Tag: maths

Questions Related to maths

State whether the given statement is true/false.
An irrational number between two numbers $\dfrac{1}{7}$ and $\dfrac{2}{7}$ is $0.1501500 15000...$ .

irrational numbers irrational numbers and their operations rational and irrational numbers real numbers (rational and irrational numbers) maths

  1. True

  2. False

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Correct Option: A
Explanation:
Let us first find the decimal forms of the given numbers as follows: 
 
$\dfrac { 1 }{ 7 } =0.\overline { 142857 } ,\dfrac { 2 }{ 7 } =0.\overline { 285714 }$

We find a number which is non-terminating non-recurring lying between them.
So, we can find infinite many such numbers. For example, $0.150150015000...$ and $0.20200200020000....$

Hence, an irrational number between two numbers $\dfrac {1}{7}$ and $\dfrac {2}{7}$ is $0.150150015000...$

Let $A$ and $B$ are two finite sets such that $n(A)=3$ and $n(B)=4$ then  the number of elements in $A\Delta B$.

maths set language different sets de morgan's law de morgan's law for set theory

  1. $2$

  2. $7$

  3. $5$

  4. can not be determined

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Correct Option: D
Explanation:

Now, we have,

$A\Delta B=(A-B)\cup(B-A)$.
But it is impossible to find the number of elements in the set $A\Delta B$ as the sets $A$ and $B$ are not given explicitly. 

$A\cup B=A\cap B$ if and only if

maths set language different sets de morgan's law de morgan's law for set theory

  1. A is an empty set

  2. B is an empty set

  3. Both A and B are empty sets

  4. Both A and B are non-empty sets

Show answer
Correct Option: C
Explanation:

Solution:- Lets assume A is an empty set and B=$\left{ a,b \right}$

Now $A\cup B=\left{ a,b \right}$  and $A\cap B=\oslash $, so in all cases other than C , the condition is not satisfied. So C is the correct answer.

If A and B be two sets such that n(A) = 15, n(B) =25, then number of possible values of $n(A\Delta B)$(symmetric difference of  A and B) is

maths set language different sets de morgan's law de morgan's law for set theory

  1. 30

  2. 16

  3. 26

  4. 40

Show answer
Correct Option: B
Explanation:
$n(A \triangle B)= n (A \cup B)- n (A \cap B)$
for $n$ (A \triangle B)$ to be max. $n (A \cap B)=0$
We know, that 
$n(A \cup B)= n (A)+ n (B)- n (A \cap B) = 15+25-0=40$
$\Rightarrow n (A \triangle B)_{max} = 40-0 =40$
For minimum value of $n (A \triangle B)$
$n (A \cup B)$ should be min, $n (A \cap B)$ should be max.
$n (A \triangle B)$ min $=25-15= 10$
So. value of 
$n (A \triangle  B)= n (A \cup B)- n(A \cap B)$ lies om the set
${10,11,12,......, 3,9,40}$
Now, when $n (A \triangle B)$ is max. i.e. when 
$n( A \cup B )=40$ & $n (A \cap B)=0$
If we decrease $n (A \cup B)$ by $1$ then $n (A \cap B)$
Will increase by $1$
$n (A \triangle B)=39-1= 38$
Similarly on for the decrease of $1$ you will get in $(A \triangle B)$ as $36$ and $30$ so on.
Hence 
Range of $n (A \triangle B)= {10,12,14,16,18,20,......,38,40}$ 
$=16$ values 

If $A=\left {x\epsilon C: x^2=1\right }$ and $B=\left {x\epsilon C: x^4=1\right }$, then $A\Delta B$ is equal to

maths set language different sets de morgan's law de morgan's law for set theory

  1. $\left {-1, 1\right }$

  2. $\left {-1, 1, i, -i\right }$

  3. $\left {-i, i\right }$

  4. None of these

Show answer
Correct Option: C
Explanation:

$x^2=1\Rightarrow x=-1, 1.\therefore A=\left {-1, 1\right }$
$x^4=1\Rightarrow x^2=-1, 1$
$\Rightarrow x=-i, i, -1, 1.\therefore B=\left {-i, i, -1, 1\right }$
$\therefore A\Delta B=(A-B)\cup (B-A)=\phi \cup \left {-i, i\right }=\left {-i, i\right }$.

Evaluate: $\dfrac{(3\dfrac{2}{3})^{2} -(2\dfrac{1}{2})^{2}}{(4\dfrac{3}{2})^{2} -(3\dfrac{1}{3})^{2}}$ $\div$ $\dfrac{3\dfrac{2}{3} -2\dfrac{1}{2}}{4\dfrac{3}{2} -3\dfrac{1}{3}}$

maths part number dividing fractions division of a fractions division of a fraction

  1. $\dfrac{35}{53}$

  2. $\dfrac{37}{53}$

  3. $\dfrac{42}{59}$

  4. $\dfrac{47}{60}$

Show answer
Correct Option: B
Explanation:

$\dfrac{(3\frac{2}{3})^2-(2\frac{1}{2})^2}{(4\frac{3}{2})^2-(3\frac{1}{3})^2} \div \dfrac{3\frac{2}{3}-2\frac{1}{2}}{4\frac{3}{2}-3\frac{1}{3}}$


$=\dfrac{(3\frac{2}{3}+2\frac{1}{2})(3\frac{2}{3}-2\frac{1}{2})}{(4\frac{3}{2}+3\frac{1}{3})(4\frac{3}{2}-3\frac{1}{3})} \div \dfrac{3\frac{2}{3}-2\frac{1}{2}}{4\frac{3}{2}-3\frac{1}{3}}$

$=\dfrac{3\frac{2}{3}+2\frac{1}{2}}{4\frac{3}{2}+3\frac{1}{3}}$

$=\dfrac{\dfrac{11}{3}+\dfrac{5}{2}}{\dfrac{11}{2}+\dfrac{10}{3}}$

$=\dfrac{\dfrac{22+15}{6}}{\dfrac{33+20}{6}}$

$=\dfrac{37}{53}$

Solve $\left[\dfrac{170}{3} +\dfrac{6}{7}\right] \div \left[\dfrac{2}{7} \times \dfrac{11}{2}\right]$

maths part number dividing fractions division of a fractions division of a fraction

  1. $\dfrac{1208}{3\times 11}$

  2. $\dfrac{1208}{11}$

  3. $\dfrac{1208}{3}$

  4. $\dfrac{1208}{9\times 11}$

Show answer
Correct Option: A
Explanation:

$\left[\dfrac{170}{3} +\dfrac{6}{7}\right] \div \left[\dfrac{2}{7} \times \dfrac{11}{2}\right]$


$=\left[ \dfrac{1190+18}{21}\right] \div \left[ \dfrac{11}{7}\right]$


$=\left[ \dfrac{1190+18}{21}\right] \times \left[ \dfrac{7}{11}\right]$

$=\left[ \dfrac{1208}{11\times 3}\right]$

Divide the difference of $\dfrac{1}{5}$ and $\dfrac{2}{7}$ by $\dfrac{2}{7}$.

maths part number dividing fractions division of a fractions division of a fraction

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

  2. $\dfrac{21}{10}$

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

  4. $\dfrac{4}{10}$

Show answer
Correct Option: C
Explanation:

Difference of  $\dfrac{1}{5}$  and $\dfrac{2}{7}$ is  


$\dfrac{2}{7} - \dfrac{1}{5} = \dfrac{2\times5 - 1\times7}{35} =\dfrac{10 - 7}{35} =\dfrac{3}{35}$

Now divide $\dfrac{3}{35}$  by $\dfrac{2}{7}$  

we have, $\dfrac{3}{35}÷\dfrac{2}{7} =\dfrac{3}{35}\times \dfrac{7}{2} =\dfrac{3}{10}$

Ans: $\dfrac{3}{10}$

Simplify $35\times 6\dfrac{1}{14}$(approximately)$=$

maths part number dividing fractions division of a fractions division of a fraction

  1. $220.5$

  2. $220$

  3. $212$

  4. $231$

Show answer
Correct Option: C
Explanation:
$6 \cfrac{1}{14} = \cfrac{14 \times 6 + 1}{14} = \cfrac{85}{14}$
$\therefore \; 35 \times 6\cfrac{1}{14} = 35 \times \cfrac{85}{14} = \cfrac{5 \times 85}{2} = \cfrac{425}{2} = 212.5 \approx 212$
Hence, 212 is the correct answer.

A ribbon of length $5\dfrac{1}{4}$m is cut in to small pieces each of length $\dfrac{3}{4}$ m number of pieces will be

maths part number dividing fractions division of a fractions division of a fraction

  1. $5$

  2. $6$

  3. $7$

  4. $8$

Show answer
Correct Option: C
Explanation:

$ \Rightarrow  $ let length l $ = 5\dfrac{1}{4}$ m 

$ l = 5+\dfrac{1}{4} = \dfrac{21}{4}m $

& Small n pieces of length $ x = \dfrac{3}{4}$m 

So, $ l = nx $

$ n = \dfrac{l}{x} = \dfrac{21}{4}\times \dfrac{4}{3} = 7 $

So, there are 7 pieces of length $ \dfrac{3}{4} $