Tag: maths

Questions Related to maths

Each piece of cardboard is $0.3\ cm$ thick. If Marie stacks $8$ pieces of cardboard on top of one another, how thick will the stack be?

  1. $2.4\ cm$

  2. $24\ cm$

  3. $2.1\ cm$

  4. $21\ cm$


Correct Option: A
Explanation:

It is given that each piece of cardboard is $0.3$ cm thick which means that one piece is $0.3$ cm thick.


If Marie stacks $8$ pieces of cardboard, then the thickness of $8$ pieces of cardboard will be:

$0.3\times 8=2.4$ cm

Hence, the stack will be $2.4$ cm thick.

If $100\times \text{ ? }=352$. Then, find the missing number.

  1. $0.352$

  2. $352$

  3. $35.2$

  4. $3.52$


Correct Option: D
Explanation:

Let $x$ be the missing number.


$100\times x=352\ \Rightarrow 100x=352\ \Rightarrow x=\dfrac { 352 }{ 100 } \ \Rightarrow x=352\times 0.01\quad \quad \quad \quad \quad \left{ \because \quad \dfrac { 1 }{ 10 } =0.1,\dfrac { 1 }{ 100 } =0.01,.... \right} \ \Rightarrow x=3.52$

Hence, the missing number is $3.52$.

Calculate $456.78\times 8$.

  1. $36542.4$

  2. $365.424$

  3. $3,654.24$

  4. $36.5424$


Correct Option: C
Explanation:
Let us first multiply the two given numbers $456.78$ and $8$ without decimal point:

$45678\times 8=365424$

Since, $456.78$ has two decimal places, therefore the answer $365424$ should also have two decimal places that is $3654.24$.

Hence, $456.78\times 8=3,654.24$.

$45.678\times \text{ ? }=1187.628$. Find $?$.

  1. $26$

  2. $27$

  3. $28$

  4. $29$


Correct Option: A
Explanation:

Let $x$ be the missing number.


$45.678\times x=1187.628\ \Rightarrow 45.678x=1187.628\ \Rightarrow \dfrac { 45678x }{ 1000 } =\dfrac { 1187628 }{ 1000 } \quad \quad \quad \quad \quad \left{ \because \quad \dfrac { 1 }{ 10 } =0.1,\dfrac { 1 }{ 100 } =0.01,.... \right} \ \Rightarrow x=\dfrac { 1187628 }{ 1000 } \times \dfrac { 1000 }{ 45678 } \ \Rightarrow x=26$

Hence, the missing number is $26$.

Multiply $43.09$ with $23$.

  1. $99.107$

  2. $991.07$

  3. $9910.7$

  4. $9.9107$


Correct Option: B
Explanation:
Let us first multiply the two given numbers $43.09$ and $23$ without decimal point:
$4309\times 23=99107$
Since, $43.09$ has two decimal places, therefore the answer $99107$ should also have two decimal places that is $991.07$.
Hence, $43.09\times 23=991.07$.

(Three rupees $5$ paise) $\times$ (five rupees $40$ paise) is equal to

  1. $Rs. 18.90$

  2. $Rs. 16.47$

  3. $Rs. 17.64$

  4. $RS. 15.37$


Correct Option: B
Explanation:

We know that 100 paise is nothing but 1 Rupee.

So $5Paise =  0.05 Rupees$
Also $40 Paise = 0.4 Rupees$
Now Three Rupees $5 Paise = 3 Rupees + 0.05 Rupees$
                                               $=3.05 Rupees$
Also Five Rupees $40 Paise= 5 Rupees + 0.40 Rupees$
                                              $= 5.40 Rupees$
Now to find $(Three\quad rupees\quad 5paise)×(fiverupees\quad 40paise)$
$3.05\times 5.40 = 16.47$

Solve the following:
$210.01\times 5$ = $1050.05$

  1. True

  2. False


Correct Option: A
Explanation:

Step 1 : $21001 x 5 = 105005$

Step 2 : Total number after decimal $= 2$
So, $210.01 \times  5 = 1050.05$

Using  identity the value obtained from the product $25.4 \times 24.6$ is

  1. $624.84$

  2. $642.84$

  3. $264.84$

  4. $62.84$


Correct Option: A
Explanation:


Given

$25.4\times24.6$

we can re-wright them as

$(25+0.4)\times (25-0.4)$

$\because \ we \ have \ an\ identity \ (a+b) (a-b)=a^2-b^2$

$a=25,\ b=0.4$


$= (25)^2-(0.4)^2$

$= 625- 0.16$

$=624.84$

$\therefore option\ A\ is\ correct.$


Without actual multiplication, the value $79.01 \times 79.01 + 2 \times 79.01 \times 20.99 + 20.99 \times 20.99$

  1. $10,009$

  2. $1000.06$

  3. $10,000$

  4. $1007$


Correct Option: C
Explanation:
$79.01\times 79.01+2\times 79.01\times 20.99+20.99\times 20.99....(1)$
$\dfrac {7901}{100}\times \dfrac {7901}{100}+2\times \dfrac {7901}{100}\times  \dfrac {2099}{100}+\dfrac {2099}{100}\times \dfrac {2099}{100}....(2)$

$7901\times 7901=(7901)^2 =(7900 +1)^2$    $\quad [\because \ 79^2=(80-1)^2\\ =80^2+1-2.80\\ =6400+1-160\\ =6241]$
$=(7900)^2 +1^2+2(7900)(1)$
$=62410000+1+15800$
$=62425801$ 

$2099\times 2099 =(2099)^2=(2100-1)^2$ $\quad [\because \ 21^2=(20+1)^2\\ =20^2+1+2.20\\ =400+40+1\\ =441]$
$=(2100)^2+1^2-2(2100)(1)$
$=4410000+1-4200$
$=4405801$

$7901\times 2099=(7900+1) (2100-1)$
$=(7900)(2100)-7900+2100-1$
$=16590000-7900+2100-1$
$=16584299\ =16584199$
from $(2)$
$\dfrac {62425801}{10000}+\dfrac {2\times 16584199}{10000}+\dfrac {4405801}{10000}$
$6242.5801+\dfrac {33168398}{10000}+440.5801$
$6242.5801+3316.8398+440.5801$
$=10,000$