Sum of temperature on 1^st, 2^nd, 3^rd and 4^th days
= (58 � 4)
= 232 degrees
Sum of temperature on 2^nd, 3^rd, 4^th and 5^th days
= (60 � 4)
= 240 degrees
Subtracting (i) from (ii), we get :
Temperature on 5^th day - Temperature on 1^st day = 8 degrees
Let temperature on 1^st and 5^th days be 7x and 8x degrees respectively.
Then,
? 8x - 7x = 8
? x = 8
? Temperature on 5^th day
= 8x
= 8 � 8
= 64 degrees

$$\eqalign{ & {\text{Average}} \cr & = \frac{{ {76 + 65 + 82 + 67 + 85} }}{5} \cr & = \frac{{375}}{5} \cr & = 75 \cr & {\text{Hence,}}\,{\text{average}} = 75 \cr} $$

Sum of weights of students after replacement
= [(52 � 45) - (48 � 5) + (54 � 5)] kg
= 2370 kg
? New average
= $$\left( {\frac{{2370}}{{45}}} \right)$$ kg
= $$52\frac{2}{3}$$ kg

$$\eqalign{ & \refore {\text{Required}}\,{\text{average}} \cr & = {\frac{{55 \times 50 + 60 \times 55 + 45 \times 60}}{{55 + 60 + 45}}} \cr & = {\frac{{2750 + 3300 + 2700}}{{160}}} \cr & = \frac{{8750}}{{160}} \cr & = 54.68 \cr} $$

The total weight of (36 + 44) Students of A and B,
= (36 � 40 + 44 � 35) kg = 2980 kg.
Therefore average weight of whole class = $$\frac{{2980}}{{80}}$$ kg
Therefore average weight = 37.25 kg

Sum of present ages of husband, wife and child
= (23 � 2 + 5 � 2) + 1
= 57 years
? Required average
= $$\left( {\frac{{57}}{3}} \right)$$
= 19 years

Weight of first box = 200 kg
Weight of third box
= 125 % of 200 kg
= 250 kg
Weight of second box
= 120% of 250 kg
= 300 kg
Weight of fourth box = 350 kg
Let weight of fifth box be x kg
Then, 70% of x = 350 kg
$$\eqalign{ & \Rightarrow x = \left( {\frac{{350 \times 100}}{{70}}} \right) \cr & \Rightarrow x = 500{\text{ kg}} \cr} $$.
Average weight of four heaviest boxes
$$\eqalign{ & {\text{ = }}\left( {\frac{{500 + 350 + 300 + 250}}{4}} \right){\text{kg}} \cr & {\text{ = 350 kg}} \cr} $$
Average weight of four lightest boxes
$$\eqalign{ & = \left( {\frac{{200 + 250 + 300 + 350}}{4}} \right){\text{kg}} \cr & = 275{\text{ kg}} \cr} $$
? Required difference
= (350 - 275)
= 75 kg

Sum of ages of three boys = 45 years
Now, (3x + 5x + 7x) = 45
? 15x = 45
? x = 3
So, age of youngest boy = 3x
= 3 � 3
= 9 years

$$\eqalign{ & {\text{Let}}\,{\text{three}}\,{\text{numbers}}\,{\text{be}}\,x,\,y,\,z. \cr & {\text{Given}}, \cr & x = 2y \cr & \Rightarrow x = 4z \cr & \Rightarrow y = 2z \cr & \Rightarrow z = z \cr & {\text{The}}\,{\text{average}}\,{\text{of}}\,{\text{reciprocal}}\,{\text{numbers}}\,{\text{is}}\,\frac{7}{{72}} \cr & \frac{{ { {\frac{1}{x}} + {\frac{1}{y}} + {\frac{1}{z}} } }}{3} = \frac{7}{{72}} \cr & \Rightarrow \frac{{ {yz + xz + xy} }}{{3xyz}} = \frac{7}{{72}} \cr & \Rightarrow \left( {14{z^2}24{z^3}} \right) = \frac{7}{{72}} \cr & \Rightarrow 504 = 84z \cr & z = 6 \cr & {\text{So}},\,x = 4z = 4 \times 6 = 24, \cr & \Rightarrow y = 2z = 2 \times 6 = 12 \cr & {\text{Thus}}\,{\text{}}\,{\text{numbers}}\,{\text{are}}\,24,\,12,\,6 \cr} $$

According to question,
$$\eqalign{ & \Rightarrow \frac{{{3^{30}} + {3^{60}} + {3^{90}}}}{3} \cr & \Rightarrow {3^{29}} + {3^{59}} + {3^{89}} \cr} $$