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

Average temperature for Wednesday, Thursday and Friday = 40� C
Total temperature = 3 � 40 = 120� C
Average temperature for Thursday, Friday and Saturday = 41� C
Total temperature = 41 � 3 = 123� C
Temperature on Saturday = 42� C
Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120;
Saturday - Wednesday = 3
Wednesday = 42 - 3 = 39� C

According to question,
Average age of a family of 5 members 3 years ago
= 17 years sum of ages of a family members
= 5 � 17
= 85 years
A baby having been born average age of family is same today.
? Sum of age of a family of 6 members
= 17 � 6
= 102 years
? Sum of age of a family of 5 members at present
= 85 + 5 � 3
= 85 + 15
= 100 years
? Age of child
= 102 - 100 = 2 years

Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.

According to question,
Average of $$\frac{{x + \frac{1}{x}}}{2} = M$$
Put x = 1
$$\eqalign{ & \refore \,\,\frac{{1 + \frac{1}{1}}}{2} = M \cr & \Rightarrow M = 1 \cr & \refore \,\,\frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{{1^2} + \frac{1}{{{1^2}}}}}{2} \cr & = 1 \cr} $$
Now check from option
Option: (C) 2M² - 1 (put M = 1)
= 2 � 1 - 1
= 1 (satisfied)

Alternate :
According to question,
$$\eqalign{ & \Rightarrow \frac{{x + \frac{1}{x}}}{2} = M \cr & \text{Squaring both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {(2M)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 4{M^2} - 2 \cr} $$
Required average
$$\eqalign{ & = \frac{{{x^2} + \frac{1}{{{x^2}}}}}{2} \cr & = \frac{{4{M^2} - 2}}{2} \cr & = 2{M^2} - 1 \cr} $$

Total cost of 10 books = Rs. 120
Total cost of 8 books = Rs. 94
? The cost of 2 books = Rs. 26
Let price of each book be x and y.
? x + y = 26 - - - - - - (1)
$$\left( {\frac{{160}}{{100}}} \right)x + y = 26$$
On Solving for y, we get
y = 10
Now, Substituting y = 10 in (1) we get,
x + 10 = 26
x = 16
So price of each book is Rs. 16 and Rs. 10 respectively.

Total cost of 10 books = Rs. 120
Total cost of 8 books = Rs. 94
? The cost of 2 books = Rs. 26
Let price of each book be x and y.
? x + y = 26 - - - - - -(1)
Given that price of 1 book is 60% more than or price
$$\eqalign{ & \left( {\frac{{160}}{{100}}} \right)y + y = 26 \cr & \Rightarrow y\left( {\frac{{160}}{{100}} + 1} \right) = 26 \cr & \Rightarrow y\left( {\frac{{160 + 100}}{{100}}} \right) = 26 \cr & \Rightarrow y = \frac{{\left( {26 \times 100} \right)}}{{260}} \cr & \Rightarrow y = 10 \cr & {\text{Substituting}}\,\,y = 10\,\,{\text{in }}{\kern 1pt} \left(1 \right){\text{we get}}, \cr & x + 10 = 26 \cr & x = 16 \cr} $$

Let original number be (10a + b)
After interchanging digits, new number becomes (10b + a)
The question states that average of 10 numbers has become 1.8 less than original average. Therefore, sum of original 10 numbers will be (10 � 1.8 = 18 more than sum of 10 numbers with digits interchanged.
10a + b = 10b + a + 18
9a - 9b = 18
a - b = 2

Total number of teachers = 10
Age of new teacher = 25 years
Age of retire teacher
= (25 + 3 � 10) years
= 55 years

Required average
$$\eqalign{ & {\text{ = }} {\frac{{67 \times 2 + 35 \times 2 + 6 \times 3}}{{2 + 2 + 3}}} \cr & {\text{ = }} {\frac{{134 + 70 + 18}}{7}} \cr & = \frac{{222}}{7} \cr & = 31\frac{5}{7}{\text{ years}} \cr} $$