The average of the squares of first ten natural numbers is-
(A) 35.5
(B) 36
(C) 37.5
(D) 38.5
Solution:
As we know that average of square of "n" natural number is $$\eqalign{ & = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} \cr & = \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr} $$ According to the question, Average of square of first ten natural number is $$\eqalign{ & = \frac{{\left( {10 + 1} \right)\left( {20 + 1} \right)}}{6} \cr & = \frac{{11 \times 21}}{6} \cr & = 38.5 \cr} $$
12.
The average of 5 consecutive odd numbers is 27. What is the product of the first and the last number?
(A) 621
(B) 667
(C) 713
(D) 725
Solution:
Let first odd number = a Five odd number = a, a + 2, a + 4, a + 6, a + 8 = 5a + 2(1 + 2 . . . 4) = 5a + 2$$\frac{{\left( 4 \right)\left( 5 \right)}}{2}$$ = 5a + 20 According to the question, $$\frac{{5a + 20}}{5}$$ = 27 a + 4 = 27 a = 23 First number = 23 Last number = 23 + 8 = 31 Product of first and last number = 23 × 31 = 713 Alternate: \[\begin{array}{*{20}{c}} {23}&{25}&{\boxed{27}}&{29}&{31} \\ \downarrow &{}& \downarrow &{}& \downarrow \\ {{\text{First Number}}}&{}&{{\text{Average}}}&{}&{{\text{Last Number}}} \end{array}\] Average is middle term if common difference between them is same. Product = 23 × 31 = 713
13.
A passenger travels from Delhi to Merut at a speed of 30 kmph and return with a speed of 60 kmph. What is the average speed?
One-fourth of certain journey is covered at the rate of 25 km/h, one-third at the rate of 30 km/h and the rest at 50 km/h. Find the average speed for the whole journey.
(A) $$\frac{{600}}{{53}}$$ km/h
(B) $$\frac{{1200}}{{53}}$$ km/h
(C) $$\frac{{1800}}{{53}}$$ km/h
(D) $$\frac{{1600}}{{53}}$$ km/h
Solution:
Let distance be 120 km Hence 30 km is covered by @25 kmph and 40 km covered by @30 kmph and rest 50 km has been covered @50 kmph Now, $$\eqalign{ & {\text{average}} = {\frac{{120}}{{{\text{total}}\,{\text{time}}\,{\text{taken}}}}} \cr & = \frac{{120}}{{\frac{{30}}{{25}} + \frac{{40}}{{30}} + \frac{{50}}{{50}}}} \cr & = \frac{{3600}}{{106}} \cr & = \frac{{1800}}{{53}}\,{\text{km/h}} \cr} $$
15.
In a class with a certain number of students, if one student weighting 50 kg is added then the average weight of the class increased by 1 kg. If one more student weighting 50 kg is added, then the average weight of the class increased by 1.5 kg over the original average. What is the original average weight (in kg) of the class?
(A) 2
(B) 4
(C) 46
(D) 47
Solution:
Let the original average weight of the class be x kg and let there be n students. Then, sum of weights of n students = (nx) kg $$\eqalign{ & \therefore \frac{{nx + 50}}{{n + 1}} = x + 1 \cr & \Rightarrow nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 50 = nx + x + n + 1 \cr & \Rightarrow x + n = 49 \cr & \Rightarrow 2x + 2n = 98.....(i) \cr & {\text{And,}} \cr & \Rightarrow \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr & \Rightarrow nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr & \Rightarrow nx + 100 = nx + 1.5n + 2x + 3 \cr & \Rightarrow 2x + 1.5n = 97.....(ii) \cr} $$ Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2 Putting n = 2 in (i), we get: x = 47
16.
The mean of 9 observation is 16. One more observation is included and the new mean becomes 17. The 10th observation is :
(A) 9
(B) 16
(C) 26
(D) 30
Solution:
According to the question mean of 9 observations is = 16 Sum of all observations is = 16 × 9 = 144 When one more observation included the new mean = 17 Sum of 10 observations = 10 × 17 = 170 ∴ 10th observation = 170 - 144 = 26
17.
The average weight of 47 balls is 4 g. if the weight of the bag (in which the balls are kept) be included; the calculated average weight per ball increases by 0.3 g. What is the weight of the bag?
(A) 14.8 g
(B) 14.4 g
(C) 15 g
(D) 18.6 g
Solution:
Answer & Solution Answer: Option E Solution: Total increased weight = 0.3 × 47 = 14.1 g
18.
The average age of Ram and his two children is 17 years and the average age of Ram's wife and the same children is 16 years. If the age of Ram is 33 years, the age of his wife is (in years) :
(A) 31 years
(B) 32 years
(C) 35 years
(D) 30 years
Solution:
According to the question,$$ = \frac{{{\text{Ram + 2C}}}}{3}$$ = 17 yearsRam + 2C = 51 years.....(i) $$ = \frac{{{\text{Wife + 2C}}}}{3}$$ = 16 years Wife + 2C = 48 years.....(ii) If Ram age = 33 years old Put this value in equation (i) ∴ 2C = 51 - 33 2C = 18 years old.....(iii) Put this value in equation (ii) Wife + 18 = 48 Wife = 48 - 18 = 30 wife = 30 years old
19.
A student obtained the following marks in percentage in his semester examination English 50, Maths 65, Statistics 70, Economics 58 and Accountancy 63. The weights of these subjects are 2, 2, 1, 1 and 1 respectively. What is the weighted arithmetic mean?