## Average questions for SSC CGL Tier 2 Solution set 28

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

Before going through these solutions you should take the test at,

* SSC CGL Tier II level Question Set 28 on Average 2*.

### Solution to 10 Average questions for SSC CGL Tier 2 Set 28 - time to solve was 12 minutes

**Problem 1.**

In a class of 50 students girls are 22 in number and they scored an average of 35 marks in a class test of full marks 50. What is average marks of the boys if the class average is 42 marks?

- 47.5
- 50
- 52.5
- 55

** Solution 1: Problem analysis and solving in mind by shortfall compensation concept**

The number of boys is,

$50-22=28$.

The shortfall of average for the girls from the class average is,

$42-35=7$.

The **shortfall in total** because of **average shortfall for 7 girls**, $7\times{22}=154$ must have been compensated by 28 boys with average marks in excess of class average of 42.

For each boy, the excess average from class average is,

$\displaystyle\frac{154}{28}=\frac{22}{4}=5.5$.

Adding this excess to the class average, the boy's average score becomes,

$42+5.5=47.5$.

You should understand the average shortfall compensation concept more clearly from the concept graphic shown.

**Answer:** Option a: 47.5.

This solution should by far be the quickest one.

**Key concepts used:** * Basic average concept -- Average shortfall compensation concept* --

**Solving in mind.****Note:** You should be clear about the use of the word, **shortfall**. For girls, **shortfall of average** from **overall average** is 7, and this **causes** a **shortfall of 154 in overall total**. We are not calculating the total here, thus saving calculation time.

To **maintain the overall total**, this shortfall of 154 must be **compensated** by an **excess in total** caused by the 28 boys.

The **excess in average of 5.5 from overall average for the 28 boys** neutralizes or compensates the shortfall caused by the girls.

The **shortfall in overall total is compensated**, which compensates in turn the overall average shortfall.

A second form of the same concept is **Average gap segmentation concept**. We'll take this opportunity to explain this concept briefly.

#### Average gap segmentation concept

It says,

In two average groups, the overall average divides the average gap in the ratio of the number of members of the two groups.

In the graphic above, the **average gap** between the **group averages** of 47.5 for the boys and 35 for the girls is,

$47.5-35=12.5$.

The **overall combined average** will be lying in this gap and **will segment it in the ratio of the number of boys and girls**. Let's see how this happens.

As **shortfall in total caused by the girls** is **compensated by excess in total caused by the boys**, from the graphic you can see that,

$154=7\times{22}=5.5\times{28}$,

Or, $\displaystyle\frac{7}{5.5}=\frac{28}{22}$.

The

total average gap of 12.5 is segmentedor divided in two portions 7 and 5.5in the ratio of 28 and 22, the number of boys and girls. 7 is shortfall for lower average of girls from overall average, and 5.5 is the excess for average of boys from overall average.

More is the number of boys, closer will be the overall average to the average of the boys.

**Care should be taken to form this ratio relation by**

Placing shortfall of lower valued average (in this case 7) and the number of members for higher valued average (in this case number of boys 28) as the two numerators in the ratio.

Depending on the nature of the problem, we would use either of these two related rich concepts for quick solution.

**Problem 2.**

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,

- 17
- 20
- 26
- 31

** Solution 2: Problem analysis and solution by basic average concept and total weight as average weight times number of items concept**

Assuming weights of A, B and C to be $a$ kg, $b$ kg and $c$ kg respectively, by the first condition,

$a+b+c=3\times{45}=135$.

By the second condition,

$a+b=2\times{40}=80$ kg, and

By the third condition,

$b+c=2\times{43}=86$ kg.

Adding the last two,

$a+2b+c=166$ kg.

Subtracting the first from the latest result,

$b=166-135=31$ kg.

**Answer:** Option d: 31 kg.

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

**Problem 3.**

If the average of six consecutive even numbers is 25, the difference between the largest and the smallest numbers is,

- 10
- 8
- 12
- 18

**Solution 3: Problem analysis and instant solution by precise problem definition and use of properties of consecutive even numbers**

The problem is formed as a problem on average, but what is really wanted? It is just the difference between the largest and the smallest numbers.

If we assume $x$ as the smallest number, the six consecutive numbers would be,

$x$, $(x+2)$, $(x+4)$, $(x+6)$, $(x+8)$ and $(x+10)$.

The required difference is simply 10. Option a is the answer.

#### Solution 3: By conventional process using concepts on average

The average of six numbers is 25, and so total of six numbers is,

$6\times{25}=150$.

With smallest of the numbers as $x$, the total of the six numbers is,

$x+(x+2)+(x+4)+(x+6)+(x+8)+(x+10)=150$,

Or, $6x+30=150$,

Or, $x=\displaystyle\frac{120}{6}=20$.

The largest number is 30 and the difference of the two, 10.

**Answer:** Option a: 10.

**Key concepts used:** ** Properties of consecutive even numbers **-- Instant solution --

*.*

**Basic average concept -- Total as product of average and number of items-- Solving in mind****Problem 4.**

The average age of 30 students in a class is 14 years 4 months. After admission of 5 new students in the class, the average becomes 13 years 9 months. The youngest one of the five new students is 9 years 11 months old. The average age of the remaining 4 new students is,

- 13 years 6 months
- 10 years 4 months
- 11 years 2 months
- 12 years 4 months

**Solution 4 : Problem analysis and strategy decision**

Thinking backwards, to get the average age of the 4 new students, you need to know either the total age of the 5 new students or the average age of them. But this evaluation will be done at the second stage after you get the average age of the five new students at the first stage.

#### Solution 4: Stage 1: Finding average age of 5 new students using Average age gap segmentation concept and average group concept

At the first stage you have effectively two goups that we call as **average groups**,

**30 out of 35**total students with**average age of 14 years 4 months**, and**5 out of same total 35 students**for whom let's assume**average age is $a$**years

Also the important information is—overall average age of 35 students is 13 years 9 months, or $13\frac{3}{4}$ years. This is a compact form for expressing age in years, but when calculating difference in two ages we'll convert the ages mentally into months.

We'll now use the **average gap segmentation concept** for age problems. It says,

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

**Note:** *See the concept explanation in solution of problem 1 above.*

In our problem then expressing age as mixed fraction in years,

$\displaystyle\frac{30}{5}=\frac{13\frac{3}{4}-a}{14\frac{1}{3}-13\frac{3}{4}}$

Or, $6\times{7\text{ months}}=13\frac{3}{4}-a$,

Or, $a=13\frac{3}{4}-3\frac{1}{2}=10\frac{1}{4}$ years.

This is a use of an advanced concept. You could have reached the same figure using basic concepts.

#### Solution 4: Finding overall average age using shortfall compensation concept

For 30 students the average age exceeds the overall average age by, $(14\frac{1}{3}-13\frac{3}{4})$ years. This is average excess for each of the 30 students. So the group of 30 students contributes to an excess to the total of,

$30\times{(14\frac{1}{3}-13\frac{3}{4})}$.

To maintain the overall average, this excess contributed by 30 students must have been compensated by equal shortfall from the total age contributed by the group of 5 new students, that is,

$5\times{(13\frac{3}{4}-a)}=30\times{(14\frac{1}{3}-13\frac{3}{4})}$,

Or, $6\times{7\text{ months}}=13\frac{3}{4}-a$,

Or, $a=13\frac{3}{4}-3\frac{1}{2}=10\frac{1}{4}$ years.

#### Solution 4: Stage 2: Finding average age of 4 new students: Applying average age gap segmentation concept a second time

Assuming average age of 4 new students other than the youngest to be $b$ years,

$\displaystyle\frac{4}{1}=\frac{10\frac{1}{4}-9\frac{11}{12}}{b-10\frac{1}{4}}$

Or, $4\times{(b-10\frac{1}{4})}=\displaystyle\frac{1}{3}$ years $=4$ months.

So just 1 month will be added to the 5 new students' average age of 10 years 3 months to get the average age of the 4 new students other than the youngest,

$10\frac{1}{4}+\frac{1}{12}=10$ years $4$ months.

**Note that**, in this second stage, *the lower average age group consisted of only the single youngest student. *

**The second point to note is**—*the average age of the 5 new students is reduced from the average age of the 4 new students other than the youngest* $b$, because of the lower age of the youngest student.

The advantage of this concept and technique is—once you are able to structure the problem suitably, the simple **segmentation relation** helps you to solve the problem quickly.

**Answer:** Option b: 10 years 4 moths.

**Key concepts used:** * Basic average concept* --

*--*

**Problem definition and structuring**

**Event sequencing -- Average age gap segmentation concept and technique -- Solving in mind**

**.****Problem 5.**

The average monthly expenditure of a family for the first 4 months is Rs.2570, for the next 3 months Rs.2490 and for the last 5 months Rs.3030. If the family saves Rs.5320 during the whole year, the average monthly income of the family during the year (in Rs.) is,

- 3580
- 3000
- 3185
- 3200

**Solution 5: Problem analysis and solution by basic concepts and shortfall compensation concept**

We will be using the simplest representation of the desired average income as,

$m=\displaystyle\frac{4\times{2570}+3\times{2490}+5\times{3030}+5320}{12}$

If we go ahead to evaluate the three products, sum up the four terms in the numerator and divide by 12 we would get the result undoubtedly. But it would be a lot of calculation.

Instead, we will equalize the durations of the last two periods of 3 months and 5 months to 8 months' average, calculations will be much less and accordingly faster.

The **technique to merge the 3 month and 5 month groups** will again be the **average gap segmentation concept**. The segmentation relation in this case is,

$\displaystyle\frac{5}{3}=\frac{a-2490}{3030-a}$,

Adding 1 to both sides of the equation,

$\displaystyle\frac{8}{3}=\frac{3030-2490}{3030-a}$,

Or, $a=3030-\displaystyle\frac{3\times{540}}{8}=3030-202.5=2827.5$.

So,

$m=\displaystyle\frac{2570+2\times{2827.5}+1330}{3}$, cancelling out factor 4 between numerator and denominator,

$=\displaystyle\frac{9555}{3}$

$=3185$.

**Answer:** Option c: 3185.

**Key concepts used:** * Basic average concept* --

**Average gap segmentation concept -- Shortfall from total concept -- Efficient simplification**

**.**#### Problem 6.

Fifteen movie theatres average 600 number of ticket sales per theatre per day. If six of the theatres close down but the total theatre attendance stays the same, then the average daily attendance per theatre among the remaining theatres is,

- 1000
- 900
- 1200
- 1100

**Solution 6 : Problem analysis and solving in mind by the concept of total as product of average and number of items**

By the basic average concepts, the desired average of attendance in 9 remaining theatres is,

$\displaystyle\frac{15\times{600}}{9}=1000$.

**Answer:** Option a: 1000.

**Key concepts used:** * Total as a product of average and number of items* -- Basic average concept

*.*

**-- Solving in mind**** Problem 7.**

In an examination, average marks obtained by the girls in a class is 85 and the average marks obtained by the boys in the same class is 87. If the girls and the boys are in the ratio 4 : 5, the average marks of the whole class approximately is,

- 85.9
- 86.5
- 86.4
- 86.1

**Solution 7: Problem analysis and quick solution by average gap segmentation concept**

The average gap between the boys' average and girls' average is, $87-2=85$ marks and this gap is divided by $4+5=9$ portions by the 4 : 5 ratio of the girls and the boys. The whole class average will then be,

$85+\displaystyle\frac{5}{9}\times{2}=86.1$ approximately.

In this solution we have used the core concept of average gap segmentation. By this concept the gap of 2 marks is divided by the whole class average in the ratio of 4 : 5 as,

$\displaystyle\frac{4}{5}=\frac{87-a}{a-85}$,

Adding 1 to both sides,

$\displaystyle\frac{9}{5}=\frac{87-85}{a-85}$

So, $a=85+\displaystyle\frac{5}{9}\times{2}=86.1$ approximately.

You get the same result.

The concept graphic is shown below.

**Effectively**, the overall average will exceed the lower average 85 by five-ninth of 2, the average gap. If you look at the overall average from the side of the higher average, it will be less than the higher average of 87 by four-ninth of 2, the average gap.

**Answer:** Option d: 86.1.

**Key concepts used:** * Basic average concepts* --

*--*

**Average gap segmentation concept**

**Solving in mind.**** Problem 8.**

The average daily income of 7 men, 11 women and 2 boys is 257.50. If the average daily income of the men is Rs.10 more than that of women and the average daily income of the women is Rs.10 more than the boys, the average daily income of a man is,

- Rs.257
- Rs.250
- Rs.265
- Rs.277.5

**Solution 8: Problem analysis and quick solution by problem definition and basic average concept**

If $x$ be the average income of men, the average income of women is $(x-10)$ and that of boys is, $(x-20)$. So the overall average is,

$257.5=\displaystyle\frac{7x+11\times{(x-10)}+2\times{(x-20)}}{20}$

Or, $20x-150=20\times{257.5}$,

Or, $x=257.5+\displaystyle\frac{150}{20}=265$.

We have reduced calculation to the minimum.

**Answer:** Option c: 265.

**Key concepts used:** ** Basic average concept **--

**Total as product of number of items and average -- Efficient simplification -- Solving in mind**

**.**** Problem 9.**

A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score of the team would have been 84. The number of points the team scored was,

- 645
- 665
- 672
- 588

**Solution 9: Problem analysis and quick solution by the simplest approach of total as the product of average and number of members**

If the best marksman scored 7 points extra the average of the team of 8 shooters would have been 84, that is their total score would have been,

$8\times{84}=672$.

Their actual total score was then,

$672-7=665$.

**Answer:** Option b: 665.

**Key concepts used:** * Basic average concept* --

**Precise problem definition -- Total score as product of average score and number of members in the group -- Solving in mind**

**.**** Problem 10.**

The average age of three boys is 15 years. If their ages are in the ratio 3 : 5 : 7, what is the age of the youngest boy?

- 8 years
- 9 years
- 8 years 3 months
- None of these

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

The number of portions in the ratio is,

$3+5+7=15$.

The total of ages being $3\times{15}=45$, value of 1 portion is,

$\displaystyle\frac{45}{15}=3$.

So the age of the youngest boy is 3 portions, that is 9 years.

**Answer:** Option b: 9 years.

**Key concepts used:** **Basic average concept -- Basic ratio concept -- Portions concept -- Solving in mind****.**

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