CBSE Sample Question Papers 2 for class 9 SA1 maths Q13 Term 1

## Add and subtract mixed fractions or mixed numbers

Add and subtract mixed fractions or mixed numbers

## NCERT Solutions for Class 10th Maths Chapter 6 Triangles Exercise 6.3 Question 4

NCERT Solutions for Class 10th Maths Chapter 6 Triangles Exercise 6.3 Question 4

## Decimal Number System

## Introduction

This system is based upon the place value and face value of a digit in a number. We have learnt that a natural number can be written as the sum of the place values of all digits of the numbers. For example

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1

Such a form of a natural number is known as its expanded form.

The expanded form of a number can also be expressed in terms of powers of 10 by using

\( {10}^{0} \) = 1, \( {10}^{1} \) = 10, \( {10}^{2} \) = 100, \( {10}^{3} \) = 1000 etc.

For example,

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1

=> 3256 = 3 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 5 x \( {10}^{1} \) + 6 x \( {10}^{0} \)

Clearly, each digit of the natural number is multiplied by \( {10}^{n} \), where n is the number of digits to its right and then they are added.

## Illustrative Examples

**Example 1:** Write the following numbers in the expanded exponential forms:

(i) 32005 (ii) 56719 (iii) 8605192 (iv) 2500132**Solution.**

(i) 32005 = 3 x \( {10}^{4} \) + 2 x \( {10}^{3} \) + 0 x \( {10}^{2} \) + 0 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

(ii) 560719 = 5 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 7 x \( {10}^{2} \) + 1 x \( {10}^{1} \) + 9 x \( {10}^{0} \)

(iii) 8605192 = 8 x \( {10}^{6} \) + 6 x \( {10}^{5} \) + 0 x \( {10}^{4} \) + 5 x \( {10}^{3} \) + 1 x \( {10}^{2} \) + 9 x \( {10}^{1} \) + 2 x \( {10}^{0} \)

(iv) 2500132 = 2 x \( {10}^{6} \) + 5 x \( {10}^{5} \) + 0 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 1 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 2 x \( {10}^{0} \)

**Example 2:** Find the number from each of the following expanded forms:

(i) 5 x \( {10}^{4} \) + 4 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

(ii) 7 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 9 x \( {10}^{0} \)

(iii) 9 x \( {10}^{5} \) + 4 x \( {10}^{2} \) + 1 x \( {10}^{1} \)

**Solution.**

(i) 5 x \( {10}^{4} \) + 4 x \( {10}^{3} \) + 2 x \( {10}^{2} \) + 3 x \( {10}^{1} \) + 5 x \( {10}^{0} \)

= 5 x 10000 + 4 x 1000 + 2 x 100 + 3 x 10 + 5 x 1

= 50000 + 4000 + 200 + 30 + 5

= 54235

(ii) 7 x \( {10}^{5} \) + 6 x \( {10}^{4} \) + 0 x \( {10}^{3} \) + 9 x \( {10}^{0} \)

= 7 x 100000 + 6 x 10000 + 0 + 9 x 1

= 700000 + 60000 + 9

= 760009

(iii) 9 x \( {10}^{5} \) + 4 x \( {10}^{2} \) + 1 x \( {10}^{1} \)

= 9 x 100000 + 4 x 100 + 1 x 10

= 900000 + 400 + 10

= 900410

## Numbers In Standard Form

## Standard Form

A number written as ( m x \(10^n\) ) is said to be in standard form if m is a decimal number such that 1 \( \le \) m \(< \)10 and n is either a positive or a negative integer.

The standard form of a number is also known as Scientific notation.

## Expressing Very Large Numbers in Standard Form

In order to write large numbers in the standard form,following steps must be followed:

* STEP I*– Obtain the number and move the decimal point to the left till you get just one digit to the left of the decimal point.

* STEP II*– Write the given number as the product of the number so obtained and \(10^n\) , where n is the number of places the decimal point has been moved to the left. If the given number is between 1 and 10, then write it as the product of the number itself and \(10^0\) .

**Illustrative Examples**

**Example 1:** Express the following numbers in the standard form:

(i) 3,90,878 (ii) 3,186,500,000 (iii) 65,950,000

**Solution.**

(i) We have,

3,90,878 = 390878.00

Clearly, the decimal point is moved through five places to obtain a number in which there is just one digit to the left of the decimal point.

Therefore, 390878.00 = 3.90878 x \(10^5\)

(ii) We have,

3,186,500,000 = 3.186500000 x \(10^9\)

= 3.1865 x \(10^9\)

(iii) We have,

65,950,000 = 65,950,000.00

= 6.5950000 x \(10^7\)

= 6.595 x \(10^7\)

**Example 2:** The distance between sun and earth is (1.496 x \( {10}^{11} \)) m and the distance between earth and moon is (3.84 x \(10^8\)) m. During solar eclipse moon comes in between earth and sun. At that time what Is the distance between moon and sun?

**Solution.** Required distance

= {(1.496 x \( {10}^{11} \)) – (3.84 x \(10^8\)) } m

= {\( \frac { 1496\times { 10 }^{ 11 } }{ { 10 }^{ 3 } } \) – (3.84 x \(10^8\))} m

= {1496 x \(10^8\)) – (3.84 x \(10^8\))} m

= {(1496 – 3.84) x \(10^8\))} m

= (1492.16 x \(10^8\)) m

Hence, the distance between moon and sun is (1492.16 x \(10^8\) ) m.

**Example 3:** Write the following numbers in the usual form:

(i) 7.54 x \(10^6\) (ii)2.514 x \(10^7\)

**Solution.** We have

(i) 7.54 x \(10^6\)

= \(\frac{754}{100}\) x \(10^6\)

= \(\frac{754 \times {10}^{6} }{{10}^{2}}\)

= 754 x \({10}^{(6-2)}\)

= (754 x \({10}^{4}\) )

= (754 x 10000) = 7540000

(ii) 2.514 x \(10^7\)

= \(\frac{2514}{1000}\) x \(10^7\)

= \(\frac{2514 \times {10}^{7} }{{10}^{3}}\)

= 2514 x \({10}^{(7-3)}\)

= (2514 x \({10}^{4}\) )

= (2514 x 10000) = 25140000

## Expressing Very Small Numbers in Standard Form

In order to write very small numbers in the standard form,following steps must be followed:

* STEP I-* Obtain the number and count the number of decimal values after the decimal point. Consider it as n.

* STEP II-* Divide the number by \( {10}^{n} \)). If the number is between 1 and 10, then write it as the product of the number itself and \( {10}^{-n} \)

**Example 1:** Write the following numbers in the standard form:

(i) 0.000000059 (ii) 0.00000000526**Solution.** We may write:

(i) 0.000000059

= \(\frac{59}{{10}^{9}}\)

= \(\frac{5.9 \times 10}{{10}^{9}}\)

= \(\frac{5.9}{{10}^{8}}\) = (5.9 x \( {10}^{-8} \))

(ii) 0.00000000526

= \(\frac{526}{{10}^{11}}\)

= \(\frac{5.26 \times 100}{{10}^{11}}\)

= \(\frac{5.26 \times {10}^{2}}{{10}^{11}}\)

= \(\frac{5.26}{{10}^{(11 – 2)}}\)

= \(\frac{5.26}{{10}^{9}}\) = (5.26 x \( {10}^{-9} \))

**Example 2:** The size of a red blood cell is 0.000007 m and that of a plant cell Is 0.00001275 m. Show that a red blood cell is half of plant cell in size.

**Solution.** We have,

Size of a red blood cell = 0.000007 m = \(\frac{7}{{10}^{6}}\) m = (7 x \( {10}^{-6} \))

Size of a plant cell

= 0.00001275 m

= \(\frac{1275}{{10}^{8}}\) m

= \(\frac{1.275 \times {10}^{3}}{{10}^{8}}\) m

= \(\frac{1.275}{{10}^{(8-3)}}\) m

= \(\frac{1.275}{{10}^{5}}\) m = (1.275 x \( {10}^{-5} \)) m

\(\frac{Size of a red blood cell}{Size of a plant cell}\)

= \(\frac{7 \times {10}^{-6}}{1.275 \times {10}^{-5}}\)

= \(\frac{7 \times {10}^{-6 + 5}}{1.275}\)

= \(\frac{7 \times {10}^{-1}}{1.275}\)

= \(\frac{7}{1.275 \times 10}\)

= \(\frac{7}{12.75}\)

= \(\frac{7}{13}\) (nearly)

= \(\frac{1}{2}\) (approximately)

Therefore, size of a red blood cell = \(\frac{1}{2}\) x (size of a plant cell)

**Example 3:** Express the following numbers in usual form:

(i) 3 x \( {10}^{-3} \) (ii) 2.34 x \( {10}^{-4} \)**Solution.** We have,

(i) 3 x \( {10}^{-3} \)

= \(\frac{3}{{10}^{3}}\)

= \(\frac{3}{1000}\) = 0.003

(ii) 2.34 x \( {10}^{-4} \)

= \(\frac{234}{100}\) x \(\frac{1}{{10}^{4}}\)

= \(\frac{234}{{10}^{2} \times {10}^{4}}\)

= \(\frac{234}{{10}^{6}}\)

= \(\frac{234}{1000000}\) = 0.000234