Shortcuts for solving Averages  Vikas Saini

Q. Tea worth rs 126 per kg and 135 rs per kg are mixed a third variety in ratio of 1:1:2.If the mixture of worth rs 153 per kg,price of third variety per kg will be ?
A. 169.5 B. 170 C.175 D.180.
Alternate approach :
126 x 1 + 135 x 1 + 2 x t = 153 x (1+1+2)
= > 126 + 135 + 2t = 612
= > 261 + 2t = 612
= > t = 175.50.Short cut :
The average of mixture here is 153.We take it as base.
1^{st} tea rate is 27 from base.2^{nd} tea rate is 18 from base.
So obviously rate of 3^{rd} tea must be higher from the base.
So 2718+2t = 0
t = +22.5.
so rate of third tea = 153 + 22.5 = 175.50.Q. There are 17 students in a class.16 students took the test,average was 77.The next day 17 students took the test and new average is 78.What was grade of 17^{th} one ?
A.78 B.80 C.91 D.94.
Alternate approach:
Grade of 17^{th} one = total marks of 17 students – total marks of 16 students
= 17 x 78 – 16 x 77
= (16+1)(77+1) – 16x77
= 16x77+16+77+116x77
=94.
Short approach :
Previous average(16 students)=77.
New average(17 students)=78.
While 17^{th} student appeared for exam,to maintain average of 77 he had to get atleast 77 marks.
(But new average is 78,it is only possible when each student will get +1 mark)
Marks of 17^{th} student = 77 + 1x17 =94.
Q. The average score of Rohit Sharma after 48 innings is 48 and in the 49^{th} innings he scored 97 runs.In the 50^{th} innings min no of runs required to increase by 2runs,than it was before 50^{th} innings ?
A. 99 B.149 C.151 D.CBD
Alternate approach :
Total score(48 innings)=48x48=2304.
After 49 innings score = 2304+97=2401.
Average after 49 innings = 2401/49=49.
Required score = (49+2)x50 – 2401 = 149.
Short approach :
Average till 48 inning = 48.
In 49^{th} inning he scored 97 runs = 48+49.
New average = (48+49)+48x48/49 = 49.
Required score in 50 to increase average by 2 = 49+2x50 = 149.
Q. The average weight of A,B,C is 45 kg.If average of A & B be 40 kg and that of B & C be 43 kg,then the weight of B?
A. 17 kg B.20 kg C. 36 kg D.31 kg
Alternate approach:
A + B + C = 3 x 45 = 135.
A + B = 2 x 40 = 80.
B + C=2 x 43 = 86.
C = 135 – 80 = 55.
A = 135 – 86 = 49.
B = 135 – 49 – 55 = 31.
Short approach :
Average of A,B,C = 45. (base)
Average of A & B = 5 (from base)
Average of B & C = 2 (from base)
Weight of B = 45 – 2x5 – 2x2 = 31.
Q.A grocer has a sale of rs 6435,rs 6927,rs 6855,rs 7230,and rs 6562 for consecutive months. How much sale must he have in the 6th month so that he gets an average sale of rs 6500.
A.4991
B.5991
C.6001
D.6991.
Solution :
alternate approach 
let 6th month he had a sale of z.
6435+6927+6855+7230+6562+z = 6 x 6500
= > 34009 + z = 39000
= > z = 4991.
Short approach :
let's take 6500 as base.
6500(65+427+355+730+62)
=6500  (1509)
= 4991.
Q. The average weight of boys in a school is 47 and girls is 42.If there are 6 more boys than girls and average weight of all class is 44.8.
How many students in class?
Alternate approach :
let no of girls = a
no of boys = a+6
47(a+6)+42a = 44.8(a+a+6)
47a+42a+282 = 44.8(2a+6)
89a + 282 = 89.6a + 268.8
0.6 a = 13.2
a = 22.
a+6 = 28.
total = 22+28 = 50.
Short approach :
average of all class=44.8.
ratio of boys & girls= (4744.8 ):(44.842)=2.2:2.8.
(2.82.2)x=6
x=10.
total students= (2.2+2.8 )x10=50.
Q.Average weight of 11 players is increased by 1kg.When one player of team weighing 55kg is replaced by new player.Find weight of new player.
Alternate approach :
average weight of each player = a.
weight of new player = k.
11a  55 + k = 11(a+1)
k  55 = 11
k = 66.
Short approach :
weight of new player=55+1x11=66.
Q. The average of 71 results is 48.If the average of first 59 results is 46 and that of the last 11 is 52.Find the 60th result.
Alternate :
71 x 48 = 59 x 46 + 11 x 52 + 60th
= > 60th value = 71 x 48  59 x 46  11 x 52.
= > 60th value = 122.
Short approach :
We take 48 as base.
59 results average is 2(from base)
11 results average is +4(from base)
So 59 x (2) + 11 x 4 = 74.
To get 60th value this value should be decrease from base.
Hence 60th value = 48(74) = 122.
Q.The average of 50 numbers is 38.If two numbers,namely 45 & 55 are discarded then what is the average of remaining numbers ?
Alternate :
Total = 50 x 38 = 1900.
two numbers discarded (45,55)
new total = 19004555 = 1800.
new average = 1800 / 48 = 37.5
Short approach :
The average of 50 numbers is 38.
so if we discarded two number then to maintain average of 38,value should be decreased by 2 x 38 = 76.
but here value decreased by 100.
So 10076 = 24.
There is 48 numbers.
so every number value will be decreased by 24/48 = 0.5
new average = 38  0.5 = 37.5.
Q. The average runs of Rohit & Dhawan is 280.The average runs of Virat & Rahane is 200.
What is the average of each of them.
Alternate approach :
Rohit + Dhawan = 2 x 280 = 560.
Virat + Rahane = 2 x 200 = 400.
Total runs = 560 + 400 = 960.
Average = 960 / 4 = 240.
Short approach :
Take average of any two
let we take Rohit & Dhawan,which is nothing but 280.(base)
now average of virat & rahane 200.
which is 80 (from base)
Average = 280  (80/2)
= 280  40
= 240.
We can solve it by take 200 too..
base = 200.
average of Rohit & Dhawan is +80(from base).
average = 200 + (80/2)
= 240.
Here divided by 2,because that is average of two person..
Q.The average weight of a class of 29 students is 40 kg.If the weight of teacher be included,the average rises by 500gm.What is the weight of the teacher ?
Alternate approach :
previous weight of class = 29 x 40 = 1160 kg.
new weight (including teacher) = 30 x 40.5 = 1215.
Teacher's weight = 1215  1160 = 55 kg.
Short approach :
base = 40.
including teacher average increased by 0.5 kg.
Teacher's weight = 40 + 30 x 0.5
= 40 + 15
= 55 kg.