## 52nd SSC CGL level Solution Set, 1st exclusively on Average

This is the 52nd solution set of 10 practice problem exercise for SSC CGL exam and 1st exclusively on topic Average.

For maximum gains, the test should be taken first, that is obvious. But more importantly, to absorb the concepts, techniques and deductive reasoning elaborated through these solutions, one must solve many problems in a systematic manner using this conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

If you have not yet taken this test you may take it by referring to the * 52nd SSC CGL question set and 1st exclusively on Average* before going through the solution.

Before going through the solutions, you may go through the basic concepts on **average**.

** Basic concepts on Average**

The average value $A$ of a set of $n$ number of values is,

$A=\displaystyle\frac{\text{Sum of }n\text{ values }S_n}{n}$.

Alternatively,

$\text{Sum of }n\text{ number of values }S_n=n\times{\text{Average of }n\text{ values }A}$.

Using these two of the most basic concepts on average, the problems on average are solved.

### 52nd solution set - 10 problems for SSC CGL exam: 1st exclusively on topic Average - answering time 15 mins

**Problem 1.**

The average height of 40 students in a class is 163 cm. When the school opened after the vacation, three students were absent and the average height of the students present was found to be 162 cm. If the heights of the two students absent were equal to each other, and the height of the third absent student was 2 cm less than either of other two, what was the height of the third absent student?

- 170 cm
- 174 cm
- 172 cm
- 176 cm

#### Solution 1 - Problem solving execution

Because of absence of 3 students, the average height reduced to 162 cm. In other words, the total height of 40 students, that is, $40\times{163}$ reduced to $37\times{162}$ by the reduction of total height of three students.

So,

$\text{Total height of 3 absent students}$

$=40\times{163}-37\times{162}$

$=(40-37)\times{162}+40$

$=3\times{162}+40$.

If we assume the height of the first two absent students whose heights are equal as, $x$ cm, the total height of these three absent students is,

$3x-2$ cm.

So,

$3x-2=3\times{162}+40$,

Or, $3x=3\times{162}+42$,

Or, $x=162+14=176$ cm.

And the height of the third absent student,

$x-2=176-2=174$ cm.

**Answer:** Option b: 174 cm.

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

**Total of averaged values concepts --***--*

**Change analysis technique***--*

**Delayed evaluation technique**

*Efficient simplification*

*.***Problem 2.**

The average weight of three men is 84 kgs. When another man, the fourth one, joins the group, the average reduces to 80 kg. Now, a fifth man with weight 3 kg more than the fourth replaces one of the first three. The average weight of these four is then 79 kg. What is the weight of the replaced man?

- 72 kg
- 75 kg
- 76 kg
- 74 kg

**Solution 2 - Problem analysis and solving**

When the fourth man joined the group of three with average weight 84 kg the average of the four reduced to 80 kg. It means the weight of the fourth man is less than 80 by 12 kgs, * the extra kgs of the first three.* So the fourth main weighed $80-12=68$ kgs.

The total weight of the first 3 was, $3\times{84}$ kg. Now the total weight of the four is, $4\times{80}$ kgs. The increase is, $80-3\times{(84-80)}=80-12=78$ kg. This is mathematical deduction to contrast the mathematical reasoning approach.

The fifth man weighed 3 kgs more than the fourth man, that is, his weight was 71 kgs.

The fifth man replacing one of the first three results in the average of the four still dropping by 1 kg to 79 kg.

This is a * second problem of the form*,

"Four persons weighed on an average 80 kgs. If one of them is replaced by another man weighing 71 kgs and the average drops to 79 kgs, what is the weight of the replaced man?"

Effectively by the replacement, the total weight reduced by 4kgs. As average of 4 was 80 and now 79 by the replacement, the * drop in total of the four* is,

$4\times{(80-79)}=4$

So the man replaced must have been weighing 4 kgs more than the man replacing him, that is, $71+4=75$ kgs.

The above deduction is by mathematical and conceptual reasoning based on change analysis technique applied to the domain of average. This approach bypasses equation formation and solving, and consequently much faster if the concepts are clear.

**Answer:** Option b : 75 kgs.

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

**Total of averaged values concepts -- Change analysis technique, we analyze the changes in the total only, bypassing calculation of the total -- Problem transformation -- Mathematical reasoning -- Conceptual reasoning.****Problem 3.**

The average of three numbers is 42. The first is twice the second and the second is twice the third. What is the difference between the largest and the smallest number?

- 54
- 48
- 50
- 46

**Solution 3 - Problem analysis**

First is twice the second and the second is twice the third. Alternatively, first is 4 times the third and second is 2 times the third.

So together the three is $4+2+1=7$ times the third total of which is $3\times{42}$.

So the third number is,

$\displaystyle\frac{3\times{42}}{7}=18$.

The first number, the largest, is then 4 times 18, that is, 72. The difference between the largest and smallest is thus, $72-18=54$.

**Answer:** Option a: 54.

**Key concepts used: ****Transitive relation concept** -- **Basic average concept -- Total of averaged values concepts -- mathematical reasoning -- conceptual reasoning -- mental maths.**

**Note:** The whole solution process having been carried out mentally, it is the fastest. If this process of solution is practiced, agility of mind increases.

**Problem 4.**

The average of 25 results is 18, of the first 12 is 14, and of last 12 is 17. What is the value of the thirteenth result?

- 68
- 78
- 87
- 79

**Solution 4 - Problem analysis and execution**

The average of 25 results is 18, and the average of the first 12 is 14. The first 12 then * creates a shortfall* of $12\times{(18-14)}=48$ from the target average of 18 (of 25 results).

Again the average of the last 12 is 17 which means a shortfall of another $12\times{(18-17)}=12$, a total shortfall from the target average 18 (of 25 results) by $48+12=60$.

This shortfall from average of 18 must be compensated by the 13th result so that the average of the 25 results is maintained at 18.

Thus the 13th result should exceed the average 18 by 60, which is $60+18=78$.

**Answer:** Option b: 78.

**Key concepts used: Basic average concept -- Shortfall concept -- Total of averaged values concepts -- mathematical reasoning -- conceptual reasoning -- mental maths.**

**Problem 5.**

If ages of three boys are in the ratio of 3 : 5 : 7 at an average of 15 years, the age of the youngest boy is,

- 8 years 3 months
- 9 years
- 8 years
- 9 years 3 months

**Solution 5 - Problem analysis and execution**

By basic ratio concepts we can assume the actual ages to be, $3n$, $5n$ and $7n$ respectively, where $n$ is the cancelled out HCF of the ratio terms. The total of these three values is, $15n=3\times{15}$, as the average is 15.

So, $n=3$ and the age of the youngest one is, $3n=9$ years.

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

**Key concepts used:** * Basic ratio concepts* --

*--*

**Total of averaged values concepts**

**Conceptual reasoning**

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

While calculating the average of ten numbers as 15 an error was made by reading 26 instead of 36 for one of the numbers. What should be the correct average?

- 16.5
- 14
- 15
- 16

**Solution 6 - Problem analysis and solving**

The error in including 26 instead of 36 reduced the total by $36-26=10$. For 10 values averaged this error then reduced the average per value by $\displaystyle\frac{1}{10}=1$. Thus the correct average would have been, $15+1=16$.

**Answer:** Option d : 16.

**Key concepts used: **

*--*

**Basic average concept***.*

**Total of averged values concepts -- Shortfall concept -- Average compensation for shortfall concept -- mathematical reasoning -- conceptual reasoning****Problem 7.**

The average age of 8 members of a committee is 39 years. A year later, a member aged 55 retires and a new member aged 39 years joins the committee in his place. What is the average age of the committee members after the change?

- 39 years
- 38 years
- 40 years
- 36 years

**Solution 7 - Problem analysis and solving**

A year later age of every member will increaee by 1 year and the average age would be $39+1=40$.

When the new member aged 39 years replaces the old member aged 55 years, $55-39=16$ years reduced from the total age. Effectively for 8 members then the * average reduced by 2 years per member,* resulting in the new average as, $40-2=38$ years.

**Answer:** Option b: 38 years.

** Key concepts used:** * Basic avrage concept -- Total of averaged values concept -- Shortfall concept -- Average compensation for shortfall concept* --

*mathematical reasoning -- conceptual reasining.*

**Problem 8.**

The average of first three numbers out of four numbers is 15, and that of last three is 16. If the last number is 19 what is the first number?

- 18
- 16
- 15
- 14

** Solution 8 - Problem analysis and solving execution**

Average of first three numbers is 15 and so the total of the three numbers is 45. The last number being 19, the total of all four numbers is 64.

As the average of last three numbers is 16 the total of these three numbers is 48. Thus the first number is, $64-48=16$.

**Answer:** Option b: 16.

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

**Total of averaged values concept -- mathematical reasoning.****Problem 9.**

The first number out of three is twice the second and half the third. If the average of the three numbers is 56, than what is the largest number?

- 50
- 75
- 96
- 48

**Solution 9 - Problem analysis**

The first number is twice the second. So the second is one-half of the first.

Again the first number is half the third. So the third number is two times the first.

Thus the total of the three numbers is,

$1+\displaystyle\frac{1}{2}+2=\displaystyle\frac{7}{2}$ times the first number.

As this total is, $3\times{\text{Average}}=3\times{56}$, the first number is,

$=3\times{56}\times{\displaystyle\frac{2}{7}}=48$

Third number is the largest number as it is two times the first and so it is, $2\times{48}=96$.

**Answer:** Option c: 96.

**Key concepts used:** **Base equalization technique, we have transformed second and third in terms of the first --****Basic average concept -- Total of averaged values concepts -- mathematical reasoning****.**

**Problem 10.**

The average (arithmetic mean) of $3^{30}$, $3^{60}$, and $3^{90}$ is,

- $3^{60}$
- $3^{29}+3^{59}+3^{89}$
- $3^{27}+3^{57}+3^{87}$
- $3^{177}$

**Solution 10 - Problem analysis and solving**

As average is sum of the three values divided by 3, the number of values, the average of the three values is,

$\displaystyle\frac{3^{30}+3^{60}+3^{90}}{3}=3^{29}+3^{59}+3^{89}$.

**Answer: **Option b: $3^{29}+3^{59}+3^{89}$.

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

**Total of averaged values concepts -- Indices concept.**### Other resources that you may find valuable

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