Subject:
MathAuthor:
emelychambersCreated:
1 year agoAnswer:
Astha borrowed approx. [tex]\text{Rs}30021[/tex].
Step-by-step explanation:
Let the sum Astha borrowed be [tex]$x$[/tex].
Then, using [tex]S.I=\frac{p*r*t}{100}[/tex], where [tex]p,r,t[/tex] are principle, rate & time respectively.
[tex]$$\begin{aligned}&\left(\frac{x \times 7 \times 2}{100}\right)+\left(\frac{x \times 6 \times 4}{100}\right)+\left(\frac{x \times 9 \times 1}{100}\right)=14110 \\&\left(\frac{14 x}{100}+\frac{24 x}{100}+\frac{9 x}{100}\right)=14110 \\\frac{47 x}{100}=14110 \\& x=\frac{14110\times 100}{47}=30021 .27\end{aligned}$$[/tex]
Hence, sum borrowed[tex]$=$ Rs 30021.27[/tex] or approximately Rs. [tex]30021[/tex].
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Author:
dámasoolson
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1Answer:
Total amount of money Astha borrowed= Rs 17000
Step-by-step explanation:
In the above question:
Given:
1. R(rate)= 7%
T(time)= 2 years
2. R(rate)= 6%
T(time)= 4 years
3. R(rate)= 9%
T(time)= 5 years
To find:
Total amount borrowed?
Solution:
Total interest= Rs 14110
We know, the formula for Simple interest as:
S.I= P×R×T/100
P= principle amount(here borrowed amount)
R= Rate of interest
T= Time
Let the total amount borrowed be 'x'
Then, total amount borrowed at the end of 11years will be;
S.I= P1×R1×T1/100+P2×R2×T2/100+P3×R3×T3/100
14110= x×7×2/100+x×6×4/100+x×9×5/100
14110= 14x/100+24x/100+45x/100(common denominator)(x is also common)
14110= 14x+24x+45x/100
14110= 83x/100
83x= 14110×100
x= 1411000/83
x= 17000
The amount of money she borrowed was Rs17000
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Author:
reillyrosario
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