Algebraic Expressions PDF

Okay, here is the markdown conversion of the provided text from the images: ### 11 Algebraic Expressions **KNOW YOUR FIELD** * Recognize variables * Frame algebraic expressions for number patterns * Evaluate an algebraic expression for a given value of the variable * Distinguish between a...

Okay, here is the markdown conversion of the provided text from the images: ### 11 Algebraic Expressions **KNOW YOUR FIELD** * Recognize variables * Frame algebraic expressions for number patterns * Evaluate an algebraic expression for a given value of the variable * Distinguish between an equation and an expression * Solve an equation * Apply the concept of algebra to the real world **NET PRACTICE** Here's a crossword puzzle: **Across** 1. Amount obtained after multiplying two or more numbers 2. Amount obtained after subtracting two numbers 3. Repeated arrangement of numbers 4. Amount obtained after dividing two numbers 5. Amount obtained after adding two numbers **Down** 2. Addition, subtraction, multiplication and division **INTRODUCTION** In previous classes, we have dealt with numbers having fixed values. We have also learnt how to perform the four basic operations +, -, ×, ÷ with these numbers. It is covered under the branch of mathematics called Arithmetic. Shyam and Siya are playing with pebbles. Study the following arrangement of pebbles made by them with ten pebbles: ``` 00 00 00 00 2×5 5×2 ``` Fig. 11.1 **REMEMBER** In Algebra, we use letters and symbols in place of unknown numbers and then perform various operations like in arithmetic. Both these arrangements represent the same rectangular number 10. Thus, it is correct to write $2 × 5 = 5 × 2$. Similarly, $3 × 4 = 4 × 3; 6 × 3 = 3 × 6; 9 × 5 = 5 × 9$ This is the commutative property of numbers. Let $x$ and $y$ represent any two counting numbers. Then we can represent the commutative property as $x × y = y × x$. Counting numbers can be represented with letters. When we substitute counting numbers with letters, we can perform all the four basic operations on them as well. Now, study the following arrangement: $2 × 4 + 2 × 5 = 2 × 9$ Thus, it can be conveniently written as $2 × 4 + 2 × 5 = 2 × (4 + 5)$. This is the distributive property of multiplication over addition. And if we replace the numbers 2, 4, 5 with letters $x, y, z$, respectively, we can generalize the distributive property for all counting numbers as follows: $x × y + x × z = x × (y + z)$ We have also dealt with even numbers and odd numbers in previous classes. The numbers that come in the times table of 2 are called even numbers. Thus, we can write: $1^{st}$ even number $= 2 × 1 = 2$, $3^{rd}$ even number $= 2 × 3 = 6$, $5^{th}$ even number $= 2 × 5 = 10$, $2^{nd}$ even number $= 2 × 2 = 4$ $4^{th}$ even number $= 2 × 4 = 8$ $n^{th}$ even number $= 2 × n = 2n$, where n is any counting number. Similarly, $1^{st}$ odd number $= 2 × 1 - 1 = 1$, $3^{rd}$ odd number $= 2 × 3 - 1 = 5$, $5^{th}$ odd number $= 2 × 5 - 1 = 9$, $2^{nd}$ odd number $= 2 × 2 - 1 = 3$ $4^{th}$ odd number $= 2 × 4 - 1 = 7$ $n^{th}$ odd number $= 2 × n - 1 = 2n - 1$, where n is any counting number **MATCHSTICK PATTERNS** Consider the matchstick pattern shown in Fig. 11.2. We can make one inverted C using three matchsticks. Now for making two inverted Cs, we need five matchsticks. Similarly, you need seven and nine matchsticks to make three and four inverted Cs, respectively. Now can you tell how many matchsticks would be there when seven inverted Cs are combined? And how many matchsticks will be there in 100 inverted Cs? For finding answer to these questions, observe the pattern. * One inverted C has 3 matchsticks. * Two inverted Cs have 3 matchsticks + 1 pair of matchsticks, that is $3 + 2 = 5$ matchsticks. * Three inverted Cs have 3 matchsticks + 2 pairs of matchsticks, that is $3 + 2 × 2 = 7$ matchsticks. * Four inverted Cs have 3 matchsticks + 3 pairs of matchsticks, that is $3 + 2 × 3 = 9$ matchsticks. Thus, seven inverted Cs will have 3 matchsticks + (7 – 1) pairs of matchsticks or 3 matchsticks + 6 pairs of matchsticks, that is $3 + 2 × 6 = 15$ matchsticks Now, hundred inverted Cs will have 3 matchsticks + 99 pairs of matchsticks, that is $3 + 2 × 99 = 201$ matchsticks. We can generalize the number of matchsticks that a number of inverted Cs will have as follows: The number of matchsticks 'n' inverted C will have 3 matchsticks + ($n - 1$) pairs of matchsticks = 3 + $2 × (n - 1) = 3 + 2n - 2 = 2n + 1$ matchsticks Thus, the number of matchsticks used = $2n + 1$, where $n$ is the number of Cs. If we have 100 repeated units, the number of matchsticks used = $2 × 100 + 1 = 201$. Now, try to make pentagons using matchsticks as shown in Fig. 11.4 and generalise the number of matchsticks required for generating $n$ number of pentagons. **MASTERSTROKE** Total number of matchsticks used to create repeated unit pattern can be generalised as follows: Number of uncommon matchsticks in the smallest repeated unit + number of units + number of common matchsticks in the smallest repeated unit. The smallest repeated unit here has 5 uncommon matchsticks and 1 common matchstick, so the general formula is $5n + 1$, where $n$ is the number of repeated units. **Solved Examples** **Example 1**: For the following matchstick pattern, fill in the table that follows: Solution: On observing the pattern, we can say that: Number of matchsticks used = $5n + 1$, where $n$ is the number of hexagons. Now, we can easily fill the table as follows: | No. of hexagons ($n$) | 1 | 2 | 3 | 4 | ? | n | | :-------------------- | :--- | :--- | :--- | :--- | :--- | :------ | | No. of matchsticks | 5x1+1=6 | 5x2+1=11 | 5x3+1=16 | 5x4+1=21 | 41 | 5n+1 | Note: We will fill in '?' while discussing equations. **Example 2**: Look at the following table and make a rule that shows relation between $a$ and $b$. Hence complete the table. Solution: Relation between $a$ and $b$ is $b = a - 3$ or $a = b + 3$ | a | 5 | 6 | 7 | 8 | 9 | 21 | n | |---|---|---|---|---|---|----|---| | b | 2 | 3 | 4 | 5 | 6 | 13 | ? | Thus, when b = 13, $a = 13 + 3 = 16$ Also, when a = 21, $b = 21 - 3 = 18$ When a = n, $b = n - 3$ **WHAT IS A VARIABLE?** You must have noticed that while making the rules for the number of matchsticks used for a pattern, we make use of letters from the English alphabet. These letters are called **variables** or **literal numbers**. In the examples discussed above, $n$, $a$ and $b$ are variables. The term 'vary' means 'to change'. So, a variable does not have a fixed value. Variables can be added, subtracted, multiplied or divided by a given number or variable, which results in an **algebraic expression**. In algebra, we use + and – signs to indicate addition and subtraction, but × and ÷ are rarely used. Multiplication of two variables is denoted by placing a dot between the variables or just writing one variable after the other with no sign of operation between them. It can also be represented by enclosing one or both variables within brackets. Thus, * The product of two variables $m$ and $n$ is represented as $m.n, mn$ or $(m)(n)$. * $6y$ represents 6 times $y$. * $x$ divided by $y$ is indicated as $\frac{x}{y}$. In Algebra, the first and foremost step is writing a statement as an algebraic expression. For example, we know that the perimeter of a square is four times its side. Now if the side of the square is $s$, then we can write Perimeter of the square = $4s$ In Algebra, an algebraic expression can represent different word phrases. For example, consider any number $a$. * $a + 5$ represents 5 added to $a$, 5 more than $a$ or $a$ increased by 5. * $a - 5$ represents 5 subtracted from $a$, 5 taken away from $a$ or $a$ decreased by 5. * $5a$ represents 5 multiplied by $a$ or 5 times $a$. * $\frac{a}{5}$ represents $a$ divided by 5, quotient of $a$ and 5 or $\frac{1}{5}$ of $a$. An algebraic expression can be evaluated only if we know value of the variable. We can find the value of all the above expressions only if the variable $a$ is given some value. For example, for $a = 15$, we have: $a + 5 = 15 + 5 = 20, \quad a - 5 = 15 - 5 = 10, \quad 5a = 5 × 15 = 75, \quad \frac{a}{5} = \frac{15}{5} = 3$ **REMEMBER** * The letters that we use for representing unknown numbers are called variables or literal numbers. * A variable does not have a fixed value while a constant has a fixed value. * An algebraic expression contains variables and constants connected by operations like +, -, ×, ÷, for example $2x + 9$. **Solved Examples** **Example 3**: Frame algebraic expressions for the following: a) 7 is added to y. b) 3 is subtracted from two times x. c) One-fourth of z is added to 6. d) z is increased by thrice y. e) Sum of x and y is subtracted from $\frac{1}{2}$. **Solution**: a) $y + 7$ b) $2x - 3$ c) $6 + \frac{z}{4}$ d) $z + 3y$ e) $\frac{1}{2} - (x + y)$ **INNINGS 11.1** 1. Observe the dot patterns shown in Fig. 11.5(i) and (ii). For each pattern: a) write down the pattern using numbers and the rule to go from one arrangement to the next. b) generalise and find out how many dots the nth arrangement will contain. 2. Study the matchstick patterns shown in Fig. 11.6. a) Write down the pattern using numbers. b) Write down the rule to go from one shape to the next. c) Write down the expression for the number of matchsticks in the nth shape. 3. Write the next number in each of the following number patterns and state the rule: a) 50, 63, 76, 89, ... b) 11, 22, 44, 88, ... c) 625, 125, 25, 5, ... d) 1, 4, 9, 16, ... 4. Write a rule to show the relationship between the two quantities in each of the following tables: | a | 16 | 19 | 21 | | r | 2 | 3 | 4 | | l | 15 | 13 | 11 | | ---- | -- | -- | -- | -- | ---- | - | - | - | -- | ---- | -- | -- | -- | | ## | 12 | 15 | 17 | | s | 3 | 5 | 7 | | m | 56 | 48 | 40 | 5. Complete the following tables using given formulas: a) | a | 5 | 6 | 7 | | - | - | - | - | | b | | | | $b = a \times a$ b) | x | 3 | 4 | 5 | | - | - | - | - | | w | | | | $w = 2(x + 1)$ 6. Taking a variable $n$, write an algebraic expression for the following: a) Ninety taken away from a number b) The sum of a number and 36 divided by 13 c) The difference between 46 and a number d) The quotient of 20 and a number e) One-third of a number increased by 62 f) Twice a number increased by 18 7. If $s$ is the side of a regular pentagon, represent the perimeter of the pentagon as an algebraic expression. Find the perimeter when $s = 5$ cm. 8. If length ($l$) of a rectangle is 2 cm less than twice its breadth ($b$). Find the expression for the length of the rectangle. Find the perimeter ($P$) of the rectangle when $b = 5$ cm. **GOOGLY** 1. The parking fee of a certain car park is as follows: For the first hour: ₹10 For every additional hour: ₹5 Frame an expression for the amount paid after $h$ hours. Also find the actual amount paid after $3\frac{1}{2}$ hours. **PRACTICAL APPLICATIONS OF ALGEBRA** It is actually quite common for an average person to perform simple algebra without realising it. Suppose, you go to a grocery store and have spent ₹50 to buy biscuits worth ₹10 a pack. Table 11.1 lists out the amount of money one should carry to buy biscuit packs and hence a formula can be generated from it. Now, cost of b packs of biscuits = 10 b Here, b represents the number of packs of biscuits. From Table 11.1, we can observe that: Cost of 1 pack of biscuit = ₹10 Cost of 2 packs of biscuit = ₹10 × 2 = ₹20 Cost of 3 packs of biscuit = ₹10 × 3 = ₹30 So, cost of 5 packs of biscuit = ₹10 × 5 = ₹50 and cost of 20 packs = ₹10 × 20 = ₹200 Now, cost of 6 packs of biscuits = ₹10 × 6 = ₹60 | No. of Biscuit Packs | Total Cost (₹) | | :------------------- | :------------- | | 1 | 10 | | 2 | 20 | | 3 | 30 | | 4 | 40 | | 5 | | | | 200 | | 6 | | Consider the following practical situations for which expressions are framed: * Jassi is twice as old as Hari. Let Hari's age be x years. Then Jassi's age is 2x years. * Amya is four years younger than her brother Arnav. What is Amya's age and Arnav's age 5 years from now? Let Arnav's age = x years. Then Amya's age = (x – 4) years In 5 years, Arnav's age will be (x + 5) years. Amya's age after 5 years = (x – 4) + 5 = (x + 1) years. * The speed of a train is 100 km/hr more than the speed of a car. Let the speed of the car be x km/hr. Then the speed of the train = (x + 100) km/hr. * Ankita and her three college friends are going to share a three-bedroom apartment. If the total rent is r, what is Ankita's share? Ankita's share = $\frac{r}{4}$ **CATCH YOUR MISTAKES** -> The following two verbal expressions give to different algebraic expressions: Twice the sum of a number and 5 is $2(x+5)$ while sum of twice a number and 5 is $2x+5.$ **Solved Examples** **Example 4**: Shruti is 4 years elder than her sister Manya. Frame an expression for Shruti's age if Manya's age is x years and fill in the following table. Solution: Let $F$ be the fund collected for the flood relief. Money collected in a week = ₹25 Money collected in 2 weeks = ₹25 × 2 So, money collected in $w$ weeks = ₹25w or ₹25w There is already ₹50 in the fund. Thus, $F = (25w + 50)$. After 4 weeks, the children will be able to collect ₹150. | Shruti's age | | 16 | | | :----------- | :- | :- | :- | | Manya's age | 8 | | 23 | | | 2 | 4 | 6 | 8 | 10 | | :------------ | :--------- | :--------- | :--------- | :--------- | :--------- | | No. of weeks | | | | | | | Fund collected | | | | | | | | 50+100=100 | 50+200=150 | 50+300=200 | 50+400=250 | 50+500=300 | Solution: Manya's age = $x$ years So, Shruti's age = $x + 4$ **Example 6**: Anu went to buy confectionery for her picnic. She took ₹500 with her and bought the following: a) 2 packets of wafers at $x$ per packet. | Shruti's age | 12 | 16 | | 27 | | :----------- | :- | :- | :- | :- | | Manya's age | 8 | 12 | | 23 | **Example 5**: In a class, children are collecting money for flood relief fund. There is already ₹50 in the fund. They are collecting ₹25 per week. Frame an expression for the money in the fund after w weeks. Also prepare a table showing the money collected after 2, 4, 6, 8 and 10 weeks. After how many weeks will they be able to collect ₹150? b) 3 packets of biscuits at $y$ per packet. c) 5 bars of chocolate at $z$ per packet. Frame an expression for the total amount spent by Anu and the amount left with her. **Solution**: Cost of 2 packets of wafers at $x$ per packet = $2x$ Cost of 3 packets of biscuits at $y$ per packet = $3y$ Cost of 5 bars of chocolate at $z$ per packet = $5z$ Now, the total amount spent by Anu = $(2x + 3y + 5z)$ The amount left with her = $[500 - (2x + 3y + 5z)]$ **INNINGS 11.2** 1. There are $y$ students who volunteered to water the school garden. The principal said that she wished she had three times as many volunteers. How many volunteers would the principal like to have? 2. Shalini has guests coming to her party. She decides to serve a *gulab jamun* as sweet dish to each guest. She already has 12 *gulab jamuns*. If she buys x *gulab jamuns* more, how many guests are coming to her place? 3. You havet toffees in your bag. Your sister has 4 times as many toffees as you do. How many toffees does your sister have? 4. Leena is 4 years younger than Jaya now. If Jaya is x years old now, frame an expression for their ages after 10 years.5. Shreya is $\frac{4}{5}$ of Reema's age. How old will Shreya be 2 years from now if Reema is $x$ years old? 5. Anwar has $p$ pencils. He wants to put these into boxes that hold 15 pencils each. He has used $x$ boxes and there are still 3 pencils left. Find an expression for the total number of pencils he has. 6. Mira is 3 years older than Jyoti. If Jyoti is $x$ years old now, frame an expression for their ages 4 years ago. 7. Lucky is 4 times as old as Ali. If Ali is x years old, what is Lucky's age 6 years hence? 8. In a school, $x$ students are split up into 3 equal groups to participate in a cricket tournament. If 4 cricketers are added in each group, frame an expression for the number of cricketers in each group now. 9. Pooja's present age is $y$ years. Find her: a) age 6 years hence. b) age 3 years ago. c) mother's age, who is 5 years elder than twice Pooja's age. d) father's age, who is 2 years elder than Pooja's mother. e) 10. Change the following statements involving algebraic expressions to statements in ordinary language: a) Sakshi is x years old. Her uncle is $(3x - 4)$ years old. b) Cost of a shampoo bottle is c. Cost of a detergent pack is $(2c + 3)$. c) Rohit scored m marks in a Maths test. Ria scored $(2m - 1)$ marks in the same test. **EQUATION** Till now, you have learnt about how to find the number of matchsticks used for making matchstick patterns. Now, how does one find the number of hexagons made given the number of matchsticks used (Fig. 11.7)? Consider the table given in Example 1. Let $n$ be the number of hexagons formed, when the number of matchsticks used is 41. Thus, $5n + 1 = 41$ . This is called an **equation**. An equation is a statement of equality of two numbers or expressions. The expression on the left hand side of the equality sign (=) is called the LHS while the expression on the right hand side of the equality sign is called the RHS. If LHS ≠ RHS, we do not get an equation. Expressions like $3y > 9$ or $3y \leq 9$ are not equations. Also, $8 – 3 = 5$ is an equation but a numerical one as it does not contain a variable. Now, let us find the number of hexagons made by using 41 matchsticks. We substitute $n = 4, 6, 8,$ etc. in the equation $5n + 1 = 41$. Putting $n = 4$, LHS =$5n + 1 = 5 × 4 + 1 = 21$ RHS = 41 Thus, LHS ≠ RHS Putting $n = 6$, LHS = $5n + 1 = 5 × 6 + 1 = 31$ RHS = 41 Thus, LHS ≠ RHS Putting $n = 8$, LHS = $5n + 1 = 5 × 8 + 1 = 41$, RHS = 41 Thus, LHS = RHS. Hence, we find that 8 hexagons will be formed when 41 matchsticks are used. Thus, any equation like $5n + 1 = 41$ is a condition on the variable and it is satisfied for a definite value of the variable. Consider Example 5 again and answer the following question: "After how many weeks will the students be able to raise a fund of ₹250?" Expression for fund collected is $F = 25w + 50$. If the fund collected is ₹250, then we get equation $25w + 50 = 250$. Looking at the table given in Example 5, we can easily say that $w = 8$ satisfies the equation. Hence, after 8 weeks the children will be able to raise a fund of ₹250. Solution of An Equation **TIPS** -> An equation involves = sign while an algebraic expression does not contain = sign. $\implies$ Interchanging the LHS and RHS of an equation does not change the equation. Thus, the equation $x + 3 = 9$ is the same as equation 9 = $x + 3$. The value of the unknown variable that satisfies an equation is called a solution of the equation or **root of the equation**. Thus, $n = 8$ is a solution of the equation $5n + 1 = 41$. We can find the solution of an equation by substituting different values of the variable. The value that satisfies the equation is the solution of the equation and the method is called the **trial and error method**. An equation is like a balanced pair of scales. If we add, subtract, multiply or divide any one of the sides of the equation we have to do the same on the other side of the equation to keep the equation balanced. Consider the equation $x + 5 = 11$. Solving One-Step Equations Using a Pan-Balance Imagine a balance with a bag of unknown counters and 5 more counters on one side of the balance and 11 on the other side. If we take away 5 counters from both sides, the balance will not get disturbed. So, we have $x + 5 - 5 = 11 - 5$ $x = 6$ Hence, $x = 6$ is the solution of the equation. This is a better and more systematic method of solving an equation and is called method of balancing an equation or inverse operation technique of solving an equation. **REMEMBER** * An equation is a statement of equality of two numbers or expressions. * We can find the solution of an equation by either of the following methods: i) By trial and error method-by substituting different values of the variable. ii) By balancing an equation by adding or subtracting the same number on both the sides, or by multiplying or dividing both sides by the same number. **RECORDS** Algebra is all about patterns. The 13th century Italian Leonardo da Pisa better known as Fibonacci was the most talented mathematician of middle age. He was known for a particular number sequence called Fibonacci sequence where sequence progressed by adding previous two terms. This number sequence is found in nature. Many species of flowering plants have number of petals in the Fibonacci sequence, the spiral arrangement of pineapples, seeds of sunflower heads, etc. all follow the same. **Solved Examples** **Example 7**: Identify which of the following are equations with a variable: a) 5=4+1 b) x=7 c) z>92 d) 15=1+3 e) 3n+12<100 f) w+4=100-w g) y-2-3=34 h) 3 (d+ 4) = 12 i) (x+24)(x-3) j) $\frac{s}{5}$ = 20 Solution: The following are equations: b) x=7 f) w+4=100-w h) 3 (d+ 4) = 12 j) $\frac{s}{5}$= 20 **Example 8**: Solve the equation by trial and error method: $2x - 4 = 6$ **Solution**: $2x - 4 = 6$ We get the values of the LHS of the given equation for different values of the variable $x$ as follows: | x | 1 | 2 | 3 | 4 | 5 | | :----- | :----- | :----- | :----- | :----- | :----- | | 2x - 4 | -2 | 0 | 2 | 4 | 6 | We find that for $x = 5$, LHS = RHS. So, $x = 5$ is the solution of the given equation. **Example 9**: Solve the following equations: a) $x - 6 = 2$ b) $x + 7 = 13$ c) $15x = 45$ d) $\frac{7x}{5}$ = 14 **Solution**: a) $x - 6 = 2$ Adding 6 to both sides, $x - 6 + 6 = 2 + 6$ or $x = 8$ So, $x = 8$ is the solution of the given equation. b) $x + 7 = 13$ Subtracting 7 from both sides, $x + 7 - 7 = 13 - 7$ or $x = 6$ So, $x = 6$ is the required solution. c) $15x = 45$ Dividing both sides by 15, $\frac{15x}{15} = \frac{45}{15}$ or $x = 3$ So, $x = 3$ is the required solution. d) $\frac{7x}{5} = 14$ Multiplying both sides by 5, $\frac{7x}{5} \times 5 = 14 \times 5$ or $7x = 70$ Now, dividing both sides by 7, $\frac{7x}{7} = \frac{70}{7}$ or $x = 10$ So, $x = 10$ is the required solution. **INNINGS 11.3** 1. Which of the following are equations? Also, identify the variable from the equation. a) $\frac{x}{12}=5$ b) $x-3>1$ c) $2z-1=3z$ d) $25y=3+4y$ e) $3+2=6-\frac{5}{5}$ f) $5p+76=3$ 2. Find whether or not the given value of the variable is the solution of the equation: a) $a-6=4, a = 10$ b) $\frac{d}{6}=18, d=18$ c) $17-v=5, v=5$ d) $3t-5=46, t=17$ 3. Find the solution of the following equations by the trial and error method: a) $x + 13 = 24$ b) $z-4=20$ c) $\frac{k}{6}=8$ d) $2n = 18$ 4. Using a variable write each of the following statements as an algebraic equation: a) A number increased by 9 is 15. b) Quotient of 39 and a number is 3. c) Five less than three times a number is 40. d) Twice the sum of a number and 5 is 48. 5. Complete the following tables and hence find the solution of the given equations: a) $x+3=7$ | x | 1 | 2 | 3 | 4 | | :----- | :----- | :----- | :----- | :----- | | x+3 | | | | | b) $\frac{n}{8}=5$ | n | 4 | 8 |16 |40 | | :----- | :----- | :----- | :----- | :----- | | $\frac{n}{8}$ | | | | | | 6. Solve the following equations by balancing the equations: a) $x-6=4$ b) $y+5=10$ c) $2m = 16$ d) $\frac{v}{2}=9$ **CHAPTER INNINGS** 1. Look at the matchstick pattern given below for letter N: Generalise a rule for the number of matchsticks required for making x number of Ns. Also, find out how many matchsticks are required for making 12 Ns. 2. Write the rule used to create the number pattern shown 250, 130, 70, 40, 25. 3. What does $\frac{m}{7}-12=23$ represent? a) One-seventh of a number is 12 less than 23. b) Seven times a number is 12 more than 23. c) One-seventh of a number is 12 more than 23. d) Seven times a number is 12 less than 23. 4. If $\frac{100a}{2}=500$, what is the value of a? 5. Reema has $12a$ flowers. Ria has 6 flowers more than Reema. How many flowers does Ria have? 6. Rahul spends $y$ daily and saves $z$ per week. What is his income for three weeks? 7. Akash hast toffees. His brother Amit has 6 toffees less than Akash. How many toffees do they have together? 8. Sandeep was $p$ years old 18 years ago. If $p = 13$, how old is Sandeep now? 9. Mr Yadav left one-third of his property to his son, one-fourth to his daughter and ₹18,00,000 for his wife. Find an expression for Mr Yadav's total property. 10. Two years ago, Suresh was