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

For the list of all aptitude articles published so far, visit mrunal.org/aptitude

## 38 Comments on “[Aptitude] Averages: More than two elements added / removed : shortcut technique explained”

hi mrunal, in the 1st eg, after arriving at the figure of 176, u said, that as the difference is 172, the lowest is 2 and highest is 174, i hav a doubt here, could it not b also like, 4 and 176, then also the difference will be 172. Then our answer also changes, i.e. the highest score.?????

Kidnly, correct me where i’m going wrong

(40*50)-(38*48) = H+L

so H+L = 176

also H-L = 172

so – + –

2 L = 4

L = 2 so H = 174 ans

thank u

@Dr.VJ,

two conditions have to be fullfilled

1. total of those two obs=176 (a+b=176)

2. difference =172. (a-b=172)

if a and b are 4 and 176 then total will be 180=first condition is not fulfilled.

thank you

Hi evryone …is it wrong to do the 1st example like this ,,

If we assume toatal of the 40 matches as ‘x’,

= then the average x/40 =50 ( as 50 is given avg)

hence X= 2000

if the lowest score is ‘y’ then the higest score will b y+172.

then the new avg will b

2000-y-(y+172)/38=48

=2000-y-y-172/38=48

1838-2y=1824

2y=14

y=7

hence the highest score must b y+172 i.e 172+7=179

could anyone tell me ver i went wrong.thanks in advance.

(2000-2y-172)\38=48

2000-2y-172=1824

176=2y+172

y=4

174 is rite ans……

Hi deeps,, …is it wrong to do the 1st example like this ,,

If we assume toatal of the 40 matches as ‘x’,

= then the average x/40 =50 ( as 50 is given avg)

hence X= 2000

if the lowest score is ‘y’ then the higest score will b y+172.

then the new avg will b

2000-y-(y+172)/38=48

=2000-y-y-172/38=48

1838-2y=1824

2y=14

y=7

hence the highest score must b y+172 i.e 172+7=179

could anyone tell me ver i went wrong.thanks in advace.

hey mrunal ….. it’s nice to read your article once again…….

here is one quick suggestion ….. please don’t waste your time on csat articles, they are Damn

easy and don’t need any explaining.

so please use that time by writing GS related stuff coz it’s very helpful !

gr8 article

Mrunal pls shift your attention to data interpretation now

Mrunal sir …could you provide an article to solve the problems based on cause and effect relation,I have one such question………..

Consider the following statements:

1. Fringe operates only when supernatural shrieks.

2.If baby reads supernatural shrieks.

3.only if baby reads does the horizon improve.

from the above four statements it may be concluded that:

a. Horizon improves only if fringe operates

b.each time supernatural shrieks horizon iproves

c.baby reads only if fringe opertaes

d.none of the above

I think the answer is “d. none of the above”.

I agree with XAERO. it shud be ‘d’.

GIVE SOME SHORTCUT USING THIS TECHNIQUE WILL CONSUME ATLEAST 5 MINUTE

this takes less than a minute once you practice enough sums.

how to do this kind of qusn..

Q: The avg weight of a grp of 20 boys was calculated to be 89.4kg and it was calculted that one weight was misread as 78kg instead 87kg. The correct avg weight is

a)88.95

b)89.25

c)89.55

d)89.85

mrunal sir how to do this?

technique is explained in this video: http://www.youtube.com/watch?v=H4iEbPdWBjM

avg=89.4 wrong mean

n=20

total=89.4*20= 1788

correct weight= +87

incorrect = -78

= 1797/20=89.85

I think calculation is wrong

thnkuu sir, avg ka concept clear ho gya:)

mrunal sir tell me wher m worng

Q. the avg age of 39 stdnts is 15 if the age of theage of the teacher be inlcuded then avg increase by 3 months find the age of the teacher?

ansr= new avg 15.3 so 39*.3=11.7

so teacher age is 15.3+11.7=27yrs

but ansr is 25,,,how

39*15=585

age of 40 persons 15yr3mth = 61/4yrs

total age of 4o person 61/4 * 40= 610yrs

610-585= 25yr

which one is correct.

@amit

teacher needs to give 3 months to 39 students each = 39 x 3 months.

and teacher should bring ((15 x 12)+3)months for himself.

so age of teacher

=(39 x 3)+((15 x 12)+3)

=300 months

=300/12

=25 years.

ya i was thinkng d same but caught in fault thnks…sir.

one more doubt

avg of a cricter whose bowling avg is 12.4 runs per wicket 5 wickts for 26 runs and there by decreases his avg by 0.4 . the number of wickets taken by him till the last match was

64 , 72 , 80 , 85

how can we use ur method here.

thnks

Use kids with money technique.

N kids had avg money 12.4

5 new kids join them.

new average comes down by 0.4 means the new comers have just below the old average.

they are added 26 Rs to the collective money pool.

as per new average they must have 5*12=60

but really have 26.

they borrow money from other kids to join this group.

so each N Kids spend 0.4 paise on 5 new kids.

60-26=34 Rs wanted.This amount contributed by N Old kids.

N*0.4=34.Solve it you will have 85.

Therefore there was N=85 old kids.i.e. senior members of the group.

thanks.

Another great article.

sir you are quants god

hi mrunal,

can we solve following type questions using money-kid method?

avg wt of 6 persons is increased by 3 kg when a person of 100 kg is replaced by another person.

what is wt of new person?

Hi to everyone any kindly help me to solve below question.

“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?”

Ans ; 80.5

But I’m not able to get above I got 95

sum of 2 persons actual weight=(62*2)+(33*0.5)=155

one person weight out of 2 person given as 60

so 155-60=95

the teacher weight is 95

Is this Right

Anybody help meeeeeeeeeeeeeeeeeeeeeeeee

(62*2)+(33*0.5)=155

the total is 140.5 not 155

The average of five positive numbers is 213. The average of the first two numbers is 233.5 and the average of last two numbers is 271. What is the third number?Any shortcut method???

thanq u mrunal sir

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 request you to explain the money problem method here sir plzz

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 request you to explain the money problem method here sir plzz

Total runs required = 282

Overall Run rate = 282/50 = 5.64

Now money Prob

10 guys – 3.2 Rs

50 guys – 5.64 rs

Sum/Contri of 40 guys = 40*5.64 + 10*2.44 (5.64-3.20) (Also , refer Reynolds Pen prob)

= 250

Therefore for 40 ovs the average should 250/40 = 6.25.

to win required=250 as 32 already scored.(3.2*10 overs)

so 250 runs has to be scored in 40 overs

which gives us average of 250/40=6.25

We have difference 87-78=9, which we need to evenly distributed in 20 people to correct average.

So, 9/20=0.45

Add this to 89.4+.45=89.85