Question

Jainil opens recurring deposit amount in the bank of rupees 400 per month how many installment does he have to pay to get a maturity of rupees 8560 if the bank pays the simple interest of 8% per annum

Answer 1

Given :

The monthly deposit amount = P = Rs 400

The maturity value = A = Rs 8560

The rate of interest = R = 8% per annum

To Find :

The number of installments

Solution :

Let The number of installments = n

∵ Interest = P × $\dfrac{n \times ( n + 1)}{2}$ × $\dfrac{R \times T}{100}$

Or,            = Rs 400 × $\dfrac{n \times ( n + 1)}{2}$ × $\dfrac{8 \times 1}{100 \times 12}$

Or,            = 2.67  × $\dfrac{n \times ( n + 1)}{2}$

Again

Maturity value = money deposit + Interest

Or,  A = Rs 400 × n +  2.67  × $\dfrac{n \times ( n + 1)}{2}$

Or,  Rs 8560 = Rs 400 n +  2.67  × $\dfrac{n \times ( n + 1)}{2}$

Or,  2.67  × $\dfrac{n \times ( n + 1)}{2}$ = 8560 - 400 n

Or,   $\dfrac{n \times ( n + 1)}{2}$ = $\dfrac{8560}{2.67}$ - 400 n

Or,  $\dfrac{n \times ( n + 1)}{2}$ = 3205.9 - 400 n

Or,  n² + n = 6411.8 - 800 n

Or, n² + 801 n - 6411.8 = 0

Solving this quadratic equation

i.e   n = $\dfrac{-801\pm \sqrt{(801)^{2}-4\times 1\times (-6411.8)}}{2\times 1}$

∴    n = 7.9 ≈ 8 , - 808.9

Number of installment = n = 8

Hence, The number of installments that have to pay is 8  . Answer

Answer 2

Use graph paper for this question,

(i) Plot the points A(2,2) and B(4,2). Reflect A and B in the x-axis to get

′ ′ respectively.Write the co-ordinates of

′ ′.

(ii) Write the equation of ′.

(iii) Name the figure

′

′.

(iv) Find the perimeter of the figure formed.