In the previous article, we saw how to approach the concepts for “Averages-> Entry/removal of One element”. (click me) Now extending that concept for Entry/removal of multiple elements.

# Situation: Removing 2 Elements

## Case: highest score in (40/38) innings? (SSC CGL)

A Batsman played 40 innings with average score of 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, then score average score of his remaining innings is 48 runs. What is the highest score of this batsman? (SSC CGL 2011)

Convert this into a Money problem

- Old gang has 40 guys and each of them got Rs.50
- Two guys leave the gang. One of them has Rs.172 more than the other guy.
- Now Average of the old gang declined by 50-48=Rs.2

Ok now let’s think about it.

- Initially everyone has Rs.50 each.
- So two guys who left the game, had total Rs.50 x 2 =Rs.100.
- We are left with 40-2=38 people.
- Two guys left the group. If they wanted to decrease the avg. of those 38 people by Rs.2. So they must have got away with 38 x 2 =Rs.76
- So the two guys had total cash of Rs.100 (theirs) + Rs.76 (they stole from others)
- =100+76=176.
- But the question says difference between highest and lowest is 172.
- And now we know that total is 176.
- Meaning lowest was 2 and highest was 174. (because 2 + 174 =176)
****** - Therefore, final answer=174 was the highest score.

******in case you’re confused:

Suppose the two observations were “a” and “b” where (a>b).

i) total of those two observations is 176 (we found this via “money problem” approach): a + b =176

ii) difference between them is 172 (already given in question): a-b =172

Solve these two equations and you get: 2a=348=> a=174

A similar question was asked in CAT-1997

## Case: Highest marks in (10 / 8) papers (CAT-1997)

The average marks of a student in 10 papers are 80. If highest and lowest scores are not considered, then the average score is 81. If the highest score is 92, then what is the lowest score?

Convert this into a Money problem.

- 10 kids have 80 rupees each.
- 2 kids run away, so 8 kids remain. (10=8-2)
- Average of remaining 8 kids is 81. (Meaning average increased by 1 (81-80=1))

Ok, now let’s think about it.

- 10 kids had 80 rupees each.
- Meaning those two kids also had 80 each
- So two kids originally had 80 x 2 =160 rupees.
- They wanted to raise the average of remaining eight kids by 1.
- So they (two kids) gave 1 rupee to each of those eight remaining kids.
- Meaning two kids gave away Rs.1 x 8kids = 8 rupees.
- So now two kids are left with 160 minus 8 =152 rupees.
- But the question says highest of those kids was 92.
- So the other kid had 152 minus 92 =Rs.60 that’s our final answer.

## Case: average price of (12/10) books (SSC CGL)

Average price of 10 books is Rs.12. Average price of 8 books is Rs.11.75.

Of the remaining two books, if one book has price 60% higher than the other book, find the price of both books.

Convert this into a Money problem. (although it is already a money problem but ‘men’ are not included).

- 10 guys had 12 rupees each.
- 2 guys left. Average goes down by 25 paisa. (12 minus 11.75=0.25)

Ok now let’s think about it.

- Originally two guys had total 2 x 12 = 24 rupees.
- They wanted to bring down the average of remaining eight guys by 0.25 paisa.
- So two guys stole 25 paisa each from the remaining eight people.
- Thus two guys stole: 8 x 0.25 =Rs.2
- Now two guys have total rupees = Rs.24 (they originally had) PLUS Rs. 2 (they stole)
- =24+2=26 rupees.

But the question says, price of one book is 60% higher than the other one.

Meaning cheap book is Rs”m” then expensive book is “1.6 times m”. **

## **side concept

Expensive book is 60% higher than cheap book.

Therefore, price of expensive book

= 100% (of cheap book) + 60% of cheap book

=160% of cheap book

=1.6 of cheap book. (because %=1 upon 100 so 160%=160/100=1.6)

=1.6m

Anyways back to the main process: We found that total is Rs.26

Therefore Rs. 26 = Cheap book + expensive book

=1.0m + 1.6m

=2.6m

If 26=2.6m, then m=26/2.6=Rs.10

And if m=10 then expensive book=1.6 x 10 = Rs.16.

So final answer is

Price of those two books are Rs.10 and Rs.16

I hope the concept of “Averages: Entry/Removal of Two elements” is clear by now.

Now let’s check **“more than two elements.”**

# Situation: More than Two elements

Raynolds produces average 4000 pens per month for the first three months. How many pens should this company produce on an average for remaining months to get yearly production average of 4375?

There are two ways to solve this problem

- Convert in Money problem
- Wine and Water mixture (alligation method)

## Approach #1: Money problem

Convert this into a money problem

- 12 guys have avg. Rs.4375 each (because 1 years= 12 months)
- 9 guys leave the gang. (because we are given avg of 3 months=12-3=9)
- 3 guys remain and their average is Rs.4000 each. (meaning average declined by 4375 minus 4000=375)
- Find the average money held by those 9 guys.

Ok now let’s think about it.

- 12 guys have avg Rs.4375 each.
- Meaning originally those 9 guys had total Rs. 4375 x 9
- They want to decrease the average of remaining 3 guys by Rs.375.
- So they steal Rs. 375 x 3 rupees from those three guys.
- Now total money with 9 guys = 4375 x 9 (they originally had) + 375 x 3 (they stole)

But the question is asking us to find the average money held by these 9 guys

=total money with 9 guys divided by 9

=(4375×9)+(375×3)/9

=Rs.4500

## Even more shortcut!

Recall what we do with “Average: errors” concept.

New average = original average plus or minus average change.

In this case original average of 9 guys =Rs.4375

And they stole Rs.375 x 3 (=increase)

So what’s the average increase= 375 x 3 / 9 =Rs.+125 (this is plus because they gained money)

And new average = original average (Rs.4375) + average change (+125)

=4375+125

=Rs.4500

## Approach #2: allegation method

If you don’t know the allegation (visual move) technique, then go through my old article (click me)

Company wants average yearly production of 4375 (this is our “middle” value)

For 3 months , they had average of 4000

That means for the remaining 9 months, average production must be higher than 4375. (let’s assume this is “m”)

Make the table.

Avg. production | 4000 | 4375 | M |

Months | 3 | 9 |

That’s it. Apply the visual move.

m-4375=3….eq1

4375-4000=9…eq2

Divide eq1 by 2 and solve it

You get m=4500 that’s our final answer.

## Case: Average run rate (10/40) overs (MAT)

- In a cricket match, team Pakistan makes 281 runs. Now team India starts batting.
- In the first ten overs, we have (bogus) run rate of 3.2 only. If we want to win this match, what must be the run rate for remaining 40 overs?

- I suggest you try the money problem method here on your own.
- Interestingly, we can solve this problem via alligiation (wine water mixture) as well!
- How?
- Since Pakistan made 281 runs, we must make 282 runs to win the game.
- And we want to do this in 50 overs (40+10)
- So what should be our “overall” runrate= 282/50=5.64
- For the first ten overs we had run rate of only 3.2. so for remaining 40 overs, we must make run rate higher than 5.64.

Run rate | 3.2 | 5.64 | M |

Weight | 10 | 40 |

That’s it, just apply the visual move!

M-5.64=10….eq1

5.64-3.2=40…eq2

Divide equation 1 by equation 2

When you solve this equation, you get m=25/4=6.25

Final answer: India must get run rate of 6.25 in the remaining 40 overs, to win the match.

# Case: Avg age of boys (18/30)

A group of 30 boys have average age of 13. Later, 18 boys, with an average of 15 years left the group. What is the average age of remaining 12 boys?

## Approach: money problem

- 30 boys had Rs.13 each
- 18 of them left, with Rs.15 each.

Let’s think about it.

- Originally those 18 boys had Rs.13 (when they were part of the gang)
- Meaning remaining 12 boys also had 12 x 13 = 156 rupees.
- Now 18 boys form a new gang with Rs.15 each. So, each of these 18 boys need Rs.2 more (15-13=2). So total they need Rs.2 x 18 = Rs.36
- They steal this money from those remaining 12 boys.
- So how much money is left with those 12 boys? = 156 they originally had MINUS Rs.36 stolen= Rs.120
- And since we are asked “average” of those 12 boys
- So we’ll divide 120 by 12=Rs.10 is our final answer.

## Approach: alligation

I’m directly constructing the table

Avg age | M | 13 | 15 |

No of boys | 12 | 18 |

Apply the visual move, you get two equations.

When you solve them, you get m=10.

Final answer: average age of remaining 12 boys is 10 years.

# Mock questions

- A class has 35 people with avg weight of 62 kilos. When we donot consider the weight of teacher and class Secretary, the average goes down to 61.5 kilos. If the weight of class Secretary is 60 kilos, how much does the teacher weight?
- A batsman made average 85 runs in 50 matches. When we donot consider his last two matches, average falls down to 82 runs. If he made 150 runs in his second last match, how many runs did he score in his last match?
- Gokuldham Society has 30 families with average annual income of 8 lakh rupees. If we ignore the families of Jethalal and Bhide, the average declines by 1.25%. If Jetalal earns 740% higher than Bhide, what is the annual income of Bhide?
- In a bookstore, there are 100 books with average price of Rs.300. Two of them are removed and average declines by 10 rupees. of the two books that were removed, if the expensive book cost Rs.1330, find the price of cheaper book.
- In a mobile store, there are 200 handsets with an average MRP of Rs.8000. When we ignore a Nokia Lumia and a Samsung Galaxy note2, the average of remaining phones goes down to Rs.7850. If Nokia Lumia costs Rs.9999, find the price of that Samsung phone.

## Answers

- 80.5kg
- 164 runs
- 2 lakh rupees. 1.25% decline in average = 0.0125×8 lakhs=10,000 rupees. Meaning Jetha+Bhide had total money of [10k x 28 + (2x8lakh)]=18.8lakhs. And Jetha+Bhide(B)=8.4B+1B=18.8lakhs. Hence Bhide(B)=2 lakhs.
- Rs.250
- Rs.35701

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