वास्तविक सङ्ख्या (Real Numbers)
Real number is a numeration system that can be placed on a number line. Real numbers are used to measure quantities such as time, mass, energy, velocity, and many more. The real numbers include natural numbers, whole numbers, rational numbers, irrational numbers, algebraic numbers, and transcendental numbers.
A real number can be thought of as a point on a number line. In this essence, to every real number, there corresponds a point on a number line. Conversely, to every point on a number line, there corresponds a unique real number.
सङख्या पद्दति (Number System)
संख्याङ्कन पद्दती (Number System) भनेको विभिन्न प्रकारका संख्याहरूको प्रतिनिधित्व गर्ने तरिका हो।
Numeration system is a systematic organization and use of numerals to represent number. For example, Hindu-Arabic numeration system. This Hindu-Arabic numeration system utilize 10 characters (digits) systematically to represent number.
- A number is a mathematical object used to count, measure, and label things. It is a theoretical concept or a mathematical abstraction or an idea in mind.
- Numbers can be expressed in many ways. Some common ways to represent number are symbol, picture, and words. These notational symbol (like word, picture) that represents a number are called numerals. Numeral are physical representation of number.
For example, \(3, III\), three, all are numerals. So, numeral is a symbol or name that stands for a number.
| Number | Numerals | |
|---|---|---|
| Nature | Abstract | Concrete |
| Essence | Idea in mind | Physical representation |
| Use | Theoretical | Different forms in practical use |
दशमलव सङ्ख्या पद्धति (Decimal System)
The numeration system developed in ancient Indian civilization by Bhaskara and Aaryabhatta around 600 AD, and later translated and disseminated by Arab civilization (al-Khwarizmi) around 1100 AD, is called the Hindu–Arabic numeration system.
This is the most scientific numeration system ever known, using an expanded system systematically to represent numbers. In this system, there are ten symbols used to represent numbers.
Characteristics of Hindu–Arabic Numeration System
- Systematic use of zero
- Expanded value system
- Large numbers can be written easily
- Based on place value system
- Ten symbols are used
- Based on “Base 10”
The number system we use in everyday life is called the Hindu–Arabic numeration system, which is a decimal system using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 to represent all numbers.
In this system, each digit represents a power of 10. To compute a number, multiply each digit value by its place value and add them all together.
For example, to represent the number 4379:
\((4 \times 1000) + (3 \times 100) + (7 \times 10) + (9 \times 1) = 4379\)
\((4 \times 1000) + (3 \times 100) + (7 \times 10) + (9 \times 1) = 4379\)
| 4 | 3 | 7 | 9 | Digit Value |
| \(10^3\) | \(10^2\) | \(10^1\) | \(10^0\) | Place Value |
| \(4 \times 1000\) | \(3 \times 100\) | \(7 \times 10\) | \(9 \times 1\) | Total value |
| 4000 | + 300 | + 70 | + 9 | = 4379 |
Number Systems and Their Properties
| Number system | Basic symbols | Base | Example |
|---|---|---|---|
| Binary | 0,1 | 2 | \(1101_2\) |
| Octal | 0–7 | 8 | \(256_8\) |
| Decimal | 0–9 | 10 | \(458_{10}\) |
| Hexadecimal | 0–9,A–F | 16 | \(2A5_{16}\) |
द्विआधार सङ्ख्या पद्दति (Binary System)
Binary numeration system भनेको base-2 मा आधारित संख्याङ्कन प्रणाली हो जसमा दुई वटा मात्र अंकहरू: 0 र 1 को प्रयोग हुन्छ। यस प्रणालीमा प्रत्येक अंकले 2 को घातलाई जनाउँछ। यो प्रणाली computing र digital electronics मा प्रयोग हुन्छ, किनभने यसले electronic switches र transistors का दुई अवस्थाहरू (on and off) सँग प्रत्यक्ष रूपमा सम्बन्ध राख्छ।
यस प्रणालीमा कुनै संख्या गणना गर्न, प्रत्येक अंकलाई त्यसको स्थान मान (place value) सँग गुणा गरिन्छ र सबैलाई जोडिन्छ। जस्तै, Binary number \(1001_2\) ले तलको संख्यालाई जनाउँछ।
\(1 × 2^3) + (0 × 2^2) + (0 × 2^1) + (1 × 2^0) = 8 + 0 + 0 + 1 = 9_10\)
| 1 | 0 | 0 | 1 | Digit Value |
|---|---|---|---|---|
| 23 | 22 | 21 | 20 | Place Value |
| 1 × 8 | 0 × 4 | 0 × 2 | 1 × 1 | Total value |
| 8 | + 0 | + 0 | + 1 | = 9 |
Group of 4 bits is known as a ‘Nibble’ and that of 8 bits as a ‘Byte’. A group of bits simultaneously processed by a digital system such as computer is known as a ‘word’ e.g. 8-bit word, 16-bit word, 32-bit word, etc. Advantages of binary number system are given below.
- Bits can be used to designate the two voltage levels in digital electronics.
- Binary arithmetic is simple compared to decimal arithmetic.
- It is also called as natural code.
Example
Convert 36 (decimal values) into binary.
The solution is
Therefore
\( 36 = 100100_2 \)
Convert 36 (decimal values) into binary.
The solution is
| 2 | 36 | 0↑ | |
| 2 | 18 | 0↑ | |
| 2 | 9 | 1↑ | |
| 2 | 4 | 0↑ | |
| 2 | 2 | 0↑ | |
| 1 | 1↑ |
\( 36 = 100100_2 \)
Exercise
Convert the following decimal values into binary.
- 8
- 17
- 11
- 70
- 132
Example
Convert the following binary values into decimal
- 0 0 1 0
- 1 1 1 0
- 1 0 1 1
- 0 0 1 0 0 0 0 1
- 0 0 0 1 1 0 1 0
- 1 0 1 0 1 0 1 0
Quinary numeration system
Quinary numeration system भनेको base-5 मा आधारित संख्याङ्कन प्रणाली हो जसमा पाँच वटा मात्र अंकहरू: 0, 1, 2, 3 र 4 को प्रयोग हुन्छ। यस प्रणालीमा प्रत्येक अंकले 5 को घातलाई जनाउँछ।
यस प्रणालीमा कुनै संख्या गणना गर्न, प्रत्येक अंकलाई त्यसको स्थान मान (place value) सँग गुणा गरिन्छ र सबैलाई जोडिन्छ। जस्तै, Quinary number\(1243_5\) ले तलको संख्यालाई जनाउँछ।
\(1 × 5^3 + 2 × 5^2 + 4 × 5^1 + 3 × 5^0 = 125 + 50 + 20 + 3 = 198\)
| 1 | 2 | 4 | 3 | Digit Value |
|---|---|---|---|---|
| 53 | 52 | 51 | 50 | Place Value |
| 1 × 125 | 2 × 25 | 4 × 5 | 3 × 1 | Total value |
| 125 | + 50 | + 20 | + 3 | = 198 |
Quinary systems were historically used in various cultures, often influenced by finger counting, where one hand represented a complete cycle of counting before moving to the next power of five.
Example
Convert 2456 (decimal values) into quinary.
The solution is
Therefore
\( 2456 = 34311_5 \)
Convert 2456 (decimal values) into quinary.
The solution is
| 5 | 2456 | 1↑ | |
| 5 | 491 | 1↑ | |
| 5 | 98 | 3↑ | |
| 5 | 19 | 4↑ | |
| 5 | 3 | 3↑ | |
| 0 |
\( 2456 = 34311_5 \)
Exercise
Convert the following decimal values into quinary.
- 45
- 568
- 2349
- 11198
- 12345
Exercise
Convert the following quinary values into decimal:
- 10234
- 1432
- 23423
- 1110043
- 33401
- 1234343
अनुपातिक सङ्ख्याहरु (Rrational Numbers)
A rational number is a real number that can be expressed in the form
\(\frac{p}{q}\)
where \( p \) and \( q \) are integers and \( q \ne 0 \). The set of all rational numbers is denoted by \( \mathbb{Q} \), so
\(\mathbb{Q} = \left\{ \frac{p}{q} \,\middle|\, p, q \in \mathbb{Z},\, q \ne 0 \right\}.\)
Rational numbers are fractions (where the numerator and denominators are integers) and repeating and terminating decimals. Fractions may not have 0 as the denominator. Some examples are \(4\) and \(\frac{22}{7}\).
अनानुपातिक सङ्ख्याहरु (Irrational Numbers)
Irrational numbers are numbers that cannot be expressed as the ratio of two integers. Examples of irrational numbers are (a) square roots of non-perfect squares, (b) \( \pi \), and (c) decimals that exist in non terminating and non-repeating pattern. Some examples of irrational numbers are \( \sqrt{2} \) and \( \pi \).
ग्रीसमा प्रसिद्ध गणितज्ञ पाइथागोरस र उनका अनुयायी पाइथागोरियनहरूले करिब 400 ईसापूर्व मा पहिलो पटक rational नभएका संख्याहरू पत्ता लगाए। यी संख्याहरूलाई irrational numbers भनियो।
NOTE:real numbers = rational numbers + irrational numbers
| Definition | Type of Decimal |
|---|---|
| Rational numbers are numbers that can be expressed as a fraction where both the numerator and the denominator are integers (with the exception of 0 as a denominator). |
Repeating \( \frac{5}{11} = 0.454545\ldots \) Terminating \( \frac{7}{8} = 0.875 \) |
| Irrational numbers are numbers that cannot be expressed as a fraction where both the numerator and the denominator are integers. |
Do not repeat or terminate First 12 digits of \( \pi \) = 3.14159265359… First 7 digits of \( \sqrt{2} \) = 1.414213… |
Locate \(\sqrt{2}\) on the number line.
SolutionConsider a square \(OABC\), with each side 1 unit in length. Then using Pythagoras theorem , we get
\(OB = \sqrt{1^2 + 1^2} = \sqrt{2}\)
Using a compass with centre O and radius OB, draw an arc intersecting the number line at the point P. Then P corresponds to \(\sqrt{2}\) on the number line.
अनुपात (Ratio)
When we compare two quantities of same kind by division, we say a ratio is formed.
अनुपात भनेको दुईवटा वस्तु, मात्रा, वा अंशहरू बीचको सम्बन्धलाई division बाट देखाउने एक तरिका हो। हामी दैनिक जीवनमा धेरै चीजहरू गणना गर्छौ र तुलना गर्छौं, जस्तै: काम गर्ने घण्टा र आराम गर्ने घण्टा। अनुपातले यस्ता दुईवटा वस्तुहरूको तुलना गर्न मद्दत गर्छ। जसमा संख्याहरुको पहिलो मात्रा अंश हो र दोस्रो मात्रा हर हो, यो भिन्न भन्दा अली फरक हुन्छ, यसमा संख्यालाई अंश वा हर दुबैमा लेख्न वा पढ्न सकिन्छ।
अनुपात भनेको दुईवटा वस्तु, मात्रा, वा अंशहरू बीचको सम्बन्धलाई division बाट देखाउने एक तरिका हो। हामी दैनिक जीवनमा धेरै चीजहरू गणना गर्छौ र तुलना गर्छौं, जस्तै: काम गर्ने घण्टा र आराम गर्ने घण्टा। अनुपातले यस्ता दुईवटा वस्तुहरूको तुलना गर्न मद्दत गर्छ। जसमा संख्याहरुको पहिलो मात्रा अंश हो र दोस्रो मात्रा हर हो, यो भिन्न भन्दा अली फरक हुन्छ, यसमा संख्यालाई अंश वा हर दुबैमा लेख्न वा पढ्न सकिन्छ।
In the ratio 3 : 2, the numbers 3 and 2 are called the terms of the ratio.
3 is called the First term or Antecedent.
2 is called the Second term or Consequent
3 is called the First term or Antecedent.
2 is called the Second term or Consequent
Ratio compares two quantities of the same kind by division: e.g., \( \frac{15}{10} = \frac{3}{2} \).
Ratio: In \( a:b \), \( a \) is antecedent, \( b \) is consequent.
Ratio
If the quantities to be compared \(a\) and \(b\) having the same unit, then the ratio of the \(a\) and \(b\) is \(\frac{a}{b}\) or \(a:b\) and the ratio of \(b\) to \(a\) is written as \(\frac{b}{a}\) or \(b:a\). Here \(a:b\) is read as \(a\) is to \(b\) and \(b:a\) is read as \(b\) is to \(a\).
If the quantities to be compared \(a\) and \(b\) having the same unit, then the ratio of the \(a\) and \(b\) is \(\frac{a}{b}\) or \(a:b\) and the ratio of \(b\) to \(a\) is written as \(\frac{b}{a}\) or \(b:a\). Here \(a:b\) is read as \(a\) is to \(b\) and \(b:a\) is read as \(b\) is to \(a\).
अनुपात (ratio) को अवधारणालाई बुझ्नको लागी तलको उदाहरण हेरौ!
Based on classroom survey data, for example: 4 boys and 5 girls, we can form different ratios as below
Based on classroom survey data, for example: 4 boys and 5 girls, we can form different ratios as below
- girls to boys \(\to\) 5 girls for every 4 boys: 5:4
- girls to total \(\to\) 5:9 students are girls
- boys to girls \(\to\) 4 boys for every 5 girls: 4:5
- boys to total \(\to\) 4:9 students are boys
- total to boys \(\to\) ratio of total to boys is 9:4
- total to girls \(\to\) ratio of total to girls is 9:5
समानुपात (Proportion)
A proportion states that two ratios are equal: \( \frac{a}{b} = \frac{c}{d} \) or \( a:b :: c:d \).
Fundamental Principle:
Product of means = Product of extremes → \( a \cdot d = b \cdot c \).
Product of means = Product of extremes → \( a \cdot d = b \cdot c \).
| Feature | Ratio | Proportion |
|---|---|---|
| Definition | Compares two quantities | Equality of two ratios |
| Symbol | : | :: or = |
| Form | Expression | Equation |
For Q.No.4(a,d) in BLE Exam
- 9 लाई पञ्चाधार सङ्ख्यामा परिवर्तन गरी लेख्नुहोस् । Convert 9 into base quinary number system. [1U]
- 4 लाई द्विआधार पद्दतिमा परिवर्तन गरी लेख्नुहोस् । Convert 4 into base two number system.[1U]
- दशमलव सङ्ख्या 435 लाई द्विआधार सङ्ख्या पद्दतिमा परिवर्तन गर्नुहोस् । Convert the decimal 435 number into binary number system. [1U]
- 101011\(_2\) लाई द्विआधार सङ्ख्या हो । उक्त सङ्ख्यालाई सरल दशमलव सङ्ख्यामा रुपान्तरण गर्नुहोस् । \(101011_2\) is a base two (binary) number. Convert the number into decimal number system. [1U]
- द्विआधार सङ्ख्या \(110011_2\) लाई दशमलव सङ्ख्या पद्दतिमा रुपान्तरण गर्नुहोस् । Convert the binary number \(110011_2\) into decimal number system. [1U]
- 512 लाई पन्चाधार सङ्ख्या पद्धतिमा रूपान्तरण गर्नुहोस् । Convert 512 into quinary number. [1U]
- 13 लाई पन्चाधार सङ्ख्या पद्धतिमा रूपान्तरण गर्नुहोस् । Convert 13 into quinary numeration system. [1U]
- \(101_{2}\) लाई दशमलव सङ्ख्यामा परिणत गर्नुहोस् । Convert \(101_{2}\) into decimal number. [1U]
- \(101_{8}\) लाई दशमलव सङ्ख्यामा रूपान्तरण गर्नुहोस् । Convert \(101_{8}\) into decimal number. [1U]
- पन्चाधार सङ्ख्या \(123_{5}\) लाई दशमलव सङ्ख्या पद्धतिमा रूपान्तरण गर्नुहोस् । Convert the quinary number \(123_{5}\) into decimal number. [1A]
- दशमलव सङ्ख्या पद्धतिमा कतिओटा अङ्कहरू हुन्छन् र ती अङ्कहरू के के हुन् ? How many digits are there in decimal (denary) number system and what are they? [1K]
- द्विआधार सङ्ख्या पद्धतिमा कतिओटा अङ्कहरू हुन्छन् र ती अङ्कहरू के के हुन् ? How many digits are there in binary number system and what are they? [1K]
- पन्चाधार सङ्ख्या पद्धतिमा कतिओटा अङ्कहरू हुन्छन् र ती अङ्कहरू के के हुन् ? How many digits are there in quinary number system and what are they? [1K]
- कस्ता सङ्ख्यालाई अपरिमेय सङ्ख्या भनिन्छ ? उदाहरणसहित लेख्नुहोस् । What type of numbers are called irational numbers? Write with example. [1K]
- तलका कुन सङ्ख्याहरू अपरिमेय सङ्ख्या हुन् ? Which of the following numbers are irrational numbers? \(\sqrt{9}, \sqrt{7}, 0.\overline{6}\). [1U]
- तलका कुन सङ्ख्याहरू अपरिमेय सङ्ख्या हुन् ? Which of the following numbers are rational numbers? \(1.414213 \dots, \sqrt{5}, 0.5\). [1U]
- सङ्ख्या 62000 लाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Write down the number 62000 in scientific notation. [1U]
- \(8.4 \times 10^4\) लाई दशमलव सङ्ख्यामा रुपान्तर गर्नुहोस् । Convert \(8.4 \times 10^4\) into decimal number. [1U]
- \(19\) लाई वैज्ञानिक सङ्केतमा Express 19 in the scientific notation. [1U]
- 0.007008 लाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Write down 0.007008 in scientific notation. [1U]
- \(3.56 \times 10^{-4}\) लाई दशमलव सङ्ख्यामा रुपान्तर गर्नुहोस् । Convert \(3.56 \times 10^{-4}\) into decimal number. [1U]
- सङ्ख्या 543000 लाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Write the decimal number 543000 in scientific notation. [1U]
- 0.0000325 लाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Write the number 0.0000325 in scientific notation. [1U]
- सङ्ख्या 235000 लाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Write number 235000 in scientific notation. [1U]
The solution is
Therefore
\( 9 = 14_5 \)
| 5 | 9 | 4↑ | |
| 1 | 1↑ |
\( 9 = 14_5 \)
The solution is
Therefore
\( 4 = 100_2 \)
| 2 | 4 | 0↑ | |
| 2 | 2 | 0↑ | |
| 1 | 1↑ |
\( 4 = 100_2 \)
The solution is
Therefore
\( 435 = 110110011_2 \)
| 2 | 435 | 1↑ | |
| 2 | 217 | 1↑ | |
| 2 | 108 | 0↑ | |
| 2 | 54 | 0↑ | |
| 2 | 27 | 1↑ | |
| 2 | 13 | 1↑ | |
| 2 | 6 | 0↑ | |
| 2 | 3 | 1↑ | |
| 1 | 1↑ |
\( 435 = 110110011_2 \)
The solution is
We expand the binary number using powers of 2:
Hence, the decimal conversion is
\(1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 32 + 0 + 8 + 0 + 2 + 1 = 43 \)
Therefore
\( 101011_2 = 43_{10} \)
We expand the binary number using powers of 2:
| 1 | 0 | 1 | 0 | 1 | 1 |
| \(2^5\) | \(2^4\) | \(2^3\) | \(2^2\) | \(2^1\) | \(2^0\) |
\(1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 32 + 0 + 8 + 0 + 2 + 1 = 43 \)
Therefore
\( 101011_2 = 43_{10} \)
The solution is
We expand the binary number using powers of 2:
Hence, the decimal conversion is
\(1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 32 + 16 + 0 + 0 + 2 + 1 = 51 \)
Therefore
\( 110011_2 = 51_{10} \)
We expand the binary number using powers of 2:
| 1 | 1 | 0 | 0 | 1 | 1 |
| \(2^5\) | \(2^4\) | \(2^3\) | \(2^2\) | \(2^1\) | \(2^0\) |
\(1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 32 + 16 + 0 + 0 + 2 + 1 = 51 \)
Therefore
\( 110011_2 = 51_{10} \)
The solution is
Therefore
\( 512 = 4022_5 \)
| 5 | 512 | 2↑ | |
| 5 | 102 | 2↑ | |
| 5 | 20 | 0↑ | |
| 4 | 4↑ |
\( 512 = 4022_5 \)
The solution is
Therefore
\( 13 = 23_5 \)
| 5 | 13 | 3↑ | |
| 2 | 2↑ |
\( 13 = 23_5 \)
The solution is
We expand the binary number using powers of 2:
Hence, the decimal conversion is
\(1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5\)
Therefore
\( 101_2 = 5_{10} \)
We expand the binary number using powers of 2:
| 1 | 0 | 1 |
| \(2^2\) | \(2^1\) | \(2^0\) |
\(1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5\)
Therefore
\( 101_2 = 5_{10} \)
The solution is
We expand the octal number using powers of 8:
Hence, the decimal conversion is
\(1 \times 8^2 + 0 \times 8^1 + 1 \times 8^0 = 64 + 0 + 1 = 65\)
Therefore
\( 101_8 = 65_{10} \)
We expand the octal number using powers of 8:
| 1 | 0 | 1 |
| \(8^2\) | \(8^1\) | \(8^0\) |
\(1 \times 8^2 + 0 \times 8^1 + 1 \times 8^0 = 64 + 0 + 1 = 65\)
Therefore
\( 101_8 = 65_{10} \)
The solution is
We expand the quinary number using powers of 5:
Hence, the decimal conversion is
\(1 \times 5^2 + 2 \times 5^1 + 3 \times 5^0 = 25 + 10 + 3 = 38\)
Therefore
\( 123_5 = 38_{10} \)
We expand the quinary number using powers of 5:
| 1 | 2 | 3 |
| \(5^2\) | \(5^1\) | \(5^0\) |
\(1 \times 5^2 + 2 \times 5^1 + 3 \times 5^0 = 25 + 10 + 3 = 38\)
Therefore
\( 123_5 = 38_{10} \)
The solution is
There are 10 digits in the decimal (denary) number system.
They are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
There are 10 digits in the decimal (denary) number system.
They are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
The solution is
There are 2 digits in the binary number system.
They are: 0 and 1.
There are 2 digits in the binary number system.
They are: 0 and 1.
The solution is
There are 5 digits in the quinary number system.
They are: 0,1,2,3 and 4.
There are 5 digits in the quinary number system.
They are: 0,1,2,3 and 4.
The numbers which cannot be expressed as \(\frac{p}{q}\) where \(p,q \in \mathbb{Z}\) and \(q \ne 0\), are called irrational numbers.
Their decimal expansions of irrational numbers are non-terminating and non-repeating.
Some example of irrational numbers are
\(\sqrt{2} = 1.414213562\ldots,\pi,e\)
Their decimal expansions of irrational numbers are non-terminating and non-repeating.
Some example of irrational numbers are
\(\sqrt{2} = 1.414213562\ldots,\pi,e\)
The solution
1. \(\sqrt{9} = 3\) → This is an integer, so it is rational number.
2. \(\sqrt{7}\)→ This cannot be written as \(\frac{p}{q}\). Hence, it is irrational number.
3. \(0.\overline{6} = 0.6666\ldots\) → This is a repeating decimal, So it is rational number.
1. \(\sqrt{9} = 3\) → This is an integer, so it is rational number.
2. \(\sqrt{7}\)→ This cannot be written as \(\frac{p}{q}\). Hence, it is irrational number.
3. \(0.\overline{6} = 0.6666\ldots\) → This is a repeating decimal, So it is rational number.
The solution
1. \(1.414213 \dots\) → This is non-terminating and non-repeating. Hence, it is an irrational number.
2. \(\sqrt{5}\) → This cannot be written as \(\frac{p}{q}\). So, it is an irrational number.
3. \(0.5 = \frac{1}{2}\) → This is a terminating decimal. So, it is a rational number.
1. \(1.414213 \dots\) → This is non-terminating and non-repeating. Hence, it is an irrational number.
2. \(\sqrt{5}\) → This cannot be written as \(\frac{p}{q}\). So, it is an irrational number.
3. \(0.5 = \frac{1}{2}\) → This is a terminating decimal. So, it is a rational number.
The solution
To express 62000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 62000 = 6.2 \times 10^4 \)
To express 62000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 62000 = 6.2 \times 10^4 \)
The solution
To convert \(8.4 \times 10^4\) into decimal number, move the decimal point 4 places to the right, which is
\( 8.4 \times 10^4 = 84000 \)
To convert \(8.4 \times 10^4\) into decimal number, move the decimal point 4 places to the right, which is
\( 8.4 \times 10^4 = 84000 \)
The solution
To express 19 in scientific notation, place the decimal after the first non-zero digit, which is
\( 19 = 1.9 \times 10^1 \)
To express 19 in scientific notation, place the decimal after the first non-zero digit, which is
\( 19 = 1.9 \times 10^1 \)
The solution
To express 0.007008 in scientific notation, move the decimal point 3 places to the right to get a number between 1 and 10, which is
\( 0.007008 = 7.008 \times 10^{-3} \)
To express 0.007008 in scientific notation, move the decimal point 3 places to the right to get a number between 1 and 10, which is
\( 0.007008 = 7.008 \times 10^{-3} \)
The solution
To convert \(3.56 \times 10^{-4}\) into decimal number, move the decimal point 4 places to the left, which is
\( 3.56 \times 10^{-4} = 0.000356 \)
To convert \(3.56 \times 10^{-4}\) into decimal number, move the decimal point 4 places to the left, which is
\( 3.56 \times 10^{-4} = 0.000356 \)
The solution
To express 543000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 543000 = 5.43 \times 10^5 \)
To express 543000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 543000 = 5.43 \times 10^5 \)
The solution
To express 0.0000325 in scientific notation, move the decimal point 5 places to the right to get a number between 1 and 10, which is
\( 0.0000325 = 3.25 \times 10^{-5} \)
To express 0.0000325 in scientific notation, move the decimal point 5 places to the right to get a number between 1 and 10, which is
\( 0.0000325 = 3.25 \times 10^{-5} \)
The solution
To express 235000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 235000 = 2.35 \times 10^5 \)
To express 235000 in scientific notation, place the decimal after the first non-zero digit, which is
\( 235000 = 2.35 \times 10^5 \)
For Q.No.4(c) in BLE Exam
- 0.0245 र 30000000 को गुणनफल पत्ता लगाउनुहोस् र गुणनफललाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Find the product of 0.0245 and 30000000 and write down result in scientific notation. [2U]
- वैज्ञानिक सङ्केतमा भएको सङ्ख्या \(1.525 \times 10^4\) लाई दशमलव सङ्ख्यामा रुपान्तरण गर्नुहोस् । Convert the number \(1.525 \times 10^4\) from scientific notation into decimal number. [2U]
- सरल गर्नुहोस् र वैज्ञानिक सङ्केतमा लेख्नुहोस्ः Simplify and write down the result in scientific notation: \(\frac{1.1 \times 10^4 \times 1.0 \times 10^2}{5.5 \times 10^4}\) [2U]
- \(612_8\) लाई पञ्चाधार सङ्ख्यामा रुपान्तरण गर्नुहोस् । Convert \(612_8\) into quinary number. [2U]
- \(8.2 \times 10^{-3}\) लाई दशमलव प्रणालीमा लेख्नुहोस् । Write \(8.2 \times 10^{-3}\) into decimal system [2U]
The solution
The multiplication is
\( 0.0245 \times 30000000 = 735000 \)
Now, scientific notation is obtained by placing the decimal after the first non-zero digit, which is
\( 735000 = 7.35 \times 10^5 \)
The multiplication is
\( 0.0245 \times 30000000 = 735000 \)
Now, scientific notation is obtained by placing the decimal after the first non-zero digit, which is
\( 735000 = 7.35 \times 10^5 \)
The solution
To convert \(1.525 \times 10^4\) into decimal number, move the decimal point 4 places to the right, which is
\( 1.525 \times 10^4 = 15250 \)
To convert \(1.525 \times 10^4\) into decimal number, move the decimal point 4 places to the right, which is
\( 1.525 \times 10^4 = 15250 \)
The solution is
\(\frac{1.1 \times 10^4 \times 1.0 \times 10^2}{5.5 \times 10^4}\)
or\( \frac{1.1 \times 1.0}{5.5} \times \frac{10^4 \times 10^2}{10^4} \)
or\( 0.2 \times 10^2 \)
Now we write in proper scientific notation (coefficient between 1 and 10), then
\( 2.0 \times 10^1\)
\(\frac{1.1 \times 10^4 \times 1.0 \times 10^2}{5.5 \times 10^4}\)
or\( \frac{1.1 \times 1.0}{5.5} \times \frac{10^4 \times 10^2}{10^4} \)
or\( 0.2 \times 10^2 \)
Now we write in proper scientific notation (coefficient between 1 and 10), then
\( 2.0 \times 10^1\)
The solution
Step 1: Convert \(612_8\) to decimal system, which is
\(6 \times 8^2 + 1 \times 8^1 + 2 \times 8^0 = 6 \times 64 + 1 \times 8 + 2 = 384 + 8 + 2 = 394\)
Step 2: Convert 394 to base 5 (quinary) by repeated division
Therefore,
\( 612_8 = 33034_5 \)
Step 1: Convert \(612_8\) to decimal system, which is
\(6 \times 8^2 + 1 \times 8^1 + 2 \times 8^0 = 6 \times 64 + 1 \times 8 + 2 = 384 + 8 + 2 = 394\)
Step 2: Convert 394 to base 5 (quinary) by repeated division
| 5 | 394 | 4↑ | |
| 5 | 78 | 3↑ | |
| 5 | 15 | 0↑ | |
| 5 | 3 | 3↑ | |
| 0 | 3↑ |
\( 612_8 = 33034_5 \)
The solution
To convert \(8.2 \times 10^{-3}\) into decimal number, move the decimal point 3 places to the left, which is
\( 8.2 \times 10^{-3} = 0.0082 \)
To convert \(8.2 \times 10^{-3}\) into decimal number, move the decimal point 3 places to the left, which is
\( 8.2 \times 10^{-3} = 0.0082 \)
For Q.No.4(c) in BLE Exam
- सरल गरि वैज्ञानिक सङ्केतमा लेख्नुहोस् । Simplify and write down the result in scientific notation. \(\frac{(9.6 \times 10^{-3}) \times (7.5 \times 10^{-2})}{2.4 \times 10^{-2}}\) [2U]
- सरल गर्नुहोस् (Simplify): \(\frac{9.8 \times 10^{-5} + 5.1 \times 10^{-6}}{2.2 \times 10^{-4}}\) [2U]
- सरल गर्नुहोस् र परिणामलाई वैज्ञानिक सङ्केतमा लेख्नुहोस् । Simplify and write down the results in scientific notation. \(\frac{(8.6 \times 10^{-3}) \times (3.9 \times 10^{-4})}{(3.6 \times 10^{-2}) \times (4.3 \times 10^{-5})}\) [2U]
- दशमलव सङ्ख्या \(0.\overline{3}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the decimal number \(0.\overline{3}\) into fraction. [2U]
- \(1.525\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert \(1.525\) into fraction.[2U]
- दशमलव सङ्ख्या \(0.\overline{31}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the decimal number \(0.\overline{31}\) into fraction.[2U]
- दशमलव सङ्ख्या \(0.4\overline{1}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the decimal number \(0.4\overline{1}\) into fraction. [2U]
- \(0.\overline{24}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the number \(0.\overline{24}\) into fraction. [2U]
- \(1.5\overline{7}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the \(1.5\overline{7}\) into a fraction. [2U]
- दशमलव सङ्ख्या \(0.16\overline{7}\) लाई भिन्नमा रुपान्तरण गर्नुहोस् । Convert the \(0.16\overline{7}\) into a fraction. [2U]
The solution
\(\frac{9.6 \times 7.5}{2.4} \times \frac{10^{-3} \times 10^{-2}}{10^{-2}}\)
or\(4 \times 7.5 \times 10^{-3}\)
or\(30 \times 10^{-3}\)
So, result = \(30 \times 10^{-3} = 3.0 \times 10^{-2}\)
\(\frac{9.6 \times 7.5}{2.4} \times \frac{10^{-3} \times 10^{-2}}{10^{-2}}\)
or\(4 \times 7.5 \times 10^{-3}\)
or\(30 \times 10^{-3}\)
So, result = \(30 \times 10^{-3} = 3.0 \times 10^{-2}\)
The solution
\(\frac{9.8 \times 10^{-5} + 5.1 \times 10^{-6}}{2.2 \times 10^{-4}}\)
or\(\frac{9.8 \times 10^{-1} + 5.1 \times 10^{-2}}{2.2 }\)
or\(\frac{0.98 + 0.051}{2.2 }\)
or\(0.4686\)
\(\frac{9.8 \times 10^{-5} + 5.1 \times 10^{-6}}{2.2 \times 10^{-4}}\)
or\(\frac{9.8 \times 10^{-1} + 5.1 \times 10^{-2}}{2.2 }\)
or\(\frac{0.98 + 0.051}{2.2 }\)
or\(0.4686\)
The solution
\(\frac{(8.6 \times 10^{-3}) \times (3.9 \times 10^{-4})}{(3.6 \times 10^{-2}) \times (4.3 \times 10^{-5})}\)
or\(\frac{8.6 \times \times 3.9 \times 10^{-7}}{3.6 \times 4.3 \times 10^{-7} }\)
or\(\frac{8.6 \times \times 3.9 }{3.6 \times 4.3 }\)
or\(\frac{2 \times 1.3 }{1.2 }\)
or\(\frac{2 \times 13 }{12 }\)
or\(\frac{13 }{6 }\approx 2.17\)
\(\frac{(8.6 \times 10^{-3}) \times (3.9 \times 10^{-4})}{(3.6 \times 10^{-2}) \times (4.3 \times 10^{-5})}\)
or\(\frac{8.6 \times \times 3.9 \times 10^{-7}}{3.6 \times 4.3 \times 10^{-7} }\)
or\(\frac{8.6 \times \times 3.9 }{3.6 \times 4.3 }\)
or\(\frac{2 \times 1.3 }{1.2 }\)
or\(\frac{2 \times 13 }{12 }\)
or\(\frac{13 }{6 }\approx 2.17\)
The solution
Let
\(x = 0.\overline{3} = 0.3333\ldots\)
Then
\(10x = 3.3333\ldots\)
Subtracting, we get
\(10x - x = 3.333\ldots - 0.333 \)
or\(9x = 3\)
or\(x=\frac{1}{3}\)
Let
\(x = 0.\overline{3} = 0.3333\ldots\)
Then
\(10x = 3.3333\ldots\)
Subtracting, we get
\(10x - x = 3.333\ldots - 0.333 \)
or\(9x = 3\)
or\(x=\frac{1}{3}\)
The solution is
\(1.525 = \frac{1525}{1000}=\frac{61}{40}\)
\(1.525 = \frac{1525}{1000}=\frac{61}{40}\)
The solution
Let
\(x = 0.\overline{31}=0.3131 \cdots\)
Then
\(100x = 31.3131\ldots\)
Subtracting, we get
\(100x - x = 31.3131\ldots - 0.3131 \cdots \)
or\(99x = 31\)
or\(x=\frac{31}{99}\)
Let
\(x = 0.\overline{31}=0.3131 \cdots\)
Then
\(100x = 31.3131\ldots\)
Subtracting, we get
\(100x - x = 31.3131\ldots - 0.3131 \cdots \)
or\(99x = 31\)
or\(x=\frac{31}{99}\)
The solution
Let
\(x = 0.4\overline{1}=0.4111 \cdots\)
or\(10x = 4.111 \cdots\)
Then
\(100x = 41.111 \ldots\)
Subtracting, we get
\(100x - 10x = 41.111 \ldots-4.1111 \ldots\)
or\(90x = 37\)
or\(x=\frac{37}{90}\)
Therefore,
\(0.4\overline{1} = \frac{37}{90}\)
Let
\(x = 0.4\overline{1}=0.4111 \cdots\)
or\(10x = 4.111 \cdots\)
Then
\(100x = 41.111 \ldots\)
Subtracting, we get
\(100x - 10x = 41.111 \ldots-4.1111 \ldots\)
or\(90x = 37\)
or\(x=\frac{37}{90}\)
Therefore,
\(0.4\overline{1} = \frac{37}{90}\)
The solution
Let
\(x = 0.\overline{24}=0.2424 \cdots\)
Then
\(100x = 24.2424\ldots\)
Subtracting, we get
\(100x - x = 24.2424\ldots - 0.2424 \cdots \)
or\(99x = 24\)
or\(x=\frac{24}{99}\)
Simplifying,
\(x=\frac{8}{33}\)
Let
\(x = 0.\overline{24}=0.2424 \cdots\)
Then
\(100x = 24.2424\ldots\)
Subtracting, we get
\(100x - x = 24.2424\ldots - 0.2424 \cdots \)
or\(99x = 24\)
or\(x=\frac{24}{99}\)
Simplifying,
\(x=\frac{8}{33}\)
The solution
Let
\(x = 1.5\overline{7}=1.5777 \cdots\)
or\(10x = 15.777 \cdots\)
Then
\(100x = 157.777 \ldots\)
Subtracting, we get
\(100x - 10x = 157.777 \ldots - 15.777 \ldots\)
or\(90x = 142\)
or\(x=\frac{142}{90}\)
Simplifying,
\(x=\frac{71}{45}\)
Therefore,
\(1.5\overline{7} = \frac{71}{45}\)
Let
\(x = 1.5\overline{7}=1.5777 \cdots\)
or\(10x = 15.777 \cdots\)
Then
\(100x = 157.777 \ldots\)
Subtracting, we get
\(100x - 10x = 157.777 \ldots - 15.777 \ldots\)
or\(90x = 142\)
or\(x=\frac{142}{90}\)
Simplifying,
\(x=\frac{71}{45}\)
Therefore,
\(1.5\overline{7} = \frac{71}{45}\)
The solution
Let
\(x = 0.16\overline{7}=0.16777 \cdots\)
or\(100x = 16.777 \cdots\)
Then
\(1000x = 167.777 \ldots\)
Subtracting, we get
\(1000x - 100x = 167.777 \ldots - 16.777 \ldots\)
or\(900x = 151\)
or\(x=\frac{151}{900}\)
Therefore,
\(0.16\overline{7} = \frac{151}{900}\)
Let
\(x = 0.16\overline{7}=0.16777 \cdots\)
or\(100x = 16.777 \cdots\)
Then
\(1000x = 167.777 \ldots\)
Subtracting, we get
\(1000x - 100x = 167.777 \ldots - 16.777 \ldots\)
or\(900x = 151\)
or\(x=\frac{151}{900}\)
Therefore,
\(0.16\overline{7} = \frac{151}{900}\)
For Q.No.4(b) in BLE Exam
- एउटा मिठाइमा दुध र चिनीको अनुपात 5:3 छ । यदि दुध 750 gm छ भने चिनीको मात्रा कति होला ? पत्ता लगाउनुहोस् । The ratio of milk and sugar in a sweet is 5:3. If milk is 750 gm, then find the quantity of sugar.[1HA]
- 1 कि.मि. र 700 मिटरको अनुपात पत्ता लगाउनुहोस् । Find the ratio of 1 km and 700 meter.[1A]
- 1000 लाई 2:3 को अनुपातमा विभाजन गर्नुहोस् । Divide 1000 in the ratio of 2:3. [1HA]
- 20 वटा स्याउहरू रमेश र उमेशलाई 3:2 को अनुपातमा बाँडफाँड गर्नुहोस् । Divide 20 apples to Ramesh and Umesh in the ratio of 3:2. [1HA]
- यदि x:y=2:5 र y:z=15:8 भए x:y:z पत्ता लगाउनुहोस् । If x:y=2:5 and y:z=15:8, then find x:y:z. [1HA]
- यदि 2:4=x:16 भए x को मान पत्ता लगाउनुहोस् । If 2:4=x:16, then find the value of x. [1HA]
- यदि 4,x र 9 निरन्तर समानुपातिक हुनुछ भने x को धनात्मक मान पत्ता लगाउनुहोस् । If 4,x and 9 are in continued proportion, find the positive value of x. [1U]
- 4,10,28 बाट प्रत्येकमा कति सङ्ख्या घटाउँदा बाँकी रहेका सङ्ख्याहरू निरन्तर समानुपातिक हुन्छन् ? What number should be subtracted from each of the numbers 4,10 and 28 so that the remainder may be in continued proportion? [1HA]
- 12 र 21 दुवैमा कति जोड्ने तिनीहरूको अनुपात 5:8 हुन्छ ? What should be added to both 12 and 21 so that they are in the ratio 5:8? [1HA]
- A ले भन्दा B ले दोब्बर र B ले भन्दा C ले दोब्बर खर्च गर्छ । जम्मा रु. 2,100 खर्च भएछ भने प्रत्येकले कति खर्च गरेछन् ? B spent double of A and C spent double of B. If they spent altogether Rs. 2100, how much did each spend? [2HA]
- यदि 7,9,x र 18 समानुपातमा भए x को मान पत्ता लगाउनुहोस् । If 7,9,x and 18 are in proportion, find the value of x. [2U]
- सुशान्त र एन्जलले रु. 600 लाई 5:7 को अनुपातमा बाँडेछ । दुवैले कति रुपियाँ पाउँछन् ? पत्ता लगाउनुहोस् । If Rs. 600 is divided to Shushant and Angel in the ratio of 5:7, then find how much money did each get? [2HA]
- 90 जना विद्यार्थी भएको कक्षामा केटा र केटीको अनुपात 4:5 छ भने केटा र केटीको सङ्ख्या पत्ता लगाउनुहोस् । In a class of 90 students, the ratio of boys to the girls is 4:5. Find the number of boys and girls. [2HA]
- दुईवटा सङ्ख्याहरू 5:8 को अनुपातमा छन् । यदि तिनीहरूको अन्तर 147 छ भने ती सङ्ख्याहरू पत्ता लगाउनुहोस् । Two numbers are in the ratio 5:8. If their difference is 147, find the numbers. [2HA]
- बाबु र आमाको हालको उमेरको अनुपात 7:5 छ । यदि 6 वर्षपछि तिनीहरूको उमेरको अनुपात 4:3 हुने छ भने तिनीहरूको हालको उमेर पत्ता लगाउनुहोस् । The ratio of the present ages of a father and mother is 7:5. If after 6 years their ages will be in the ratio of 4:3, find their present ages.[2HA]
According to the question
ratio of milk and sugar in a sweet = 5:3
Therefore
quantity of milk=5x
quantity of sugar=3x
Also given that
milk = 750 gm
or5x = 750
orx =150
Hence, the quantity of sugar is
sugar=3x=3 \(\times\)150 = 450 gm
ratio of milk and sugar in a sweet = 5:3
Therefore
quantity of milk=5x
quantity of sugar=3x
Also given that
milk = 750 gm
or5x = 750
orx =150
Hence, the quantity of sugar is
sugar=3x=3 \(\times\)150 = 450 gm
According to the question
1 km = 1000 meters
So, the ratio of is.
Ratio =\(\dfrac{ 1000}{700}\)
orRatio =\(\dfrac{ 10}{7}\)
1 km = 1000 meters
So, the ratio of is.
Ratio =\(\dfrac{ 1000}{700}\)
orRatio =\(\dfrac{ 10}{7}\)
According to the question
Total amount = 1000
Total Ratio = 2+ 3=5
Therefore,
First part = \( \frac{2}{5} \times 1000 = 400 \)
Second part = \( \frac{3}{5} \times 1000 = 600 \)
Total amount = 1000
Total Ratio = 2+ 3=5
Therefore,
First part = \( \frac{2}{5} \times 1000 = 400 \)
Second part = \( \frac{3}{5} \times 1000 = 600 \)
According to the question
Total apples = 20
Total Ratio = 3 + 2 = 5
Therefore,
Ramesh's share = \( \frac{3}{5} \times 20 = 12 \) apple
Umesh's share = \( \frac{2}{5} \times 20 = 8 \) apple
Total apples = 20
Total Ratio = 3 + 2 = 5
Therefore,
Ramesh's share = \( \frac{3}{5} \times 20 = 12 \) apple
Umesh's share = \( \frac{2}{5} \times 20 = 8 \) apple
According to the question
x : y = 2 : 5
y : z = 15 : 8
To combine the ratios, make the value of y same in both.
LCM of 5 and 15 is 15.
Now, the combine ratio is
x:y and y:z
or2:5 and 15:8
or6:15 and 15:8
Now, combining ratio is
x : y : z = 6 : 15 : 8
x : y = 2 : 5
y : z = 15 : 8
To combine the ratios, make the value of y same in both.
LCM of 5 and 15 is 15.
Now, the combine ratio is
x:y and y:z
or2:5 and 15:8
or6:15 and 15:8
Now, combining ratio is
x : y : z = 6 : 15 : 8
According to the question
2 : 4 = x : 16
We can write this proportion as:
\( \frac{2}{4} = \frac{x}{16} \)
or\( 4x=32 \)
or\( x=8\)
2 : 4 = x : 16
We can write this proportion as:
\( \frac{2}{4} = \frac{x}{16} \)
or\( 4x=32 \)
or\( x=8\)
According to the question,
4, x, 9 are in continued proportion.
So, we have:
4 : x = x : 9
or\( \dfrac{4}{x} = \dfrac{x}{9} \)
or\(x^2 =36\)
or\(x = 6\)
4, x, 9 are in continued proportion.
So, we have:
4 : x = x : 9
or\( \dfrac{4}{x} = \dfrac{x}{9} \)
or\(x^2 =36\)
or\(x = 6\)
Let the number to be subtracted be \( x \)
Then the new numbers are:
\( 4 - x,\ 10 - x,\ 28 - x \)
Since they are in continued proportion,
\( (4 - x) : (10 - x) = (10 - x) : (28 - x) \)
or\( \frac{4 - x}{10 - x} = \frac{10 - x}{28 - x} \)
or\( (4 - x)(28 - x) = (10 - x)^2 \)
or\( 4×28 - 4x - 28x + x^2 = 100 - 20x + x^2 \)
or\( 112 - 32x + x^2 = 100 - 20x + x^2 \)
or\( 112 - 32x = 100 - 20x \)
or\( 112 - 100 = 32x - 20x \)
or\( 12 = 12x \)
or\( x = 1 \)
Then the new numbers are:
\( 4 - x,\ 10 - x,\ 28 - x \)
Since they are in continued proportion,
\( (4 - x) : (10 - x) = (10 - x) : (28 - x) \)
or\( \frac{4 - x}{10 - x} = \frac{10 - x}{28 - x} \)
or\( (4 - x)(28 - x) = (10 - x)^2 \)
or\( 4×28 - 4x - 28x + x^2 = 100 - 20x + x^2 \)
or\( 112 - 32x + x^2 = 100 - 20x + x^2 \)
or\( 112 - 32x = 100 - 20x \)
or\( 112 - 100 = 32x - 20x \)
or\( 12 = 12x \)
or\( x = 1 \)
Let the number to be added be \( x \)
Then the new numbers are:
\( 12 + x,\ 21 + x \)
According to the question,
\( (12 + x) : (21 + x) = 5 : 8 \)
or\( \dfrac{12 + x}{21 + x} = \dfrac{5}{8} \)
or\( 8(12 + x) = 5(21 + x) \)
or\( 96 + 8x = 105 + 5x \)
or\( 8x - 5x = 105 - 96 \)
or\( 3x = 9 \)
or\( x = 3 \)
Then the new numbers are:
\( 12 + x,\ 21 + x \)
According to the question,
\( (12 + x) : (21 + x) = 5 : 8 \)
or\( \dfrac{12 + x}{21 + x} = \dfrac{5}{8} \)
or\( 8(12 + x) = 5(21 + x) \)
or\( 96 + 8x = 105 + 5x \)
or\( 8x - 5x = 105 - 96 \)
or\( 3x = 9 \)
or\( x = 3 \)
Let the amount spent by A be \( x \)
Then,
B spent = \( 2x \)
C spent = \( 4x \)
Total amount spent = Rs. 2100, thus
\( x + 2x + 4x = 2100 \)
or\( 7x = 2100 \)
or\( x = 300 \)
Therefore,
A spent = \( x = \) Rs. 300
B spent = \( 2x = \) Rs. 600
C spent = \( 4x = \) Rs. 1200
Then,
B spent = \( 2x \)
C spent = \( 4x \)
Total amount spent = Rs. 2100, thus
\( x + 2x + 4x = 2100 \)
or\( 7x = 2100 \)
or\( x = 300 \)
Therefore,
A spent = \( x = \) Rs. 300
B spent = \( 2x = \) Rs. 600
C spent = \( 4x = \) Rs. 1200
According to the question,
7, 9, x, 18 are in proportion
So,
7 : 9 = x : 18
or\( \frac{7}{9} = \frac{x}{18} \)
Cross-multiplying:
\( 9x = 7 \times 18 \)
\( 9x = 126 \)
\( x = \frac{126}{9} = 14 \)
7, 9, x, 18 are in proportion
So,
7 : 9 = x : 18
or\( \frac{7}{9} = \frac{x}{18} \)
Cross-multiplying:
\( 9x = 7 \times 18 \)
\( 9x = 126 \)
\( x = \frac{126}{9} = 14 \)
According to the question,
Total amount = Rs. 600
Ratio = 5 : 7
Sum of ratio parts = 5 + 7 = 12
Therefore,
Shushant's share = \( \frac{5}{12} \times 600 = 250 \)
Angel's share = \( \frac{7}{12} \times 600 = 350 \)
Total amount = Rs. 600
Ratio = 5 : 7
Sum of ratio parts = 5 + 7 = 12
Therefore,
Shushant's share = \( \frac{5}{12} \times 600 = 250 \)
Angel's share = \( \frac{7}{12} \times 600 = 350 \)
According to the question,
Total number of students = 90
Ratio of boys to girls = 4 : 5
Sum of ratio parts = 4 + 5 = 9
Therefore,
Number of boys = \( \frac{4}{9} \times 90 = 40 \)
Number of girls = \( \frac{5}{9} \times 90 = 50 \)
Total number of students = 90
Ratio of boys to girls = 4 : 5
Sum of ratio parts = 4 + 5 = 9
Therefore,
Number of boys = \( \frac{4}{9} \times 90 = 40 \)
Number of girls = \( \frac{5}{9} \times 90 = 50 \)
Let the two numbers be \( 5x \) and \( 8x \)
According to the question,
Difference = 147
or\( 8x - 5x = 147 \)
or\( 3x = 147 \)
or\( x = 49 \)
Therefore,
First number = \( 5x = 5 \times 49 = 245 \)
Second number = \( 8x = 8 \times 49 = 392 \)
According to the question,
Difference = 147
or\( 8x - 5x = 147 \)
or\( 3x = 147 \)
or\( x = 49 \)
Therefore,
First number = \( 5x = 5 \times 49 = 245 \)
Second number = \( 8x = 8 \times 49 = 392 \)
Let the present ages of father and mother be \( 7x \) and \( 5x \) years respectively.
After 6 years,
Father's age = \( 7x + 6 \)
Mother's age = \( 5x + 6 \)
According to the question,
\( \frac{7x + 6}{5x + 6} = \frac{4}{3} \)
Cross-multiplying:
\( 3(7x + 6) = 4(5x + 6) \)
\( 21x + 18 = 20x + 24 \)
\( 21x - 20x = 24 - 18 \)
\( x = 6 \)
Therefore,
Father's present age = \( 7x = 7 \times 6 = 42 \) years
Mother's present age = \( 5x = 5 \times 6 = 30 \) years
After 6 years,
Father's age = \( 7x + 6 \)
Mother's age = \( 5x + 6 \)
According to the question,
\( \frac{7x + 6}{5x + 6} = \frac{4}{3} \)
Cross-multiplying:
\( 3(7x + 6) = 4(5x + 6) \)
\( 21x + 18 = 20x + 24 \)
\( 21x - 20x = 24 - 18 \)
\( x = 6 \)
Therefore,
Father's present age = \( 7x = 7 \times 6 = 42 \) years
Mother's present age = \( 5x = 5 \times 6 = 30 \) years
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