Question

in how much time will ₹2000 give C.I. of ₹163.20, if the rate of interest is 4% per annum?​

Answer 1

${\huge{\underline{\large{\mathbb{\red{KABIR}}}}}}$

We have,

Rate = 4 % per annum

CI = ₹163.20

Principal (P) = Rs 2000

By using the formula,

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

163.20 = 2000[(1 + 4/100)n – 1]

163.20 = 2000[(1.04)n -1]

163.20 = 2000 × (1.04)n – 2000

163.20 + 2000 = 2000 × (1.04)n

2163.2 = 2000 × (1.04)n

(1.04)n = 2163.2/2000

(1.04)n = 1.0816

(1.04)n = (1.04)2

So on comparing both the sides, n = 2

Therefore,

Time required is 2 years.

Hopes this helps you ✌☺

Answer 2

Answer:

We have,

We have,Rate = 4 % per annum

We have,Rate = 4 % per annumCI = ₹163.20

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000(1.04)n = 1.0816

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000(1.04)n = 1.0816(1.04)n = (1.04)2

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000(1.04)n = 1.0816(1.04)n = (1.04)2So on comparing both the sides, n = 2

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000(1.04)n = 1.0816(1.04)n = (1.04)2So on comparing both the sides, n = 2Therefore,

We have,Rate = 4 % per annumCI = ₹163.20Principal (P) = Rs 2000By using the formula,CI = P [(1 + R/100)n – 1]Substituting the values, we have163.20 = 2000[(1 + 4/100)n – 1]163.20 = 2000[(1.04)n -1]163.20 = 2000 × (1.04)n – 2000163.20 + 2000 = 2000 × (1.04)n2163.2 = 2000 × (1.04)n(1.04)n = 2163.2/2000(1.04)n = 1.0816(1.04)n = (1.04)2So on comparing both the sides, n = 2Therefore,Time required is 2 years.