NCERT Solutions for Class 8 help students understand algebraic expressions, variables, equations, and mathematical patterns through simple explanations and practice questions.
NCERT Solutions for Class 8 provide step-by-step methods and clear explanations that help students solve algebraic problems confidently and understand concepts more effectively.
Algebra develops logical thinking and problem-solving skills. It also forms the foundation for advanced mathematical topics taught in higher classes.
The chapter includes questions related to variables, expressions, equations, number patterns, and basic algebraic operations that help students strengthen their understanding of algebra.
Yes, NCERT Solutions for Class 8 help students practice different algebraic methods regularly, improving both accuracy and calculation speed.
Students can improve by practicing equations daily, understanding formulas carefully, revising solved examples, and using NCERT Solutions for Class 8 for proper guidance.
Yes, the chapter explains how algebra is used in everyday situations such as finding unknown values, solving practical problems, and identifying mathematical relationships.
Yes, NCERT Solutions for Class 8 provide complete solutions for all textbook exercises, examples, and practice questions from the latest NCERT textbook.
Regular revision helps students remember formulas, improve logical reasoning, and solve algebraic questions more confidently during exams.
Yes, NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 6 are prepared according to the latest CBSE and NCERT syllabus guidelines.
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NCERT Solutions for Class 8 Maths Ganith Prakash 2 Chapter 6 Algebra Play
In Chapter 6, Algebra Play, students learn how to play around with algebra and experience both the logical side of mathematics and the fun side! This chapter teaches students to convert verbal puzzles and “Think of a Number” tricks into algebraic equations, which shows them that variables are more than just boxes to hold numbers; they are powerful ways to help see and explore the relationships between numbers.
Students can develop their ability to think mathematically by mastering some key concepts, which provide a bridge between arithmetic and more advanced disciplines such as science. NCERT Solutions for Class 8 are one way in which students develop this ability because they present students with a clear and logical approach to breaking down the more complex methods they encounter in order to complete different algebraic modelling examples.
Access detailed and easy-to-understand NCERT Class 8 Maths Ganith Prakash 2 Chapter 6 Solutions. Download the free PDF to practice important questions, understand concepts clearly, and prepare effectively for exams.
Chapter 6 : Algebra Play
2.0Key Concepts covered in Class 8 Maths : Algebra Play
This chapter focuses on applying algebraic modeling to solve recreational and investigative problems.
Modeling 'Think of a Number' Tricks: Learning to represent steps like "Double it" (2x) or "Add four" (2x+4) to show why results are predictable.
Number Pyramids and Grids: Using variables to find the relationship between the base numbers and the final sum in tiered structures.
Maximizing Products: Investigating how the arrangement of digits impacts the total value when forming numbers to multiply.
Algebraic Justification: Using algebraic expressions to prove mathematical properties and divisibility tricks that appear like "magic."
Modeling Word Scenarios: Solving logic-based stories, such as the "Genie and the Coins" puzzle, by setting up and solving equations.
3.0NCERT Solutions for Class 8 Maths Chapter 6 : All Exercises
Exercise 6.1 This exercise focuses on modeling "Think of a Number" tricks. Students translate verbal instructions into algebraic expressions to prove why certain results are always predictable, regardless of the initial starting number chosen.
Exercise 6.2 In this exercise, the number pyramids and magic grids are explored Investigating them enables students to determine which variables can represent cells and how to find general summation rules and prove that the occurrences of certain patterns exist by expanding mathematically and legislatively by means of algebraic processes.
Exercise 6.3 In the final exercise, students will study algebraic models from a recreational nature completing the last assignment and completing difficult tasks (i.e., finding the largest number that can be formed given a set of digits; determining if a number is divisible by another; using algebra to prove their solution is correct.)
4.0Detailed Class 8 Maths Chapter 6 Algebra Play - NCERT Solutions
Figure it out-01
1. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.
4
13
8
7
11
3
10
14
25
Sol. We know that,
(1) Given, bottom row:
4
13
8
a=4b=13c=8
The number in the topmost row
=a+2b+c=4+2×13+8=4+26+8=38
(2) Given, bottom row:
7
11
3
a=7b=11c=3
The number in the topmost row =7+2×11+3
=7+22+3=32
(3) Given, bottom row:
10
14
25
a=10,b=14,c=25
The number in the topmost row
=10+2×14+25=10+28+25=63
2. Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.
Sol. Let a,b,c,d,e,f,g,h,i, and j be the elements of the pyramid.
Thus, the expression for the top row is (a+3b+3c+d).
3. Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.
8
19
21
13
7
18
19
6
9
7
5
11
:---
:---
:---
:---
Recall the Virahānka-Fibonacci number sequence 1,2,3,5,… where each number is the sum of the two numbers before it.
Sol. (1) If a,b,c and d are the bottom row, then the expression of the topmost row of the pyramid is a+3b+3c+d.
Given, bottom row;
8
19
21
13
Here, a=8,b=19,c=21, and d=13
∴ The number in the topmost row =
a+3b+3c+d=8+3(19)+3(21)+13=8+57+63+13=141
Thus, the number in the topmost row is 141.
(2) Given, bottom row:
7
18
19
6
Here, a=7,b=18,c=19 and d=6
∴ The number in the topmost row =
a+3b+3c+d=7+3(18)+3(19)+6=7+54+57+6=124
Thus, the number in the topmost row is 124 .
(3) Given, bottom row:
9
7
5
11
Here, a=9,b=7,c=5, and d=11
∴ The number in the topmost row
=a+3b+3c+d=9+3(7)+3(5)+11=9+21+15+11=56
Thus, the number in the topmost row is 56 .
4. If the first three Virahānka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahānka-Fibonacci numbers?
Sol. We know that the first three VirahankaFibonacci number sequence =1,2,3
Here, the bottom row
1
2
3
Let a,b, and c be the missing numbers.
b=1+2=3c=2+3=5
and a=b+c=3+5=8
The complete pyramid is:
The numbers appear in the grid
=1,2,3,5,8
∴ The number at the top =8
Yes, 1, 2, 3, 5, 8 are Virahanka-Fibonacci numbers.
5. What can you say about the numbers in the pyramid and the number at the top in the following cases?
(i) The first four Virahānka-Fibonacci numbers are written in the bottom row of a four-row pyramid.
(ii) The first 29 Virahānka-Fibonacci numbers are written in the bottom row of a 29 -row pyramid.
Sol. (i) We know that, The first four VirahankaFibonacci numbers =1,2,3,5
Here, the bottom row
1
2
3
5
Let a,b,c,d,e, and f be the missing numbers.
d=1+2=3e=2+3=5f=3+5=8b=d+e=3+5=8c=e+f=5+8=13 and a=b+c=8+13=21
The numbers in the pyramid are 1,2, 3,5,8,13,21,…
We can say that the numbers are a Virahanka-Fibonacci sequence.
∴ The number at the top =21 Total number in bottom rowpyramid = 4
Sequence of number at the top =2×4−1=7 th
Fibonacci numbers =21
(ii) From the above solution, we get The number at the top =2× (Number of Virahāṅka-Fibonacci numbers present at the bottom) - 1
If the bottom row of an n row pyramid contains the first n Virahānka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?
Sol. When the bottom row uses the first n Virahanka-Fibonacci numbers, every number that appears anywhere in the grid will be a Virahanka-Fibonacci number.
∴ The number at the top of the pyramid =(2n−1)th Virahanka-Fibonacci number.
Figure it out-02
1. Fill the digits 1,3 , and 7 in □□×□ to make the largest product possible.
Sol. There are six ways to place three digits:
The six choices are:
3 7 ×1 and 7 3 ×1
1 7 ×3 and 7 1 ×3
1 3 ×7 and 3 1 ×7
In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.
7 □ 3 □ 1 , □ 7 □ 1 × □ 3 and 33× □ 7
It is clear that □ 7 □ 1 × □ 3 is greater than □ 7 □ 3 × □ 1 , so we only need to compare □ 7 □ 1 × □ 3 and □ 3 □ 1 × □ 7
71×3=7×10×3+1×331×7=3×10×7+1×7
Thus, the first term in both expressions is equal.
The second term shows that □ 3 □ 1 × □ 7
=217 is the largest.
2. Fill the digits 3, 5, and 9 in □□×□ to make the largest product possible.
Sol. There are six ways to place three digits: We can fill the first box with 3,5 , or 9 . For each of these choices, we have 2 ways of filling the remaining 2 digits.
The six choices are:
5
9
×
3
and
9
5
×
3
3
9
×
5
and
9
3
×
5
3
5
×
9
and
5
3
×
9
In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.
95× □ 3 , 9 □ 3 × □ 5 53× □ 9
It is clear that □ 9 □ 3 × □ 5 is greater than ____ 9 □ 5 × □ 3 , so we only need to compare □ 9 □ 3 × □ 5 and □ 5 □ 3 × □ 9
Let us expand these two:
93×5=9×10×5+3×553×9=5×10×9+3×9
Thus, the first term in both expressions is equal.
The second term shows that □ 5 □ 3 × □ 9 =477 is the largest.
Figure it out-03
1.
In the trick given above, what is the quotient when you divide by 9 ? Is there a relationship between the two numbers and the quotient?
Sol. Let ab be the two-digit number. (b>a)∴ba>ab
The difference is (10b+a)−(10a+b)=10b+a−10a−b=9b−9a=9(b−a), is divisible by 9 .
When 9(b−a) is divided by 9 , then the quotient is ( b−a ).
The quotient is equal to the difference between the digits of a two-digits number.
2.
In the trick given above, instead of finding the difference of the two 2 -digit numbers, find their sum. What will happen? For example:
We start with 31. After reversing we get 13 . Adding 31 and 13 , we get 44 .
We start with 28. After reversing we get 82 . Adding 28 and 82 , we get 110 .
We start with 12. After reversing we get 21 . Adding 12 and 21 , we get 33 .
Observe that all these numbers are divisible by 11 . Is this always true? Can we justify this claim using algebra?
Sol. 44, 110, 33 are divisible by 11 .
Yes, it is always true.
Using Algebra
Original number =10a+b
Reversed number =10b+a
Sum =10a+b+10b+a=11(a+b)
Hence, the sum is always divisible by 11 .
3. Consider any 3-digit number, say abc (100a +10b+c ). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37 . Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]
Sol. abc=100a+10b+cbca=100b+10c+acab=100c+10a+b
Sum of abc+bca+cab=111a+111b+111c=111(a+b+c)=37×3(a+b+c), is always divisible by 37 .
111=1+1+1=3, is always divisible by 3 .
For example:
Consider a number 153.
Other two numbers = 531 and 315
Sum =153+531+315=999999=37×27, which is divisible by 37 .
999=9+9+9=27, which is also divisible by 3 .
4. Consider any 3-digit number, say abc. Make it a 6 -digit number by repeating the digits, that is abcabc. Divide this number by 7 , then by 11 , and finally by 13 . What do you get? Try this with other numbers.
Figure out why it works. [Hint: Multiply 7, 11 and 13.]
Sol. Given that abc is a 3 -digit number.
abc=100a+10b+c
Make it a 6-digit number = abcabc
=100000a+10000b+1000c+100a+10b+c=100100a+10010b+1001c=1001(100a+10b+c)1001=7×11×13∴abcabc=1001(100a+10b+c), is divisible by 7,11 , and 13 .
Consider 836 which is a 3 -digit number.
Make it 6-digit number
=836836=1001×836∴836836 is divisible by 7,11 , and 13 .
This works because 1001=7×11×13 and repeating a 3 -digit number creates a multiple of 1001.
5. There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1 . Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2 . Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?
Sol. Let x be the initial number of flowers, and k be the equal number of flowers placed in each of the three shrines.
In shrine 1, the remaining flowers =2x−k
In shrine 2, the remaining flowers
=2(2x−k)−k=4x−2k−k=4x−3k
In shrine 3 , the remaining flowers
=2(4x−3k)−k=8x−6k−k=8x−7k∴8x−7k=0⇒8x=7k⇒x=7k/8
For the minimum possible number of flowers, we use the smallest positive integer k , which is k=8.
∴x=8(7×8)=7
Thus, the person started with 7 flowers and placed 8 flowers in each shrine.
So, for different multiple values of k=8 i.e. 16,24,32,…, we can get different number of flowers to start with and different number of flowers to be put in each shrine.
6. A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150 . How many horses and how many hens are on the farm?
Can you solve this without letternumbers? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]
Sol. Method-1: Using Algebra
Let x and y be the number of horses and hens, respectively.
According to the questions,
x+y=55
And, 4x+2y=150⇒2x+y=75
Subtracting (i) from (ii), we get
2x+y−x−y=75−55⇒x=20
Putting x=20 in equation (i), we get
20+y=55⇒y=55−20=35
Thus, the number of horses =20 and the number of hens =35.
Method-2: (without letter numbers)
If all 55 animals were hens
Total legs would be 55×2=110 legs
But actual legs = 150
Difference =150−110=40 legs
Each time we replace a hen with a horse.
We remove 2 legs (hen) and add 4 legs (horse).
Net increase = 2 legs
So, the extra 40 legs come from horses.
Number of horses needed =40÷2=20
Number of hens =55−20=35
7. A mother is 5 times her daughter's age. In 6 years, the mother will be 3 times her daughter's age. How old is the daughter now?
Sol. Let the present age of the daughter
= x years
and the present age of her mother
= y years
According to the question,
5(x)=y⇒5x=y
In 6 years,
3(x+6)=y+6⇒3x+18=y+6⇒3x+18−6=y⇒3x+12=y
From equations (i) and (ii), we get
3x+12=5x⇒5x−3x=12⇒2x=12⇒x=6
The present age of the daughter =6 years
8. Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, "You have twice as many cows as I do". Naina says, "That's true, but if I gave you three of my cows, we would each have the same number of cows". How many cows do Gauri and Naina have?
Sol. Let x and y be the number of cows of Gauri and Naina.
According to the question, 2x=y
Also, x+3=y−3x−y=−3−3=−6
By putting the value of y in eq. (ii)
x−2x=−6⇒−x=−6
or x=6
Putting x=6 in equation (i), we get,
y=2×6=12
Thus, Gauri and Naina have 6 and 12 cows, respectively.
9. I run a small dosa cart, and my expenses are as follows:
Rent for the dosa cart is ₹ 5000 per day.
The cost of making one dosa (including all the ingredients and fuel) is ₹ 10 .
(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹ 2000?
(ii) If my customers are willing to pay only ₹ 50 for a dosa, how many dosas should I aim to sell in a day to make a profit of ₹ 2000?
Sol. Given, rent for the dosa cart =₹5000/ day .
The total cost of making one dosa =₹10
(i) Given, Number of dosas = 100
∴ The cost of making 100 dosas =100×₹10=₹1000
Total cost price = Rent for the dosa cart + The cost of making 100 dosas
=₹5000+₹1000=₹6000
Profit = ₹ 2000
∴ Total selling price
= ₹ 6000 + ₹ 2000= ₹ 8000
The selling price of one dosa
=8000/100=₹80
(ii) Let n be the number of dosa.
Then total cost price =n×₹10+₹
5000
Total selling price =n×₹50
Profit = ₹ 2000
S.P = C.P + profit
⇒50n=10n+5000+2000⇒50n−10n=5000+2000⇒40n=7000⇒n=7000/40
n = 175
So, i should sell 175 dosa to make a profit of ₹2000.
10. Evaluate the following sequence of fractions:
31,(5+7)(1+3),(7+9+11)(1+3+5)
What do you observe? Can you explain why this happens?
[Hint: Recall what you know about the sum of the first n odd numbers.]
Sol. 31=31,
5+71+3=124=31,
7+9+111+3+5=279=31
Thus, the given sequences are equivalent fractions.
We know that the sum of the first n odd numbers is n2.
Numerators:
1=12=11+3=22=41+3+5=32=9
11. Karim was taking a nap under a tree. He had a dream about a magical lamp and a genie. He heard a voice saying, "I have come to serve you, Oh master". He woke up and to his surprise, it was a genie! Do you want to make money?", asked the genie. Karim nodded dumbly in bewilderment. The genie continued, "Do you see the banyan tree over there? All you have to do is go around it once. The money in your pocket will double".
Karim immediately started towards the tree, only to be stopped by the genie. "One moment!", said the genie.
Since I am bringing you great riches, you should share some of your gains with me.
You must give me 8 coins each time you go around the tree.
Thinking that was a trifling amount, Karim readily agreed.
He went around the tree once. Just as the genie had said, the number of coins in his pocket doubled! He gave 8 coins to the genie. He made another round. Again, the number of coins doubled. He gave 8 more coins to the genie. He went around the tree for the third time. The number of coins doubled again, but to his horror, he was left with only 8 coins, exactly the number of coins he owed the genie!
As Karim began to wonder how the genie tricked him, the genie let out a loud laugh and disappeared.
(i) How many coins did Karim initially have?
(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?
(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim's coins?
Sol. (i) Let Karim have n coins initially.
After 1st round, Number of coins left with Karim =2n−8
After 2nd round, Number of coins left with Karim
=2(2n−8)−8=4n−18−8=4n−24
After 3rd round, Number of coins left with Karim
=2(4n−24)−8=8n−48−8=8n−56
But after 3rd round, Karim will be left with no coins i.e. 8n−56=0⇒8n=56⇒n=7
(ii) Let c be the cost per round i.e. number of coins to be given to genie.
Starting with 7 coins:
After round 1 : 2(7) - c = 14-c
For Karim to make money, his ending amount (14−c) must be greater than 7 .
14−c>7⇒ 14-7>c
⇒c<7
To actually make a profit, Karim must agree to a cost of 6 coins or less per round.
(iii) Let Karim's starting number of coins =n Let the genie's cost per round =c
After 1st round, Number of coins left with Karim =2n−c
After 2nd round, Number of coins left with Karim
=2(2n−c)−c=4n−2c−c=4n−3c
After 3rd round, Number of coins left with Karim
=2(4n−3c)−c=8n−6c−c=8n−7c
For the Genie to get all coins,
8n−7c=0⇒7c=8n⇒c=78n
The genie should charge 78n coins per round, where n should be multiple of 7 .
(Such as c=8,16,24,… for the values of n=7,14,21,… )
5.0Key Features and Benefits of Class 8 Maths Ganith Prakash 2 Chapter 6
Gamified Learning Experience: By focusing on "tricks and puzzles," the solutions make algebra approachable and entertaining, reducing "math anxiety" for many students.
Emphasis on Logical Proofs: Students learn not just that a trick works, but why it works, which is the foundational skill for advanced proofs in higher education.
Critical Thinking and Invention: The chapter encourages students to invent their own puzzles, fostering creativity and a deeper ownership of mathematical concepts.
Pattern Recognition: Investigations into number grids and pyramids help students see structure in data, an essential skill for computer science and coding.
Scientific and Modeling Tools: The solutions highlight that algebra is an "indispensable tool" used across all areas of science to model real-world scenarios.
Exam and Olympiad Advantage: This investigative approach to algebra is a hallmark of CBSE competency-based questions and is highly relevant for competitive Olympiads like the IMO
