Which is heavier: $\frac{12}{15}$ of 500 grams or $\frac{3}{20}$ of 4 kg?Solution:

Question

Accepted Answer

Solution:
When one of the numbers being multiplied is between 0 and 1, the product is less than the other number.
E.g., 0 < $\frac{1}{2}$ < 1, and $\frac{1}{2}$ × 100 = 50 < 100 When one of the numbers being multiplied is greater than 1, the product is greater than the other number.
E.g., 1$\frac{1}{2}$ > 1 and $1 \frac{1}{2} \times \frac{1}{4}=\frac{3}{2} \times \frac{1}{4}=\frac{3}{8}$
Now, $\frac{1}{4}=\frac{2}{8}$
⇒ $\frac{1}{4}<\frac{3}{8}$ 8.2 Division of Fractions Dividend, Divisor, and the Quotient NCERT In-Text Questions (Pages 189-190) When do you think the quotient is less than the dividend, and when is it greater than the dividend?
Is there a similar relationship between the divisor and the quotient?
Use your understanding of such relationships in multiplication to answer the questions above.
Solution:
When the divisor is between 0 and 1, the quotient is greater than the dividend.
E.g. $\frac{1}{2} \div \frac{1}{3}=\frac{1}{2} \times \frac{3}{1}=\frac{3}{2}=1 \frac{1}{2}$ and $\frac{1}{2}$ < 1$\frac{1}{2}$
When the divisor is greater than 1, the quotient is less than the dividend.
E.g. $\frac{1}{5} \div 2=\frac{1}{5} \times \frac{1}{2}=\frac{1}{10}$ and $\frac{1}{10}<\frac{1}{5}$.
When the divisor is 1, the quotient is equal to the dividend.
E.g. $\frac{3}{5} \div 1=\frac{3}{5}$
There is no similar relationship between the divisor and the quotient. 8.3 Some Problems Involving Fractions NCERT In-Text Questions (Page 191) Example 5.
This problem was posed by Chaturveda Prithudakasvami (c. 860 CE) in his commentary on Brahmagupta’s book Brahmasphutasiddhanta.
Four fountains fill a cistern. The first fountain can fill the cistern in a day. The second can fill it in half a day. The third can fill it in a quarter of a day. The fourth can fill the cistern in one-fifth of a day. If they all flow together, in how much time will they fill the cistern?
Let us solve this problem step by step.
In a day, the number of times- 
The first fountain will fill the cistern in 1 ÷ 1 = 1
The second fountain will fil the cistern is 1 ÷ $\frac{1}{2}$ = _____
The third fountain will fil the cistern is 1 ÷ $\frac{1}{4}$ = _____
The fourth fountain will fil the cistern is 1 ÷ $\frac{1}{5}$ = _____
The number of times the four fountains together will fill the cistern in a day is ___ + ____ + ___ + ____ = 12.
 Solution:
In a day, the number of times- 
The first fountain will fill the cistern in 1 ÷ 1 = 1
The second fountain will fill the cistern is 1 ÷ $\frac{1}{2}$ = 1 × $\frac{2}{1}$ = 4
The third fountain will fill the cistern is 1 ÷ $\frac{1}{4}$ = 1 × $\frac{4}{1}$ = 4
The fourth fountain will fill the cistern, is third fountain will fill the cistern is 1 ÷ $\frac{1}{5}$ = 1 × $\frac{5}{1}$ = 5
The number of times the four fountains together will fill the cistern in a day is 1 + 2 + 4 + 5 = 12.
 Fractional Relations NCERT In-Text Questions (Page 193) In each of the figures given below, find the fraction of the big square that the shaded region occupies.
[Working with Fractions Class 7 Solutions Ganita Prakash Maths Chapter 8 Page 193 Q1]
Solution:
Consider the following images
[Working with Fractions Class 7 Solutions Ganita Prakash Maths Chapter 8 Page 193 Q1.1]
We can see in image (i) that the top right square occupies $\frac{1}{4}$ of the area of the whole square.
Now, draw similar squares in the other three corners of the whole square as shown in Figure II.
From fig (ii), it is clear that the shaded region is $\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)$ of the top right square, that is $\frac{3}{2}$ of the top right square. But the top right square occupies $\frac{1}{4}$ of the area of the whole square.
Therefore, the shaded region occupies $\frac{3}{2} \times \frac{1}{4}$ of the area of the whole square, that is $\frac{3}{2} \times \frac{1}{4}=\frac{3}{8}$ of the area of whole square.
Now, consider the second given images (Fig. IV)
[Working with Fractions Class 7 Solutions Ganita Prakash Maths Chapter 8 Page 193 Q1.2]
From figure (iii), it is clear that the top left square shown by the bold line occupies $\frac{1}{4}$ of the area of the whole square.
Now, from figure (iv) it is clear that the top left square is divided into 8 identical triangles, out of which two are the shaded triangles.
So, the shaded region occupies $\frac{2}{8}$ of the top left square. But, the top left square occupies $\frac{1}{4}$ of the area of the whole.
Therefore the shaded region occupies $\frac{2}{8} \times \frac{1}{4}$ of the area of the whole square.
That is, $\frac{2}{8} \times \frac{1}{4}=\frac{1}{16}$
Thus, the shaded region occupies $\frac{1}{16}$ of the area of the whole square. A Dramma-tic Donation NCERT In-Text Questions (Page 194) If we assume 1 gold dinar = 12 silver drammas, 1 silver dramma = 4 copper panas, 1 copper pana = 6 mashakas, and 1 pana = 30 cowrie shells,
1 copper pana = $\frac{1}{48}$ gold dinar $\left(\frac{1}{12} \times \frac{1}{4}\right)$
1 cowrie shell = ___________ copper panas
1 cowrie shell = ___________ gold dinar.
Solution:
1 copper pana = $\frac{1}{48}$ gold dinar $\left(\frac{1}{12} \times \frac{1}{4}\right)$
1 cowrie shell = $\frac{1}{30}$ copper panas
1 cowrie shell = $\frac{1}{48} \times \frac{1}{30}$ gold dinar = $\frac{1}{1440}$ gold dinar Figure it Out (Page 196-198)