7th Standard Syllabus & Materials
7th Standard
TN 7th Tamil рокро░рпБро╡роорпН -1 роЗропро▓рпН 1 - роЕроорпБродродрпНродрооро┐ро┤рпН - роТройрпНро▒ро▓рпНро▓ роЗро░рогрпНроЯро▓рпНро▓ Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th Tamil рокро░рпБро╡роорпН -1 роЗропро▓рпН 1 - роЕроорпБродродрпНродрооро┐ро┤рпН - роОроЩрпНроХро│рпН родрооро┐ро┤рпН Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th English T1 - Poem - Your Space Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th English T1 - Prose - A Prayer to the Teacher Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th English T1 - Poem - The Listeners Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th English T1 - Prose - The Wind on Haunted Hill Important Questions And Answers Study Material - QB365 Set A
Published on: 04/11/2019
Term 2 Algebra
Download Tamil Nadu 7th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
Subtract x3 − x2 + x + 3 from 3x3 − 2x2 − 7x + 6 and find the degree.
2.
Find the degree of the following expressions.
12xyz − 3x3y2z + z8
3.
Find the degree of the following expressions.
- 4xy2z3
4.
Simplify by using the law of exponents.
613 x 4813 ÷ 1213
5.
Simplify using power rule of exponents.
(26)2 x (24)7
6.
Simplify using power rule of exponents.
(83)4
7.
Simplify using quotient rule of exponents.
\(\frac { { 6 }^{ 4 } }{ { 6 }^{ 0 } } \)
8.
Simplify using quotient rule of exponents.
\(\frac { { 2 }^{ 8 }\times { 3 }^{ 5 }\times { 5 }^{ 4 } }{ { 3 }^{ 3 }\times { 5 }^{ 3 }\times { 2 }^{ 4 } } \)
9.
Simplify using Product Rule of exponents.
25 × 32 × 625 × 64
10.
Simplify using Product Rule of exponents.
33 x 32 x 34
11.
Which is greater 34 or 43?
12.
Find the value of 23 + 32
13.
Find the value of
(−7)2
14.
Express the following numbers in exponential form with the given base:
243, base 3.
15.
Express 729 in exponential form.
1.
This can be written as (3x3 − 2x2 − 7x + 6)− (x3 − x2 + x + 3)
When there is a –ve sign before the brackets, it can be removed by changing the sign of every term inside the bracket.
(3x3 − 2x2 − 7x + 6)−(x3 − x2 + x + 3) = 3x3 − 2x2 − 7x + 6 − x3 + x2 −x−3
= (3x3 − x3 )+ (−2x2 + x2 )+ (−7x − x)+ (6 − 3)
= x3 (3 −1)+ x2 (−2 +1)+ x (−7 −1)+ (6 − 3)
= 2x3 − x2 − 8x + 3
Hence, the degree of the expression is 3.
2.
The terms of the given expression are 12xyz, 3x3 y2z, z8
Degree of each of the terms : 3, 6, 8
Terms with highest degree : z8.
Therefore, degree of the expression is 8.
3.
In −4xy2z3, the sum of powers of x, y and z is 6 (that is, 1 + 2 + 3). Thus, the degree of the expression is 6.
4.
613 x 4813 ÷ 1213 = 613 x (4813 ÷ 1213) [BIDMAS]
= 613x\(\left( \frac { 48 }{ 12 } \right) ^{ 13 }\) [since \(\frac { { a }^{ m } }{ { b }^{ m } } =\left( \frac { a }{ b } \right) ^{ m }\)]
= 613 x 413
= (6 x 4)13 [Since, am x bm = (a x b)m]
= (24)13
5.
(26)2 x (24)7 = 26 x 2 x 24 x 7 [since (am)n = amxn]
= 212 x 228
=212 + 28 = 240 [since am x an = am+n]
6.
(83)4 = 83 x 4 = 812 [since (am)n = am x n]
7.
\(\frac { { 6 }^{ 4 } }{ { 6 }^{ 0 } } \) = 64-0 = 64 (or) \(\frac { { 6 }^{ 4 } }{ { 6 }^{ 0 } } \)=\(\frac { { 6 }^{ 4 } }{ 1 } \) = 64 [since 60 = 1]
8.
\(\frac { { 2 }^{ 8 }\times { 3 }^{ 5 }\times { 5 }^{ 4 } }{ { 3 }^{ 3 }\times { 5 }^{ 3 }\times { 2 }^{ 4 } } \) = \(\frac { { 2 }^{ 8 } }{ { 2 }^{ 4 } } \times \frac { { 3 }^{ 5 } }{ { 3 }^{ 3 } } \times \frac { { 5 }^{ 4 } }{ { 5 }^{ 3 } } \) [grouping exponential numbers with the same base]
= 28-4 x 35-3 x 54-3 = 24 x 32 x 51 \(\left[ \frac { { a }^{ m } }{ { a }^{ n } } ={ a }^{ m-n } \right] \)
9.
25 x 32 x 625 x 64 = (5 x 5) x (2 x 2 x 2 x 2 x 2) x (5 x 5 x 5 x 5) x (2 x 2 x 2 x 2 x 2 x 2)
= 52 x 25 x 54 x 26
= (52 x 54 ) x (25 x 26 ) [grouping exponential numbers with the same base]
= 52+4 x 25+6 = 56 x 211
10.
33 x 32 x 34 = 33+2 x 34 = 35 x 34
= 35+4 = 39
11.
34 = 3 x 3 x 3 x 3 = 81
43 = 4 x 4 x 4 = 64
81> 64 gives 34 > 43
Therefore, 34 is greater.
12.
23 + 32 = (2 x 2 x 2) + (3 × 3)
= 8 + 9 = 17
13.
(−7)2 = (−7) x (−7) = 49
14.
243 = 3 x 3 x 3 x 3 x 3 = 35
15.
Dividing by 3, we get
729 = 3 x 3 x 3 x 3 x 3 x 3 = 36
Also, 729 = 9 × 9 × 9 = 93
7th Standard Syllabus & Materials
7th Standard
TN 7th English T1 - Supplementary - On Monday Morning Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th English T1 - Prose - Eidgah Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th Social Science T1 - ECO - Production Important Questions And Answers Study Material - QB365 Set A
NEW7th Standard
TN 7th Social Science T1 - GEO - Interior of the Earth Important Questions And Answers Study Material - QB365 Set A
Tamilnadu Stateboard 7th Standard Subjects
Tamilnadu Stateboard Standards
