Kerala Syllabus Class 7 Mathematics - Unit 12 Algebra - Questions and Answers | Teaching Manual 

Questions and Answers for Class 7 Mathematics - Unit 12 Algebra - Study Notes | Text Books Solution STD 7 - Maths: Unit 12 Algebra - Questions and Answers | ഗണിതം - ബീജഗണിതം - ചോദ്യോത്തരങ്ങൾ 

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Kerala Syllabus STD 7 Maths: Unit 12 Algebra - Textbook Solutions

♦ Textbook Activities (Textbook Page No: 166)
Numbers and algebra

♦ Now try these problems:
(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.
Answer:
Take the consecutive natural numbers 5, 6 and 7.
Their sum will be: 
5 + 6 + 7 = 18 = 3 x 6 = 3 x middle number
Also consider the consecutive natural numbers 27, 28 and 29
Their sum will be
27 + 28 + 29 = 84 = 3 x 28 = 3 x middle number

(ii) Explain why this relation holds for any three consecutive natural numbers.
Answer:
Any three consecutive natural numbers will have the first number 1 less than the middle number and the last number 1 more than the middle number. Therefore, adding the three numbers is equivalent to adding the middle number three times.
For example, if we take the numbers 27, 28 and 29:
27 = 28 −1
29 = 28 + 1
27 + 28 + 29 = (28 − 1) + 28 + (28 + 1)
                    = (28 + 28 + 28) + 1 − 1
                    = 28 x 3 = 3 x 28
                    = 3 x middle number

(iii) Write this relation, first in ordinary language, and then using algebra.
Answer:
The sum of three consecutive natural numbers is three times the middle number.
Let's write it using algebra:
Let the middle number be x. Then the first number will be 1 subtracted from x, that is, x - 1, and the last number will be 1 added to x, that is, x + 1.
∴ (x − 1) + x + (x + 1) = (x + x + x) + 1 − 1 = 3x
We can also write it as follows:
If x, y and z are three consecutive natural numbers, then x + y + z = 3y.

(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:
1 + 4 = 5        2 + 3 = 5 
2 + 5 = 7        3 + 4 = 7
3 + 6 = 9        4 + 5 = 9
4 + 7 = 11      5 + 6 = 11
 
(i) Explain why such sums are equal for any four consecutive natural numbers
Answer:
Let's take four consecutive natural numbers.
For example, 27, 28, 29, and 30.
27 + 30 = 57 ; 28 + 29 = 57
In these consecutive natural numbers, the first number is 1 subtracted from the second number.
27 = 28 − 1
And the fourth number is 1 added to the third number.
30 = 29 + 1
Therefore, the sum of the first and the last number will be equal to the sum of the numbers in the middle.
27 + 30 = (28 − 1) + (29 + 1) = 28 + 29 + 1 − 1 
ie, 27 + 30 = 28 + 29

(ii) Write this relation using algebra.
Answer:
Take x, x + 1, x + 2 and x + 3 as the four consecutive natural numbers.
Sum of the first and last numbers = x + (x + 3) = 2x + 3
Sum of the second and the third numbers = (x + 1) + (x + 2) = 2x + 3
ie, x +(x + 3) = 2x + 3
(x + 1) + (x + 2) = 2x + 3
This can also be written like this:
If x, y, z and w are four consecutive natural numbers, then x + w = y + z.

(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:
Let's take four consecutive natural numbers.
For example, 4, 5, 6 and 7.
Sum of the first two numbers = 4 + 5 = 9
Sum of the last two numbers = 6 + 7 = 13
In these consecutive numbers, the third number is 2 added to the first number.
4 + 2 = 6
The fourth number is 2 added to the second number.
5 + 2 = 7
Therefore, the sum of the third and the fourth numbers will be 4 more than the sum of the first and the second numbers.
4 + 2 = 6
5 + 2 = 7
4 + 5 + 2 + 2 = 6 + 7
ie, (4 + 5) + 4 = 6 + 7
Algebraic form of the relation
Take the four consecutive natural numbers as x, x + 1, x + 2 and x + 3.
Sum of the first and the second number = x + (x + 1) = 2x + 1
Sum of the third and the fourth number = (x + 2) + (x + 3) = 2x + 5 = (2x + 1) + 4
The second sum is 4 more than the first sum.
This can also be written as:
Let x, y, z and w be four consecutive natural numbers.
Then x + y + 4 = z + w
Now, among four consecutive natural numbers, let's examine the relation between the sum of the first and third numbers and the sum of the second and the fourth number.
For example, let's take the numbers 4, 5, 6 and 7.
Here,
Sum of the first and third numbers = 4 + 6 = 10
Sum of the second and fourth numbers = 5 + 7 = 12
In these four numbers, the second number is 1 added to the first number.
4 + 1 = 5
The fourth number is 1 added to the third number.
6 + 1 = 7
Therefore, the sum of the second and the fourthnumber is 2 more than the sum of the first and the third number.
4 + 1 = 5;   6 + 1 = 7
4 + 6 + 1 + 1 = 5 + 7
ie, (4 + 6) + 2 = 5 + 7
Algebraic form:
Take the four consecutive natural numbers as x, x + 1, x + 2 and x + 3
Sum of the first and third number = x + (x + 2) = 2x +2
Sum of the second and fourth number = (x + 1) + (x + 3) = 2x + 4 = (2x + 2) + 2
The second sum is 2 more than the first sum.
We can also write  this as:
Let x, y, z and w be four consecutive natural numbers, then x + y + 2 = y + w.
♦ Textbook Activities (Textbook Page No: 171, 172, 173)
Number curiosities

♦ Now try these problems:
(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:
The topmost number is 16 times the middle number, 10 in the bottom row.
(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:
Take x as the middle number in the bottom row.
The numbers at the left side of x are (x − 2) and (x − 4), and the numbers at the right side of x are (x + 2) and x + 4.
Numbers on the bottom row (first row) in the tower are as follows.
Using these numbers, we can find the numbers in the next row.
(x − 2) + (x − 4) = 2x − 6
(x - 2) + x = 2x − 2
x + (x + 2) = 2x + 2
(x + 2) + (x + 4) = 2x + 6
Numbers in the second row from the bottom
Using these numbers, we can find the numbers in the next row.
(2x − 6) + (2x − 2) = 4x − 8
(2x − 2) + (2x + 2) = 4x
2x + 2 + 2x + 6 = 4x + 8
Numbers in the third row
Using these numbers, we can find the numbers in the next row.
(4x − 8) + 4x = 8x − 8
4x + (4x + 8) = 8x + 8
Numbers in the fourth row
8x − 8 8x + 8
(8x − 8) + (8x + 8) = 16 x
Numbers in the fifth row
   16x    
This number is 16 times the middle number in the bottom row

(iii) Is there any method to determine this without writing all the numbers?
Answer:
The numbers in the third row are 4 times the middle three numbers in the bottom row. The top number is 4 times the middle number in the third row.
For example, take 1, 3, 5, 7 and 9
(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:
(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:
(iii) Explain the reason for this using algebra
Answer:
Numbers in the first row
  x     y   x − y   x  
Numbers in the second row
x + y   x   2x − y
Numbers in the third row
2x − y 3x + y
Numbers in the fourth row
  5x   
Here we take x = 10
Then the number in the topmost is 5 X 10 = 50

(iv)  Fill in the empty cells of this tower:
(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 picture. 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:
In each square, the sum of the numbers at the left and the right is twice the middle number.

(ii) The relation between the middle number and the sum of the numbers on its top and bottom. 
Answer:
In each square, the sum of the numbers at the top and the bottom is twice the middle number.

(iii) The relation between the middle number and the sums of the pairs of numbers diagonally on its top and bottom.
Answer:
In each square, the sum of the numbers at the opposite corners is twice the middle number.

(iv) The relation between the middle number and sum of all the numbers in the square
Answer:
In each square, the sum of all the numbers is nine times the middle number.
Explain all this using algebra
If we take the middle number as x.
x − 11  x − 10 x − 9
x − 1      x   x + 1
x + 9  x  + 10 x + 11
i) The sum of the numbers on the left and rights side of the middle number = x − 1 + x + 1 = 2x
ii) The sum of the numbers at the top and bottom of the middle number = x − 10 + x + 10 = 2x
iii) The sum of the numbers at the opposite corners of the middle number = x − 9 + x + 9 = 2x
iv) The sum of all the numbers in each square = (x − 11) + (x − 10) + (x − 9) + (x − 1) + x + (x + 1) + (x + 9) + (x + 10) + (x + 11) = 9x
 
(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.
Answer:
Using algebra, we have
   x  x + 1
x + 7  x + 9
x + x + 1 + x + 7 + x + 8 = 4x + 16 = 4(x + 4)
The sum of all four numbers is a multiple of 4.

(ii) Explain using algebra, the relation between this sum and the smallest number in the square.
Answer:
The sum 4x + 16 is 16 added to 4 times the smallest number x in the square. 

Multiple and remainder
Any even number can be written in the form 2n, where n is one of the numbers 0,1, 2, 3,...
Any odd number can be written in the form 2n + 1, where n is one of the numbers 0, 1, 2, 3,...
 
♦ Textbook Activities (Textbook Page No: 176)
♦ Now try these problems:
(1) Add a number leaving a remainder of 1 on division by 3, and a number leaving a remainder of 2 on division by 3. Explain using algebra, why the sum of any two such numbers is divisible by 3
Answer:
Let's take 4 as the number that leaves a remainder of 1 when divided by 3, and 11 as the number which leaves a remainder of 2 when divided by 3.
Sum = 4 + 11 = 15, which can be divided by 3 without leaving a remainder.
Algebraic explanation:
Let the number that leaves a remainder of 1 when divided by 3 be 3n + 1
Let the number that leaves a remainder of 2 when divided by 3 be 3n + 2
Their sum = (3n + 1) + (3n + 2) = 3(2n + 1)
Here (2n + 1) is a natural number.
Therefore, 3(2n + 1) will always be a multiple of 3.

(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?
Answer:
When divided by 11, the remainder will be 1,

(ii) Write the general algebraic form of all these numbers
Answer:
11n + 1 (n = 1, 2, 3, ...)

(iii) Is 100 among these numbers? What about 1000?
Answer:
100 = 11 X 9 + 1. Therefore, 100 will be there.
1000 = 999 + 1
Here, the number 999 is not a multiple of 3, therefore 1000 will not be in the number sequence.

(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?
Answer:
When divided by 11, the remainder will be 10.

(ii)  Write the general algebraic form of all these numbers.
Answer:
11n + 10 (n = 1, 2, 3, ...)

(iii) Is 100 among these numbers? What about 1000?
Answer:
100 90 + 10.
Here, the number 90 is not a multiple of 11, therefore 100 will not be in the number sequence.
1000 will be in the number sequence.
♦ Textbook Activities (Textbook Page No: 180)
Digits and numbers
♦ Now try these problems:

(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:
For example: 35 + 53 = 88
Two-digit number = 10m + n
Reversing the digits gives 10n + m
Their sum = 10m + n + 10n + m = 10m + m + n + 10n
                 = 11m + 11n = 11(m + n)
This is 11 times of (m + n). That is, 11 times the sum of the digits m and n.

(2) From a two-digit number, subtract the sum of the digits. Explain using algebra why all such differences are multiples of 9.
Answer:
For example: 25 − (2 + 5) = 25 − 7 = 18
Two-digit number = 10m + n
Some of the digits = m + n
Their difference = (10m + n) − (m + n)
                         = 10m + n − m − n = 9m
Here, 9m is a multiple of 9.

(3) (i) Write the algebraic form of all three-digit numbers.
Answer:
General form of three digit numbers = 100x + 10y + z

(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.
Answer:
General form of three digit numbers = 100x + 10y + z
Digits reversed = 100z + 10y + x
Their difference = (100x + 10y + z) − (100z + 10y + x)
                         = 100x + 10y + z − 100z − 10y − x
                         = 100x − x + 10y − 10y − 100z + z
                         = (100x − x) + (10y − 10y) − (100z − z)
                         = 99x + 0 − 99z = 99x − 99z
                         = 99(x − z)
This is a multiple 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:
General form of three digit numbers = 100x + 10y + z
Sum of digits = (x + y + z)
Difference = (100x + 10y + z) − (x + y + z)
                         = 100x + 10y + z − x − y − z
                         = (100x − x) + (10y − y)+ (z − z)
                         = 99x + 9y + 0 = 99x + 9y
                         = 9(11x + y)
This is a multiple of 9.