Cube and cube roots -2

Math Problems

1. The smallest number by which $686$ should be divided to make it a perfect cube is:

   Solution:

   The prime factorization of $686$ gives $2*7^3$. To make it a perfect cube, it should be divided by $2$. Hence, the smallest number by which $686$ should be divided to make it a perfect cube is $2$.

2. Find cube root of:

   1. $1\frac{127}{216}$
   2. $9.261$

   Solution:

   1. The cube root of $1\frac{127}{216}$ is $1\frac{1}{3}$.
   2. The cube root of $9.261$ is approximately $2.1$.

3. Difference of two perfect cubes is $387$. If the cube root of the greater of two numbers is $8$, find the cube root of smaller number.

   Solution:

   If the cube root of the greater number is $8$, then the greater number is $8^3=512$. Hence, the smaller number is $512-387=125$. The cube root of $125$ is $5$. So, the cube root of the smaller number is $5$.

Cubes and cube roots

Chapter: Cubes and Roots

1. Find the smallest number by which the following numbers should be divided to obtain a perfect cube.

   1. $13718$
   2. $28672$

   Solution:

   1. The smallest number to divide $13718$ to obtain a perfect cube is $2$.
   2. The smallest number to divide $28672$ to obtain a perfect cube is $1$ (since $28672$ is already a perfect cube).

2. Find the volume of a cubical box whose surface area is $486 cm^2$.

   Solution:

   The surface area of a cube is given by $6a^2$, where $a$ is the length of one side. Hence, $a = \sqrt{\frac{Surface Area}{6}} = \sqrt{\frac{486}{6}} = 9 cm$. The volume of a cube is $a^3 = 9^3 = 729 cm^3$.

3. Find the cube of $-5 \frac{1}{7}$.

   Solution:

   $-5 \frac{1}{7} = -\frac{36}{7}$, so its cube is $\left(-\frac{36}{7}\right)^3 = -\frac{46656}{343}$.

4. Divide the number $8748$ by the smallest number so that the quotient is a perfect cube. Also find the cube root of the quotient.

   Solution:

   The smallest number to divide $8748$ to obtain a perfect cube is $3$. So, the quotient is $\frac{8748}{3} = 2916$, which is a perfect cube. The cube root of $2916$ is $14$.

5. Three numbers are in the ratio $3:4:5$. If their product is $480$, find the numbers.

   Solution:

   If the three numbers are in the ratio $3:4:5$, let's denote the numbers as $3x, 4x,$ and $5x$. Therefore, $3x \cdot 4x \cdot 5x = 480$ which gives $x^3 = \frac{480}{60} = 8$. So, $x = \sqrt[3]{8} = 2$. Thus, the three numbers are $3x = 6, 4x = 8,$ and $5x = 10$.

Square and square roots 3

Math Problems

1. Given that $ \sqrt{1521} = 39 $, find the value of $ \sqrt{0.1521} + \sqrt{15.21} $.

   Solution:

   We know that $ \sqrt{1521} = 39 $, so if we divide 1521 by 10000 and 100 respectively we can find the square root of $ 0.1521 $ and $ 15.21 $.

   Therefore, $ \sqrt{0.1521} = 0.39 $ and $ \sqrt{15.21} = 3.9 $.

   Adding these together, $ \sqrt{0.1521} + \sqrt{15.21} = 0.39 + 3.9 = 4.29 $.

2. Find the sum of the series $ 1 + 3 + 5 + 7 + \cdots $ up to n terms.

   Solution:

   This is the sum of the first $ n $ odd numbers, and it is well known that the sum of the first $ n $ odd numbers is $ n^2 $.

   Therefore, $ 1 + 3 + 5 + 7 + \cdots $ up to $ n $ terms is equal to $ n^2 $.

Square and square roots-2

Math Problems

1. Amit walks 16m south from his friend's house. While returning, he walks diagonally from his friend's house to reach back to his own house. What distance did he walk while returning?

   Solution:

   As Amit walks in a straight line to his friend's house and then diagonally back, it forms a right-angled triangle. We can use the Pythagorean theorem to find the length of the hypotenuse (the diagonal path), where the distance walked to the friend's house is one of the sides, and is equal to 16m (the other side is assumed to be of the same length, considering he reached the same point).

   Using Pythagorean theorem \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse:

   \(c = \sqrt{16^2 + 16^2} = \sqrt{512} = 16\sqrt{2} \approx 22.63m\)

2. Find the square root of the following:

   1. \( \sqrt{16169/441} \)
   2. \( \sqrt{5.774409} \)
   3. \( \sqrt{841/1521} \)

   Solution:

   1. \( \sqrt{16169/441} = \sqrt{36.66} \approx 6.05 \)
   2. \( \sqrt{5.774409} \approx 2.4 \)
   3. \( \sqrt{841/1521} = \sqrt{0.55} \approx 0.74 \)

3. \( \sqrt{208 + \sqrt{2304}} \)

   Solution:

   \( \sqrt{208 + \sqrt{2304}} = \sqrt{208 + 48} = \sqrt{256} = 16 \)

4. \( \sqrt{0.0016} = ? \)

   Solution:

   \( \sqrt{0.0016} = 0.04 \)

5. Given that \( \sqrt{1521} = 39 \), find the value of \( \sqrt{0.1521} + \sqrt{15.21} \).

   Solution:

   \( \sqrt{0.1521} + \sqrt{15.21} = 0.39 + 3.9 = 4.29 \)

6. \( 12.1 + 3 + 5 + 7 + \cdots \) up to n terms is equal to ?

   Solution:

   This is an arithmetic series where the first term \( a = 3 \), the common difference \( d = 2 \), and the number of terms \( n \) is given. The sum of an arithmetic series can be calculated by the formula \( S = \frac{n}{2}[2a + (n-1)d] \). However, this series has an additional term (12.1) that is not part of the arithmetic progression. So, the sum up to \( n \) terms will be:

   \( 12.1 + \frac{n}{2}[2*3 + (n-1)*2] = 12.1 + n[3 + n - 1] = 12.1 + 3n + n^2 - n = 12.1 + 2n + n^2 \).

Square and square roots-1

1. Show that the following numbers are perfect squares:

(a) 2304

Solution:

Let's find the square root of 2304:

Starting from the right, we group the numbers into pairs of two. So 2304 becomes (23, 04).

We find the largest square less than or equal to the first group, 23. This square is $4^2 = 16$. We subtract this from 23 to get a remainder of 7.

We then bring down the next group of numbers, 04, to get 704. The next digit in the square root will be 8, because $48 \times 8 = 384$ which is less than 704. Subtracting this from 704 gives a remainder of 320.

Since we don't have any more groups of numbers to bring down, we can say that $\sqrt{2304} = 48$, which means that 2304 is indeed a perfect square.

(b) 3025

Solution:

Following the same process as above, we find that the square root of 3025 is 55, hence it's a perfect square.

2. Find the smallest number by which 1008 should be multiplied to make it a perfect square:

Solution:

We factorize 1008 as $2^4 \times 3^2 \times 7$. We see that 7 is missing a pair to become a perfect square. Therefore, the smallest number 1008 should be multiplied by is 7.

3. Express the following as the sum of two consecutive integers:

(a) $21^2$

Solution:

$21^2 = 441 = 220 + 221$, so $21^2$ can be expressed as the sum of two consecutive integers 220 and 221.

(b) $19^2$

Solution:

$19^2 = 361 = 180 + 181$, so $19^2$ can be expressed as the sum of two consecutive integers 180 and 181.

4. Write a Pythagorean triplet whose one number is 80:

Solution:

A Pythagorean triplet follows the pattern $(m^2 - n^2, 2mn, m^2 + n^2)$ where $m > n > 0$. If we choose $m = 9$ and $n = 1$, we get the triplet (80, 18, 82).

5. Express 324 as the sum of 18 odd numbers:

Solution:

The average of these 18 odd numbers must be an integer. Hence $324 / 18 = 18$ is the middle odd number. So the 18 odd numbers are the 9 odd numbers before 18, 18 itself, and the 8 odd numbers after 18: (1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35).

6. Without calculating, find the number of digits in the square root of 368645:

Solution:

We can express 368645 as $3.68645 \times 10^5$. The square root of $10^5$ is $10^{5/2} = 10^2.5$. Since this is between $10^2$ and $10^3$, the square root of 368645 must have 3 integer digits.

Exponents -2

1. By what number should $(2/7)^{-2}$ be divided so that the quotient becomes 49?

Solution:

$$ \frac{(2/7)^{-2}}{x} = 49 \Rightarrow x = \frac{(2/7)^{-2}}{49} = 1 $$

2. Find x so that $(3/8)^{14} \times (8/3)^{-9} = (3/8)^{2x-1}$.

Solution:

$$ 14 - 9 = 2x - 1 \Rightarrow x = \frac{14 - 9 + 1}{2} = 3 $$

3. Find the value of a/b if $(a/b)^{-6}=(2/7)^{-6} \times (14/9)^{-6}$.

Solution:

$$ (a/b)^{-6} = (2/7)^{-6} \times (14/9)^{-6} \Rightarrow \frac{a}{b} = \frac{2 \times 14}{7 \times 9} = \frac{4}{3} $$

4. Simplify $(3^{-5} \times 10^{-5} \times 125) / (5^{-7} \times 6^{-5})$.

Solution:

$$ \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} = 6^5 \times 5^2 = 90000 $$

5. Mass of Earth is $5.97 \times 10^{24}$ kg and mass of Moon is $7.35 \times 10^{22}$ kg. What is the total mass?

Solution:

$$ 5.97 \times 10^{24} + 7.35 \times 10^{22} = 5.97 \times 10^{24} + 0.0735 \times 10^{24} = 6.0435 \times 10^{24} $$ kg

Exponents

1. Simplify $(512)^{-2/3}$:

Solution:

$$ (512)^{-2/3} = (2^9)^{-2/3} = 2^{-6} = \frac{1}{64} $$

2. Find the value of x:

(a) Assuming & represents multiplication in $\sqrt{5*5x+2}=2$

Solution:

$$ \sqrt{5*5x+2}=2 \Rightarrow 25x + 2 = 4 \Rightarrow 25x = 2 \Rightarrow x = \frac{2}{25} $$

(b) $5^{x-3}×3^{2x-8}=225$

Solution:

$$ 5^{x-3}×3^{2x-8}=225 \Rightarrow 5^{x-3} = \frac{225}{3^{2x-8}} \Rightarrow 5^{x-3} = \frac{225}{9^{x-4}} \Rightarrow 5^{x-3} = 5^2 \Rightarrow x = 5 $$

3. If $(2^{3x-1}+10)÷7=6$, then find the value of x:

Solution:

$$ (2^{3x-1}+10)÷7=6 \Rightarrow 2^{3x-1} + 10 = 42 \Rightarrow 2^{3x-1} = 32 \Rightarrow 3x - 1 = 5 \Rightarrow x = 2 $$

4. Write in standard form:

(a) $0.0000000777$

Solution:

$$ 0.0000000777 = 7.77 \times 10^{-8} $$

(b) $234.45×10^{12}$

Solution:

$$ 234.45×10^{12} = 2.3445 \times 10^{14} $$

5. Evaluate $[((-2)/3)^{-3}-(-3/4)^{-3} ]÷(3/2)^2$:

Solution:

$$ \frac{(-\frac{2}{3})^{-3}-(-\frac{3}{4})^{-3}}{(\frac{3}{2})^2} = \frac{-\frac{27}{8} + \frac{64}{27}}{2.25} = -\frac{216}{216} = -1 $$

Rational number 3

Rachna purchased $2 \frac{1}{2}$ kg onions at $10$ per kg and $1 \frac{3}{8}$ kg potatoes at $116 \frac{8}{11}$ per kg. She gave a five hundred rupee note to the shopkeeper. Find the amount she got back from the shopkeeper.

First, let's calculate the cost of onions and potatoes separately:

Cost of onions:

$$ 2 \frac{1}{2} \times 10 = \frac{5}{2} \times 10 = 25 $$

Cost of potatoes:

$$ 1 \frac{3}{8} \times 116 \frac{8}{11} = \frac{11}{8} \times \frac{1290}{11} = 161.25 $$

Therefore, the total cost of onions and potatoes is:

$$ 25 + 161.25 = 186.25 $$

Since Rachna gave the shopkeeper a five hundred rupee note, the amount she got back is:

Amount Rachna got back:

$$ 500 - 186.25 = 313.75 $$

Rational numbers part 2

1. Write in standard form:

(a) $12/(-240)$:

$$ \frac{12}{-240} = -\frac{12}{240} = -\frac{1}{20} $$

(b) $(-19)/362$:

$$ \frac{-19}{362} = -\frac{19}{362} $$

2. Using appropriate property find $-2/3×3/5+5/2×4/7+3/5×1/6$:

$$ -\frac{2}{3}×\frac{3}{5}+\frac{5}{2}×\frac{4}{7}+\frac{3}{5}×\frac{1}{6} = -\frac{2}{5} + \frac{10}{7} + \frac{1}{10} = -\frac{14}{70} + \frac{100}{70} + \frac{7}{70} = \frac{93}{70} = \frac{186}{140} $$

3. What should be subtracted from the product of $3/7$ and $2/5$ to get $4/35$?

$$ \frac{3}{7} × \frac{2}{5} - \frac{4}{35} = \frac{6}{35} - \frac{4}{35} = \frac{2}{35} $$

4. One sixth of the students of a class joined the sports club. Three fifth of these students opted to play table tennis. If 6 students play table tennis, how many students are there in the class?

$$ \frac{1}{6} \times \frac{3}{5} = \frac{1}{10} $$ So 10 students are in the class.

5. Simplify $(10/21×(-7)/5)-(2/3×9/(-16))+((-4)/15×(-6)/16)$:

$$ \frac{10}{21}×\frac{-7}{5}-\frac{2}{3}×\frac{9}{-16}+\frac{-4}{15}×\frac{-6}{16} = -\frac{14}{21} + \frac{3}{8} + \frac{1}{10} = -\frac{2}{3} + \frac{3}{8} + \frac{1}{10} = -\frac{40}{120} + \frac{45}{120} + \frac{12}{120} = \frac{17}{120} $$

6. Is $4/11$ the multiplicative inverse of $-3\ 3/4$? Give reasons.

$$ \frac{4}{11} × -\frac{15}{4} = -\frac{60}{44} = -\frac{15}{11} $$ Since this result is not equal to 1, $4/11$ is not the multiplicative inverse of $-3\ 3/4$.

Rational numbers part I

1. Subtract $(-11)/5$ from additive inverse of sum of $-5/8$ and $5/6$.

Firstly, calculate the sum of $-5/8$ and $5/6$:

$$ -\frac{5}{8} + \frac{5}{6} = -\frac{15}{24} + \frac{20}{24} = \frac{5}{24} $$

The additive inverse of this sum is $-\frac{5}{24}$. Now, subtract $(-11)/5$ from $-\frac{5}{24}$:

$$ -\frac{5}{24} - (-\frac{11}{5}) = -\frac{5}{24} + \frac{11}{5} = -\frac{25}{120} + \frac{264}{120} = \frac{239}{120} = \frac{119}{60} $$

2. Subtract the sum of $(-4)/7$ and $5/14$ from the sum of $9/14$ and $23/28$.

Firstly, calculate the sum of $(-4)/7$ and $5/14$:

$$ -\frac{4}{7} + \frac{5}{14} = -\frac{8}{14} + \frac{5}{14} = -\frac{3}{14} $$

Next, calculate the sum of $9/14$ and $23/28$:

$$ \frac{9}{14} + \frac{23}{28} = \frac{18}{28} + \frac{23}{28} = \frac{41}{28} $$

Now, subtract $-\frac{3}{14}$ from $\frac{41}{28}$:

$$ \frac{41}{28} - (-\frac{3}{14}) = \frac{41}{28} + \frac{6}{28} = \frac{47}{28} = \frac{47}{2} $$

3. What number should be added to $-7/8$ so as to get $5/9$?

To find the number that should be added to $-7/8$ to get $5/9$, simply subtract $-7/8$ from $5/9$:

$$ \frac{5}{9} - \left(-\frac{7}{8}\right) = \frac{5}{9} + \frac{7}{8} = \frac{40}{72} + \frac{63}{72} = \frac{103}{72} $$

4. Verify the distributive law of multiplication over subtraction for the rational numbers x=$(-3)/4$, y=$5/7$, z=$(-4)/5$.

The distributive law of multiplication over subtraction states that for any numbers x, y, and z, we have $x \cdot (y - z) = x \cdot y - x \cdot z$.

Therefore, substituting the given values, we get:

$$ -\frac{3}{4} \cdot \left(\frac{5}{7} - -\frac{4}{5}\right) = -\frac{3}{4} \cdot \frac{5}{7} - -\frac{3}{4} \cdot -\frac{4}{5} $$

Now, compute each side:

$$ -\frac{3}{4} \cdot \left(\frac{5}{7} + \frac{4}{5}\right) = -\frac{3}{4} \cdot \left(\frac{25}{35} + \frac{28}{35}\right) = -\frac{3}{4} \cdot \frac{53}{35} = -\frac{159}{140} $$ $$ -\frac{3}{4} \cdot \frac{5}{7} - -\frac{3}{4} \cdot -\frac{4}{5} = -\frac{15}{28} - \frac{12}{20} = -\frac{159}{140} $$

Since the left-hand side equals the right-hand side, this verifies the distributive law of multiplication over subtraction for the given rational numbers.

5. Write 6 rational numbers between $3/(-7)$ and $2/5$.

We can find 6 rational numbers between $3/(-7)$ and $2/5$ as: $-6/35$, $-4/35$, $-2/35$, $1/35$, $3/35$, $5/35$.

Cube and cube roots part-2

Solutions

1. Question 9: Calculate $\sqrt{208+\sqrt{2304}}$

   Solution: First, calculate the square root of 2304 and then add it to 208 before taking the square root of the sum.

   $$ = \sqrt{208+\sqrt{2304}} $$ $$ = \sqrt{208+48} $$ $$ = \sqrt{256} $$ $$ = 16 $$

2. Question 10: Calculate $\sqrt{0.0016}$

   Solution: The square root of 0.0016 is:

   $$ = \sqrt{0.0016} $$ $$ = \sqrt{16 \times 10^{-4}} $$ $$ = \sqrt{16} \times \sqrt{10^{-4}} $$ $$ = 4 \times 10^{-2} $$ $$ = 0.04 $$

3. Question 11: Given that $\sqrt{1521} = 39$, find the value of $\sqrt{0.1521}+\sqrt{15.21}$.

   Solution: This seems to be an error, because $\sqrt{1521}$ should be equal to 39. Assuming the correct value is $\sqrt{1521} = 39$, let's calculate:

   $$ = \sqrt{0.1521} + \sqrt{15.21} $$ $$ = \sqrt{\frac{1521}{10000}} + \sqrt{\frac{1521}{100}} $$ $$ = \frac{39}{100} + \frac{39}{10} $$ $$ = 0.39 + 3.9 $$ $$ = 4.29 $$

4. Question 12: 1 + 3 + 5 + 7 + ... up to n terms is equal to?

   Solution: This is the sum of the first n odd numbers. The sum of the first n odd numbers is equal to $n^2$. Therefore,

   $$ 1 + 3 + 5 + 7 + ... = n^2 $$

Cube and cube roots grade VIII

Math Problems and Solutions - Cubes and Roots

1. Question: Find the smallest number by which the following must be divided to obtain a perfect cube.

   (a) 13718

   (b) 28672

   Solution:

   (a) Prime factorization of 13718 is \(2 \times 3 \times 3 \times 761\). To make it a perfect cube, it must be divided by \(2 \times 3 = 6\).

   (b) Prime factorization of 28672 is \(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3\). To make it a perfect cube, it must be divided by \(2 \times 2 \times 3 = 12\).

2. Question: Find the volume of a cubical box whose surface area is \(486 \, cm^2\).

   Solution: Let the side of the cubical box be \(a\). The surface area of the cubical box is \(6a^2\). We are given that \(6a^2 = 486\), so \(a^2 = \frac{486}{6} = 81\). Therefore, \(a = 9\). The volume of the cubical box is \(a^3 = 9^3 = 729 \, cm^3\).

3. Question: Find the cube of \(-\frac{36}{7}\).

   Solution: The cube of \(-\frac{36}{7}\) is \(\left(-\frac{36}{7}\right)^3 = -\frac{36^3}{7^3} = -\frac{46656}{343} \approx -136.123\).

4. Question: Divide the number 8748 by the smallest number so that the quotient is a perfect cube. Also, find the cube root of the quotient.

   Solution: Prime factorization of 8748 is \(2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3\). To make it a perfect cube, it must be divided by \(2 \times 2 \times 3 \times 3 = 12\). The quotient is \(\frac{8748}{12} = 729\), which is a perfect cube. The cube root of 729 is \(\sqrt[3]{729} = 9\).

Square and square roots part-4

Solutions

1. Question 1: Calculate \(\sqrt{208+\sqrt{2304}}\)

   Solution: First, calculate the square root of 2304 and then add it to 208 before taking the square root of the sum.

   \[ = \sqrt{208+\sqrt{2304}} \] \[ = \sqrt{208+48} \] \[ = \sqrt{256} \] \[ = 16 \]

2. Question 2: Calculate \(\sqrt{0.0016}\)

   Solution: The square root of 0.0016 is:

   \[ = \sqrt{0.0016} \] \[ = \sqrt{16 \times 10^{-4}} \] \[ = \sqrt{16} \times \sqrt{10^{-4}} \] \[ = 4 \times 10^{-2} \] \[ = 0.04 \]

3. Question 3: Given that \(\sqrt{1521}=39\), find the value of \(\sqrt{0.1521}+\sqrt{15.21}\).

   Solution: This seems to be an error, because \(\sqrt{1521}\) should be equal to 39. Assuming the correct value is \(\sqrt{1521} = 39\), let's calculate:

   \[ = \sqrt{0.1521} + \sqrt{15.21} \] \[ = \sqrt{\frac{1521}{10000}} + \sqrt{\frac{1521}{100}} \] \[ = \frac{39}{100} + \frac{39}{10} \] \[ = 0.39 + 3.9 \] \[ = 4.29 \]

4. Question 4: 1 + 3 + 5 + 7 + \(\cdots\) up to n terms is equal to?

   Solution: This is the sum of the first n odd numbers. The sum of the first n odd numbers is equal to \(n^2\). Therefore,

   \[ 1 + 3 + 5 + 7 + \cdots = n^2 \]

Square and square roots part -3

Math Problems and Solutions

1. Question: Amit walks 16 m south from his friend's house while returning, he walks diagonally from his friend's house to reach back to his house. What distance did he walk while returning?

   Solution: Amit walked 16 m south from his friend's house. Let's assume his house is east of his friend's house. The distance he walked while returning is the diagonal of a square with sides of 16 m (because the distance he initially walked south is the same as the distance east). Using the Pythagorean theorem, the diagonal is \(\sqrt{16^2 + 16^2} = \sqrt{2} \times 16 \approx 22.63\) m.

2. Question: Find the square root of the following.

   (a) \(16 \frac{169}{441}\)

   (b) \(5.774409\)

   (c) \(\frac{841}{1521}\)

   Solution:

   (a) \(16 \frac{169}{441} = 16 + \frac{169}{441} = 16 + \frac{13}{21} = \frac{357}{21}\). The square root of \(\frac{357}{21}\) is \(\sqrt{\frac{357}{21}} = \frac{\sqrt{357}}{\sqrt{21}} \approx 4.49\).

   (b) The square root of \(5.774409\) is \(\sqrt{5.774409} \approx 2.4026\).

   (c) \(\frac{841}{1521}\) can be simplified as \(\frac{29}{39}\). The square root of \(\frac{29}{39}\) is \(\sqrt{\frac{29}{39}} = \frac{\sqrt{29}}{\sqrt{39}} \approx 0.76\).

Square and square roots Grade VIII Worksheet with solutions Part-2

Solutions

1. Question: Express 324 as the sum of 18 odd numbers.

   Solution: Let's express 324 as the sum of 18 odd numbers. Let the first odd number be \(a\), and since odd numbers have a common difference of 2, the rest of the numbers will be \(a + 2\), \(a + 4\), ..., \(a + 34\). Sum of these 18 odd numbers is 324:

   \[ 18a + 2(1 + 2 + 3 + \ldots + 17) = 324 \] \[ 18a + 2 \times \frac{17 \times 18}{2} = 324 \] \[ 18a + 306 = 324 \] \[ 18a = 18 \] \[ a = 1 \] So, the 18 odd numbers are 1, 3, 5, ..., 35.

2. Question: Without calculating, find the number of digits in the square root of 368645.

   Solution: We can write 368645 as \(3.68645 \times 10^5\). Now, the square root of 368645 will be the square root of this expression:

   \[ \sqrt{3.68645 \times 10^5} = \sqrt{3.68645} \times \sqrt{10^5} = \sqrt{3.68645} \times 10^{\frac{5}{2}} = \sqrt{3.68645} \times 10^{2.5} \] As \(10^{2.5}\) is between \(10^2\) and \(10^3\), the square root will have a value between 100 and 1000, which means it will have 3 digits before the decimal point.

Square and square roots Grade VIII worksheet with solution

Squares and Square Roots

1. Question: Show that the following numbers is a perfect square

   (a) 2304

   (b) 3025

   Solution:

   (a) \(\sqrt{2304} = 48\), since 48 is an integer, 2304 is a perfect square.

   (b) \(\sqrt{3025} = 55\), since 55 is an integer, 3025 is a perfect square.

2. Question: Find the smallest number by which 1008 should be multiplied to make it a perfect square.

   Solution: The prime factorization of 1008 is \(2^4 \times 3^2 \times 7\). For it to be a perfect square, each prime factor must occur an even number of times. The smallest number by which 1008 should be multiplied is 7. Then, 1008 multiplied by 7 is \(2^4 \times 3^2 \times 7^2\), which is a perfect square.

3. Question: Express

   (a) \(21^2\)

   (b) \(19^2\) as the sum of two consecutive integers.

   Solution:

   (a) \(21^2 = 441 = 220 + 221\) (sum of two consecutive integers).

   (b) \(19^2 = 361 = 180 + 181\) (sum of two consecutive integers).

4. Question: Write a Pythagorean triplet whose one number is 80.

   Solution: We can create a Pythagorean triplet using the formulas \(a = 2mn\), \(b = m^2 - n^2\), and \(c = m^2 + n^2\), where \(m\) and \(n\) are positive integers with \(m > n\). For \(a = 80\), let's choose \(m = 9\) and \(n = 1\). Then, \(b = 9^2 - 1^2 = 80\) and \(c = 9^2 + 1^2 = 82\). So, one such Pythagorean triplet is \(80, 80, 82\).

5. Question: Express 324 as sum

Exponents Grade VIII Worksheet with solutions

Solutions

1. Question: By what number should \( \left(\frac{2}{7}\right)^{-2} \) be divided so that the quotient becomes 49?

   Solution: Let's call the number by which we should divide \(x\). Then the equation becomes \(\frac{\left(\frac{2}{7}\right)^{-2}}{x} = 49\). Simplifying, we get \(x = \frac{1}{49}\).

2. Question: Find x so that \( \left(\frac{3}{8}\right)^{14} \times \left(\frac{8}{3}\right)^{-9} = \left(\frac{3}{8}\right)^{2x-1} \).

   Solution: We can equate the powers because the bases are equal. This gives us \(14 - 9 = 2x - 1\). Solving for \(x\), we find \(x = 12\).

3. Question: Find the value of a/b if \( \left(\frac{a}{b}\right)^{-6} = \left(\frac{2}{7}\right)^{-6} \times \left(\frac{14}{9}\right)^{-6} \).

   Solution: Since the left-hand side and the right-hand side are equal, we can equate the bases which gives us \( \frac{a}{b} = \frac{2}{7} \times \frac{14}{9} = \frac{4}{3} \). So, \( a = 4 \) and \( b = 3 \).

4. Question: Simplify \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \).

   Solution: This simplifies to \( \frac{125 \times 5^2 \times 6^5}{3^5} = \frac{3125 \times 7776}{243} = 100000 \).

5. Question: The mass of Earth is \( 5.97 \times 10^{24} \) kg and the mass of the Moon is \( 7.35 \times 10^{22} \) kg. What is the total mass?

   Solution: The total mass is the sum of the mass of the Earth and the Moon, which is \( 5.97 \times 10^{24} + 7.35 \times 10^{22} = 5.97 \times 10^{24} + 0.0735 \times 10^{24} = 6.0435 \times 10^{24} \) kg.

Rational Number and Exponents Grade VIII with solutions

Worksheet

1. Question: Rachna purchased \(2 \frac{1}{2}\) kg onions at ₹10 per kg and \(1 \frac{3}{8}\) kg potatoes at ₹\(16 \frac{8}{11}\) per kg. She gave a five hundred rupee note to the shopkeeper. Find the amount she got back from the shopkeeper.

   Solution: Total cost of onions = \(2 \frac{1}{2}\) kg x ₹10 = ₹25

   Total cost of potatoes = \(1 \frac{3}{8}\) kg x ₹\(16 \frac{8}{11}\) = ₹\(1 \frac{3}{8}\) x \(16.7273\) ≈ ₹23.64

   Total cost = ₹25 + ₹23.64 ≈ ₹48.64

   Amount Rachna got back = ₹500 - ₹48.64 = ₹451.36

2. Question: Simplify \((512)^{\left(-\frac{2}{3}\right)}\).

   Solution: \((512)^{\left(-\frac{2}{3}\right)} = \left(\sqrt[3]{512}\right)^{-2} = (8^{-2}) = \frac{1}{64}\).

3. Question: Find the value of \(x\).

   (a) \( \sqrt[3]{5x + 2} = 2 \)

   Solution: \( 5x + 2 = 2^3 \Rightarrow 5x + 2 = 8 \Rightarrow 5x = 6 \Rightarrow x = \frac{6}{5} \).

   (b) \(5^{x-1} \times 3^{2x-5} - 225\)

   Solution: It appears that the equation is incomplete. Please check the equation again.

4. Question: Write in standard form

   (a) 0.0000000777

   Solution: \(7.77 \times 10^{-8}\).

   (b) \(234.45 \times 10^{12}\)

   Solution: \(2.3445 \times 10^{14}\).

5. Question: Evaluate \(\left[\left(\frac{-2}{3}\right)^{-2} - \left(\frac{-3}{4}\right)^{-1} \right] \times \left(\frac{3}{2}\right)^2\).

   Solution: \(\left[\left(\frac{-2}{3}\right)^{-2} - \left(\frac{-3}{4}\right)^{-1} \right] \times \left(\frac{3}{2}\right)^2 = \left[\left(\frac{3}{2}\right)^2 - \left(\frac{4}{3}\right) \right] \times \frac{9}{4} = \left[\frac{9}{4} - \frac{4}{3} \right] \times \frac{9}{4} = \left(\frac{27}{12} - \frac{16}{12}\right) \times \frac{9}{4} = \frac{11}{4} \times \frac{9}{4} = \frac{99}{16}\).

Rational Number- Grade VIII Part -I

Rational Numbers Worksheet

1. Question: Subtract \( \frac{-11}{5} \) from additive inverse of sum of \( \frac{-5}{8} \) and \( \frac{5}{6} \).

   Solution: \( -\left(\frac{-5}{8} + \frac{5}{6}\right) - \frac{-11}{5} = -\left(\frac{-15 + 20}{24}\right) + \frac{11}{5} = -\left(\frac{5}{24}\right) + \frac{11}{5} = -\frac{5}{24} + \frac{132}{24} = \frac{127}{24} \).

2. Question: Subtract the sum of \( \frac{-4}{7} \) and \( \frac{5}{14} \) from the sum of \( \frac{9}{14} \) and \( \frac{23}{28} \).

   Solution: \( \left(\frac{9}{14} + \frac{23}{28}\right) - \left(\frac{-4}{7} + \frac{5}{14}\right) = \frac{18 + 46}{28} - \left(\frac{-8 + 5}{14}\right) = \frac{64}{28} - \left(\frac{-3}{14}\right) = \frac{64}{28} + \frac{3}{14} = \frac{71}{14} \).

3. Question: What number should be added to \( \frac{-7}{8} \) so as to get \( \frac{5}{9} \)?

   Solution: Let the number be \( x \). Therefore, \( x + \frac{-7}{8} = \frac{5}{9} \). Solving for \( x \), \( x = \frac{5}{9} + \frac{7}{8} = \frac{40 + 63}{72} = \frac{103}{72} \).

4. Question: Verify the distributive law of multiplication over subtraction for the rational numbers \( x = \frac{-3}{4} \), \( y = \frac{5}{7} \), \( z = \frac{-4}{5} \).

   Solution: \( x(y - z) = x \cdot y - x \cdot z = \frac{-3}{4} \cdot \frac{5}{7} - \frac{-3}{4} \cdot \frac{-4}{5} = \frac{-15}{28} + \frac{12}{20} = \frac{-15}{28} + \frac{42}{28} = \frac{27}{28} \).

5. Question: Write 6 rational numbers between \( \frac{3}{-7} \) and \( \frac{2}{5} \).

   Solution: Six rational numbers between \( \frac{3}{-7} \) and \( \frac{2}{5} \) are \( \frac{-2}{7} \), \( \frac{-1}{7} \), \( 0 \), \( \frac{1}{5} \), \( \frac{2}{7} \), and \( \frac{3}{7} \).

6. Question: Write in standard form (a) \( \frac{12}{-240} \) (b) \( \frac{-19}{362} \).

   Solution: (a) \( \frac{12}{-240} = \frac{-1}{20} \) (b) \( \frac{-19}{362} \) is already in standard form.

7. Question: Using appropriate property find \( \frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} \times \frac{4}{7} + \frac{3}{5} \times \frac{1}{6} \).

   Solution: \( \frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} \times \frac{4}{7} + \frac{3}{5} \times \frac{1}{6} = \frac{-2}{5} + \frac{10}{7} + \frac{1}{10} = \frac{-14}{35} + \frac{50}{35} + \frac{7}{35} = \frac{43}{35} \).

8. Question: What should be subtracted from the product of \( \frac{3}{7} \) and \( \frac{2}{5} \) to get \( \frac{4}{35} \)?

   Solution: Let the number be \( x \). Therefore, \( \frac{3}{7} \times \frac{2}{5} - x = \frac{4}{35} \). Solving for \( x \), \( x = \frac{3}{7} \times \frac{2}{5} - \frac{4}{35} = \frac{6}{35} - \frac{4}{35} = \frac{2}{35} \).

9. Question: One sixth of the students of a class joined the sports club. Three fifth of these students opted to play table tennis. If 6 students play table tennis, how many students are there in the class?

   Solution: Let the total number of students be \( x \). Therefore, \( \frac{1}{6}x \times \frac{3}{5} = 6 \). Solving for \( x \), \( x = \frac{6 \times 6 \times 5}{3} = 60 \). So, there are 60 students in the class.

10. Question: Simplify \( \left(\frac{10}{21} \times \frac{-7}{5}\right) - \left(\frac{2}{3} \times \frac{9}{-16}\right) + \left(\frac{-4}{15} \times \frac{-6}{16}\right) \).

    Solution: \( \left(\frac{10}{21} \times \frac{-7}{5}\right) - \left(\frac{2}{3} \times \frac{9}{-16}\right) + \left(\frac{-4}{15} \times \frac{-6}{16}\right) = \frac{-70}{105} + \frac{3}{8} + \frac{1}{10} = \frac{-56}{105} + \frac{42}{105} + \frac{21}{105} = \frac{7}{105} \).

11. Question: Is \( \frac{4}{11} \) the multiplication inverse of \( -3 \frac{3}{4} \)? Give reasons.

    Solution: \( -3 \frac{3}{4} \) is equal to \( -\frac{15}{4} \) and its multiplicative inverse is \( -\frac{4}{15} \). Since \( -\frac{4}{15} \) is not equal to \( \frac{4}{11} \), \( \frac{4}{11} \) is not the multiplicative inverse of \( -3 \frac{3}{4} \).

Test

Self Assessment Test 6

Based on Chapter 6

Time: 1 hour

M.M. 25

Note: Q.1-2 carry 1 mark each, Q.3-5 carry 2 marks each, Q.6-8 carry 3 marks each and Q.9-10 carry 4 marks each.

1. Question: Express \( \frac{5}{12} \) as a percentage.

   Solution: \( \frac{5}{12} = \frac{5}{12} \times 100\% = 41.67\% \).

2. Question: Simplify \( \sqrt[3]{64} \).

   Solution: \( \sqrt[3]{64} = 4 \).

3. Question: Simplify \( \frac{4x^2 - 9}{2x + 3} \).

   Solution: \( \frac{4x^2 - 9}{2x + 3} = \frac{(2x - 3)(2x + 3)}{2x + 3} = 2x - 3 \).

4. Question: If \( f(x) = x^2 + 2x - 3 \), find \( f(-1) \).

   Solution: Substituting \( x = -1 \) into \( f(x) \), we get \( f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \).

5. Question: Solve the inequality \( 2x - 3 < 7 \).

   Solution: Adding 3 to both sides, we get \( 2x < 10 \). Dividing by 2, we get \( x < 5 \).

6. Question: Find the slope of the line passing through the points (3, 4) and (6, -2).

   Solution: The slope is \( \frac{-2 - 4}{6 - 3} = \frac{-6}{3} = -2 \).

7. Question: Evaluate \( \frac{2^3 \times 3^2}{6} \).

   Solution: \( \frac{2^3 \times 3^2}{6} = \frac{8 \times 9}{6} = 12 \).

8. Question: Solve for \( x \) in the equation \( 4x - 7 = 3x + 8 \).

   Solution: Subtracting \( 3x \) from both sides, we get \( x - 7 = 8 \). Adding 7 to both sides, we get \( x = 15 \).

9. Question: Factorize \( x^2 - 6x + 9 \).

   Solution: \( x^2 - 6x + 9 = (x-3)^2 \).

10. Question: Find the area of a triangle with sides of length 5, 12, and 13 units.

    Solution: Using Heron's formula, let \( s = \frac{5 + 12 + 13}{2} = 15 \). The area is \( \sqrt{15(15-5)(15-12)(15-13)} = 30 \) square units.

Linear equation in two variables grade IX -Worksheet 2 with solutions

Self Assessment Test 5

Based on Chapter 5

Time: 1 hour

M.M. 25

Note: Q.1-2 carry 1 mark each, Q.3-5 carry 2 marks each, Q.6-8 carry 3 marks each and Q.9-10 carry 4 marks each.

1. Question: Simplify \(3a^2b \times 2ab^2\).

   Solution: \(3a^2b \times 2ab^2 = 3 \times 2 \times a^2 \times a \times b \times b^2 = 6a^3b^3\).

2. Question: Factorize \(x^2 - 6x + 9\).

   Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).

3. Question: Solve the quadratic equation \(x^2 - 5x + 6 = 0\).

   Solution: Factoring the equation, we get \((x-2)(x-3) = 0\). Therefore, \(x = 2\) or \(x = 3\).

4. Question: Simplify \(\frac{a^2 - b^2}{a^2 + b^2}\) and express in the form \(a + b\).

   Solution: \(\frac{a^2 - b^2}{a^2 + b^2} = \frac{(a+b)(a-b)}{a^2 + b^2} = \frac{a+b}{a^2/b^2 + 1} = a + b - \frac{2b^2}{a^2 + b^2}\).

5. Question: Solve for \(x\): \(3x + 4 = 2x - 5\).

   Solution: Subtracting \(2x\) from both sides, we get \(x + 4 = -5\). Subtracting 4 from both sides, we get \(x = -9\).

6. Question: If \(f(x) = 3x^2 - 4x + 5\), find \(f(2)\).

   Solution: Substituting \(x = 2\) into \(f(x)\), we get \(f(2) = 3(2)^2 - 4(2) + 5 = 3(4) - 8 + 5 = 12 - 8 + 5 = 9\).

7. Question: Find the zeros of the quadratic function \(f(x) = x^2 - 4\).

   Solution: The zeros are the x-values that make \(f(x) = 0\), so we solve the equation \(x^2 - 4 = 0\). Factoring, we get \((x-2)(x+2) = 0\), so \(x = 2\) or \(x = -2\).

8. Question: Simplify \(\frac{3x^3 - 2x^2 + 4x - 3}{x - 1}\).

   Solution: Using polynomial division, we find that \(\frac{3x^3 - 2x^2 + 4x - 3}{x - 1} = 3x^2 + x + 5 - \frac{8}{x - 1}\).

9. Question: Factorize \(2x^2 - 5x - 3\).

   Solution: \(2x^2 - 5x - 3 = (2x+1)(x-3)\).

10. Question: Solve the system of equations: \(x + y = 5\) and \(x - y = 1\).

    Solution: Adding the two equations, we get \(2x = 6\), or \(x = 3\). Substituting into \(x + y = 5\), we get \(3 + y = 5\), or \(y = 2\). Therefore, the solution is \(x = 3\), \(y = 2\).

Linear equation in two variable Grade IX Worksheet with solutions

Self Assessment Test 7

Based on Chapter 7

Time: 1 hour

M.M. 25

Note: Q.1-2 carry 1 mark each, Q.3-5 carry 2 marks each, Q.6-8 carry 3 marks each and Q.9-10 carry 4 marks each.

1. Question: Simplify \(\sqrt{49}\).

   Solution: \(\sqrt{49} = 7\).

2. Question: Convert \(\frac{3}{4}\) to a decimal.

   Solution: \(\frac{3}{4} = 0.75\).

3. Question: If \(f(x) = 2x + 5\), find \(f(3)\).

   Solution: \(f(3) = 2(3) + 5 = 6 + 5 = 11\).

4. Question: Solve for \(x\): \(3x - 7 = 8\).

   Solution: \(3x - 7 = 8 \Rightarrow 3x = 15 \Rightarrow x = 5\).

5. Question: Simplify: \(3(x + 4) - 2(2x - 1)\).

   Solution: \(3(x + 4) - 2(2x - 1) = 3x + 12 - 4x + 2 = -x + 14\).

6. Question: Factorize: \(x^2 - 6x + 9\).

   Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).

7. Question: Solve the system of equations for \(x\) and \(y\): \(x + y = 7\), \(x - y = 3\).

   Solution: Adding the two equations, \(2x = 10 \Rightarrow x = 5\). Substituting into the first equation, \(5 + y = 7 \Rightarrow y = 2\). So, \(x = 5\) and \(y = 2\).

8. Question: Find the area of a triangle with sides of length 6, 8, and 10 units.

   Solution: Using Heron's formula, \(s = \frac{6+8+10}{2} = 12\), \(Area = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \cdot 6 \cdot 4 \cdot 2} = 24\).

9. Question: If the circumference of a circle is 44 cm, find its radius.

   Solution: Circumference = \(2 \pi r = 44\), \(\Rightarrow r = \frac{44}{2 \pi} \approx 7\).

10. Question: Solve the inequality: \(3x + 5 > 14\).

    Solution: \(3x + 5 > 14 \Rightarrow 3x > 9 \Rightarrow x > 3\).

Coordinate Geometry Grade IX Worksheet with solutions

Self Assessment Test 4 - Solutions

Based on Chapter 4

1. Question: Simplify \(3x + 5x\).

   Solution: \(3x + 5x = 8x\).

2. Question: Solve for \(x\): \(5x - 3 = 12\).

   Solution: \(5x - 3 = 12 \Rightarrow 5x = 12 + 3 \Rightarrow 5x = 15 \Rightarrow x = 3\).

3. Question: Find the area of a rectangle with length 5 units and width 3 units.

   Solution: Area of a rectangle = length × width = 5 × 3 = 15 square units.

4. Question: Simplify: \(\frac{6x^2 + 9x}{3x}\).

   Solution: \(\frac{6x^2 + 9x}{3x} = 2x + 3\).

5. Question: Factorize: \(x^2 - 6x + 9\).

   Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).

6. Question: Find the slope of the line passing through the points (2, 3) and (4, 7).

   Solution: Slope = \(\frac{7-3}{4-2} = \frac{4}{2} = 2\).

7. Question: Solve the system of equations: \(2x + 3y = 8\) and \(x - 4y = -6\).

   Solution: Solving the system, we find that \(x = 2\) and \(y = 2\).

8. Question: Evaluate: \(2^{3} \times 3^{-1}\).

   Solution: \(2^{3} \times 3^{-1} = 8 \times \frac{1}{3} = \frac{8}{3}\).

9. Question: The following table shows the number of books sold by a bookstore over 6 months. Find the average number of books sold per month.

   Month	Jan	Feb	Mar	Apr	May	Jun
   Books Sold	20	24	22	26	25	27

   Solution: Average number of books sold per month = \(\frac{20 + 24 + 22 + 26 + 25 + 27}{6} = \frac{144}{6} = 24\).

10. Question: Find the sum of the infinite geometric series: \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\).

    Solution: This is a geometric series with first term \(a = \frac{1}{3}\) and common ratio \(r = \frac{1}{3}\). The sum of an infinite geometric series is \(\frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\).

Coordinate Geometry Grade IX Worksheet Solution

Self Assessment Test 4 - Solutions

Based on Chapter 4

1. Question: Simplify \(3x + 5x\).

   Solution: \(3x + 5x = 8x\).

2. Question: Solve for \(x\): \(5x - 3 = 12\).

   Solution: \(5x - 3 = 12 \Rightarrow 5x = 12 + 3 \Rightarrow 5x = 15 \Rightarrow x = 3\).

3. Question: Find the area of a rectangle with length 5 units and width 3 units.

   Solution: Area of a rectangle = length × width = 5 × 3 = 15 square units.

4. Question: Simplify: \(\frac{6x^2 + 9x}{3x}\).

   Solution: \(\frac{6x^2 + 9x}{3x} = 2x + 3\).

5. Question: Factorize: \(x^2 - 6x + 9\).

   Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).

6. Question: Find the slope of the line passing through the points (2, 3) and (4, 7).

   Solution: Slope = \(\frac{7-3}{4-2} = \frac{4}{2} = 2\).

7. Question: Solve the system of equations: \(2x + 3y = 8\) and \(x - 4y = -6\).

   Solution: Solving the system, we find that \(x = 2\) and \(y = 2\).

8. Question: Evaluate: \(2^{3} \times 3^{-1}\).

   Solution: \(2^{3} \times 3^{-1} = 8 \times \frac{1}{3} = \frac{8}{3}\).

9. Question: The following table shows the number of books sold by a bookstore over 6 months. Find the average number of books sold per month.

   Month	Jan	Feb	Mar	Apr	May	Jun
   Books Sold	20	24	22	26	25	27

   Solution: Average number of books sold per month = \(\frac{20 + 24 + 22 + 26 + 25 + 27}{6} = \frac{144}{6} = 24\).

10. Question: Find the sum of the infinite geometric series: \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\).

    Solution: This is a geometric series with first term \(a = \frac{1}{3}\) and common ratio \(r = \frac{1}{3}\). The sum of an infinite geometric series is \(\frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\).

Coordinate Geometry Grade IX Worksheet Solution

Self Assessment Test 3 - Solutions

Based on Chapter 3

1. Question: What are the signs of a point in 2nd quadrant?

   Solution: In the 2nd quadrant, the x-coordinate is negative and the y-coordinate is positive.

2. Question: What is the distance of the point \(A(-2,-3)\) from y-axis?

   Solution: The distance of a point \((x, y)\) from the y-axis is equal to the absolute value of its x-coordinate. So, the distance of point \(A(-2, -3)\) from y-axis is \(|-2| = 2\) units.

3. Question: Plot the points \(P(2,-1)\), \(Q(0,5)\), \(C(3,4)\) and \(D(-3,-5)\) in the cartesian plane.

   Solution: To plot these points on the cartesian plane, plot each point by counting spaces: Right and up for positive coordinates, left and down for negative coordinates. For example, to plot P(2, -1), move 2 spaces to the right and 1 space down from the origin.

4. Question: A point \(R\) is at a distance of 3 units from the x-axis and 7 units from y-axis and is in 3rd quadrant. Write the coordinates of \(R\).

   Solution: In the 3rd quadrant, both x and y coordinates are negative. Therefore, the coordinates of point \(R\) are \((-7, -3)\).

5. Question: Which of the following points lie in quadrant IV? \(L(3,-5)\), \(M(5,-3)\), \(N(-2,-1)\), \(P(1,2)\), \(Q(2,-7)\) and \(R(-1,-1)\)

   Solution: In the 4th quadrant, x is positive and y is negative. Therefore, points \(L(3, -5)\), \(M(5, -3)\), and \(Q(2, -7)\) lie in the 4th quadrant.

6. Question: Plot the points \(A(2,0)\), \(B(0,2)\), \(C(-2,0)\) and \(D(0,-2)\). What type of figure is ABCD ?

   Solution: After plotting these points on the cartesian plane, it can be observed that ABCD forms a square.

7. Question: Find the values of \(x\) and \(y\) if \((3x-7,2y+11)=(2x-3,2-y)\)

   Solution: From \(3x - 7 = 2x - 3\), we can find \(x = 4\). From \(2y + 11 = 2 - y\), we can find \(y = -3\).

8. Question: The points \(E(2,4)\), \(F(-5,4)\) and \(H(2,-3)\) are the vertices of a square EFGH. Plot the points on the graph paper and hence, find the coordinates of \(G\).

   Solution: The coordinates of \(G\) are \((-5, -3)\) as it will be diagonally opposite to \(E\) and have the same x-coordinate as \(F\) and the same y-coordinate as \(H\).

9. Question: Plot the points \((-4,-10)\),\((-2,-4)\),\((0,2)\),\((2,8)\) and \((4,14)\) and show that they are collinear.

   Solution: These points are collinear because the differences in the x-coordinates and y-coordinates are proportional: \(\frac{y_2-y_1}{x_2-x_1} = \frac{y_3-y_2}{x_3-x_2} = \ldots\) for collinear points. You can verify this by plotting them on graph paper.

10. Question: The following table gives the cost of different number of toys.

    \(x\) (number of toys)	2	4	6	8	10	12	15
    \(y\) (cost in ₹)	20	40	60	80	100	120	150

    Plot the points (as ordered pairs) and join them.

    Solution: Plot each point from the table as an ordered pair (x, y) on graph paper, such as (2, 20), (4, 40), etc., and join them to see the linear relationship between the number of toys and the total cost.

Coordinate Geomtery Grade IX

Self Assessment Test 3

Based on Chapter 3

Time: 1 hour

M.M. 25

Note: Q. 1-2 carry 1 mark each, Q. 3 - 5 carry 2 marks each, Q. 6-8 carry 3 marks each and Q. 9-10 carry 4 marks each.

1. What are the signs of a point in 2nd quadrant?

2. What is the distance of the point \(A(-2,-3)\) from y-axis?

3. Plot the points \(P(2,-1)\), \(Q(0,5)\), \(C(3,4)\) and \(D(-3,-5)\) in the cartesian plane.

4. A point \(R\) is at a distance of 3 units from the x-axis and 7 units from y-axis and is in 3rd quadrant. Write the coordinates of \(R\).

5. Which of the following points lie in quadrant IV? \(L(3,-5)\), \(M(5,-3)\), \(N(-2,-1)\), \(P(1,2)\), \(Q(2,-7)\) and \(R(-1,-1)\)

6. Plot the points \(A(2,0)\), \(B(0,2)\), \(C(-2,0)\) and \(D(0,-2)\). What type of figure is ABCD ?

7. Find the values of \(x\) and \(y\) if \((3x-7,2y+11)=(2x-3,2-y)\)

8. The points \(E(2,4)\), \(F(-5,4)\) and \(H(2,-3)\) are the vertices of a square EFGH. Plot the points on the graph paper and hence, find the coordinates of \(G\).

9. Plot the points \((-4,-10)\),\((-2,-4)\),\((0,2)\),\((2,8)\) and \((4,14)\) and show that they are collinear.

10. The following table gives the cost of different number of toys.

    \(x\) (number of toys)	2	4	6	8	10	12	15
    \(y\) (cost in ₹)	20	40	60	80	100	120	150

    Plot the points (as ordered pairs) and join them.

Number System Grade IX - 3

Self Assessment Test 3

Based on Chapter 1

Time: 1 hour

M.M. 25

Note: Q. 1 - 2 carry 1 mark each, Q. 3 - 5 carry 2 marks each, Q. 6-8 carry 3 marks each and Q.9 - 10 carry 4 marks each.

1. Find the product: $(3-\sqrt{5})(3+\sqrt{5})$.

2. Express $0.3\overline{47}$ (recurring decimal) in the form $\frac{p}{q}$, where p and q are integers and $q \neq 0$.

3. Find a point corresponding to $\frac{5}{2} + \sqrt{3}$ on the number line.

4. Rationalise the denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$ and subtract it from $\sqrt{5}-\sqrt{2}$.

5. Evaluate:

   (i) $(32768)^{\frac{1}{5}}$

   (ii) $\left(\frac{-1}{27}\right)^{-\frac{1}{3}}$

6. Simplify: $8\sqrt{18} + 4\sqrt{72} - 5\sqrt{32} + 3\sqrt{98} + 2\sqrt{50}$.

7. Find two irrational numbers in decimal form between $\sqrt{2}$ and $2$.

8. If $\sqrt{5}=2.236$ and $\sqrt{20}=4.472$, find the value of $\frac{49}{\sqrt{125}-\sqrt{45}}$.

9. If $\frac{(\sqrt{2}-3)}{(\sqrt{2}+3)} - \frac{(\sqrt{2}+3)}{(\sqrt{2}-3)} = e + f\sqrt{2}$, find e and f where e and f are rational numbers.

10. If $z = 5 + 2\sqrt{6}$, find:

    (i) $\sqrt{z} + \frac{1}{\sqrt{z}}$

    (ii) $\sqrt{z} - \frac{1}{\sqrt{z}}$

Number System Grade IX

 This is a self assessment based on new syllabus

Self Assessment Test 3 (Based on Chapter 1) Time: 1 hour M.M. 25 Note: Q. 1 - 2 carry 1 mark each, Q. 3 - 5 carry 2 marks each, Q. 6-8 carry 3 marks each and Q.9 - 10 carry 4 marks each.

1. Find the product: $(4-\sqrt{7})(4+\sqrt{7})$.

2. Express $0.4\overline{56}$ (recurring decimal) in the form $\frac{p}{q}$, where p and q are integers and $q \neq 0$.

3. Find a point corresponding to $\frac{7}{3} + \sqrt{2}$ on the number line.

4. Rationalise the denominator of $\frac{1}{\sqrt{7}+\sqrt{3}}$ and subtract it from $\sqrt{7}-\sqrt{3}$.

5. Evaluate: (i) $(59049)^{\frac{1}{5}}$ (ii) $\left(\frac{-1}{64}\right)^{-\frac{1}{3}}$

6. Simplify: $10\sqrt{50} + 3\sqrt{98} - 7\sqrt{200} + 2\sqrt{32} + 4\sqrt{12}$.

7. Find two irrational numbers in decimal form between $\sqrt{7}$ and $3$.

8. If $\sqrt{8}=2.828$ and $\sqrt{18}=4.243$, find the value of $\frac{64}{\sqrt{144}-\sqrt{72}}$.

9. If $\frac{(\sqrt{3}-4)}{(\sqrt{3}+4)} - \frac{(\sqrt{3}+4)}{(\sqrt{3}-4)} = e + f\sqrt{3}$, find e and f where e and f are rational numbers.

10. If $z = 8 + 3\sqrt{5}$, find: (i) $\sqrt{z} + \frac{1}{\sqrt{z}}$ (ii) $\sqrt{z} - \frac{1}{\sqrt{z}}$