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Algebraic Expressions and Identities Test - 22

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Algebraic Expressions and Identities Test - 22
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Weekly Quiz Competition
  • Question 1
    1 / -0
    The value of $$(2.3 a^{5}b^{2})\times (1.2a^{2}b^{4})$$, when $$a = 1$$ and $$b = 0.5$$ is ________.
    Solution
    Given expression is $$(2.3a^5b^2)\times (1.2a^2b^4)$$ $$ = 2.76 \times a^{7}b^{6} $$
    Also given $$a=1, b=0.5$$
    Subtituting the values, we get 
    $$  2.76 \times 1^{7}\times (0.5)^{6} $$
    $$ = 2.76 \times (0.015625) $$
    $$=0 .043125$$
  • Question 2
    1 / -0
    Simplify: $$(l + m)^2-  4lm$$
    Solution
         $$(l+m)^2-4lm$$
    $$=l^2+2lm+m^2-4lm$$
    $$=l^2-2lm+m^2$$
    $$=(l-m)^2$$
  • Question 3
    1 / -0
    Obtain the product of: $$2, 4y, 8y^2, 16y^3$$
    Solution
    $$2, 4y, 8y^2, 16y^3$$
    $$=2\times 4y\times 8y^2\times 16y^3$$
    $$=1024y^6$$
  • Question 4
    1 / -0
    Find the errors and correct the given expression.
    $$(z + 5)^2 = z^2 + 25$$
    Solution
    LH.S=$$(z+5)^2=z^2+2(z)(5)+5^2      .....................   [(a+b)^2=a^2+2ab+b^2]$$
    $$=z^2+10z+25 \neq$$ R.H.S

    The correct option is C.
  • Question 5
    1 / -0
    Find the product of given expressions:
    $$a, 2b, 3c, 6abc$$
    Solution
    $$a, 2b, 3c, 6abc$$
    $$=a\times 2b\times 3c\times 6abc$$
    $$=36a^2b^2c^2$$
  • Question 6
    1 / -0
    Multiply
    $$(5-2x)$$ and $$ (3 + x)$$
    Solution
    $$(5 -2x)\times (3 + x)$$
    $$=5(3 + x)-2x(3 + x)$$
    $$=15+5x-6x-2x^2$$
    $$=15-x-2x^2$$
  • Question 7
    1 / -0
    Obtain the product of: $$a, a^2, a^3$$
    Solution
    Given monomials are $$a,  a^2, a^3$$
    Product of the given monomials $$=a\times a^2\times a^3$$
    We know that, $$a^m\times a^n=a^{m+n}$$
    $$\therefore \ a\times a^2\times a^3=a^{1+2+3}$$
    $$=a^6$$

    Hence, the product of the given terms is $$a^6$$.
  • Question 8
    1 / -0
    Find the product. $$\left ( \dfrac{2}{3}xy \right ) \times \left ( \dfrac{-9}{10}x^2 y^2 \right )$$
    Solution
    $$\left ( \frac{2}{3}xy \right ) \times \left ( \frac{-9}{10}x^2 y^2 \right )$$
    $$=-\frac{3}{5}x^3 y^3$$
  • Question 9
    1 / -0
    The coefficient of $$x$$ in the product $$(x-1)(1-2x)$$ is:
    Solution
    We solve the given product $$(x-1)(1-2x)$$ as follows:
    $$(x-1)(1-2x)=(x\times 1)+\left[ x\times (-2x) \right] +(-1\times 1)+\left[ (-1)\times (-2x) \right]$$
    $$ =x-2x^{ 2 }-1+2x$$
    $$=-2x^{ 2 }+3x-1$$
    From the above calculations, we observe that the coefficient of $$x^2$$ is $$-2$$ and coefficient of $$x$$ is $$3$$.
    Hence, the coefficient of $$x$$ is $$3$$.
  • Question 10
    1 / -0
    Find the product of $$\left ( -\dfrac{10}{3}pq^3 \right ) \times \left ( \dfrac{6}{5}p^3 q \right )$$
    Solution
    $$\left ( -\frac{10}{3}pq^3 \right ) \times \left ( \frac{6}{5}p^3 q \right )$$
    $$=-2pq^3\times 2p^3q$$
    $$=-4p^4q^4$$
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