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CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 3 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Logarithm And Antilogarithm
Quiz 3
The value of $$(2^0 - 3^0) \times 4^2$$ is---
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$$16$$
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$$0$$
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$$-16$$
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None of these
Explanation
$$(2^0 - 3^0) \times 4^2\\ = (1 - 1) \times 4^2\\ = 0\times 16\\=0$$
If $$\displaystyle \log x=n$$ then 2n is equal to
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$$\displaystyle \log\left ( x^{2} \right )$$
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$$\displaystyle \left ( \log x \right )^{2}$$
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$$\displaystyle \log \left ( x+2 \right )$$
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$$\displaystyle \log 2x$$
Explanation
Given , $$\log x = n $$
We know that $$\log x^{2} = 2 \log x$$
$$\Rightarrow 2n = 2 \log x = \log x^{2}$$
Or $$ \log x^{2} = 2 \log x = 2n$$.
$$a^0=1$$ is true for all $$a$$ except
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$$-1$$
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negative integers
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0
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1
Explanation
The correct answer is C
A. $${ (-1 )}^{ 0 }=1$$
B. $${ (-2) }^{ 0 }=1$$
C. $${ 0 }^{ 0 }$$ is not defined.
D. $${ 1 }^{ 0 }=1$$
If $$\displaystyle \log _{x}y=100\: and\: \log _{2}x=10 $$ then the value of y is
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$$\displaystyle 2^{1000}$$
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$$\displaystyle 2^{100}$$
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$$\displaystyle 2^{2000}$$
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$$\displaystyle 2^{10000}$$
Explanation
$$\because log_2 x=10$$
$$\therefore x=2^{10}$$
Also, $$log _x y=100$$
$$\therefore y=x^{100}$$
Now, put the value of x then
$$y=2^{1000}$$
If $$\displaystyle x=\frac{y}{(1+a)^p}$$, then $$p$$ is equal to
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$$\displaystyle\frac{\displaystyle\log_e{\left(\frac{y}{x}\right)}}{\log_e{(1+a)}}$$
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$$\displaystyle\log{\left\{\frac{y}{x(1+a)}\right\}}$$
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$$\displaystyle\log{\left\{\frac{y-x}{1+a}\right\}}$$
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$$\displaystyle\frac{\log{y}}{\log{\{x(1+a)\}}}$$
Explanation
$$\displaystyle\because x=\frac{y}{(1+a)^p}$$
$$\displaystyle\therefore(1+a)^p=\frac{y}{x}$$
$$\displaystyle\therefore p\log_e{(1+a)}=\log_e{\frac{y}{x}}$$
$$\displaystyle\therefore p=\frac{\displaystyle\log_e{\left(\frac{y}{x}\right)}}{\log_e{(1+a)}}$$
Which of the following expresses zero law of exponents?
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$$0^{x} = 0$$
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$$x^{0} = 1$$
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$$x^{1} = x$$
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None of the above
Explanation
According to the zero law of exponents,
any non - zero number raised to the power of zero is equal to $$1$$.
$$\therefore x^{0} = 1$$, where $$x\neq 0$$
So, option $$B$$ is correct.
$$4\times 4^{10}$$ is represented as:
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$$4^{40}$$
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$$4^{10}$$
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$$4^{11}$$
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$$16^{10}$$
Explanation
$$4\times 4^{10} = 4^{1 + 10}$$
$$= 4^{11}$$
So, option $$C$$ is correct.
In simplified form, $$((3)^{0} + (5)^{0})^{0}$$ is equal to:
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$$2$$
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$$1$$
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$$0$$
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$$8$$
Explanation
$$(3^{0} + 5^{0})^{0} = (1 + 1)^{0}$$
$$= (2)^{0}$$
$$= 1$$
So, option $$B$$ is correct.
Find the value of $$(6)^{0} - (10)^{0}$$
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$$-4$$
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$$2$$
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$$1$$
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$$0$$
Explanation
$$(6)^0 - (10)^0 = 1-1 = 0$$
So, option $$D$$ is correct.
The expression $$((2)^{0} + (3)^{0} + (5)^{0})^{0}$$ is equal to _____
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$$3$$
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$$1$$
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$$10$$
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$$0$$
Explanation
$$((2)^{0} + (3)^{0} + (5)^{0})^{0} = (1 + 1 + 1)^{0}$$
$$= 3^{0}$$
$$= 1$$
So, option $$B$$ is the correct answer.
The value of $$21^{0}$$ is _____.
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$$0$$
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$$21$$
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$$1$$
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$$-21$$
Explanation
$$21^{0} = 1$$
$$\because a^{0} = 1$$
This is the law for zero exponent.
So, option $$C$$ is correct.
Which of the following expresses product law of exponents?
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$$(a^{m})^{n} = a^{mn}$$
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$$a^{m}\times a^{n} = a^{m+n}$$
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$$a^{m} \div a^{n} = a^{m-n}$$
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$$a^{m}\div b^{m} = \left (\dfrac {a}{b}\right )^{m}$$
Explanation
The product law of exponents states that,
when multiplying two powers that have the same base,
we can add their exponents.
$$\therefore a^{m}\times a^{n} = a^{m+n}$$
So, option $$B$$ is correct.
The value of $$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$$ is _____
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$$2$$
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$$4$$
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$$1$$
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$$0$$
Explanation
$$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$$
$$ = (1 + 1)\div (1 + 1)$$
$$= 2\div 2$$
$$= 1$$
So, option $$C$$ is correct.
The value of $$(100)^{0}$$ is _____.
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$$1$$
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$$0$$
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$$100$$
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$$1000$$
Explanation
According to the law of power of $$0$$,
$$(100)^{0} = 1$$.
So, option $$A$$ is correct.
Evaluate: $$({(10)^0} + (12)^{0})\times (18)^{0}$$
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$$1$$
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$$0$$
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$$2$$
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$$18$$
Explanation
$$((10)^{0} + (12)^{0})\times (18)^{0} = (1 + 1) \times 1$$
$$= (2)\times 1$$
$$= 2$$
So. option $$C$$ is correct.
Given that $$2^h\times 2^3 = 2^9$$, find the value of h.
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$$3$$
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$$6$$
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$$8$$
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$$12$$
Explanation
$$2^{h}\times 2^{3}=2^{9}$$
$$\Rightarrow 2^{h+3}=2^{9}$$
$$\Rightarrow h+3=9$$
$$\Rightarrow \boxed{h=6}$$
Find the correct expression, if $$\log _{ c }{ a } =x$$.
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$${ x }^{ c }=a$$
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$${ a }^{ x }=c$$
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$${ c }^{ a }=x$$
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$${ c }^{ x }=a$$
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None of these
Explanation
Given, $$\log _c a=x$$
Taking antilog, we get
$$c^x = a$$.
Which one of the following is the value of $$(101)^{0}$$?
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$$0$$
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$$101$$
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$$1010$$
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$$1$$
Explanation
$$101^{0} = 1$$
So, option $$D$$ is correct.
To simplify the following expression correctly, what must be done with the exponents?
$${ { 5 }^{ a }\times { 5 }^{ b }\times }{ 5 }^{ c }$$
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Add the exponents.
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Multiply the exponents.
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Divide the exponents
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Subtract the exponents.
Explanation
Since the base is same, the exponents can be added (Product rule of exponents).
$$\therefore 5^a \times 5^b \times 5^c = 5^{(a+b+c)}$$
Hence, option $$A$$ is correct.
The value of $$(10)^{0} \times (20)^{0}\times (30)^{0}$$ is:
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$$1$$
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$$3$$
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$$60$$
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$$0$$
Explanation
$$(10)^{0} \times (20)^{0} \times (30)^{0} = 1\times 1\times 1 = 1$$
So, option $$A$$ is correct.
Which of the following is equal to $$1$$?
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$$(3)^{0} + (3)^{0}$$
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$$(3)^{0} - (3)^{0}$$
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$$(3)^{0}$$
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$$(3^{3})^{0} + (3^{0})^{3}$$
Explanation
$$\textbf{Step 1: Simplify the following expressions.}$$
$$(a)\ 3^0+3^0=1+1=2\neq 1$$
$$(b)\ 3^0-3^0=1-1=0\neq 1$$
$$(c)\ 3^0=1$$
$$(d)\ (3^3)^0+(3^0)^3=9^0+1^3=1+1=2\neq 1$$
$$\textbf{Hence, Option C is correct.}$$
Simplify and give reasons:
$${(-2)}^{7}$$
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$$-128$$
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$$128$$
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$$-28$$
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None of these
Explanation
$$(-2)^7$$
$$=(-1*2)^7$$
$$=(-1)^7(2)^7$$ $$\because (ab)^m=a^m*b^m$$
$$=-1*2^7$$
$$=-2^7$$
$$=-128$$
$${(1000)}^{9}\div {10}^{24}=$$?
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$$10000$$
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$$1000$$
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$$100$$
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$$10$$
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None of the above
Explanation
Given Exp.=$$\cfrac { { \left( 1000 \right) }^{ 9 } }{ { 10 }^{ 24 } } =\cfrac { { \left( { 10 }^{ 3 } \right) }^{ 9 } }{ { 10 }^{ 24 } } =\cfrac { { 10 }^{ 27 } }{ { 10 }^{ 24 } } ={ 10 }^{ (27-24) }={ 10 }^{ 3 }=1000$$
Simplify and give reasons:
$${(-3)}^{-4}$$
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$$\cfrac{1}{81}$$
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$$\cfrac{-1}{81}$$
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$$\cfrac{3}{81}$$
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None of these
Explanation
$$(-3)^{-4}$$
$$=(-1*3)^{-4}$$
$$=(-1)^{-4}(3)^{-4}$$ $$\because (ab)^m=a^m*b^m$$
$$=1*3^{-4}$$
$$=3^{-4}$$ $$\because a^{-m}=\dfrac1{a^m}$$
$$=(\dfrac13)^4$$
$$=\dfrac1{81}$$
Simplify:
$$\left( { 4 }^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }$$
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$$\cfrac{1}{2}$$
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$$\cfrac{1}{4}$$
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$$\cfrac{2}{2}$$
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None of these
Explanation
we know,
$$a^m*b^m = (ab)^m$$
so,
$$4^{-1} * 3^{-1} = 12^{-1}$$
we know,
$$\dfrac{a^{-m}}{b^{-m}}=(\dfrac{a}{b})^{-m}=(\dfrac{b}{a})^{m}$$
so,
$$\dfrac{12^{-1}}{6^{-1}}=2^{-1}=\dfrac{1}{2}$$
Simplify the following:
$${ (-2) }^{ 7 }\times { (-2) }^{ 3 }\times { (-2) }^{ 4 }$$
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$${(-2)}^{14}$$
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$${(-2)}^{24}$$
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$${(-2)}^{54}$$
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None of these
Explanation
we know,
$$a^{m}*a^{n}=a^{m+n}$$
$$\implies$$
$$a^{m}*a^{n}*a^{p}=a^{m+n+p}$$
so,
$${ (-2) }^{ 7 }\times { (-2) }^{ 3 }\times { (-2) }^{ 4 }$$
$$=(-2)^{7+3+4}$$
$$=(-2)^{14}$$
Simplify and give reasons:
$$\cfrac { { 3 }^{ -2 } }{ 3 } \times \left( { 3 }^{ 0 }-{ 3 }^{ -1 } \right) $$
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$$\cfrac{2}{81}$$
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$$\cfrac{1}{81}$$
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$$\cfrac{4}{81}$$
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None of these
The logarithmic form of $${5}^{2}=25$$ is
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$$\log _{ 5 }{ 2 } =25$$
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$$\log _{ 2 }{ 5 } =25$$
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$$\log _{ 5 }{ 25 } =2$$
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$$\log _{ 25 }{ 5 } =2$$
Explanation
$$5^2=25$$
Taking log with base $$5$$ both sides, we get
$$\log_55^2=\log_525$$
$$\Rightarrow \log_525=2\log_55$$
$$\Rightarrow \log_525=2$$ $$(\log_aa=1)$$
Hence, C is the correct option.
Exponential form of $$\log_{4}8 = x$$ is _____
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$$x^{8} = 4$$
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$$x^{4} = 8$$
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$$4^{x} = 8$$
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$$8^{x} = 4$$
Explanation
We have, $$\log_{4}{8}=x$$
If $$ \log_{b}{a}=x$$ then $$a={b}^{x}$$
So, $${4}^{x}=8$$ is the correct answer.
Option $$C$$ is the correct answer.
The exponential form of $$\log _{ 2 }{ 16 } =4$$ is
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$${2}^{4}=16$$
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$${4}^{2}=16$$
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$${2}^{16}=4$$
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$${4}^{16}=2$$
Explanation
Since, $$\log_2 16=4$$
$$\implies \log $$ of $$16$$ to the base $$2$$ is $$4$$
$$\Rightarrow 16=2^4$$
Hence, A is the correct option.
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