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

According to question,
Average of x number is y²
? Sum of x number is = xy²
Average of y number is = x²
? Sum of y number is = yx²
Average of all number is
= $$\frac{{{xy^2} + {yx^2}}}{{x + y}}$$
= $$\frac{{xy\left( {x + y} \right)}}{{x + y}}$$
= xy

Since month begins with a Sunday, so re will be five Sundays in month.
? Required average
$$\eqalign{ & = \left( {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \right) \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$

According to question,
Wrong marks = 42 + 74 = 116
Correct marks = 56 + 32 = 88
Difference = 116 - 88 = 28 marks
? This difference effect 14 students = $$\frac{28}{14}$$ = 2
And, incorrect average = 71
? Correct average = 71 - 2 = 69

According to question,
Total weekly emoluments workers = Rs. 1534
Number of workers
= $$\frac{1534}{118}$$
= 13

Average
$$ = \left( {\frac{{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23}}{9}} \right)$$
$$\eqalign{ & = \frac{{100}}{9} \cr & = 11\frac{1}{9} \cr} $$

Total age of 2 players = 18 + 20 = 38 years
Increased years = 2 × 11 = 22 months
Age of new players
= 38 years + 22 months
= 39 years 10 months
Average = 19 years 11 months

Alternate
Let total age of 11 players = 11x
Then sum of age of new players = y years
According to question,
$$\eqalign{ & \Rightarrow 11x + y - 18 - 20 = 11\left( {x + \frac{1}{6}} \right) \cr & \Rightarrow 11x + y - 38 = 11x + \frac{{11}}{6} \cr & \Rightarrow y = \frac{{11}}{6} + 38 \cr & \Rightarrow y = \frac{{239}}{6} \cr} $$
? Average
$$\eqalign{ & = \frac{{239}}{{2 \times 6}} \cr & = \frac{{239}}{{12}} \cr} $$
= 19 years 11 months

Number of men
= $$\frac{1}{5}$$ × 120
= 24
Number of women
= $$\frac{1}{4}$$ × 120
= 30
Number of children
= 120 - (24 + 30)
= 66
Average age of men = 60 years
Average age of children
= $$\frac{1}{4}$$ × 60
= 15 years
Average age of women
= $$\frac{5}{6}$$ × 60
= 50 years
? Avarage age of group
$$\eqalign{ & = \left( {\frac{{60 \times 24 + 50 \times 30 + 15 \times 66}}{{120}}} \right){\text{ years}} \cr & = \left( {\frac{{3930}}{{120}}} \right){\text{ years}} \cr & = 32.75{\text{ years}} \cr} $$

Weight of teacher
= (35.4 × 25 - 35 × 24) kg
= 45 kg

Let five consecutive odd numbers are 1, 3, 5, 7, 9
Here, a = 1, b = 3, c = 5, d = 7, e = 9
According to question,
Average
$$\eqalign{ & = \frac{{1 + 2 + 5 + 7 + 9}}{5} \cr & = \frac{{25}}{5} \cr & = 5 \cr} $$
Now check option
Option (D) a + 4
Here a = 1
1 + 4 = 5 satisfy