CBSE Class 8 Maths Exemplar Chapter 9: Comparing Quantities NCERT Solutions

Solution:

Let the principal sum be P.

Scenario 1: Simple Interest

The sum doubles in 2 years, so the Amount (A) = 2P.

The Simple Interest (SI) = A - P = 2P - P = P.

Using the formula for Simple Interest: $$SI\; =\; \backslash frac\{P\; \backslash times\; r\; \backslash times\; T\}\{100\}$$

Substituting the values: $P\; =\; \backslash frac\{P\; \backslash times\; r\; \backslash times\; 2\}\{100\}$

Dividing both sides by P (assuming P ≠ 0): $1\; =\; \backslash frac\{2r\}\{100\}$

Solving for r: $r\; =\; \backslash frac\{100\}\{2\}\; =\; 50$

So, the simple interest rate is r = 50%.

Scenario 2: Compound Interest

The sum doubles in 2 years, so the Amount (A) = 2P.

Using the formula for Amount with Compound Interest: $$A\; =\; P\; \backslash left(\; 1\; +\; \backslash frac\{R\}\{100\}\; \backslash right)^\{T\}$$

Substituting the values: $2P\; =\; P\; \backslash left(\; 1\; +\; \backslash frac\{R\}\{100\}\; \backslash right)^\{2\}$

Dividing both sides by P (assuming P ≠ 0): $2\; =\; \backslash left(\; 1\; +\; \backslash frac\{R\}\{100\}\; \backslash right)^\{2\}$

Taking the square root of both sides: $\backslash sqrt\{2\}\; =\; 1\; +\; \backslash frac\{R\}\{100\}$

Solving for R: $\backslash frac\{R\}\{100\}\; =\; \backslash sqrt\{2\}\; -\; 1$

$R\; =\; 100\; (\backslash sqrt\{2\}\; -\; 1)$

We know that $\backslash sqrt\{2\}\; \backslash approx\; 1.414$.

$R\; \backslash approx\; 100\; (1.414\; -\; 1)$

$R\; \backslash approx\; 100\; (0.414)$

$R\; \backslash approx\; 41.4$

So, the compound interest rate is approximately R = 41.4%.

Comparison:

We found r = 50% and R ≈ 41.4%.

Comparing the two rates, we see that R (41.4%) is less than r (50%).

Therefore, R < r.

Answer: (b) R < r

Solution:

Given:

Principal (P) = Rs. 50,000

Rate of Interest (R) = 4% per annum

Time (T) = 2 years

The formula for the Amount (A) when interest is compounded annually is:

$$A\; =\; P\; \backslash left(\; 1\; +\; \backslash frac\{R\}\{100\}\; \backslash right)^\{T\}$$

Substitute the given values into the formula:

$$A\; =\; 50000\; \backslash left(\; 1\; +\; \backslash frac\{4\}\{100\}\; \backslash right)^\{2\}$$

Simplify the fraction inside the parenthesis:

$$A\; =\; 50000\; \backslash left(\; 1\; +\; \backslash frac\{1\}\{25\}\; \backslash right)^\{2\}$$

$$A\; =\; 50000\; \backslash left(\; \backslash frac\{25\; +\; 1\}\{25\}\; \backslash right)^\{2\}$$

$$A\; =\; 50000\; \backslash left(\; \backslash frac\{26\}\{25\}\; \backslash right)^\{2\}$$

Calculate the square of the fraction:

$$A\; =\; 50000\; \backslash times\; \backslash frac\{26\; \backslash times\; 26\}\{25\; \backslash times\; 25\}$$

$$A\; =\; 50000\; \backslash times\; \backslash frac\{676\}\{625\}$$

Now, perform the multiplication and division:

$A\; =\; \backslash frac\{50000\; \backslash times\; 676\}\{625\}$

We can simplify this calculation. Note that $50000\; /\; 625\; =\; 80$.

$A\; =\; 80\; \backslash times\; 676$

$A\; =\; 54080$

So, the total amount after 2 years is Rs. 54,080.

The Compound Interest (CI) is the difference between the Amount and the Principal:

CI = A - P

CI = 54080 - 50000

CI = Rs. 4080

Answer: The compound interest is Rs. 4,080.

Solution:

Given:

Marked Price (MP) = Rs. 1,200

Discount Percentage = 12%

The discount is calculated on the marked price.

Discount Amount = Discount Percentage of Marked Price

$$Discount\; Amount\; =\; \backslash frac\{12\}\{100\}\; \backslash times\; 1200$$

Calculate the discount amount:

$$Discount\; Amount\; =\; 12\; \backslash times\; 12$$

$Discount\; Amount\; =\; 144$

So, the discount is Rs. 144.

The Selling Price (SP) is calculated by subtracting the discount amount from the marked price:

SP = Marked Price - Discount Amount

SP = 1200 - 144

SP = Rs. 1056

Answer: The selling price of the article is Rs. 1,056.