GSEB Class 6 Maths Solutions Chapter 11 Algebra Exercise 11.1

Question 1. Find the rule which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(a) A pattern of letter T as T
(b) A pattern of letter Z as Z
(c) A pattern of letter U as U
(d) A pattern of letter V as V
(e) A pattern of letter E as E
(f) A pattern of letter S as S
(g) A pattern of letter A as A
Answer:
(a) For \( n = 1 \), it needs 2 or \( 2 \times n \) matchsticks.
For \( n = 2 \), it needs 4 or \( 2 \times n \) matchsticks.
For \( n = 3 \), it needs 6 or \( 2 \times n \) matchsticks.
For \( n = 4 \), it needs 8 or \( 2 \times n \) matchsticks.
Therefore, the rule is \( 2n \).
(b) A number of matchsticks is found as follows:
For \( n = 1 \), it is 3 or \( 3 \times n \).
For \( n = 2 \), it is 6 or \( 3 \times n \).
For \( n = 3 \), it is 9 or \( 3 \times n \).
For \( n = 4 \), it is 12 or \( 3 \times n \).
Therefore, the general rule is \( 3n \).
(c) The number of matchsticks is found as:
For \( n = 1 \), it is 3 or \( 3 \times n \).
For \( n = 2 \), it is 6 or \( 3 \times n \).
For \( n = 3 \), it is 9 or \( 3 \times n \).
For \( n = 4 \), it is 12 or \( 3 \times n \).
So, the rule is \( 3n \).
(d) The number of matchsticks is calculated:
For \( n = 1 \), it is 2 or \( 2 \times n \).
For \( n = 2 \), it is 4 or \( 2 \times n \).
For \( n = 3 \), it is 6 or \( 2 \times n \).
For \( n = 4 \), it is 8 or \( 2 \times n \).
Hence, the rule is \( 2n \).
(e) The number of matchsticks is determined:
For \( n = 1 \), it is 5 or \( 5 \times n \).
For \( n = 2 \), it is 10 or \( 5 \times n \).
For \( n = 3 \), it is 15 or \( 5 \times n \).
Thus, the rule is \( 5n \).
(f) The number of matchsticks is calculated:
For \( n = 1 \), it is 5 or \( 5 \times n \).
For \( n = 2 \), it is 10 or \( 5 \times n \).
For \( n = 3 \), it is 15 or \( 5 \times n \).
Therefore, the rule is \( 5n \).
(g) The number of matchsticks is given as:
For \( n = 1 \), it is 6 or \( 6 \times n \).
For \( n = 2 \), it is 12 or \( 6 \times n \).
For \( n = 3 \), it is 18 or \( 6 \times n \).
Thus, the rule is \( 6n \).
In simple words: For each pattern, count how many matchsticks are used for a single unit (n=1), then for two units (n=2), and so on. Find a formula that connects the number of units (n) to the total matchsticks. This formula is the rule.

Exam Tip: Always analyze the progression for the first few values of 'n' to correctly identify if the rule is linear (like \( an \)) or has a constant term (like \( an + b \)).

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?
Answer: The rules for finding the number of matchsticks are: The letter L uses \( 2n \). The letter C uses \( 3n \). The letter V uses \( 2n \). The letter U uses \( 3n \). The letter T uses \( 2n \). The letter F uses \( 4n \). The letters L, V, and T have the same rule. This happens because each of these patterns needs 2 matchsticks for one basic unit (when \( n = 1 \)).
In simple words: Letters L, V, and T all follow the same rule, which is "2 times n" (\( 2n \)). This is because each of these shapes is made up of 2 matchsticks when you make just one of them.

Exam Tip: When comparing rules, identify the base number of matchsticks needed for the simplest form (n=1) of each pattern, and see which ones match.

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.)
Answer: Let \( n \) represent the number of rows. There are 5 cadets in each row. So, for \( n = 1 \) row, there are 5 cadets or \( 5 \times n \) cadets. For \( n = 2 \) rows, there are 10 cadets or \( 5 \times n \) cadets. For \( n = 3 \) rows, there are 15 cadets or \( 5 \times n \) cadets. The rule for finding the total number of cadets in \( n \) rows is \( 5n \).
In simple words: If there are 5 cadets in every row, and you have 'n' rows, you just multiply 5 by 'n' to find the total number of cadets. So, the rule is \( 5n \).

Exam Tip: Remember to clearly define your variable (n for number of rows) and show the pattern of calculation for different values of n before stating the final rule.

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.)
Answer: The number of mangoes in one box is 50. Therefore, to determine the total number of mangoes: When \( b = 1 \) box, there are 50 mangoes, which is \( 50 \times 1 = 50 \times b \). When \( b = 2 \) boxes, there are 100 mangoes, which is \( 50 \times 2 = 50 \times b \). When \( b = 3 \) boxes, there are 150 mangoes, which is \( 50 \times 3 = 50 \times b \). Thus, the total quantity of mangoes is \( 50b \).
In simple words: If one box has 50 mangoes, then for 'b' number of boxes, you just multiply 50 by 'b'. So, the total mangoes are \( 50b \).

Exam Tip: This is a direct multiplication problem. Ensure you use the specified variable 'b' in your final algebraic expression.

Question 5. The teacher distributes 5 pencils per student. Can you tell me how many pencils are needed, given the number of students? (Use s for the number of students.)
Answer: Let \( s \) denote the number of students. Each student receives 5 pencils. So, when \( s = 1 \) student, the number of pencils is 5 or \( 5 \times 1 = 5 \times s \). When \( s = 2 \) students, it is 10 pencils or \( 5 \times 2 = 5 \times s \). When \( s = 3 \) students, it is 15 pencils or \( 5 \times 3 = 5 \times s \). Therefore, the total number of pencils required is \( 5s \).
In simple words: Each student gets 5 pencils. If there are 's' students, you multiply 5 by 's' to find out how many pencils are needed. The answer is \( 5s \).

Exam Tip: Always make sure your final expression clearly uses the variable given in the question, 's' in this case.

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.)
Answer: Let \( t \) represent the flying time in minutes. The bird flies 1 kilometer in one minute. So, the distance covered: When \( t = 1 \) minute, it is 1 km or \( 1 \times 1 \) km \( = 1 \times t \) km. When \( t = 2 \) minutes, it is 2 km or \( 1 \times 2 \) km \( = 1 \times t \) km. When \( t = 3 \) minutes, it is 3 km or \( 1 \times 3 \) km \( = 1 \times t \) km. Hence, the distance covered in \( t \) minutes is \( t \) km.
In simple words: The bird flies 1 km every minute. So, if it flies for 't' minutes, the distance covered will be 't' kilometers.

Exam Tip: When the rate is 1 unit per unit of time, the total amount is simply equal to the time variable itself.

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?
Answer: Let \( r \) denote the number of rows. There are 9 dots in each row. So, the total number of dots: For \( r = 1 \) row, there are 9 dots or \( 9 \times 1 = 9 \times r \) dots. For \( r = 2 \) rows, there are 18 dots or \( 9 \times 2 = 9 \times r \) dots. For \( r = 3 \) rows, there are 27 dots or \( 9 \times 3 = 9 \times r \) dots. Therefore, the total number of dots is \( 9r \). Now, if \( r = 8 \), we have \( 9 \times 8 = 72 \) dots. If \( r = 10 \), we have \( 9 \times 10 = 90 \) dots.
In simple words: For 'r' rows with 9 dots in each, the total dots are \( 9r \). If there are 8 rows, that's \( 9 \times 8 = 72 \) dots. If there are 10 rows, that's \( 9 \times 10 = 90 \) dots.

Exam Tip: This question requires two steps: first find the general rule, then use that rule to calculate for specific values of 'r'.

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.
Answer: Let's assume Radha's age is \( x \) years. Leela is 4 years younger than Radha. This means Leela's age is Radha's age minus 4 years. So, Leela's age \( = (x - 4) \) years.
In simple words: If Radha is 'x' years old, and Leela is 4 years younger, then Leela's age is \( x - 4 \) years.

Exam Tip: Younger means subtraction. Always set up the age of the unknown person (Leela) in terms of the known variable (Radha's age, x).

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 I, how many laddus did she make?
Answer: The number of laddus remaining is 5. The number of laddus given away is \( l \). Therefore, the total number of laddus made by the mother is the sum of those given away and those remaining, which is \( l + 5 \).
In simple words: To find the total laddus made, you add the laddus given away (l) to the laddus that are left (5). So, she made \( l + 5 \) laddus.

Exam Tip: Total items are always the sum of what was used/given away and what remains. This is a basic addition problem.

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 is taken to be x, what is the number of oranges in the larger box?
Answer: The number of oranges in a large box is \( 2x \). This comes from filling two smaller boxes, where each small box has \( x \) oranges, plus 10 additional oranges that are left over. Since \( x \) stands for the number of oranges in a smaller box, the number of oranges in a large box can be written as \( 2x + 10 \).
In simple words: A large box holds enough oranges to fill two small boxes (which is \( 2 \times x \)) and still has 10 oranges left over. So, the large box has \( 2x + 10 \) oranges.

Exam Tip: Break down word problems into smaller parts. "Fill two smaller boxes" means \( 2 \times x \), and "10 remain outside" means add 10.

Question 11.
(a) Look at the following matchstick pattern of squares (Fig. I). The squares are not separate. Two neighboring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)
(b) Fig. II gives a matchstick pattern of triangles. As in Q. 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.
Answer:
(a) Let \( n \) represent the number of squares. For (a), the number of matchsticks is found as follows:
When \( n = 1 \), it is 4, or \( (3 \times 1) + 1 = 3n + 1 \).
When \( n = 2 \), it is 7, or \( (3 \times 2) + 1 = 3n + 1 \).
When \( n = 3 \), it is 10, or \( (3 \times 3) + 1 = 3n + 1 \).
When \( n = 4 \), it is 13, or \( (3 \times 4) + 1 = 3n + 1 \).
Therefore, the rule needed is \( 3n + 1 \).
(b) For (b), let \( n \) represent the number of triangles. The number of matchsticks is calculated:
When \( n = 1 \), it is 3, or \( (2 \times 1) + 1 = 2n + 1 \).
When \( n = 2 \), it is 5, or \( (2 \times 2) + 1 = 2n + 1 \).
When \( n = 3 \), it is 7, or \( (2 \times 3) + 1 = 2n + 1 \).
When \( n = 4 \), it is 9, or \( (2 \times 4) + 1 = 2n + 1 \).
Hence, the rule needed is \( 2n + 1 \).
In simple words: For square patterns, each new square adds 3 matchsticks, plus 1 for the first one. So, it's \( 3n + 1 \). For triangle patterns, each new triangle adds 2 matchsticks, plus 1 for the first one. So, it's \( 2n + 1 \).

Exam Tip: Observe the incremental change in matchsticks as 'n' increases. This increment will often be the coefficient of 'n' in your rule. The remaining constant is the matchsticks in the first shape that don't increase.