COMPOUND INTEREST / 26 POINT
CLASS IX. QUESTION 13 IN THE PIC...
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1
heya !!
here's your answer:
(i) Rate of depreciation = (720 - 648/720)*(100) = (72/720)*100 = 10%
__________________________
(ii) Let the original cost be x.
Rate of depreciation 10%
Then, x - (10x/100) = 720
=>(100x - 10x)/100 = 720
=>90x/100 = 720
=>x = 720 * 10/9 = Rs 800
__________________________
(iii) Value at the end of third year = 648 - (10* 648/100) = 648 - 64.8 = Rs 583.2
◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
hope helped!!☺☺
here's your answer:
(i) Rate of depreciation = (720 - 648/720)*(100) = (72/720)*100 = 10%
__________________________
(ii) Let the original cost be x.
Rate of depreciation 10%
Then, x - (10x/100) = 720
=>(100x - 10x)/100 = 720
=>90x/100 = 720
=>x = 720 * 10/9 = Rs 800
__________________________
(iii) Value at the end of third year = 648 - (10* 648/100) = 648 - 64.8 = Rs 583.2
◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
hope helped!!☺☺
Answered by
0
heya !!
here's your answer:
(i) Rate of depreciation = (720 - 648/720)*(100) = (72/720)*100 = 10%
__________________________
(ii) Let the original cost be x.
Rate of depreciation 10%
Then, x - (10x/100) = 720
=>(100x - 10x)/100 = 720
=>90x/100 = 720
=>x = 720 * 10/9 = Rs 800
__________________________
(iii) Value at the end of third year = 648 - (10* 648/100) = 648 - 64.8 = Rs 583.2
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