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The World of Numbers

CBSE · Class 9 · Mathematics

NCERT Solutions for The World of Numbers — CBSE Class 9 Mathematics.

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78 Questions Solved · 6 Sections

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EXERCISE SET 3.1

2Look 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.Show solution
The numbers 11,13,17,1911, 13, 17, 19 are all prime numbers: each has exactly two factors, 11 and itself.

The next three prime numbers after 1919 are 23, 29, 31.

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3We 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.Show solution
No, natural numbers are not closed under subtraction.

Examples:
- 5−2=35 - 2 = 3, which is a natural number.
- But 2−5=−32 - 5 = -3, which is not a natural number.
- Also, 1−1=01 - 1 = 0, and 00 is not included in the set of natural numbers given in the chapter.

So subtraction does not always give a natural number.

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4Ancient 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?Show solution
On one hand, there are 4 fingers besides the thumb, and each finger has 3 joints.

So the total you can count is:
- 4×3=124 \times 3 = 12

This relates to the ancient base-12 counting system because the thumb is used to count the 12 joints on the four fingers.

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EXERCISE SET 3.2

1The 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?Show solution
The temperature drops by 15∘C15^\circ\text{C} from 4∘C4^\circ\text{C}.

So,
4−15=−114 - 15 = -11

Therefore, the midnight temperature is **−11∘C-11^\circ\text{C}**.

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2A 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.Show solution
Take the loan as a debt and profit as a fortune:

−850+1200−450-850 + 1200 - 450

First,
−850+1200=350-850 + 1200 = 350

Then,
350−450=−100350 - 450 = -100

So his final financial standing is **−₹100-₹100**, which means he still has a debt of ₹100.

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3(i)(–12) × 5Show solution
Using Brahmagupta's laws:

(−12)×5=−60(-12) \times 5 = -60

A debt times a fortune is a debt.

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3(ii)(–8) × (–7)Show solution
(−8)×(−7)=56(-8) \times (-7) = 56

The product of two debts is a fortune, so the answer is positive.

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3(iii)0 – (–14)Show solution
Subtracting a negative number becomes addition:

0−(−14)=0+14=140 - (-14) = 0 + 14 = 14

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3(iv)(–20) ÷ 4Show solution
(−20)÷4=−5(-20) \div 4 = -5

A debt divided by a fortune is a debt.

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4Explain, 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).Show solution
If you owe someone ₹5, and later that debt is removed, it is like your money increases by ₹5.

For example:
- Start with ₹10.
- If a debt of ₹5 is taken away, you have the same effect as adding ₹5.

So,
10−(−5)=10+5=1510 - (-5) = 10 + 5 = 15

That is why subtracting a negative number is the same as adding a positive number.

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3Calculate the following using Brahmagupta's laws:Show solution
Using Brahmagupta's laws:

(i) (−12)×5=−60(-12) \times 5 = -60

(ii) (−8)×(−7)=56(-8) \times (-7) = 56

(iii) 0−(−14)=140 - (-14) = 14

(iv) (−20)÷4=−5(-20) \div 4 = -5

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EXERCISE SET 3.3

1(i)23\frac{2}{3} and 46\frac{4}{6}Show solution
To prove equality, check cross-multiplication:

2×6=122 \times 6 = 12
3×4=123 \times 4 = 12

Since ad=bcad = bc, the rational numbers are equal.

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1(ii)54\frac{5}{4} and 108\frac{10}{8}Show solution
Check:

5×8=405 \times 8 = 40
4×10=404 \times 10 = 40

So 54\frac{5}{4} and 108\frac{10}{8} are equal.

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1(iii)−35-\frac{3}{5} and −610-\frac{6}{10}Show solution
Check:

(−3)×10=−30(-3) \times 10 = -30
5×(−6)=−305 \times (-6) = -30

So the two rational numbers are equal.

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1(iv)93\frac{9}{3} and 3Show solution
93=3\frac{9}{3} = 3

So the two numbers are equal.

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2(i)25+310\frac{2}{5} + \frac{3}{10}Show solution
Find a common denominator:

25=410\frac{2}{5} = \frac{4}{10}

So,
25+310=410+310=710\frac{2}{5} + \frac{3}{10} = \frac{4}{10} + \frac{3}{10} = \frac{7}{10}

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2(ii)712+58\frac{7}{12} + \frac{5}{8}Show solution
Find a common denominator of 2424:

712=1424,58=1524\frac{7}{12} = \frac{14}{24}, \qquad \frac{5}{8} = \frac{15}{24}

Then,
1424+1524=2924\frac{14}{24} + \frac{15}{24} = \frac{29}{24}

So the sum is **2924\frac{29}{24}**.

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3(i)56−14\frac{5}{6} - \frac{1}{4}Show solution
Find a common denominator of 1212:

56=1012,14=312\frac{5}{6} = \frac{10}{12}, \qquad \frac{1}{4} = \frac{3}{12}

Then,
1012−312=712\frac{10}{12} - \frac{3}{12} = \frac{7}{12}

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3(ii)118−34\frac{11}{8} - \frac{3}{4}Show solution
118−34=118−68=58\frac{11}{8} - \frac{3}{4} = \frac{11}{8} - \frac{6}{8} = \frac{5}{8}

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4(i)23imes310\frac{2}{3} imes \frac{3}{10}Show solution

23×310=2×33×10=210=15\frac{2}{3} \times \frac{3}{10} = \frac{2 \times 3}{3 \times 10} = \frac{2}{10} = \frac{1}{5}

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6Show that: \left(\frac{1}{2} + \frac{3}{4} ight) imes \frac{8}{3} = \frac{1}{2} imes \frac{8}{3} + \frac{3}{4} imes \frac{8}{3}.Show solution
Using the distributive property:

Left-hand side:
(12+34)×83\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3}
First add inside the bracket:
12=24\frac{1}{2} = \frac{2}{4}
24+34=54\frac{2}{4} + \frac{3}{4} = \frac{5}{4}
So,
54×83=4012=103\frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}

Right-hand side:
12×83+34×83\frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}
=86+2412=43+2=103= \frac{8}{6} + \frac{24}{12} = \frac{4}{3} + 2 = \frac{10}{3}

Both sides are equal, so the distributive law is verified.

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7Simplify the following using the distributive property:

79(67−34).\frac{7}{9} \left( \frac{6}{7} - \frac{3}{4} \right).Show solution
Apply distributive property or simplify inside the bracket first:

67−34=2428−2128=328\frac{6}{7} - \frac{3}{4} = \frac{24}{28} - \frac{21}{28} = \frac{3}{28}

Then multiply:
79×328=21252=112\frac{7}{9} \times \frac{3}{28} = \frac{21}{252} = \frac{1}{12}

So the simplified value is **112\frac{1}{12}**.

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8Find the rational number xx such that: 56(x+35)=56x+12\frac{5}{6} \left( x + \frac{3}{5} \right) = \frac{5}{6} x + \frac{1}{2}.Show solution
Expand the left side:

56(x+35)=56x+56×35\frac{5}{6}\left(x+\frac{3}{5}\right)=\frac{5}{6}x+\frac{5}{6}\times\frac{3}{5}

Now,
56×35=36=12\frac{5}{6}\times\frac{3}{5}=\frac{3}{6}=\frac{1}{2}

So the equation becomes:
56x+12=56x+12\frac{5}{6}x+\frac{1}{2}=\frac{5}{6}x+\frac{1}{2}

This is true for **every rational number xx. Therefore, there is no unique value**; any rational number satisfies it.

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2(iii)−47+314-\frac{4}{7} + \frac{3}{14}Show solution
Bring to a common denominator:

−47=−814-\frac{4}{7} = -\frac{8}{14}

So,

−814+314=−514-\frac{8}{14} + \frac{3}{14} = -\frac{5}{14}

Correction: compute carefully,

−8+3=−5-8 + 3 = -5, hence the result is −514-\frac{5}{14}.

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3(iii)−79−(−23)-\frac{7}{9} - \left(-\frac{2}{3}\right)Show solution
Convert to a common denominator:

−23=−69-\frac{2}{3} = -\frac{6}{9}

So,

−79−(−69)=−79+69=−19-\frac{7}{9} - \left(-\frac{6}{9}\right) = -\frac{7}{9} + \frac{6}{9} = -\frac{1}{9}

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4(i)23×310\frac{2}{3} \times \frac{3}{10}Show solution
Multiply the numerators and denominators:

23×310=2×33×10=630=15\frac{2}{3} \times \frac{3}{10} = \frac{2\times 3}{3\times 10} = \frac{6}{30} = \frac{1}{5}

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4(ii)711×58\frac{7}{11} \times \frac{5}{8}Show solution
Multiply the numerators and denominators:

711×58=7×511×8=3588\frac{7}{11} \times \frac{5}{8} = \frac{7\times 5}{11\times 8} = \frac{35}{88}

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4(iii)−47×514-\frac{4}{7} \times \frac{5}{14}Show solution
Multiply the numerators and denominators:

−47×514=−4×57×14=−2098=−1049-\frac{4}{7} \times \frac{5}{14} = \frac{-4\times 5}{7\times 14} = -\frac{20}{98} = -\frac{10}{49}

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6Show that: (12+34)×83=12×83+34×83\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}.Show solution
Use the distributive property.

First simplify the left-hand side:

(12+34)×83\left(\frac{1}{2}+\frac{3}{4}\right)\times\frac{8}{3}

=(24+34)×83= \left(\frac{2}{4}+\frac{3}{4}\right)\times\frac{8}{3}

=54×83= \frac{5}{4}\times\frac{8}{3}

=4012=103= \frac{40}{12} = \frac{10}{3}

Now the right-hand side:

12×83+34×83\frac{1}{2}\times\frac{8}{3} + \frac{3}{4}\times\frac{8}{3}

=86+2412= \frac{8}{6} + \frac{24}{12}

=43+2= \frac{4}{3} + 2

=43+63=103= \frac{4}{3} + \frac{6}{3} = \frac{10}{3}

Since both sides are equal to 103\frac{10}{3}, the statement is proved.

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7Simplify the following using the distributive property:Show solution
Apply the distributive property:

79(67−34)\frac{7}{9}\left(\frac{6}{7}-\frac{3}{4}\right)

First find the bracket:

67−34=24−2128=328\frac{6}{7}-\frac{3}{4} = \frac{24-21}{28} = \frac{3}{28}

Now multiply:

79×328=21252=112\frac{7}{9}\times\frac{3}{28} = \frac{21}{252} = \frac{1}{12}

So the simplified value is 112\frac{1}{12}.

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1Prove that the following rational numbers are equal:Show solution
Check equality by cross-multiplication:

(i) 23\frac{2}{3} and 46\frac{4}{6}:
2×6=122\times 6 = 12 and 3×4=123\times 4 = 12, so they are equal.

(ii) 54\frac{5}{4} and 108\frac{10}{8}:
5×8=405\times 8 = 40 and 4×10=404\times 10 = 40, so they are equal.

(iii) −35-\frac{3}{5} and −610-\frac{6}{10}:
(−3)×10=−30(-3)\times 10 = -30 and 5×(−6)=−305\times (-6) = -30, so they are equal.

(iv) 93\frac{9}{3} and 33:
93=3\frac{9}{3} = 3, so they are equal.

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2Find the sum:Show solution
Add the fractions with a common denominator:

25+310=410+310=710\frac{2}{5}+\frac{3}{10} = \frac{4}{10}+\frac{3}{10} = \frac{7}{10}

712+58=1424+1524=2924\frac{7}{12}+\frac{5}{8} = \frac{14}{24}+\frac{15}{24} = \frac{29}{24}

−47+314=−814+314=−514-\frac{4}{7}+\frac{3}{14} = -\frac{8}{14}+\frac{3}{14} = -\frac{5}{14}

The sum is not a single value because the question set contains several parts; for the first part, the answer is 710\frac{7}{10}.

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3Find the difference:Show solution
Find each difference:

(i) 56−14=1012−312=712\frac{5}{6}-\frac{1}{4} = \frac{10}{12}-\frac{3}{12} = \frac{7}{12}

(ii) 118−34=118−68=58\frac{11}{8}-\frac{3}{4} = \frac{11}{8}-\frac{6}{8} = \frac{5}{8}

(iii) −79−(−23)=−79+69=−19-\frac{7}{9}-\left(-\frac{2}{3}\right)= -\frac{7}{9}+\frac{6}{9} = -\frac{1}{9}

For the first part, the answer is 712\frac{7}{12}.

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4Find the product:Show solution
Find each product:

(i) 23×310=15\frac{2}{3}\times\frac{3}{10}=\frac{1}{5}

(ii) 711×58=3588\frac{7}{11}\times\frac{5}{8}=\frac{35}{88}

(iii) −47×514=−1049-\frac{4}{7}\times\frac{5}{14}=-\frac{10}{49}

For the first part, the answer is 15\frac{1}{5}.

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5Find the quotient:Show solution
Find each quotient:

(i) 23÷310=23×103=209\frac{2}{3}\div\frac{3}{10}=\frac{2}{3}\times\frac{10}{3}=\frac{20}{9}

(ii) 711÷58=711×85=5655\frac{7}{11}\div\frac{5}{8}=\frac{7}{11}\times\frac{8}{5}=\frac{56}{55}

(iii) −47÷514=−47×145=−85-\frac{4}{7}\div\frac{5}{14}= -\frac{4}{7}\times\frac{14}{5}= -\frac{8}{5}

For the first part, the answer is 209\frac{20}{9}.

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EXERCISE SET 3.4

1Represent the rational numbers 23\frac{2}{3}, −54-\frac{5}{4} and 1121\frac{1}{2} on a single number line.Show solution

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2Find three distinct rational numbers that lie strictly between −12-\frac{1}{2} and 14\frac{1}{4}.Show solution
Any three rational numbers strictly between −12-\frac{1}{2} and 14\frac{1}{4} will do.

For example:

−14, 0, 18-\frac{1}{4},\ 0,\ \frac{1}{8}

Each of these is greater than −12-\frac{1}{2} and less than 14\frac{1}{4}.

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3Simplify the expression: \left(-\frac{1}{4} ight) + \left(\frac{5}{12} ight).Show solution
Take a common denominator of 12:

−14=−312-\frac{1}{4} = -\frac{3}{12}

So,

−312+512=212=16-\frac{3}{12} + \frac{5}{12} = \frac{2}{12} = \frac{1}{6}

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4A tailor has 153415\frac{3}{4} metres of fine silk. If making one kurta requires 2142\frac{1}{4} metres of silk, exactly how many kurtas can he make?Show solution
Convert to improper fractions:

1534=63415\frac{3}{4} = \frac{63}{4}

214=942\frac{1}{4} = \frac{9}{4}

Now divide:

634÷94=634×49=639=7\frac{63}{4} \div \frac{9}{4} = \frac{63}{4}\times\frac{4}{9} = \frac{63}{9} = 7

So he can make 7 kurtas.

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5Find three rational numbers between 3.1415 and 3.1416.
6Can you think of other way(s) to find a rational number between any two rational numbers?
3Simplify the expression: (−14)+(512)\left(-\frac{1}{4}\right) + \left(\frac{5}{12}\right).

EXERCISE SET 3.5

1Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 720\frac{7}{20}, 415\frac{4}{15} and 13250\frac{13}{250}. Then check your answers
2Perform the long division for 113\frac{1}{13}. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 213\frac{2}{13}? Now compute 313\frac{3}{13}, 413\frac{4}{13}, etc. What do you notice?
3Classify the following numbers as rational or irrational:
4The number 0.9‾0.\overline{9} (which means 0.99999 ...) is a rational number. Using algebra (let x=0.9‾x = 0.\overline{9}, multiply by 10, and subtract), explain why 0.9‾0.\overline{9} is exactly equal to 1.
5We have seen that the repeating block of 17\frac{1}{7} is a cyclic number. Try to find more numbers (nn) whose reciprocals \left(\frac{1}{n} ight) produce decimals with repeating blocks that are cyclic.
5We have seen that the repeating block of 17\frac{1}{7} is a cyclic number. Try to find more numbers (nn) whose reciprocals (1n)\left(\frac{1}{n}\right) produce decimals with repeating blocks that are cyclic.

END-OF-CHAPTER EXERCISES

1(i)350\frac{3}{50}
2Prove that qrt5qrt{5} is an irrational number.
3(i)12.6
3(ii)0.0120
3(iii)3.052‾3.0\overline{52}
3(iv)1.235‾1.2\overline{35}
3(v)0.23‾0.\overline{23}
3(vi)2.05‾2.0\overline{5}
3(vii)2.125‾2.12\overline{5}
3(viii)3.125‾3.12\overline{5}
3(ix)2.162‾5‾2.\overline{162}\overline{5}
4(i)0.532
4(ii)1.15‾1.1\overline{5}
5Find 6 rational numbers between 3 and 4.
6Find 5 rational numbers between 25\frac{2}{5} and 35\frac{3}{5}.
7Find 5 rational numbers between \fracrac16\fracrac{1}{6} and 25\frac{2}{5}.
8If x3+x5=1615\frac{x}{3} + \frac{x}{5} = \frac{16}{15}, find the rational number xx.
9Let aa and bb be two non-zero rational numbers such that a+1b=0a + \frac{1}{b} = 0. Without assigning any numerical values, determine whether abab is positive or negative. Justify your answer.
10A 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 p104\frac{p}{10^4}, where pp 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 242^4 or 545^4? Give reasons.
11Without performing division, determine whether the decimal expansion of 18125\frac{18}{125} is terminating or non-terminating. If it terminates, state the number of decimal places.
12A rational number in its lowest form has denominator 23imes52^3 imes 5. How many decimal places will its decimal expansion have? Explain your answer.
13Let a=712a = \frac{7}{12} and b=56b = \frac{5}{6}. Express both aa and bb in the form k1m\frac{k_1}{m} and k2m\frac{k_2}{m} where k1,k2k_1, k_2 and mm are integers and k2−k1>6k_2 - k_1 > 6. Using the same denominator mm, write exactly five distinct rational numbers lying between aa and bb keeping an integer numerator. Explain why the condition k2−k1>n+1k_2 - k_1 > n + 1 is necessary to find nn such rational numbers between the two rational numbers aa and bb using this method.
14Three rational numbers x,y,zx, y, z satisfy x+y+z=0x + y + z = 0 and xy+yz+zx=0xy + yz + zx = 0. Show that all the rational numbers x,y,zx, y, z must be simultaneously zero.
15Show that the rational number (a+b)2\frac{(a+b)}{2} lies between the rational numbers aa and bb.
2Prove that 5\sqrt{5} is an irrational number.
7Find 5 rational numbers between 16\frac{1}{6} and 25\frac{2}{5}.
12A rational number in its lowest form has denominator 23×52^3 \times 5. How many decimal places will its decimal expansion have? Explain your answer.
1(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:
3Convert the following decimal numbers in the form of pq\frac{p}{q}.
4Locate the following rational numbers on the number line.

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