Tag: multiplication of decimals

Questions Related to multiplication of decimals

(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 $(2)$
$\dfrac {62425801}{10000}+\dfrac {2\times 16584199}{10000}+\dfrac {4405801}{10000}$
$6242.5801+\dfrac {33168398}{10000}+440.5801$
$6242.5801+3316.8398+440.5801$
$=10,000$

Find the product
$\displaystyle 0.123\times 100$

  1. $1.23$

  2. $123$

  3. $12.3$

  4. $1230$


Correct Option: C
Explanation:
To find $0.123\times 100$
Since, $100$ has $2$ zeros and $0.123$ has decimal point after $3$ digits
So, the product will have decimal point after $1$ digit from left

Therefore, value of $0.123\times 100$ is $12.3$.

Product of $78.12$ and $1.5$ is :

  1. $117.81$

  2. $117.18$

  3. $117.32$

  4. $117.80$


Correct Option: B
Explanation:

The product of $78.12$ and $1.5$ is

 $78.12 \times 1.5=117.18$
Hence, the answer is $117.18$.

Product of $\displaystyle 3.92\times 0.1\times 0.0\times 6.3$ is:

  1. $0.392$

  2. $0.1176$

  3. $0$

  4. $6.3$


Correct Option: C
Explanation:

We know if we multiply $0$ by any number then result will be zero.

So the value of $ 3.92\times .01\times 0.0\times 6.3=0$
Hence, the answer is $0$.

Find the product
$\displaystyle 7.854\times 10$

  1. $785.4$

  2. $78.54$

  3. $7854$

  4. $78540$


Correct Option: B
Explanation:
To find $7.854\times 10$
Since, $10$ has $1$ zero and $7.854$ has decimal point after $3$ digits
So, the product will have decimal point after $2$ digits from left

Therefore, value of $7.854\times 10$ is $78.54$.