Exercise 6.3 covers number reversals, divisibility patterns, cyclic numbers, and repeated three-digit numbers.
The exercise examines the difference and sum of the original number and its reverse.
Algebraic expressions are used to show that the difference is a multiple of 9.
The sum can be expressed as 11 times the sum of its digits.
They are numbers formed by cycling the digits of a three-digit number, such as abc, bca, and cab.
Algebra is used to express their sum as a multiple of both 37 and 3.
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Algebra Play Exercise 6.3 Class 8 Maths – Chapter 6 NCERT Solutions
The detailed NCERT Solutions for Class 8 Maths Chapter 6, Algebra Play, Exercise 6.3 are available here. Curated by the ALLEN expert team in line with the latest updated syllabus, these solutions cover the exercise questions with detailed working to support students throughout their practice and preparationExercise 6.3 looks at patterns with two-digit numbers, including reversing numbers and looking at their sums and differences. It also contains algebraic proofs for divisibility by 9 and 11, and discusses cyclic three-digit numbers, repeating three-digit numbers and the patterns of divisibility that they follow.
1.0Download Class 8 Maths Chapter 6 Ex 6.3 NCERT Solutions PDF
Download the detailed Class 8 Maths Chapter 6 Exercise 6.3 NCERT Solutions PDF for step-by-step explanations and convenient revision.
Chapter 6 : Algebra Play Exercise 6.3
2.0Detailed NCERT Solutions for Algebra Play Exercise 6.3 Class 8 Maths (Ganita Prakash Term 2)
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 valuesof n=7,14,21,… )
3.0Key Concepts of Chapter 6 Algebra Play Exercise 6.3
Reversing the digits of a two-digit number.
Finding the difference between a two-digit number and its reverse.
Using algebra to show divisibility by 9.
Adding a two-digit number to its reverse and showing divisibility by 11.
Cycling the digits of a three-digit number and finding the resulting sum.
Using algebra to show that the sum of cyclic three-digit numbers is divisible by 37 and 3.
Understanding the divisibility of repeated three-digit numbers by 7, 11, and 13.
4.0Benefits of NCERT Solutions Class 8 Maths Chapter 6 Exercise 6.3
Finds the difference between a two-digit number and its reverse.
Uses algebraic expressions to establish divisibility by 9.
Determines the sum of a two-digit number and its reverse.
Uses algebra to establish divisibility by 11.
Forms cyclic three-digit numbers and calculates their sum.
Establishes the divisibility of cyclic three-digit number sums by 37 and 3.
Explains why repeated three-digit numbers are divisible by 7, 11, and 13.
