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CBSE Questions for Class 8 Maths Algebraic Expressions And Identities Quiz 9 - MCQExams.com
CBSE
Class 8 Maths
Algebraic Expressions And Identities
Quiz 9
The product of $$(x - 1)(2x - 3)$$ is _____
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$$2x^{2} - 5x - 3$$
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$$2x^{2} - 5x + 3$$
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$$2x^{2} + 5x - 3$$
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$$2x^{2} + 5x + 3$$
Explanation
Expanding: $$(x-1)\times(2x-3)$$
$$=2x^{2}-2x-3x+3$$
After combining the like terms we get:
$$=2x^{2}-5x+3$$
Find the product of $$\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\right)$$ and $$\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\right)$$.
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$$\dfrac{1}{9}y^4-\dfrac{1}{4}x^4$$
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$$\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$$
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$$4x^2-9y^4$$
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$$\dfrac{1}{4}x^4+\dfrac{1}{9}y^4$$
Explanation
Given, product of
$$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)$$ and $$ \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$
i.e.
$$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$
.
We know, the identity $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$\Bigl(\dfrac{1}{2}x^2-\dfrac{1}{3}y^2\Bigr)\times \Bigl(\dfrac{1}{2}x^2+\dfrac{1}{3}y^2\Bigr)$$
$$=\Bigl(\dfrac{1}{2}x^2\Bigr)^2-\Bigr(\dfrac{1}{3}y^2\Bigl)^2$$
$$=\dfrac{1}{4}x^4-\dfrac{1}{9}y^4$$.
Therefore, option $$B$$ is correct.
The value of $$(-5{x}^{2}y) \times (-\dfrac{2}{3} x{y}^{2}z) \times (\dfrac{8}{15} xy{z}^{2}) \times (-\dfrac{1}{4} z)$$ is ______.
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$$-\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$$
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$$\dfrac{4}{9} {x}^{4}{y}^{4}{z}^{4}$$
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$$-\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$
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$$\dfrac{4}{9} {x}^{3}{y}^{3}{z}^{3}$$
Explanation
We have,
$$\quad \quad \left( -5{ x }^{ 2 }y \right) \times \left( -\dfrac { 2 }{ 3 } x{ y }^{ 2 }z \right) \times \left( \dfrac { 8 }{ 15 } xy{ z }^{ 2 } \right) \times \left( -\dfrac { 1 }{ 4 } z \right) \\ =\left( -5\times \left( -\dfrac { 2 }{ 3 } \right) \times \dfrac { 8 }{ 15 } \times \left( -\dfrac { 1 }{ 4 } \right) \right) \times \left( { x }^{ 2 }y\quad \times \quad x{ y }^{ 2 }z\quad \times \quad xy{ z }^{ 2 }\quad \times \quad z \right) \quad \\ =\left( -\dfrac { 2\times 2 }{ 3\times 3 } \right) \times \left( \left( { x }^{ 2 }\times x\times x \right) \times \left( y\times { y }^{ 2 }\times y \right) \times \left( z\times { z }^{ 2 }\times z \right) \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \times \left( { x }^{ 4 } \right) \times \left( { y }^{ 4 } \right) \times \left( { z }^{ 4 } \right) \\ =\left( -\dfrac { 4 }{ 9 } \right) \left( { x }^{ 4 } \right) \left( { y }^{ 4 } \right) \left( { z }^{ 4 } \right)$$
Answer is A
The value of $$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$ is:
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$$2$$
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$$3$$
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$$5$$
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$$1$$
Explanation
Given, $$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$.
We know, $$(a+b)(a-b)=a^2-b^2$$.
Then,
$$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$
$$=\displaystyle 5^{\tfrac {1}{2}\times2} - 3^{\tfrac {1}{2}\times2}$$
$$=\displaystyle 5^{1} - 3^{1}$$
$$=\displaystyle 5- 3$$
$$=\displaystyle 2$$.
Hence, option $$A$$ is correct.
Using identities, evaluate:
$$71^2$$.
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$$5041$$
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$$3041$$
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$$6041$$
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$$2041$$
Explanation
Given, $$71^2$$,
i.e.
$$71^2$$
$$=(70+1)^2$$
.
We know,
$$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$$.
Then,
$$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$$
Hence, $$71^2=5041$$.
Therefore, option $$A$$ is correct.
Find the product $$24x^2(1-2x)$$ and evaluate if for $$x=2$$
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$$288$$
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$$384$$
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$$-288$$
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$$-384$$
Explanation
Let us multiply the given monomial and binomial as shown below:
$$24{ x }^{ 2 }(1-2x)\\ =\left( 24{ x }^{ 2 }\times 1 \right) -\left( 24{ x }^{ 2 }\times 2x \right) \\ =24{ x }^{ 2 }-48{ x }^{ 3 }$$
Now, let us substitute $$x=2$$ in
$$24{ x }^{ 2 }-48{ x }^{ 3 }$$ as follows:
$$24{ (2) }^{ 2 }-48{ (2) }^{ 3 }=(24\times 4)-(48\times 8)=96-384=-288$$
Hence,
$$24{ x }^{ 2 }(1-2x)=-288$$ for $$x=2$$.
The value of $${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$$, corrected to three decimal places is:
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$$2.001$$
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$$2.004$$
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$$2.006$$
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$$1.995$$
Explanation
Given,
$$(1.02)^2+(0.98)^2$$.
We know,
$$(a+b)^2= a^2+2ab+b^2$$.
Then,
$$(1.02)^2+(0.98)^2$$
$$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$$)
$$=(1.02+0.98)^2-1.9992$$
$$=(2)^2-1.9992$$
$$=4-1.9992$$
$$=2.0008$$.
When corrected to three decimal digit,
$$2.0008$$
$$=2.001$$.
Therefore, option $$A$$ is correct.
Factorise : $${ (ax+by) }^{ 2 }+{ (2bx-2ay) }^{ 2 }-6abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+2{ b }^{ 2 }{ x }^{ 2 }+2{ a }^{ 2 }{ y }^{ 2 }+6abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }-12abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }+4abxy$$
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$${ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }+4{ b }^{ 2 }{ x }^{ 2 }+4{ a }^{ 2 }{ y }^{ 2 }+6abxy$$
Explanation
$$\left( ax+by\right)^2+\left( 2bx-2ay\right)^2-babxy$$
Here we will use the following identity,
$$\Rightarrow \left( m+n\right)^2=m^2+n^2+2mn$$
and $$\left(m-n\right)^2=m^2+n^2-2mn$$
Using this we have :
$$= \left(ax\right)^2+\left(by\right)^2+2axby+4\left(bx\right)^2+4\left(ay\right)^2-8axby-6abxy$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-14axby+2axby$$
$$=a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby$$
Hence, the answer is $$a^2x^2+b^2y^2+4b^2x^2+4a^2y^2-12axby.$$
If $$xy=20$$ and $$(x+y)^{2}=70$$, then $$x^{2}+y^{2}$$ is equal to
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$$25$$
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$$35$$
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$$30$$
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$$50$$
Explanation
$$(x+y)^2=x^2+y^2+2xy$$
It is given that $$(x+y)^2=70$$ and $$xy=20$$
$$\Rightarrow 70=x^2+y^2+(2\times 20)$$
$$x^2+y^2=70-40$$
$$=30$$
The product of $$\left( 7a-8b \right) $$ and $$\left( 7a-8b \right) $$ is _______________.
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$${ 14a }^{ 2 }-112ab+{ 16b }^{ 2 }$$
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$${ 49a }^{ 2 }+112ab+{ 64b }^{ 2 }$$
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$${ 49a }^{ 2 }-112ab+{ 64b }^{ 2 }$$
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$${ 49a }^{ 2 }-112ab-{ 64b }^{ 2 }$$
Explanation
We know that $$(x-y)^2 = x^2 -2xy + y^2$$
So,
$$(7a-8b)(7a-8b)$$
$$= (7a)^2-2(7a)(8b)+(8b)^2$$
$$=49a^2-112ab+64b^2$$
Thus, C is the correct answer.
Product of the following monomials $$4p$$, $$- 7q^{3}$$, $$-7pq$$ is
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$$ 196 p^{2}q^{4}$$
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$$ 196 pq^{4}$$
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$$-196p^{2}q^{4}$$
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$$ 196 p^{2}q^{3}$$
Explanation
Given monomials are : $$4p$$, $$- 7q^{3}$$, $$-7pq$$
$$\therefore$$ Product $$=(4p)\times(- 7q^{3})\times(-7pq)=4\times (-7)\times (-7)\times p\times q^{3} \times pq =196p^{2}q^{4}$$
Hence, option (A) is correct.
The value of $$(a +b)^{2} + (a-b)^{2}$$ is
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$$ 2a + 2b $$
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$$ 2a - 2b$$
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$$ 2a^{2} + 2b^{2}$$
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$$ 2a^{2} - 2b^{2}$$
Explanation
$$\textbf{Step-1: Apply the relevant formula & simplify.}$$
$$\text{We have,}$$
$$(a + b)^{2} + (a - b)^{2}$$
$$\text{Now,}$$
$$(a+b)^2 = a^2+2ab+b^2$$ $$......(i)$$
$$(a-b)^2= a^2-2ab+b^2$$ $$......(ii)$$
$$\textbf{Step-2: Add the above expressions to get the result}$$
$$(a + b)^{2} + (a - b)^{2}$$
$$a^2+2ab+b^2 + a^2-2ab+b^2$$
$$=2a^2 +2b^2$$
$$\textbf{Hence, option - C is the answer}$$
State whether the statement is true (T) or false (F).
The product of one negative and one positive term is a negative term.
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True
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False
Explanation
The given statement is true.
The product of one negative and a positive term is always a negative term.
For example, $$(-2)\times3=-6$$
or, $$-a\times b = -ab$$
What is the product of $$ (x+2)(x-3) ? $$
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$$ 2 x-6 $$
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$$ 3 x-2 $$
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$$ x^{2}-x-6 $$
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$$ x^{2}-6 x $$
Explanation
$$ (x+2)(x-3) $$
$$x(x-3)+2(x-3)$$
$$x^2-3x+2x-6$$
$$ x^{2}-x-6 $$
Simplify the following expression:
$$(3x^2 * 7x^7)+ (2y^3 * 9y^{12})$$
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$$21x^{14} + 18y{26}$$
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$$10x^{9} + 11y{15}$$
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$$21x^{14} + 18y{15}$$
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$$21x^{9} + 18y{15}$$
If
$$x + y = 2013$$ and $$\frac{1}{x}+\frac{1}{y}=2013$$,
what is the value of xy?
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$$\frac{1}{2013}$$
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4026
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0
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1
Find the product of $$(p+2q)(p+3q)$$ when $$p=1$$, $$q=0$$.
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
$$p\times p+2q\times p+p\times 3q+2q\times 3q$$
$$=p^2+6q^2+5pq$$
at $$p=1, q=0$$
product $$= 1+0+0=1$$
Simplify the following expression:
$$(2x^4y^7m^2z) ^{TM} (5x^2y^3m^8)$$
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$$10x^6y^9m^{10}z $$
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$$7x^6y^{10}m^{10}z$$
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$$10x^5y^{10}m^{10}z$$
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$$10x^6y^{10}m^{10}z$$
$$Let\,x = \sqrt {3 - \sqrt 5 } \,and\,y = \sqrt {3 + \sqrt 5 } .$$ If the value of expression $$x - y + 2{x^2}y + 2x{y^2} - {x^4}y + x{y^4}$$ can be expressed in the form $$\sqrt p + \sqrt q $$ $$where\,p,\,q \in N,$$ then $$(p+q)$$ has the value equal
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$$ 448 $$
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$$610$$
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$$510$$
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$$540$$
Explanation
Multiply:$$\frac{2}{3},xy,6{x^2}{y^2}\,\,and\,\,\frac{1}{4}{y^3}\,\,is\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,} $$
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$${x^3}{y^3}$$
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$$\frac{2}{3}{x^3}{y^6}$$
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$${x^3}{y^6}$$
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$$\frac{2}{3}$$
Multiply: $$4x, - 3y\,\,and\,\, - 2xy$$
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$$24xy$$
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$$24{x^2}{y^2}$$
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$$ - 24{x^2}{y^2}$$
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$$ - 24xy$$
Multiply: $$3{x^2},2xy,4x{y^2}$$
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$$24{x^4}{y^3}$$
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$$12x{y^2}$$
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$$24{x^2}{y^2}$$
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$$24xy$$
Find the value of $$\dfrac { 10.5 \times 10.5 - 15 \times 10.5 + 7.5 \times 7.5 } { 10.5 - 7.5 }.$$
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$$17$$
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$$3$$
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$$6$$
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$$22$$
The product of a monomial and a binomial is a
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Monomial
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binomial
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Trinomial
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None of these
If $$A = \left[ \overset { -4 }{ 3 } \quad \overset { -1 }{ 1 } \right]$$ , then the determinant of the matrix $$(A^{2016} - 2A^{2015} - A^{2014})$$ is :
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$$2014$$
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$$- 175$$
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$$2016$$
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$$-25$$
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