Practice Average

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A batsman scores \[30, 40\] and \[50\] runs in three consecutive matches. What is his average score?

\[20\]

\[40\]

\[60\]

\[80\]

Average score of the batsman

\[= \frac{Total\ score}{Total\ matches}\]

\[=\frac{\left(30+40+50\right)}{3}\]

\[=\frac{120}{3}\]

\[=40\]

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What is the average of the first five even natural numbers?

\[10\]

\[8\]

\[6\]

\[4\]

Average of the first five even natural numbers

\[=\frac{\left(2+4+6+8+10\right)}{5}\]

\[=\frac{30}{5}\]

\[=6\]

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The average marks obtained by \[200\] students in a certain examination is \[45\]. Find the total marks.

\[300\]

\[500\]

\[700\]

\[900\]

Total marks \[=\] Average marks \[\times\] Number of students

\[=200\times45=900\]

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Total temperature for the month of September is \[840°C\]. If the average temperature of that month is \[28°C\], find of how many days is the month of September.

\[28\]

\[30\]

\[31\]

Can't be determined

Number of days in the month of September

\[=\frac{Total\ temperarure}{Average\ temperature}\]

\[=\frac{840}{28}\]

\[=30\] days

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The average of the marks obtained in an examination by \[8\] students was \[51\] and by \[9\] other students was \[68\]. The average marks of all \[17\] students was :

Sum of total number of \[8\] students in exam

\[= 8 × 51 = 408\]

Sum of total number of \[9\] students in exam

\[= 9 × 68 = 612\]

\[\therefore\] Required average

\[=\frac{\left(408+612\right)}{17}\]

\[=\frac{1020}{17}\]

\[=60\]

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If the average marks of three batches of \[55, 60\] and \[45\] students respectively is \[50, 55\] and \[60,\] then the average marks of all the students is

\[40.52\]

\[54.68\]

\[62.84\]

\[74.96\]

The required average marks

\[=\frac{\left(55\times50\right)+\left(60\times55\right)+\left(45\times60\right)}{55+60+45}\]

\[=\frac{2750+3300+2700}{160}\]

\[=\frac{8750}{160}\]

\[=54.68\]

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If the average weight of \[6\] students is \[50 kg\]; that of \[2\] students is \[51 kg\]; and that of other \[2\] students is \[55 kg\]; then the average weight of all students is

\[51.2\]

\[50.4\]

\[49.6\]

\[48.4\]

Required average weight

\[=\frac{\left(55\times6\right)+\left(51\times2\right)+\left(55\times2\right)}{10}\]

\[=\frac{300+102+110}{10}\]

\[=\frac{512}{10}\]

\[=51.2\]

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\[12 kg\] of rice costing \[Rs.30\] per \[kg\] is mixed with \[8 kg\] of rice costing \[Rs.40\] per \[kg\].The average per kg price of mixed rice is

\[Rs.28\]

\[Rs.30\]

\[Rs.32\]

\[Rs.34\]

Total cost price of \[20kg\] of mixed rice

\[=(12 × 30 + 8 × 40)\]

\[= Rs.680\]

\[\therefore\] Average per kg price

\[=\frac{680}{20}\]

\[= 34\]

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The average of \[30\] numbers is \[40\] and that of other \[40\] numbers is \[30\]. The average of all the numbers is

\[10\]

\[20\]

\[40\]

\[60\]

Average of all numbers

\[=\frac{\left(30\times40\right)+\left(40\times30\right)}{60}\]

\[=\frac{1200+1200}{60}\]

\[=\frac{2400}{60}\]

\[=40\]

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\[4\] boys and \[3\] girls spent \[Rs.120\] on the average, of which boys spent \[Rs.150\] on the average. Then the average amount spent by the girls is

\[Rs.50\]

\[Rs.60\]

\[Rs.70\]

\[Rs.80\]

Total expenditure

\[= 120 × 7\]

\[= Rs. 840\]

Total expenditure of \[4\] boys

\[= 150 × 4\]

\[= Rs.600\]

Total expenditure of \[3\] girls

\[= 840 – 600\]

\[= Rs.240\]

\[\therefore\] Their average expenditure

\[=\frac{240}{3}\]

\[=Rs.80\]

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The average of \[8\] numbers is \[27\]. If each of the numbers is multiplied by \[8\], find the average of new set of numbers.

\[182\]

\[186\]

\[212\]

\[216\]

If each number is multiplied by \[8\], their average gets multiplied by \[8\].

\[\therefore\] Required average

\[= 8 × 27\]

\[= 216\]

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The average of all the numbers between \[6\] and \[50\] which are divisible by \[5\] is

\[24.1\]

\[25.1\]

\[26.5\]

\[27.5\]

Numbers are :

\[10, 15, 20, 25, 30, 35, 40, 45\]

Sum \[= 220\]

Average \[=\frac{220}{8}\]

\[=27.5\]

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The average of \[1, 3, 5, 7, 9, 11 ......\] to \[25\] terms is

100

Sum of first \[n\] odd natural numbers

\[= n^2\]

Their average \[=\frac{n^2}{n}=n\]

\[\therefore\] Required average \[= 25\]

because \[n = 25\]

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The average of \[2, 4, 6, 8, 10, 12 ......\] to \[25\] terms is

\[22\]

\[24\]

\[26\]

\[28\]

Sum of first \[n\] even natural numbers

\[= n(n+1)\]

Their average \[=\frac{n(n+1)}{n}=n+1\]

\[\therefore\] Required average \[= 25+1=26\]

because \[n = 25\]

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The mean high temperature of the first four days of a week is \[25°C\] whereas the mean of the last four days is \[25.5°C\]. If the mean temperature of the whole week is \[25.2°C\] then the temperature on the \[4th\] day is

\[25.2°C\]

\[25.4°C\]

\[25.6°C\]

\[25.8°C\]

Temperature on \[4th\] day

\[= (4 × 25 + 4 × 25.5–25.2 × 7)°C\]

\[= (100 + 102– 176.4)°C\]

\[= 25.6°C\]

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Six friends have an average height of \[167 cms\]. A boy with height \[162 cm\] leaves the group. Find the new average height.

\[166 cm\]

\[167 cm\]

\[168 cm\]

\[169 cm\]

Total height of \[5\] friends

\[= (6 × 167 – 162) cm.\]

\[= (1002 – 162) cm.\]

\[= 840 cm.\]

\[\therefore\] Required average

\[=\frac{840}{5}\]

\[= 168 cm.\]

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The average weight of \[A, B\] and \[C\] is \[45 kg\]. If the average weight of \[A\] and \[B\] be \[40 kg\] and that of \[B\] and \[C\] be \[43 kg\], then the weight (in \[kg\]) of \[B\] is

\[27kg\]

\[28kg\]

\[29kg\]

\[31kg\]

Weight of \[B\]

\[= (A + B)’s\] weight \[ + (B + C)’s\] weight \[– (A + B + C)’s\] weight

\[= 40 × 2 + 43 × 2 – 45 × 3\]

\[= 80 + 86 – 135\]

\[= 166 – 135\]

\[= 31 kg.\]

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The average weight of \[8\] persons increases by \[2.5 kg\] when a new person comes in place of one of them weighing \[65 kg\]. The weight of the new person is

\[81kg\]

\[83kg\]

\[85kg\]

\[87kg\]

Weight of new person

\[= (65 + 8 × 2.5) kg\]

\[= (65 + 20) kg\]

\[= 85 kg\]

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The average of \[25\] results is \[20\]. The average of first \[12\] results is \[15\] and that of the last \[12\] results is \[18\]. Then, the \[13th\] result is

\[100\]

\[102\]

\[104\]

\[106\]

Value of \[13th\] result

\[= 25 × 20 – 12 × 15 + 12 × 18\]

\[= 500 – 180 – 216\]

\[= 104\]

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The average of \[20\] numbers is \[12\]. The average of the first \[12\] numbers is \[11\] and that of the next \[7\] numbers is \[10\]. The last number is :

\[32\]

\[34\]

\[36\]

\[38\]

Last number

\[=\] Sum of \[20\] numbers \[–\] sum of first \[12\] numbers \[–\] sum of next \[7\] numbers

\[= 20 × 12 – 11 × 12 – 7 × 10\]

\[= 240 – 132 – 70\]

\[= 38\]