Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 7 Maths Chapter 7 Solutions Shorthand Math
Class 7 Maths Chapter 7 Shorthand Math Questions and Answers Kerala State Syllabus
Shorthand Math Class 7 Questions and Answers Kerala Syllabus
Page 96
Question 1.
Write the following statements using the language of algebra.
1) Zero added to any number gives the same number.
2) Zero subtracted from any number gives the same number.
3) Any number subtracted from the same number gives zero.
4) Any number multiplied by zero gives zero.
5) Any number divided by the same number gives 1.
6) Twice a number added to the number makes three times the number.
7) Twice a number subtracted from thrice the number gives the number.
8) A number added to a number, and then the added number subtracted gives the original number.
Answer:
Let n be the number.
1) n + 0 = n
2) n – 0 = n
3) n – n = 0
4) n × 0 = 0
6) n + 2n = 3n
7) 3n – 2n = n
8) Let m be the other number, n + m – m = n
Page 98
Question 1.
Do the following problems mentally:
(i) 49 + 125 + 75
(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
(iii) 15.5 + 0.25 + 0.75
(iv) 38 + 27
(v) 136 + 64
Answer:
(i) 49 + 125 + 75 = 49 + 100 + 25 + 75
= 49 + 100+ 100
= 49 + 200
= 249
(ii) 3\(\frac{1}{2}\) + 8\(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{7}{2}+\frac{36}{4}\)
= \(\frac{7}{2}+\frac{18}{2}\)
= \(\frac{25}{2}\)
= 12\(\frac{1}{2}\)
(iii) 15.5 + 0.25 + 0.75 = 15.5 + 1.0
= 16.5
(iv) 38 + 27 = 38 + 2 + 25
= 40 + 25
= 65
(v) 136 + 64 = 136 + 4 + 60
= 140 + 60
= 200
Question 2.
Do the following proble mentally:
(i) (135 – 73) – 27
(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\)
(iii) (298 – 4.5) – 3.5
(iv) 78 – 29
(v) 140 – 51
Answer:
(i) (135 – 73) – 27 = 135 – (73 + 27)
= 135 – 100
= 35
(ii) (37 – 1\(\frac{1}{2}\)) – \(\frac{1}{2}\) = (37 – 1.5) – 0.5
= 37 – (1.5 + 0.5)
= 37 – 2
= 35
(iii) (298 – 4.5) – 3.5
= 298 – (4.5 + 3.5)
= 298 – 8
= 290
(iv) 78 – 29 = 78 – (30 – 1)
= 78 – 30 + 1
= 48 + 1
= 49
(v) 140 – 51 = 140 – (50 + 1)
= 140 – 50 – 1
= 90 – 1
= 89
Page 99
Do the following problems mentally:
(i) (136 + 29) – 19
(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{4}\)
(iii) (298 + 14.5) – 12.5
(iv) 23 + (35 – 18)
(v) 65 + 98
Answer:
(i) (136 + 29) – 19 = 136 + (29 – 19)
= 136 + 10
= 146
(ii) (3\(\frac{1}{2}\) + 5\(\frac{3}{4}\)) – 2\(\frac{1}{2}\) = (3.5 + 5.75) – 2.25
= 3.5+ (5.75 – 2.25)
= 3.5 + 3.5
= 7
(iii) (298 + 14.5) – 12.5 = 298 + (14.5 – 12.5)
= 298 + 2
= 300
(iv) 23 + (35 – 18) = (23 + 35) – 18
= 58 – 18
= 40
(v) 65 + 98 = 65 + (100 – 2)
= (65 + 100) – 2
= 165 – 2
= 163
Page 101
Do the following problems mentally:
(i) (135 – 73) + 23
(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\)
(iii) (19 – 6.5) + 2.5
(iv) 135 – (35 – 18)
(v) 240 – (40 – 13)
Answer:
(i) (135 -73) + 23 = 135 -(73 – 23)
= 135 – 50
= 85
(ii) (38 – 8\(\frac{1}{2}\)) + \(\frac{1}{2}\) = (38 – 8.5) + 0.5
= 38 – (8.5 – 0.5)
= 38 – 8
= 30
(iii) (19 – 6.5) + 2.5 = 19 – (6.5 – 2.5)
= 19 – 4
= 15
(iv) 135 – (35 – 18) = (135 – 35) + 18
= 100 + 18
= 118
(v) 240 – (40 – 13) = (240 – 40) + 13
= 200 + 13
= 213
Page 102
Do the following problems mentally:
(i) 103 × 15
(ii) 98 × 25
(iii) (63 × 12) + (37 × 12)
(iv) (65 × 11) – (55 × 11)
(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
Answer:
(i) 103 × 15 = (100+ 3) × 15
= (100 × 15) +(3 × 15)
= 1500 + 45
=1545
(ii) 98 × 25 = (100 – 2) × 25
= (100 × 25)-(2 × 25)
= 2500 – 50
= 2450
(iii) (63 × 12) + (37 × 12)
= (63 + 37) × 12
= 100 × 12
= 1200
(iv) (65 × 11) – (55 × 11)
= (65 – 55) × 11
= 10 × 11
= 110
(v) (15 × \(\frac{3}{4}\)) + (5 × \(\frac{3}{4}\))
= \(\frac{3}{4}\) × (15 + 5)
= \(\frac{3}{2}\) × 15
= \(\frac{45}{2}\)
(vi) (5\(\frac{1}{2}\) × 23) – (4\(\frac{1}{2}\) × 23)
= 23 × (5\(\frac{1}{2}\) + 4\(\frac{1}{2}\))
= 23 × (5.5 + 4.5)
= 23 × 10
= 230
Class 7 Maths Chapter 7 Kerala Syllabus Shorthand Math Questions and Answers
Question 1.
Write the following statements using the language of algebra.
i. When 15 is added to a number, then it is equal to three times that number.
ii. If 25 is added to three times a number, the result is 70.
iii. One third of a number when added to 1 gives 15
Answer:
i. n + 15 = 3n
ii. 3n + 25 = 70
iii. 1 + \(\frac{n}{3}\) = 15
Question 2.
Find 24 + 16 + 34
Answer:
24 + 16 + 34 = 24 + (16 + 34)
= 24 + 50
= 74
Question 3.
Find 79 – 52 – 18
Answer:
79 – 52 – 18 = 79 – (52 + 18)
= 79 – 70
= 9
Question 4.
Find 3 × 13 + 3 × 7
Answer:
3 × 13 + 3 × 7 = 3(13 + 7)
= 3 × 20
= 60
Question 5.
Find 7 × 48
Answer:
7 × 48 = 7(50 – 2)
= 350 – 14
= 336
Class 7 Maths Chapter 7 Notes Kerala Syllabus Shorthand Math
Algebra is one of the main branches of mathematics. In algebra, we make use of letters to solve mathematics. This chapter introduces some basic concepts of algebra. Following are the main topics discussed in this chapter.
Numbers and letters
- The method of stating relations between measures or numbers using letters is called algebra.
- In algebra, we write products without multiplication signs.
- In products with numbers and letters, we write the number first in algebra
- We can use any letter to denote the number or the measure in algebra
- We write division as a fraction in algebra
- We need to use more than one letter when we talk about several numbers in algebra
One by one and altogether
- While adding three numbers in a row, we can use the following result.
(x + y) + z = x + (y + z), for any three numbers x, y, z. - While subtracting three numbers in a row, we can use the following result.
(x – y) – z — x – (y + z) for any three numbers x, y, z.
Addition and Subtraction
Adding a larger number to the given number and then subtracting a smaller number gives the same result as adding the difference between, the larger number and the smaller number to the given number. That is,
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z
Subtracting a larger number from the given number and then adding a smaller number gives the same result as subtracting the difference between the larger number and the smaller number from the given number. That is,
(x – y) + z = x – (y – z), for any x, y,-z with y > z
We must use brackets to show the order of operations clearly.
Addition and subtraction and then multiplication
- Multiplying a sum by the given number gives the same result as multiplying each number in the
sum separately by the given number and then adding them. That is,
(x + y)z = xz + yz, for any three numbers x, y and z - Multiplying a difference by a number gives the same result as multiplying each number in the difference and then subtracting them. That is,
(x – y)z = xz – yz, for any three numbers x, y, z
Numbers And Letters
The method of stating relations between measures or numbers using letters is called algebra. Eg:
Consider the following fact: “A number added to itself is twice the number.”
In the language of math, we write it as follows; a number + the same number = 2 × number Let’s denote the number by n.
Then, n + n = 2n for any number n, which is the algebraic representation of the given fact.
Some features of algebra:
- Write products without multiplication sign.
Eg:
3 × n can be written as 3n. - In products with numbers and letters, write the number first.
Eg;
5 × m is written as 5m, not m5. - We can use any letter to denote the number or the measure.
Eg:
Consider the fact; “A number added to itself is twice the number.”
- If we denote the number as n, the algebraic form is n + n = 2n.
- If we denote the number as x, the algebraic form is x + x = 2x.
We write division as a fraction.
Eg:
Consider the fact: “Five times a number divided by five gives that number.”
If we denote the number as n, the algebraic form is \(\frac{5n}{5}\) = n.
We need to use more than one letter when we talk about several numbers.
Eg:
Consider the fact: If we add a number to another number and then subtract the original number, we get the added number.
Let n be the original number and m be the added number. Then the fact becomes; n + m – n = m
The letters that we use to represent numbers or measures in algebra are generally know as variables
One By One And Altogether
Adding two numbers one after another to a number, or adding their sum, gives the same result. Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.
Eg:
Consider the numbers 1, 2 and 3.
(1 + 2) + 3 = 3 + 3 = 6
1 + (2 + 3) = 1 + 5 = 6
There are situations where adding the sum of these numbers is easier than adding one after another. .
Eg:
15 + 28 + 2 =15 + 30
= 45
There are situations where adding one after another is easier than adding the sum.
Eg:
25 + 18 = 25 + 5 + 13
= 30 + 13
= 43
Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z.
Eg:
Consider the numbers 10, 7 and 2.
(10 – 7) – 2 = 3 – 2 = 1
10 – (7 + 2) = 10 – 9 = 1
There are situations where subtracting the sum is easier than subtracting the numbers one after the other.
Eg:
35 – 17 – 3 = 35 – (17 + 3)
= 35 – 20
= 15
There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
Eg:
500 – 201 = 500 – 200 – 1
= 300 – 1
= 299
Addition And Subtraction
(x + y) – z = x + (y – z), for any three numbers x, y, z with y > z
Eg:
Consider the numbers 5, 4 and 3.
(5 + 4) – 3 = 9 – 3 = 6
5 + (4 – 3) = 5 + 1 = 6
Therefore, (5 + 4) – 3 = 5 + (4 – 3)
Sometimes, it is more convenient to apply it in the reverse.
Eg:
25 + 99 = 25 + (100 – 1)
= (25 + 100) – 1
= 125 – 1
= 124
(x – y) + z = x – (y – z), for any x, y, z with y > z Eg:
Consider the numbers 10, 7 and 4.
(10 – 7) + 4 = 3 + 4 = 7
10 – (7 – 4) = 10 – 3 = 7
Therefore, (10 – 7) + 4 = 10 – (7 – 4)
We must use brackets to show the order of operations clearly.
Explanation:
Adding 7 and 4 and then adding 2 gives 13.
Adding 4 and 2 first and then adding this to 7 also gives 13.
That is, (7 + 4) + 2 = 7 + (4 + 2)
So, we may write this sum as 7 + 4 + 2 without brackets.
But if we subtract 4 from 7 and then subtract 2, we get 1.
That is, (7 – 4) – 2 = 1. Whereas if we
subtract 2 from 4 and then subtract it from 7, we get 5.
That is 7 – (4 – 2) = 5.
So, if we just write 7 – 4 – 2, the answer will be different depending on which operation we do first.
So, we must use brackets to show the order of operations clearly.
Addition And Subtraction And Then Multiplication
(x + y)z = xz + yz, for any three numbers x, y and z Eg:
Consider the numbers 1, 2, 3.
(1 + 2) × 3 = 3 × 3 = 9
(1 × 3) + (2 × 3) = 3 + 6 = 9
Therefore, (1 + 2) × 3 = (1 × 3) + (2 × 3)
Reading these in reverse is also useful in some problems. That is, xz + yz = (x + y)z
Eg:
32 + 56 =(4 × 8)+ (7 × 8)
= 8 × (4 + 7)
= 8 × 11
= 88
(x – y)z = xz – yz, for any three numbers x, y, z
Eg:
Consider the numbers 7, 5 and 3. .
(7 – 5) × 3 = 2 × 3 = 6
(7 × 3) – (5 × 3) = 21 – 15 = 6
Therefore, (7 – 5) × 3 = (7 × 3) – (5 × 3)
Reading these in reverse is also useful in some problems, That is, xz – yz = (x – y)z
Eg:
(\(\frac{1}{2}\) × 35) – (\(\frac{1}{2}\) × 15) = \(\frac{1}{2}\) (35 – 15)
= \(\frac{1}{2}\) × 20
= 10
The method of stating relations between measures or numbers using letters is called algebra.
Features of algebra:
- Write products without multiplication sign.
- In products with numbers and letters, write the number first.
- We can use any letter to denote the number or the measure.
- We write division as a fraction.
- We need to use more than one letter when we talk about several numbers.
Adding two numbers one after another to a number, or adding their sum, gives the same result.
Algebraically, (x + y) + z = x + (y + z), for any three numbers x, y, z.
- There are situations where adding the sum of these numbers is easier than adding one after another.
- There are situations where adding one after another is easier than adding the sum.
Subtracting two numbers one after the other from a number or subtracting the sum of these two numbers from the first number both give the same result. Algebraically,
(x – y) – z = x – (y + z) for any three numbers x, y, z. ,
There are situations where subtracting the numbers one after the other is easier than subtracting the sum.
(x + y) – zfc x + (y – z), for any three numbers x, y, z with y > z
Sometimes, it is more convenient to apply it in the reverse. That is, x + (y – z) = (x + y) – z
- We must use brackets to show the order of operations clearly.
- (x – y) + z = x – (y – z), for any x, y, z with y > z
- (x + y)z = xz + yz, for any three numbers x, y, z
Reading these in reverse is also useful in some problems. That is,
xz + yz = (x + y)z ,
- (x – y)z = xz – yz, for any three numbers x, y, z
- Reading these in reverse is also useful in some problems. That is, xz – yz = (x – y)z