/100 =72
so x=7200/6
=RS. 1200

Answer :- 1200 (option :- 5)

Like? Yes (10) | No (10)

BHANU KUMAR 10 years AGO

let the sum be s and rate be r%p.a.
then s*(r+3)*2/100-s*r*2/100=72
6s=7200
s=1200 (OPTION 5)

Like? Yes (2) | No (2)

PAWAN KUMAR 10 years AGO

let the sum be Rs x and rate of interest be y% p.a.
Then,
x*(y+3)*2/100 - x*y*2/100 =72
=>x*(y+3) - x*y =72*100/2 =3600
=>3x=3600
=> x=1200. Ans.

Like? Yes (1) | No (1)

Asha Praveena 10 years AGO

Assume sum =rs. X
rate=r%
we know that simple interest =PTR/100

( X*(r+3)*2)/100-(X*r*2)/100=72

2Xr+6x-2Xr=7200

6X=7200

X=Rs.1200

X is nothing but sum=Rs.1200

Like? Yes (2) | No (2)

Sumit Ranjan 10 years AGO

rate=y%,principal=p
in 1st case,s.i=2py/100
in 2nd case,s.i=2p(y+3)/100
so,2p(y+3)/100-2py/100=72
we will get,,p=Rs.1200

Like? Yes (1) | No (1)

Surya Rajitha 10 years AGO

let principle p, time t=2, rate is r and interest is i

2*p*r/100 = i

2*p*(r+3)/100 = i+72

2*p*r=100i
2*p*r + 6*p = 100i +7200

6*p = 7200

p=1200

Like? Yes (2) | No (2)

chinatamani chinna reddeppa 10 years AGO

(p*t*(r+3)/100)-(p*t*r/100)=72
p*t*r+3p*t-p*t*r=7200
3*p*t=7200
3p=3600
p=1200

Like? Yes (2) | No (2)

Vinayaka 10 years AGO

p*2*(r+3)/100-PTR/100=72

Like? Yes (1) | No (1)

vignesh waran A 10 years AGO

SI = PNR/100 formula

Like? Yes (1) | No (1)

Aranya Ghosh 10 years AGO

(x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

Like? Yes (1) | No (1)

Anish Mathur 10 years AGO

sum=rs.x; rate = r% then, (x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

Like? Yes (1) | No (1)

SAURADIPTA SADHU 10 years AGO

Let the sum be x
For extra 3% on x the interest is Rs 72 for 2 years
For 1 year it would be Rs 36
So, x*3/100=36
So, x= Rs.1200

Like? Yes (2) | No (2)

mahesh 10 years AGO

ans 1200

Like? Yes (1) | No (1)

AMAN KUMAR 10 years AGO

p*(r+3)*2/100-p*r*2/100=72
hence p=1200

Like? Yes (1) | No (1)

Liti 10 years AGO

{(p*2(r+3))(2pr)}100= 72
6p=7200
p=1200

Like? Yes (1) | No (1)

HARISHCHANDRA RAIKWAR 10 years AGO

p*2*x/100 + 72 = p*2*(x+3)/100
solving for p we get
6p=7200
p=1200

Like? Yes (1) | No (1)

Deepanwita 10 years AGO

by use of simultaneous equations.
(x*r*2)/100 +72=(x*(r_+3)*2)/100
x=1200

Like? Yes (1) | No (1)

RAJIB BANERJEE 10 years AGO

Let,Principle=p,Rate=r
So,1st time SI =(p*r*2)/100
Now,Rate=r+3
SI = {p*(r+3)*2}/100
So,{p*(r+3)*2}/100 - (p*r*2)/100=72
->6*p/100 =72
->p=1200

Like? Yes (1) | No (1)

Ritika 10 years AGO

Let x be the principal
x*3*2÷100=72
So,x=1200

Like? Yes (1) | No (1)

k priyanka 10 years AGO

sum=rs.x; rate = r% then, (x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

Like? Yes (1) | No (1)

Rudradeb MAhato 10 years AGO

find out urself

Like? Yes (2) | No (2)

Manikumar R 10 years AGO

pt(r+2)/100-ptr/100=72
p*2*3/100=72
p=1200

Like? Yes (1) | No (1)

BISWAJIT NATH 10 years AGO

((P2r)/100)+72=(P2(r+3))/100;
p=Rs.1200

Like? Yes (1) | No (1)

MRITUNJAY KUMAR 10 years AGO

sum*0.03*2=72
sum=1200.

Like? Yes (2) | No (2)

NUGGU SINDHURI 10 years AGO

I= PTR/100

72=P×2×3/100
P=1200

Like? Yes (1) | No (1)

manish kumar 10 years AGO

if p=principal,t=time,and r=rate of interesr
so
interst(i)=p*t*r/100
72=(p*t*r/100)-p*t*(r+3)/100
p=1200

Like? Yes (1) | No (1)

Amrita priyadarshi 10 years AGO

sum=x
rate = r%
then, (x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

Like? Yes (1) | No (1)

sonam 10 years AGO

[(s*2*r)/100]+72=[s*2*(r+3)]/100
7200+2sr=2sr+6s
6s=7200
s=1200

Like? Yes (1) | No (1)

Niraj Ojha 10 years AGO

3*x*2/100=72
So ans 1200

Like? Yes (1) | No (1)

Santosh Kumar Senapati 10 years AGO

sum=72*100/6

Like? Yes (1) | No (1)

Kushagra Tiwari 10 years AGO

sum=rs.x;
rate = r% then,
(x*(r+3)*2)/100- (x*r*2)/100=72

on solving this

x=1200

which is in 5th option

Like? Yes (1) | No (1)

Abhishek Vishwakarma 10 years AGO

sum=rs.x;
rate = r% then,
(x*(r+3)*2)/100- (x*r*2)/100=72

on solving this

x=1200

ans is ---(5)

Like? Yes (1) | No (1)

Ashish Shende 10 years AGO

sum=rs.x;
rate = r% then,
(x*(r+3)*2)/100- (x*r*2)/100=72

on solving this

x=1200

ans is ---(5)

Like? Yes (1) | No (1)

Rupak Roy 10 years AGO

suppose the amount he deposited is 'x',and the interest is y%.
then
100rs..1year..interest is y
for x amount and for 2years the interest is 2xy/100
now,if the interest was 3% higher then
100 rs..1 year..interest is (y+3)
for x amount and for 2 years the interest is 2(y+3)x/100
now ,
2(y+3)x/100 - 2xy/100 =72
solving this equation we get x=1200

Like? Yes (1) | No (1)

pramita raha 10 years AGO

I=PTR/100 WHERE I=TOTAL INTEREST P=SUM , T=TIME,R=INTEREST RATE

I=PTR/100 WHERE T=2
I=2PR/100;

IF THE SUM HAD BEEN PUT 3%HIGHER RATE INTEREST WILL RS.72 MORE
HERE I=(I+72), R=(R+3) ,T=2;
I+72=P2(R+3)/100
=>2PR/100+72=2PR/100+6P/100;
=>6P/100=72
=>P=1200

Like? Yes (1) | No (1)

souharda 10 years AGO

100x=sum, rate =r,r+3
n=2
2*100x/100(r+3-r)=72
2*3x=72 x=12 100x=1200

Like? Yes (1) | No (1)

Nagar Bansri 10 years AGO

rate =3% , time=2 yr, si=72
so,
si=(P*R*T)/100
72=(P*3*2)/100
(72*100)/(3*2)=P
SP, P=1200

Like? Yes (1) | No (1)

Mani Shankar 10 years AGO

(PxRx2)/100 + 72 = (Px(R+3)x2)/100
=> P = 1200

Like? Yes (1) | No (1)

yagnik pandya 10 years AGO

I=72
R= 3%
N=2
P=?
According to simple interest equation I = PRN/100.
We get 1200.

Like? Yes (1) | No (1)

prince kachhela 10 years AGO

let the principle amount is X and rate of interest is R%
then
X*2*(R+3)/100-X*2*R/100=72
so X=1200

Like? Yes (1) | No (1)

AVINASH GAURAV 10 years AGO

Let sum be 'x' and rate be 'r'.
({x*2*r}/100 )+72 = (x*2*{r+3})/100.
72=(6*x)/100.
x=1200.

Like? Yes (1) | No (1)

vinay agarwal 10 years AGO

let P be the sum.and X be the interest rate.
(2*(X+3)*P/100) - (2*X*P/100)=72
THEREFORE 6P/100=72
P=RS 1200

Like? Yes (1) | No (1)

T.VIGNESWARA REDDY 10 years AGO

intrest = p*2*r/100 for first time so second time intrest is p*2*(r+3)/100 so difference is given as 72 p*6/100=72 so p=1200

Like? Yes (1) | No (1)

Lakshmi Narmada 10 years AGO

(npr)/100=x
np(r+3%)=x+72
solving we get n=1200

Like? Yes (1) | No (1)

Aravinth kumar 10 years AGO

sum=rs.x; rate = r%
(x*(r+3)*2)/100- (x*r*2)/100=72
x=1200

Like? Yes (1) | No (1)

Sagar Laxmikantrao Gomte 10 years AGO

Let the SI for 2 years and rate R be Rs. y and sum be Rs. x
So,
y=(x*R*2)/100.................eq(1)

now if rate increases by 3% then SI increases by Rs. 72
hence,
y+72 = (x*(R+3)*2)/100
=> y = (2x*(R+3)-7200)/100
=> (2*x*R)/100 = ((2*x*R)+6*x-7200)/100

=> 6*x = 7200
=> x = Rs. 1200

Like? Yes (1) | No (1)

RAJESH KUMAR 10 years AGO

(1200*2*3)/100 =72

Like? Yes (1) | No (1)

prem kumar 10 years AGO

from 1st condition x=p*2*r/100 and from 2 nd condition x+72=p*2*(r+3)/100 by solving we get p=1200

Like? Yes (1) | No (1)

rakesh kumar 10 years AGO

(p*r*2)/100+72=(p*(r+3r/100)*2)/100
p=1200

Like? Yes (1) | No (1)

Yogesh 10 years AGO

Let Sum=x
rate=r%
then
(x*r*2/100)-(x*(r+3)*2/100)=72
3*x*r/50=72
by solving this we got 1200rs...

Like? Yes (1) | No (1)

Vikas Bhardwaj 10 years AGO

(3*2*p)/100=720

Like? Yes (1) | No (1)

Divya 10 years AGO

sum=p
rate=R
so interest=(P*R*T)/100
then
from given
P*(R+3)*3/100=((p*r*3)/100)+75
by solving sum is 1200
answr is 1200

Like? Yes (1) | No (1)

T P Niveditha 10 years AGO

P*2*(r+3)/100-p*2*r/100=72

Like? Yes (1) | No (1)

Kapil Shirsath 10 years AGO

(P*N*(R+3))/100 - P*N*R/100

solving above equation we get 6P=7200

so p=1200

Like? Yes (1) | No (1)

shubham varshney 10 years AGO

si=pr2/100
si+72=p(r+3)2/100
on subtracting both equations we get p=1200

Like? Yes (1) | No (1)

Rajamohan Raju 10 years AGO

S = 2PR/100 ---------------------------------(1)
S+72 =2P(R+3)/100 ------------------------(2)

Substituting 1 in 2 then we get S = 1200

Like? Yes (1) | No (1)

Navneet Kaur 10 years AGO

P*(r+3)*2/100-P*r*2/100=72
p*r*2+p*6-p*r*2=72*100
p*6=7200
p=1200

Like? Yes (1) | No (1)

myvizhi 10 years AGO

sum=rs.x; rate = r% then, (x*(r+3)*2)/100- (x*r*2)/100=72
6x/100=72;x=1200

Like? Yes (1) | No (1)

Sunke Sinddhuu 10 years AGO

(x*r*2)/100-(x*(r+3)*2)/100=72

on solving we get x=1200/-

Like? Yes (1) | No (1)

paaredi muthyam reddy 10 years AGO

((p*2*(r+3))/100)-((p*2*r)/100)=72

solve it then we will get p=1200

Like? Yes (1) | No (1)

Ramesh Sriram 10 years AGO

lets take amount=p, interest=x, rate=r
x=p*2*r/100
for raised rate
x+72=p*2*(r+3)/100
x+72=(2*p*r/100)+6p/100
x+72=x+6*p/100
p=1200

Like? Yes (1) | No (1)

Ravi Kumar Kodari 10 years AGO

((p*2*(r+3))/100)-((p*2*r)/100)=72

solve it then we will get p=1200

Like? Yes (1) | No (1)

RAVI RANJAN 10 years AGO

3% in 2 year is 72
so in 1 year 36
if 3% =36 then 100% =1200

Like? Yes (1) | No (1)

priya 10 years AGO

SI=PRT/100
let P=x
R=r%
then, (x*(r+3)*2)/100- (x*r*2)/100=72
[(x*2)/100](r+3-r)=72
6x=7200
=> x=1200

Like? Yes (1) | No (1)

Tania Mondal 10 years AGO

Let sum= x and rate of interest=y
now,
[x*(r+3)*2/100]-[x*r*2/100]=72
2xr+6x-2xr=7200
x=1200
So sum=1200rs

Like? Yes (1) | No (1)

Mohit Aggarwal 10 years AGO

72 means 36 in a year.
x * 3% = 36
x=1200

Like? Yes (1) | No (1)

shashikanth 10 years AGO

p*2*r=100x
p*2*(r+3)=100*(x+72)
solving we get p=1200

Like? Yes (1) | No (1)

RIPON DEB 10 years AGO

let the interest be r
so {2*p(r+3)}/100-(2*p*r)/100=72
p=1200

Like? Yes (1) | No (1)

AMIT KUMAR SHAW 10 years AGO

Let sum was=x
rate of interest=y%
simple interest for 2years at y%=(x*y*2)/100
simple interest for 2years at (y+3)%=x*(y+3)*2/100
Hence,
{x*(y+3)*2/100}-{(x*y*2)/100}=72
or,{(2xy+6x)/100}-2xy/100=72
or,6x/100=72
or,x=1200
Hence,The sum is =1200

Like? Yes (1) | No (1)

Dipak Singh 10 years AGO

let, x = principal ; y = rate of interest
2x(y/100)+72=2[x({y+3)/100}]
therefore,x=1200

Like? Yes (1) | No (1)

prakash kumar 10 years AGO

S.I.= (p*r*t)/1000
(p*(r+3)*2)/100 - (p*r*2)/100 = 72,
p/100(2r+6-2r) = 72,
p = 1200

Like? Yes (1) | No (1)

gouthamdaneti 10 years AGO

I=p*r*t
I is interest
p is principle
t is time period

Like? Yes (1) | No (1)

ananthavalli 10 years AGO

(p*(R+3)*2)/100-(P*R*2)/100=72
2PR+6P-2PR=7200
6P=7200
so P=1200 The sum is Rs.1200
so option 5) is correct

Like? Yes (1) | No (1)

Arijit Acharyya 10 years AGO

S.I(1)=p*r*2/100------(1)
S.I(2)=p*(r+3)*2/100-----(2)
S.I(2)-S.I(1)=72
p(2*r+6-2*r)=7200
6*p=7200
p=1200
option 5)

Like? Yes (1) | No (1)

sandeep yendamuri 10 years AGO

intelligence

Like? Yes (1) | No (1)

suma sindhu 10 years AGO

SI=PTR/100
Let, x=p(2)(r)/100.................(1)
x+72=p(2)(r+3)/100.............(2)
sub(1) in (2)
72=p(2)(3)/100
p=Rs.1200

Like? Yes (1) | No (1)

swarupa nuggu 10 years AGO

1200

Like? Yes (1) | No (1)

shubhankar 10 years AGO

1200

Like? Yes (1) | No (1)

yogesh keshari 10 years AGO

Si=pr2/100
Si+72=p(r+3)2/100
Solve for p

Like? Yes (1) | No (1)

kundan kumar 10 years AGO

(p*r*2/100)+72=(p*(r+3)*2)/100
by solving
6p=7200
p=1200

Like? Yes (1) | No (1)

vibhu raj gupta 10 years AGO

it's given that from the new rate the intrest is increased by 72, so the
p*(r+3)*2/100=p*r*2/100 +72
by solving this we get principal is 1200.

Like? Yes (1) | No (1)

Udai Pratap Singh 10 years AGO

simple interest=sum*x*2/100 where x=certain rate

modified simple interest=sum*(x+3)*2/100

modified simple inertest - simple interest=72
sum*(x+3)*2/100 -sum*x*2/100=72
sum=1200

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Rachit Agarwal 10 years AGO

72=(x*3*2)/100
x=1200

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Devendra Marghade 10 years AGO

If sum=x, rate=r, then
Simple Interest wih rate 'r' for 2 years=(x*2*r)/100 ---(i)
Simple Interest with rate 'r+3' for 2 years=x*2*(r+3)/100 --(ii)
Given, [(x*2*r)/100]+72 =x*2*(r+3)/100
On solving, x=1200

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Vinay Sipani 10 years AGO

let 'P' be sum and 'r' be rate.

we have
72 + (P*2*r)/100 = P*2*(r+3)/100
=> 7200 + 2Pr = 2Pr + 6P
=> P = Rs. 1200

Like? Yes (2) | No (3)

KRISHNA CHANDRA UPADHYAY 10 years AGO

sum= Rs. x; rate = r% then, (x*(r+3)*2)/100- (x*r*2)/100=72

x=1200.
hence option (5) is correct.

Like? Yes (2) | No (3)

ABHINAV THAKUR 10 years AGO

Since the sum was put on simple interest, so 3% interest for 2 years=6% interest. Now, 6% of sum= Rs. 72 => Sum= Rs. 1200.

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RAKESH 10 years AGO

p*(r+3)*2 - p*r*2 = 72*100
=> 2pr + 6p - 2rp = 7200
=> 6p = 7200
=> p = 1200

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guddu 10 years AGO

P*(R+3)*2/100-P*R*2/100=72
P=1200

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DIPAK TIWARI 10 years AGO

Let sum=X ₹ and rate=Y%
{X*(Y+3)*2}/100 - {X*Y*2}/100 =72₹
(2XY+6X)/100 - 2XY/100 =72₹
6X/100=72₹
X=1200₹
Option 5

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ANKUR ANAND 10 years AGO

let x be the required sum
and y be the rate
(x*y*2)/100 + 72=(x*(y+3)*2)/100
=> 2xy/100 +72=2xy/100 + 6x/100
comparing two sides we get
6x/100=72
=> x=1200

correct answer is 5) Rs. 1200

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Nitin Jaiswal 10 years AGO

Let the sum be P, the interest be I and the interest rate be R%.
Then, P * R * 2 / 100 = I
and P * (R+3) * 2 / 100 = I + 72
Subtracting the above two equations we get P = 1200

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Tark Patel 10 years AGO

sum=x , rate=r , year=n
-> ( ( x * r * n ) / 100 ) - ( ( x * (r+3) * n ) / 100 ) = 72
-> ( ( x * r * 2 ) / 100 ) - ( ( x * (r+3) * 2 ) / 100 ) = 72
-> ( ( 6 * x ) / 100 ) = 72
-> x = 1200

Answer is Rs. 1200

Like? Yes (1) | No (2)

naina 10 years AGO

S.I.=p*r*t
72=x*(r+3)*2/100-x*r*2/100
where x is the sum
=>x=1200

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Ravi Bhalani 10 years AGO

sum = P,
Origional rate=R
Then,
(P*(R+3)*2/100)-(P*R*2/100)=72
=>P=1200 Rs.

Like? Yes (1) | No (2)

HARDIP BHALANI 10 years AGO

sum = P,
Origional rate=R
Then,
(P*(R+3)*2/100)-(P*R*2/100)=72
=>P=1200 Rs.

Like? Yes (1) | No (2)

Harshad 10 years AGO

sum = P,
Origional rate=R
Then,
(P*(R+3)*2/100)-(P*R*2/100)=72
=>P=1200 Rs.

Like? Yes (1) | No (2)

RAJEEV KUMAR 10 years AGO

sum=rs.x; rate = r% then, (x*(r+3)*2)/100- (x*r*2)/100=72

x=1200

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piyush ranjan behera 10 years AGO

p(r+3)2/100= 72+p2r/100
pr2=6p=7200+ p2r
p=7200/6= 1200

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Narendra Jadon 10 years AGO

SI = PRT / 100

in 2nd case SI = [P x (R + 3) x T ]/100
Difference is of 72
Hence:
[P x (R + 3) x T ]/100 - PxRxT/100 = 72
3 x P x 2 = 7200
P = 1200
Hence sum is 1200
Option 5

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Saraswathy 10 years AGO

2pr/100 = x, --->(1)

2p(r+3)/100 = x+72 --->(2)

(2) - (1)

2pr + 6p - 2pr/100 = 72
6p = 7200
p = 1200

Ans : 1200 (Option 5)

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Anshu sharma 10 years AGO

P×(r+3)×2/100-p×r×t/100=72
p=1200.ans

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bismita kalyani 10 years AGO

LET SUM= P RATE = R
SO, {P*(R=3)*2/100}-{P*R*2/100} = 72
THEN, => 2PR + 6P - 2PR = 7200
=> 6P=7200
=> P = 7200/6 = 1200
SO, ANSWER IS Rs. 1200

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nageshwarnk 10 years AGO

1200

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TINKU MITTAL 10 years AGO

let sum is M and rate is a% and interest is b%
then (M x a x 2) /100=b
now, (M x(a+3) x 2)/100=b+72
by deviding both, we have 24a=b
by putting this value in 1st eq. we get, M=1200
that's the answer

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MANU 10 years AGO

I=2XR/100,,WHERE R=RATE,X=AMOUNT
AND I+72=2X(R+3)/100
SOLVING WE GET X=1200

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krishna kanth 10 years AGO

(p*2*R)/100+72=(p*2*(R+3))/100
2PR+7200=2PR+6P
7200=6p
p=1200

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Satyaki 10 years AGO

Let ‘x’ be the sum and ‘r’ the rate percent per annum.
Then, x (r + 3) × 2/100 – x × r × 2/100 = 72
solving, we get answer is : 1200

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RANDHIR KUMAR 10 years AGO

difeerrence of simple interest

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gurnor 10 years AGO

simple interest=(p*r*t)/100
case1:
s.i.=p*r*2/100
case2:
s.i.=p*(r+3)*2 /100
as given:
p*r*2 /100 +72= p*(r+3)*2 /100
3p/50=72
p=1200

Like? Yes (1) | No (2)

Satya 10 years AGO

Total interest payable by the end of 2 years is 2xP/100.
By taking x% interest per annum
The rate is 3% higher.
That means Total interst would be (x+3)% per annum
Total interest payable = 2P*(x+3)/100
Difference amount = 2P*(x+3)/100 - 2xP/100 = 72
6P/100 = 72
P = (72*100)/6 = 1200

Like? Yes (1) | No (2)

Saurabh Kumar 10 years AGO