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Calculation using standard form - class-X

Description: calculation using standard form
Number of Questions: 39
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Tags: repeated multiplication square and square root maths fundamentals powers and exponents fractions and standard forms indices basic algebra exponents and powers numbers real numbers
Attempted 0/39 Correct 0 Score 0

Evaluate the product of $2.3\times 10^4$ and $3\times 10^3$.

  1. $6.9\times 10^4$

  2. $6.9\times 10^7$

  3. $6.9\times 10^8$

  4. $6.9\times 10^6$


Correct Option: B
Explanation:

$(2.3\times 10^4) \times (3 \times 10^3)$
$\Rightarrow (2.3 \times 3) \times(10^{3+4})$

$\therefore 6.9 \times 10^7$
Ans- Option $B$.

Evaluate $\dfrac{6.3\times 10^3}{3\times 10^7}$

  1. $2\times 10^{-3}$

  2. $2.1\times 10^{-3}$

  3. $2.1\times 10^{-4}$

  4. $2\times 10^{-5}$


Correct Option: C
Explanation:

$\dfrac{6.3\times 10^{3}}{3\times 10^7}$


$\Rightarrow \dfrac{6.3}{3} \times 10^{3-7}$


$\therefore 2.1\times 10^{-4}$

Ans-Option $C$.

Solve:

$\cfrac{2.3^{n+1} + 7.3^{n-1}}{3^{n+2}-2 \left ( \cfrac{1}{3} \right )^{1-n}} $

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: B
Explanation:

We have,

$\cfrac{2.3^{n+1} + 7.3^{n-1}}{3^{n+2}-2 \left ( \cfrac{1}{3} \right )^{1-n}}  $

$\Rightarrow \cfrac{6.3^n+ \dfrac{7}{3}.3^n}{9.3^n-\dfrac{2}{3}. 3^n}  $

$\Rightarrow \cfrac{18.3^n+ 7.3^n}{27.3^n-2. 3^n}  $

$\Rightarrow \cfrac{25.3^n}{25.3^n}  $

$\Rightarrow 1  $

Hence, this is the answer.

The average distance between the Sun and a certain planet is approximately $\displaystyle 2.3\times { 10 }^{ 14 }$ inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately $\displaystyle 3.9\times { 10 }^{ 4 }$ inches )

  1. $\displaystyle 7.1\times { 10 }^{ 8 }$

  2. $\displaystyle 5.9\times { 10 }^{ 9 }$

  3. $\displaystyle 1.6\times { 10 }^{ 10 }$

  4. $\displaystyle 1.6\times { 10 }^{ 111 }$

  5. $\displaystyle 5.9\times { 10 }^{ 11 }$


Correct Option: B
Explanation:

The average distance in KM will be $\frac { 2.3\times { 10 }^{ 14 } }{ 3.9\times { 10 }^{ 4 } } =0.59\times { 10 }^{ 10 }=5.9\times { 10 }^{ 9 }\quad km$

So correct answer will be option B

Simplify: $\dfrac{1.8\times 10^{11}}{2\times 10^4}$

  1. $9\times 10^7$

  2. $0.9\times 10^6$

  3. $0.9\times 10^5$

  4. $9\times 10^6$


Correct Option: D
Explanation:

$\dfrac{1.8\times 10^{11}}{2\times 10^4}$


$= \dfrac{1.8}{2} \times 10^{11-4}$


$=0.9 \times 10^7$

$= 9\times 10^6$


Ans-Option $D$.

Express $(3000)^2\times (20)^3$ in scientific notation:

  1. $7.2\times 10^{10}$

  2. $3.6\times 10^{8}$

  3. $6\times 10^9$

  4. $4.8\times 10^{12}$


Correct Option: A
Explanation:

${ 3000 }^{ 2 }\times { 20 }^{ 3 }=(9\times { 10 }^{ 6 })\times (8\times { 10 }^{ 3 })\ =72\times { 10 }^{ 9 }=7.2\times { 10 }^{ 10 }$

So correct answer will be option A

Write the number of significant digits in:
$0.97 \times 10^{-2}$

  1. $1$

  2. $5$

  3. $2$

  4. $3$


Correct Option: C
Explanation:

$0.97\times 10^{-2}$
There are $2$ significant figures $9$ and $7$. When a number is  written in scientific notation, only significant figures are placed into the numerical portion.

$225$ can be expressed as

  1. $5\times 3^2$

  2. $5^2\times 3$

  3. $5^2\times 3^2$

  4. $5^3\times 3$


Correct Option: C
Explanation:

$ 225 = 25 \times 9$
$ 225 = (5\times 5)  \times (3 \times 3)$

$ 225 = 5^2 \times 3^2$
Option C is correct

Which of the following is greater than 1000.01 ?

  1. 0.00010001 x 10$\displaystyle ^{7}$

  2. 0.00001 x 10$\displaystyle ^{8}$

  3. 1.1 x 10$\displaystyle ^{2}$

  4. 1.00001 x 10$\displaystyle ^{3}$


Correct Option: A
Explanation:

A $\Rightarrow \displaystyle 0.00010001\times 10^{7}=1000.1 $
B $\Rightarrow \displaystyle 0.00001\times 10^{8}=1000 $
C $\Rightarrow \displaystyle 1.1\times 10^{2}=110 $
D $\Rightarrow \displaystyle 1.00001\times 10^{3}=1000.01 $

We observe that $\displaystyle 1000.1> 1000.01 $
i.e., $\displaystyle 0.00010001\times 10^{7}> 1000.01 $

What is the smallest integer n for which $\displaystyle { 25 }^{ n }>{ 5 }^{ 12 }$?

  1. $6$

  2. $7$

  3. $8$

  4. $9$

  5. 10


Correct Option: B
Explanation:

${25}^{n}>{5}^{12}$

${({5}^{2})}^{n}>{5}^{12}$
${{5}^{2n}}>{5}^{12}$
Comparing the power
$2n>12$
$n>\dfrac{12}{2}$
$n>6$
Smallest integer greater than $6$ will be $7$

If $4^{n-2} + 4^{2} = 32$, then what is the value of $n$?

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: B
Explanation:

Given, ${ 4 }^{ n-2 }+{ 4 }^{ 2 }=32$
${ 4 }^{ n-2 }=32-{ 4 }^{ 2 }$
${ 4 }^{ n-2 }=32-16$
${ 4 }^{ n-2 }=16$
${ 4 }^{ n-2 }={ 4 }^{ 2 }$
$n-2=2$ ..... (as the bases are equal)
$n=4$

If $\sqrt { { 2 }^{ x } } =16$, then $x=$

  1. $8$

  2. $4$

  3. $2$

  4. $10$


Correct Option: A
Explanation:

$\sqrt { { 2 }^{ x } } =16\ \Rightarrow { (2) }^{ \dfrac { x }{ 2 }  }={ 2 }^{ 4 }$


Exponents should be equal.

$ \dfrac { x }{ 2 } =4\Rightarrow x=8$

If $2^x - {2^{x - 1}} = 4$ then $x^x$ is equals to 

  1. $7$

  2. $3$

  3. $27$

  4. None of these


Correct Option: C
Explanation:
$2^{x}-2^{x-1}=4$

$2^{x}-2^{x}2^{-1}=4$

$2^{x}\left (1-\dfrac{1}{2}\right )=2^{x-1}=4=2^2$

$\therefore$ $x-1=2$

$\text{from this we get }$  $x=3$

$x^{x}=3^{3}=27$

If $3^x=5^y=7^z=105$, then $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ is equal to?

  1. $1$

  2. $0$

  3. $-1$

  4. None


Correct Option: A
Explanation:

$x = \cfrac{\log 105}{\log 3}, y = \cfrac{\log 105}{\log 5} , z = \cfrac{\log 105}{\log 7}$

$\cfrac{1}{x} + \cfrac{1}{y} + \cfrac{1}{z} = \cfrac{\log 3 + \log 5 + \log 7}{\log 105} = \cfrac{\log 105}{\log 105} = 1$

Which is greatest among following $2^{156},\ 4^{79},\ 128^{23}$ and $8^{54}$?

  1. $4^{79}$

  2. $128^{23}$

  3. $2^{156}$

  4. $8^{54}$


Correct Option: D
Explanation:
$2^{156}\rightarrow 2^{156}$
$4^{79}\rightarrow {2^{2\times 79}}\rightarrow 2^{158}$
$128^{23}\rightarrow 2^{7\times 23}\rightarrow 2^{161}$
$8^{54}\rightarrow 2^{3\times 54}\rightarrow 2^{162}$
$8^{54}$ is greatest
D is correct.

If $ { 9 }^{ x-1 }={ 3 }^{ 2x-1 }-486 $,then the value of x is:

  1. $\dfrac{7}{2}$

  2. 4

  3. 1

  4. 0


Correct Option: A
Explanation:
$9^{x-1}=3^{2x-1}-486$
$3^{2x-1}=9^{x-1}t486$
$\dfrac{3^{2x}}{3}=3^{2x-2}+486$
$3^{2x}=\dfrac{3(3^{2x})}{3^{z}}t(486)3$
$3^{2q}=3^{2x-1}+1458$
$2177=729+1458$
$3^{7}=3^{6}+1458$
$2x=7$
$x=\dfrac{7}{2}$

$For \ a \  natural \  number \  n , \  2n(n-1)!\leqslant  n^n, if$

  1. $n<2$

  2. $n>2$

  3. $n\ge2$

  4. Never


Correct Option: C

Which of the following represents the given expression?
$a^2b^3\times 2ab^2$ ?

  1. $2a^3b^4$

  2. $2a^3b^5$

  3. $2ab$

  4. $a^3b^5$


Correct Option: B
Explanation:
We know that if a number say $b$ is multiplied three times. That is, $b\times b\times b$ can be written as $b^3$.
=> $b^{3}$ = $b\times b\times b$
Similarly,
$a^{2}b^3$ = $a \times a\times b\times b\times b$
$2ab^{2}$ = $ 2 \times a\times b\times b$.
Thus, 
$a^{2}b^{3}\times 2ab^{2} = a \times a\times b\times b\times b \times 2 \times a\times b\times b $.
                      $= 2 \times a \times a \times a\times b\times b\times b\times b\times b$.
                      $ = 2a^{3}b^{5}$.

If $\displaystyle 2^{2^{3}}=j, 2^{3^{2}}=k, 3^{2^{2}}=\varphi ,$ then 

  1. $k = 2j$

  2. $j < k$

  3. $\displaystyle \varphi < k$

  4. All of these


Correct Option: D
Explanation:

  $j=2^{2{^3}}$, $k=2^{3^{2}}$, $\varphi=3^{2^{2}}$

$\Rightarrow$  $j=256,\,\,k=512,\,\,\varphi=81$
$\Rightarrow$  Option A says $k=2j$, which is true because $j=256$, so $k=2\times 256=512$
$\Rightarrow$  Option B says $j<k$, which is true because value of $j$ is $255$ and value of $k$ is $512$
$\Rightarrow$   Option C says $\varphi <k$, which is true because value of $\varphi$ is $81$ and value of $k$ is $512$

If $\displaystyle a^{m}=b^{m}$ and $(m > 0)$, then which of the following options could be true:

  1. $a = -b$

  2. $a  + b= 0$

  3. $2a - b = 0$

  4. $\displaystyle \dfrac{a^{2}}{b^{2}}=1$


Correct Option: D
Explanation:

Given $a^m=b^m$


Dividing by $b^m$

$\dfrac{a^m}{b^m}=1$
${\left(\dfrac{a}{b}\right)}^m=1$

Only Option D satisfies this answer.

Use an appropriate comparison symbol $0.00000998$ ______ $0.0000116$.

  1. $<$

  2. $>$

  3. $=$

  4. None of these


Correct Option: A
Explanation:

$\displaystyle 0\cdot 00000998= 9\cdot 89\times 10^{-6}$
$\displaystyle 0\cdot 0000116= 1\cdot 16\times 10^{-5}$
Thus $\displaystyle 9\cdot 89\times 10^{-6}< 1\cdot 16\times 10^{-5}$

$0.000008$ _______ $0.000016$

  1. is half of

  2. is double than

  3. is one-fourth of

  4. is one-third of


Correct Option: A
Explanation:

$\displaystyle 0\cdot 000008= 8\times 10^{-6}$


$\displaystyle 0\cdot 000016= 1\cdot 6\times 10^{-5}$

Now
$\displaystyle \frac{8\times 10^{-6}}{1\cdot 6\times 10^{-5}}= \frac{8}{1\cdot 6}\times 10^{-1}= 5\times 10^{-1}=0\cdot 5$

$\displaystyle \therefore 8\times 10^{-6}$ is half of $\displaystyle 1\cdot 6\times 10^{-5}$

The thickness of paper is $0.004$ m and that of another paper is $0.008$ m. Compare their sizes.

  1. Paper-1 is double thicker than paper-2.

  2. Thickness of paper-1 is half than paper-2.

  3. Thickness of paper-1 is one-fourth than paper-2.

  4. Paper-1 and paper-2 are both equally thick.


Correct Option: B
Explanation:

$\displaystyle 0\cdot 004= 4\times 10^{-3}$ m


$\displaystyle 0\cdot 008=8\times 10^{-3}$

Now
$\displaystyle \frac{0\cdot 004}{0\cdot 008}= \frac{4\times 10^{-3}}{8\times 10^{-3}}= \frac{4}{8}= \frac{1}{2}$

$\displaystyle \therefore $ Thickness of paper-1 is half than paper-2

Compare the folllowing:

$0.000000038$ _______ $\displaystyle 3\cdot 8\times 10^{-8}$

  1. $<$

  2. $>$

  3. $=$

  4. None of these


Correct Option: C
Explanation:

$\displaystyle 0\cdot 000000038= 3\cdot 8\times 10^{-8}$

Use an appropriate comparison symbol $0.0000486$ _____ $0.00000387$.

  1. $<$

  2. $>$

  3. $=$

  4. None of these


Correct Option: B
Explanation:

$\displaystyle 0.0000486=4\cdot 86\times 10^{-5}$
$\displaystyle 0.00000387=3\cdot 87\times 10^{-6}$
Thus
$\displaystyle 4\cdot 86\times 10^{-5}> 3\cdot 87\times 10^{-6}$

If $2^{p + 2} + 2^{p + 1} = 96$, then find the value of $ p$.

  1. $1$

  2. $2$

  3. $3$

  4. $4$

  5. $5$


Correct Option: D
Explanation:

As given, $2^{p+2}+2^{p+1}=96$
Rewriting the left hand side of equation using $x^a.x^b=x^{a+b}$.
$2^p.2^2+2^p.2^1=96$
$2^p(2^2+2^1)=96$
$2^p\times6=96$
$2^p=16=2^4$
$p=4$
Hence, option D is correct.

If $8^x = 16^{x-1}$, find $x $.

  1. $\dfrac{1}{8}$

  2. $\dfrac{1}{2}$

  3. $2$

  4. $4$


Correct Option: D
Explanation:

Given, $8^x=(16)^{x-1}$

Left hand side, $8^x=(2^3)^x=2^{3x}$
Right hand side $16^{x-1}=(2^4)^{x-1}=2^{4(x-1)}=2^{(4x-4)}$
Equating both sides, we get
$2^{3x}=2^{{4x-4}}$
As bases are equal, then powers must be equal, so
$\Rightarrow 3x=4x-4$
$ \Rightarrow x=4$

If $a^{b} = 4  -ab$ and $b^{a} = 1$, where $a$ and $b$ are positive integers, find $a$.

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: C
Explanation:
Plug in real numbers for $a$ and $b$. Since it isn’t clear what numbers to plug in to satisfy the first equation, look at the second equation instead. First, realize that  $a$  cannot be  $0$  since a is a positive integer.
Since, $a\neq0$, so the only way to get $b^a=1$ is if $b=1 $(As $1$ to any power is $1$).
Plugging $b=1$ in to the first equation,
$a^b=4-ab$
$a^1=4-a\times1$
$a=4-a$
$a=2$
Hence, option C is correct.

If $9^n = 27^{n+1}$, then calculate the value of $2^n $.

  1. $-\dfrac{10}{3}$

  2. $-\dfrac{8}{3}$

  3. $-\dfrac{3}{8}$

  4. $\dfrac{1}{8}$

  5. $\dfrac{3}{8}$


Correct Option: D
Explanation:

Given ${ 9 }^{ n }={ 27 }^{ n+1 }$, which implies
$ { 3 }^{ 2n }={ 3 }^{ 3n+3 }$
Now compare powers, we get
$2n = 3n+3$ , which implies $n = -3$
Therefore ${ 2 }^{ n }={ 2 }^{ -3 }=\dfrac { 1 }{ 8 } $

If $4^{2x + 2} = 64$, then calculate the value of $x $.

  1. $\dfrac {1}{2}$

  2. $1$

  3. $\dfrac {3}{2}$

  4. $2$


Correct Option: A
Explanation:

Given is $4^{2x+2}=64$

LHS: 
$\Rightarrow 4^{ 2x+2 }=(2^{ 2 })^{ 2x+2 }\ \Rightarrow { 2 }^{ 4x+4 }$
RHS: 
$\Rightarrow 64=2^6$

Now, LHS $=$ RHS
$\Rightarrow { 2 }^{ 4x+4 }={ 2 }^{ 6 }$
As bases are equal, so powers must be equal,
$\Rightarrow 6=4x+4$
$\Rightarrow 4x=2\ \Rightarrow x=\dfrac { 1 }{ 2 } $

If $x$ is a positive integer satisfying $x^7=k$ and $x^9=m$, which of the following must be equal to $x^{11}$?

  1. $\cfrac{m^2}{k}$

  2. $m^2-k$

  3. $m^2-7$

  4. $2k-\cfrac{m}{3}$

  5. $k+4$


Correct Option: A
Explanation:
Given, $x^7=k, x^9=m$
$x^{11}=\dfrac{x^{18}}{x^{7}}$ 
Put the given values, we get
$x^{11}=\dfrac{m^{2}}{k}$

If $n$ and $k$ are positive integers and $8^n=2^k$, what is the value of $\dfrac{n}{k}$?

  1. $\dfrac{1}{4}$

  2. $\dfrac{1}{3}$

  3. $\dfrac{1}{2}$

  4. $3$


Correct Option: B
Explanation:

Given 

${8}^{n}$ $=$ ${2}^{k}$
As $'8'$ can be written as $'$${2}^{3}$$'$,
${({2}^{3})}^{n}$ $=$ ${2}^{k}$
${2}^{3 \space \times \space n}$ $=$ ${2}^{k}$
${2}^{3n}$ $=$ ${2}^{k}$
Equating the exponents as the bases are same, we get
$3n$ $=$ $k$
Rearranging the terms, we get
$\dfrac {n}{k}$ $=$ $\dfrac {1}{3}$
Therefore, the value of $\dfrac {n}{k}=\dfrac {1}{3}$.

If $7^n\times 7^3 = 7^{12}$, what is the value of $n$?

  1. $2$

  2. $4$

  3. $9$

  4. $15$

  5. $36$


Correct Option: C
Explanation:

Given, ${7}^{n}$ $\times$ ${7}^{3}$ $=$ ${7}^{12}$

Rearranging terms, we get
${7}^{n}$ $=$ $\dfrac {{7}^{12}}{{7}^{3}}$
We know that
$\dfrac {{x}^{a}}{{x}^{b}}$ $=$ ${x}^{a \space - \space b}$
Hence, ${7}^{n}$ $=$ ${7}^{12 \space - \space 3}$
$\therefore {7}^{n}$ $=$ ${7}^{9}$
Again rearranging terms, we get
$n$ $=$ $\log _{7}{{7}^{9}}$
We know that
$\log _{x}{{x}^{a}}$ $=$ $a$$\log _{x}{x}$ and $\log _{x}{x}$ $=$ $1$
Hence, $n$ $=$ $9$$\log _{7}{7}$
$\therefore n$ $=$ $9$ $\times$ $1$
$\therefore n$ $=$ $9$
Therefore, the value of $'n'$ is $'9'$.

Which of the following has the greatest value?

  1. $(6^{2} \times 6)^{4}$

  2. $(36)^{5}$

  3. $(36^{2} \times 6^{3})^{2}$

  4. $(216)^{4}$

  5. $(6^{4})^{4}$


Correct Option: E
Explanation:

  1. ${ ({ 6 }^{ 2 }\times 6) }^{ 4 }={ { (6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  2. ${ 36 }^{ 5 }={ ({ 6 }^{ 2 }) }^{ 5 }={ 6 }^{ 10 }$
  3. ${ ({ 36 }^{ 2 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 4 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 7 }) }^{ 2 }={ 6 }^{ 14 }$
  4. ${ 216 }^{ 4 }={ ({ 6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  5. ${ ({ 6 }^{ 4 }) }^{ 4 }={ 6 }^{ 16 }$
  • Therefore option $E$ has maximum value

If $3^{n} = n^{6}$, find the value of $ n^{18} $

  1. $3^{n} n^{3}$

  2. $3^{n} n^{12}$

  3. $9^{n}$

  4. $3^{12n}$


Correct Option: B
Explanation:
Given, $3^n=n^6$
We need to find the value of $n^{18}$
$\therefore {n}^{18}= n^6. n^{12}=3^n.n^{12}$
$\therefore n^{6+12}=3^n. n^{12}$
$\therefore n^{18}= {3}^{n}{n}^{12}$

If $64^{x} = 4^{x^{2} - 4}$, then find the value of $x$.

  1. $x = 4$ or $x = -1$

  2. $x = -4$ or $x = 1$

  3. $x = 10$

  4. $x = \sqrt {20}$

  5. $x = 3$


Correct Option: A
Explanation:
  • ${ 64 }^{ x }={ 4 }^{ 3x }={ 4 }^{ { x }^{ 2 }-4 }$ , by equating powers , we get,
  • ${ x }^{ 2 }-4 = 3x$ , which implies ${ x }^{ 2 }-3x-4 = 0$
  • $\Rightarrow x^2-4x+x-4=0$
  • $\Rightarrow x(x-4)+1(x-4)= 0 $
  • $\Rightarrow (x-4)(x+1)=0$
  • The roots are $x=4,-1$

If $2^{3x - 2} = 16$, then calculate the value of $x $.

  1. $\dfrac {1}{2}$

  2. $1$

  3. $2$

  4. $\dfrac {3}{2}$


Correct Option: C
Explanation:

Given, ${ 2 }^{ 3x-2 }=16$
$\Rightarrow { 2 }^{ 3x-2 }=16={ 2 }^{ 4 }$

As bases are equal, there powers must be equal.
$\therefore 3x-2=4$
$\therefore 3x=6$
$\therefore x=2$

Place the following list of numbers with the given labels in order of greatest to least.
$F={({10}^{10})}^{10}$


$G={10}^{10}$ ${10}^{10}$

$H=\cfrac{{10}^{100}}{{10}^{10}}$

$I=100$

  1. $F,G,H,I$

  2. $G,F,H,I$

  3. $F,H,G,I$

  4. $H,F,G,I$


Correct Option: C
Explanation:

$F={({10}^{10})}^{10}=10^{10\times 10}=10^{100}$


$G={10}^{10}$ ${10}^{10}=10^{10+10}=10^{20}$

$H=\cfrac{{10}^{100}}{{10}^{10}}=10^{100-10}=10^{90}$

$I=100$

Clearly $F>H>G>I$

If $5^{k^2}(25^{2k})(625) = 25\sqrt{5}$ and $k < -1$, find the value of $k$.

  1. $-3.581$

  2. $-3.162$

  3. $-2.613$

  4. $-1.581$

  5. $-0.419$


Correct Option: A
Explanation:

Given, ${ 5 }^{ { k }^{ 2 } }({ 25 }^{ 2k })(625)=25\sqrt { 5 } $
${ 5 }^{ { k }^{ 2 }+4k+4 }={ 5 }^{ \tfrac 52 }$ 

By comparing powers, we get
${ k }^{ 2 }+4k+4=\cfrac 52$ which implies ${ 2k }^{ 2 }+8k+3=0$
$\Rightarrow k = -3.581$           ...(given $k<-1$)

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