Please !!!!help me to solve Q.no-1.I will mark as BRAINLIST.
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Answers
Answered by
2
HII BUDDY
HERE'S THE ANSWER..
SEE IN THIS PICTURE...........
so, amount received after 6 months =63600.
.................................
NEXT....
see in 2 picture.....
so, amount received after 1 year =67416..
HOPE IT HELPS YOU.....
PLS MARK IT AS BRAINLIEST ANSWER....
IF IT HELPS YOU...
HERE'S THE ANSWER..
SEE IN THIS PICTURE...........
so, amount received after 6 months =63600.
.................................
NEXT....
see in 2 picture.....
so, amount received after 1 year =67416..
HOPE IT HELPS YOU.....
PLS MARK IT AS BRAINLIEST ANSWER....
IF IT HELPS YOU...
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PROMIS:
are you Indian
Answered by
1
Hi there!
Here's the answer:
When interest is compounded Halfyearly
![Amount =P[1 + \frac{\frac{R}{2}}{100}]^{2n} Amount =P[1 + \frac{\frac{R}{2}}{100}]^{2n}](https://tex.z-dn.net/?f=Amount+%3DP%5B1+%2B+%5Cfrac%7B%5Cfrac%7BR%7D%7B2%7D%7D%7B100%7D%5D%5E%7B2n%7D)
Where
P = Principal
Rate = R% per annum
Time = n years
Now,
Given,
(i) n = 6 months =
year
2n = 1
P = 60000
R = 12
![Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{1} Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{1}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B%5Cfrac%7B12%7D%7B2%7D%7D%7B100%7D%5D%5E%7B1%7D)
=>![Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{1} Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{1}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B%5Cfrac%7B12%7D%7B2%7D%7D%7B100%7D%5D%5E%7B1%7D)
=>![Amount =60000[1 + \frac{6}{100}] Amount =60000[1 + \frac{6}{100}]](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B6%7D%7B100%7D%5D)
=>![Amount =60000[1 + 0.06] Amount =60000[1 + 0.06]](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+0.06%5D)
=>![Amount = 60000[1.06] Amount = 60000[1.06]](https://tex.z-dn.net/?f=Amount+%3D+60000%5B1.06%5D)
=> Amount = 63600
•°• Amount after 6 months = 63600₹
(ii) (i) n = 1 year
=> 2n = 2
P = 60000
R = 12
![Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{2} Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{2}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B%5Cfrac%7B12%7D%7B2%7D%7D%7B100%7D%5D%5E%7B2%7D)
=>![Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{2} Amount =60000[1 + \frac{\frac{12}{2}}{100}]^{2}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B%5Cfrac%7B12%7D%7B2%7D%7D%7B100%7D%5D%5E%7B2%7D)
=>![Amount =60000[1 + \frac{6}{100}]^{2} Amount =60000[1 + \frac{6}{100}]^{2}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+%5Cfrac%7B6%7D%7B100%7D%5D%5E%7B2%7D)
=>![Amount =60000[1 + 0.06]^{2} Amount =60000[1 + 0.06]^{2}](https://tex.z-dn.net/?f=Amount+%3D60000%5B1+%2B+0.06%5D%5E%7B2%7D)
=>![Amount = 60000[1.06]^{2} Amount = 60000[1.06]^{2}](https://tex.z-dn.net/?f=Amount+%3D+60000%5B1.06%5D%5E%7B2%7D)
=>![Amount = 60000[1.1236] Amount = 60000[1.1236]](https://tex.z-dn.net/?f=Amount+%3D+60000%5B1.1236%5D)
=> Amount = 67416.
•°• Amount after 1 year = 67416₹
Hope it helps
Here's the answer:
When interest is compounded Halfyearly
Where
P = Principal
Rate = R% per annum
Time = n years
Now,
Given,
(i) n = 6 months =
2n = 1
P = 60000
R = 12
=>
=>
=>
=>
=> Amount = 63600
•°• Amount after 6 months = 63600₹
(ii) (i) n = 1 year
=> 2n = 2
P = 60000
R = 12
=>
=>
=>
=>
=>
=> Amount = 67416.
•°• Amount after 1 year = 67416₹
Hope it helps
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