Q: The taxi fare in a city is as follows: For the first kilometre, the fare is \$8\$ and for the subsequent distance it is \$5\$ per km. Taking the distance covered as \$x\$ km and total fare as Rs.\$y\$, write a linear equation for this information and draw its graph.

\$x\$\$0\$ \$ 1\$\$ 2\$ \$y\$\$3\$\$ 8\$ \$ 13\$ Taxi fare for first kilometer \$=\$ Rs. \$8\$Taxi fare for subsequent distance \$=\$ Rs. \$5\$Total distance covered \$= x\$Total fare \$= y\$Since the fare for first kilometer \$=\$ Rs.\$ 8\$According to problem,Fare for \$(x – 1)\$ kilometer \$=\$ \$5(x-1)\$So, the total fare \$y = 5(x-1) + 8\$\$\Rightarrow y = 5(x-1) + 8\$\$\Rightarrow y = 5x – 5 + 8\$\$\Rightarrow y = 5x + 3\$\$Hence\$, \$y = 5x + 3\$ \$\text{is the required linear equation.}\$Now the equation is\$y = 5x + 3\$ ...(1)Now, putting the value \$x = 0\$ in (1)\$y = 5 \times 0 + 3\$\$y = 0 + 3 = 3 \$ So the solution is \$(0, 3)\$Putting the value \$x = 1\$ in (1)\$y = 5 \times 1 + 3\$\$y = 5 + 3 = 8\$. So the solution is \$(1, 8)\$Putting the value \$x = 2\$ in (1)\$y = 5 \times 2 + 3\$\$y = 10 + 3 = 13\$. So the solution is \$(2, 13)\$

Q: Sum of the digits of a two digit number is \$9\$. When we interchange the digit it is found that the resulting new number is greater than the original number by \$27\$. What is the two digit number

Let the digits at tens place and ones place: \$x\$ and \$9 - x\$ respectively.\$\therefore\$ original number \$= 10 x + (9 - x)\$\$= 9x + 9\$Now Interchange the digits: Digit at ones place and tens place: \$x\$ and \$9 - x\$ respectively.\$\therefore\$ New number: \$10 (9 - x) + x\$\$= 90 - 10x + x\$\$= 90 - 9x\$AS per the questionNew number \$=\$ Original number \$+ 27\$\$90 - 9x = 9x + 9 + 27\$\$90 - 9x = 9x + 36\$\$18x = 54\$\$x = \dfrac {54}{18}\$\$x = 3\$Digit at tens place \$\Rightarrow 3\$ and one's place : \$6\$\$\therefore\$ Two digit number: \$36\$

Q: A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.

Let the number be \$x\$. Then,\$x+12=160\times \dfrac{1}{x}\$\$x^{2}+12x-160=0\$\$(x+20)(x-8)=0\$\$x=-20,8\$Therefore, the required number is \$8.\$

Q: Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score ?

Let the Rahul scored \$x\$ runsGiven that Sachin scored twice as many runs as Rahul.Therefore, The runs scored by Sachin is \$2x\$Given that Their total runs fell two short of double century.Therefore, \$x+2x=200-2\$\$\implies 3x=198\$\$\implies x=66\$Therefore, the runs scored by Rahul is \$66\$ and the runs scored by Sachin is \$2\times 66=132\$

Q: One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two digit number and add the resulting number to the original number, you get \$88\$. What is the original number?

Let the digit in units place be \$x\$. Then, digit in ten's place \$= 3x\$So, original number \$= 10(3x) + x = 31x\$On interchanging the digits, new number \$= 10x + 3x = 13x\$According to the given condition,\$ {31x + 13x = 88} \$\$\Rightarrow 44x=88\\ \Rightarrow x=2\$So, the original number is \$31x\$ or \$31 \times 2 = 62\$.On the other hand, if we consider the digit in ten's place as \$x\$, then the digit in the unit's place will be \$3x\$.So, original number \$= 10(x) + 3x = 13x\$On interchanging the digits, new number \$= 10(3x) + x = 31x\$According to the given condition,\$ {13x + 31x = 88} \$\$\Rightarrow 44x=88\\ \Rightarrow x=2\$So, the original number is \$13x\$ or \$13 \times 2 = 26\$.Here, both answers are correct, as \$26 + 62 = 88\$.

Q: Simplify \$3x(4x-5)+3\$ and find its value for \$x=3\$.

\$3x(4x-5)+3=12x^2-15x+3\$\$x=3\$ gives the value \$12(9)-15\times 3 +3\$ \$=108-45+3\$ \$=66\$

Q: The perimeter of a rectangular swimming pool is \$154\ m\$. Its length is \$2\ m\$ more than twice its breadth. What are the length and the breadth of the pool?

The perimeter of rectangular swimming pool is \$154\ m\$ and its length is \$2\$ meter more than twice its breadth.\$\textbf{Step 1: Perimeter of rectangle}\$ \$\boldsymbol{=2\times (Length +Breadth)}\$ Let the breadth of swimming pool be \$x\ m,\$ then its length will be \$(2 x + 2)\ m\$ Perimeter of swimming pool \$= 154\$ \$\Rightarrow 2\times (2x+2+x)=154\$ \$\Rightarrow 2x+2+x=\dfrac{154}{2}\$ \$\Rightarrow 3x+2=77\$ \$\Rightarrow 3x=77-2\$ \$\Rightarrow 3x=75\$ \$\Rightarrow x=\dfrac{75}{3}\$ \$\Rightarrow x=25\$ So, Length of swimming pool \$=2\times 25+2\$ \$=50+2\$ \$=52\ m\$\$\textbf{Hence, the length of swimming pool is 52}\$ \$\boldsymbol{m}\$ \$\textbf{and its breadth if 25}\$ \$\boldsymbol {m.}\$

Q: Laxmi's father is \$49\$ years old. He is \$4\$ years older than three time Laxmi's age. What is Laxmi's age?

\${\textbf{Step - 1: Writing mathematical equation according to question}}\$ \${\text{Let Laxmi's age be x,}}\$ \${\text{Then according to question, }}\$ \$\text{Laxmi's father is 49 years old. He is 4 years older than three time Laxmi's age}\$ \${\text{49 = 4 + 3x}}\$ \${\textbf{Step - 2: Finding Laxmi's age}}\$ \$ \Rightarrow {\text{ 49 = 4 + 3x}}\$ \$ \Rightarrow {\text{ 45 = 3x}}\$ \$ \Rightarrow {\text{ x = 15}}\$ \${\textbf{Hence, Laxmi's age is 15 years}}{\text{.}}\$

Q: Give expressions for the following cases.(a) \$7\$ added to \$p\$(b) \$7\$ subtracted from\$p\$(c) \$p\$ multiply by \$7\$(d) \$p\$ divide by \$7\$(e) \$7\$ subtracted from \$-m\$(f) \$-p\$ multiply by \$5\$(g) \$-5\$ divided by \$5\$(h) \$p\$ multiply by \$-5\$

\$(a) 7+p\$\$(b) p-7\$\$(c) 7\times p\$\$(d)p/7\$\$(e) -m-7\$\$(f)-5p\$\$(g)\frac{-5}{6}\$\$(h) -5p\$

Question 1

The taxi fare in a city is as follows: For the first kilometre, the fare is $8$ and for the subsequent distance it is $5$ per km. Taking the distance covered as $x$ km and total fare as Rs.$y$, write a linear equation for this information and draw its graph.

Accepted Answer

$x$$0$ $ 1$$ 2$ $y$$3$$ 8$ $ 13$ Taxi fare for first kilometer $=$ Rs. $8$Taxi fare for subsequent distance $=$ Rs. $5$Total distance covered $= x$Total fare $= y$Since the fare for first kilometer $=$ Rs.$ 8$According to problem,Fare for $(x &#8211; 1)$ kilometer $=$ $5(x-1)$So, the total fare $y = 5(x-1) + 8$$\Rightarrow y = 5(x-1) + 8$$\Rightarrow y = 5x &#8211; 5 + 8$$\Rightarrow y = 5x + 3$$Hence$, $y = 5x + 3$ $\text{is the required linear equation.}$Now the equation is$y = 5x + 3$    ...(1)Now, putting the value $x = 0$ in (1)$y = 5 \times 0 + 3$$y = 0 + 3 = 3 $ So the solution is $(0, 3)$Putting the value $x = 1$ in (1)$y = 5 \times 1 + 3$$y = 5 + 3 = 8$. So the solution is $(1, 8)$Putting the value $x = 2$ in (1)$y = 5 \times 2 + 3$$y = 10 + 3 = 13$. So the solution is $(2, 13)$

Question 2

Sum of the digits of a two digit number is $9$. When we interchange the digit it is found that the resulting new number is greater than the original number by $27$. What is the two digit number

Accepted Answer

Let the digits at tens place and ones place: $x$ and $9 - x$ respectively.$\therefore$ original number $= 10 x + (9 - x)$$= 9x + 9$Now Interchange the digits: Digit at ones place and tens place: $x$ and $9 - x$ respectively.$\therefore$ New number: $10 (9 - x) + x$$= 90 - 10x + x$$= 90 - 9x$AS per the questionNew number $=$ Original number $+ 27$$90 - 9x = 9x + 9 + 27$$90 - 9x = 9x + 36$$18x = 54$$x = \dfrac {54}{18}$$x = 3$Digit at tens place $\Rightarrow 3$ and one's place : $6$$\therefore$ Two digit number: $36$

Question 3

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 75 questions and secures 125 marks, the number of questions he attempted correctly, is

Accepted Answer

If the number of correct answer be x, thenthe number of incorrect answer is $\displaystyle (75- x)$.$\displaystyle \therefore \quad 4x-\left( 75-x \right) =125\quad or\quad x=40$

Question 4

A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.

Accepted Answer

Let the number be $x$. Then,$x+12=160\times \dfrac{1}{x}$$x^{2}+12x-160=0$$(x+20)(x-8)=0$$x=-20,8$Therefore, the required number is $8.$

Question 5

Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score ?

Accepted Answer

Let the Rahul scored $x$ runsGiven that Sachin scored twice as many runs as Rahul.Therefore, The runs scored by Sachin is $2x$Given that Their total runs fell two short of double century.Therefore, $x+2x=200-2$$\implies 3x=198$$\implies x=66$Therefore, the runs scored by Rahul is $66$ and the runs scored by Sachin is $2\times 66=132$

Question 6

One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two digit number and add the resulting number to the original number, you get $88$. What is the original number?

Accepted Answer

Let the digit in units place be $x$. Then, digit in ten's place $= 3x$So, original number $= 10(3x) + x = 31x$On interchanging the digits, new number $= 10x + 3x = 13x$According to the given condition,$ {31x + 13x = 88} $$\Rightarrow 44x=88\ \Rightarrow x=2$So, the original number is $31x$ or $31 \times 2 = 62$.On the other hand, if we consider the digit in ten's place as $x$, then the digit in the unit's place will be $3x$.So, original number $= 10(x) + 3x = 13x$On interchanging the digits, new number $= 10(3x) + x = 31x$According to the given condition,$ {13x + 31x = 88} $$\Rightarrow 44x=88\ \Rightarrow x=2$So, the original number is $13x$ or $13 \times 2 = 26$.Here, both answers are correct, as $26 + 62 = 88$.

Question 7

Simplify $3x(4x-5)+3$ and find its value for $x=3$.

Accepted Answer

$3x(4x-5)+3=12x^2-15x+3$$x=3$ gives the value $12(9)-15\times 3 +3$                                          $=108-45+3$                                          $=66$

Question 8

The perimeter of a rectangular swimming pool is $154\ m$. Its length is $2\ m$ more than twice its breadth. What are the length and the breadth of the pool?

Accepted Answer

The perimeter of rectangular swimming pool is $154\ m$ and its length is $2$ meter more than twice its breadth.$\textbf{Step 1: Perimeter of rectangle}$ $\boldsymbol{=2\times (Length +Breadth)}$                 Let the breadth of swimming pool be $x\ m,$ then its length will be $(2 x + 2)\ m$                Perimeter of swimming pool $= 154$                $\Rightarrow 2\times (2x+2+x)=154$                $\Rightarrow 2x+2+x=\dfrac{154}{2}$                $\Rightarrow 3x+2=77$                $\Rightarrow 3x=77-2$                $\Rightarrow 3x=75$                $\Rightarrow x=\dfrac{75}{3}$                $\Rightarrow x=25$                So, Length of swimming pool $=2\times 25+2$                                                                  $=50+2$                                                                  $=52\ m$$\textbf{Hence, the length of swimming pool is 52}$ $\boldsymbol{m}$ $\textbf{and its breadth if 25}$ $\boldsymbol {m.}$

Question 9

Laxmi's father is $49$ years old. He is $4$ years older than three time Laxmi's age. What is Laxmi's age?

Accepted Answer

${\textbf{Step  - 1: Writing mathematical equation according&#10;to question}}$&#10;&#10;                    ${\text{Let&#10;Laxmi's age be x,}}$&#10;&#10;                    ${\text{Then&#10;according to question, }}$                    $\text{Laxmi's father is 49 years old. He is 4 years older than three time Laxmi's age}$&#10;&#10;                    ${\text{49  = &#10;4  +  3x}}$&#10;&#10;${\textbf{Step  - 2: Finding Laxmi's age}}$&#10;&#10;                    $&#10;\Rightarrow {\text{ 49  =  4 &#10;+  3x}}$&#10;&#10;                    $&#10;\Rightarrow {\text{ 45  =  3x}}$&#10;&#10;                    $&#10;\Rightarrow {\text{ x  =  15}}$&#10;&#10;${\textbf{Hence, Laxmi's age is 15 years}}{\text{.}}$

Question 10

Give expressions for the following cases.(a)  $7$ added to $p$(b)  $7$ subtracted from$p$(c)  $p$ multiply by $7$(d)  $p$ divide by $7$(e)  $7$ subtracted from $-m$(f)   $-p$ multiply by $5$(g)  $-5$ divided by $5$(h)   $p$ multiply by $-5$

Accepted Answer

$(a) 7+p$$(b) p-7$$(c) 7\times p$$(d)p/7$$(e) -m-7$$(f)-5p$$(g)\frac{-5}{6}$$(h) -5p$