Playing With Numbers - Class 8 Maths - Extra Questions

Consider the given fraction.

$$\dfrac{8}{9}$$

According to the question

$$ =\dfrac{8}{9}\times \dfrac{4}{4} $$

$$ =\dfrac{32}{36} $$

Hence, this is the answer.

We have,

$${{\left( \dfrac{625}{81} \right)}^{-\frac{3}{4}}}\times \left[ {{\left( \dfrac{4}{25} \right)}^{\frac{-3}{2}}}\div {{\left( \dfrac{2}{3} \right)}^{-3}} \right]$$

$$ ={{\left( \dfrac{81}{625} \right)}^{\frac{3}{4}}}\times \left[ {{\left( \dfrac{25}{4} \right)}^{\frac{3}{2}}}\div {{\left( \dfrac{3}{2} \right)}^{3}} \right] $$

$$ ={{\left( \dfrac{{{3}^{4}}}{{{5}^{4}}} \right)}^{\frac{3}{4}}}\times \left[ \left( \dfrac{25}{4} \right)\times \left( \dfrac{5}{2} \right)\times {{\left( \dfrac{2}{3} \right)}^{3}} \right] $$

$$ ={{\left( \dfrac{3}{5} \right)}^{3}}\times \left[ \left( \dfrac{125}{8} \right)\times {{\left( \dfrac{2}{3} \right)}^{3}} \right] $$

$$ =\left( \dfrac{27}{125} \right)\times \left( \dfrac{125}{27} \right) $$

$$ =1 $$

Hence, this is the answer.

$$i) 26,04,783$$

two million six hundred four thousand seven hundred eighty-three

$$ii) 5,901,080$$

five million nine hundred one thousand eighty

$$iii) 70,051,900$$

seventy million fifty-one thousand nine hundred

$$iv} 17,25,09,250$$

one hundred seventy-two million five hundred nine thousand two hundred fifty

We have,

$$0.46\div 10$$

$$ =\dfrac{0.46}{10} $$

$$ =0.046 $$

Hence, this is the answer.

We have,

$$ \dfrac{8m-1}{2m+3}=2 $$

$$ 8m-1=4m+6 $$

$$ 4m=7 $$

$$ m=\dfrac{7}{4} $$

Hence, this is the answer.

$$160$$ students are required to seat $$=40$$ benches

$$1$$ students are required to seat $$=\dfrac{40}{160}$$

 Therefore,

$$240$$ students are required to seat 

$$=\dfrac{40}{160}\times 240$$ benches

$$=60$$ benches.

 Hence, this is the answer.

$$\textbf{Step -1: Draw the lines for the given numbers.}$$

                  $$\text{For adding }2+2,$$

                  $$\text{Draw two lines}+\text{Draw two lines}$$

                  $$(||)+(||)$$

$$\textbf{Step -2: Count all the drawn lines.}$$

                  $$\text{Number of drawn lines}=4$$

$$\textbf{Therefore, }\mathbf{2+2=4.}$$

Let the dfraction is $$\dfrac{x}{y}$$

Then, according to given question

$$y=x+3\,\,\,\,.......\,\,\,\,\left( 1 \right)$$

Again,

$$ \dfrac{x}{y}+\dfrac{y}{x}=2\dfrac{9}{10} $$

$$ \dfrac{{{x}^{2}}+{{y}^{2}}}{xy}=\dfrac{29}{10} $$

$$ 10{{x}^{2}}+10{{y}^{2}}=29xy\,\,\,\,\,.......\,\,\,\,\left( 2 \right) $$

Put the value of y in equation (2).

$$ 10{{x}^{2}}+10{{\left( x+3 \right)}^{2}}=29x\left( x+3 \right) $$

$$ \Rightarrow 10{{x}^{2}}+10\left( {{x}^{2}}+9+6x \right)\,=29{{x}^{2}}+87x $$

$$ \Rightarrow 10{{x}^{2}}+10{{x}^{2}}+90+60x=29{{x}^{2}}+87x $$

$$ \Rightarrow 20{{x}^{2}}-29{{x}^{2}}+60x-87x+90=0 $$

$$ \Rightarrow -9{{x}^{2}}-27x+90=0 $$

$$ \Rightarrow -9\left( {{x}^{2}}+3x-10 \right)=0 $$

$$ \Rightarrow {{x}^{2}}+3x-10=0 $$

$$ \Rightarrow \left( x+5 \right)\left( x-2 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,by\,\,\,\,factorize $$

$$ \Rightarrow x+5=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-2=0 $$

$$ \Rightarrow x=-5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\,\,\,\,\,\,\,\,\,\, $$

Put  $$x=2$$               In equation (1)

$$y=2+3=5$$

fraction $$\dfrac{x}{y}=\dfrac{2}{5}$$  $$ans.$$

$$\textbf{Step-1: Assume the digits to be equal to some variables & then apply all given information's }$$

                 $$\text{Let the digit in the unit place be y and digit in the ten's place be x.}$$

                 $$\therefore$$  $$\text{Two digit number = 10x+y}$$

                 $$\text{According to the given question,}$$

                 $$\text{The sum of a two digit number and the number obtained by interchanging}$$ 

                 $$\text{the digits of the number is 121}$$

                 $$ \left( 10x+y \right)+\left( 10y+x \right)=121 $$

                 $$ \Rightarrow 11x+11y=121 $$

                 $$ \Rightarrow x+y=11\,\,\,\,\,........\,\,\,\left( 1 \right) $$

                 $$\text{Again, we know that, digits of the number differ by 3}$$

                 $$x-y=3\,\,\,........\,\,\,\,\left( 2 \right)$$

$$\textbf{Step-2: Simplify unknown.}$$

                 $$\text{By equation (1) and (2) , we get}$$

                 $$x=7$$

                 $$\text{Put in the equation (1) and we get,}$$

                 $$ 7+y=11 $$

                 $$ y=4 $$

                 $$\text{So, the number is}$$

                 $$ =10\times 7+4 $$

                 $$ =74 $$

$$\textbf{Hence, the number is 74.}$$

Let the two digit number is $$10x+y$$ whose digits is $$x\,\And \,y$$.

Then, According to given question,

$$ x\times y=14 $$

$$ y=\dfrac{14}{x}\,\,\,\,.........\,\,\,\left( 1 \right) $$

Now again, according to given question,

$$ 4\times x=2y-6 $$

$$ \Rightarrow 4x-2y+6=0\,\,\,\,........\,\,\,\left( 2 \right) $$

Put the value of $$y$$ by equation (1) in to (2), we get,

$$ 4x-2\times \dfrac{14}{x}+6=0 $$

$$ \Rightarrow 4{{x}^{2}}-28+6x=0 $$

$$ \Rightarrow 4{{x}^{2}}+6x-28=0 $$

$$ \Rightarrow 2{{x}^{2}}+3x-14=0 $$

$$ \Rightarrow 2{{x}^{2}}+\left( 7-4 \right)x-14=0 $$

$$ \Rightarrow 2{{x}^{2}}+7x-4x-14=0 $$

$$ \Rightarrow 2{{x}^{2}}-4x+7x-14=0 $$

$$ \Rightarrow 2x\left( x-2 \right)+7\left( x-2 \right)=0 $$

$$ \Rightarrow \left( x-2 \right)\left( 2x+7 \right)=0 $$

$$ \Rightarrow x-2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x+7=0 $$

$$ \Rightarrow x=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=-\dfrac{7}{2} $$

Put the value of x in equation (1) and we get,

$$ y=\dfrac{14}{x} $$

$$ y=\dfrac{14}{2}=7 $$

Then the number is

$$ =10x+y $$

$$ =10\times 2+7 $$

$$ =27 $$

Hence, this is the answer.

Step 1: Write the sum of digits of number as the first equation 

Given: $$2791A$$ is divisible by $$9$$

$$⇒$$ Sum of the digits of a number $$2791A$$ must be divisible by $$9$$

$$⇒ 2 + 7 + 9 + 1 + A = 19 + A$$ must be divisible by $$9$$

Substitute $$A$$ value from $$0$$ to $$9$$ by trial-error method

We get

$$A = 8$$ as $$27918$$ is divisible by $$9.$$

$$\Rightarrow 19 + 8 = 27$$ is divisible by $$9.$$

Hence, $$A$$ is $$8.$$

$$  {\textbf{Step -1: Open the bracket and simplify it.}} $$

                 $$  \left( {3 + \sqrt {23} } \right) - \sqrt {23} $$$$  =  3 + \sqrt {23}  - \sqrt {23}  $$

                                                  $$   = 3 $$

$$ {\textbf{Hence, the result is 3}}{\text{.}} $$ 

The last digit of $$2340 $$  is $$0$$

This indicates that one of the numbers must end in with $$5$$ and the other number must be even.

Let the missing digit be $$x$$

The number $$x5$$ has a value $$10x+5$$

The number $$5x$$  has a value $$50+x$$

The product is $$2340$$

$$(10x+5)(50+x)=2430$$

$$500x+10x^2+250+5x=2430$$ 

$$  10x^2+505x-2180=0 $$ ...... (divide both side by $$5$$)

Using ac method:

$$2\times 436=872$$

Find the factor of $$872$$ which differ by $$101$$

$$1,2,4,8........109,218,432,872$$

$$\Rightarrow (2x+109)(x-4)=0$$

By solving equation we have

$$x=-54.5 $$ ..... (ignore as $$x$$ is an integer)

$$\Rightarrow x=4$$

If  $$x=4,$$ then the numbers are $$45$$ and $$54$$.

Let the tenth digit be $$10x$$

Then, unit's unit $$=x+2$$

Number $$=10x+(x+2)=11x+2$$

Sum of digits $$=x+x+2= 2x+2$$

$$(11x+2)(2x+2)=144$$

$$\Rightarrow 11x^2 +26x-140=0$$

$$\Rightarrow 11x^2 +13x-70=0$$

$$\Rightarrow (x-2)(11x+35)=0$$

$$\Rightarrow x=2$$

Hence the required number is $$(11\times2)+2=24$$.