Average — Part 1 MCQs with Explanations

Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings. In the 10 th inning, he scored 100 runs then average became (x+8). And he scored (x + 8) × 10 runs in 10 innings. Now, $$\eqalign{ & \Rightarrow 9x + 100 = 10 \times \left( {x + 8} \right) \cr & {\text{or}},\,9x + 100 = 10x + 80 \cr & {\text{or}},\,x = 100 - 80 \cr & {\text{or}},\,x = 20 \cr & {\text{New}}\,{\text{average}} = \left( {x + 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 28\,{\text{runs}} \cr} $$

Average temperature for Wednesday, Thursday and Friday = 40° C Total temperature = 3 × 40 = 120° C Average temperature for Thursday, Friday and Saturday = 41° C Total temperature = 41 × 3 = 123° C Temperature on Saturday = 42° C Now, (Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120; Saturday - Wednesday = 3 Wednesday = 42 - 3 = 39° C

$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{{\text{total}}\,{\text{sum}}\,{\text{of}}\,{\text{multiple}}\,{\text{of}}\,9}}{5}} \cr & = {\frac{{9 + 18 + 27 + 36 + 45}}{5}} \cr & = 27 \cr} $$ Note that, average of 9 and 45 is also 27. And average of 18 and 36 is also 27.

$$\eqalign{ & {\text{Average speed}}, \cr & = \frac{{ {2 \times x \times y} }}{{ {x + y} }} \cr & = \frac{{ {2 \times 100 \times 150} }}{{ {100 + 150} }} \cr & = \frac{{ {200 \times 150} }}{{250}} \cr & = 120\,{\text{km/hr}} \cr} $$

1 st Method: Average of 1 st n natural number is given by $$ = \frac{{ {\frac{{ {{\text{n}} \times \left( {{\text{n}} + 1} \right)} }}{2}} }}{{\text{n}}}$$ Average of 1 st 97 natural number is given by $$\eqalign{ & {\frac{{ {\frac{{ {97 \times \left( {97 + 1} \right)} }}{2}} }}{{97}}} \cr & = 49 \cr} $$ 2 st Method: These numbers are in AP series so average. $$\eqalign{ & = \frac{{ {{\text{sum of corresponding term}}} }}{2} \cr & = \frac{{ {1 + 97} }}{2} = 49 \cr & {\text{or, }}\frac{{ {2 + 96} }}{2} = 49 \cr & {\text{or, }}\frac{{ {3 + 95} }}{2} = 49 \cr & {\text{And so on}}{\text{.}} \cr} $$

Sum of ages of three boys = 45 years Now, (3x + 5x + 7x) = 45 ⇒ 15x = 45 ⇒ x = 3 So, age of youngest boy = 3x = 3 × 3 = 9 years

Let N be the no. of persons in the group. Required number of person is given by; Member in group × aged increased = difference of replacement N × 5 = 38 - 18 Or, 5N = 20 Or, N = 4

Member in group × age increased = difference of replacement Or, 8 × 1 = new comer - man going out Or, new comer = 8 + 60 Or, new comer = 68 kg.

Sum of 1 st 12 results = 12 × 14 Sum of last 12 results = 12 × 17 13 th result = x (let) Now, 12 × 14 + 12 × 17 + x = 25 × 18 Or, x = 78

$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{total}}\,{\text{time}}}} \cr & {\text{Total}}\,{\text{Distance}} \cr & = {16 + 20 + 10} \cr & = 46\,{\text{km}} \cr & {\text{Time}}\,{\text{taken}} \cr & = {\frac{{16}}{{20}}} + {\frac{{20}}{{40}}} + {\frac{{10}}{{15}}} \cr & = {\frac{{4}}{{5}}} + {\frac{{1}}{{2}}} + {\frac{{2}}{{3}}} \cr & = {\frac{{24 + 15 + 20}}{{30}}} \cr & = \frac{{59}}{{30}} \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{46 \times 30}}{{59}} \cr & = 23\frac{{23}}{{59}}\,{\text{km/hr}} \cr} $$

$$\eqalign{ & {\text{Total}}\,{\text{salary}}\,{\text{of}}\,660\,{\text{workers}} \cr & = 660 \times 380 \cr & = Rs.\,250800 \cr & {\text{If}}\,{\text{other}}\,{\text{workers}}\,{\text{be}}\,x;\,{\text{then}}, \cr & = \left[ {\left( {660 - x} \right) \times 2100} \right] + 340x \cr & = 250800 \cr & Or,\,1386000 - 2100x + 340x = 250800 \cr & 1760x = 1135200 \cr & {\text{Hence}},\,x = \frac{{1135200}}{{1760}} = 645 \cr & {\text{Number}}\,{\text{of}}\,{\text{other}}\,{\text{workers}} = 645 \cr} $$

$$\eqalign{ & {\text{Correct average}} \cr & = 120 + \left( {\frac{{33 - 13}}{4}} \right) \cr & = 120 + 5 \cr & = 125 \cr} $$ Solve while reading method: Average given is 120. Difference of 33 and 13 is 20. That means 20 must be added to total. Then average of is and so must be added to average, i.e. Correct average = 120 + 5 = 125

$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,{\text{pass}}\,{\text{student}}\,{\text{}}\,{\text{first,}}\,{\text{second}}\,{\text{and}}\,{\text{third}}\,{\text{class}}, \cr & = \left( {70\% \,{\text{of}}\,10} \right) + \left( {60\% \,{\text{of}}\,15} \right) + \left( {80\% \,{\text{of}}\,25} \right) \cr & = 7 + 9 + 20 = 36 \cr & {\text{Total}}\,{\text{students}}\,{\text{appeared}}, \cr & = 10 + 15 + 25 = 50 \cr & {\text{Pass}}\,{\text{average}}, \cr & = 36 \times \frac{{100}}{{50}} = 72\% \cr} $$

Required average speed, $$ = {\frac{{\left( {3 \times 40 \times 30 \times 15} \right)}}{{ {\left( {40 \times 30} \right) + \left( {40 \times 15} \right) + \left( {30 \times 15} \right)} }}} $$ $$ = 24\,{\text{km/hr}}$$ Alternatively Time taken to traveled $$\frac{1}{3}$$ distance of journey with speed 40 kmph, $$ = \frac{{ {\frac{1}{3}} }}{{40}} = \frac{1}{{120}}$$ Time taken to traveled $$\frac{1}{3}$$ distance of journey with speed 30 kmph, $$ = \frac{{ {\frac{1}{3}} }}{{30}} = \frac{1}{{90}}$$ Time taken to traveled $$\frac{1}{3}$$ distance of journey with speed 15 kmph, $$\eqalign{ & = \frac{{ {\frac{1}{3}} }}{{15}} = \frac{1}{{45}} \cr & {\text{Total time taken}} \cr & = {\frac{1}{{120}}} + {\frac{1}{{90}}} + {\frac{1}{{45}}} \cr & = \frac{{45}}{{1080}} \cr & {\text{Average speed}} \cr & = \frac{{{\text{Total distance traveled}}}}{{{\text{Total time taken}}}} \cr & = \frac{1}{{ {\frac{{45}}{{1080}}} }} \cr & = 24\,{\text{kmph}} \cr} $$

Total present age of A, B, C and D = (45 × 4) + (4 × 5) = 200 years Total age present age of A, B, C, D and E = 49 × 5 = 245 years So, Age of E = 45 years

Total of 80 numbers = 80 × 42 = 3360 Now, total of 85 numbers = 85 × 45 = 3825 Hence, sum of 5 numbers = 3825 - 3360 = 465 Average of five numbers = $$\frac{{465}}{5}$$ = 93 Alternatively, Solve while reading method: Average of 80 number was 42 When 5 more numbers are added average become 45 that means 3 is given to each number to make their average. 45 and 5 numbers also keep 45 as to maintain the entire average. So, the sum of five numbers = 240 + 225 = 465 Hence, average of five numbers = 93

Let 100 be the original population. 1 st year's population increased = 30% So, Population after first year = (100 + 30% of 100) = 130 Population in second year increases by 40%, Then Population = (130 + 40% of 130) = 182 The final population become 182 which was originally at 100. It means there is 82% increment in the population in 2 years So, Average increment = $$\frac{{82}}{2}$$ = 41% Mind Calculation Method: Increase in population is given by, 100 == 30% $$ \uparrow $$ ⇒ 130 == 40% $$ \uparrow $$ ⇒ 182 Hence, average increase = $$\frac{{82}}{2}$$ = 41%

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, $$\eqalign{ & {\text{average}} = {\frac{{120}}{{{\text{total}}\,{\text{time}}\,{\text{taken}}}}} \cr & = \frac{{120}}{{\frac{{30}}{{25}} + \frac{{40}}{{30}} + \frac{{50}}{{50}}}} \cr & = \frac{{3600}}{{106}} \cr & = \frac{{1800}}{{53}}\,{\text{km/h}} \cr} $$

Part of the runs scored in the 87 th innings will go towards increasing the average of the first 86 innings to the new average and remaining part of the runs will go towards maintaining the new average for the 87 th innings. The only constraint in this problem is that there is increase in average by a whole number of runs. This is possible for all three options.

Step 1: Find the weight of the third box. The third box is 25% heavier than the first box (200 kg). 25% of 200 kg = (25/100) * 200 kg = 50 kg Weight of the third box = 200 kg + 50 kg = 250 kg Step 2: Find the weight of the second box. The second box is 20% heavier than the third box (250 kg). 20% of 250 kg = (20/100) * 250 kg = 50 kg Weight of the second box = 250 kg + 50 kg = 300 kg Step 3: Find the weight of the fifth box. The fourth box (350 kg) is 30% lighter than the fifth box. This means the fourth box is 70% of the fifth box's weight. Let the weight of the fifth box be x. 0.7x = 350 kg x = 350 kg / 0.7 = 500 kg Step 4: Arrange the weights in ascending order. The weights are: 200 kg, 250 kg, 300 kg, 350 kg, 500 kg Step 5: Calculate the average weight of the four heaviest boxes. Four heaviest boxes: 250 kg, 300 kg, 350 kg, 500 kg Total weight = 250 + 300 + 350 + 500 = 1400 kg Average weight = 1400 kg / 4 = 350 kg Step 6: Calculate the average weight of the four lightest boxes. Four lightest boxes: 200 kg, 250 kg, 300 kg, 350 kg Total weight = 200 + 250 + 300 + 350 = 1100 kg Average weight = 1100 kg / 4 = 275 kg Step 7: Find the difference in the averages. Difference = 350 kg - 275 kg = 75 kg Therefore, the correct option is B: 75 kg

The number of pass candidates are 2 + 6 + 18 + 40 = 66 out of total 100. Hence, Pass percentage = 66%

Let average expenditure of 19 people be x then, 19x = 13 × 79 + 6 × (x + 4) or, 19x = 13 × 79 + 6x + 24 or, x = 80.84 So, total money spent = 80.84 × 19 = Rs. 1536.07

Total increased weight = 0.3 × 47 = 14.1 g

Let he scored x runs in final innings Now, 37 × 20 + x = 40 × 21 x = 840 - 740 x = 100

$$\eqalign{ & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{first}}\,{\text{three}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{3000}}{{100}}} \right)}^2} + 10} \right] \times 3 \cr & = Rs.\,2710 \cr & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{next}}\,{\text{five}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{5000}}{{100}}} \right)}^2} + 10} \right] \times 5 \cr & = Rs.\,12550 \cr & {\text{Bonus}}\,{\text{for}}\,{\text{the}}\,{\text{next}}\,{\text{four}}\,{\text{month}}, \cr & = \left[ {{{\left( {\frac{{8000}}{{100}}} \right)}^2} + 10} \right] \times 4 \cr & = Rs.\,25640 \cr & {\text{Total}}\,{\text{earning}}\,{\text{as}}\,{\text{bonus}}\,{\text{for}}\,{\text{whole}}\,{\text{year}}, \cr & = 2710 + 12550 + 25640 \cr & = Rs.\,40900 \cr & {\text{His}}\,{\text{average}}\,{\text{bonus}} \cr & = \frac{{40900}}{{12}} = Rs.\,3410 \cr & {\text{Thus}}\,{\text{his}}\,{\text{average}}\,{\text{earning}}\,{\text{for}}\,{\text{whole}}\,{\text{year}}, \cr & = 3410 + 200 \cr & = Rs.\,3610 \cr} $$

The weight is increasing in form of GP so the total weight of eggs in the end of r th will be n r k.

$$\eqalign{ & {\text{Average}}\,{\text{speed}}, \cr & = {\frac{{{\text{Total}}\,{\text{Distance}}}}{{{\text{Total}}\,{\text{time}}}}} \cr & = {\frac{{ {200 + 300 + 500} }}{{ { {\frac{{200}}{{40}}} + {\frac{{300}}{{20}}} + {\frac{{500}}{{10}}} } }}} \cr & = \frac{{1000}}{{70}} \cr & = 14{\frac{2}{7}}\, \text{kmph} \cr} $$

A + B + C + D + E = 40 × 5 = 200 A + B = 35 × 2 = 70 C + D = 42 × 2 = 84 Therefore, E = (A + B + C + D + E) - (A + B + C + D) E = 200 - 70 - 84 E = 46 years

Let number of boys are x and then number of girls = (100 - x) Thus, 50x + (100 - x) × 40 = 46 × 100 → x = 60 Number of boys = 60

Let the number of students who got the job of A, B and C categories is a, b and c respectively, Then the total salary, $$ = {\frac{{\left\{ {26\left( {a + b} \right) + 44\left( {b + c} \right) + 34\left( {c + a} \right)} \right\}}}{{2\left( {a + b + c} \right)}}} $$ $$\eqalign{ & = {\frac{{\left( {60a + 70b + 78c} \right)}}{{2\left( {a + b + c} \right)}}} \cr & = \frac{{\left[ {30\left( {a + b + c} \right) + \left( {5b + 9c} \right)} \right]}}{{\left( {a + b + c} \right)}} \cr} $$ = 30 + some positive value Thus, the minimum salary must be Rs. 30 lakh and the maximum salary can not exceed 44, which is the highest of the three

$$\eqalign{ & \frac{{\left[ {\left( {10 \times 50} \right) + \left( {50 \times 10} \right) + \left( {20 \times 25} \right) + \left( {25 \times 20} \right)} \right]}}{4} \cr & = 500\,{\text{paise}} \cr} $$

$$\eqalign{ & A + B + C = 12000 \times 3 \cr & B + C + D = 15000 \times 3 \cr & \to D - A = 3000 \times 3 = 9000 \cr & {\text{Also}},\,D = 2A \cr & {\text{Then}},\,D = 18000\,{\text{and}}\,A = 9000 \cr & {\text{Therefore}}, \cr & {\text{Average}}\,{\text{salary}}\,{\text{of}}\,B\,{\text{and}}\,C, \cr & = \frac{{ {45000 - 18000} }}{2} \cr & = 13500 \cr} $$

Average Speed $$\eqalign{ & = \frac{{2 \times 60 \times 30}}{{60 + 30}} \cr & = \frac{{3600}}{{90}} \cr & = 40\,{\text{kmph}} \cr} $$

Average of 20 students = 12 years Total age of 20 students = 20 × 12 = 240 years When teacher included average become 13 years Now, total age 20 students and teacher = 13 × 21 = 273 years ∴ Age of teacher = 273 - 240 = 33 years

The total wages earned during the 15 days that the worker worked , = 15 × 90 = Rs. 1350 The total wages earned during the first 7 days = 7 × 87 = Rs. 609 The total wages earned during the last 7 days = 7 × 92 = Rs. 644 Total wages earned during the 15 days, = wages during first 7 days + wage on 8 th day + wages during the last 7 days. 1350 = 609 + wage on 8 th day + 644 wage on 8 th day = 1350 - 609 - 644 = Rs. 97

Let X be the number of employees. We are given that 40% of the employees are workers. Now, 40% of X is $$\frac{{40}}{{100}}$$ × X = 0.4X Hence, the number of workers is $$\frac{{2{\text{X}}}}{5}$$ All the remaining employees are executives, so the number of executives equals, (The number of Employees) $$ - $$ (The number of Workers) = X $$ - $$ $$\frac{{2{\text{X}}}}{5}$$ = $$\frac{{3{\text{X}}}}{5}$$ The annual income of each worker is Rs. 390. Hence, the total annual income of all workers together is = $$\frac{{2{\text{X}}}}{5}$$ × 390 = 156X Also, the annual income of each executive is Rs.420. Hence, the total income {of all the executives together is, = $$\frac{{3{\text{X}}}}{5}$$ × 420 = 252X Hence, the total income of the employees is, = 156X + 252X = 408X The average income of all the employees together = 408

Total cost of 10 books = Rs. 120 Total cost of 8 books = Rs. 94 ⇒ The cost of 2 books = Rs. 26 Let the price of each book be x and y. ⇒ x + y = 26 - - - - - - (1) $$\left( {\frac{{160}}{{100}}} \right)x + y = 26$$ On Solving for y, we get y = 10 Now, Substituting y = 10 in (1) we get, x + 10 = 26 x = 16 So the price of each book is Rs. 16 and Rs. 10 respectively.

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{three}}\,{\text{numbers}}\,{\text{be}}\,x,\,y,\,{\text{and}}\,z.\, \cr & {\text{We}}\,{\text{are}}\,{\text{given}}\,{\text{that}}\, \cr & \frac{{ {x + y} }}{2} = 2 \cr & \frac{{ {y + z} }}{2} = 3 \cr & \frac{{ {x + z} }}{2} = 4 \cr & {\text{Adding}}\,{\text{three}}\,{\text{equations}}, \cr & \frac{{ {x + y} }}{2} + \frac{{ {y + z} }}{2} + \frac{{ {x + z} }}{2} = 2 + 3 + 4 \cr & x + y + z = 9 \cr & {\text{The}}\,{\text{average}}\,{\text{of}}\,{\text{the}}\,{\text{three}}\,{\text{numbers}}\,{\text{is}}, \cr & \frac{{ {x + y + z} }}{3} = \frac{9}{3} = 3 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{three}}\,{\text{numbers}}\,{\text{be}}\,x,\,y,\,z. \cr & {\text{Given}}, \cr & x = 2y \cr & \Rightarrow x = 4z \cr & \Rightarrow y = 2z \cr & \Rightarrow z = z \cr & {\text{The}}\,{\text{average}}\,{\text{of}}\,{\text{reciprocal}}\,{\text{numbers}}\,{\text{is}}\,\frac{7}{{72}} \cr & \frac{{ { {\frac{1}{x}} + {\frac{1}{y}} + {\frac{1}{z}} } }}{3} = \frac{7}{{72}} \cr & \Rightarrow \frac{{ {yz + xz + xy} }}{{3xyz}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{2z \times z + 4z \times z + 4z \times 2z}}{{3\left( {4z \times 2z \times z} \right)}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{2{z^2} + 4{z^2} + 8{z^2}}}{{3 \times 8{z^3}}} = \frac{7}{{72}} \cr & \Rightarrow \frac{{14{z^2}}}{{24{z^3}}} = \frac{7}{{72}} \cr & \Rightarrow 504 = 84z \cr & z = 6 \cr & {\text{So}},\,x = 4z = 4 \times 6 = 24, \cr & \Rightarrow y = 2z = 2 \times 6 = 12 \cr & {\text{Thus}}\,{\text{the}}\,{\text{numbers}}\,{\text{are}}\,24,\,12,\,6 \cr} $$

The total salary amount = 15 × 600 = 9000 The salary of the exiting manager = 720 Therefore, the salary of 12 workers and the remaining 2 managers, = 9000 − 720 = 8280 When a new manager joins, the new average salary drops to Rs. 580 for the total team of 15 of them The total salary for the 15 people i.e., 12 workers, 2 old managers and 1 new manager =580 × 15 = 8700 Therefore, the salary of the new manager is, 9000 - 8700 = 300 less than that of the old manager who left the company, which is equal to 720 - 300 = 420 Alternatively , The average salary dropped by Rs. 20 for 15 of them. Therefore, the overall salary has dropped by 15 × 20=300 Therefore, the new manager's salary should be Rs. 300 less than that of the old manager, = 720 - 300 = 420