Question 1

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?

Accepted Answer

Given: 15 ingots for every 2 bags of spices.

To find: Number of copper ingots for 12 bags of spices.

Step 1: Find the number of ingots per bag = 15/2 = 7.5 ingots per bag.
Step 2: For 12 bags, total ingots = 7.5 × 12 = 90 ingots.

Therefore, the merchant will leave with 90 copper ingots. — The ratio of ingots to bags is 15:2, so for each bag, the merchant gets 7.5 ingots. Multiplying by 12 bags gives 90 ingots.

Question 2

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.

Accepted Answer

The numbers 11, 13, 17, and 19 are all prime numbers. They are consecutive prime numbers starting from 11.

The next three prime numbers after 19 are 23, 29, and 31.

Therefore, the next three numbers in the pattern are 23, 29, and 31. — All given numbers are prime numbers (numbers greater than 1 that have no divisors other than 1 and themselves). The sequence lists consecutive primes starting from 11.

Question 3

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.

Accepted Answer

Natural numbers are not closed under subtraction.

Example 1: 5 and 3 are natural numbers.
5 - 3 = 2, which is a natural number.

Example 2: 3 and 5 are natural numbers.
3 - 5 = -2, which is not a natural number.

Since subtraction of two natural numbers can result in a number that is not natural (like a negative number), natural numbers are not closed under subtraction. — Closure under an operation means the result of the operation on members of the set is always in the set. Since subtraction can produce negative numbers which are not natural numbers, natural numbers are not closed under subtraction.

Question 4

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?

Accepted Answer

Each finger (except the thumb) has 3 joints. There are 4 fingers (index, middle, ring, little) on one hand.

Number of joints counted = 4 fingers × 3 joints each = 12 joints.

The thumb is used to point to each joint to count.

This method allows counting up to 12 on one hand, which relates to the ancient base-12 (duodecimal) counting system used by Indians and other ancient cultures.

Thus, the finger-joint counting method naturally supports the base-12 system. — Counting finger joints with the thumb gives a total of 12 countable units on one hand, which aligns with the base-12 system historically used in India.

Question 5

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?

Accepted Answer

Given noon temperature = 4 °C
Temperature drop = 15 °C
Midnight temperature = Noon temperature - Temperature drop = 4 - 15 = -11 °C
So, the midnight temperature is -11 °C. — Since the temperature drops by 15 °C from 4 °C, subtract 15 from 4 to get the midnight temperature: 4 - 15 = -11 °C.

Question 6

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.

Accepted Answer

Let the loan (debt) be represented as -850 (since it is a debt).
Profit is +1200.
Loss is -450.
Equation: -850 + 1200 - 450
Calculate:
-850 + 1200 = 350
350 - 450 = -100
Final financial standing is -₹100, meaning he still owes ₹100. — Debt is negative, profit positive, loss negative. Adding these: (-850) + 1200 - 450 = 350 - 450 = -100. So, the trader is in debt by ₹100 after all transactions.

Question 7

Calculate the following using Brahmagupta's laws:

(i) (–12) × 5
(ii) (–8) × (–7)
(iii) 0 – (–14)
(iv) (–20) ÷ 4

Accepted Answer

(i) (–12) × 5 = –60
Since negative × positive = negative.

(ii) (–8) × (–7) = 56
Since negative × negative = positive.

(iii) 0 – (–14) = 0 + 14 = 14
Subtracting a negative number is same as adding a positive number.

(iv) (–20) ÷ 4 = –5
Negative divided by positive is negative. — Apply Brahmagupta's rules:
- Negative × Positive = Negative
- Negative × Negative = Positive
- Subtracting a negative number equals adding the positive
- Negative ÷ Positive = Negative
Calculate each accordingly.

Question 8

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).

Accepted Answer

Subtracting a negative number means removing a debt, which increases the amount you have. For example, if you have ₹10 and someone forgives a debt of ₹5 (which is like subtracting –5), your total money increases to ₹15.
Mathematically, 10 – (–5) = 10 + 5 = 15.
Thus, subtracting a negative number is equivalent to adding a positive number. — In real life, if you owe money (negative), and that debt is removed (subtracting a negative), your net amount increases. This is why subtracting a negative number adds to the total.

Question 9

While adding or subtracting two rational numbers having different denominators, how will you make the denominators equal?

Accepted Answer

To add or subtract two rational numbers with different denominators, we find the Least Common Denominator (LCD) of the two denominators. Then, convert each fraction to an equivalent fraction with the LCD as the denominator by multiplying numerator and denominator appropriately. After this, add or subtract the numerators and keep the denominator same.

For example, to add $ \frac{a}{b} + \frac{c}{d} $, find LCD of b and d, say L. Then,
$ \frac{a}{b} = \frac{a \times (L/b)}{L} $ and $ \frac{c}{d} = \frac{c \times (L/d)}{L} $.
Now add: $ \frac{a \times (L/b) + c \times (L/d)}{L} $. — Making denominators equal allows direct addition or subtraction of numerators. The LCD is the smallest number divisible by both denominators, ensuring equivalent fractions.

Question 10

Verify the distributive law for rational numbers.

Accepted Answer

The distributive law states that for rational numbers p, q, and r:

p(q + r) = pq + pr

Verification:
Let p = $ \frac{a}{b} $, q = $ \frac{c}{d} $, r = $ \frac{e}{f} $, where a,b,c,d,e,f are integers and denominators non-zero.

Calculate left side:
$ p(q + r) = \frac{a}{b} \times \left( \frac{c}{d} + \frac{e}{f} \right) = \frac{a}{b} \times \frac{cf + ed}{df} = \frac{a(cf + ed)}{bdf} $

Calculate right side:
$ pq + pr = \frac{a}{b} \times \frac{c}{d} + \frac{a}{b} \times \frac{e}{f} = \frac{ac}{bd} + \frac{ae}{bf} = \frac{acf}{bdf} + \frac{ade}{bdf} = \frac{a(cf + ed)}{bdf} $

Since left side equals right side, distributive law holds for rational numbers. — By expressing rational numbers as fractions and performing algebraic operations, both sides of the distributive law expression simplify to the same fraction, confirming the law.

Question 11

Prove that the following rational numbers are equal:

(i)  $\frac{2}{3}$  and  $\frac{4}{6}$

(ii)  $\frac{5}{4}$  and  $\frac{10}{8}$

(iii)  $-\frac{3}{5}$  and  $-\frac{6}{10}$

(iv)  $\frac{9}{3}$  and 3

Accepted Answer

Solution:
(i) $\frac{2}{3} = \frac{4}{6}$
Multiply numerator and denominator of $\frac{2}{3}$ by 2:
$\frac{2 \times 2}{3 \times 2} = \frac{4}{6}$.
Hence, they are equal.

(ii) $\frac{5}{4} = \frac{10}{8}$
Multiply numerator and denominator of $\frac{5}{4}$ by 2:
$\frac{5 \times 2}{4 \times 2} = \frac{10}{8}$.
Hence, they are equal.

(iii) $-\frac{3}{5} = -\frac{6}{10}$
Multiply numerator and denominator of $-\frac{3}{5}$ by 2:
$-\frac{3 \times 2}{5 \times 2} = -\frac{6}{10}$.
Hence, they are equal.

(iv) $\frac{9}{3} = 3$
Divide numerator by denominator:
$9 \div 3 = 3$.
Hence, they are equal. — To prove two rational numbers are equal, we check if their cross products are equal or if one can be obtained by multiplying numerator and denominator of the other by the same number.

(i) $2 \times 6 = 12$ and $3 \times 4 = 12$, so equal.
(ii) $5 \times 8 = 40$ and $4 \times 10 = 40$, so equal.
(iii) $-3 \times 10 = -30$ and $5 \times -6 = -30$, so equal.
(iv) $9 \div 3 = 3$, so equal.

Question 12

Find the sum:

(i)  $\frac{2}{5} + \frac{3}{10}$

(ii)  $\frac{7}{12} + \frac{5}{8}$

(iii)  $-\frac{4}{7} + \frac{3}{14}$

Accepted Answer

Solution:
(i) $\frac{2}{5} + \frac{3}{10}$
LCM of 5 and 10 is 10.
$\frac{2}{5} = \frac{4}{10}$, so sum = $\frac{4}{10} + \frac{3}{10} = \frac{7}{10}$.

(ii) $\frac{7}{12} + \frac{5}{8}$
LCM of 12 and 8 is 24.
$\frac{7}{12} = \frac{14}{24}, \frac{5}{8} = \frac{15}{24}$
Sum = $\frac{14}{24} + \frac{15}{24} = \frac{29}{24} = 1 \frac{5}{24}$.

(iii) $-\frac{4}{7} + \frac{3}{14}$
LCM of 7 and 14 is 14.
$-\frac{4}{7} = -\frac{8}{14}$
Sum = $-\frac{8}{14} + \frac{3}{14} = -\frac{5}{14}$. — Add rational numbers by finding the LCM of denominators, converting to equivalent fractions, then adding numerators.

(i) LCM(5,10)=10, convert $\frac{2}{5}$ to $\frac{4}{10}$, sum $\frac{4}{10} + \frac{3}{10} = \frac{7}{10}$.
(ii) LCM(12,8)=24, convert $\frac{7}{12} = \frac{14}{24}$, $\frac{5}{8} = \frac{15}{24}$, sum $\frac{29}{24}$.
(iii) LCM(7,14)=14, convert $-\frac{4}{7} = -\frac{8}{14}$, sum $-\frac{5}{14}$.

Question 13

Find the difference:

(i)  $\frac{5}{6} - \frac{1}{4}$

(ii)  $\frac{11}{8} - \frac{3}{4}$

(iii)  $-\frac{7}{9} - \left(-\frac{2}{3}\right)$

Accepted Answer

Solution:
(i) $\frac{5}{6} - \frac{1}{4}$
LCM of 6 and 4 is 12.
$\frac{5}{6} = \frac{10}{12}, \frac{1}{4} = \frac{3}{12}$
Difference = $\frac{10}{12} - \frac{3}{12} = \frac{7}{12}$.

(ii) $\frac{11}{8} - \frac{3}{4}$
LCM of 8 and 4 is 8.
$\frac{3}{4} = \frac{6}{8}$
Difference = $\frac{11}{8} - \frac{6}{8} = \frac{5}{8}$.

(iii) $-\frac{7}{9} - \left(-\frac{2}{3}\right) = -\frac{7}{9} + \frac{2}{3}$
LCM of 9 and 3 is 9.
$\frac{2}{3} = \frac{6}{9}$
Sum = $-\frac{7}{9} + \frac{6}{9} = -\frac{1}{9}$. — Subtract rational numbers by finding LCM of denominators, converting to equivalent fractions, then subtracting numerators.

(i) LCM(6,4)=12, $\frac{5}{6} = \frac{10}{12}$, $\frac{1}{4} = \frac{3}{12}$, difference $\frac{7}{12}$.
(ii) LCM(8,4)=8, $\frac{3}{4} = \frac{6}{8}$, difference $\frac{5}{8}$.
(iii) $-\frac{7}{9} - (-\frac{2}{3}) = -\frac{7}{9} + \frac{6}{9} = -\frac{1}{9}$.

Question 14

Find the product:

(i)  $\frac{2}{3} \times \frac{3}{10}$

(ii)  $\frac{7}{11} \times \frac{5}{8}$

(iii)  $-\frac{4}{7} \times \frac{5}{14}$

Accepted Answer

Solution:
(i) $\frac{2}{3} \times \frac{3}{10} = \frac{2 \times 3}{3 \times 10} = \frac{6}{30} = \frac{1}{5}$.

(ii) $\frac{7}{11} \times \frac{5}{8} = \frac{7 \times 5}{11 \times 8} = \frac{35}{88}$.

(iii) $-\frac{4}{7} \times \frac{5}{14} = -\frac{4 \times 5}{7 \times 14} = -\frac{20}{98} = -\frac{10}{49}$. — Multiply numerators and denominators directly and simplify.

(i) $2 \times 3 = 6$, $3 \times 10 = 30$, simplify $\frac{6}{30} = \frac{1}{5}$.
(ii) $7 \times 5 = 35$, $11 \times 8 = 88$, fraction $\frac{35}{88}$.
(iii) $-4 \times 5 = -20$, $7 \times 14 = 98$, simplify $-\frac{20}{98} = -\frac{10}{49}$.

Question 15

Find the quotient:

(i)  $\frac{2}{3} \div \frac{3}{10}$

(ii)  $\frac{7}{11} \div \frac{5}{8}$

(iii)  $-\frac{4}{7} \div \frac{5}{14}$

Accepted Answer

Solution:
(i) $\frac{2}{3} \div \frac{3}{10} = \frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$.

(ii) $\frac{7}{11} \div \frac{5}{8} = \frac{7}{11} \times \frac{8}{5} = \frac{56}{55}$.

(iii) $-\frac{4}{7} \div \frac{5}{14} = -\frac{4}{7} \times \frac{14}{5} = -\frac{56}{35} = -\frac{8}{5}$. — Divide rational numbers by multiplying the first by the reciprocal of the second.

(i) Reciprocal of $\frac{3}{10}$ is $\frac{10}{3}$, multiply $\frac{2}{3} \times \frac{10}{3} = \frac{20}{9}$.
(ii) Reciprocal of $\frac{5}{8}$ is $\frac{8}{5}$, multiply $\frac{7}{11} \times \frac{8}{5} = \frac{56}{55}$.
(iii) Reciprocal of $\frac{5}{14}$ is $\frac{14}{5}$, multiply $-\frac{4}{7} \times \frac{14}{5} = -\frac{56}{35} = -\frac{8}{5}$.

The Dawn of Mathematics: The Human Need to Count

The Dawn of Mathematics: The Human Need to Count — Study Notes

3.1. THE DAWN OF MATHEMATICS: THE HUMAN NEED TO COUNT

3.1.1 A History Written in Bone

3.1.2 The Indian Context: Trade and Astronomy

Practice Questions — The Dawn of Mathematics: The Human Need to Count

All 8 Chapters in Mathematics