Answer
Verified
458.1k+ views
Hint: First, evaluate the amount which is charged on the extra minutes. The total bill amount is the sum of the fixed charge per month and the amount she charged for extra minutes.
Complete step-by-step answer:
We are given that Divya pays Rs. $145$per month for her mobile phone service and $75$ paise for each extra minute she talks over the allowed number of minutes in the monthly plan.
Let the number of extra minutes is $x$.
First, convert $75$ paise into rupees.
We know that $100$ paise is equal to Rs. $1$
Therefore, $75$ paise is equal to Rs. $0.75$.
Now, extra minutes are $x$ and the amount charged on each extra minute is Rs. $0.75$. Therefore, the total extra amount will be the product of the total number of extra minutes and charge of each minute.
That is, $0.75x$
Since, the bill includes the total amount, therefore, Rs. $178$is equal to the sum of Rs. $145$and total extra amount charged.
That is, $0.75x + 145 = 178$
Simplify the equation to evaluate the value of $x$ by taking constants at one side.
$
\Rightarrow 0.75x = 178 - 145 \\
\Rightarrow 0.75x = 33 \\
$
Divide both sides by $0.75$.
$
\Rightarrow x = \dfrac{{33}}{{0.75}} \\
\Rightarrow x = 44 \\
$
$\therefore$, $44$ extra minutes, she used her phone beyond the allowed time. Hence, option (C) is correct.
Note:
We can solve this question directly by option.
The difference in the bill and she pays per month is Rs. $178 - 145 = 33$ and charge for each minute is Rs.$0.75$.
Therefore, that option will be our answer whose product and $0.75$is equal to $33$.
Option-(C)
$44 \times 0.75 = 33$
Hence, option (C) is correct.
Complete step-by-step answer:
We are given that Divya pays Rs. $145$per month for her mobile phone service and $75$ paise for each extra minute she talks over the allowed number of minutes in the monthly plan.
Let the number of extra minutes is $x$.
First, convert $75$ paise into rupees.
We know that $100$ paise is equal to Rs. $1$
Therefore, $75$ paise is equal to Rs. $0.75$.
Now, extra minutes are $x$ and the amount charged on each extra minute is Rs. $0.75$. Therefore, the total extra amount will be the product of the total number of extra minutes and charge of each minute.
That is, $0.75x$
Since, the bill includes the total amount, therefore, Rs. $178$is equal to the sum of Rs. $145$and total extra amount charged.
That is, $0.75x + 145 = 178$
Simplify the equation to evaluate the value of $x$ by taking constants at one side.
$
\Rightarrow 0.75x = 178 - 145 \\
\Rightarrow 0.75x = 33 \\
$
Divide both sides by $0.75$.
$
\Rightarrow x = \dfrac{{33}}{{0.75}} \\
\Rightarrow x = 44 \\
$
$\therefore$, $44$ extra minutes, she used her phone beyond the allowed time. Hence, option (C) is correct.
Note:
We can solve this question directly by option.
The difference in the bill and she pays per month is Rs. $178 - 145 = 33$ and charge for each minute is Rs.$0.75$.
Therefore, that option will be our answer whose product and $0.75$is equal to $33$.
Option-(C)
$44 \times 0.75 = 33$
Hence, option (C) is correct.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Write the difference between order and molecularity class 11 maths CBSE
A rainbow has circular shape because A The earth is class 11 physics CBSE
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What are noble gases Why are they also called inert class 11 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between calcination and roasting class 11 chemistry CBSE