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SCERT Class 7 Maths Chapter 12 Solutions Algebra
Class 7 Maths Chapter 12 Algebra Questions and Answers Kerala State Syllabus
Algebra Class 7 Questions and Answers Kerala Syllabus
Page 166
Now try these problems:
Question 1.
Take some sets of three consecutive natural numbers and add all three.
i) Check if the sum has any relation with any one of the three numbers added.
ii) Explain why this relation holds for any three consecutive natural numbers.
iii) Write this relation, first in ordinary language, and then using algebra.
Answer:
Let’s take three examples for three consecutive natural numbers and add them together:
1. 21 + 22 + 23 = 66
2. 34 + 35 + 36= 105
3. 78 + 79 + 80 = 237
i) For the consecutive natural numbers 21, 22, 23.
21 + 22 + 23 = 66 and 3 × 22 = 66
For the consecutive natural numbers 34, 35, 36.
34 + 35 + 36= 105 and 3 × 35 = 105
For the consecutive natural numbers 78, 79, 80.
78 + 79 + 80 = 237 and
3 ×79 = 237
That is, the sum of three consecutive numbers is always three times the middle number.
ii) Consider the three consecutive numbers 35, 36, and 37. These can be expressed in relation to the middle term as 36 – 1, 36, and 36 + 1.
Thus, the sum of these three numbers is:
(36 – 1) + 36 + (36 + 1) = 108
On simplifying this expression, we get (36 – 1) + 36 + (36 + 1) = 3 × 36 = 108
This is because in the addition of the three numbers, the 1 and -1 cancel each other out, resulting in a sum that is three times the middle number,
iii) In ordinary language, we can say it as:
The sum of any three consecutive natural numbers is three times the middle number.
Algebraically, if the three consecutive numbers are x – 1, x, and x + 1, the sum is:
(x – 1) + x + (x + 1) = 3x, for any natural numbers x
Question 2.
For some sets of four consecutive natural numbers, the sum of the first and the last, and the sum of the middle two, are shown separately below:
i) Explain why such sums are equal for any four consecutive natural numbers.
ii) Write this relation using algebra.
Answer:
i) Consider the four consecutive natural numbers as 4, 5, 6 and 7
The sum of the first and last numbers is:
4 + 7 = 4 + (4 + 3) = (2 × 4) + 3 = 11
The sum of the middle two numbers is:
5 + 6 = (4 + 1) + (4 + 2) = (2 × 4) + 3 = 11
When all three numbers are expressed in terms of the first number, we observe that the sum of the first and the last numbers, as well as the sum of the middle two terms, is equal to the sum of twice the first number and three.
ii) Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3
Then, the sum of the first and last numbers is:
n + (n + 3) = 2n + 3
The sum of the middle two numbers is:
(n + 1) + (n + 2) = 2n + 3
That is,
n + (n + 3) = (n + 1) + (n + 2) = 2n + 3, for any natural number n
Question 3.
For four consecutive natural numbers, is there any relation between the sum of the first two numbers and the last two numbers? Explain the reason for this relation and write the relation using algebra. What about the sum of the first and the third numbers and the sum of the second and the fourth? .
Answer:
Consider the four consecutive natural numbers as 5,6,7,8.
The sum of the first two numbers is:
5 + 6 = 5 + (5 +1) = (2 × 5) + 1 = 11
The sum of the last two numbers is:
7 + 8 = (5 + 2) + (5 + 3) = (2 × 5) + 5 = 15
Taking the difference between these two sums, that is
[(2 × 5) + 5] – [(2 × 5) + 1] = 15 – 11 = 4
When all three numbers are expressed in terms of the first number, we observe that the sum of the first two numbers equals twice the first number plus 1, and the sum of the last two numbers equals twice the first number plus 5.
The difference between these sums is a constant value of 4.
Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3 then
The sum of the first two numbers is:
n + (n + 1) = 2n + 1
The sum of the last two numbers is:
(n + 2) + (n + 3) = 2n + 5
Taking the difference between these two sums:
(2n + 5) – (2n + 1) = 4
That is, the sum of the first two numbers and the sum of the last two numbers differ by 4.This can be expressed as:
[n + (n + 1)] — [(n + 2) + (n + 3)] = (2n + 5) — (2n + 1) = 4, for any natural number n.
Now, we will examine the sums of the first and third numbers and the second and fourth numbers using the four consecutive natural numbers 5,6,7,8
The sum of the first and third numbers is:
5 + 7 = 5 + (5 + 2) = (2 × 5) + 2 = 12
The sum of the second and fourth numbers is:
6 + 8 = (5 + 1) + (5 + 3) = (2 × 5) + 4 = 14
Taking the difference between these two sums:
[(2 × 5) +4] – [(2 × 5) + 2] = 14 – 12 = 2
When all three numbers are expressed in terms of the first number, we observe that the sum of the first and third number equals twice the first number plus 2, and the sum of the second and fourth numbers equals twice the first number plus 4.
The difference between these sums is a constant value of 2.
Algebraically, if the four consecutive natural numbers are n, n + 1, n + 2, n + 3 then
The sum of the first and third numbers is:
n + (n + 2) = 2n + 2
The sum of the second and fourth numbers is:
(n + 1) + (n + 3) = 2n + 4
Taking the difference between these two sums:
(2n + 4) – (2n + 2) = 2
That is, the sum of the first and third numbers and the sum of the second and fourth numbers differ by 2. This can be expressed as:
[(n + 1) + (n + 3)] — [n + (n + 2)] = (2n + 4) — (2n + 2) = 2, for any natural number n.
Page 171
Question 1.
The bottom row of a five-storey number tower (like the three-storey and four-storey towers we have discussed) is shown below:
i) Before actually writing down the other numbers, see if you can guess how many times 10 is the topmost number. Check whether your guess is correct by filling in the other numbers
Answer:
Yes, here the topmost number is 16 times of 10.
ii) Explain using algebra that if we start with any five numbers equally apart and make a five- storey tower like this, then the topmost number is a fixed multiple of the middle number.
Answer:
Let’s start with x and add y each time to get the next four numbers. First line:
x, x + y, x + 2y, x + 3y, x + 4 y
Second line:
x + (x + y) = 2x + y .
(x + y) + (x + 2 y) = 2x + 3 y
(.x + 2y) + (x + 3 y) = 2x + 5y
(x + 3 y) + (x + 4 y) = 2x + 7 y
Third line: .
(2x + y) + (2.x + 3 y) = 4x + 4y
(2x + 3 y) + (2x + 5y) = 4x + 8y
(2x + 5 y) + (2x + 7 y) = 4x + 12 y
Fourth line:
(4x + 4 y) + (4x + 8y) = 8x + 12y
(4x + 8 y) + (4x + 12 y) = 8x + 20 y
At the Top:
(8x + 12y) + (8x + 20y) = 16x + 32y
From this we can say that starting with five numbers equally apart, we can construct a tower with the top number being 16 times the middle number of the first line.
iii) Is there any method to determine this without writing all the numbers?
Answer:
We can also find the top number just after writing the second line.
Here the numbers in the second line is:
2.x + y, 2x + 3y, 2x + 5y, 2x + 7y
The sum of the two middle numbers is
(2x + 3y) + (2x + 5y) = 4x + 8 y
And four times of this is:
4 x (4x + 8y) = (4 x 4x) + (4 x 8y) = 16x + 32y
So, the top number of a four-story tower must be four times the sum of the two middle number of the first line.
Question 2.
We can make number towers starting with any set of numbers, not necessarily equally apart. For example, look at this tower:
(i) Fill in the empty cells of the tower below:
Answer:
ii) In a tower like this, keep the 10’s where they are and change the second number of the bottom row to some number other than 1 or 2 and compute the other numbers. Is the topmost number still 50?
Answer:
Yes, here also the topmost number is 50 itself,
iii) Explain the reason for this using algebra.
Answer:
We can explain this using algebra,
Let’s consider x as the second number of the bottom row. So the tower will look like this:
First line will be:
First number : 10
Second number : x
Third number :10 – x
Fourth number: 10
Second line:
First number : 10 + x = 10 + x
Second number: 10
Third number : (10 – x) + 10 = 20 – x
Third line:
First number : (10 + x) + 10 = 20 + x
Second number : 10 + (20 – x) = 30 – x
At the top:
(20 + x) + (30 – x) = 50
From this we can find that the sum of middle numbers of the bottom line is equal to 10.
So whatever the number in the place of x the topmost number will be 50 which means 5 times of 10.
iv) Fill in the empty cells of this tower:
Answer:
Question 3.
Write the numbers from 1 to 100 in rows and columns as in the first picture below. Then draw some 9-cell squares in it, as in the second pictWe. Mark the middle number of each such square also:
Find the following for each of the squares:
i) The relation between the middle number and the sum of the numbers on its left and right.
Answer:
First Square
Middle number = 14
Left number = 13
Sum of left and right = 13 +
Relation between the middle and sum = \(\frac{28}{2}\) = 14
Second Square
Middle number = 63
Left number = 62
Sum of left and right = 62 +
Relation between the middle and sum = \(\frac{126}{2}\) = 63
Third Square
Middle number = 78
Left number = 77
Sum of left and right = 77 + 79 = 156
Relation between the middle and sum = \(\frac{156}{2}\) = 78
Let’s explain this using algebra,
Let the middle number be x and its left and right numbers be x – 1 and x + 1 respectively,
The sum of number on the left and right of the middle number = (x – 1) + (x + 1) = 2x
Thus, relation between the middle and sum = \(\frac{2x}{2}\) = x
This means the middle number is equal to the average of sum of its left and right numbers.
ii) The relation between the middle number and the sum of the numbers on its top and bottom.
Answer:
First Square
Middle number = 14
Top number = 4
Bottom number = 24
Sum of top and bottom = 4 + 24 = 28
Relation between the middle and sum = \(\frac{28}{2}\) =14
Second Square
Middle number = 63
Top number = 53
Bottom number = 73
Sum of top and bottom = 53 + 73 = 126
Relation between the middle and sum = \(\frac{126}{2}\) = 63
Third Square
Middle number = 78
Top number = 68
Bottom number = 88
Sum of top and bottom = 68 + 88 = 156
Relation between the middle and sum = \(\frac{156}{2}\) =78
Let’s explain this using algebra,
Let the middle number be x and its top and bottom numbers be x – 10 and x + 10 respectively,
The sum of number on the top and bottom of the middle number = (x – 10) + (x + 10) = 2x
Thus relation between the middle and sum = \(\frac{(x-10)+(x+10)}{2}=\frac{2 x}{2}\) = x
That is the middle number is equal to the average of sum of its top and bottom numbers.
iii) The relation between the middle number and the sums of the pairs of numbers diagonally on its top and bottom.
Answer:
First Square
Middle number =14
Sum of first pair of diagonal numbers = 3 + 25 = 28
Sum of second pair of diagonal numbers = 5 + 23 = 28
Sums of both diagonal pairs = 28 + 28 = 56
Relation between the middle and sum = \(\frac{56}{4}\) = 14
Second Square
Middle number = 63
Sum of first pair of diagonal numbers = 52 + 74 = 126
Sum of second pair of diagonal numbers = 54 + 72 = 126
Sums of both diagonal pairs = 126 + 126 = 252
Relation between the middle and sum = \(\frac{252}{4}\)= 63
Third Square
Middle number = 78
Sum of first pair of diagonal numbers = 67 +89 = 156
Sum of second pair of diagonal numbers = 69 + 87 = 156
Sums of both diagonal pairs= 156 + 156 = 312
Relation between the middle and sum = \(\frac{312}{4}\) = 78
Let’s explain this using algebra,
Let take the first number as x.
Then we can write the rest of the number as
so, the sum of first pair of diagonal numbers = x + x + 22
= 2x + 22
sum of second pair of diagonal numbers =x + 20 + x + 2
= 2x + 22
Thus, the sum of both diagonal pairs = 2x + 22 + 2x + 22
= 4x + 44
= 4 (x + 11)
Taking on fourth of the sum,that’s
\(\frac{4(x+11)}{4}\) = x + 11
That is , the middle number in equal to the one fourth of the sum of it’s diagonal pairs.
iv) The relation between the middle number and sum of all the numbers in the square Explain all this using algebra.
Answer:
First Square
Middle number =14
Sum of all numbers in the square
= 3 + 4 + 5 + 13 + 14 + 15 + 23 + 24 + 25 = 126
Relation between the middle and sum = \(\frac{56}{4}\) = 14
Second Square
Middle number = 63
Sum of all numbers in the square
= 52 + 53 + 54 + 62 + 63 + 64 + 72 + 73 + 74 = 567
Relation between the middle and sum = \(\frac{252}{4}\) = 63
Third Square
Middle number = 78
Sum of all numbers in the square
= 67 + 68 + 69 + 77 + 78 + 79 + 87 + 88 + 89
= 702
Relation between the middle and sum = \(\frac{312}{4}\) = 78
Let’s explain this using algebra,
Let take the first number as x.
Then we can write the rest of the number as x x + 1 x + 2
Thus the sum of all number in the square
= x + x + 1 + x + 2 + x + 10 + x + 11 + x + 12 + x + 20 + x + 21 + x + 22
= 9x + 99
= 9(x + 11)
Taking 1/9 th of the sum = \(\frac{9(x+11)}{9}\) = x + 11
That in the sum of all number in equal to the 9 times the middle number
Question 4.
On the calendar of any month, draw 4-cell squares at various positions.
i) Explain using algebra, why the sum of all four numbers in such a square is a multiple of 4.
ii) Explain using algebra, the relation between this sum and the smallest number in the square.
Answer:
Let’s consider each square
Sum of all four numbers in first square =3 + 4 + 10 + 11 = 28, which is a multiple of 4.
Sum of all four numbers in second square =13 + 14 + 20 + 21 = 68, which is a multiple of 4.
Sum of all four numbers in third square =23 + 24 + 30 + 31 = 108, which is a multiple of 4.
Let’s explain reason for this using algebra,
Consider x as the first number in the square.
Then what will be the other numbers?
Then we can form any four number in the square of the calendar as:
So, the sum of these four number will be:
= x + (x + 1) + (x + 7) + (x + 8)
= 4x + 16
= 4(x + 4)
That is that the sum of four numbers in a 2×2 square on a calendar is always a multiple of 4,
Let the smalles number be x the sum of all the four number in a square = 4x + 16 = 4 (x + 4)
That is by adding 4 to the smallest number and then taking four times that sum gives the total of all numbers in the square.
Page 176
Now try these problems:
Question 1.
Add a number leaving remainder 1 on division by 3, and a number leaving remainder 2 on division by 3. Explain, using algebra, why the sum of any two such numbers is divisible by 3
Answer:
Consider the number 7, which leaves a remainder 1 on division by 3. Thus, we can write 7 as:
7 = (2 × 3) + 1
Consider the number 17, which leaves a remainder 2 on division by 3. Thus, we can write 17 as:
17 = (5 × 3) + 2
Adding 7 and 17 we get
7 + 17 = 24 = 3 × 4
That is, there sum is divisible by 3.
Using algebra, let’s see why the sum of any two numbers is divisible by 3.
For that, consider the two numbers as 3n + 1 (remainder 1 when divided by 3) and 3m + 2 (remainder 2 when divided by 3), where m and n are natural numbers.
Then their sum is,
(3m + 1) + (3n + 2) = 3 m + 3n + 3 = 3(m + n + 1)
When m + n + 1 = p, then
(3m + 1) + (3n + 2) = 3p
where p is a natural number.
That is, the sum of these two numbers leaves no reminder. Thus, we can say that the sum of a number leaves add a number leaving remainder 1 on division by 3, and a number leaving remainder 2 on division by 3 division by 3 will leaves no reminder.
Question 2.
The numbers 12, 23, 34,… are got by starting with 12 and adding 11 again and again.
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The reminder will be 1. Thus, the number can be written as:
12 = (1 × 11)+ 1
23 = (2 × 11) + 1
34 = (3 × 11) + 1
ii) Algebraically we can write in the form of 1 In + 1 where n is one of the numbers 0, 1, 2, 3….
iii) For the number 100 we can write it as:
100 = (9 × 11) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10
Question 3.
The numbers 21,32, 43… are got by starting with 21 and adding 11 again and again
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers.
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The reminder will be 10. Thus, the number can be written as:
21 =(1 × 11) + 10
32 = (2 × 11)+ 10
43 = (3 × 11)+ 10
ii) Algebraically, we can write in the form of 11n + 10 where n is one of the numbers 0, 1, 2, 3….
iii) For the number 100 we can write it as:
100 = (9 × 1) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10
Page 180
Now try these problems:
Question 1.
Add any two-digit number and the number got by reversing the digits. Explain using algebra, why all such sums are multiples of 11.
Answer:
Consider three examples for any two-digit number and the number got by reversing the digits, then 23 + 32 = 55
35 + 53 = 88
47 + 74 = 121
Using algebra, let’s check this.
Take the larger two numbers as 10m + n.
On reversing the digit, we get the reversed number as 10n + m.
We can now determine the sum between two-digit number and its reverse as follows:
(10m + n) + (10n + m) = 10m + n + 10n + m
= 11m + 11n = 11(m + n)
Thus, we get the sum as a multiple of 11 where m + n is the sum of the digits of the numbers.
Question 2.
From a two-digit number, subtract the sum of the digits. Explain using algebra why all such differences are multiples of 9?
Answer:
Consider a two-digit number 53,
Where the sum of the digits = 5 + 3 = 8.
Now, subtract the sum of the digits from the two-digit number, that is: 53 – 8 = 45
Using algebra, let’s check this.
Take the two-digit number as 10a + b, where the sum of the digits is a + b.
When the two-digit number is subtracted from the sum of the digits, then
(10a + b) – (a + b) = 10a + b – a – b = 9a
That is, the difference is always a multiple of 9.
Question 3.
i) Write the algebraic form of all three-digit numbers.
ii) Take any three-digit number and the number got by reversing its digits. Subtract the smaller from the larger. Explain using algebra, why all such differences are multiples of 99.
iii) From a three-digit number, subtract the sum of the digits. Explain using algebra why all such differences are multiples of 9.
Answer:
i) Consider the three-digit number as xyz, where x is in the hundreds place, y is in the tens place, and z is in the one’s place. Thus, the value of the three-digit number can be expressed as:
100x + 10y + z
ii) Consider the three-digit number 531; on reversing the digits, we get 135.
Then subtract the smaller from the larger we get, 531 – 135 = 396
On dividing 396 ÷ 99 = 4,
so 396 is a multiple of 99.
Let’s explain this using algebra;
Take the three-digit number as:
100a + 10b + c
On reversing the digit, we can write it as:
100c + 10b + a
So, the difference between these two numbers is:
(100a + 10b + c) – (100c + 10b + a) = 100a – a + 10b – 10b + c – 100c
= 99a – 99c = 99(a – c)
The difference between any three-digit number and its reverse is always a multiple of 99 because the algebraic form of the difference is 99(a – c), which is clearly divisible by 99.
iii) Consider the three-digit number 352
Sum of digits: 3 + 5 + 2 = 10
Difference: 352 – 10 = 342
On dividing, 342 ÷ 9 = 38, so 342 is a multiple of 9.
Let’s explain this using algebra;
For that, consider the three-digit number
100a + 10b + c
The sum of the digits of this number is:
a + b + c
On subtracting the three-digit number from the sum of the digits of this number
(100a + 10b + c) – (a + b + c) = 100a + 10b + c – a – b – c
= (100a – a) + (10b – b) + (c – c)
= 99a + 9b
= 9(11a + b)
Intext Questions And Answers
Question 1.
What if we take the starting numbers as x — 2, x, x + 2 instead?
Answer:
First line is x – 2, x, x + 2
Second line:
(x — 2) + x = 2x – 2 x + (x + 2) = 2x + 2
At the top:
(2x – 2) + (2x + 2) = 4x
So, starting with three numbers equally apart, we can construct a tower with the top number being four times the middle number.
For example,
Now consider given four-story towers,
The first tower starts with the four consecutive numbers 6, 7, 8, 9.
While in second 9, 12, 15, 18, was obtained by repeatedly adding 3 to 9 from the beginning.
In both of this cases we can see that the top number is equal to four times the sum of the two middle numbers of the first line.
Question 2.
Make some more towers like this, starting with other number ?
Answer:
Let’s write this using algebra:
Start with x and add y each time to get the next three numbers.
First line:
x, x + y, x + 2y, x + 3y
Second line:
x + (x + y) = 2x + y
(x + y) + (x + 2y) = 2x + 3y
(x + 2y) + (x + 3y) = 2x + 5y
Third line:
(2x + y) + (2x + 3 y) = 4x + 4y
(2x + 3 y) + (2x + 5y) = 4x + 8y
At the top:
(4x + 4y) + (4x + 8y) = 8x + 12y
The sum of the two middle numbers is
(x + y) + (x + 2y) = 2x + 3y
And four times of this is: ‘
4 × (2x + 3y) = (4 × 2x) + (4 × 3y) = 8x + 12y
Thus we can conclude that the top number equal to the four times the sum of the two middle number of the first line.
Now let’s see how can be find the top number just after writing the second line. .
The numbers in the second line are ‘
2x + y, 2x + 3y, 2x + 5y
Let’s rewrite it like this:
2x + y
2x + 3y = (2x + y) + 2y
2x + 5y = (2x + 3y) + 2y
The three numbers are at an equal distance of 2y apart.
So here also, the top number of a three-story tower must be four times the middle number 2x + 3y.
Question 3.
In the same way, can you explain using algebra, why the sum of an even number and an odd number is an odd number?
Answer:Consider the even number as 2m and the odd number as 2n + 1, where m and n are either 0 or some natural numbers.
Then their sum is,
2m + (2n + 1) = 2m + 2n +1 = 2(m + n) + 1
When m + n = p, then
2m + (2n + 1) = 2p + 1
where p is a natural number.
Here, 2p is an even number, as it is two times a natural number. Adding 1 to an even number results in an odd number.
Therefore, we can conclude that the sum of an even number 2m and an odd number 2n + 1 is always an odd number.
Now we can explore this based on the remainders obtained when dividing by 3.
| Set of numbers |
Specialty |
Algebraic form |
| 0, 3, 6, 9,… |
Remainder 0 on division by 3 |
3n (n = 0,1,2,3 …) |
| 1, 4, 7, 10,… |
Remainder 1 on division by 3 |
3n + 1 (n = 0,1,2,3 …) |
| 2, 5, 8, 11,… |
Remainder 2 on division by 3 |
3n + 2 (n = 0,1,2,3 …) |
Class 7 Maths Chapter 12 Kerala Syllabus Algebra Questions and Answers
Question 1.
Find the algebraic expression for the numbers got by adding 10 repeatedly to 9.
Answer:
9 + 10 = 19
19+ 10 = 29
29 + 10 = 39
Thus, the numbers are 19, 29, 39….
Using algebra, we can express it as 10n + 9.
Question 2.
Take a set of five consecutive natural numbers and add all five.
i) Check if the sum has any relation with any one of the five numbers added.
ii) Explain why this relation holds for any five consecutive natural numbers.
iii) Write this relation, first in ordinary language and then using algebra.
Answer:
Let’s take an example of consecutive natural numbers and add them together:
34 + 35 + 36 + 37 + 38 = 180
i) For the consecutive natural numbers 34, 35, 36, 37, 38.
34 + 35 + 36 + 37 + 38 = 180 and 5 × 36 = 180
That is, the sum of five consecutive numbers is always five times the middle number.
ii) Consider the five consecutive numbers 34, 35, 36, 37 and 38. These can be expressed in relation to the middle term as 36 – 2, 36 – 1, 36, and 36 + 1, 36 + 2.
Thus, the sum of these five numbers is:
(36 – 2) + (36 – 1) + 36 + (36 + 1) + (36 + 2) = 180
On simplifying this expression, we get
(36 – 2) + (36 – 1) + 36 + (36 + 1) + (36 + 2) = 5 × 36 = 180
This is because, in the addition of the five numbers the 2, 1 and -1,-2 cancel each other out, resulting in a sum that is five times the middle number.
iii) In ordinary language, we can say it as:
The sum of any five consecutive natural numbers is five times the middle number. Algebraically, if the five consecutive numbers are x – 2, x — 1, x, and x + 1, x + 2 the sum is:
(x – 2) + (x – 1) + x + (x + 1) + (x + 2) = 5x, for any natural numbers x
Question 3.
The numbers 25,36, 47… are got by starting with 25 and adding 11 again and again
i) What is the remainder if any such number is divided by 11?
ii) Write the general algebraic form of all these numbers.
iii) Is 100 among these numbers? What about 1000?
Answer:
i) The remainder will be 3. Thus, the number can be written as:
25 = (2 × 11) + 3
36 = (3 × 11) + 3
47 = (4 × 11) + 3
ii) Algebraically we can write in the form of 11n + 3 where n is one of the numbers 0, 1, 2, 3….
iii) For the number 100 we can write it as: 100 = (9 × 11) + 1
For the number 1000 we can write it as:
1000 = (90 × 11) + 10
Question 4.
Consider a 5 storey pyramid start with numbers one and adding 2 repeatedly to 1 from the beginning the first line of the pyramid and so on. Explain the relations you find using algebra.
Answer:
Algebraic form of pyramid:
The middle number in the bottom row = x + 4
Top most number = 16(x + 4) = 16x + 64
That is the top most number is 16 times the middle number in the bottom row.
Question 5.
a) Three number pyramids are given here. Find the relation between the first and second pyramid and complete the third pyramid accordingly.
Answer:
b) Which will be the starting number to get 100 as the topmost number?
Answer:
Beginning with 10, the top most number be 100.
c) Find the relation between the topmost number and the first number of the bottom row. Write the algebraic form of it.
Answer:
The first number of the bottom can be taken as x.
Then the rest of the numbers became 2x, 3x, 4x respectively.
For the second row x + 2x = 3x 3x + 4x = 7x
Top most number is 3x + 7x = 10x
From this we can say that the top most number will be 10 times of the first number of the bottom row.
Class 7 Maths Chapter 12 Notes Kerala Syllabus Algebra
In this chapter, we will explore the exciting world of algebra, looking at different ways to express mathematical ideas about measurements and numbers more simply. Algebra serves as a powerful tool that allows us to represent relationships and patterns using symbols and letters, making complex ideas easier to understand.
- The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and’ one. We can represent the statements in algebraic form as follows: x + (x + 1) = 2x + 1, for any natural number x.
- For any three consecutive natural numbers, the sum of the first and the last, is equal to twice the middle number. We can represent the statements in algebraic form as follows:
(x – 1) + (x + 1) = 2x, for any natural number x.
- Even numbers are those numbers that are divisible by 2. That is, any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,… .
- Odd numbers are those that leave a remainder 1 on division by 2. That is, any odd number can be written in the form 2n + 1, where n is one of the numbers 0,1,2,3,…
- When arranging all two-digit numbers in rows and columns, we can express the general algebraic form of any two-digit number as: 10 m + n (m = 1,2, …,9; n = 0,1,2, …,9)
- When the digits of a two-digit number are reversed and the smaller number is subtracted from the larger, the resulting difference is always a multiple of 9. This difference is given by m – n, where m and n are the digits of the original number.
Throughout this chapter, we will uncover algebraic forms for various sets of consecutive numbers, enhancing our understanding of how algebra can simplify and clarify numerical relationships.
Numbers And Algebra
Let’s look at more ways to write facts about measurements and numbers in shorthand using algebra.
First, let’s examine the case of two consecutive numbers. Take 156 and 157 as an example. Let’s check if adding 156 and 157 gives the same result as adding 1 to 2 times 156.
That is,
156 + 157 = 313
(2 × 156) + 1 = 313
Here, the number 157 in the first operation is not there in the second operation. If we can write 157 as:
156 + 1 = 157
Then,
156 + 157 = 156 + (156 + 1)
That is,
156 + (156 + 1) = (156 + 156) + 1
= (2 × 156) + 1
In general, we can write it as:
The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and one.
Let’s examine how this can be expressed in algebraic terms.
Let x and y be two consecutive natural numbers, with x as the smaller number and y as the larger number. Using x we can denote the next natural number as x + 1. That is, add 1 to the x.
So, let’s get started like this:
- x is a natural number
- The next natural number is x + 1
- Adding these two gives the number x + (x + 1)
Now, let’s examine the number obtained by adding one to twice the smaller number:
- The smaller number is x
- Twice this is 2x
- 1 added to this is 2x + 1
Thus, we can express the statements above in algebraic form as:
x + (x + 1) = 2x + 1, for any natural number x
This result holds true not only for natural numbers but also for fractions.
That is, it is true for all numbers. The only thing is, instead of two consecutive numbers, we should say, a number and one more than the number.
Hence, we can say
The sum of a number and one more than it, is equal to sum of twice the number and one.
We already know that:
(x + y) + z = x + (y + z) for any numbers x,y,z .
In reverse, we can read as
x + (y + z) = (x + y) + z
Taking y as x and z as 1, we get
x + (x + 1) = (x + x) + 1
That is,
x + (x + 1) = 2x + 1
Now, let’s examine the case of three consecutive natural numbers. Take 54, 55, 56 as an example.
Let’s check if 54 + 56 and 2 × 55 give the same result.
That is,
54 + 56 = 110
2 × 55 = 110
Now we can link 54 and 56 to 55 as:
54 = 55 + 56 = 55 + 1
Thus, we can write
54 + 56 = (55 – 1) + (55 + 1)
Now evaluate the operation done in (55 – 1) + (55 + 1), one by one:
- Two 55’s are added
- 1 is added
- 1 is subtracted
That is,
54 + 56 = (55 – 1) + (55 + 1)
= (2 × 55) + 1 – 1
= 2 × 55
In general, we can write it as:
For any three consecutive natural numbers, the sum of the first and the last, is equal to twice the middle number.
Let’s examine how this can be expressed in algebraic terms.
Here, the operation is described using the middle number. Therefore, let’s represent it as x.
- Middle number is x
- The first number is 1 subtracted from x, that is x – 1
- The last number is 1 added to x, that is x + 1 So, the above result as an equation is
(x – 1) + (x + 1) = 2x
As said earlier, the operations in (x – 1) + (x + 1) means adding two x’s, then adding 1, and then subtracting 1; in effect, just adding two x’s. And this means multiplying x by 2.
(x – 1) + (x + 1) = 2x, for any natural number x
We can also state it like this: ‘
x + z = 2y for any consecutive natural numbers x, y, z.
Multiple And Remainder
We know that for any two natural numbers, one number can be expressed as the sum of a multiple of the other number plus a remainder, using division.
Consider two natural numbers, 7 and 3 On dividing 7 by 3, we can write:
7 = (3 × 2) + 1
On dividing 3 by 7, we can write:
3 = (0 × 7) + 3
Let’s consider the numbers that can be divided by 2 without a remainder.
For example,
2 = 1 × 2
4 = 2 × 2
6 = 3 × 2
These numbers are called even numbers.
Since
0 = 0 × 2
That is, zero is also an even number.
In general,
Even numbers are those numbers which are divisible by 2.
We can express this algebraically as:
Any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,…
Now let consider the numbers that leave a remainder when divided by 2.
For example,
1 = (0 × 2) + 1
3 = (1 × 2) + 1 ‘
5 = (2 × 2) + 1 .
These numbers are called odd numbers.
In general,
Odd numbers are those that leave a remainder 1 on division by 2.
We can express this algebraically as:
Any odd number can be written in the form 2n + 1, where n is one of the numbers 0, 1, 2, 3,…
Using algebra, let’s demonstrate that the sum of two even numbers is also an even number. Consider two even numbers 2m and 2n, where m and n are either 0 or some natural numbers.
Then their sum is,
2m + 2n = 2(m + n)
When m + n = p, then
2m + 2n = 2p
where p is either 0 or some natural number.
Therefore, 2p is an even number.
Using algebra, now let’s demonstrate that the sum of two odd numbers is also an even number. Consider two even numbers 2m + 1 and 2n + 1, where m and n are either 0 or some natural numbers. Then their sum is,
(2m + 1) + (2 n + 1) = 2m + 2n + 2
= 2 (m + n + 1)
When m + n + 1 = p, then
(2m + 1) + (2 n + 1) = 2 p
where p is a natural number.
Therefore, 2p is an even number.
Number Curiosities
Let’s write any three consecutive natural numbers.in a line:
3 4 5
Add the adjacent pair of numbers and write on the top:
Again add these two number and write it op the top:
Now we can create some more set of consecutive numbers and create tower like this:
Here, the number at the top is 4 times the middle number of the three we start with
using algebra we can explain the reason.
For that let’s start by taking the middle number as x.
What about the numbers on the left and right?
But we know the numbers before and after the middle number in three consecutive numbers are one less and one more.
So the bottom line will be x – 1, x, x + 1.
Now we can find the numbers on above that:
Then the number at the top will be:
Digits And Numbers
We can now arrange all two-digit numbers in rows and columns as follows:
The general algebraic expression for the first row can be written as:
10 + n (n = 0, 1,2,3 )
For the second row, it can be written as:
20 + n (n = 0, 1,2,3 )
Thus, the algebraic form of numbers in each row can be represented as:
where n represents the digit in the one’s place in each of these numbers.
If we write the digit in the ten’s place as m, then all these numbers can be written in the algebraic form 10m + n. Thus, the general algebraic form of all two-digit numbers is
10m+ n (m = 1, 2,…………..9 n = 0,1,2, …,9)
Now let’s look at a problem where we can take any two-digit number and the two-digit number got by reversing its digits, then subtract the smaller from the larger.
For example:
32 – 23 = 9
42 – 24-= 18
Using algebra, let’s check this.
Take the larger two numbers as 10m +n. On reversing the digit, we get the reversed number as lOn + m.
We can now determine the difference between a two-digit number and its reverse as follows:
(10m + n) – (10n + m) = (10m + n — 10n) – m
= (10m – 10n + n) – m
= (10m – 9 n) – m – 10m – m – 9n
= 9m -9n
= 9(m – n)
Thus, we get the difference as a multiple of 9 where m – n is the difference of the digits of the numbers.
The sum of two consecutive natural numbers is equal to the sum of twice the smaller number and one. We can represent the statements in algebraic form as follows:
x + (x + 1) = 2x + 1, for any natural number x.
- For any three consecutive natural numbers, the sum of the first and the last is equal to twice the middle number. We can represent the statements in algebraic form as follows:
(x – 1) + (x + 1) = 2x, for any natural number x.
- Even numbers are those numbers that are divisible by 2. That is, any even number can be written in the form 2n, where n is one of the numbers 0, 1, 2, 3,…
- Odd numbers are those that leave a remainder of 1 on division by 2. That is, any odd number can be written in the form 2n + 1, where n is one of the numbers 0, 1, 2, 3,…
- When arranging all two-digit numbers in rows and columns, we can express the general algebraic form of any two-digit number as:
10m + n (m = 1, 2,…,9; n = 0, 1, 2,…,9)
- When the digits of a two-digit number are reversed and the smaller number is subtracted from the larger, the resulting difference is always a multiple of 9. This difference is given by m – n, where m and n are the digits of the original number.
