# Quant Boosters - Vikas Saini - Set 1

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source -

• Q1) 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.

• 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.
1st tea rate is -27 from base.2nd tea rate is -18 from base.
So obviously rate of 3rd tea must be higher from the base.
So -27-18+2t = 0
t = +22.5.
so rate of third tea = 153 + 22.5 = 175.50.

• Q2) 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 17th one ?
a. 78
b. 80
c. 91
d. 94

• Grade of 17th 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+1-16x77
= 94.

Short approach :-
Previous average(16 students) = 77.
New average(17 students) = 78.
While 17th 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 17th student = 77 + 1 x 17 = 94.

• Q3) The average score of Rohit Sharma after 48 innings is 48 and in the 49th innings he scored 97 runs. In the 50th innings, what is the minimum runs required to increase the average by 2 runs than it was before 50th innings ?
a. 99
b. 149
c. 151
d. CBD

• 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 49th 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

• Q4) The average weight of A,B and 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

• 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.

• Q5) 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.

• 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.

• Q6) 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 are in the class?

• 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= (47-44.8 ):(44.8-42)=2.2:2.8.
(2.8-2.2)x=6
x=10.
total students= (2.2+2.8 )x10=50

• Q7) 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

• 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 + 1 x 11 = 66.

• Q8) The average of 71 results is 48. If the average of first 59 results is 46 and that of the last 11 is 52 then find the 60th result.

• 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.

• Q9) The average of 50 numbers is 38. If two numbers, namely 45 & 55 are discarded then what is the average of remaining numbers ?

• Total = 50 x 38 = 1900.
two numbers discarded (45,55)
new total = 1900-45-55 = 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 100-76 = 24.
There is 48 numbers.
so every number value will be decreased by 24/48 = 0.5
new average = 38 - 0.5 = 37.5.

• Q10) The average runs of Rohit & Dhawan is 280. The average runs of Virat & Rahane is 200. What is the average of each of them.

61

63

61

46

61

43

61

54