Shortcut - Average
Common increase or decrease for all the elements. If al the elements in a series are increased or decreased or multiplied or divided by a certain number, the old average should also be added or subtracted or multiplied or divided respectively to get the new average.
Question:
Average age of a family of four members is 34. What is the average age of the family after 4 years?
Answer:Since every ones age is increased by four in 4 years, the average will also increase by 4
The new average = 34 + 4 = 38
Proof: Total age of the family before 4 years = 34 x 4 = 136
Total increased age for four members = 4 x 4 = 16
New total age = 136 + 16 = 152
New average = 152/4 = 38
Shortcut – Average
Adding or removing or replacing an element. $A_{1}N_{1} - A_{2}N_{2}$
Question 1:
Average age of 5 students was 18. After adding teacher’s age the average became 20. What is the age of the teacher?
Answer := (18 x 5) - (20 x 6) = 30
Question 2:
Average age of 5 students and a teacher was 21. After removing the teacher’s age the average became 18. What is the age of the teacher?
Answer:= (21 x 6) - (18 x 5) = 36
Question 3:
Average weight of 40 bags is 20kg. After removing 1 bag the average became 19.5. What is the weight of removed bag?
Answer:= (20 x 40) - (19.5 x 39) = 39.5
Question 4:
Average weight of 40 bags is 20kg. After replacing a bag by another bag the average became 39. What is the difference between the weights of those two bags?
Answer:= (20 x 40) ~ (20 x 39) = 20
Shortcut – Average
Weighted Average $A_{0}=\dfrac{A_{1}N_{1}+A_{2}N_{2}+.....+A_{n}N_{n}}{N_{1}+N_{2}+...+N_ {n}}$
Question:
Average marks scored by students in three classes are 40, 50, and 60. If there are 10, 20 and 30 students in the classes respectively, find the overall average of the three classes.
Answer:A1 = 40; A2 = 50; A3 = 60
N1 = 10; N2 = 20; N3 = 30
Substitute in the above formula. We get,
=[(40x10) + (50x20) + (60x30)]/[10+20+30]
=[400 + 1000 + 1800]/60
= 53.33
Shortcut - Average
Finding the middle subject mark, if the middle subject overlaps.
$M=A_{1}N_{1}+A_{2}N_{2}-A_{0}N_{0}$
Question:
Average marks scored by Sai in 11 subjects is 60. Average marks scored by her in first 6 subjects is 50 and average marks scored by her in last 6 subjects is 62. Find the mark scored by Sai in 6th subject.
Answer:$A_{0} = 60; N_{0} = 11$
$A_{1} = 50; N_{1} = 6$
$A_{2} = 62; N_{2} = 6$
Substituting the values in the formula, we get
M = 50x6 + 62x6 – 60x11
= 300 + 372 – 660
= 12
Marks scored by Sai in 6th subject = 12
Shortcut - Average
Finding the middle subject mark if the middle subject is left out.
$M=A_{0}N_{0}-A_{1}N_{1}-A_{2}N_{2}$
Question:
The average marks scored by Shradha in 9 subjects is 75. The average marks in first 4 subjects is 69 and average marks in last 4 subjects is 78. Find the marks scored by her in 5th subject.
Answer:
$A_{0} = 75; N_{0} = 9$
$A_{1}= 69; N_{1} = 4$
$A_{2} = 78;N_{2} = 4$
Substituting the above values in the formula, we get
M = 75x9 – 69x4 – 78x4
= 675 – 276 – 312
= 87
Marks scored by Sradha in 5th subject = 87
Properties of average:
1.Average = sum/N
2.Avg=$\dfrac{a_{1}+a_{2}+a_{3}+.......+a_{n}}{n}$
3.Suppose if we increase every quantity by x,then the effect on average is
Newavg=$\dfrac{(a_{1}+x)+(a_{2}+x)+(a_{3}+x)+..+(a_{n}+x)}{n}$
=$\dfrac{(a_{1}+a_{2}+a_{3}+..+a_{n} )}+{nx} $
[Note:If x is added for n times the value will become nx]
=$\dfrac{(a_{1}+a_{2}+a_{3}+..+a_{n} )}{n}+\dfrac{nx}{n}$
New avg=A+x From this we know that if we increase every quantity by x ,then the new average will be an old average (+) plus x.
Likewise,if we subtract every quantity by x,then the new average will be an old average minus (-) x.
Likewise,if we multiply every quantity by x,then the new average will be an old average multiply (*)by x.
Sum of n terms of an AP =$\dfrac{n}{2}[2a+(n-1)d$
=$\dfrac{n}{2}[a+a+(n-1)d$
=$\dfrac{n}{2}[1st term +last term]$[Where,a is the 1st term and a+(n-1)d is the last term
From this we find the average Avg=$\dfrac{Sum}{N}$ where n=no of terms
=$\dfrac{\dfrac{n}{2}[1st term +last term]}{n}$
=$\dfrac{1st term +Last term}{2}$
For example:
Find the average of 2,4,6,8,10.
Formula:
$\dfrac{\dfrac{n}{2}[1st term +last term]}{n}$
=$\dfrac{2+10}{2}$=6
We can also says that in question if they have given odd number of values then the mid term will be the answer of average.
To find average or change in average from a set of values
Question:
The average of a batsman in 16 innings is 36. In the next innings, he is scoring 70 runs. What will be his new average?
Solution:Formula:
New average=((old sum+ new score)/(total number of innings)
New average= ((16 ×36)+70)/((16+1)) = 38
Shortcut tricks:
Step 1: Take the difference between the new score and the old average = 70 – 36= 34
Step 2: This is 34 extra runs which is spread over 17 innings. So, the innings average will increase by 34/17 = 2
Step 3: Hence, the average increases by => 36+2 = 38.
To find new value when average is given
Question:
The average age of 29 students is 18. If the age of the teacher is also included the average age of the class becomes 18.2. Find the age of the teacher?
Solution: Conventional Method:Let the average age of the teacher = x
(29 × 18 + x × 1)/30
Solving for x, we get x = 24.
Shortcut tricks:
Step 1: Calculate the change in average = 18.2 – 18 = 0.2.
This change in 0.2 is reflected over a sample size of 30.
Step 2:The new age is increased by 30 × 0.2 = 6 years above the average i.e. 18 + 6 = 24; which is the age of the teacher.
Question:
Find the average of 20,30,40,50,60,70,80,90,100 if each number is increased by 5.Find the new average
By the property 3,we can simply say the mid value is 60 and each number is increased by 5 then the average will also increased by 5.so,60+5=65 will be the answer.
Question:
The students attended classes from monday to saturday are consecutive integers in increasing order and the average monday to wednesday is 50 find the average from monday to friday.
Solution:Monday =n
Tuesday =n+1
Wednesday =n+2
Thursday =n+3
Friday =n+4
saturday =n+5
Avg from monday to wednesday =$\dfrac{n+(n+1)+(n+2)}{3}=50$
=>$\dfrac{3n+3}{3}$=50
=>3n=150-3
=>n=147/3=49
Avg from monday to friday =$\dfrac{n+(n+1)+(n+2)+(n+3) +(n+4)}{5}$
=5n+10/5=n+2=49+2=51
Question:
Method of deviation:Find the average of 2004,1998,1996,2005,2007
2004=>2000+4
1998=>2000-2
1996=>2000-4
2005=>2000+5
2007=>2000+7
--------------------
(5* 2000 +10)/5
=2000+2=2002
Question:
Find average of 58,57,62,65
Here the reference is 60
Deviation from reference :
58 -2
57 -3
62 +2
65 +5
----------
2/4
reference +2/4=>reference +1/2=60+0.5=60.50
average =60.5
Simplification -large, complex numerical expression into a simpler form by performing various mathematical operations, in accordance with the BODMAS rule.
Learn squares and cubes of number:
Squares (12 to 302):• 12 - 1
• 22 - 4
• 32 - 9
• 42 - 16
• 52 - 25
• 62 - 36
• 72 - 49
• 82 - 64
• 92 - 81
• 102 -100
• 112 -121
• 122 -144
• 132 - 169
• 142 - 196
• 152 - 225
• 162 - 256
• 172 - 289
• 182 - 324
• 192 - 361
• 202 - 400
• 212 - 441
• 222 - 484
• 232 - 529
• 242 - 576
• 252 - 625
• 262 - 676
• 272 - 729
• 282 – 784
• 292 – 841
• 302 - 900
Cubes (13 to 153):
• 13 - 1
• 23- 8
• 33 - 27
• 43 - 64
• 53 - 125
• 63 - 216
• 73 - 343
• 83 - 512
• 93 - 729
• 103 - 1000
• 113 - 1331
• 123 - 1728
• 133 - 2197
• 143 - 2477
• 153 - 3375
Example 1: 212 / 49 × 6
Solution:From the above question if we know the square value of 212, then this question will be easily solved
STEP 1: 212 = 441
STEP 2: 441/49= 9
STEP 3: 9×6 = 54
STEP 4: Hence the answer for above series is 54
2) REMEMBER FREQUENTLY ASKED FRACTION VALUES
• 5% = 0.05
• 6 ¼ % = 0.0625
• 10% = 0.1
• 12 ½ = 0.125
• 16 × (2/3)% = 0.166
• 20 % = 0.2
• 25 % = 0.25
• 33 × (1/3)%= 0.33
• 40 % = 0.4
• 50% = 0.5
• 60% = 0.6
• 66 × (2/3) =0.66
• 75 %= 0.75
• 80 %= 0.8
• 90 % = 0.9
• 100% = 1
• 125 % = 1.25
• 150% = 1.5
• 200 % = 2
• 250 % =2.5
EXAMPLE 2: 60% of 250 +25% of 600
STEP 1: Know the values of 60% =0.6 and 25 % = 0.25
STEP 2: Now directly multiply 0.6×250 + 0.25×600
STEP 3: 0.6×250= 150
0.25×600=150
STEP 4: 150+ 150 = 300
STEP 5: Hence the answer for above series is 300
3) Solve mixed fraction – Multiplication EXAMPLE 3: 2×(3/5) × 8×(1/3) + 7 ½ × 2× (2/3)
STEP 1: 2×(3/5) × 8×(1/3) = (13/5) × (25/3) = 65/3
STEP 2: + 7 ½ ×2×(2/3)= 43/6 × 12/5 = 86/5
STEP 3: 65/3 + 86/5 = 38×(15/13)
STEP 4: hence the answer for above series is 38×(15/13)
4) Solve Mixed Fraction addition Example 4: 19×(3/5) + 23×(2/3) – 24×(1/5)
STEP 1: Take all the whole number outside the bracket i.e. 19+23 -24 = 18
STEP 2: Add fractions within bracket 18×[(3/5) + (2/3) – (1/5)] = 18(16/15)
STEP 3: Hence the answer for above series is 18(16/15)
Example 5: (?)2 +18×12= 62 ×5×2
STEP 1: Multiply 18 × 12 = 216
STEP 2: Square of 6 = 36
STEP 3: Multiply 36 ×5×2= 360
STEP 4: (X)2 +216 = 360
STEP 5: (X)2 = 360-216 = 144
STEP 6: Therefore X = 12