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Question Number 44222 by Necxx last updated on 23/Sep/18

A man borrowed x at 5% per annum  simple interest.He paid 198000   after 2years.Calculate _   a)the value of x  b)simple interest

$${A}\:{man}\:{borrowed}\:{x}\:{at}\:\mathrm{5\%}\:{per}\:{annum} \\ $$$${simple}\:{interest}.{He}\:{paid}\:\mathrm{198000}\: \\ $$$${after}\:\mathrm{2}{years}.{Calculate}\:_{} \\ $$$$\left.{a}\right){the}\:{value}\:{of}\:{x} \\ $$$$\left.{b}\right){simple}\:{interest} \\ $$

Answered by olj55336@awsoo.com last updated on 24/Sep/18

taken 100+5+5.25 pay full amount  198000/110.25×100=179591.84

$${taken}\:\mathrm{100}+\mathrm{5}+\mathrm{5}.\mathrm{25}\:{pay}\:{full}\:{amount} \\ $$$$\mathrm{198000}/\mathrm{110}.\mathrm{25}×\mathrm{100}=\mathrm{179591}.\mathrm{84} \\ $$$$ \\ $$

Commented by 12344 last updated on 24/Sep/18

no no no

$$\mathrm{no}\:\mathrm{no}\:\mathrm{no} \\ $$

Answered by 12344 last updated on 24/Sep/18

I=((PRT)/(100))  cross multiply  I×100=PRT  100I=PRT  where P=x.R=5. T=2. I=?  100I=x×5×2  100I=x×10  100I=10x  ((100I)/(10))=((10x)/(10))  10I=x...............(1)       BUT  x+I=198000.........(2)           substitute (1) into(2)  10I+I=198000  11I=198000  ((11I)/(11))=((198000)/(11))  I=18000    but  10I=x  10(18000)=x  180000=x  therefore  a) x=180000  b)simple interest(I)=18000

$$\mathrm{I}=\frac{\mathrm{PRT}}{\mathrm{100}} \\ $$$$\mathrm{cross}\:\mathrm{multiply} \\ $$$$\mathrm{I}×\mathrm{100}=\mathrm{PRT} \\ $$$$\mathrm{100I}=\mathrm{PRT} \\ $$$$\mathrm{where}\:\mathrm{P}={x}.\mathrm{R}=\mathrm{5}.\:\mathrm{T}=\mathrm{2}.\:\mathrm{I}=? \\ $$$$\mathrm{100I}={x}×\mathrm{5}×\mathrm{2} \\ $$$$\mathrm{100I}=\mathrm{x}×\mathrm{10} \\ $$$$\mathrm{100I}=\mathrm{10}{x} \\ $$$$\frac{\mathrm{100I}}{\mathrm{10}}=\frac{\mathrm{10}{x}}{\mathrm{10}} \\ $$$$\mathrm{10I}={x}...............\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:{B}\mathrm{UT} \\ $$$${x}+\mathrm{I}=\mathrm{198000}.........\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{substitute}\:\left(\mathrm{1}\right)\:\mathrm{into}\left(\mathrm{2}\right) \\ $$$$\mathrm{10I}+\mathrm{I}=\mathrm{198000} \\ $$$$\mathrm{11I}=\mathrm{198000} \\ $$$$\frac{\mathrm{11I}}{\mathrm{11}}=\frac{\mathrm{198000}}{\mathrm{11}} \\ $$$$\mathrm{I}=\mathrm{18000} \\ $$$$\:\:\mathrm{bu}{t} \\ $$$$\mathrm{10I}={x} \\ $$$$\mathrm{10}\left(\mathrm{18000}\right)={x} \\ $$$$\mathrm{180000}={x} \\ $$$${therefore} \\ $$$$\left.\mathrm{a}\right)\:{x}=\mathrm{180000} \\ $$$$\left.\mathrm{b}\right)\mathrm{simple}\:\mathrm{interest}\left(\mathrm{I}\right)=\mathrm{18000} \\ $$

Commented by Necxx last updated on 24/Sep/18

Thank you so much sir

$${Thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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