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Published on: 16/06/2021
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Questions + Answers key
Take MCQ Maths Test1.
Simplify and find the degree of the expression
(4m2 + 3n)− (3m + 9n2 )− (3m2 − 6n2 ) + (5m− n)
2.
Subtract x3 − x2 + x + 3 from 3x3 − 2x2 − 7x + 6 and find the degree.
3.
Add the expressions 4x2 + 3xy + 9y2 and 2x2 − 9xy + 6y2 and find the degree.
4.
Find the degree of the following expressions.
3a3b4 −16c6 + 9b2c5 + 7
5.
Find the degree of the following expressions.
- 4xy2z3
1.
(4m2 + 3n) − (3m+ 9n2 ) − (3m2 − 6n2 ) + (5m−n)
= 4m2 + 3n − 3m − 9n2 − 3m2 + 6n2 + 5m − n
= (4m2 − 3m2 ) + (3n − n) + (−3m+ 5m) + (−9n2 + 6n2)
= m2 + 2n + 2m − 3n2
Hence, the degree of the expression is 2.
2.
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.
3.
This can be written as (4x2 + 3xy + 9y2 ) + (2x2 − 9xy + 6y2)
Let us group the like terms, thus we have
(4x2 + 2x2 )+ (3xy − 9xy)+ (9y2 + 6y2 ) = x2 (4 + 2)+ xy (3 − 9) + y2 (9 + 6)
= 6x2 − 6xy +15y2
Thus, the degree of the expression is 2.
4.
The terms of the given expression are 3a3b4, −16c6, 9b2c5, 7
Degree of each of the terms: 7, 6, 7, 0
Terms with highest degree: 3a3b4, 9b2c5
Therefore, degree of the expression is 7.
5.
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.
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Tamilnadu Stateboard 7th Standard Subjects
Tamilnadu Stateboard Standards
