The World of Numbers Class 9 Important Questions Maths Chapter 3

During revision, students quickly go through Class 9 Maths Important Questions and Ganita Manjari Class 9 Maths Chapter 3 The World of Numbers Important Questions with Solutions for clarity.

Class 9 The World of Numbers Important Questions

Important Questions of The World of Numbers

Very Short Answer Type Questions

Question 1.
Are natural numbers closed under subtraction? Give an example to justify your answer.
Solution:
Natural numbers are not closed under subtraction as subtraction of two natural numbers may not result in a natural number.
For example, 7 – 12 = -5
Here, 7 and 12 are both natural numbers, but -5 is not a natural number.

Question 2.
Find 4 rational numbers between 4 and 5.
Solution:
4 rational numbers between 4 and 5 are 4.2, 4.4, 4.6, and 4.8.
There may be many more possible answers to this question.

Question 3.
Write two irrational numbers between 1 and 2.
Solution:
We can write two irrational numbers between 1 and 2 as √2 and √3.

Question 4.
Write two irrational numbers between 0 and 1.
Solution:
Two irrational numbers between 0 and 1 can be written as 0.10100100010000… and 0.505505550….
There may be many more irrational numbers between 0 and 1.

Question 5.
Write an irrational number and a rational number between √2 and √3.
Solution:
Here, √2 = 1.4142… and √3 = 1.732…
So, we can write an irrational number between √2 and √3 as 1.515515551… and a rational number between √2 and √3 as 1.5.
There may be many more alternative answers possible.

Question 6.
How many maximum number of digits possible in the repeating block of the decimal representation of \(\frac {1}{7}\)?
Solution:
There can be at most (n – 1) digits in the repeating block of the rational number \(\frac {1}{n}\).
So, there will be at most 6 digits in the decimal representation of \(\frac {1}{7}\).

Question 7.
An investor gains ₹ 350 in the stock market on Monday and loses ₹ 520 on Tuesday. Write these as integers and find the net gain or loss.
Solution:
Profit on Monday = ₹ +350
Loss on Tuesday = ₹ -520
350 + (-520) = 350 – 520 = -170
∴ Net loss = ₹ 170

Question 8.
Using Brahmagupta’s rule, calculate:
(i) (-10) × 5
(ii) (-9) × (-7)
Solution:
(i) (-10) × 5 = -50 …..[debt × fortune = debt]
(ii) (-9) × (-7) = +63 …..[debt × debt = fortune]

Question 9.
Verify: Is the set of integers closed under subtraction? Give two examples to justify.
Solution:
Yes, integers are closed under subtraction.
Example 1: 3 – 8 = -5, which is an integer.
Example 2: (-4) – (-9) = -4 + 9 = 5, which is an integer.
∴ The difference of any two integers is always an integer.

Question 10.
Find the sum: \(\frac{7}{10}+\frac{3}{5}\)
Solution:
\(\frac{7}{10}+\frac{3}{5}=\frac{7}{10}+\frac{3 \times 2}{5 \times 2}\) = \(\frac{7}{10}+\frac{6}{10}\)
[LCM of 10 and 5 is 10]
∴ \(\frac{7}{10}+\frac{3}{5}=\frac{13}{10}\)

Question 11.
Find the difference: \(\frac{9}{14}-\frac{2}{7}\)
Solution:
\(\frac{9}{14}-\frac{2}{7}=\frac{9}{14}-\frac{2 \times 2}{7 \times 2}\) = \(\frac{9}{14}-\frac{4}{14}\)
[LCM of 14 and 7 is 14]
∴ \(\frac{9}{14}-\frac{2}{7}=\frac{5}{14}\)

Question 12.
Find the product: \(\frac{4}{9} \times \frac{3}{8}\)
Solution:
\(\frac{4}{9} \times \frac{3}{8}=\frac{4 \times 3}{9 \times 8}=\frac{12}{72}=\frac{1}{6}\)

Question 13.
Find the quotient: \(\frac{5}{12} \div \frac{5}{6}\)
Solution:
\(\frac{5}{12} \div \frac{5}{6}=\frac{5}{12} \times \frac{6}{5}\) = \(\frac{6}{12}=\frac{1}{2}\)

Question 14.
Find five rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\).
Solution:
\(\frac{3}{5}=\frac{30}{50}\) and \(\frac{4}{5}=\frac{40}{50}\)
∴ Five rational numbers between \(\frac {3}{5}\) and \(\frac {4}{5}\) are:
\(\frac{31}{50}, \frac{32}{50}, \frac{33}{50}, \frac{34}{50}\) and \(\frac {35}{50}\).

Question 15.
Represent \(\frac {3}{5}\) on the number line.
Solution:
\(\frac {3}{5}\) lies between 0 and 1.
∴ Divide the interval between 0 and 1 into 5 equal parts and move three parts to the right of 0.

Question 16.
Determine whether the following rational numbers have terminating or non-terminating repeating decimal expansions without performing long division:
(i) \(\frac {329}{400}\)
(ii) \(\frac {7}{18}\)
Solution:
(i) \(\frac {329}{400}\)
Prime factorisation of denominator:
400 = 24 × 52
Since the denominator has only prime factors 2 and 5.
∴ The decimal expansion is terminating.
(ii) \(\frac {7}{18}\)
18 = 2 × 9
Since the denominator has a prime factor other than 2 and 5.
∴ The decimal expansion is a non-terminating repeating.

Question 17.
Convert \(0 . \overline{6}\) into the form \(\frac {p}{q}\).
Solution:
Let x = \(0 . \overline{6}\) ……(i)
Since one digit repeats, multiplying both sides by 10, we get
10x = \(6 . \overline{6}\) ……(ii)
Subtracting (i) from (ii), we get
10x – x = \(6 . \overline{6}-0 . \overline{6}\)
⇒ 9x = 6
⇒ x = \(\frac{6}{9}=\frac{2}{3}\)

Question 18.
Convert \(0 . \overline{45}\) into the form \(\frac {p}{q}\).
Solution:
Let x = \(0 . \overline{45}\) …..(i)
Since two digits repeat, multiplying both sides by 100, we get
100x = \(45 . \overline{45}\) ……(ii)
Subtracting (i) from (ii), we get
100x – x = \(45 . \overline{45}-0 . \overline{45}\)
⇒ 99x = 45
⇒ x = \(\frac{45}{99}=\frac{5}{11}\)

Question 19.
Convert \(0.1 \overline{6}\) into the form \(\frac {p}{q}\).
Solution:
Let x = \(0.1 \overline{6}\)
Since one digit is non-repeating, multiplying by 10, we get
10x = \(1 . \overline{6}\) …..(i)
Now, multiplying by 10, we get
100x = \(16 . \overline{6}\) …..(ii)
Subtracting (i) from (ii), we get
100x – 10x = \(16 . \overline{6}-1 . \overline{6}\)
⇒ 90x = 15
⇒ x = \(\frac{15}{90}=\frac{1}{6}\)

Class 9 Maths The World of Numbers Important Questions

Short Answer Type Questions

Question 1.
A shopkeeper sells biscuit packs in carton boxes. If there are 30 biscuit packs in 5 cartons, then how many biscuit packs will be there in 12 cartons?
Solution:
Number of biscuit packs in 5 cartons = 30
So, number of biscuit packs in 1 carton = \(\frac {30}{5}\) = 6
Therefore, number of biscuit packs in 12 cartons = 12 × 6 = 72.
Hence, there will be 72 biscuit packs.

Question 2.
A merchant in a port city of Lothal of the Harappan Civilization is exchanging bags of spices for copper ingots. He receives 18 ingots for every 3 bags of spices. If he brings 20 bags of spices to the market, how many copper ingots will he leave with?
Solution:
Number of ingots received for 3 bags of spices = 18
So, number of ingots received for 1 bag of spices = \(\frac {18}{3}\) = 6
Therefore, number of ingots received for 20 bags of spices = 20 × 6 = 120.
Hence, the merchant will leave with 120 ingots.
An ingot means a solid block of metal usually in the shape of a brick.

Question 3.
Find six rational numbers between \(\frac {1}{4}\) and \(\frac {1}{5}\).
Solution:
To get six rational numbers between \(\frac {1}{4}\) and \(\frac {1}{5}\), we need to make the denominators of \(\frac {1}{4}\) and \(\frac {1}{5}\) same.
To do that, we take the L.C.M. of 4 and 5, which is 20.
To get 20 in the denominator of \(\frac {1}{4}\), we multiply its numerator and denominator both by 5, to get
\(\frac{1}{4}=\frac{1 \times 5}{4 \times 5}=\frac{5}{20}\)
Similarly, we write \(\frac {4}{20}\) for \(\frac {1}{5}\).
To get 6 rational numbers between \(\frac {4}{20}\) and \(\frac {5}{20}\), we multiply both the rational numbers by 7 in the numerator as well as in the denominator.
So, we have the two given rational numbers as \(\frac{4 \times 7}{20 \times 7}=\frac{28}{140}\) and \(\frac{5 \times 7}{20 \times 7}=\frac{35}{140}\).
Now we are in a position to write six rational numbers between the given rational numbers as:
\(\frac{29}{140}, \frac{30}{140}, \frac{31}{140}, \frac{32}{140}, \frac{33}{140}, \frac{34}{140}\)
Or simply, \(\frac{29}{140}, \frac{3}{14}, \frac{31}{140}, \frac{8}{35}, \frac{33}{140}, \frac{17}{70}\)

Question 4.
If \(\frac{x}{4}+\frac{x}{5}=\frac{9}{10}\), then find the value of x.
Solution:
Here, \(\frac{x}{4}+\frac{x}{5}=\frac{9}{10}\)
⇒ \(\frac{(5 x \times 4 x)}{20}=\frac{9}{10}\)
⇒ \(\frac{9 x}{20}=\frac{9}{10}\)
⇒ x = \(\frac{9}{10} \times \frac{20}{9}\)
⇒ x = 2

Question 5.
In the ancient port city of Lothal, a Dealer trades spices for copper ingots at the rate of 15 ingots per 2 bags.
(i) If he starts with 10 bags and keeps 4 for personal use, how many ingots does he receive?
(ii) If he then trades back 30 ingots for spices at the same rate, how many bags does he get?
Solution:
(i) Bags available for trading = 10 – 4 = 6
Now, 2 bags have 15 ingots
So, 6 bags will have = \(\frac {15}{2}\) × 6 ingots
= 15 × 3
= 45
Thus, ingots received = 45
(ii) Given: 15 ingots = 2 bags
1 ingot = \(\frac {2}{15}\)
∴ Number of bags having 30 ingots = \(\frac {2}{15}\) × 30
= 2 × 2
= 4
Thus the dealer gets 4 bags of spices.

Question 6.
A submarine starts at sea level (0 m). It dives 250 m below sea level, then rises 80 m, and then dives another 120 m.
(i) Write the position after each movement.
(ii) If it now needs to surface (reach 0 m), how many metres must it ascend?
(iii) If the same journey is repeated 3 times in a day, what is the total downward distance covered?
Solution:
(i) The submarine is at sea level, i.e., at 0 m.
It dives 250 m below sea level.
0 – 250 = -250 m
∴ Position after first dive is -250 m.
Now, it rises 80 m.
-250 + 80 = -170 m
∴ Position after rising is -170 m.
Again, it dives 120 m.
-170 – 120 = -290 m
∴ Position after second dive is -290 m.
(ii) To reach sea level (0 m), it must ascend:
0 – (-290) = 290 m
(iii) Total downward distance in one journey:
250 + 120 = 370 m
The journey is repeated 3 times:
3 × 370 = 1110 m
∴ The total downward distance covered is 1110 m.

Question 7.
A student claims: \(\frac{4}{9} \div \frac{2}{3}=\frac{2}{6}\)
Check whether the statement is correct. If not, find the correct answer.
Solution:
\(\frac{4}{9} \div \frac{2}{3}=\frac{4}{9} \times \frac{3}{2}\) = \(\frac{12}{18}=\frac{2}{3}\)
Student’s answer: \(\frac {2}{6}\) which is equal to \(\frac {1}{3}\)
\(\frac{2}{3} \neq \frac{2}{6}\)
The student’s answer is incorrect.
The correct answer is \(\frac {2}{3}\).

Question 8.
Pooja reads \(\frac {2}{5}\)th of a book on Monday and \(\frac {1}{4}\)th on Tuesday. Find the fraction of the book still left unread.
Solution:
Total part read = \(\frac{2}{5}+\frac{1}{4}\)
L.C.M. of 5 and 4 is 20.
∴ \(\frac{8}{20}+\frac{5}{20}=\frac{13}{20}\)
Whole book = 1 = \(\frac {20}{20}\)
Unread part = \(\frac{20}{20}-\frac{13}{20}=\frac{7}{20}\)
\(\frac {7}{20}\)th part of the book is unread.

Question 9.
Using distributive property, simplify: \(\frac{5}{7}\left(\frac{7}{10}+\frac{14}{15}\right)\)
Solution:
Using the distributive property:
\(\frac{5}{7}\left(\frac{7}{10}+\frac{14}{15}\right)=\frac{5}{7} \times \frac{7}{10}+\frac{5}{7} \times \frac{14}{15}=\frac{1}{2}+\frac{2}{3}\)
L.C.M. of 2 and 3 is 6.
∴ \(\frac{1}{2}+\frac{2}{3}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}\)
∴ \(\frac{5}{7}\left(\frac{7}{10}+\frac{14}{15}\right)=\frac{7}{6}\)

Question 10.
Using distributive property, verify: \(\frac{4}{7}\left(\frac{7}{8}-\frac{1}{2}\right)=\frac{3}{14}\)
Solution:
L.H.S = \(\frac{4}{7}\left(\frac{7}{8}-\frac{1}{2}\right)=\frac{4}{7} \times \frac{7}{8}-\frac{4}{7} \times \frac{1}{2}=\frac{1}{2}-\frac{2}{7}\)
L.C.M. of 2 and 7 is 14.
∴ \(\frac{1}{2}-\frac{2}{7}=\frac{7}{14}-\frac{4}{14}=\frac{3}{14}\) = R.H.S.
∴ \(\frac{4}{7}\left(\frac{7}{8}-\frac{1}{2}\right)=\frac{3}{14}\)

Question 11.
Find three rational numbers between 0.1 and 0.11.
Solution:
0.1 = \(\frac{1}{10}=\frac{1 \times 100}{10 \times 100}=\frac{100}{1000}\)
0.11 = \(\frac{11}{100}=\frac{11 \times 10}{100 \times 10}=\frac{110}{1000}\)
∴ Three rational numbers between 0.1 and 0.11 are \(\frac{101}{1000}, \frac{102}{1000}\), and \(\frac {103}{1000}\)
i.e., 0.101, 0.102, and 0.103.

Question 12.
A shopkeeper has 18 metres of cloth. If one school uniform requires 2\(\frac {1}{4}\) metres of cloth, exactly how many uniforms can be stitched?
Solution:
Measure of cloth required to stitch one uniform = 2\(\frac {1}{4}\) = \(\frac {9}{4}\) m
Total measure of cloth = 18 m
∴ Required number of uniforms = 18 ÷ \(\frac {9}{4}\)
= 18 × \(\frac {4}{9}\)
= 8
∴ Exactly 8 uniforms can be stitched from 18 m of cloth.

Question 13.
Represent \(\frac {5}{3}\) and \(-\frac {7}{4}\) on a single number line.
Solution:
(i) \(\frac{5}{3}=1 \frac{2}{3}\), it lies between 1 and 2.
Divide the interval between 1 and 2 into 3 equal parts and move two parts to the right of 1.
(ii) \(-\frac{7}{4}=-1 \frac{3}{4}\), it lies between -2 and -1.
Divide the interval between -2 and -1 into 4 equal parts and move three parts to the left of -1.

Question 14.
Classify the following numbers as rational or irrational:
(i) √5
(ii) \(-\frac {11}{4}\)
(iii) 1.2727…
Solution:
(i) √5 is irrational because 5 is not a perfect square, so √5 cannot be expressed in \(\frac {p}{q}\) form.
(ii) \(-\frac {11}{4}\) = -2.75, which is a rational number.
(iii) 1.272727…
Since the number is repeating, it is a rational number.

Question 15.
Convert \(2.35 \overline{7}\) into the form \(\frac {p}{q}\).
Solution:
Let x = \(2.35 \overline{7}\)
Since two digits are non-repeating, multiplying by 100, we get
100x = \(235 . \overline{7}\) ……(i)
Since one digit repeats, multiplying by 10, we get
1000x = \(2357 . \overline{7}\) ……(ii)
Subtracting (i) from (ii), we get
1000x – 100x = \(2357 . \overline{7}-235 . \overline{7}\)
⇒ 900x = 2122
⇒ x = \(\frac{2122}{900}=\frac{1061}{450}\)

Question 16.
Convert \(2.45 \overline{37}\) into the form \(\frac {p}{q}\).
Solution:
Let x = \(2.45 \overline{37}\)
Since two digits are non-repeating, multiplying by 100, we get
100x = \(245 . \overline{37}\) ….(i)
Since two digits are repeating, multiplying by 100, we get
10000x = \(24537 . \overline{37}\) ……(ii)
Subtracting (i) from (ii), we get
10000x – 100x = \(24537 . \overline{37}-245 . \overline{37}\)
⇒ 9900x = 24292
⇒ x = \(\frac{24292}{9900}=\frac{6073}{2475}\)

Question 17.
Determine whether \(\frac {72}{625}\) has a terminating decimal expansion or a non-terminating recurring decimal expansion. If it is terminating, then find its decimal form.
Solution:
\(\frac {72}{625}\)
625 = 5⁴
Since the denominator contains only the prime factor 5, the decimal expansion is terminating.
Now, to find its decimal form:
Multiplying numerator and denominator by 2⁴
∴ \(\frac{72}{625}=\frac{72 \times 2^4}{5^4 \times 2^4}=\frac{72 \times 16}{625 \times 16}=\frac{1152}{10,000}\) = 0.1152

The World of Numbers Important Questions Class 9

Long Answer Type Questions

Question 1.
Prove that √3 is an irrational number. Give all necessary steps.
Solution:
We will use the method of contradiction to prove the irrationality of √3.
Step 1: Assumption: Let us assume that √3 is a rational number.
Therefore, it can be written as a fraction \(\frac {p}{q}\), q ≠ 0, in its simplest form,
i.e., p and q are coprime integers.
Hence, √3 = \(\frac {p}{q}\) …….(i)
Step 2: Squaring both sides of equation (i),
we have 3 = \(\frac{p^2}{q^2}\) …….(ii)
Step 3: Multiplying both sides of equation (ii) by q², we obtain
\(3 \times q^2=\frac{p^2}{q^2} \times q^2\)
⇒ 3q² = p² …..(iii)
Step 4: Deduction for p: Because p² is equal to 3 times some integer, p² must have a factor of 3.
If a prime number is a factor of the square of a number, then it is a factor of that number also.
So, 3 is a factor of p.
Let us take p = 3k, for some integer k.
Step 5: Substituting p = 3k in equation (iii), we get
3q² = (3k)²
⇒ 3q² = 9k²
⇒ q² = 3k²
Step 6: Deduction for q: Because q² is equal to 3 times some integer, q² must have a factor of 3.
If a prime number is a factor of the square of a number, then it is a factor of that number also.
So, 3 is a factor of q.
Step 7: The Contradiction: We deduced that p and q have a common factor 3.
This means they are not coprime integers as we had assumed them in step 1.
So, our assumption is wrong, and it is a contradiction.
Therefore, we cannot write √3 as a fraction.
Hence, √3 is irrational.

Question 2.
Construct a line segment of length √2 and mark its position on the number line.
Solution:
We follow the following steps to draw and mark √2 on the number line:
Step 1: On the number line, we measure OA = 1 unit and draw a perpendicular on OA through A.

Step 2: On this perpendicular line, we mark the point B such that AB = 1 unit and join the origin to B.
Clearly, OB = √2 units.
Step 3: With O as centre and OB as radius, with a compass we draw an arc which intersects the number line at P.
Clearly, OP = √2 units.
Hence, P represents the irrational number √2.

Question 3.
Express \(0.3+0 . \overline{7}+0.3 \overline{6}\) in the form of \(\frac {p}{q}\) (q ≠ 0), where p and q are integers.
Solution:
Let x = 0.3 = \(\frac {3}{10}\)
Let y = \(0 . \overline{7}\) ……(i)
Multiplying both sides by 10, we get
10y = \(7 . \overline{7}\) …..(ii)
Subtracting (i) from (ii), we get
10y – y = \(7 . \overline{7}-0 . \overline{7}\)
⇒ 9y = 7
⇒ y = \(\frac {7}{9}\)
Let z = \(0.3 \overline{6}\)
Multiplying both sides by 10, we get
10z = \(3 . \overline{6}\) …..(iii)
Again, multiplying by 10, we get
100z = \(36 . \overline{6}\) ……(iv)
Subtracting (iii) from (iv), we get
100z – 10z = \(36 . \overline{6}-3 . \overline{6}\)
⇒ 90z = 33
⇒ z = \(\frac{33}{90}=\frac{11}{30}\)
∴ \(0.3+0 . \overline{7}+0.3 \overline{6}=\frac{3}{10}+\frac{7}{9}+\frac{11}{30}\)
= \(\frac{27+70+33}{90}\)
= \(\frac {130}{90}\)
= \(\frac {13}{9}\)

Class 9 Maths The World of Numbers Important Questions

Case-Study Based Questions

Question 1.
The temperature in the high-altitude desert of Ladakh drops significantly at night. The drop in temperature may be as high as 25 degrees as compared to the day temperature on some days.

Based on the information given above, answer the following questions:
(i) On Monday, the temperature was recorded as 4°C at noon. By midnight, it dropped by 12°C. What is the midnight temperature?
(ii) On Tuesday, the temperature at noon was recorded as 7°C. What was the rise in temperature as compared to the midnight of last night?
(iii) On Tuesday, the temperature dropped to -12°C at midnight. What was the drop in temperature on Tuesday?
(iv) Find the temperature difference at midnight of Monday and Tuesday.
Solution:
(i) On Monday, temperature at noon = 4 °C
Drop in temperature by midnight = 12 °C
At midnight, temperature = 4°C – 12 °C = -8 °C

(ii) On Tuesday, temperature at noon = 7 °C
Temperature at midnight of last night = -8 °C
Rise in temperature = |-8 °C – 7 °C| = |-15 °C| = 15 °C.

(iii) On Tuesday, temperature at noon = 7 °C
Temperature at midnight = -12 °C
Drop in temperature = |-12 °C – 7 °C| = |-19 °C| = 19 °C.
(iv) On Monday, temperature at midnight = -8 °C
On Tuesday, temperature at midnight = -12 °C
Difference in temperature = |-12 °C – (-8) °C|
= |-12 °C + 8 °C|
= |-4 °C|
= 4 °C

Question 2.

In a video game tournament, a player earns bonus points and loses points due to penalties. Positive integers represent points gained and negative integers represent points lost due to penalties. His performance over 5 rounds is recorded as follows:

The tournament organiser wants to calculate the player’s overall performance. Answer the following questions.
(i) Find the player’s net score after all 5 rounds.
(ii) If only the positive scores are counted for the “Best Player Award”, how many bonus points did the player earn?
(iii) (a) Identify the points lost due to penalties and find the total penalty points lost.
OR
(b) If the player gives the same performance in another set of 5 rounds, what will be his total score for the tournament?
Solution:
(i) Net score = 12 + (-4) + 18 + (-7) + 9
= 12 – 4 + 18 – 7 + 9
= 28
∴ Net score after 5 rounds = 28
(ii) Total bonus points = 12 + 18 + 9 = 39
(iii) (a) Penalty points are -4 and -7.
∴ Total penalty points lost = 4 + 7 = 11
OR
(b) Net score after 5 rounds = 28
If the same performance is repeated: 28 × 2 = 56
∴ Total score for the tournament = 56

Question 3.
During the annual “Catalyst Quest” at school, students must unlock the final laboratory safety vault by solving rational number challenges related to chemical mixtures. Team Alpha receives the following task cards. Read the information carefully and answer the questions that follow.
Clue 1: To initiate the reaction, the team collects \(\frac {5}{6}\) litres of Reagent A and \(\frac {3}{4}\) litres of Reagent B in a mixing beaker.
Clue 2: To stabilise the mixture, the team carefully pours out \(\frac {2}{3}\) litres of the solution into a neutralising tank.
Clue 3: The remaining liquid in the beaker is shared equally between 2 laboratory groups who reached the final round.
Final Bonus Clue:
To unlock the equipment shed, the team must simplify the following expression representing the pressure change using the distributive property: \(\frac{4}{5}\left(\frac{3}{8}+\frac{5}{6}\right)\)

(i) Find the total quantity of solution collected in the mixing beaker initially.
(ii) Find the quantity of solution left after pouring out \(\frac {2}{3}\) litres into the neutralising tank.
(iii) (a) Final Sharing Challenge
Before unlocking the laboratory safety vault, Team Alpha must share the remaining solution equally between 2 laboratory groups. Calculate the quantity of solution received by each group.
OR
(b) Bonus Lock Challenge
To unlock the final laboratory safety vault, the team must solve the bonus expression shown on the control panel: \(\frac{4}{5}\left(\frac{3}{8}+\frac{5}{6}\right)\)
Simplify it using the distributive property.
Solution:
(i) Total quantity of solution collected initially = \(\frac{5}{6}+\frac{3}{4}\)
LCM of 6 and 4 is 12.
∴ \(\frac{5}{6}+\frac{3}{4}=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
∴ Total quantity of solution collected = \(\frac {19}{12}\) litres
(ii) Quantity of solution left after pouring out \(\frac {2}{3}\) litres = \(\frac{19}{12}-\frac{2}{3}\) = \(\frac{19}{12}-\frac{8}{12}\)
[L.C.M of 12 and 3 is 12]
∴ Quantity of solution left = \(\frac {11}{12}\) litres
(iii) (a) Remaining solution = \(\frac {11}{12}\)
∴ Solution received by each group = \(\frac {11}{12}\) ÷ 2
= \(\frac{11}{12} \times \frac{1}{2}\)
= \(\frac {11}{24}\)
∴ Quantity of solution received by each group = \(\frac {11}{24}\) litres
OR
(a) \(\frac{4}{5}\left(\frac{3}{8}+\frac{5}{6}\right)=\frac{4}{5} \times \frac{3}{8}+\frac{4}{5} \times \frac{5}{6}\)
= \(\frac{12}{40}+\frac{20}{30}\)
= \(\frac{3}{10}+\frac{2}{3}\)
L.C.M of 10 and 3 is 30.
∴ \(\frac{3}{10}+\frac{2}{3}=\frac{9}{30}+\frac{20}{30}=\frac{29}{30}\)

Question 4.
A bakery owner maintains a weekly stock record of liquid ingredients used for preparing cakes. Some ingredient quantities are recorded in decimal form, including terminating and repeating decimals.
On Monday, the quantity of chocolate syrup used was 0.5 litre.
On Tuesday, the quantity of vanilla essence used was \(0 . \overline{3}\) litre.
On Wednesday, the quantity of strawberry syrup used was \(0.4 \overline{5}\) litre.
To prepare the final stock report, the owner must convert these decimal values into rational numbers in the form \(\frac {p}{q}\), where p and q are integers and q ≠ 0.

(i) The owner wants to record the chocolate syrup quantity in fraction form. Express 0.5 in the form \(\frac {p}{q}\).
(ii) For the vanilla essence entry, convert \(0 . \overline{3}\) into the form \(\frac {p}{q}\).
(iii) (a) For the strawberry syrup stock entry, express \(0.4 \overline{5}\) in the form \(\frac {p}{q}\).
OR
(b) The owner wants to find the combined quantity of chocolate syrup and vanilla essence used. Express 0.5 + \(0 . \overline{3}\) in the form \(\frac {p}{q}\).
Solution:
(i) 0.5 = \(\frac{5}{10}=\frac{1}{2}\)
The quantity of chocolate syrup = \(\frac {1}{2}\) litres.
(ii) Let x = \(0 . \overline{3}\) …..(i)
Multiplying both sides by 10, we get
10x = \(3 . \overline{3}\) ……(ii)
Subtracting (i) from (ii), we get
10x – x = \(3 . \overline{3}-0 . \overline{3}\)
⇒ 9x = 3
⇒ x = \(\frac{3}{9}=\frac{1}{3}\)
∴ The quantity of vanilla essence is \(\frac {1}{3}\) litres.
(iii) (a) Let y = \(0.4 \overline{5}\)
Multiplying both sides by 10, we get
10y = \(4 . \overline{5}\) …..(i)
Again, multiplying by 10, we get
100y = \(45 . \overline{5}\) ……(ii)
Subtracting (i) from (ii), we get
100y – 10y = \(45 . \overline{5}-4 . \overline{5}\)
⇒ 90y = 41
⇒ y = \(\frac {41}{90}\)
∴ \(0.4 \overline{5}=\frac{41}{90}\)
∴ The quantity of strawberry syrup = \(\frac {41}{90}\) litres.
OR
(b) 0.5 = \(\frac {1}{2}\)
\(0 . \overline{3}=\frac{1}{3}\)
L.C.M. of 2 and 3 is 6.
∴ \(0.5+0 . \overline{3}=\frac{1}{2}+\frac{1}{3}\)
= \(\frac{3}{6}+\frac{2}{6}\)
= \(\frac {5}{6}\)
∴ Combined quantity of chocolate syrup and vanilla essence = \(\frac {5}{6}\) litres.

Self Assessment

Very Short Answer Type Questions

Question 1.
Aarya went to a shop. She purchased 12 packets of pens containing 10 pens in each packet. How many pens will she have in total?
Answer:
120

Question 2.
There is a rectangle. The rectangle has length 4 cm and width 3 cm. By cutting its length and width into parts of 1cm each, how many squares of area 1 cm2 can be made?
Answer:
12

Question 3.
How many maximum number of digits possible in the repeating block of the decimal representation of \(\frac {1}{11}\)?
Answer:
10

Short Answer Type Questions

Question 1.
Find two rational numbers between \(\frac {2}{3}\) and \(\frac {5}{7}\).
Answer:
\(\frac{43}{63}, \frac{44}{63}\)

Question 2.
Write two irrational numbers between 7 and 8.
Answer:
7.454454445…, 7.878878887…

Question 3.
Find the value of x, if \(\frac{x}{2}\left(\frac{3}{2}-\frac{1}{2}\right)\) = 10.
Answer:
20

Long Answer Type Questions

Question 1.
Represent √5 on the number line.
Answer:

Question 2.
Prove that √5 is irrational.
Answer:
Use the method of Contradiction.

Case Study Based Questions

Question 1.
On average, the temperature inside the Earth increases by approximately 1 °C for every 32 m to 40 m. This rate of increase is known as the geothermal gradient. At a place there is an underground mine. The geothermal gradient at this place is 1 °C for each 35 m. On a particular day, the temperature at the ground level is 29 °C. On the other hand, the temperature decreases with the increase in height from the ground level by an average of 1 °C for every 165 m.

Based on the information given above, answer the following questions:
(i) In this mine, if there is a point A, which is below 70 m, what will be the temperature at A?
(ii) In this mine, there is another point B below 1 km and 50 m. What will be the temperature at B?
(iii) A point C is 330 m above the Earth’s surface. What is the difference in temperature between A and C?
(iv) What is the difference between B and C?
Answer:
(i) 31 °C
(ii) 59 °C
(iii) 58 °C
(iv) 86 °C

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