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Kerala State Syllabus 7th Standard Maths Solutions Chapter 3 Unchanging Relations
Unchanging Relations Text Book Questions and Answers
Measures and relations Textbook Page No. 36
We can draw squares of different sizes. The perimeter and area change with the length of sides. But in all the squares, the perimeter is always four times the length of side; and the area is the product of the length of side by
itself.
There are many such situations where even though the measures change, certain relations between them do not change. For example, if we take several objects made of iron, their volume and weight will be different. But if we divide the weight by the volume, we get the same number, 7.8. This is called the density of iron. Similarly if we do this for objects made of copper, weight divided by volume gives 8.9. This is the density of copper.
In such contexts, the unchanging relation between measures is denoted using letters.
For example, if we write the weight of an object made of iron as w and the volume as y, then we can write
w = 7.8v
If we take copper instead of iron, the relation will be
w = 8.9v
In general, if the weight of an object is w, its volume is y and if it is made of a material of density d, then their relationship can be written as
w = dv
Measures and numbers
Numbers were invented by Man to indicate and compare various measurements. For example, instead of saying, “a large group of people”, if we say, “a group of hundred people”, one gets a clear picture. Or instead of saying “walked some distance”, we can be more precise and say, “walked two and a half kilometers”.
Length, weight, and time are measured directly using instruments; area, volume, and density are not directly measured but are got by computations. For that, we need operations with numbers. For example, to find the volume of a rectangular block we must measure its length, width, and height and calculate the product.
In course of time, men started thinking about operations with pure numbers without linking them to measurement. For example, after discovering that to compute the perimeter of a rectangle, instead of measuring all four sides and adding, we need only measure two different sides and then double their sum they went on to discover the number principle that instead of multiplying two numbers separated by two and adding, we need only multiply their sum by two.
Much later, we started writing this using letters as
2x + 2y = 2(x + y)
Number theory Textbook Page No. 38
We have seen that general properties of operations on numbers can be stated in shorthand, using letters.
For example, the fact that
zero is added to any number gives the same number.
Can be shortened to
x + 0 = x, for any number x
Similarly, the fact that
To find the sum of two numbers, they can be added in any order can be written
x + y = y + x, for all number x and y.
Simple ideas like these which are easily recognized need not be shortened in this manner.
But look at this:
Adding to a number, one more than itself, and adding one to double the number gives the same result.
It is much more convenient to shorten the elaborate statement as
x + (x + 1) = 2x + 1, for any number x.
Here we must note an important thing. It is easy to remember such shorthand formulas, but to use them in time, we must know their meaning clearly.
Folding umbrellas are easy to carry; but if we don’t know how to open them, we are sure to get wet.
Two operations, one result
Adding one to twice a number is an arithmetical operation and the number got by doing it is the result. For example, doing this on 3 gives 7, and on 10 gives 21.
Adding one to a number and then adding the sum to the original number is another operation. For example, this operation on 4 gives 4 + (4 + 1) = 9. These two operations done on the same number give the same result. It is this fact that we write in algebra as
x + (x + 1) = 2x + 1
In this, the short form x + (x + 1) on the left side means, adding a number to one more than the number. The form 2x + 1 on the right means, doubling a number and adding one to it. The equal sign says that these two operations lead to the same result.
Similarly, we can write the operation of doubling two numbers separately and adding 2x + 2y in algebra. The algebraic form of the operations of doubling the sum of two numbers is 2 (x + y). The general principle that these two operations on the same pairs of numbers give the same result is written
2x + 2y = 2(x + y) for all numbers x, y.
Many general properties of numbers have a similar form, starting that apparently different operations giving the same results.
Arithmetic and algebra Textbook Page No. 40
Studies on numbers are generally known as arithmetic, and stating them using letters is algebra.
In arithmetic, the operations of adding 3 and 7 are written as 3 + 7. The sum, or the result of the operation, is 10. And we write,
3 + 7 = 10
combining the operation and the result.
In algebra, the operation of adding two numbers can be written x + y. How do we write the sum? We cannot find it without knowing the actual numbers added. So we can only write x + y for the sum also.
But the fact that
Any number added to itself is double the number that can be written in algebra as x + x = 2x, for all numbers x.
Note that this is not a general principle, but is the very definition of multiplication by 2.
Difference of difference
Finding the sum of three numbers is very natural. So three is actually no need to keep in mind the algebraic expression
(x + y) + z = x + (y + z)
Its only use is that it sometimes makes computation easier. For example, in calculating 29 + 37 + 63, if we can quickly see 37 + 63 = 100, then the total can be easily formed as 129. (To add in the given order may require pen and paper).
But we must be careful with subtraction.
The meaning of
(10 – 3) – 2
is, subtract 3 from 10 and then subtract 2 from the resulting 7; that is, the final result is 5.
What about this?
10 – (3 – 2)
First, subtract 2 from 3 to get 1; then subtract this 1 from 10, to get 9.
In other words, these operations give different results. But the result of both (10 – 3) – 2 and 10 – (3 + 2) is 5. We must keep in mind the general result
(x – y) – z = x – (y + z)
and it’s meaning:
instead of subtracting two numbers one after another, we need only subtract their sum.
Theory and practice Textbook Page No. 42
To do 25 + 20 – 15, we can first add and then subtract to get 45 – 15 = 30; or we can first do the subtraction and then addition to get 25 + 5 = 30.
But in 25 + 10 – 15, it is easy to see that we cannot do the subtraction first.
In operations like this with actual numbers, it is easy to see that certain operations cannot be done. But when we write them in algebra, we must also state the conditions which make them true.
That is why in writing
(x + y) – z = x + (y – z) we also write the condition y > z.
Less and more
Look at these differences:
10 – 9 = 1
10 – 8 = 2
10 – 7 = 3
10 – 6 = 4
When we subtract less, we get more, right?
How much more?
When one is subtracted, we get one more; when two are subtracted, we get two more.
In general
When less is subtracted, the result is more; whatever less is subtracted, that much more is the result.
Let’s write this in algebra. First note that if x and y are two numbers, then y subtracted from x is x – y.
Now taking another number z, the number y – z is z less than y; so x – (y – z) is z more than x – y.
Thus,
x – (y – z) = (x – y) + z
Sum and difference Textbook Page No. 44
What happens when we add the sum and difference of two numbers?
The difference is the smaller number subtracted from the larger; the sum is the small number added to the larger.
For example, taking the numbers 7 and 3, the sum is 7 + 3 and the difference is 7 – 3. Without reducing these to 10 and 4, if we write their sum, we get
(7 + 3) + (7 – 3)
Here the larger number 7 is added twice; the smaller 3 is added once and then subtracted.
So that net result is 7 + 7 = 14.
In other words, by changing the order of operations, we find
(7 + 3) + (7 – 3) = (7 + 7) + (3 – 3) = 14
It is this fact that we write as the general algebraic rule,
(x + y) + (x – y) = (x + x) + (y – y) = 2x
Different ways
Look at this problem:
The price of a book and pen together is 16 rupees. The price of the book is 10 rupees more than that of the pen. What is the price of each?
If we forget about books and pens and look at the prices as mere numbers, the problem is like this:
The sum of two numbers is 16 and their difference is 10. What are the numbers?
Double the larger number is 16 + 10 = 26; so the larger number is 13. And then the smaller number is 16 – 13 = 3. Thus the price of the book is 13 rupees and the price of the pen is 3 rupees.
There is another way. Instead of a book and pen, suppose two books are bought. Since the price of the book is 10 rupees more, we would have to give 10 more rupees; that is, 16 + 10 = 26 rupees.
This is the price of two books. So, the price of one book is 13 rupees.
Calendar math Textbook Page No. 46
Take any month’s calendar and mark four numbers making a square:
In the picture, the sum of the chosen numbers is 8 + 9 + 15 + 16 = 48. Divide this by 4 and subtract 4. We get the first number 8, right?
Why does this happen?
Answer:
If we take the first number as x, the numbers marked are
Their sum is
x + (x + 1) + (x + 7)(x + 8) = 4x + 16
We can write this as
4x + 16 = (4 × x) + (4 × 4)
= 4(x + 4)
This is adding 4 to the first number and then multiplying by 4.
So to get back the first number we need to divide by 4 and then subtract 4.
Another trick
Take any month’s calender and mark nine numbers in a square, instead of four.
Their sum in the picture above is 144. And it is 9 times the middle number 16.
Mark other numbers like this and check whether this happens everytime.
To see why this is so, take the middle number as x. We can fill in the other numbers like this.
In this, if we match up pairs like x – 8, x + 8, we can see without any computation that the sum is 9x. That is 9 times the middle number.
Measures and numbers Textbook Page No. 36
One side of a square is 3 centimeters. What is its perimeter?
What about a square of side 5 centimeters?
The perimeter of any square is four times the length of a side, right? And don’t you remember how we wrote it in short form using letters?
If we write s for the length of a side of a square and p for the perimeter, then we can write
p = 4 × s
We also know that when we write such a relation between numbers using letters, we don’t write the multiplication symbol × (and why). So, we write the relation between the length of side s and perimeter p of a square as
p = 4s
What about a rectangle, instead of a square?
If we know the lengths of two unequal sides, how do we find out the perimeter?
If we denote the length of each side as l and b and the perimeter as p, how do we write the relation between p, l, and b?
How do we write the relation between the lengths of sides of a rectangle and its area using letters?
Answer: Given a square figure with a side of 3cm.
Next, the side of 5cm square is, P = 4 x 5 = 20 sq.cm
The perimeter of a square is 4 x S.
The length of two unequal sides is called a rectangle. By using the formula, we can find the perimeter.
The formula for the perimeter of a rectangle is P = 2(l+b). This is a relationship between p, l, and b.
The area of a rectangle is A = l x b.
Number relations
Look at these sums:
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7
We add two consecutive natural numbers.
Now look at these:
(2 × 1) + 1 = 3
(2 × 2) + 1 = 5
(2 × 3) + 1 = 7
We double natural numbers and add 1.
How come we end up with the same numbers?
Let’s take a natural number and do the first operation. For example, if we start with 7, the next natural number is 8; and the sum
7 + 8 = 15
Suppose we write the 8 here as 7 + 1? We see that
7 + 7 + 1 = (2 × 7) + 1 = 15
This we can do for any natural number instead of 7.
If we take any natural number and add the next natural number, or if we double the first number and add 1, we get the same number as the result.
Can we do this only with natural numbers?
Answer: Let us consider that, the natural number is x.
Add the next natural number as , x+1 i.e., x+(x+1).
If we double the first number is 2x.
Add one to the first number i.e, 2x+1
We get the same result which this equation.
The equation is x+(x+1) = 2x+1.
Consider the x value as 1. Then the equation is,
1+(1+1) = 2(1) + 1
1+(2) = 2+1
3 =3.
Observed that, we get the same values.
For example let’s take the fraction, half. There is no meaning in saying the next fraction; but we can say, the number got by adding one to it. That is, half and one make one and a half. The half we started with, together with the one and a half we got now makes two.
On the other hand, doubling half makes one; and one added to it makes 2. That is,
\(\frac{1}{2}\) + 1\(\frac{1}{2}\) = \(\frac{1}{2}\) + (\(\frac{1}{2}\) + 1) = (2 × \(\frac{1}{2}\)) + 1
This computation is right, whatever fraction we start with. So we can extend the fact given above.
If we add to any number, one more than itself, or if we double the number and add one to it, we get the same number as the result.
This general property of numbers can be denoted in a shorter form using letters. For this, let’s write x for the number we start with. One added to it is x + 1; and the sum of these is x + (x + 1). Next, x doubled is written as 2x; and one added to this is 2x + 1. So the general property we have discovered can be written thus;
x + (x + 1) = 2x + 1, for every number x.
This mathematical shorthand for writing number-related facts using letters is called algebra.
Let’s look at a simple example. Suppose, to one number we add another and then subtract the added number. What do we have now? The first number back, right?
If we take the first number as x and the added number (which is later subtracted) as y, then we can write in algebra what happens as
(x +y) – y = x, for all numbers x, y.
Note that this is a general principle, which applies to all numbers. Certain properties hold only for specific numbers; for example both 2 + 2 and 2 x 2 give 4. But x + x = x x x is not a general principle (if we replace 3 for 2, it becomes incorrect).
Now do each of the following operations on several numbers and describe the results in a different form. Write each such relation in ordinary language. Then write it as an algebraic expression, using letters.
- Add to a number, two more than the number.
- Add one to a number and subtract two.
- From a number, subtract another, and then add twice the subtracted number.
- Add to a number the double of itself.
- Add two consecutive natural numbers and find the number, one less than this.
- Add two consecutive odd numbers and subtract the even number in their middle.
- To a number, add another and subtract the first.
- To a number, add another, and then add this sum to the first number.
- Subtract two times a number from five times the number.
- Add two times a number with three times the same number.
However, we add
What is 38 + 25 + 75?
We can add in the given order:
38 + 25 = 63
63 + 75 = 138
We can also add like this:
25 + 75 = 100
38 + 100 = 138
We don’t need pen and paper to do it the second way. Now try this:
29 + \(\frac{1}{3}\) + \(\frac{2}{3}\)
To do this easily, which two numbers would you add first?
What do we see from these two sums?
Answer:
To find the sum of three numbers, either we can find the sum of the first two and add this to the third, or we can find the sum of the last two and add this to the first. This we can state in another form
Instead of adding to one number, two numbers one after another, we need only add their sum.
We can show the order of operations using brackets. For example, we can write the first sum like this:
(38 + 25) + 75 = 38 + (25 + 75)
And the second sum like this
(29 + \(\frac{1}{3}\)) + \(\frac{2}{3}\) = 29 + (\(\frac{1}{3}\) + \(\frac{2}{3}\))
So we can write the general principle of adding three numbers using algebra:
(x + y) + z = x + (y + z), for all number x, y, z.
Now suppose we went to calculate 36 + 25 + 64.
Isn’t it easier to add 36 and 64 first?
First write 25 + 64 as 64 + 25 and then write the full sum as (36 + 64) + 25.
That is, the addition of numbers can be done in any order.
Now try to find these sums in your head:
• 49 + 125 + 75
Answer: 249
Explanation: Given the numbers 49, 125, and 75.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (49+125)+75 = 49+(125+75)
174+75 = 49+(200)
249 = 249
• 347 + 63 + 37
Answer: 447
Explanation: Given the numbers 347, 63, and 37.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (347+63)+37 = 347+(63+37)
410+37 = 347+(100)
447= 447
• 88 + 72 + 12
Answer: 172
Explanation: Given the numbers 88, 72, and 12.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (88+72)+12 = 88+(72+12)
160+12 = 88+(84)
172 = 172
• \(\frac{1}{4}\) + 1 \(\frac{3}{4}\) + 2
Answer: 4
Explanation: Given the numbers 1/4, (1)(3/4), and 2.
Now, we will find the sum value.
1/4+ 7/4+2.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the value is (1/4+7/4)+2 = 1/4+(7/4+2)
8/4+2 = 1/4+(15/4)
2+2 = 16/4
4 = 4
• 15.5 + 0.25 + 0.75
Answer: 16.5
Explanation: Given the numbers 15.5, 0.25, and 0.75.
Now, we will find the sum value.
Using the principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values is (15.5+0.25)+ 0.75 = 15.5+(0.25+0.75)
15.30+0.75 = 15.5+(1)
16.5 = 16.5
• 8.2 + 3.6 + 6.4
Answer: 18.2
Explanation: Given the numbers 8.2, 3.6, and 6.4.
Now, we will find the sum value.
The principle for these numbers is,
(x+y)+z = x+(y+z)
So, the values are (8.2+3.6)+6.4= 8.2+(3.6+6.4)
11.8+6.4 = 8.2+(10.0)
18.2 = 18.2
Addition and subtraction Textbook Page No. 40
We have seen the general principle of adding three numbers.
What if we subtract repeatedly, instead of adding?
Answer: Instead of adding, if we subtract three numbers we get the difference value.
Look at this problem:
Unni had 500 rupees with him and gave 100 rupees to Appu. Sometime later, Abu borrowed 50 rupees from him. Now, how much money does Unni have?
After the loan to Appu, he had
500 – 150 = 350 rupees
And after lending money to Abu, he had
350 – 50 = 300 rupees
We can also think in a different way. The total money he lent is
150 + 50 = 200 rupees
So what he finally has is
500 – 200 = 300 rupees
In other words, whether we do (500 – 150) – 50 or 500 – (150 + 50), we get the same number as the result. Similarly, can you do this in your head?
218 – 20 – 80
How can we state what we have seen here as a general principle?
Answer: The general principle for subtracting 3 numbers is (x-y)-z = x-(y+z), then we get an equal value (L.H.S = R.H.S).
Instead of subtracting from one number, two numbers one after another, we need only subtract the sum of these two numbers.
And in algebra?
(x – y) – z = x – (y + z), for all numbers x, y, z.
Instead of adding or subtracting two numbers in succession, suppose we add one number and subtract another.
Look at this problem.
There were 38 children when the class started. 5 came in late. Sometime later, 3 went to attend the Math Club meeting. How many are in the class now?
Let’s do this in the order of events. When 5 more joined, there were
38 + 5 = 43
And when 3 children left, there were 43 – 3 = 40
If instead, we look at the events together, we can compute like this: 5 children came in and 3 others left. So the final increase in number is only
5 – 3 = 2
In the beginning, there were 38 children.
So the total number now is
38 + 2 = 40
Thus, instead of adding one number and then subtracting another, we can subtract the second number from the first and add. For example,
(108 + 25) – 15 = 108 + (25 – 15) = 118
We should be a bit careful here. To compute like this, the number subtracted should be less than the number added. For example, look at this problem:
25 + 10 – 15
To do this, we cannot first subtract 15 from 10.
In algebra
(x + y) – z = x + (y – z) , for all numbers x, y, z with y > z.
Using these ideas, try to do these problems without pen and paper:
• (135 – 73) – 27
Answer: 35
Explanation: Given the numbers 135, 73, and 27.
Now, we will find the difference value.
The principle for this expression is,
(x-y)-z = x-(y+z)
So, the values are (135-73)-27= 13-(73+27)
62-27 = 135-(100)
35 = 35
The value is 35.
• (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\)
Answer: 35
Explanation: Given the numbers 37, (1)(1/2), and 1/2.
Now, we will find the difference value.
The value of (1)(1/2) in fraction form is 3/2
The principle for this given expression is,
(x-y)-z = x-(y+z)
So, the values are (37-3/2)-1/2= 37-(3/2+1/2)
(71-1)/2 = 37-(4/2)
70/2 = (74-4)/2
35 = 35
• (298 – 4.5) – 3.5
Answer: 290
Explanation: Given the numbers 298, 4.5, and 3.5.
Now, we will find the difference value.
The principle for this given expression is,
(x-y)-z = x-(y+z)
So, the values are (298-4.5)-3.5
293.5 – 3.5
290.
• (128 + 79) – 29
Answer: 178
Explanation: Given the numbers 128, 79, and 29.
Now, we will find the value.
The principle for a given expression is, (x+y)-z where y>z.
So, the values are (128+79)-29
207 – 29
178.
• (298 + 4.5) – 3.5
Answer: 299
Explanation: Given the numbers 298, 4.5, and 3.5.
Now, we will find the value.
The principle for this expression is, (x+y)-z where y>z.
So, the values are (298+4.5)-3.5
302.5 – 3.5
299
• (149 + 3\(\frac{1}{2}\)) – 2 \(\frac{1}{2}\)
Answer: 150
Explanation: Given the numbers 149, 3(1/2), and (2)1/2.
Now, we will find the value.
The principle for this expression is, (x+y)-z where y>z.
So, the values are (149+ 7/2)-(2)1/2
149 + 7/2 -5/2
149+2/2
149+1 = 150.
Subtracting and adding
Look at this problem.
Gopu had 110 rupees in his savings box. He took out 15 rupees to buy a pen. He got a pen for 10 rupees. He
returned 5 rupees to the box. How much is in the box now?
Let’s first compute in the order of what Gopu did. When he took out 15 rupees, the box had
110 – 15 = 95 rupees
Since he put back 5 rupees, the box now has 95 + 5 = 100 rupees
After all, this have happened, we can also think like this: he took out 15 rupees and put back 5 rupees; so the actual decrease in the box is only
15 – 5 = 10 rupees
So the box now has
110 – 10 = 100 rupees
Writing the first computation as (110 – 15) + 5 and the second as 110 – (15 – 5), what we see is that (110 – 15) + 5 = 110 – (15 – 5)
That is, instead of subtracting a number and then adding another, we need only subtract the difference of the second from the first. For example,
(29 – 17) + 7 = 29 – (17 – 7) = 19
Can we do this in all problems of subtracting and then adding? For example, can we rewrite
(29 – 7) + 17
in this manner?
So, the change of operation is written in algebra as
(x – y) + z = x – (y – z) , for all numbers x, y, z with y > z
Now use this idea to calculate these without pen and paper.
• (135 – 73) + 23
Answer: 85
Explanation: Given the numbers 135, 73, and 23.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values (135-73)+23
62 +23
85.
• (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\)
Answer: 30
Explanation: Given the numbers 38, 8(1/2), and 1/2.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values (38-17/2)+1/2
76 -17/2 +1/2
59/2+1/2
60/2
30.
• (19 – 6.5) + 5.5
Answer: 18
Explanation: Given the numbers 19, 6.5, and 5.5.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values (19-6.5)+5.5
12.5 +5.5
18
• (135 – 6.5) + 5.5
Answer: 134
Explanation: Given the numbers 135, 6.5, and 5.5.
Now, we have to find the value.
The principle for these equation is, (x-y)+z = x-(y-z).
So, after substituting the values (135-6.5)+5.5
128.5 +5.5
134.
• 135 – (35 – 18)
Answer: 118
Explanation: Given the numbers 135, 35, and 18.
Now, we have to find the value.
The principle for these equation is, x-(y-z) = (x-y)+z.
So, after substituting the values 135-(35-18)
135-(17)
118
• 4.2 – (3.2 – 2.3)
Answer: 3.3
Explanation: Given the numbers 4.2, 3.2, and 2.3.
Now, we have to find the value.
The principle for these equation is, x-(y-z) = (x-y)+z).
So, after substituting the values 4.2 -(3.2-2.3)
4.2 – 0.9
3.3
Sums and differences Textbook Page No. 44
Athulya often teases her classmates with her new discoveries. Today she had a new trick. “Think of any two numbers and give me their sum and difference; I can tell you the numbers.”
“Sum is 10 and difference is 2”, Rahim started.
“Easy! numbers are 6 and 4”, Athulya said.
“Sum 16, difference 5”, mischievous Jessy challenged. Athulya thought for a moment and said,
“Nice try! Numbers are 10 \(\frac{1}{2}\) and 5\(\frac{1}{2}\) ”
How did Athulya find the numbers?
How do we find two numbers, using their sum and difference?
Let’s take the numbers x and y. Then the sum is x + y. If we take the larger number as x, the difference is x – y. Using these, we have to find x and y.
To find x from x + y, we need only subtract y:
(x + y) – y = x
But we don’t know y.
What if we add x again?
(x + y) – y + x = x + x = 2x
Subtracting y and then adding x is the same as adding x and then subtracting y, right?
(x + y) + (x – y) = 2x
What does this mean?
Answer: Adding sum and difference gives twice the larger number.
For example, Rahim’s sum is 10 and the difference is 2. Their sum is 12. This is twice his larger number. So the larger number is 6, and his smaller number is 10 – 6 = 4.
Now let’s look at what Jessy said: sum 16, difference 5; their sum is 21. So the larger number is half of this and so it is 10 \(\frac{1}{2}\) ; the smaller is 16 – 10 \(\frac{1}{2}\) = 5 \(\frac{1}{2}\).
Do you get Athulya’s trick?
We can see another thing here. Subtract the difference from the sum:
(x + y) – (x – y) =(x+y)-x+y
= x + y – x + y = x – x + y + y = 2y
What does this mean?
Answer: Subtracting the difference from the sum gives twice the smaller number.
For example, in Rahim’s case, the sum is 10, and the difference is 2. So, double the smaller number is 10 – 2 = 8, and so the smaller number is half of this, which is 4.
The sum and difference of some pairs of numbers are given below. Can you find the numbers?
- Sum 12 difference 8
Answer: 2
Explanation: Given that, the sum is 12 and the difference is 8.
The smaller double number is 12 – 8 = 4.
The smaller number is half of this, which is 2.
- Sum 140, difference 80
Answer: 30
Explanation: Given that, the sum is 140 and the difference is 80.
The smaller double number is 140 – 80 = 60.
The smaller number is half of this, which is 30.
- Sum 23, difference 11
Answer: 6
Explanation: Given that, the sum is 23 and the difference is 11.
The smaller double number is 23 – 11 = 12.
The smaller number is half of this, which is 6.
- Sum 20, difference 5
Answer: 7.5
Explanation: Given that, the sum is 20 and the difference is 5.
The smaller double number is 20 – 5 = 15.
The smaller number is half of this, which is 7.5.
Addition and multiplication
We have seen that twice a number is added to thrice the number gives five times the number. (The last problem in the section Number relations)
What is the algebraic form of this statement?
2x + 3x = 5x, for every number x.
This we can state in another manner.
Instead of multiplying a number by 2 and 3 separately and adding, we need only multiply this number by 5.
For example,
(2 × 16) + (3 × 16) = 5 × 16 = 80
In this, suppose we take other numbers instead of 2 and 3?
Look at this problem:
In a math conference, two rooms are used for discussion. In one room there are 40 people and in the other, 35. At tea-time each is to be given 2 biscuits. How many biscuits are needed?
In the first room, we need
40 × 2 = 80
And in the second,
35 × 2 = 70
So altogether, we need
80 + 70 = 150
We can also think like this: The total number of people is
40 + 35 = 75
So the number of biscuits needed is
75 × 2 = 150
What do we see here? Instead of multiplying 2 by 40 and 35 separately and then adding, we need only multiply their sum 75 by 2.
This can be done in multiplication by fractions also. For example, if we add half of 4 and half of 6, we get 2 + 3 = 5; half of the sum 10 is also 5.
What general principle do we see in all these?
Multiplying two numbers by a number separately and adding give the same result as multiplying their sum by the number.
How do we write this in algebra?
xz + yz = (x + y) z, for all numbers x, y, z
What about subtraction?
Multiplying two numbers by a number separately and subtracting give the same result as multiplying their difference by the number.
In algebra,
xz – yz = (x – y) z, for all numbers x, y, z
Now try these problems.
• (63 × 12) + (37 × 12)
Answer: 120
Explanation: Given the expression is (63 x 12) + (37 x 12)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(63 x 12) + (37 x 12) = (63+37) x 12
100 x 12 = 120.
So, the final value is 120.
• (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
Answer: 15
Explanation: Given the expression is (15 x 3/4) + (5 x 3/4)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(15 x 3/4) + (5 x 3/4) = (15+5) x 3/4
20 x 3/4 = 5 x3 = 15
So, the final value is 15.
• (\(\frac{1}{3}\) × 20) + (\(\frac{2}{3}\) × 20)
Answer: 20
Explanation: Given the expression is (1/3 x 20) + (2/3 x 20)
The expression is in the form of xz + yz.
The algebraic form of xz + yz = (x+y)z.
Based on this, the given expression is,
(1/3 x 20) + (2/3 x 20) = (1/3+2/3) x 20
(3/3) x 20 = 1 x 20= 20
Hence, the final value is 20.
• (65 × 11) – (55 × 11)
Answer: 110
Explanation: Given the expression is (65 x 11) – (55 x 11)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(65 x 11) – (55 x 11) = (65-55) x 11
10 x 11 = 110
Therefore, the value is 110.
• (2\(\frac{1}{2}\) × 23) – (1\(\frac{1}{2}\) × 23)
Answer: 23
Explanation: Given the expression is (2(1/2) x 23) – (1(1/2) x 23)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(2(1/2) x 23) – (1(1/2) x 23) = (5/2-3/2) x 23
2/2 x 23 = 1 x 23 = 23.
Therefore, the value is 23
• (13.5 × 40) – (3.5 × 40)
Answer: 400
Explanation: Given the expression is (13.5 x 40) – (3.5 x 40)
The expression is in the form of xz – yz.
The algebraic form of xz – yz = (x-y)z.
Based on this, the given expression is,
(13.5 x 40) – (3.5 x 40) = (13.5-3.5) x 40
10 x 40 = 400
Therefore, the value is 400.
Let’s do it! Textbook Page No. 47
Question 1.
From the square below, take any 9 numbers making a square. Find the relation between their sum and the number in the middle of the square. Justify this relation using algebra.
Now try this with 25 numbers in a square.
Answer: (i) Given the square, which consists of 1 to 36 numbers.
First, take 9 numbers in given square box numbers.
The taken 9 numbers are below,
Now, we will find the sum of those 9 numbers.
Then the sum is, 9+10+11+15+16+17+21+22+23 = 144.
Now, the middle number is 16.
So, 9 times the middle number is 16 x 9 = 144 (Because the table consists of 9 numbers).
The relation between the middle number and the sum is, that the sum of 9 numbers in the given square is 9 times the middle number.
(ii) Now, try this with 25 numbers in a square box.
The 25 numbers are,
Now, we will find the sum of 25 numbers.
Then the sum is, 1+2+3+4+5+7+8+9+10+11+13+14+15+16+17+19+20+21+22+23+25+26+27+28+29= 375.
Now, the middle number is 15.
So, 25 times the middle number is 25 x 15 = 375 (Because the table consists of 25 numbers).
Therefore the relation between the middle number and the sum, that is the sum of 25 numbers in the given square box is 25 times the middle number.