Average problems for SSC CGL Tier 2 Solution set 27

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Solution to average problems for SSC CGL Tier 2 Set 27

Average problems with solutions for SSC CGL Set 27

Learn to solve 10 average problems for SSC CGL Tier 2 set 27 in 12 minutes using basic concepts and problem solving techniques. But take the test first.

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SSC CGL Tier II level Question Set 27 on Average 1.

Solution to 10 average problems for SSC CGL Tier 2 set 27 - time to solve was 12 mins

Problem 1.

The arithmetic mean of the following numbers, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, and 7, 7, 7, 7, 7, 7, 7 is,

Solution 1: Problem analysis, Problem definition and solving in mind by pattern discovery

By definition the arithmetic mean or average of the given series of numbers is,

$A=\displaystyle\frac{\text{Sum of numbers}}{\text{Number of numbers}}$

A quick examination of the series of numbers identifies the sum as,

$S=1^2+2^2+3^2+4^2+5^2+6^2+7^2$.

There are 1 number of 1, 2 numbers of 2's, 3 numbers of 3's and so on. When summed up, 2 times 2 is $2^2$, 3 times 3 is $3^2$ and so on.

This is the all important pattern discovered for solving the problem quickly.

Instead of using any formula, we consider it faster to enumerate and sum up the squares of first 7 natural numbers as,

$S=1+4+9+16+25+36+49=140$.

Again, the number of numbers in the denominators is the sum of 1 number of 1, 2 numbers of 2, 3 numbers of 3 and so on. In other words, the number of numbers is,

$N=1+2+3+4+5+6+7=4\times{7}=28$.

The average of first 7 numbers is the middle number (as 7 is odd) 4, so that the total is, 28.

The desired arithmetic mean is then,

$\displaystyle\frac{140}{28}=5$.

Answer: Option a: 5.

Key concepts used: Basic average concept -- Key pattern identification -- Natural number average and summation --Solving in mind.

Problem 2.

The mean of 100 items was 46. Later it was found that an item 43 was misread as 34 and a second item 16 was misread as 61. It was also found that the number of items was 90 instead of 100. Then the correct mean must be,

52.7
50.7
50
52

Solution 2: Problem analysis and solution by basic average concept and shortfall and excess amount concept

By the first statement, the total of the items calculated was,

$S=100\times{46}=4600$, assuming that at least the calculation is correctly done even if the observations were erroneous.

Because of the first misread the less amount added has been,

$E_1=43-34=9$.

And by the second misread an excess amount added has been,

$E_2=63-16=45$.

Together by the two misreads then effectively an excess amount added has been,

$E=45-9=36$.

The actual total would then be 4600 less 36, that is,

$4600-36=4564$,

And the correct mean of 90 items would be,

$\displaystyle\frac{4564}{90}=\frac{456.4}{9}=50.7$.

Answer: Option b: 50.7.

Key concepts used: Basic average concept -- Total of items as product of average and number of items -- Shortfall and excess amount concept -- Erroneous reading problem -- Solving in mind.

Problem 3.

In a school with 600 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the students in the school is 11 years and 9 months, then the number of girls in the school is,

Solution 3: Problem analysis and solution by rich concept of proportionate average gap segmentation

In a conventional solution, you can assume number of girls as $x$, so that number of boys becomes $(600-x)$ and the total age relation would be,

$600\times{11.75}=11x+12(600-x)$,

Or, $x=600\times(12-11.75)=150$.

This method is mechanical and doesn't really convey the rich concept embedded in the problem.

Let us explain what we mean by this.

The problem involves two groups of members with average age for each group given. The 12 years average age of the boys is larger than the average age 11 years of the girls. The gap of the two averages is 1 year. These are facts. Now comes first part of the concept.

If the number of girls were equal to that of boys, what would be the average of the two groups taken together? It would simply be the mid-point of the average gap, that is, 11.5 years.

The more the number of girls increases in comparison to the number of boys, how would the overall average age change? It will simply move closer to 11 years, the average age of the girls. In the same way, if the number of boys exceeds the number of girls more and more, the overall average age would move closer to 12 years, the average age of the boys. This makes sense, isn't it? Up to this point it is believable.

But unless we are able to measure how much nearer to 12 years the overall average age moves when the number of boys exceeds the number of girls more and more, we won't be able to use this concept in practice.

Finally then the last part of concept—the change in overall average age from its mid-point 11.5 years of the average gap (when number of members of both groups are equal) is proportional to change in number of members of any of the two groups.

More precisely,

The overall average age divides the average age gap in the ratio of the number of members of the two groups.

As the average age 11.75 years divides the gap of 1 year in the ratio 3 : 1 in favor of the boys, number of boys is 3 times that of number of girls which is, being 1 portion out of total 4, must be one-fourth of 600, that is 150.

Practically without any calculation, this wholly conceptual method gives the solution in maximum 15 seconds.

Answer: Option c: 150.

Key concepts used: Basic average concept -- Rich average concept -- Proportionate average gap segmentation concept -- Solving in mind.

Problem 4.

Five players of a team are weighed consecutively and their average weight calculated after each player is weighed. If the average weight increases by 1 kg each time, how much heavier is the last player than the first one?

4 kg
5 kg
20 kg
8 kg

Solution 4 : Problem analysis and solving in mind by precise event sequencing and problem definition

This is a problem where if you can follow the events of 5 weighings one after the other and define the core of the problem, solution comes very quickly.

Question is, do we need to carefully track each of the weighings? We don't, because difference between weights of the last and the first player weights is only wanted.

Let's assume, the weight of the first player be $x$ kg. So after the first player is weighed, the average weight is $x$ kg and after the last player is weighed, the average weight is $(x+4)$ kg for the team of 5 players. The total weight of 5 players is then,

$5(x+4)=5x+20$ kg.

Now we just need to know the total weight after weighing of the first 4 players, which must be,

$4(x+3)=4x+12$ kg, as average weight of the first players was $(x+3)$ kg

And the difference of these two totals is the weight of the fifth player, which is,

$(x+8)$ kgs, that is 8 kgs more than the first player.

Easy, isn't it?

Answer: Option d: 8 kg.

Key concepts used: Basic average concept -- Total weight is product of average weight and number of members -- Event sequencing -- Problem definition -- Solving in mind.

Problem 5.

A librarian purchased 60 story books for her library. But later she found that she could get 4 extra books by spending Rs.336 more and then the overall average price per book would be reduced by Rs.1. The previous average price per book (in Rs.) was,

Solution 5: Problem analysis and solving in mind by total price as product of average price and total number of items concept

As overall average price per book reduced by Rs.1 when 4 more books were purchased for Rs.336, assuming original average price as Rs.$x$,

$64\times{(x-1)}=60x+336$,

Or, $4x=336+64=400$,

Or, $x=\displaystyle\frac{400}{4}=100$.

Conceptually, as overall average price of 64 books reduced by Rs.1 by purchase of the 4 extra books at Rs.336, if this amount of Rs.336 were Rs.64 more, that is Rs.400, the old average would have remained unchanged.

For 4 items total amount of Rs.400 means, old average was Rs.100.

This is use of shortfall from total concept.

Answer: Option a: 100.

Key concepts used: Basic average concept -- Total price as the product of average price and number of items concept -- Shortfall from total concept.

Problem 6.

In a team of 10 persons, 9 persons spent Rs.40 each for their meals and the remaining one spent Rs.9 more than the average expenditure of all the 10 persons. The total expenditure for their meals (in Rs.) was,

510
610
410
310.

Solution 6 : Problem analysis and solving in mind by excess distribution on member average

For first 9 members average expenditure was Rs.40.

The 10th member made an expenditure in excess of Rs.9 from 10 member final average. This extra amount if distributed equally to each of the first 9 members assuming their original expenditure as Rs.40 each, the expenditure for each increases by Rs.1 to Rs.41 which is then the final average.

The total expenditure for 10 persons is then,

$10\times{41}=410$.

Answer: Option c: 410.

Key concepts used: Total as a product of average and number of members -- Distribution of amount excess of final average among other members equally to reach final average -- Excess from average distribution technique -- Average as actual normalized equal value for each member concept -- Solving in mind.

Problem 7.

Of the three numbers whose average is 22, the first is, $\displaystyle\frac{3}{8}$th of the sum of the other two. What is the first number?

Solution 7: Problem analysis and quick solution by basic average concepts

As average of the three numbers is 22, their total is 66.

If $x$ be the first number, sum of the other two numbers is,

$66-x$.

So by the given condition as the first number is three-eighth of the sum of the other two,

$x=\displaystyle\frac{3}{8}(66-x)$

Or, $\displaystyle\frac{11}{3}x=66$,

Or, $x=18$.

Answer: Option b: 18.

Key concepts used: Basic average concepts -- Total as product of number of items and average -- Solving in mind.

Problem 8.

The average of three consecutive odd numbers is 52 more than $\displaystyle\frac{1}{3}$rd of the largest one. What is the smallest of the three numbers?

Solution 8: Problem analysis and quick solution by odd number pattern concept

Average of three consecutive odd numbers is the middle number as each number is more than the previous number by 2.

So, the total of the three numbers is,

$3\times{M}=3\times{(52+\frac{1}{3}L)}=156+(M+2)$, where $M$ is the middle number and $L=M+2$ is the largest number.

So $2M=158$,

Or, $S=M-2=79-2=77$, where $S$ is the smallest number.

Answer: Option d: 77.

Key concepts used: Basic average concept -- Total as product of number of items and average -- Properties of consecutive odd numbers -- Solving in mind.

Problem 9.

The average of six numbers is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. The average of the remaining two numbers is,

Solution 9: Problem analysis and quick solution by average group concept

This is an ideal problem to apply average group concept. We are not to find the values of the remaining two numbers, but their average, while averages of other two pairs of numbers are given. For all practical purposes then we can consider each pair of numbers as a single group so that six numbers consist of 3 groups each having two numbers.

At the second step of abstraction of average group concept, you can very well then consider the six numbers as effectively three numbers with average of these three as 3.95 and the value of the first two as, 3.4 and 3.85, the average values themselves.

This is possible because of the basic mechanism of effective value equalization of all members of a group of members by averaging.

So the value of the third group of pair of numbers which is their average is,

$3\times{3.95}-(3.4+3.85)$.

3.85 is 0.10 short and 3.4 is 0.55 short from 3.95, a total of 0.65 short. The 3rd average must make up this total shortfall from 3.95. Adding this shortfall of 0.65 to 3.95 the average of 3rd pair of numbers is then 4.6.

The conventional calculations are simple and can be carried out with ease, but we have highlighted the mental process of going through two stages of application of concepts,

First, average group concept by abstraction, and
Second, shortfall compensation concept.

Answer: Option a: 4.6.

Key concepts used: Basic average concept -- Abstraction -- Average group concept -- Shortfall compensation concept -- Solving in mind -- Solving in mind.

Problem 10.

What is the average of all numbers between 100 and 200 which are divisible by 13?

143.5
149.5
147.5
145.5

Solution 10: Problem analysis and quick solution using arithmetic progression series properties of the consecutive target numbers

We have to first identify mentally what are the target numbers whose average we have to find out.

In this process of precisely identifying the target numbers, the first number in the series we identify as,

$8\times{13}=104$.

As in a 100, there are 7 multiples of 13, there will be $7+1=8$ multiples of 13 between 100 and 200 because of the additional one number of starting multiple of 104.

These eight numbers form a series of 8 numbers each more than the previous by 13. 104 is the starting number of the series and 195 the ending number.

As the difference between two consecutive numbers in this series is same, the average of the first 7 numbers of the series will be the 4th multiple, that is,

$104+3\times{13}=143$.

The difference between the eighth and last number 195 and first seven number average of 143, that is, 52 if divided by 8 and added to 143, you would get the average of eight numbers as,

$143+\displaystyle\frac{(195-143)}{8}=149.5$.

This is distribution of excess from average for the last member of the eight member set of numbers.

Alternatively, we could have calculated the total of the eight numbers in the AP series and divided by 8 to get the same result. But that is remembering and using formulas. We prefer to adopt faster concept driven approach without explicitly dealing with the eight numbers.

Answer: Option b: 149.5.

Key concepts used: Basic average concept -- Multiples concept -- AP series concept -- Average of an AP series—concept of average as middle number for an AP series with odd number of members -- Excess from average distribution technique -- Solving in mind.

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SSC CGL Tier II level Solution Set 27, Average 1