In chapter 11, we are going to deal with algebra. We will learn how variables help us form equations and solve various mathematical problems. We will solve some matchstick pattern questions to understand the concept better.
In this chapter, we will create various shapes and letters using matchsticks. If you arrange matchsticks to form the shape of a letter or a geometric figure, you can learn about numbers and calculations more interactively.
A ‘variable’ is a fundamental concept in algebra. It’s like a placeholder or a blank space in mathematics. It can be filled with different values. A variable represents a letter (like x, y, n) that stands for a number we don’t know yet or can change. For example, if we say that x is the number of matchsticks needed to make a shape, x can be any number depending on the shape we choose.
Variables are used in various situations. For example, if Radha has x number of apples and gets 5 more, then she has x + 5 apples. In this case, x represents the initial number of apples. Similarly, if a pattern requires 3 matchsticks for the first shape, 6 for the second, 9 for the third, and so on, we can use a variable (let’s say n) to describe the total number of matchsticks needed for any number of shapes. The rule could be written as 3n, where n is the number of shapes.
NCERT Solutions for Class 6 Maths Exercise 11.1 Chapter 11 Algebra
Question 1. Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
(a) A pattern of letter T
(b) A pattern of letter Z
(c) A pattern of letter U
(d) A pattern of letter V
(e) A pattern of letter E
(f) A pattern of letter S
(g) A pattern of letter A
Solution
(a) A pattern of letter ‘T’
The letter ‘T’ typically requires 2 matchsticks to form the top horizontal line and 1 matchstick for the vertical line.
Thus, a single ‘T’ uses 3 matchsticks. For multiple ‘T’s, the pattern would be 3, 6, 9, 12, etc.
3 × 1, 3 × 2, 3 × 3, 3 × 4, and so on.
This can be written as 3 × n, where n is the number of ‘T’s.
The required pattern is 3 × n = 3n.
(b) A pattern of letter ‘Z’
The letter ‘Z’ requires 2 matchsticks for the top and bottom horizontal lines and 1 matchstick for the diagonal line, totaling 3 matchsticks for one ‘Z’.
For multiple ‘Z’s, the pattern would be 3, 6, 9, 12, etc., which is 3 × 1, 3 × 2, 3 × 3, 3 × 4, and so on.
This can be written as 3 × n.
The required pattern is 3 × n = 3n.
(c) A pattern of letter ‘U’
The letter ‘U’ can be formed using 2 vertical matchsticks for the sides and 1 horizontal matchstick for the bottom, totaling 3 matchsticks.
For multiple ‘U’s, it follows the same pattern as ‘T’ and ‘Z’.
The required pattern is 3 × n = 3n.
(d) A pattern of letter ‘V’
A ‘V’ typically requires 2 matchsticks, one for each diagonal line. For multiple ‘V’s, the pattern would be 2, 4, 6, 8, etc., which is 2 × 1, 2 × 2, 2 × 3, 2 × 4, and so on.
This can be written as 2 × n.
The required pattern is 2 × n = 2n.
(e) A pattern of letter ‘E’
The letter ‘E’ typically requires 4 matchsticks – 1 for the vertical line and 3 for the horizontal lines.
For multiple ‘E’s, the pattern would be 4, 8, 12, 16, etc., which is 4 × 1, 4 × 2, 4 × 3, 4 × 4, and so on.
This can be written as 4 × n.
The required pattern is 4 × n = 4n.
(f) A pattern of letter ‘S’
The letter ‘S’ can be formed with 5 matchsticks – 2 for the top and bottom curves and 1 for the middle part.
For multiple ‘S’s, the pattern would be 5, 10, 15, 20, etc., which is 5 × 1, 5 × 2, 5 × 3, 5 × 4, and so on.
This can be written as 5 × n.
The required pattern is 5 × n = 5n.
(g) A pattern of letter ‘A’
The letter ‘A’ typically requires 3 matchsticks – 2 for the diagonal lines and 1 for the horizontal line.
For multiple ‘A’s, the pattern would be 3, 6, 9, 12, etc., which is 3 × 1, 3 × 2, 3 × 3, 3 × 4, and so on.
This can be written as 3 × n.
The required pattern is 3 × n = 3n.
Question 2. We already know the rule for the pattern of letters L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution
The rule for creating the letter L with matchsticks involves using 2 matchsticks: one for the vertical line and one for the horizontal line at the bottom.
So, the rule for multiple L’s would be 2 × n, where n is the number of L’s.
Now, let’s compare this with the rules we derived for T, Z, U, V, E, S, and A:
T: Requires 3 matchsticks (rule: 3n).
Z: Requires 3 matchsticks (rule: 3n).
U: Requires 3 matchsticks (rule: 3n).
V: Requires 2 matchsticks (rule: 2n).
E: Requires 4 matchsticks (rule: 4n).
S: Requires 5 matchsticks (rule: 5n).
A: Requires 3 matchsticks (rule: 3n).
From this comparison, we can see that the letter V has the same rule as the letter L, which is 2n.
This is because both L and V require only 2 matchsticks to be formed.
Both L and V are composed of two lines (one vertical and one slanted for V, one vertical and one horizontal for L), making their matchstick requirements and patterns identical in terms of quantity.
Question 3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution
The number of cadets in each row is 5. If there are n rows, then the total number of cadets is the number of cadets in a row multiplied by the number of rows.
Therefore, the total number of cadets = 5 cadets/row × n rows.
Total number of cadets = 5n.
Question 4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution
Each box contains 50 mangoes. If there are b boxes, then the total number of mangoes is the number of mangoes in one box multiplied by the number of boxes.
Therefore, the total number of mangoes = 50 mangoes/box × b boxes.
Total number of mangoes = 50b.
Question 5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution
Each student receives 5 pencils. If there are s students, the total number of pencils needed is the number of pencils per student multiplied by the total number of students.
Therefore, the total number of pencils = 5 pencils/student × s students.
Total number of pencils = 5s.
Question 6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution
The bird flies 1 kilometer in one minute. If the bird flies for t minutes, then the total distance covered by the bird is the distance flown in one minute multiplied by the total flying time in minutes.
Therefore, the total distance covered = 1 kilometer/minute × t minutes.
Total distance covered = 1t kilometers or t kilometers.
Question 7. Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution
Each row contains 9 dots. If there are r rows, then the total number of dots is the number of dots in a row multiplied by the number of rows.
Therefore, the total number of dots = 9 dots/row × r rows.
Total number of dots = 9r.
For 8 rows: Total dots = 9 dots/row × 8 rows = 72 dots.
For 10 rows: Total dots = 9 dots/row × 10 rows = 90 dots.
Question 8. Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution
Let Radha’s age be x years. Since Leela is 4 years younger than Radha, Leela’s age is Radha’s age minus 4 years.
Therefore, Leela’s age = Radha’s age – 4 years = x years – 4 years.
Leela’s age = x – 4 years.
Question 9. Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution
Let the number of laddus given away be l. There are still 5 laddus remaining. Therefore, the total number of laddus made is the sum of the laddus given away and the laddus remaining.
Total laddus made = laddus given away + laddus remaining = l + 5.
Total laddus made = l + 5.
Question 10. Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution
Let the number of oranges in each smaller box be x. When a large box is emptied, it fills two smaller boxes and there are still 10 oranges remaining. Therefore, the number of oranges in the large box is twice the number of oranges in a small box plus 10.
Total oranges in a large box = (oranges in a small box × 2) + 10 oranges = (x × 2) + 10.
Total oranges in a large box = 2x + 10.
Additional Worksheet Questions for Exercise 11.1 Algebra for Class 6
- A gardener plants 6 trees in each row of a garden. How many trees are there in total for r rows?
- Each student in a class receives 4 notebooks. How many notebooks are needed for s students?
- A baker bakes 12 cupcakes in one batch. How many cupcakes does the baker make in b batches?
- A parking lot has space for 8 cars in each row. How many cars can it hold if there are n rows?
- In a bookshelf, each shelf can hold 15 books. If there are k shelves, how many books can the bookshelf hold in total?
- A cyclist covers 20 kilometers in one hour. How far can the cyclist travel in h hours?
- A tailor uses 2 meters of fabric to make a shirt. How much fabric is needed to make m shirts?
- An artist uses 7 colors to paint a picture. If the artist paints p pictures, how many colors are used in total?
- A farmer harvests 30 apples from each apple tree. How many apples does the farmer harvest from t trees?
- A construction site receives 50 bricks in one delivery. How many bricks does the site receive after d deliveries?
Answers
- Total trees = 6r
- Total notebooks = 4s
- Total cupcakes = 12b
- Total cars = 8n
- Total books = 15k
- Total distance = 20h kilometers
- Total fabric = 2m meters
- Total colors used = 7p
- Total apples = 30t
- Total bricks = 50d
