Average — Part 2 MCQs with Explanations

Let a, b, c, d, and e be the five positive numbers in the decreasing order of size such that e is the smallest number. We are given that the average of the five numbers is 25. Hence, we have the equation $$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + {\text{e}}}}{5} = 25$$ a + b + c + d + e = 125 ----------- (1) by multiplying by 5. The smallest number in a set is at least less than the average of the numbers in the set if at least one number is different. For example, the average of 1, 2, and 3 is 2, and the smallest number in the set 1 is less than the average 2. Hence, we have the inequality 0 < e < 25 0 > -e > -25 by multiplying both sides of the inequality by -1 and flipping the directions of the inequalities. Adding this inequality to equation (1) yields 0 + 125 > (a + b + c + d + e) + (-e) > 125 - 25 125 > (a + b + c + d) > 100 125 > (a + b + c + d + 0) > 100 by adding by 0 25 > $$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + 0}}{5}$$ ⇒ 20 by dividing the inequality by 5 25 > The average of numbers a, b, c, d and 0 > 20 Hence, x equals (Average of the numbers a, b, c, d and e) – (Average of the numbers a, b, c, and d) = 25 − (A number between 20 and 25) ⇒ A number less than 5 Hence, x is less than 5

Let a, b, and c be the annual incomes of Ramesh, Suresh, and Pratap, respectively. Now, we are given that The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. Hence, $$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 3800 ⇒ a + b = 2 × 3800 = 7600 The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800. Hence, $$\frac{{{\text{b}} + {\text{c}}}}{2}$$ = 4800 ⇒ b + c = 2 × 4800 = 9600 The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. Hence, $$\frac{{{\text{c}} + {\text{a}}}}{2}$$ = 5800 ⇒ c + a = 2 × 5800 = 11,600 Adding these three equations yields: (a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600 2a + 2b + 2c = 28,800 a + b + c = 14,400 The average of the incomes of the three equals the sum of the incomes divided by 3, $$\eqalign{ & \frac{{{\text{a}} + {\text{b}} + {\text{c}}}}{3} \cr & = \frac{{14,400}}{3} \cr & = {\text{Rs}}{\text{.}}\,4800 \cr} $$

Let Arun's weight by X kg. According to Arun, 65 < X < 72 According to Arun's brother, 60 < X < 70 According to Arun's mother, X $$ \leqslant $$ 68 The values satisfying all the above conditions are 66, 67 and 68. Required average, $$\eqalign{ & = \frac{{66 + 67 + 68}}{3} \cr & = \frac{{201}}{3} \cr & = 67\,{\text{kg}}{\text{.}} \cr} $$

Let the original number be (10a + b) After interchanging the digits, the new number becomes (10b + a) The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be (10 × 1.8 = 18 more than the sum of the 10 numbers with the digits interchanged. 10a + b = 10b + a + 18 9a - 9b = 18 a - b = 2

First five multiples of three are: 3, 6, 9, 12, 15. $$\eqalign{ & {\text{Average}} = \frac{{ {3 + 6 + 9 + 12 + 15} }}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{45}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9 \cr} $$ Alternatively , Average = $$\frac{{ {3 + 15} }}{2} = 9$$

The total weight of (36 + 44) Students of A and B, = (36 × 40 + 44 × 35) kg = 2980 kg. Therefore average weight of the whole class = $$\frac{{2980}}{{80}}$$ kg Therefore average weight = 37.25 kg

The required average speed given by the formula, $$\eqalign{ & {\text{Average speed}} \cr & = {\frac{{2xy}}{{ {x + y} }}} \,\,{\text{km/hr}} \cr & {\text{Where}}, \cr & x = 84{\kern 1pt} \,{\text{kmph}} \cr & y = 56\,{\kern 1pt} {\text{kmph}} \cr & {\text{Average speed}} \cr & = {\frac{{ {2 \times 84 \times 56} }}{{ {84 + 56} }}} \cr & = 67.2\,{\kern 1pt} {\text{kmph}} \cr} $$

Total sum of 48 numbers, = (50 × 30) – (35 +40) = 1500 – 75 = 1425 Average = $$\frac{{1425}}{{48}}$$ = 29.68

Total sum of last 4 matches, = (10 × 38.9) - (6 × 42) = 389 - 252 = 137 Average = $$\frac{{137}}{4}$$ = 34.25

Let the average after 17 th innings = x Then average after 16 th innings = (x - 3) Therefore 16 × (x - 3) + 87 = 17x Therefore x = 39

Let the original average expenditure be Rs. x then, 42(x - 1) - 35x = 42 ⇒ 7x = 84 ⇒ x = 12 Therefore original expenditure, = Rs. (35 × 12) = Rs. 420

Let the average expenditure of all the nine be x Then, 12 × 8 + (x + 8) = 9x Therefore, x = 13 Total money spent, = 9x = Rs. (9 × 13) = Rs. 117

$$\eqalign{ & {\text{Average}} \cr & = \frac{{ {76 + 65 + 82 + 67 + 85} }}{5} \cr & = \frac{{375}}{5} \cr & = 75 \cr & {\text{Hence,}}\,{\text{average}} = 75 \cr} $$

Average after 11 innings = 36 Required number of runs, = ( 36 × 11) - (32 × 10) = 396 - 320 = 76

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°

Let the average weight of the 59 students be X Therefore, the total weight of the 59 of them will be 59X The questions states that when the weight of this student who left is added, the total weight of the class = 59X + 45 When this student is also included, the average weight decreases by 0.2 kgs 59X + $$\frac{{45}}{{60}}$$ = X - 0.2 => 59X + 45 = 60X - 12 => 45 + 12 = 60X - 59X => X = 57

Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24°. ⇒ a – b = 24 Since the average of the two angles is 54°, we have $$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 54 Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields, $$ {\frac{{\left\{ {{\text{a}} + \left( {{\text{a}} - 24} \right)} \right\}}}{2}} = 54$$ 2a − 24 = 54 × 2 2a − 24 = 108 2a = 108 + 24 2a = 132 a = 66 Also, b = a − 24 = 66 − 24 = 42 Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is 180°, a + b + c = 180° Putting the previous results into the equation yields 66 + 42 + c = 180° Solving for c yields c = 72° Hence, the greatest of the three angles a, b and c is c, which equal.

At present the total age of the family = 5 × 20 =100 The total age of the family at the time of the birth of the youngest member, = 100 - 10 - (10 × 4) = 50 Therefore, average age of the family at the time of birth of the youngest member, $$ = \frac{{50}}{4} = 12.5$$

The average of the elements in the original set S is: $$\eqalign{ & \frac{{0 + 2 + 4 + 5 + 9}}{5} \cr & = \frac{{20}}{5} \cr & = 4 \cr} $$ If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}. The average of the elements is, $$\eqalign{ & \frac{{0 + 2 + 5 + 9}}{4} \cr & = \frac{{16}}{4} \cr & = 4 \cr} $$

Average cost of 5 apples and 4 mangoes = Rs. 36 Total cost = 36 × 9 = 324 Average cost of 7 apples and 8 mangoes = 48 Total cost = 48 × 15 = 720 Total cost of 12 apples and 12 mangoes = 324 + 720 = 1044 Therefore, cost of 24 apples and 24 mangoes = 1044 × 2 = 2088

In this question, sum of numbers is provided, net required sum (i.e. A + B+ C+ D + E) cannot be calculated by the given data. Therefore the answer is Data inadequate.

Let 10 numbers be x1, x2, x3, . . . . . . . x10 According to question average of these 10 numbers is 10 $$ \Rightarrow \frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}} = x$$ Now if each number is increased by 10%, then new average, say y. $$y = \frac{{ {1.1x1 + 1.1x2 + 1.1x3 + .... + 1.1x10} }}{{10}}$$ $${\kern 1pt} \Rightarrow y = 1.1 \times {\frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}}} $$ $$\eqalign{ & \Rightarrow y = 1.1x \cr & \Rightarrow y\,{\text{is }}10\% \,{\text{increased}} \cr} $$

Let age of the baby is x. 3 years ago total age of the family = 5 × 17 = 85 years Total age of the 5 member at present time = 85 + 3 × 5 = 100 years Total age of the family at present time including baby, = 100 + X The average of the family including baby at present time, = 17 years $$\frac{{100 + {\text{x}}}}{6} = 17$$ 100 + X = 102 X = 102 - 100 = 2 years

Let the rest workers = x Now, According to question, (7 + x) × 8000 = 12000 × 7 + 6000x 56000 + 8000x = 84000 + 6000x 2000x = 28000 x = 14 So total number of worker = 14 + 7 = 21

$$\eqalign{ & {\text{Required run rate}} \cr & = {\frac{{282 - \left( {3.2 \times 10} \right)}}{{40}}} \cr & = \frac{{250}}{{40}} \cr & = 6.25 \cr} $$

$$\eqalign{ & {\text{Required average}} \cr & {\text{ = }} {\frac{{67 \times 2 + 35 \times 2 + 6 \times 3}}{{2 + 2 + 3}}} \cr & = {\frac{{134 + 70 + 18}}{7}} \cr & = \frac{{222}}{7} \cr & = 31\frac{5}{7}{\text{years}} \cr} $$

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009 Therefore Required sale = Rs. [(6500 × 6) - 34009] = Rs. (39000 - 34009) = Rs. 4991

Average of 20 numbers = 0 ∴ Sum of 20 numbers (0 x 20) = 0 It is quite possible that 19 of these numbers may be positive and if their sum is a then 20 th number is (-a)

Total weight increased = (8 x 2.5) kg = 20 kg. Weight of new person = (65 + 20) kg = 85 kg.

Let the average age of the whole team by x years ∴ 11x - (26 + 29) = 9(x - 1) ⇒ 11x - 9x = 46 ⇒ 2x = 46 ⇒ x = 23 So, average age of the team is 23 years

Let P, Q and R represent their respective monthly incomes. Then, we have: P + Q = (5050 x 2) = 10100 .... (i) Q + R = (6250 x 2) = 12500 .... (ii) P + R = (5200 x 2) = 10400 .... (iii) Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv) Subtracting (ii) from (iv), we get P = 4000 Therefore P's monthly income = Rs. 4000

Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years ∴ Husband's present age = (90 - 50) years = 40 years

Total quantity of petrol consumed in 3 years $$\eqalign{ & = \left( {\frac{{4000}}{{7.50}} + \frac{{4000}}{8} + \frac{{4000}}{{8.50}}} \right){\text{litres}} \cr & = 4000\left( {\frac{2}{{15}} + \frac{1}{8} + \frac{2}{{17}}} \right){\text{litres}} \cr & = {\frac{{76700}}{{51}}} {\text{ litres}} \cr & {\text{Total}}\,{\text{amount}}\,{\text{spent}} \cr & = Rs.\,\left( {3 \times 4000} \right) \cr & = Rs.\,12000 \cr & \therefore {\text{Average}}\,{\text{Cost}} \cr & = Rs.\, {\frac{{12000 \times 51}}{{76700}}} \cr & = Rs.\,\frac{{6120}}{{767}} \cr & = Rs.\,7.98 \cr} $$

Let Arun's weight by X kg. According to Arun, 65 < X < 72 According to Arun's brother, 60 < X < 70 According to Arun's mother, X < 68 The values satisfying all the above conditions are 66, 67 and 68 $$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{66 + 67 + 68}}{3}} \cr & = {\frac{{201}}{3}} \cr & = 67\,kg. \cr} $$

Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 x 3) = 135 .... (i) A + B = (40 x 2) = 80 .... (ii) B + C = (43 x 2) = 86 ....(iii) Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv) Subtracting (i) from (iv), we get : B = 31 Therefore B's weight = 31 kg.

$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{50.25 \times 16 + 45.15 \times 8}}{{16 + 8}}} \cr & = {\frac{{804 + 361.20}}{{24}}} \cr & = \frac{{1165.20}}{{24}} \cr & = 48.55 {\text{ kg.}} \cr} $$

Since the month begins with a Sunday, to there will be five Sundays in the month. $$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$

$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{55 \times 50 + 60 \times 55 + 45 \times 60}}{{55 + 60 + 45}}} \cr & = {\frac{{2750 + 3300 + 2700}}{{160}}} \cr & = \frac{{8750}}{{160}} \cr & = 54.68 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{there}}\,{\text{be}}\,x\,{\text{pupils}}\,{\text{in}}\,{\text{the}}\,{\text{class}}{\text{.}} \cr & {\text{Total}}\,{\text{increase}}\,{\text{in}}\,{\text{marks}} \cr & = {x \times \frac{1}{2}} = \frac{x}{2} \cr & \therefore \frac{x}{2} = \left( {83 - 63} \right) \cr & \Rightarrow \frac{x}{2} = 20 \cr & \Rightarrow x = 40 \cr} $$

Total marks of six boys = 48 + 59 + 87 + 37 + 78 + 57 = 366 Required Average = $$\frac{{366}}{6}$$ = 61