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Published on: 18/09/2019
Algebraic Expressions and Identities
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Questions + Answers key
Take MCQ Mathematics Test
1.
Using identities, evaluate (99)2
2.
Using identities, evaluate (102)2
3.
Simplify (a + 7)(a2 + 3a + 5)
4.
Multiply (a2 + 2b2) and (5a - 3b)
5.
Multiply (-3x) and (-5y + 2)
6.
Find the product of 4xy, 5x2y2 and 633y3
7.
Find the product of 2x, Sy and 7z
8.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively (3mn,4np)
9.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively (4x,3x2)
10.
Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively (20x2,5y2)
11.
Add the following:
ab - bc, bc - ca, ca - ab
12.
Identify the terms, their coefficients for each of the following expressions,
\(\frac { X }{ 2 } +\frac { Y }{ 2 } -XY\)
13.
Identify the terms, their coefficients for each of the following expressions,
3- pq+ qr- rp
14.
Subtract 13a -18ab+ 3b+16 from 22a+ 30ab+ 6b+ 8.
15.
Show on the number line z - 3.
1.
9801
2.
10404
3.
a3 + 10a2 + 26a + 35
4.
(a2 + 2b2) x (5a – 3b) = a2 (5a – 3b) + 2b2 x (5a – 3b)
= 5a3 – 3a2b + 10ab2 – 6b3
5.
15xy - 6x
6.
120 x6y6
7.
70 xyz
8.
We have,length = 3mn and breadth = 4np
∴Area of rectangle = Length x Breadth
=(3mn) x (4np) = (3 x 4)ｘ(mn) x (np)
= 12 x m x (n x n)ｘp = 12mn2p
9.
We have,4x and breadth = 3x2
∴Area of rectangle = LengthｘBreadth
= (4x) x (3x2) = (4 x 3)ｘ(x ｘ x2)
= 12 x (x3 ) = 12 x3
10.
We have,length = 20x2 and breadth = 5y2
∴ Area of rectangle = LengthｘBreadth
= (20x2) x (5y2) = (20 x 5) x (x2 x y2) =100x2y2
11.
We have,ab - bc, bc - ca, ca - ab
On writing the given expressions in separate rows with like terms one below the other, we get
ab - bc
+bc - ca
-ab +ca
_____________
0 +0 +0
_____________
Thus, the sum of the expression is 0.
12.
We have\(\frac { X }{ 2 } +\frac { Y }{ 2 } -XY\)
| Terms | \(\frac { X }{ 2 } \) | \(\frac { Y }{ 2 } \) | -XY |
| Coefficients | \(\frac { 1 }{ 2 } \) | \(\frac { 1 }{ 2 } \) | -1 |
13.
We have, 3 - pq + qr - rp
| Terms | 3 | -pq | qr | -rp |
| Coefficients | 3 | -1 | 1 | -1 |
14.
22a + 30ab + 6b + 8
13a -18ab + 3b + 16
(-) (+) (-) (-)
__________________
9a + 48ab + 3b - 8
___________________
15.
We have, (z - 3)
For given expression (z - 3), the variable z has a position Z on the number line. Z may be anywhere on the number line but the value of(z - 3) is given by position P, 3 units to the left of Z.
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CBSE 8th Standard CBSE Subjects
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