Cake given to Naina =\(1\frac { 1 }{ 2 } \) piece
Cake given to Najma =\(1\frac { 1 }{ 3 } \) piece
Total cake given to both of them = \(1\frac { 1 }{ 2 } +1\frac { 1 }{ 3 } =\frac { 1\times 2+1 }{ 2 } +\frac { 1\times 3+1 }{ 3 } =\frac { 3 }{ 2 } +\frac { 4 }{ 3 } \)
LCM of 2 and 3 = 6
∴  \(\frac { 3 }{ 2 } +\frac { 4 }{ 3 } =\frac { 3\times 3 }{ 2\times 3 } +\frac { 4\times 2 }{ 3\times 2 } \) =\(\frac { 9 }{ 8 } +\frac { 8 }{ 6 } =\frac { 9+8 }{ 6 } =\frac { 17 }{ 6 } \)
Hence, total cake given to both of them =\(\frac { 17 }{ 6 } \) piece

Jitesh

\(\frac { 36 }{ 63 } ,\frac { 35 }{ 63 } ,\frac { 42 }{ 63 } \)

LCM of 9 and 5 = 45
Now,\(\frac { 5 }{ 9 } =\frac { 5\times 5 }{ 9\times 5 } =\frac { 25 }{ 45 } \)  and \(\frac { 4 }{ 5 } =\frac { 4\times 9 }{ 5\times 9 } =\frac { 36 }{ 45 } \)
Thus, both fractions are like fractions but \(\frac { 25 }{ 45 } \neq \frac { 36 }{ 45 } \)
∴  \(\frac { 5 }{ 9 } \neq \frac { 4 }{ 5 } \)

\(\frac { 2 }{ 11 } ,\frac { 2 }{ 21 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 7 } ,\frac { 2 }{ 8 } ,\frac { 2 }{ 15 } \)
Here, descending order of denominators
 =  21>15>11> 9>8>7
∴ Ascending order of fractions =  \(\frac { 2 }{ 11 } ,\frac { 2 }{ 15 } ,\frac { 2 }{ 11 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 8 } ,\frac { 2 }{ 7 } \)and descending order of fractions =\(\frac { 2 }{ 7 } ,\frac { 2 }{ 8 } ,\frac { 2 }{ 9 } ,\frac { 2 }{ 11 } ,\frac { 2 }{ 15 } ,\frac { 2 }{ 21 } \)

No

\(\frac { 4 }{ 9 } \)

\(7\frac { 3 }{ 4 } =7+\frac { 3 }{ 4 } =\frac { 7\times 4+3 }{ 4 } =\frac { 31 }{ 4 } \)

Possible pairs of numerator and denominator which add upto 10 are (0,10), (1,9), (2, 8), (3, 7) and (4,6)
Now, proper fractions = \(\frac { 0 }{ 10 } ,\frac { 1 }{ 9 } ,\frac { 2 }{ 8 } ,\frac { 3 }{ 7 } \) and \({4\over 6}.\)
There are five fractions. So, we can make five such fractions.

15

In the given figure, four parts are not equal.
∴ Fraction is not equal to 1/4.

Total number of equal parts = 7; Shaded parts = 3
 ∴ Fraction representing the shaded portion = \(\frac { 3 }{ 7 } \)

We know that, one day = 24 h
\(\therefore\) The required fraction = \(\frac { 15\ h }{ 24\ h } =\frac { 15 }{ 14 } =\frac { 5 }{ 8 } \)

Here, the given figure is divided into 4 equal parts and 3 parts out of 4 parts are shaded.
∴ The required fraction = \(\frac { 3 }{ 4 } \)

Sunil purchased juice on Monday =12\(\frac{1}{2}\)L
Juice purchased on Tuesday = 14\(\frac{3}{4}\)L
∴Total juice purchased = 12 \(\frac{1}{2}\)+ 14\(\frac{3}{4}\)  =\(\frac{25}{2}+\frac{59}{4}\)
= \(\frac{50+59}{4}=\frac{109}{4}=27\frac{1}{4}L\)
So, he purchased 27\(\frac{1}{4}\)L of juice in two days.

Cloth required for skirt = \(1\frac{3}{4}\)m
Cloth required for scarf = \(\frac{1}{2}\)m
∴ Total cloth = \((1\frac{3}{4}+\frac{1}{2})m= \frac{7}{4}+\frac{1}{2}\)
\(\frac{7+2}{4}=\frac{9}{4}\)        [∵ LCM of 2 and 4 is 4]
=\(2\frac{1}{4}m\)
so, he must buy \(2\frac{1}{4}m\) cloth.

We have \(\Box{ } -\frac { 5 }{ 8 } =\frac { 1 }{ 4 } \)
Here\(\frac { 5 }{ 8 } \) is subtracted from missing fraction to get\(\frac { 1 }{ 4 } .\) This means addition of \(\frac { 5 }{ 8 } \) and \(\frac { 1 }{ 4 } \) gives the missing fraction.
∴ Missing fraction = \(\frac { 1 }{ 4 } +\frac { 5 }{ 8 } \)
⇒ \(\frac { 1 }{ 4 } =\frac { 1\times 2 }{ 4\times 2 } =\frac { 2 }{ 8 } \)
Now, \(\frac { 1 }{ 4 } +\frac { 5 }{ 8 } =\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =\frac { 2+5 }{ 8 } =\frac { 7 }{ 8 } \)
Hence  \(\boxed { \frac { 7 }{ 8 } } -\frac { 5 }{ 8 } =\frac { 1 }{ 4 } \)

Here, fraction of pencils used by Ramesh = \(\frac { 10 }{ 20 } \)
Fraction of pencils used by Sheelu = \(\frac { 25 }{ 10 } \)
Fraction of pencils used by Jamaal = \(\frac { 40 }{ 80 } \)
Now,  \(\frac { 10 }{ 20 } =\frac { 10\div 10 }{ 20\div 10 } =\frac { 1 }{ 2 } \) [∵ HCF of 10 and 20 is 10]
\(\frac { 25 }{ 50 } =\frac { 25\div 25 }{ 50\div 25 } =\frac { 1 }{ 2 } \)    [∵ HCF of 25 and 50 is 25]
and  \(\frac { 40 }{ 80 } =\frac { 40\div 40 }{ 80\div 40 } =\frac { 1 }{ 2 } \)   [∵ HCF of 40 and 80 is 40]
Thus \(\frac { 10 }{ 10 } =\frac { 25 }{ 50 } =\frac { 40 }{ 80 } =\frac { 1 }{ 2 } \)
Hence, each has used up an equal fraction of his/her pencils

Simplest form of the fraction \(\frac { 84 }{ 98 } \) is \(\frac { 6 }{ 7 } \)

Total number of sandwiches = 2; Total number of boys = 3.
(a) Here, Arya has to divide 2 sandwiches into 3 persons. So, he will divide each sandwich into three equal parts and give one part of each sandwich to each of them Thus, each person gets 2/3 sandwiches.
(b) One sandwich is to be divided among three boys.
∴ Each boy will get = 1/3 part of a sandwich.

\(\frac{2}{15}\)

(a) Riya's saving = \(\frac{1}{3}\) of her salary
Remaining salary \(=1-{1\over 3}={2\over 3}\)
She spends = \(\frac{1}{2}\) of \(\frac{2}{3}\)\(=\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
She gives for charity=\(\frac { 2 }{ 3 } -\frac { 1 }{ 3 } =\frac { 2-1 }{ 3 } =\frac { 1 }{ 3 } \)
∴ The fraction she gives for charity=\(\frac{1}{3}\)
(b) The values depict from this are:
Helpfulness, caring for poor people and cooperation. 

Given, quantity of CNG saved = 33000 tonne
Quantity of diesel saved = 3300 tonne
Quantity of petrol saved = 21000 tonne
(i) The fraction of the quantity of diesel saved to the quantity a petrol saved=\(\frac{3300}{21000}\)
\(=\frac { 33 }{ 210 } =\frac { 33\div3 }{ 210\div3 }\)  [\(\because\) HCF of 33 and 210 is 3]
\(=\frac { 11 }{ 70 } \)
(ii) The fraction of the quantity of diesel to the quantity of CNG saved =\(\frac{3300}{21000}\)\(=\frac { 33 }{ 330 } =\frac { 33\div33 }{ 330\div33 } =\frac { 1 }{ 10 } \) 

Here, a watermelon is divided into 16 parts.
∴ Each part =\(\frac { 1 }{ 16 } \)
I ate 7 parts out of 16 parts i.e. \(\frac { 7 }{ 16 } \) part
My friend ate 4 parts out of 16 parts i.e\(\frac { 4 }{ 16 } \) part
Now, the part which we eat together
=\(\frac { 7 }{ 16 } +\frac { 4 }{ 16 } =\frac { 7+4 }{ 16 } =\frac { 11 }{ 16 } \)  part
Now, difference between our parts = \(\frac { 7 }{ 16 } -\frac { 4 }{ 16 } =\frac { 7-4 }{ 16 } =\frac { 3 }{ 16 } \)
So, I ate\(\frac { 3 }{ 16 } \) part of watermelon more than my friend. Now, remaining part of the watermelon = 1- \(\frac { 11 }{ 16 } -\frac { 16 }{ 16 } -\frac { 11 }{ 16 } \)
 [∵ \(\frac { 16 }{ 16 } \)and 1 are equivalent fractions]
= \(\frac { 16-11 }{ 16 } =\frac { 5 }{ 16 } \)

(i) Total number of equal parts = 2;
Number of shaded parts = 1
∴ Fraction of shaded portion = \(\frac { Number\ of\ shaded\ parts\ }{ Total\ number\ of\ equal\ parts } =\frac { 1 }{ 2 } \)
(ii) Total number of equal parts = 4;
Number of shaded parts = 2
∴ Fraction of shaded portion = 2/4
(iii) Total number of equal parts = 6;
Number of shaded parts = 3
∴ Fraction of shaded portion = 3/6
(iv) Total number of equal parts = 8;
Number of shaded parts = 4
∴ Fraction of shaded portion = 4/8
Now, \(\frac { 2 }{ 4 } =\frac { 2\div 2 }{ 4\div 2 } =\frac { 1 }{ 2 } \)  [∵ HCF of 2 and 4 is 2]
\(\frac { 3 }{ 6 } =\frac { 3\div 3 }{ 6\div 3 } =\frac { 1 }{ 2 } \)  [∵ HCF of 3 and 6 is 3]
\(\frac { 4 }{ 8 } =\frac { 4\div 4 }{ 8\div 4 } =\frac { 1 }{ 2 } \) [∵ HCF of 4 and 8 is 4]
Since all fraction in simple form are same.
∴ \(\frac { 1 }{ 2 } =\frac { 2 }{ 4 } =\frac { 3 }{ 6 } =\frac { 4 }{ 8 } \) i.e the fractions are equivalent.