At what rate per cent per year will a sum double itself in $11.111$ years ?

9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

11

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $9$ %
Given Amount $= 2 P, T = 11.111$
Amount $= S I + P \Rightarrow 2 P = S I + P$
$S I = P$
We know that
$S I = \frac{P R T}{100} = P$
$R = \frac{100}{T} = \frac{100}{11.111} = 9$
$∴$ At rate $9$ % pa, sum will double itself.

Suggest Corrections

Similar questions

Q. At what rate per cent per year will a sum double itself in

5.55

years ?

Q. At what rate per cent per year will a sum double itself in

6 \frac{1}{4}

years ?

Q. At what rate per cent per annum simple interest will a sum double itself in

10

years?

(i) At what rate per cent per annum will Rs. 630 produce an interest of Rs. 126 in 4 years?
(ii) At what rate per cent per year will a sum double itself in $6 \frac{1}{4}$ years?

Q. At what per cent per annum will a sum double itself in

8

years?

Join BYJU'S Learning Program

At what rate per cent per year will a sum double itself in 11.111 years ?

The correct option is D 9%Given Amount=2P,T=11.111Amount=SI+P⇒2P=SI+PSI=PWe know thatSI=PRT100=PR=100T=10011.111=9∴At rate 9% pa, sum will double itself.

At what rate per cent per year will a sum double itself in $11.111$ years ?

The correct option is D $9$ %
Given Amount $= 2 P, T = 11.111$
Amount $= S I + P \Rightarrow 2 P = S I + P$
$S I = P$
We know that
$S I = \frac{P R T}{100} = P$
$R = \frac{100}{T} = \frac{100}{11.111} = 9$
$∴$ At rate $9$ % pa, sum will double itself.