Math, asked by ARSH214, 8 months ago

What sum of money invested at 7.5% p.a simple interest for 2 years produces twice as much interest as ₹9600 in 3 years and 6 months at 10% p.a simple interest ?

Answers

Answered by Anonymous
22

SI for 3 yrs = 1200

SI for 1 yr = 400

ie CI for 1st yr = 400

now

CI for 2 yrs = 832

therefore

CI for 2nd yr = (832-400) = 432

since difference between CI for two consecutive yrs is interest for 1 yr on Rs 400

therefore

P= 400 , SI =(432-400) = 32 , T =1

R = 32 × 100

---------- = 8%

400 × 1

now

since SI for 1 yr is 400

therefore

P = SI × 100 400 × 100

----------- = ------------- = 5000

R × T 8 × 1

now to find difference between CI and SI for 3 yrs

we have SI for 3 yrs = 1200

to find CI for 3 yrs.

ise formula

CI = P { (1+R/100)^n - 1}

CI = 5000 { ( 1 + 8/100)^3 - 1 }

= 1298.56

thus

difference of CI and SI

= 98.56

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Answered by HariesRam
17

1st case given

P=9600

R =10%

T =3 years or7/2,years

Soln

SI = PXRXT/100

PUTTING THE GIVEN VALUES

SI =9600X7/2X10/100

=96X35=RS3360/-

NOW A/Q

NEW SI is 2 x 3360

=6720/-

Again for new SI

6720,=,PXRXT/100

6720=PX75/10X100X2

6720X100/15=P

THEREFOR P=44800/- IS THE REQUIRED ANS.

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Ram

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