The average of 33 numbers is 74. The average of the first 17 numbers is 72.8 and that of the last 17 numbers is 77.2. If the 17th number is
excluded, then what will be the average of the remaining numbers (correct to one decimal place)?
(A) 72.9
(B) 73.4
(C) 71.6
(D) 70.8
Solution:
Average of 33 members = 74 Sum of numbers = 33 × 74 = 2442 Sum of the first 17 numbers = 1237.6 Sum of the last 17 numbers = 17 × 77.2 = 1312.4 17th number = 1237.6 + 1312.4 - 2442 = 108 Average of 32 numbers $$ = \frac{{2442 - 108}}{{32}} = 72.9$$
72.
The average of 30 numbers is 40 and that of other 40 numbers is 30. The average of all the numbers is :
(A) $$34\frac{2}{7}$$
(B) 35
(C) 34
(D) 34.5
Solution:
According to the question,Average of 30 numbers is = 40Sum of 30 numbers is = 40 × 30 = 1200Average of 40 numbers is = 30Sum of 40 numbers is = 40 × 30 = 1200Total average = $$\frac{1200 + 1200}{70}$$= $$\frac{2400}{70}$$ = $$34\frac{2}{7}$$
73.
The average of the first five multiples of 9 is:
(A) 20
(B) 27
(C) 28
(D) 30
Solution:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{{\text{total}}\,{\text{sum}}\,{\text{of}}\,{\text{multiple}}\,{\text{of}}\,9}}{5}} \cr & = {\frac{{9 + 18 + 27 + 36 + 45}}{5}} \cr & = 27 \cr} $$ Note that, average of 9 and 45 is also 27. And average of 18 and 36 is also 27.
74.
The average age of a husband and wife at the time of their marriage was 25 years. A son was born to them two years after their marriage. The present average age of all three of them is 24 years. How many years is it since the couple got married?
(A) 5 years
(B) 6 years
(C) 8 years
(D) 9 years
Solution:
Sum of the ages of husband and wife at the time of their marriage = (25 × 2) yrs = 50 yrs Sum of the ages of husband and wife when their son was born = (50 + 2 × 2) yrs = 54 yrs Sum of the ages of husband, wife and son at present = (24 × 3) yrs = 72 yrs ∴ Age of son = $$\frac{{\left( {72 - 54} \right)}}{3}$$ = $$\frac{{18}}{3}$$ = 6 years Hence, the couple got married (6 + 2) = 8 years ago
75.
The average of marks in Mathematics for 5 students was found to do 50. Later, it was discovered that in the case of one student the marks 48 were misread as 84. The correct average is :
(A) 40.2
(B) 40.8
(C) 42.8
(D) 48.2
Solution:
According to the question, Correct average $$\eqalign{ & = \frac{{5 \times 50 + 48 - 84}}{5} \cr & = \frac{{250 - 36}}{5} \cr & = \frac{{214}}{5} \cr} $$ = 42.8
76.
A student finds the average of 10, 2 digits numbers. If the digits of one of the numbers interchanged, the average increases by 3.6. The difference between the digits of the 2 digits number is :
(A) 4
(B) 3
(C) 2
(D) 5
Solution:
According to the question, Let as consider by mistake he writes 10th number with its digits interchanged ∴ $$\frac{{10x + y - \left( {10y + x} \right)}}{{10}} = 3.6$$ ∴ In these remaining nine numbers are same and they cancel out $$\eqalign{ & \Rightarrow \frac{{10x + y - \left( {10y + x} \right)}}{{10}} = 3.6 \cr & \Rightarrow 9x - 9y = 36 \cr & \Rightarrow x - y = 4 \cr} $$
77.
An elevator can carry maximum of 16 passengers with an average weight of 80 kg. However, four boys more than the maximum carrying capacity of the elevator entered it making the average weight as 86 kg and overloading the elevator. What is the average weight of those four boys ?
(A) 112 kg
(B) 108 kg
(C) 110 kg
(D) 98 kg
Solution:
Passengers Weight Total wight 16 × 80 = 1280 20 × 86 = 1720 Weight of 4 boys = 440 Average weight of 4 boys = $$\frac{440}{4}$$ = 110 kg
78.
The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
The average age of 30 students of a class is 14 years 4 months. After admission of 5 new students in the class the average becomes 13 years 9 months. The youngest one of the five new students is 9 years 11 months old. The average age of the remaining 4 new students is :
(A) 10 years 4 months
(B) 12 years 4 months
(C) 11 years 2 months
(D) 13 years 6 months
Solution:
According to the question, Total age of 30 students = 30 × (14 years 4 months)= 30 × $$14\frac{1}{3}$$ = $$\frac{30 × 43}{3}$$ = 430 years Total age of (30 + 5) students = 35 (13 years 9 months) = 35 × $$13\frac{3}{4}$$ = $$\frac{1925}{4}$$ years Total age of 5 students = $$\frac{1925}{4}$$ - 430 = $$\frac{205}{4}$$ = 51 years 3 months ∴ One of the new five student is = 9 years 11 months old ⇒ Remaining 4 students age = $$\frac{{{\text{41 years 4 months}}}}{4}$$ = 10 years 4 months
80.
The average temperature on Wednesday, Thursday and Friday was 25ÃÂÃÂÃÂð. The average temperature on Thursday, Friday and Saturday was 24ÃÂÃÂÃÂð. If the temperature on Saturday was 27ÃÂÃÂÃÂð, what was the temperature on Wednesday?
(A) 24°
(B) 21°
(C) 27°
(D) 30°
Solution:
Total temperature on Wednesday, Thursday and Friday was 25 × 3 = 75° Total temperature on Thursday, Friday and Saturday was 24 × 3 = 72° Hence, difference between the temperature on Wednesday and Saturday = 3° If Saturday temperature =27°, then Wednesday's temperature = 27 + 3 = 30°