Shortcut tricks in Average
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 New avg=$\dfrac{(a_{1}+x)+(a_{2}+x)+...+(a_{n}+x)}{n}$ =$\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.
4.Arithmetic progression: a,a+d,a+2d,a+3d......a+(n-1)d
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.
Shortcut 1:To find average or change in average from a set of values
Q1: 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)
Conventional Method :
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.
Shortcut 2:Shortcut to find new value when average is given
Q2: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.
Q3: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.
Q4: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
Q5: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
Q6: 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