The World of Numbers Class 9 Chapter 3 (Ganita Manjari I)

A merchant in the port city of Lothal is exchanging bags of spices for copper ingots. He receives 15 ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with?
Given that he receives 15 ingots for every 2 bags of spices
Thus,
Cost of 2 bags of spices = 15 ingots

Look at the sequence of numbers on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern.
The Ishango bone has numbers 11, 13, 17, 19
All these are prime numbers

We know that Natural Numbers are closed under addition
(the sum of any two natural numbers is always a natural number). Are they closed under subtraction? Provide a couple of examples to justify your answer
Closure is when an operation (eg: "adding") on members of a set (eg: natural numbers) always makes a member of the same set.

Ancient Indians used the joints of their fingers to count, a practice still seen today. Each finger has 3 joints, and the thumb is used to count them. How many can you count on one hand? How does this relate to the ancient base-12 counting systems?
Thus, we can count till 12 in one hand

The temperature in the high-altitude desert of Ladakh is recorded as 4 °C at noon. By midnight, it drops by 15 °C. What is the midnight
temperature?
Initial temperature = 4 °C
Temperature drop at midnight = 15 °C

A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an equation using integers and calculate his final financial standing.
Let’s look at the signs of debt, fortune and loss
Debt is negative
Profit is positive
Loss is negative

(i) Calculate the following using Brahmagupta’s laws:
(i) (–12) × 5
Since Positive × Negative = Negative

Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number
(e.g., 10 – (–5) = 15).
5. THE PRODUCT OF TWO DEBTS IS A FORTUNE (Negative * Negative $=$ Positive) *The trickiest one.

(i) Prove that the following rational numbers are equal:
(i) 2/3 and 4/6
To check if they are equal, we make them equal and cross-multiply

(i) Find the sum:
(i) 2/5 +3/10
2/5 +3/10
= (𝟐 × 𝟏𝟎 + 𝟑 × 𝟓)/(𝟓 × 𝟏𝟎)
= (20 + 15)/50
= 𝟑𝟓/𝟓𝟎
Simplifying
= 35/50
= 𝟕/𝟏𝟎
Note:
We could directly take
LCM of 5 and 10 = 10

(i) Find the difference:
(i) 5/6 −1/4
5/6 −1/4
= (𝟓 × 𝟒 − 𝟏 × 𝟔)/(𝟔 × 𝟒)
= (20 − 6)/24
= 𝟏𝟒/𝟏𝟐
Simplifying
= 14/24
= 𝟕/𝟏𝟐
Note:
We could directly take
LCM of 6 and 4 = 12

(i) Find the product:
(i) 2/3×3/10
2/3×3/10
= (𝟐 × 𝟑)/(𝟑 × 𝟏𝟎)
= (2 × 1)/(1 × 10)
= 2/10
= 𝟏/𝟓

(i)
Find the quotient:
(i) 2/3÷3/10
2/3÷3/10
= 𝟐/𝟑 ×𝟏𝟎/𝟑
= (2 × 10)/(3 × 3)
= 𝟐𝟎/𝟗
Ex 3.3, 5 (ii)
Find the quotient:
(ii) 7/11÷5/8
7/11÷5/8
= 𝟕/𝟏𝟏 ×𝟖/𝟓
= (7 × 8)/(11 × 5)
= 𝟓𝟔/𝟓𝟓
Ex 3.3, 5 (iii)
Find the quotient:
(iii) −4/7÷5/14
−4/7÷5/14
= −𝟒/𝟕 ×𝟏𝟒/𝟓
= (−4 × 14)/(7 × 5)
= (−4 × 2)/(1 × 5)
= (−𝟖)/𝟓

Show that: (1/2+3/4)×8/3=1/2×8/3+3/4×8/3.
To prove this, we solve the Left Hand Side (LHS) and Right Hand Side (RHS) separately and prove that they are equal

Simplify the following using the distributive property:
7/9 (6/7−3/4).
In Distributive property, we distribute the outside term to both terms in the product
Now,
𝟕/𝟗 (𝟔/𝟕−𝟑/𝟒)=7/9 × (6/7−3/4)
=𝟕/𝟗 ×𝟔/𝟕−𝟕/𝟗 ×𝟑/𝟒
=(7 × 6)/(9 × 7)−(7 × 3)/(9 × 4)
=(1 × 6)/(9 × 1)−(7 × 1)/(3 × 4)
=6/9−7/12
=𝟐/𝟑−𝟕/𝟏𝟐
Since LCM of 3 and 12 is 12
=2/3 ×4/4−7/12
=8/12−7/12
=(8 − 7)/12
=𝟏/𝟏𝟐

Find the rational number 𝑥 such that:
5/6 (𝑥+3/5)=5/6 𝑥+1/2.
Using Distributive property on let hand side
We distribute the outside term to both terms in the product

Represent the rational numbers 2/3,−5/4 and 1 1/2 on a single number line.
To represent all of them in one number line, we make their denominators same
Our rational numbers are
𝟐/𝟑,
(−𝟓)/𝟒 and
1 1/2=1+1/2=(1 × 2 + 1)/2=(2 + 1)/2=𝟑/𝟐

Find three distinct rational numbers that lie strictly between (−1)/2 and 1/4.
Since numbers don’t have the same denominator
We make denominator same

Simplify the expression: (−1/4)+(5/12).
Since denominators of both fractions are different
Frist, we make denominators same

A tailor has 15 3/4 metres of fine silk. If making one kurta requires 2 1/4 metres of silk, exactly how many kurtas can he make?
First, we convert both mixed fractions into normal fraction

Find three rational numbers between 3.1415 and 3.1416.
To find numbers between two decimals,
We add another 0 behind decimal point and write down the gap numbers

Can you think of other way(s) to find a rational number between
any two rational numbers?
We found rational numbers beween any two numbers by Common denominator & Scaling method

Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.
A rational number 𝑝/𝑞 to be terminating decimal
We check the prime factorisation of denominator
If Denominator = q = 2n 5m
where n ≥ 0 & m ≥ 0
Then, our fraction is terminating

Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13? Now compute 3/13, 4/13, etc. What do you notice?
Let’s divide 𝟏/𝟏𝟑
We notice that
1/13 = 0.(𝟎𝟕𝟔𝟗𝟐𝟑) ̅

(i) Classify the following numbers as rational or irrational:
(i) √81
Simplifying our number
√𝟖𝟏 = √(9^2 )
= 9

The number 0.9 ̅ (which means 0.99999 … ) is a rational number.
Using algebra (let x = 0.9 ̅, multiply by 10, and subtract), explain
why 0.9 ̅ is exactly equal to 1.
Let
x = 0.999…

We have seen that the repeating block of 1/7 is a cyclic number.
Try to find more numbers (n) whose reciprocals (1/𝑛) produce decimals with repeating blocks that are cyclic.
We saw that 1/7 produces the cyclic block 142857.
Since 7 is a prime number, lets try others

(i) Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(i) 3/50
Thus,
3/50=𝟎.𝟎𝟔
So, it is terminating decimal expansion
Question 1 (ii) Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:
(ii) 2/9
Thus,
2/9=0.222…=𝟎.𝟐 ̅
So, it is non-terminating repeating decimal expansion

Prove that √5 is an irrational number.
We have to prove √5 is irrational
Let us assume the opposite,
i.e., √𝟓 is rational

(i) Convert the following decimal numbers in the form of 𝑝/𝑞.
(i) 12.6
Now,
12.6 = 126/10
= 𝟔𝟑/𝟓
Question 3 (ii) Convert the following decimal numbers in the form of 𝑝/𝑞.
(ii) 0.0120
Now,
0.0120 = 0120/10000
= 120/10000
= 𝟏𝟐/𝟏𝟎𝟎𝟎
= 6/500
= 𝟑/𝟐𝟓𝟎
Question 3 (iii) Convert the following decimal numbers in the form of 𝑝/𝑞.
(iii) 3.0(52) ̅
Let
x = 3.0525252…..

(i) Locate the following rational numbers on the number line.
(i) 0.532
To plot 0.532, we just magnify our number line step-by-step
Using successive magnification
Zoom 1 (Tenths): 0.532 lies between 0 and 1. We zoom into that space and divide it into 10 parts. Our number lies right between 0.5 and 0.6.
Zoom 2 (Hundredths): We take the tiny segment between 0.5 and 0.6 and magnify it, dividing it into 10 smaller parts. 0.532 lies between 0.53 and 0.54.
Zoom 3 (Thousandths): Finally, we magnify the space between 0.53 and 0.54. We count two ticks over to accurately plot our point at exactly 0.532.

Find 6 rational numbers between 3 and 4.
As we have to find 6 rational numbers, we multiply the numbers by 𝟕/𝟕
So, our numbers become
3 = 3 × 7/7
& 4 = 4 × 7/7

Find 5 rational numbers between 2/5 and 3/5.
As we have to find 5 rational numbers, we multiply the numbers by 𝟔/𝟔
So, our numbers become
2/5 = 2/5 × 6/6
3/5 = 3/5 × 6/6

Find 5 rational numbers between 1/6 and 2/5.
First, we make the numbers have same denominator
Now,
Common denominator = LCM of 6 & 5
= 2 × 3 × 5
= 30

If 𝑥/3+𝑥/5=16/15, find the rational number 𝑥.
Now,
𝑥/3+𝑥/5=16/15
𝒙 × (𝟏/𝟑+𝟏/𝟓)=𝟏𝟔/𝟏𝟓
𝑥 × ((1 × 5 + 3 × 1)/(3 × 5))=16/15
𝑥 × ((5 + 3)/15)=16/15
𝒙 ×𝟖/𝟏𝟓=𝟏𝟔/𝟏𝟓
𝑥 =16/15 ×15/8

Let a and b be two non-zero rational numbers such that 𝑎+1/𝑏=0
Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
Given that 𝒂+𝟏/𝒃=𝟎, and a ≠ 0, b ≠ 0

A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form 𝑝/(104 ), where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 24 or 54? Give reasons.
Let’s do it in two parts

Without performing division, determine whether the decimal
expansion of 18/(125 ) is terminating or non-terminating. If it terminates, state the number of decimal places.
A rational number 𝑝/𝑞 to be terminating decimal
We check the prime factorisation of denominator
If Denominator = q = 2n 5m
where n ≥ 0 & m ≥ 0
Then, our fraction is terminating

A rational number in its lowest form has denominator 23 × 5. How
many decimal places will its decimal expansion have? Explain your answer
To find the exact decimal value of a fraction without long division, we try to turn the denominator into a perfect power of 10

Let 𝑎=7/(12 ) and 𝑏=5/(6 ) Express both a and b in the form 𝑘_1/(𝑚 ) and 𝑘_2/(𝑚 ) where 𝑘_1, 𝑘_2 and 𝑚 and are integers and 𝑘_2−𝑘_1>6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition 𝑘_2−𝑘_1>𝑛+1 is necessary to find n such rational
numbers between the two rational numbers a and b using this
method.
Simplifying the question
We need to find 5 rational numbers between 7/(12 ) and 5/(6 )
First, the question says make denominator common
Then, it asks to explain the condition 𝑘_2−𝑘_1>𝑛+1
Let’s do this
Since numbers 7/(12 ) and 5/(6 ) don’t have the same denominator
We make denominator same

Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
We have an identity with 3 variables, it is
(𝒙+𝒚+𝒛)^𝟐=𝒙^𝟐+𝒚^𝟐+𝒛^𝟐+𝟐(𝒙𝒚+𝒚𝒛+𝒛𝒙)

Show that the rational number ((𝑎+𝑏))/(2 ) lies between the rational
numbers a and b.
This is asking us to prove that the average of two numbers will always sit exactly between them on the number line.
We can prove this using a little bit of algebral

Find the lengths of the hypotenuses of all the right triangles in
Fig. 3.14 which is referred to as the square root spiral.
The Square Root Spiral is a beautiful geometric trick for drawing irrational lengths using only a ruler and the Pythagorean theorem (a2 + b2 = c2)