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CAT 2025 Lesson : Averages - Common Types

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6. Common Types of Average Questions

6.1 Age based questions

With no changes to a group of n people,
1) Average age of the group will increase by 1\bm{1} year for every year that passes by.
2) Total age of the group will increase by n\bm{n} years for every year that passes by.

Example 13

The average age of a family of 6 members, that include Mary, John and Oliver, was 4545 in 20102010. Mary and John are married to each other and had a baby in 20142014. In 20162016, Oliver died at the age of 6262. What was the average age of the family in 20182018, if nobody else had a baby or died during this period?
(11) 3939            (22) 4141            (33) 4343            (44) 4545           

Solution

Average age of the 66 members in 2014=45+4=492014 = 45 + 4 = 49

Total age of the 66 members in 2014=49×6=2942014 = 49 \times 6 = 294

With a just-born baby, total age of 7\bm{7} members in 2014\bm{2014} remains the same at 294\bm{294}.

The total age would have increased to 294+7×2=308294 + 7 \times 2 = 308 = 308 in 20162016.

As a person aged 6262 died, average in 2016=308−626=412016 = \dfrac{308 - 62}{6} = 41

Average age of the members in 2018=41+2=432018 = 41 + 2 = 43

Answer: 4343


6.2 Consecutive Integers

For nn consecutive integers x1,x2,...,xnx_{1}, x_{2}, ... , x_{n},

11) As they are in AP, Average of these integers, x‾=x1+xn2\overline{x} = \dfrac{x_{1} + x_{n}}{2}
∴ Averages of consecutive integers will be in multiples of 0.5\bm{0.5}

22) If the smallest term x1x_{1} is removed, new average =x‾+0.5= \overline{x} + 0.5

33) If the largest term xnx_{n} is removed, new average =x‾−0.5= \overline{x} - 0.5

∴ New average when an integer is removed is between x‾−0.5\overline{x} - 0.5 and x‾+0.5\overline{x} + 0.5

Example 14

A set of consecutive positive integers beginning with 11 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is 3571735 \dfrac{7}{17}. What was the number erased?
[CAT 2001]
(11) 77            (22) 88            (33) 99            (44) None of these           

Solution

Average of a set of consecutive natural number is always a multiple of 0.50.5. When one number is removed from this set, the average increases or decreases by a maximum of 0.50.5 as explained above.

When a number is removed, the average becomes 3571735 \dfrac{7}{17}, which is between 3535 and 35.535.5. Therefore, the initial average should have been either 35\bm{35} or 35.5\bm{35.5}.

Denominator for the new average has 1717. Therefore, the new number of terms is a multiple of 17\bm{17}.

Let n be the initial number of terms on the board.
Average of nn consecutive natural numbers =n+12= \dfrac{n + 1}{2}

If the initial average =35= 35, then n=69\bm{n = 69}. And, if initial average 35.5,n=7035.5, \bm{n = 70}. When the number of terms is reduced by 11, we get the number of terms, which should be a multiple of 1717. This is possible only when the initial number of terms n=69\bm{n = 69} and the new number of terms is 68(17×4)68 (17 \times 4).

∴ Initially, there were 6969 integers on the board and the average was 35.35.
Sum of 6969 integers =35×69=2415= 35 \times 69 = \bm{2415}

Sum of 6868 integers =35717×68=60217×68=2408= 35 \dfrac{7}{17} \times 68 = \dfrac{602}{17} \times 68 = \bm{2408}

The number erased =2415−2408=7= 2415 - 2408 = 7

Answer: (1) 77


6.3 Items joining or leaving a group

When 11 or more items join or leave a group,

New Average =New Sum of ValuesNew Number of Items= \dfrac{\text{New Sum of Values}}{\text{New Number of Items}}

Alternatively, we can assume that the items joining or leaving the group have the same value as the initial average and add or subtract the relative impact. This is shown in the following example's alternative solution.

Example 15

The average weight of 6363 students in a class is 7777 kg. When a new student joined the class, this average increased to 7878 kg. What is the weight of the new student?

Solution

Let the weight of the new student be xx.

Initial Weight =63×77=4851= 63 \times 77 = 4851

New average =4851+x63+1=78= \dfrac{4851 + x}{63 + 1} = 78

⇒ 4851+x4851 + x = 49924992

⇒ xx = 141141

Alternatively (Recommended Method)

Average of 6363 students in the class =77= 77 kg

Average remains unchanged, if the new student's weight was 77 kg.

However, the average increases by 1\bm{1} kg for the new total of 64\bm{64} students. This means
11) The new student brought in additional weight
22) Additional weight =1×64== 1 \times 64 = 64\bm{64} kg

New student's weight == Assumed Weight ++ Additional Weight
=77+64=141= 77 + 64 = 141

Answer: 141141 kg

Note: Calculation in the second method is minimal.



Example 16

The average weight of 5656 students in a class is 4343 kg. When 22 students left the class, the new average increased to 43.543.5 kg. What is the average weight of the 22 students who left the group?

Solution

Average of 5656 students in the class =43= 43 kg

Average remains unchanged if the 22 students had weighed 4343 kg each, a total of 86\bm{86} kg.

However, when they left, the average increases by 0.5\bm{0.5} kg for the new total of 54\bm{54} students. This means
11) The students who left took away less weight (or weighed less than the average)
22) Reduced weight =0.5×54== 0.5 \times 54 = 27\bm{27} kg

Weight of students who left == Assumed Weight −- Reduced Weight
=86−27=59= 86 - 27 = 59

Average weight of the 22 students =592=29.5= \dfrac{59}{2} = 29.5 kg

Answer: 29.529.5 kg

Note: The solution includes the thought process, which might make it long. But the calculation is minimal.


Example 17

Average runs scored by a player is the total runs divided by the number of matches played by him/her. Smriti had average of 3030 runs in the first 2424 matches. After the next 44 matches, her average reduced to 28.528.5 runs. What is the average of the runs she scored in her last 44 matches?

Solution

Average of 2424 matches =30= 30

Average remains unchanged if she scored 3030 runs in each of the next 44 matches, totalling to 120\bm{120} runs

However, average decreases by 1.5\bm{1.5} runs for the new number of 28\bm{28} matches.
11) She scored less runs in the additional matches
22) Reduced Runs =1.5×28=42= 1.5 \times 28 = \bm{42} runs

Run in last 44 matches =120−42=78= 120 - 42 = 78

Average in last 44 matches =784=19.5= \dfrac{78}{4} = 19.5

Answer: 19.519.5 runs


6.4 Replacement

When one or more items leave a group and the same number of items join the group, the number of items remain unchanged. In these questions, we can only find the difference between the values of items leaving the group and those joining the group. We, however, need additional information to find their exact values.

Example 18

The average height of 11 players in a team is 178 cm. When Joel leaves the team and Jeff joins, the average height reduces by 3 cm.Which of the following is true?
(11) Joel is shorter than Jeff by 3030 cm           

(22) Joel is shorter than Jeff by 3333 cm           

(33) Joel is taller than Jeff by 3030 cm           

(44) Joel is taller than Jeff by 3333 cm           

Solution

Total height of 1010 players and Joel =178×11=1958= 178 \times 11 = 1958
Total height of 1010 players and Jeff =(178−3)×11=1925= (178 - 3) \times 11 = 1925

∴ Joel is taller than Jeff by 1958−1925=331958 - 1925 = 33 cm

Alternatively (Recommended Method)

With Joel leaving and Jeff joining, the number of players is unchanged at 1111 players.

As the average height reduces, Jeff's height is lower than Joel.

Impact on total height is 33 cm for 1111 players.

∴ Joel is taller than Jeff by 3×11=33\bm{3 \times 11 = 33} cm

Answer: (44) Joel is taller than Jeff by 3333 cm


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