original length of the iron rod=175.65 % increase=6(1/3)%×175.65 =((19)/3)×(1/(100))×175.65 =((19×175.65)/(3×100))=((3337.35)/(300))=11.1245 new length=original length+increased length =175.65+11.1245 =186.7745cm solution by CASIO.....


Question and Answers Forum

All Questions Topic List
Logarithms Questions
Previous in All Question Next in All Question
Previous in Logarithms Next in Logarithms
Question Number 145437 by Saiki last updated on 04/Jul/21

$o r i g i n a l l e n g t h o f t h e i r o n r o d = 175.65$ $% i n c r e a s e = 6 \frac{1}{3} % \times 175.65$ $= \frac{19}{3} \times \frac{1}{100} \times 175.65$ $= \frac{19 \times 175.65}{3 \times 100} = \frac{3337.35}{300} = 11.1245$ $n e w l e n g t h = o r i g i n a l l e n g t h + i n c r e a s e d l e n g t h$ $= 175.65 + 11.1245$ $= 186.7745 c m$ $s o l u t i o n b y C A S I O . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum