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MATHS [4 marks sum]

Algebraic Expressions · ICSE STD 8 (ICSE - English Medium). 18 questions.

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[4 marks sum]

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18 questions · timed · auto-graded

Question 14 Marks
Simplify: $7x + 4 \{x^2 \div (5x \div 10)\} − 3 \{2 − x^3 \div (3x^2 \div x)\}$
Answer
$ 7 x+4\left\{x^2 \div(5 x \div 10)\right\}-3\left\{2-x^3 \div\left(3 x^2 \div x\right)\right\}$
$ =7 x+4\left\{x^2 \div\left(\frac{5 x}{10}\right)\right\}-3\left\{2-x^3 \div\left(\frac{3 x^2}{x}\right)\right\}$
$ =7 x+4\left\{x^2 \div \frac{x}{2}\right\}-3\left\{2-x^3 \div 3 x\right\}$
$ =7 x+4\left\{x^2 \times \frac{2}{x}\right\}-3\left\{2-\frac{x^3}{3 x}\right\}$
$ =7 x+8 x-6+x^2$
$ =x^2+15 x-6$
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Question 24 Marks
Simplify : $3 x-[4 x-\overline{3 x-5 y}-3\{2 x-(3 x-\overline{2 x-3 y})\}]$
Answer
$ 3 x-[4 x-\overline{3 x-5 y}-3\{2 x-(3 x-\overline{2 x-3 y})\}]$
$ =3 x-[4 x-3 x+5-3\{2 x-(3 x-2 x+3 y)\}]$
$ =3 x-[4 x-3 x+5 y-3\{2 x-(x+3 y)\}]$
$ =3 x-[4 x-3 x+5 y-3\{2 x-x-3 y\}]$
$ =3 x-[x+5 y-6 x+3 x+9 y]$
$ =3 x-[-2 x+14 y]$
$ =3 x+2 x-14 y$
$ =5 x-14 y$
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Question 34 Marks
Simplify : $2[6+4\{m-6(7-\overline{n+p})+q\}]$
Answer
$ 2[6+4\{m-6(7-\overline{n+p})+q\}]$
$ =2[6+4\{m-6(7-n-p)+q\}]$
$ =2[6+4\{m-42+6 n+6 p+q\}]$
$ =2[6+4 m-168+24 n+24 p+4 q]$
$ =2[4 m+24 n+24 p+4 q-162]$
$ =8 m+48 n+48 p+8 q-324$
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Question 44 Marks
Divide $x^6 – y^6$ by the product of $x^2 + xy + y^2$ and $x – y.$
Answer
Product of $\left(x^2+x y+y^2\right)$ and $(x-y)$
$ =(x-y)\left(x^2+x y+y^2\right) $
$ =x\left(x^2+x y+y^2\right)-y\left(x^2+x y+y^2\right) $
$ =x^3+x^2 y+x y^2-x^2 y-x y^2-y^3 $
$ =x^3-y^3$
Now, $\left(x^6-y^6\right) \div\left(x^3-y^3\right)$
$=x^3+y^3$
$ \left.x^3-y^3\right) \overline{x^6-y^6}\left(x^3+y^3\right. $
$ x^6-x^3 y^3 $
$ -+ $
$ x^3 y^3-y^6 $
$ x^3 y^3-y^6 $
$ -\quad+ $
$ \times $
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Question 54 Marks
The area of a rectangle is $x^3 – 8x + 7$ and one of its sides is $x – 1$. Find the length of the adjacent side.
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Question 64 Marks
Find the quotient and the remainder when : $6x^2 + x − 15$ is divided by $3x +5$. verify your answer.
Answer
Image
Verification :
Dividend = Quotient $\times$ Divisor + Reminder
$= (2x − 3) (3x + 5) + 0$
$= 6x^2 + 10x − 9x − 15 + 0$
$= 6x^2 + x − 15$
which is given
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Question 104 Marks
Multiply $-4xy^3$ and $6x^2y$ and verify your result for $x = 2$ and $y= 1.$
Answer
$(−4xy^3) \times (6x^2y) = (−4 \times 6) (x\ x\ x^2) (y^3 x\ y)$
$= −24x^3y^4$
For $x = 2$ and $y = 1$
$(−4xy^3) \times (6x^2y) = (−4 \times 2 \times 1^3) \times (6 \times 2^2 \times 1)$
$= (−8) \times 24 = −192$
And, $−24x^3y^4 = −24 \times 2^3 \times 1^4$
$= −24 \times 8 \times 1 = −192$
$\therefore$ For $x = 2$ and $y = 1,$ it is verified that
$(−4xy^3) \times (6x^2y) = −24x^3y^4$
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Question 114 Marks
Simplify : $(5 – x) (6 – 5x) (2 -x).$
Answer
$(5 – x) (6 – 5x) (2 -x)$
$[(5 – x) (6 – 5x)] (2 -x)$
$[(5(6 – 5x) − x (6 − 5x)] (2 -x)$
$[30 − 25x − 6x + 5x^2] (2 − x)$
$(5x^2 − 31x + 30) (2 − x)$
$2(5x^2 − 31x + 30) − x (5x^2 − 31x + 30)$
$10x^2 − 62x + 60 − 5x^3 + 31x^2 − 30x$
$−5x^3 + 10x^2 + 31x^2 − 62x − 30x + 60$
$−5x^3+ 41x^2 − 92x + 60$
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Question 124 Marks
Multiply and then verify : $−3x^2y^2$ and $(x – 2y)$ for $x = 1$ and $y = 2.$
Answer
$(−3x^2y^2) \times (x – 2y)$
$= (−3x^2y^2) \times (x) − (−3x^2y^2)(2y)$
$= −3x^2y^2 + 6x^2y^3$
$= 6x^2y^3 − 3x^3y^2$
For $x = 1$ and $y = 2$
$(−3x^2y^2) \times (x – 2y)$
$= (−3 \times 1^2\times 2^2) \times (1 − 2 \times 2)$
$= (6 \times 1 \times 8) − (3 \times 1 \times 4)$
$= 48 − 12$
$= 36$
$\therefore$ For $x = 1$ and $y = 2,$ it is verified that,
$(−3x^2y^2) \times (x – 2y)$
$= 6x^2y^3 − 3x^3y^2$
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Question 134 Marks
Multiply: $2m^2 − 3m − 1$ and $4m^2 − m − 1$
Answer
$(2m^2 − 3m − 1) (4m^2 − m − 1)$
$= 2m^2(4m^2 −  m −  1) − 3m(4m^2 −  m − 1) −1(4m^2 − m −1)$
$= 8m^4 − 2m^3− 2m^2 − 12m^3 + 3m^2 + 3m −  4m^2+ m + 1$
$= 8m^4 − 14m^3 − 6m^2 + 3m^2 + 4m + 1$
$= 8m^4 − 14m^3 − 3m^2 + 4m + 1$ 
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Question 144 Marks
Separate monomials, binomials, trinomials and polynomials from the following algebraic expressions :$8 − 3x, xy^2, 3y^2 − 5y + 8, 9x − 3x^2 + 15x^3 − 7,3x \times 5y, 3x ÷ 5y, 2y ÷ 7 + 3x − 7$ and $4 − ax^2 + bx + y$
Answer
Monomials are : $xy^2, 3x \times 5y, 3x \div 5y;$
Bionomials are: $8 − 3x$
Trinomials are: $3y^2 − 5y + 8, 2y \div 7 + 3x − 7$​​​​​​​
Polynomials are: $8 − 3x, 3y^2 − 5y + 8, 9x − 3x^2 + 15x^3 − 7, 2y \div 7 + 3x −7, 4 − ax^2 + bx + y$
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Question 154 Marks
The product of two numbers$-$is $16x^4 – 1.$ If one number is $2x – 1,$ find the other.
Answer
Product of two numbers $=16 x^4-1$
One number $=2 x-1$
Then the second number $=\frac{16 \mathrm{x}^4-1}{2 \mathrm{x}-1}$
$=8 x^3+4 x^2+2 x+1$
Image
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Question 164 Marks
Find the quotient and the remainder when : $3x^4 + 6x^3 − 6x^2 + 2x − 7$ is divided by $x − 3.$ verify your answer.
Answer
Image
$\therefore$ Quotient $= 3x^3 + 15x^2 + 39x + 119$ and reminder $= 350$
Verification :
Dividend $=$ Quotient $\times$ Divisor $+$ Reminder
$= (3x^3 + 15x^2 + 39x + 119) (x − 3) + 350$
$= 3x^4 + 15x^3 + 39x^2 + 119x − 9x^3 − 45x^2 − 117x − 357 + 350$
$= 3x^4 + 6x^3 − 6x^2 + 2x − 7$
which is given
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Question 174 Marks
Find the quotient and the remainder when : $a^3− 5a^2 + 8a + 15$ is divided by $a + 1$. verify your answer.
Answer
Image
$\therefore $ Quotient $= a^2 − 6a + 14$ and reminder $= 1$
Verification :
Dividiend $=$ Quotient $\times $ Divisor $+$ Reminder
$= (a^2 − 6a + 14) \times (a + 1) + 1$
$= a^3 − 6a^2 + 14a + a^2 − 6a + 14 + 1$
$= a^3 − 5a^2 + 8a + 15$
which is given
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Question 184 Marks
If $x = 6a + 86 + 9c ; y = 2b – 3a – 6c$ and $z = c – b + 3a $ ; find :$(i)\ x + y + z\ (ii)\ x – y + z\ (iii) \ 2x – y – 3z\ (iv) \ 3y – 2z – 5x$
Answer
$(i) x = 6a + 8b + 9c$
$y = − 3a + 2b − 6c$
$z = +3a − b + c$
Adding $x + y + z = 6a + 9b + 4c$
$(ii) x − y + z = (6a + 8b + 9c) − (2b − 3a − 6c) + (c − b + 3a)$
$= 6a + 8b + 9c − 2b + 3a + 6c + c − b + 3a$
$= 6a + 3a + 3a + 8b − 2b − b + 9c + 6c + c$
$= 12a + 5b + 16c$
$(iii) 2x − y − 3z = 2(6a + 8b + 9c) − (2b − 3a − 6c) − 3(c −b + 3a)$
$= 12a + 16b + 18c − 2b + 3a + 6c − 3c +3b − 9a$
$= 12a + 3a − 9a + 16b + 3b − 2b + 18c + 6c −3c$
$= 6a + 17b + 21c$
$(iv) 3y − 2z − 5x = 3(2b − 3a − 6c) − 2(c − b + 3a) − 5(6a + 8b + 9c)$
$= 6b − 9a − 18c − 2c + 2b − 6a − 30a − 40b − 45c$
$= − 9a − 6a − 30a + 6b + 2b − 40b − 18c − 2c − 45c$
$= − 45a − 32b − 65c$
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[4 marks sum] - MATHS STD 8 Questions - Vidyadip