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Chapter 9

Percent and Percentage — Statement I-II Type Questions

Class - 7 Concise Mathematics Selina



Statement I-II Type Questions

Question 16

Statement 1: x is 20% more than y, then y is 20% less than x.

Statement 2: % change = (Decrease (or increase) in the valueOriginal value×100)\left(\dfrac{\text{Decrease (or increase) in the value}}{\text{Original value}} \times 100\right)%

Which of the following options is correct?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

Let y = 100.

x is 20% more than y

x=100+20100×100\Rightarrow x = 100 + \dfrac{20}{100} \times 100

= 100 + 20

= 120

Now, y is less than x by 120 − 100 = 20.

Percentage by which y is less than x

=(20120×100)= \left(\dfrac{20}{120} \times 100\right)%

=2000120= \dfrac{2000}{120}%

=503= \dfrac{50}{3}%

=1623= 16\dfrac{2}{3}%

Since y is 162316\dfrac{2}{3}% less than x and not 20% less than x, Statement 1 is false.

We know that,

% change = (Decrease (or increase) in the valueOriginal value×100)\left(\dfrac{\text{Decrease (or increase) in the value}}{\text{Original value}} \times 100\right)%

This is the correct formula. So, Statement 2 is true.

Statement 1 is false and Statement 2 is true.

Hence, option 4 is the correct option.

Question 17

Statement 1: 2 min 24 seconds is 4% of 1 hour.

Statement 2: While comparing two quantities, they must be of the same kind and in the same units and required % = (Any QuantityAnother Quantity×100)\left(\dfrac{\text{Any Quantity}}{\text{Another Quantity}} \times 100\right)%

Which of the following options is correct?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and statement 2 is false.

  4. Statement 1 is false, and statement 2 is true.

Answer

We know, 1 hour = 60 × 60 seconds = 3600 seconds.

4% of 1 hour

=4100×3600 seconds=14400100 seconds=144 seconds= \dfrac{4}{100} \times 3600 \text{ seconds}\\[1em] = \dfrac{14400}{100} \text{ seconds}\\[1em] = 144 \text{ seconds}

Now, 144 seconds = (120 + 24) seconds = 2 min 24 seconds.

Since 4% of 1 hour = 2 min 24 seconds, Statement 1 is true.

We know that,

While comparing two quantities, they must be of the same kind and in the same units, and

required % = (Any QuantityAnother Quantity×100)\left(\dfrac{\text{Any Quantity}}{\text{Another Quantity}} \times 100\right)%

This is correct. So, Statement 2 is true.

Both Statement 1 and Statement 2 are true.

Hence, option 1 is the correct option.

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