The key fact about aver­ages is short: sum = num­ber of obser­va­tions × aver­age. Almost every aver­age prob­lem is solved by turn­ing the aver­ages into sums, adding or tak­ing away those sums, and then divid­ing again if needed. This les­son works through the com­mon kinds of prob­lem: ages, parts of a list, vis­i­tors in a month, cricket scores and run rates, and one per­son replac­ing another.

Ages

Exam­ple 1: Before the youngest was born

The aver­age age of a fam­ily of 6 mem­bers is 22 years. The youngest mem­ber is 7 years old. What was the aver­age age of the fam­ily when the youngest mem­ber was born?

Sum of the 6 ages now =6×22=132= 6 \times 22 = 132 years.

The youngest is 7, so the other 5 mem­bers add up to 132−7=125132 - 7 = 125 years.

When the youngest was born, 7 years ago, each of those 5 was 7 years younger. Their sum then was

125−7×5=125−35=90125 - 7 \times 5 = 125 - 35 = 90

At that time the fam­ily had only 5 mem­bers. So the aver­age age then was

905=18 years\displaystyle \frac{90}{5} = 18 \text{ years}

Parts of a list

Exam­ple 2: The miss­ing mid­dle num­ber

The aver­age of 9 num­bers is 30. The aver­age of the first 5 num­bers is 25, and the aver­age of the last 3 num­bers is 35. What is the 6th num­ber?

x1+x2+⋯+x9=9×30=270x_1 + x_2 + \cdots + x_9 = 9 \times 30 = 270

x1+x2+x3+x4+x5=5×25=125x_1 + x_2 + x_3 + x_4 + x_5 = 5 \times 25 = 125

x7+x8+x9=3×35=105x_7 + x_8 + x_9 = 3 \times 35 = 105

The first 5, the 6th, and the last 3 make up all 9:

125+x6+105=270x6=270−230=40\begin{aligned} 125 + x_6 + 105 &= 270 \\ x_6 &= 270 - 230 = 40 \end{aligned}

The 6th num­ber is 40.

Exam­ple 3: The rest of the group

The aver­age age of 30 girls is 13 years. The aver­age age of the first 18 girls is 15 years. Find the aver­age age of the remain­ing 12 girls.

Sum of all 30 ages =30×13=390= 30 \times 13 = 390.

Sum of the first 18 ages =18×15=270= 18 \times 15 = 270.

Sum of the other 12 ages =390−270=120= 390 - 270 = 120.

average of the 12=12012=10 years\displaystyle \text{average of the 12} = \frac{120}{12} = 10 \text{ years}

Exam­ple 4: The last two num­bers

The sum of five num­bers is 555. The aver­age of the first two is 75 and the third num­ber is 115. Find the aver­age of the last two num­bers.

x1+x2=2×75=150x_1 + x_2 = 2 \times 75 = 150

150+115+x4+x5=555x4+x5=555−265=290\begin{aligned} 150 + 115 + x_4 + x_5 &= 555 \\ x_4 + x_5 &= 555 - 265 = 290 \end{aligned}

x4+x52=2902=145\displaystyle \frac{x_4 + x_5}{2} = \frac{290}{2} = 145

Exam­ple 5: A num­ber counted twice

The aver­age of 9 obser­va­tions is 9. The aver­age of the first 5 is 10, and the aver­age of the last 5 is 8. Find the 5th obser­va­tion.

x1+x2+⋯+x9=9×9=81(1)x_1 + x_2 + \cdots + x_9 = 9 \times 9 = 81 \quad (1)

x1+x2+x3+x4+x5=5×10=50(2)x_1 + x_2 + x_3 + x_4 + x_5 = 5 \times 10 = 50 \quad (2)

x5+x6+x7+x8+x9=5×8=40(3)x_5 + x_6 + x_7 + x_8 + x_9 = 5 \times 8 = 40 \quad (3)

The "first 5" and the "last 5" over­lap: both con­tain x5x_5. Add (2) and (3). You get all nine obser­va­tions, with x5x_5 counted twice:

(x1+x2+⋯+x9)+x5=50+40=9081+x5=90x5=9\begin{aligned} (x_1 + x_2 + \cdots + x_9) + x_5 &= 50 + 40 = 90 \\ 81 + x_5 &= 90 \\ x_5 &= 9 \end{aligned}

Vis­i­tors in a month

Exam­ple 6

A library has an aver­age of 510 vis­i­tors on Sun­days and 240 on other days. Find the aver­age num­ber of vis­i­tors per day in a month of 30 days that begins with a Sun­day.

The month starts on a Sun­day, so the Sun­days fall on days 1, 8, 15, 22 and 29. That is 5 Sun­days, and 30−5=2530 - 5 = 25 other days.

Vis­i­tors on Sun­days =5×510=2550= 5 \times 510 = 2550.

Vis­i­tors on other days =25×240=6000= 25 \times 240 = 6000.

average per day=2550+600030=855030=285\displaystyle \text{average per day} = \frac{2550 + 6000}{30} = \frac{8550}{30} = 285

Cricket scores

Exam­ple 7: Rais­ing the aver­age

A bats­man's aver­age over 10 innings is 32. How many runs must he score in the next innings to raise his aver­age by 4?

Runs in 10 innings =10×32=320= 10 \times 32 = 320.

After 11 innings he wants an aver­age of 32+4=3632 + 4 = 36, so a total of 11×36=39611 \times 36 = 396.

runs needed=396−320=76\text{runs needed} = 396 - 320 = 76

Exam­ple 8: The 16th innings

A bats­man scores 64 runs in his 16th innings, and this raises his aver­age by 3. Find his aver­age after the 16th innings.

Let his aver­age after 15 innings be PP. Then his total after 15 innings is 15P15P.

After the 16th innings the total is 15P+6415P + 64, and the aver­age is P+3P + 3:

15P+6416=P+315P+64=16P+4864−48=16P−15PP=16\displaystyle \begin{aligned} \frac{15P + 64}{16} &= P + 3 \\ 15P + 64 &= 16P + 48 \\ 64 - 48 &= 16P - 15P \\ P &= 16 \end{aligned}

His aver­age after the 16th innings is 16+3=1916 + 3 = 19.

Exam­ple 9: Run rate

In the first 10 overs of a cricket match, the run rate was only 3.2. What run rate is needed in the remain­ing 40 overs to reach a tar­get of 282 runs?

The run rate is the aver­age num­ber of runs per over.

Runs in the first 10 overs =3.2×10=32= 3.2 \times 10 = 32.

Runs still needed =282−32=250= 282 - 32 = 250.

run rate needed=25040=6.25\displaystyle \text{run rate needed} = \frac{250}{40} = 6.25

One per­son replaced by another

Exam­ple 10

The aver­age weight of 8 peo­ple goes up by 2.5 kg when a new per­son comes in place of one of them who weighs 65 kg. What is the weight of the new per­son?

The total weight of the 8 peo­ple goes up by 8×2.5=208 \times 2.5 = 20 kg.

The only change is that 65 kg went out and the new per­son came in. So the new per­son weighs 20 kg more than 65 kg:

N=65+20=85 kgN = 65 + 20 = 85 \text{ kg}

With let­ters: if the old aver­age is PP, the old total is 8P8P. The new total is 8P−65+N8P - 65 + N, and it equals 8(P+2.5)=8P+208(P + 2.5) = 8P + 20. So N−65=20N - 65 = 20 and N=85N = 85.

Prac­tice

  1. The aver­age of 7 num­bers is 20. The aver­age of the first 4 is 17. Find the aver­age of the last 3.
  2. The aver­age of 11 num­bers is 50. The aver­age of the first 6 is 49 and the aver­age of the last 6 is 52. Find the 6th num­ber.
  3. A bats­man scores 87 runs in his 17th innings and raises his aver­age by 3. Find his aver­age after the 17th innings.
  4. The aver­age age of a fam­ily of 5 is 24 years. The youngest mem­ber is 4 years old. What was the aver­age age of the fam­ily when the youngest was born?
  5. The aver­age weight of 10 peo­ple goes up by 1.5 kg when a per­son weigh­ing 60 kg is replaced by a new per­son. Find the weight of the new per­son.
  6. In the first 20 overs, the run rate was 4.5. What run rate is needed in the remain­ing 30 overs to reach a tar­get of 300?
  7. A park has an aver­age of 300 vis­i­tors on Sun­days and 150 on other days. Find the aver­age num­ber of vis­i­tors per day in a month of 30 days that begins with a Sun­day.
  8. A play­er's aver­age over 12 innings is 40. How many runs must he score in the 13th innings to raise his aver­age to 42?

Answers

Show answers
  1. Total =140= 140 and first 4 =68= 68. Last 3 =72= 72, so their aver­age is 24.
  2. 6×49+6×52−11×50=294+312−550=566 \times 49 + 6 \times 52 - 11 \times 50 = 294 + 312 - 550 = 56.
  3. 16a+87=17(a+3)16a + 87 = 17(a + 3), so a=36a = 36. The new aver­age is 39.
  4. Sum =120= 120. The other 4 add to 116. Four years ago they added to 116−16=100116 - 16 = 100, so the aver­age was 25 years.
  5. The total goes up by 10×1.5=1510 \times 1.5 = 15 kg, so the new per­son weighs 60+15=7560 + 15 = 75 kg.
  6. Runs so far =90= 90. Needed =210= 210 in 30 overs, so the run rate is 7.
  7. 5 Sun­days and 25 other days. 5×300+25×15030=525030=175\displaystyle \frac{5 \times 300 + 25 \times 150}{30} = \frac{5250}{30} = 175.
  8. Needed total =13×42=546= 13 \times 42 = 546. Runs so far =480= 480. He needs 66 runs.