
Reena has enough money to buy 10 kg potatoes worth \[{\rm{Rs\,18/kg}}\] with the same money. How much kilograms of potatoes can she buy if the price of potatoes increases to \[{\rm{Rs\,20/kg}}\]?
Answer
575.1k+ views
Hint: Here, we will first find the amount that Reena had by multiplying the rate with the quantity. Then by using the amount she had and the increased price of potato, we will find the quantity of the kilograms she can buy for the same amount.
Complete step-by-step answer:
We are given that Reena has enough money to buy 10 kg potatoes worth \[{\rm{Rs\,18/kg}}\].
Now, we will find the money that Reena has with her at first.
We can find the money that Reena has by multiplying the quantity of the potatoes she bought by the rate of the potatoes for a kilogram.
Since Reena bought 10 kg potatoes worth \[{\rm{Rs\, 18/kg}}\], we get
Amount of money \[ = \] Quantity of Potatoes\[ \times \] Rate of potatoes for a kilogram
\[ \Rightarrow \] Amount of money \[ = 10 \times 18\]
Multiplying the terms, we get
\[ \Rightarrow \] Amount of money \[ = {\rm{Rs}}.180\]
So, Reena has Rs.180 to buy 10 kg potatoes worth \[{\rm{Rs \,18/kg}}\].
Now, we will find the quantity of kilograms of potatoes she can buy if the price of potatoes increases to \[{\rm{Rs\,20/kg}}\].
Amount of money \[ = \] Quantity of Potatoes \[ \times \] Rate of potatoes for a kilogram
We have found that Reena has Rs. 180.
Let \[x\] be the quantity of the kilograms.
\[ \Rightarrow 180 = x \times 20\]
By dividing the equation, we get
\[ \Rightarrow x = \dfrac{{180}}{{20}}\]
Dividing 180 by 20, we get
\[ \Rightarrow x = 9{\rm{kg}}\]
Thus, 9 kilograms of potatoes can be bought by Reena.
Therefore, 9 kilograms of potatoes can be bought if the price of potatoes increases to \[{\rm{Rs\,20/kg}}\] with the same amount of money.
Note: We know that multiplication and division are inverse to each other. If we want to know the quantity that has to be given for the given amount, then the total amount has to be divided by the rate. If we want to find the total amount, then the quantity has to be multiplied with the rate, to find the amount of money. In simple words, if we want to find a particular, then it has to be divided and if we want to find the whole, then it has to be multiplied.
Complete step-by-step answer:
We are given that Reena has enough money to buy 10 kg potatoes worth \[{\rm{Rs\,18/kg}}\].
Now, we will find the money that Reena has with her at first.
We can find the money that Reena has by multiplying the quantity of the potatoes she bought by the rate of the potatoes for a kilogram.
Since Reena bought 10 kg potatoes worth \[{\rm{Rs\, 18/kg}}\], we get
Amount of money \[ = \] Quantity of Potatoes\[ \times \] Rate of potatoes for a kilogram
\[ \Rightarrow \] Amount of money \[ = 10 \times 18\]
Multiplying the terms, we get
\[ \Rightarrow \] Amount of money \[ = {\rm{Rs}}.180\]
So, Reena has Rs.180 to buy 10 kg potatoes worth \[{\rm{Rs \,18/kg}}\].
Now, we will find the quantity of kilograms of potatoes she can buy if the price of potatoes increases to \[{\rm{Rs\,20/kg}}\].
Amount of money \[ = \] Quantity of Potatoes \[ \times \] Rate of potatoes for a kilogram
We have found that Reena has Rs. 180.
Let \[x\] be the quantity of the kilograms.
\[ \Rightarrow 180 = x \times 20\]
By dividing the equation, we get
\[ \Rightarrow x = \dfrac{{180}}{{20}}\]
Dividing 180 by 20, we get
\[ \Rightarrow x = 9{\rm{kg}}\]
Thus, 9 kilograms of potatoes can be bought by Reena.
Therefore, 9 kilograms of potatoes can be bought if the price of potatoes increases to \[{\rm{Rs\,20/kg}}\] with the same amount of money.
Note: We know that multiplication and division are inverse to each other. If we want to know the quantity that has to be given for the given amount, then the total amount has to be divided by the rate. If we want to find the total amount, then the quantity has to be multiplied with the rate, to find the amount of money. In simple words, if we want to find a particular, then it has to be divided and if we want to find the whole, then it has to be multiplied.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 7 Social Science: Engaging Questions & Answers for Success

Master Class 7 Science: Engaging Questions & Answers for Success

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Convert 200 Million dollars in rupees class 7 maths CBSE

Write a letter to your aunt thanking her for the birthday class 7 english CBSE

How did Douglas overcome his fear of water class 7 english CBSE

What are the controls affecting the climate of Ind class 7 social science CBSE

The founder of Jainism was A Rishabhadev B Neminath class 7 social science CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE


