Question

Ron borrows $ 8000 at 12% per annum simple interest and Mike borrows $9100 at 10% per annum simple interest. In how many years will their borrowed amounts(debt) be equal ?

a)18
b)20
c)22
c)24

Answer 1

we can use Arithmetic Sequence and its Sum...

Answer 2

how ?

Answer 3

the 1st item for Ron, will be the 1st year of his debt, that is...
p1=8000(1.12)=8960
p2=8960+960=9920
p3=8960+960(2)=10880
our common d=960=8000(0.12)=the interest per annum...

Answer 4

so the sum of debt for Roy (let it be SRoy) at t years would be...
\[SRoy=\frac{t}{2}(2(8960)+960(t-1))\]

Answer 5

we'll do the same for Mike... and equate the sum of their debts to solve for t years...

Answer 6

got the idea @moli1993 ?

Answer 7

the sequence for Mike will be as follows:
p0=9100
p1=9100(1.10)=10010
p2=9100(1.20)=10010+910=10920
the common d=910=9100(0.10)=interest for Mike...

Answer 8

same formula applies for the sum of Mike's debt for t years...\[SMike=\frac{t}{2}(2(10010)+910(t-1))\]

Answer 9

Equating SRoy=SMike will allow us to solve t years...

Answer 10

ohk u r applying arithmetic progression i.e. sum=n/2(2a+(n-1)d)

Answer 11

yup...

Answer 12

that's the way i see it...

Answer 13

ohh cool but i cudnt understand why are u taking
8960 instead of 8000 in the formula ?
and vice versa for 9100?

Answer 14

that's the start of their payment if they'll pay for 1st year of their borrowed money...

Answer 15

ummm i calculated but the answer is not matching with any of the options :(

Answer 16

it's coming 43

Answer 17

maybe the others have different idea...

Answer 18

i'm thinking also it it's a geometric progression but can't see a common ratio....

Answer 19

can u do it using the formula
simple interest= principle*rate*time/100

Answer 20

time will just cancel once you equate the two...

Answer 21

@ganeshie8
can u help us out !

Answer 22

yes time has cancelled out, the inner one is remaining and t=43 is coming on solving this equation

Answer 23

hmmm... let us try equation of a line... simple interest is a linear function... let us use the slope-intercept formula...\[y=mx+b\]Let's have some definitions:\[y=S(t)=principal+interest\]\[m=principal\times rate\]\[b=principal\]\[t=time~(years)\]The equation for Roy's scheme will be: \[S(t)=960t+8000~~~~(1)\]For Mike's scheme it will be:\[S(t)=910t+9100~~~~(2)\]Equating (1) and (2) will yield\[960t+8000=910t+9100\]\[(960-910)t=9100-8000\]\[50t=1100\]\[t=\frac{1100}{50}\]\[t=22~years\]

Answer 24

... so it is letter c... \(\ddot\smile\)