KnowledgeBoat Logo
|

Mathematics

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then the cost of machine becomes :

  1. ₹ x×120100×75100x \times \dfrac{120}{100} \times \dfrac{75}{100}

  2. ₹ x×(120100)2×75100x \times \Big(\dfrac{120}{100}\Big)^2 \times \dfrac{75}{100}

  3. ₹ x×(120100)2×(75100)2x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2

  4. ₹ x×(80100)2×(75100)2x \times \Big(\dfrac{80}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2

Compound Interest

5 Likes

Answer

According to question :

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years.

⇒x×(1+20100)2×(1−25100)2⇒x×(120100)2×(75100)2.\Rightarrow x \times \Big(1 + \dfrac{20}{100}\Big)^2 \times \Big(1 - \dfrac{25}{100}\Big)^2 \\[1em] \Rightarrow x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2.

Hence, Option 3 is the correct option.

Answered By

2 Likes

Related Questions