AVERAGE SET 2:

1. The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is:
   A.51.4
   B.52.6
   C.56.1
   D.55.3
   E.None of these
   
   Answer – A (51.4)
   Explanation – Let the required mean score be a
   Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100
   1600 + 775 + 55a = 5200
   55a = 2825
   a = 51.4
   
   
2. The average of a non-zero number and its square is 5 times the number. The number is:
   A.0 , 7
   B.0 , 6
   C.5 , 7
   D.0 , 9
   E.None of these
   
   Answer – D (0 , 9)
   Explanation – Let the number be x.
   Then,
   (x + x^2) / 2 = 5x
   x^2 – 9x = 0
   x (x – 9) = 0
   x = 0 or x = 9.
   
   
3. If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of a^2, b^2, c^2 is:
   A.3 M x M
   B.3 M
   C.9 M
   D.9 M x M
   E.None of these
   
   Answer – A (3 M x M)
   Explanation –
   We have :
   (a + b + c) / 3 = M or (a + b + c) = 3M.
   Now, (a + b + c)^2 = (3M)^2 = 9M^2
   a^2 + b^2 + c^2 + 2 (ab + bc + ca) = 9M^2
   a^2 + b^2 + c^2 = 9M^2
   Required mean =
   (a^2 + b^2 + c^2)/3= 9M^2/ 3 = 3 M^ 2
   
   
4. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one them weighing 65 kg. What might be the weight of the new person?
   A.65 kg
   B.70 kg
   C.85 kg
   D.92 kg
   E.None of these
   
   Answer – C (85 kg)
   Explanation – Total weight increased = (8 x 2.5) kg = 20 kg
   Weight of new person = (65 + 20) kg = 85 kg
   
   
5. The average age of the boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is:
   A.15
   B.16
   C.17
   D.Data inadequate
   E.None of these
   
   Answer – D (Data inadequate)
   Explanation – Clearly, to find the average, we ought to know the number of boys, girls or students in the class, neither of which has been given
   So, the data provided is inadequate
   
   
6. A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must he make in his next innings so as to raise his average to 24?
   A.44
   B.45
   C.49
   D.48
   E.None of these
   
   Answer – C (49)
   Explanation – Total of 10 innings = 21.5 x 10 = 215
   Suppose he needs a score of a in 11th innings; then average in 11 innings =
   (215 + a) / 11= 24
   or, a = 264-215 = 49
   
   
7. 1/3 rd of certain journey is covered at the rate of 25 km/hr, 1/4 th at the rate of 30 km/hr and rest at 50 km/hr. Find the averae speed for the whole journey?
   A.31 1/3 km/hr
   B.30 1/3 km/hr
   C.33 1/3 km/hr
   D.34 1/3 km/hr
   E.None of these
   
   Answer – C (33 1/3)
   Explanation – Let the journey by a km. Then a/3 km at the speed of 25 km/hr and a/4 km at 30 km/hr and the rest distance ( a- a/3 –a/4 ) = 5/12 x a at the speed of 50 km/hr.
   Total time taken during the journey of a km
   
   
   
8. The average of six numbers is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. What is the average of the remaining two numbers?
   A.4.2
   B.4.6
   C.5.1
   D.5.6
   
   Answer – B (4.6)
   Explanation – Sum of the remaining two numbers = (3.95 x 6) – [(3.4 x 2) + (3.85 x 2)]
   = 23.70 – (6.8 + 7.7) = 23.70 – 14.5 = 9.20
   Average = 9.2/2 = 4.6
   
   
   
9. The average salary of the entire staff in a office is Rs 120 per month. The average salary of officers is Rs 460 and that of non officer is Rs 110. If the number of officer is 15, then find the number of non-officer in the office?
   A.450
   B.550
   C.510
   D.520
   E.None of these
   
   Answer – C (510)
   Explanation – Let the required number of non-officers = a
   Then, 110a + 460 x 15 = 120 (15 + a)
   or, 120a – 110a = 450 x 15 – 120 x 15 = 15 (460 – 120)
   or, 10a = 15 x 340; a = 15 x 34 = 510
   
   
10. Find the average of first 20 multiple of 7?
    A.71.5
    B.73.5
    C.75.2
    D.76.6
    E.None of these
    
    Answer – B (73.5)
    Explanation –
    Average = 7 (1+2+3+…….+20)/ 20= 7 x 20 x 21/ 20×2= 73.5
    
11.