What is the value of the equator?​

Step-by-step explanation:

answer is given above

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Given,

The student's ages are in A.P.

The common difference  = 3 months.

The younger student's age = 12.

The total age = 375yrs.

To Find,

The count of students in the group.

Solution,

Given that,

The ages of students are in A.P.

The difference between their ages = 3months.

3/12 = 1/4yrs

The total of their age = 375yrs.

Younger one's age =12yrs.

In A.P, the sum of the terms

S = n/2 (2a +(n-1)d)

a =12.

S = n/2 (2*12+ (n-1)1/4)

S = 375

n/2(2*12+(n-1)1/4) = 375

n/2(24+(n-1)1/4) = 375

n(24+(n-1)1/4) = 750

n(24*4+(n-1)) = 750 * 4

n(96+(n-1)) = 3000

96n + [tex]n^{2}[/tex] -n = 3000

95n + [tex]n^{2}[/tex] = 3000

[tex]n^{2}[/tex] + 95n -3000 = 0

Let's factorise it.

[tex]n^{2}[/tex] + 120n -25n -3000 = 0

n(n+120) -25(n +120) = 0

So,

n = -120 or 25

Hence, 25is the total number of students in the group.

Answer:

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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Answer:

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

#SPJ2

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