Test Level 3: Averages - 1 - CAT MCQ

Go with the options:
If 60 is the heaviest weight, to make 121, the second heaviest weight must be 61, which is not possible.
If 62 is the heaviest weight, the second heaviest weight will be (121 - 62) = 59
The third heaviest will be (120 - 62) = 58
Now, the lightest weight will be 112 - 58 = 54 ( Second least sum = Smallest + Third)
So, the second lightest weight must be 110 - 54 = 56
So, the weights are 54, 56, 58, 59 and 62.
We can make all the ten pairs from these.

Let x litres of the first mixture be mixed with y litres of the second mixture.
Milk Water

Since the third vessel contains half milk and half water;

16x + 3y = 2x + 15y
14x = 12y
x/y = 12/14 = 6/7
Hence, y = 7 /(7+ 6) of 26 gallons = 7/13 x 26 = 14 gallons

Total temperature for (M + T + W) = 3 × 55 = 165°C
Total temperature for (T + W + Th) = 3 × 60 = 180°C
On subtracting, we get 
M + T + W = 165

Th – M = 15
56 – M = 15
Temperature on Monday = 41°C

Let x consecutive numbers be a + 1, a + 2, ……… a + x.

Next y consecutive numbers be a + x + 1, a + x + 2, …….. a + x + y.

2x + 1 + y = x + 1 + 12
x + y = 12

Alternative method:
Let 3 consecutive numbers be 1, 2 and 3.
Average, 

Next numbers be 4, 5, 6 ….
Average = n + 6 = 8
∴ Next numbers must be 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Hence, y must be 9.
border=0 x + y = 3 + 9 = 12

Average weight of 12 drums = 60 kg
Sum of weights = 12 × 60 = 720 kg
60 = H – S_v ...(1) [Where H and S_v are the weights of the two heaviest drums]

And, 1/2(60) = (Total weight of remaining 10 oil drums)/10
300 = Total weight of remaining 10 oil drums …(2)

It means that the sum of the remaining two drums is 420 kg.
H + S_v = 420 ……(3)
H – S_v = 60 ……(4)
2H = 480
H = 240 kg

Total number of students = 120
Maximum marks of the exam = 100
∵ The fractional marking is not allowed, so there are 101 students who can get different marks from 0 to 100, remaining 19 students can also get the marks from 0 to 100.
∵ No three students can get the same marks, so there are 19 pairs of students who are awarded the same marks.

Average earning for the entire year = $1800
Total earnings in the year = $21,600
According to the question:
5 × 1500 + 2000 + 2500 + 2475 + 2325 + 0 + 2x + x = 21600 (where x is his earnings in December)
⇒ x = 1600
Earnings in November = 2x = $3200

Let the total distance covered be x km.

30% of 60 = 18
If x more games were played by the team and all were won continuously, then
18 + x = 0.50(60 + x)
0.5x = 12
x = 24
Hence, 24 more games have to be won in order to attain this average.

Option (c) is correct, since you need to transfer out whatever you got into A, in order to keep the value of A’s average at the minimum.