In the afternoon, a student read 100 pages at the rate of 60 pages/h. In the evening, when she was tired, she read 100 more pages at the rate of 40 pages/h . What was her average rate of reading in pages per hour?

Let the monthly salary of a man = ? N
? The annual salary = ? 12N

According to the question,
Annually spends of a man = 7 x 1694.70 + 5 x 1810.50
= ? 20915.40

? His monthly spends = 20915.40/12 = ? 1742.95

And monthly saving = 3084.60/12 = ? 257.05
? His monthly salary
= 1742.95 + 257.05
= ? 2000

Total temperature of first and fourth days= (4 x 48 - 2 x 34)°C
= (192 - 68)°C= 124°C

Now, according to the question

Temperature of the first day = (9/20) x 124 = 55.8°C
Temperature of fourth day = (11/20) x 124 = 68.2°C

If three equal distances are covered at different spees of A, B and C km/h, respectively, then average speed during whole journey
= 3ABC/(AB + BC + CA) km/h
? Required average speed
= (3 x 15 x 10 x 5) / (15 x 10 + 10 x 5 + 15 x 5)
= (3 x 15 x 10 x 5)/(150 + 50 + 75)
= (3 x 15 x 10 x 5)/275
= 8.18 km/h

Monthly income of 4 persons = 7350 x 4 = ? 29400
Monthly income of 3 person (excluding the dead person) = 6500 x 3 = ? 19500
? Monthly income of dead person = 29400 - 19500 = ? 9900

Given that total number of students = 150 = ( n₁+ n₂) and
Mean weight of 150 students = 60 kg.
The mean weight of boys a₁= 70 kg and
The mean weight of girls a₂ = 55 kg
Let n₁ and n₂ be the number of boys and girl, respectively.
? n₁ + n₂ = 150 ...(i)
11 n₁ + 11 n₂ = 1650 ...(ii)
Average weight = (n₁ a₁ + n₂ a₂) / (n₁ + n₂)
? 60 = (70 n₁ + 55n₂)/150
? 70 n₁ + 55n₂ = 9000
? 14n₁ + 11n₂ =1800 ...(iii)
On subtracting Eq. (ii) from Eq. (iii), we get
3n₁ = 150 ? n₁ = 50
Hence, required number of boys is 50.

According to the average speed formula,
P = 1/4, Q =3/5, R = 3/20
x = 8 km/h, y = 6 km/h, z = 10km/h
? Required average speed = 1 / [(P/x) + (Q/y) + (R/z)]
=1 / [(1/32) + (1/10) + (3/200)]
= 1/[(25 + 80 + 12)/800]
= 800/117
= 6.83 km/h

Here, P = 9 km/h, Q = 25 km/h, R = 30 km
x = 3 km/h, y = 5 km/h and z = 10 km/h
? Required average speed
= [P + Q + R] / [(P/x) + (Q/y) + (R/z)] = [9+25 + 30] / [(9/3) + (25/5) + (30/10)]
= 64/(3 + 5 + 3)
= 64/11
= 5⁹/_{11 km/h}

Let temperature on sunday be x° C.
(20° + 18.6° + 17° + 21.4° + 19° + 19.8° + x) / 7 = 19.5°
? 115.8 + x = 19.5 x 7
? x = 136.5 - 11.58
? x = 20.7°C

According to the question
(P + T + R) / 3 = 54¹/₃
? P + T + R = (163/3) x 3
? P + T + R = 163 ............(i)
Also (T + F + G)/3 = 53
? T + G + F = 159 ..........(ii)
Clearly, from the two Eqs. (i) and (ii), the problem cannot be solved.

Let the minimum score = x
? Maximum score = x + 100
? x + (x + 100) = 30 x 42 - 40 x 28
? 2x + 100 = 1260 - 1120 = 140
? 2x = 140 - 100 = 40
? x = 20
Hence, the maximum score = x + 100
= 20 + 100 = 120