CBSE Class 9 Maths Ganita Manjari Part 1 Ch 03 The World of Numbers Short Answer Questions with Answers

2 Mark Questions

Question 1. Virat lives in a residential flat. He parks his car and, through the lift, moves from floor No. −2 to floor No. 5. He says that his total movement is 8 floors. By calculation, show whether he is correct or not.
Answer:
Movement = final floor − starting floor = \( 5 - (-2) = 5 + 2 = 7 \) floors.
Counting the steps: −2 → −1 → 0 → 1 → 2 → 3 → 4 → 5 gives 7 moves.
\( \therefore \) Virat is not correct. He moved 7 floors, not 8.
In simple words: From −2 up to 5 is 7 steps; people often wrongly count the starting floor too.

Exam Tip: Use final − initial and remember that subtracting a negative number means adding.

Question 2. A trader records the following transactions: ₹200 earned, ₹350 lost and ₹150 earned again. Determine the final balance and state its implication.
Answer:
Take earnings as positive and losses as negative.
Final balance = \( 200 + (-350) + 150 = 350 - 350 = 0 \)
\( \therefore \) The final balance is ₹0: the trader has neither profit nor loss overall.
In simple words: He earned ₹350 in total and lost ₹350, so he ends where he started.

Exam Tip: Always state what the sign of the final answer means in the real situation.

Question 3. Using the number 35, illustrate how zero behaves differently under addition and multiplication.
Answer:
Addition: \( 35 + 0 = 35 \). Adding zero leaves the number unchanged, so 0 is the additive identity.
Multiplication: \( 35 \times 0 = 0 \). Multiplying by zero always gives zero; this is the zero property of multiplication.
So zero keeps the number the same in addition but turns it into zero in multiplication.
In simple words: Adding 0 changes nothing, but multiplying by 0 makes everything 0.

Exam Tip: Do not confuse the additive identity (0) with the multiplicative identity (1).

Question 4. Goutam is converting 0.125 into \( \frac{p}{q} \) form. Show his steps and get the answer.
Answer:
\( 0.125 = \frac{125}{1000} \) (three decimal places, so divide by 1000)
Divide numerator and denominator by 125: \( \frac{125 \div 125}{1000 \div 125} = \frac{1}{8} \)
\( \therefore 0.125 = \frac{1}{8} \)
In simple words: Write the decimal over 1000 and cancel down to lowest terms.

Exam Tip: Always reduce the fraction to its lowest terms for full marks.

Question 5. A right triangle has both perpendicular sides of length 1 unit. Find the length of the hypotenuse and explain how this result is connected to irrational numbers.
Answer:
By the Baudhāyana–Pythagoras theorem: \( h^2 = 1^2 + 1^2 = 2 \Rightarrow h = \sqrt{2} \) units.
Since 2 is not a perfect square, \( \sqrt{2} \) cannot be written as \( \frac{p}{q} \); it is irrational.
So a simple triangle with whole-number sides can have a hypotenuse whose length is an irrational number, \( \sqrt{2} \approx 1.414… \). This is also how \( \sqrt{2} \) can be marked exactly on a number line.
In simple words: A triangle with two sides of 1 has a slanting side of √2, a length that cannot be written as a fraction.

Exam Tip: Mention that √2 is irrational because 2 is not a perfect square.

3 Mark Questions

Question 6. A person moves on a number line from −3 to 5 and then back to 1.
(a) Calculate the total distance covered.
(b) Determine the net displacement.
(c) Explain the difference between distance and displacement.
Answer:
(a) From −3 to 5: \( 5 - (-3) = 8 \) units. From 5 back to 1: \( 5 - 1 = 4 \) units. Total distance = 8 + 4 = 12 units.
(b) Net displacement = final − initial = \( 1 - (-3) = \) 4 units (in the positive direction).
(c) Distance is the total length of the path travelled and is never negative. Displacement is the change from the starting point to the finishing point, and its sign shows the direction.
In simple words: The person walked 12 units in all, but ended up only 4 units from where they began.

Exam Tip: For displacement, ignore the middle of the journey and use only the start and end points.

Question 7. A container holds \( \frac{9}{10} \) litre of water. If \( \frac{2}{5} \) litre of water is removed,
(a) determine the remaining quantity.
(b) express it in simplest form.
Answer:
(a) Remaining = \( \frac{9}{10} - \frac{2}{5} = \frac{9}{10} - \frac{4}{10} = \frac{5}{10} \) litre
(b) \( \frac{5}{10} = \frac{1}{2} \) litre = 0.5 litre
In simple words: Convert both fractions to tenths, subtract, then simplify to one-half litre.

Exam Tip: Make the denominators equal before subtracting fractions.

Question 8. Examine whether the decimal expansion of \( \frac{7}{40} \) is terminating or non-terminating.
(a) Express the denominator in its prime factorised form.
(b) Verify whether it satisfies the condition for a terminating decimal.
(c) State the final conclusion regarding the nature of its decimal expansion.
Answer:
(a) \( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5 \)
(b) A fraction in lowest terms has a terminating decimal if its denominator has no prime factors other than 2 and 5. Here \( \frac{7}{40} \) is in lowest terms and 40 has only 2 and 5, so the condition is satisfied.
(c) The decimal expansion is terminating: \( \frac{7}{40} = 0.175 \).
In simple words: 40 is made only of 2s and a 5, so 7/40 ends after a few digits: 0.175.

Exam Tip: Check that the fraction is in lowest terms before applying the 2-and-5 test.

Question 9. Convert the mixed recurring decimal \( 0.2\overline{3} \) into the form \( \frac{p}{q} \). Show the complete method used to obtain the fractional form.
Answer:
Let \( x = 0.2333… \) … (1)
Multiply by 10 (to move the non-repeating digit): \( 10x = 2.333… \) … (2)
Multiply by 100: \( 100x = 23.333… \) … (3)
(3) − (2): \( 90x = 21 \Rightarrow x = \frac{21}{90} = \frac{7}{30} \)
\( \therefore 0.2\overline{3} = \frac{7}{30} \)
In simple words: Make two numbers with the same repeating tail and subtract them so the tail disappears.

Exam Tip: Check: 7 ÷ 30 = 0.2333…, which matches.

Question 10. Alexa tries to simplify the expression \( \frac{3}{4} \times \frac{2}{5} + \frac{1}{2} \). She claims that the result is an irrational number. Is she correct? Show the steps of solution.
Answer:
\( \frac{3}{4} \times \frac{2}{5} = \frac{6}{20} = \frac{3}{10} \)
\( \frac{3}{10} + \frac{1}{2} = \frac{3}{10} + \frac{5}{10} = \frac{8}{10} = \frac{4}{5} \)
\( \therefore \) The result is \( \frac{4}{5} \), which is rational. Alexa is not correct.
Rational numbers are closed under multiplication and addition, so this result had to be rational.
In simple words: Multiplying and adding fractions always gives another fraction, so the answer 4/5 is rational.

Exam Tip: Follow the order of operations: multiply first, then add.

Question 11. One number is \( \sqrt{49} \) and another number is \( \sqrt{10} \).
(a) Evaluate the expressions.
(b) Classify each number as rational or irrational.
(c) Compare the nature of these numbers with appropriate justification.
Answer:
(a) \( \sqrt{49} = 7 \) (since \( 7^2 = 49 \)); \( \sqrt{10} \approx 3.162… \)
(b) \( \sqrt{49} = 7 \) is rational; \( \sqrt{10} \) is irrational.
(c) 49 is a perfect square, so its square root is the integer 7, which can be written as \( \frac{7}{1} \). 10 is not a perfect square, so \( \sqrt{10} \) has a non-terminating, non-repeating decimal and cannot be written as \( \frac{p}{q} \).
In simple words: A square root is rational only if the number under it is a perfect square.

Exam Tip: Use the words 'perfect square' in your justification; that is the key idea examiners look for.