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Numerical Ability
Simple & Compound Interest
Two equal amounts are deposited in two bank at same interest rate 12% p.a. for 7 years and 4.5 years respectively. If the different between the interest is Rs. 585 then what was each amount?
1) Rs. 1650
2) Rs. 1750
3) Rs. 1850
4) Rs. 1950
5) Rs. 2050
Read Solution (Total 9)
-
- According to the question Sum ,P is same R= 12% in both case
Then Interest of Rs.585 will be the interest of the difference of time period= (7-4.5)= 2.5 years
Now, P = 100 x SI/R*T = 100×585/12×2.5
P = 100×585/30 = 1950 - 8 years agoHelpfull: Yes(12) No(0)
- 4)Rs. 1950
- 8 years agoHelpfull: Yes(3) No(5)
- let x be the principle amount.
(x*7*12)/100 - (x*4.5*12)/100=585
therefore,x=1950
- 8 years agoHelpfull: Yes(1) No(0)
- Acc to question Sum ,P is same R= 12% in both case
Then Interest of Rs.585 will be the interest of the difference of time period= (7-4.5)= 2.5 years
Now, P = 100xSI/RT = 100×585/12×2.5
P = 100×585//30 = 1950 - 8 years agoHelpfull: Yes(0) No(1)
- 585=p*12*7/100 - p*12*4.5/100
solving for p, we get 1950 - 8 years agoHelpfull: Yes(0) No(0)
- let the amount be'x'.
interest of 12%for 7,4.5 yrs.
(x*12*7)/100*(x*12*4.5)/100=585
ans:1950 - 8 years agoHelpfull: Yes(0) No(0)
- option 4 i.e 1950
p*12*7 - p*12*4.5 = 58500
on solving this we get p=1950 - 8 years agoHelpfull: Yes(0) No(0)
- Here difference of interest=585
N=7 R=12
N2=4.5 R2=12
ie SI1 - SI2=585
SI1=( P * N * R ) / 100
SI2=( P * N2 * R2 ) / 100
SI1 - SI2=( P * N * R ) / 100-( P * N2 * R2 )/100
Taking common term P/100 outside we get
SI1 - SI2=P( N * R - N2 * R2 )/100
585 =P(7 * 12 - 4.5 * 12 ) / 100
=P ( 84 - 54 ) /100
585 * 100 = 30 P
58500 / 30= P
Therefore answer is
Option 4) 1950
585=P/100 - 8 years agoHelpfull: Yes(0) No(0)
- we know simple interest si=(ptr)/100.
in first case let p=x,t=7 and r=12
in second case p=x and given t=4.5,r=12
so (x*7*12)/100=(x*4.5*12)/100
=>30x=585*100
=>x=1950 - 8 years agoHelpfull: Yes(0) No(1)
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