# PE 2 - Average | Arithmetic - Average

**PE 2 - Average | Arithmetic - Average**

There are five students whose average weight is 62 kgs. Then, four new students join the group whose weights are (a + 2), (a – 2), (a + 5) and 2a kgs more than the original average of the group. If the average weight of the group is now 67 kgs, find the value of a.

- (a)
8

- (b)
7

- (c)
6

- (d)
5

Answer: Option A

**Explanation** :

Total weight of 5 students = 5 × 62 = 310 kgs

Total weight of 4 new students = (62 + a + 2) + (62 + a - 2) + (62 + a + 5) + (62 + 2a) = 62 × 4 + 5a + 5

Average weight of all the nine students = (5 × 62 + 62 × 4 + 5a + 5)/9 = 67

⇒ 9 × 62 + 5a + 5 = 67 × 9

⇒ a = 8

**Alternately**,

Because of these 4 new students, the average of now 9 students increases by 5.

∴ Extra weight of these 4 students = 9 × 5 = 45

⇒ (a + 2) + (a - 2) + (a + 5) + 2a = 45

⇒ 5a = 40

⇒ a = 8

Hence, option (a).

Workspace:

**PE 2 - Average | Arithmetic - Average**

Harsh scored a total of 308 points in 28 basketball games. Karan played 10 fewer games than Harsh and his average score was 1.5 points per game higher than Harsh's average score. How many points in total did Karan score?

Answer: 225

**Explanation** :

If Harsh scored 308 points in 28 games, this implies that his average is 308/28 = 11 points per game.

∴ Karan must then have averaged 12.5 points in each of the 18 games he played.

⇒ Karan’s total score = 12.5 × 18 = 225 points

Hence, 225.

Workspace:

**Answer the next 4 questions based on the information given below.**

The following table shows some incomplete information about the runs scored by four batsmen in four innings.

**Additional information:**

- Average runs scored in the 1
^{st}innings for all the players were 101.25. - Rohit's average runs for all 4 innings were 1.5 times the runs scored by Dhawan in 1
^{st}innings. - Virat's average runs were 108.75 in all four innings.
- Total runs scored by Pant in all the four innings were 450.
- The runs scored in 2
^{nd}innings were 15 less than the runs scored in 1^{st}innings for all the players. - Rohit and Pant scored the same runs in 4
^{th}innings. - Sum of runs scored by all the players in the 4
^{th}innings is 405.

**PE 2 - Average | Arithmetic - Average**

The average runs scored by each player, per innings, are (rounded off to the nearest integer).

Answer: 96

**Explanation** :

(i) Average runs scored in the 1^{st} innings = 101.25

Total score = 101.25 × 4 = 405

(ii) Average runs of Virat = 108.75

Total runs of Virat = 108.75 × 4 = 435

So, the runs made by Virat in 3^{rd} innings = 435 - 120 - 105 - 120 = 90

(iii) Total score in the 4^{th} innings = 405

Now, let Rohit’s, Pant’s and Dhawan’s scores in 4th innings be x, x and y.

⇒ 2x + y + 120 = 405

⇒ 2x + y = 285 …(1)

Let Dhawan’s and Pant’s scores in 1^{st} innings be p and q, respectively.

⇒ p + q + 120 + 90 = 405

⇒ p + q = 195 …(2)

From 2: 6p - x = 255 …(3)

From 5: Dhawan’s and Pant’s scores in 2^{nd} innings are p - 10 and q - 10, respectively.

⇒ Dhawan’s total runs = 330

⇒ p + p - 15 + 90 + y = 330

⇒ 2p + y = 255 …(4)

Solving (1) and (4), we get 2x - 2p = 30 …(5)

Solving (3) and (4), we get x = 69 and p = 54

From (1) and (2),

y = 147 and q = 141

Thus,

The average runs scored by each player, per innings = 1026/16 = 96.1875.

Hence, 96.

Workspace:

**PE 2 - Average | Arithmetic - Average**

Who scored the lowest runs in the 1st innings?

- (a)
Virat

- (b)
Rohit

- (c)
Dhawan

- (d)
Pant

- (e)
Cannot be determined

Answer: Option C

**Explanation** :

Consider the solution to the first question of this set.

Lowest runs were scored in 1st innings by Dhawan (i.e., 54).

Hence, option (c).

Workspace:

**PE 2 - Average | Arithmetic - Average**

Who scored the highest total runs in all innings?

- (a)
Virat

- (b)
Rohit

- (c)
Dhawan

- (d)
Pant

- (e)
Cannot be determined

Answer: Option D

**Explanation** :

Consider the solution to the first question of this set.

Highest total runs were scored by Pant (i.e., 450).

Hence, option (d).

Workspace:

**PE 2 - Average | Arithmetic - Average**

Who scored the least in the 2^{nd} innings?

- (a)
Virat

- (b)
Rohit

- (c)
Dhawan

- (d)
Pant

- (e)
Cannot be determined

Answer: Option C

**Explanation** :

Consider the solution to the first question of this set.

Lowest runs were scored in 2^{nd} innings by Dhawan.

Hence, option (c).

Workspace:

**PE 2 - Average | Arithmetic - Average**

A baseball team, which played 60 games, won 30% of its played games. After a phenomenal winning streak, this team raised its winning percentage to 50%. What is the least numbers of games the team must have additionally played in order to attain this average?

- (a)
18

- (b)
20

- (c)
24

- (d)
26

Answer: Option C

**Explanation** :

The team won 30% out of 60 games.

∴ Number of games won initially = 30% of 60 = 18.

If x more games were played by the team and all were won continuously, then

18 + x = 50% of (60 + x)

⇒ 0.5x = 12

⇒ x = 24

∴ 24 more games were played in order to attain this average.

Hence, option (c).

Workspace:

**PE 2 - Average | Arithmetic - Average**

Avinash was asked to calculate the arithmetic mean of ten positive integers, each of which had two digits. By mistake, he interchanged the two digits, say a and b, in one of these ten integers i.e. ab. As a result, his answer for the arithmetic mean was 2.7 more than what it should have been. Then, b - a is equal to

- (a)
1

- (b)
2

- (c)
-2

- (d)
3

Answer: Option D

**Explanation** :

Since the average differs by 2.7, the difference in sum will be 2.7 × 10 = 27.

Let the original number is ab.

Let the sum of the rest nine terms be x.

Sum of all 10 numbers including ‘ab’ = x + 10a + b

Sum of all 10 numbers including reverse number ‘ba’ = x + 10b + a

Difference in sum = (x + 10b + a) - (x + 10a + b) = 27

⇒ 9(b - a) = 27

⇒ (b - a) = 3

**Alternately**,

The difference in total due to reversing the number = (10b + a) - (10a + b) = 9(b - a)

Since average increases by 2.7, the total increases by 2.7 × 10 = 27.

⇒ 9(b - a) = 27

⇒ (b - a) = 3

Hence, option (d).

Workspace:

**PE 2 - Average | Arithmetic - Average**

The average of x consecutive integers is n and the average of the next y consecutive integers is n + 6. Find the relation between x and y.

- (a)
x + y = 12

- (b)
x - y = 6

- (c)
x = y

- (d)
2x - y = 6

Answer: Option A

**Explanation** :

Let the 'x' consecutive numbers be a + 1, a + 2, ……… a + x.

Average = $\frac{(a+1+a+2+a+3+\cdots +a+x)}{x}$ = $\frac{\mathrm{a}+1+\mathrm{a}+\mathrm{x}}{2}$ = a + $\frac{x+1}{2}$ = n

Next y consecutive numbers be a + x + 1, a + x + 2, …….. a + x + y.

Average = a + x + $\frac{1+y}{2}$ = n + 6

∴ a + x + $\frac{1+y}{2}$ = a + $\frac{x+1}{2}$ + 6

⇒ 2x + 1 + y = x + 1 + 12

⇒ x + y = 12

**Alternately**,

Let 3 consecutive numbers be 1, 2 and 3.

Average n = $\frac{1+2+3}{3}$ = 2

Next numbers will be 4, 5, 6 ….

The average these numbers should be (n + 6) = 8

∴ Next numbers must be 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Hence, y must be 9.

⇒ x + y = 3 + 9 = 12

Hence, option (a).

Workspace:

**PE 2 - Average | Arithmetic - Average**

A group of students decided to buy speakers for a party worth Rs. 600. But at the last moment, five students backed out, so each of the remaining had to pay Rs. x more than what they had planned. If ten students had backed out, each of the remaining would have had to pay Rs. 10 more than what they had planned. Find the value of x and total number of students.

- (a)
Rs. 6, 50

- (b)
Rs. 4, 30

- (c)
Rs. 5, 30

- (d)
Rs. 4, 20

Answer: Option B

**Explanation** :

Let the total number of students be 'n'.

Cost of the speaker = Rs. 600

Let the price to be paid by each student initially be 'y'.

So, n × y = 600 …(1)

Now, five students backed out, and average price increased by 'x' i.e.

⇒ (n - 5) × (y + x) = 600 …(2)

Now, if ten students had backed out, average price would have increased by Rs. 10 i.e.

⇒ (n - 10) × (y + 10) = 600 …(3)

Solving the three equations:

From (1) and (3):

$\left(\frac{600}{y}-10\right)\left(y+10\right)$ = 600

y^{2} + 10y – 600 = 0

y = $\frac{-10\pm \sqrt{100+2400}}{2}$ = 20

⇒ y = 20

⇒ n = 600/20 = 30 (# students)

∴ x = 600/(30 - 5) - 20 = 4 [From (2)]

x = 4

Hence, option (b).

Workspace:

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