Get the simplified Class 8 Maths Extra Questions Part 2 Chapter 6 Algebra Play Class 8 Extra Questions and Answers with complete explanation.
Class 8 Algebra Play Extra Questions
Class 8 Maths Chapter 6 Algebra Play Extra Questions
Class 8 Maths Chapter 6 Extra Questions – Algebra Play Extra Questions Class 8
Very Short Answer Questions
Question 1.
What is a variable In algebra?
Answer:
A variable is a letter used to represent an unknown number.
Question 2.
In a number pyramid, how is each number formed?
Answer:
Each number is the sum of the two numbers directly below it.
Question 3.
What is the algebraic expression for a number increased by 7?
Answer:
Let x be the unknown number.
Then, the number is increased by 7 i.e. x + 7.
Question 4.
If the bottom row of a pyramid is a, b, c, what is the top number?
Answer:
If the bottom row of a pyramid is a, b, c.
Then, the top number = (a + b) + (b + c)
= a + 2b + c
Question 5.
Which digits 3,6 and 9 should be used as the multiplier to get the largest product?
Answer:
We know that if digits p<q<r, the largest product is always formed as qpxr.
Therefore, the largest digit should be the multiplier
i. e. 9.
Question 6.
A two-digit number is ab. Write its reverse.
Answer:
A two-digit number is ab.
Then, its reverse = ba = 10b + a
Question 7.
The difference of a two-digit number and its reverse is divisible by which number?
Answer:
Let the two digit number be ab i.e. 10 a + band its reverse
= ba = 10b+a
∴ Required difference = 10b + a – 10a – b = 9b – 9a = 9(b – a)
Therefore, the difference of a number and its reverse is divisible by 9.
Question 8.
The sum of a two-digit number and its reverse is divisible by which number?
Answer:
Let the two-digit number be ab i.e. 10a + b and its reverse = ba – 10 b + a
Required sum = 10a + b + 10b + a
= 11a + 11b = 11(a + b)
Therefore, the sum of a number and its reverse is divisible by 11.
Short Answer Type Questions
Question 1.
If a person thinks of a number, doubles it, adds 4 and divides the result by 2, what must he subtract from this value to always get a result of 2? Show your work.
Answer:
Think of a number: x Double it: 2x
Add 4 : 2x + 4 Divide by two : x + 2
Subtract the original number x + 2 – x = 2
Hence, original number is subtracted to get 2.
Question 2.
What is the Virahanka-Fibonacci sequence?
Answer:
The Virahanka-Fibonacci sequence is a mathematical series, where each number is the sum of the two preceding ones.
The Fibonacci sequence is 1, 2, 3, 5, 8, 13, ……………
Question 3.
Represent the sum of four numbers in a 2 × 2 calendar grid, algebraically, where a is the top-left number.
Answer:
Let the top-left number be a.
Then, the grid will be
Now, the sum of 4 numbers in 2 × 2 grid
= a + a + 1 + a + 7 + a + 8
= 4a + 16
Question 4.
If the sum of four numbers in a 2 × 2 calendar grid is 36, what is the top-left number?
Answer:
Let the top-left number be a.
We know that the sum of all numbers in 2 x 2 calendar grid = 4a+ 16
∴ 4a + 16 = 36
⇒ 4a = 36 – 16
⇒ a = \(\frac{20}{4}\) = 5
Hence, top-left number is 5.
Question 5.
Using the digits 1,3 and 7 once each, what is the largest possible product of a two-digit number and a one-digit number?
Answer:
We know that to get the largest product from three numbers, the largest digit should be the multiplier and the other two digits should be arranged in decreasing order to form the multiplicand.
The given digits are 1, 3 and 7.
Here, the largest digit is 7, so it is the multiplier.
The remaining digits are 3 and 1, which in decreasing order form the number 31.
The expression is 31 × 7 i.e. 217.
Hence, the largest product is achieved with the expression 31 × 7 and the result is 217.
Question 6.
Why is the difference between a two-digit number ab and its reverse ba always divisible by 9?
Answer:
Let the two-digit number be ab.
When it is reversed, the new number is ba.
If b > a then ba > ab.
So, the difference = 10b + a – (10a + b)
= 10 b + a – 10a – b = 9b – 9a = 9(b – a) and if a > b then ab > ba.
So, the difference = 10a +b – (10b + a)
= 9a – 9b =9(a – b)
Hence, the difference is always divisible by 9.
Question 7.
Show that the number formed by repeating a three-digit number twice like 456456 is always divisible by 7, 11 and 13.
Answer:
The 6-digit number 456456 can be written as
= 400000 + 50000 + 6000 + 400 + 50 + 6
= 400400 + 50050 + 6006 = 1001[400 +50 + 6]
= 7 × 11 × 13 × 456
When divided sequentially by 7, 11 and 13, the result is the original 3-digit number.
Long Answer Type Questions
Question 1.
Find the dates if the final answer is 1066.
Answer:
Let the month be M and day be D.
We know that 1066 = 100M + 165 + D
⇒ 1066 – 165 = 100M + D
⇒ 901= 100M + D
Here, the hundreds and thousands digits represents the month (M) and the last two digits represent the day (D).
Hence, the date is the 1st September.
Question 2.
Fill this number pyramid.
Answer:
The bottom row is 5, 11, 17, 9.
The first number in the second row is the sum of the first two numbers in the bottom row i.e. 5 + 11= 16.
The second number in the second row is the sum of the middle numbers in the bottom row i.e. 11 + 17 = 28.
Similarly,
The third number in the second row = 17 + 9 = 26
The first number of the third row = 16 + 28 = 44
The second number of the third row = 28 + 26 = 54
The topmost row number = 44 + 54 = 98
Hence, the pyramid is
Question 3.
The bottom row of a number pyramid contains the following three consecutive Fibonacci numbers.
(i) 13, 21, 34
(ii) 34, 55, 89
Find the number at the top of the pyramid.
Answer:
We know that if the bottom row of a number pyramid is a, h, c and the expression for the nubmer at the top row of the number pyramid is a + 2b + c.
(i) Here, a = 13, b = 21 and c = 34
Number at top row of the pyramid = 13 + 2(21) + 34
= 13 + 42 + 34 = 89
(ii) Here, a = 34, b = 55 and c = 89
Number at the top row of the pyramid = 34+ 2(55) + 89
= 34 + 110 + 89 = 233
Question 4.
There are 3 enhanted wells. When coins are dropped into any well, the number of coins becomes three times. A traveller has some coins. Fie drops all the coins into the first well and then puts some coins into pocket 1. He then drops the remaining coins into the second well and puts the some coins into pocket 2. Finally, he drops the remaining coin into the third well and puts all of them into pocket 3.
If the number of coins in each pocket is equal.
How many coins did the traveller start with? How many coins are in each pocket?
Answer:
Let x be the initial number of coins and y be the equal number of coins in each pocket.
After the first well, the number of coins becomes 3x. After placing coins in pocket 1, the remaining coins are 3x – y.
The remaining coins are dropped in the second well, so the number of coins
3(3x – y) = 9x – 3y
After puts y coins in pocket 2, the remaining coins = 9x – 3y – y = 9x – 4y
The remaining coins are dropped in the third well, so the number of coins
= 3(9x – 4y)
= 27x – 12 y
The traveller places all of these coins in pocket 3.
So, this quantity must equal to y.
27x – 12y = y
⇒ 21 x = 13y
⇒ \(\frac{x}{y}=\frac{13}{27}\)
∴ The smallest positive integer solution is x = 13 and y = 27.
Hence, the person started with 13 coins and puts 27 coins in each pocket.
Question 5.
A genie doubles Karim’s money every round but takes 4 coins each time. After 3 rounds, Karim has 20 coins left. How many coins did he have initially?
Answer:
Let the number of coins Karim had initially be x.
If money doubles and 4 coins are taken, so after first round the remaining coins = 2x – 4
Similarly, after second round, the remaining coins
= 2(2x – 4) – 4
= 4x – 8 – 4 = 4x – 12
and after third round, the remaining coins
= 2(4x – 12) – 4
= 8x – 24 – 4 = 8x – 28
If after third round, Karim has 20 coins.
∴ 8x – 28 = 20
⇒ 8x = 48
⇒ x = \(\frac{48}{8}\) = 6
Hence, the number of coins Karim had initially is 6.
Question 6.
In farm, there are hens and goats.
The total number of heads is 18 and total number of legs is 52. Find the number of hens and goats.
Answer:
Given, the total number of heads of animals = 18
If all 18 animals were hens, each having 2 legs.
∴ The total number of legs = 18 × 2 = 36
Also, given the total number of legs = 52
∴ Remaining legs = 52 – 36 = 16
Each goat has 4 legs, which is 2 more than a hen.
This difference of 16 legs is due to the goat.
∴ The number of goat = \(\frac{16}{2}\) = 8
So, the number of hens = 18 – 8 = 10
Hence, there are 10 hens and 8 goats in the farm.
Question 7.
A juice seller spends ₹ 120 every day and earns ₹ 20 per glass sold. How many glasses must he sell to make a profit of ₹ 180?
Answer:
Given, a juice seller spends ₹ 120 every day and cost of per glass = ₹ 20
Let x be the number of glasses sold.
∴ Income = 20x
and profit = Income – Expense
⇒ 180 = 20x – 120
⇒ 20x = 180 + 120
Hence, 15 glasses must he sell to make a profit of ₹ 180.
Skill Based Questions
Question 1.
Designing Your Own “Think of a Number’’.
Trick You want a trick that always ends with the answer 7.
- Think of a number.
- Do some operations.
- Final answer is always 7.
(a) Propose a sequence of atleast 4 arithmetic steps that always results in 7.
(b) Justify using algebra.
Answer:
Do yourself
Question 2.
In a 4 row pyramid, the bottom row is a, b, c, d.
(a) Derive an expression for the top number in terms of a, b, c, d.
Answer:
a + 3b + 3c +d
(b) Check your expression using a simple example (e.g. 1,2,3,4).
Answer:
Do yourself
Question 3.
On an August 2025 calendar, a 2 × 2 block sums to 52.
(a) Let the top-left number be a. Write the sum of the four numbers in terms of a.
Answer:
4a + 16
(b) Find the four numbers.
Answer:
9, 10, 16, 17
Question 4.
Using digits 1,3 and 7 exactly once, fill __ __ × __ to get the largest product.
(a) Try to reason structurally (without checking all six possibilities), which arrangement is best.
Answer:
Do yourself
(b) Then, compute and confirm.
Answer:
217
Question 5.
There are 3 shrines, each with a magical pond that doubles the flowers dipped. A person starts with xflowers.
- Dip in pond 1 double → place k flowers in shrine 1.
- Dip the remaining flowers in pond 2 → double → place k flowers in shrine 2.
- Dip the remaining flowers in pond 3 → double → place all remaining in shrine 3, again k flowers.
Find the initial number of flowers and value of k.
Answer:
8 flowers, k = 8 ⇒ x = 7
Case Study Based Question
Question 1.
Aditi is playing a game with her friend using a calendar. She asks her friend to choose any 3 × 3 square of dates from the current month.
She claims that if her friend tells her the sum of all 9 numbers in that square, she can instantly tell her the middle number and every other date in that square. According to given information answers the following questions.
(i) If the middle number of the square is represented by x, write an expression for the number directly above it and the number directly below it.
Answer:
The number directly above is x – 7 and the number directly below is x + 7.
(ii) Show algebraically, why the sum of any 3 × 3 calendar square is always 9 times the middle number?
Answer:
Let the middle number be x.
The nine numbers are:
Row 1: (x – 8), (x – 7), (x – 6)
Row 2: (x – 1), x, (x + 1)
Row 3: (x + 6), (x + 7), (x + 8)
So, the sum of all numbers
= x – 8 + x – 7 + x – 6 + x – 1 + x + x + 1 + x + 6 + x + 7 + x + 8
= 9x
(iii) If the sum of all 9 numbers in the 3 × 3 square is 135, what is the middle number?
Answer:
According to the question, 9x = 135
x = \(\frac{135}{9}\) = 15