my sister brought my a mathematics question to help solve, but i was unable to solve it, after so many trials. fellow naira lander, the solution is yours. thanks!


Q. A car was bought for 2,400,000 naira, every year it loses 10% of its value, when will it be worth 1,200,000 naira?

7th year. but this question too cheap now

10% of the initial cost price is 240000 naira .so the second year I.e after 1 year. the value drops to 2,400,000 -240,000= 2,160,000 naira.
the new value of d car is 2,160,000. so the 3rd year it becomes 2,160,000 - (10% of 2,160,000=216000) = 1944000. and so on . I busy now I for continue

Omo.. Will it ever be worth that amount?

by 4th year .it will worth (1944000-194400) =1,749,600naira. by 5th year it will worth 1,749,600- 174,960= 1574640. then 6th year the car worth will be 1417176. 7th year it will worth 1,275,459 naira. by the end of the
8th year it will be 1275459-127545= 1147914. but there should be a mathematical direct formula for the calculation

6.57....
Approximately 7years

---

Formular; V=A(1-r)^x

V=Cost of car after "x" years of depreciation (N1,200,000)
A=Actual cost of car (2,400,000)
r=rate of depreciation (10/100=0.10)

1,200,000=2,400,000(1-0.10)^x
1,200,000/2,400,000=0.9^x
1/2=0.9^x
0.5=0.9^x
Using logarithm to solve for x
Log 0.5=Log 0.9^x
Log 0.5=xLog 0.9
x=Log 0.5/Log 0.9
x=-0.3010/-0.0458
x=6.57years or 7years approximately
QED!

ifoundmyperfect:
6.57....
Approximately 7years

---

Formular; V=A(1-r)^x

V=Cost of car after "x" years of depreciation (N1,200,000)
A=Actual cost of car (2,400,000)
r=rate of depreciation (10/100=0.10)

1,200,000=2,400,000(1-0.10)^x
1,200,000/2,400,000=0.9^x
1/2=0.9^x
0.5=0.9^x
Using logarithm to solve for x
Log 0.5=Log 0.9^x
Log 0.5=xLog 0.9
x=Log 0.5/Log 0.9
x=-0.3010/-0.0458
x=6.57years or 7years approximately
QED!

excellent!!!! u r absolutely correct! the formular u gave was the correct approach.

thanksssss! i really appreciate.

demsid:
by 4th year .it will worth (1944000-194400) =1,749,600naira. by 5th year it will worth 1,749,600- 174,960= 1574640. then 6th year the car worth will be 1417176. 7th year it will worth 1,275,459 naira. by the end of the
8th year it will be 1275459-127545= 1147914. but there should be a mathematical direct formula for the calculation

u gat a nice work there! that's the same approach i av been working on, not giving rit result!

Kessyy2k4:
excellent!!!! u r absolutely correct! the formular u gave was the correct approach.

thanksssss! i really appreciate.

Glad I could be of help...

hello fellas, here is another related problem.,.....


Q. A sum of money was made for an investment at the 5.5% compound interest. for how long will it become exactly double of the money invested?.


thanks

ifoundmyperfect:


Glad I could be of help...

thanks brother!......here is another problem......



Q. A sum of money was made for an investment at the 5.5% compound interest. for how long will it become exactly double of the money invested?.


thanks

ifoundmyperfect:


Glad I could be of help...

still waiting for the solution bro!

Kessyy2k4:
still waiting for the solution bro!

The compound interest formular we wld be adopting is;
FV=PV x (1+r)^t

where;
FV= Future value
PV= Present value
r = Interest rate (5.5/100= 0.055)
t = period

---

We are trying to find "t" given an interest rate of 5.5%
Hence, it was also stated that at this period "t", the Future value, FV will be twice the present value, PV
representing that mathematically, it simply;
FV= 2PV

so now lets proceed;

FV = PV x (1+r)^t
2PV = PV x (1+0.055)^t
2PV/PV = (1.055)^t
2 = 1.055^t
Using log to solve for t
Log 2 = Log 1.055^t
Log 2 = t Log 1.055
t = Log 2/Log 1.055
t = 0.3010/0.0233
t = 12.9
approx 13years
Q.E.D!!!

ifoundmyperfect:


The compound interest formular we wld be adopting is;
FV=PV x (1+r)^t

where;
FV= Future value
PV= Present value
r = Interest rate (5.5/100= 0.055)
t = period

---

We are trying to find "t" given an interest rate of 5.5%
Hence, it was also stated that at this period "t", the Future value, FV will be twice the present value, PV
representing that mathematically, it simply;
FV= 2PV

so now lets proceed;

FV = PV x (1+r)^t
2PV = PV x (1+0.055)^t
2PV/PV = (1.055)^t
2 = 1.055^t
Using log to solve for t
Log 2 = Log 1.055^t
Log 2 = t Log 1.055
t = Log 2/Log 1.055
t = 0.3010/0.0233
t = 12.9
approx 13years
Q.E.D!!!

that's incredible bro! u r 100% correct! pls. where did u graduated, course & school. I'm really proud of u & ur school ooo

Kessyy2k4:
that's incredible bro! u r 100% correct! pls. where did u graduated, course & school. I'm really proud of u & ur school ooo

Graduate of Geology for a South Western State University...
Reservoir Geologist by profession...

Thanks for your kind words...

ifoundmyperfect:
6.57....
Approximately 7years

---

Formular; V=A(1-r)^x

V=Cost of car after "x" years of depreciation (N1,200,000)
A=Actual cost of car (2,400,000)
r=rate of depreciation (10/100=0.10)

1,200,000=2,400,000(1-0.10)^x
1,200,000/2,400,000=0.9^x
1/2=0.9^x
0.5=0.9^x
Using logarithm to solve for x
Log 0.5=Log 0.9^x
Log 0.5=xLog 0.9
x=Log 0.5/Log 0.9
x=-0.3010/-0.0458
x=6.57years or 7years approximately
QED!

Dis is lovely...............I cut cap for u

ifoundmyperfect:

pls sir I'm here again with another question.

Q. Aina has 1800 naira to spend on cake.if the price of each cake decreases by 20naira, then she can buy three(3) more of them, find the price of each cake.




thanks Sir!

Kessyy2k4:
pls sir I'm here again with another question.

Q. Aina has 1800 naira to spend on cake.if the price of each cake decreases by 20naira, then she can buy three(3) more of them, find the price of each cake.




thanks Sir!

. it's. 1800/3=600 for each cake.....
this implies that each cake sells at rate of 620, nd d question states that the price is being reduced by 20. which price will now be 620-20=600 for each

Harpheez50:
. it's. 1800/3=600 for each cake.....
this implies that each cake sells at rate of 620, nd d question states that the price is being reduced by 20. which price will now be 620-20=600 for each

u are korrect! the same answer i arrived at, but the actual answer from the text book was "120" i think the given answer was wrong......thksss! bst regard!

The textbook answer is correct....Reduce the word problem into two equations....xy=1800......(1) (y-20)(x+3)=1800......(2) .........,Solve simultaneously to get y as the price and x as the number of cakes........