AMCAT Simple and Compound Interest Quiz-1

Solution:
Let the first part be x then second part be 850-x.
\frac{(x*6*2)}{100} = \frac{[(850-x)*12*4]}{100}
x= 850*4 -4x
5x=850*4
x=680
Then second part 850-680=Rs 170.

Solution:
Let x = equal installment at the end of one year( rate% annually) .
Now 1st year,
P =550,
Interest = \frac{(550*20*1)}{100} = 110.
Now, at the beginning of 2nd year,
P =550 + 110 – x
Interest at the end of 2nd year,
= \frac{[(660 – x) *20*1]}{100} = 132 –x/5.
Hence, the total installment,
2x = 550 + 110 + 132 – x/5
2x + x/5= 792
x = 360.

Solution:
1525-1300= 225 for 1.5 yrs (3.5-2)
So for one yr \frac{225}{1.5}= 150
Then for 2 yrs interest is 150+150=300
Then principal 1300-300=1000.
Now (\frac{150}{1000})*100= 15%

Solution:
CI =[4000 *((1 + \frac{10}{100})^{2})– 4000]
=4000*\frac{11}{10}*frac{11}{10} – 4000
=Rs840.
Then Sum in SI = 420 (ie \frac{840}{2})
= \frac{(P*3*8)}{100}
=Rs1750.

Solution:
15,000 *(1+\frac{20}{100})^{2}[ (1+\frac{25}{100}) –(1+\frac{}{100}20/100)]
15,000*\frac{120}{100}*\frac{120}{100}* [\frac{125}{100}-\frac{120}{100}]
15000*\frac{144}{100}(\frac{5}{100})
150*144*\frac{5}{100}=1080

Solution:
[P(1+\frac{r}{100})^{2}/[P(1+r*\frac{1}{100})]=\frac{6}{5}
1+\frac{r}{100}=\frac{}{100}
\frac{r}{100}=\frac{1}{5}
r=20%

Solution:
Let the sum be Rs. 100. Then,
S.I. for first 6 months = \frac{100*10*1}{100*2}
=Rs5
S.I. for last 6 months = \frac{105*10*1}{100* 2}
=Rs5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%

Solution:
Gain in 2 yrs =[(3000*25/4*2/100)-(3000*2*5/100)]
=375-300=75.
Gain in 1yr=75/2=37.5

Solution:
SI-CI for 3 yrs =\frac{Pr^{2}}{100^{3}*(300+r)}
640=\frac{P*20^{2}}{100^{3}*320}
640=(P*20*\frac{20}{100}*100*100)*320
P=Rs5000

Solution:
Let the principal be Rs x.
Then the SI =\frac{4x}{25}
Rate of interest = Time
r= \frac{(100*\frac{4}{25x})}{x*r}
r^{2}=\frac{400}{25}
r=\frac{20}{5}=4%