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Sia ? 6 years, 6 months ago
Compounded half yearly.
P = Rs. 80000
R = 10% per annum
{tex}\frac{{10}}{2}\% {/tex} per half year
{tex}n = 1\frac{1}{2}{/tex} years
{tex} = 1\frac{1}{2} \times 2{/tex} half years
= 3 half years
{tex}\therefore A = P{\left( {1 + \frac{R}{{100}}} \right)^n} = 80000{\left( {1 + \frac{5}{{100}}} \right)^3}{/tex}
{tex} = 80000{\left( {1 + \frac{1}{{20}}} \right)^3} = 80000{\left( {\frac{{21}}{{20}}} \right)^3}{/tex}
{tex} = 80000 \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}}{/tex}
= Rs. 92610
This is the required amount
Difference in amounts = Rs. 92610 – Rs. 92400 = Rs. 210
Hence, the difference in amounts is Rs. 210.
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Sia ? 6 years, 6 months ago
Radius={tex}\frac {Diameter}{2}{/tex}={tex}\frac {56}{2}=28{/tex}
Height =2.25m
We have to convert radius in m
Therefore, {tex}\frac {28}{100}m=0.28m{/tex}
T.S.A = 2πr (r + h) = 2{tex}\times{/tex}{tex}\frac {22}{7}{/tex}{tex}\times{/tex}0.28{tex}\times{/tex}(0.28+2.25)=1.76{tex}\times{/tex}2.53m2=4.4528m2
Cost of metal = (4.4528{tex}\times{/tex}80) = 356.22
Yogita Ingle 6 years, 6 months ago
Radius=Diameter/2 = 56/2cm = 28cm
Height = 2.25m
We have to convert radius in m
Therefore, 28/100m = 0.28m
T.S.A = 2πr (r+h) =2*22/7*0.28*(0.28+2.25) =1.76*2.53m2 =4.4528m2
Cost of metal=₹ (4.4528*80)= ₹ 356.22
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Yogita Ingle 6 years, 6 months ago
Let ones digit number be y
and tens digit number be x
where, x>y
according to question
x=6y....... (1)
and x +y = 9
6y +y=9
7y= 9
y=9/7
putting the value of y in equation (1)
x=6*9/7
x=54/7
so x=54/7 and y=9/7
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Diya Joshi 6 years, 6 months ago
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