Arun Sharma Test: Averages- 1 - CAT MCQ

Let the age of the teacher be T.

The initial total age of the class (including the teacher) is (30+1)*14 = 434 years.

If we exclude the teacher, the total age of the 30 students is 30*(14-0.5) = 405 years.

So the teacher's age, T, must be:

434 - 405 = 29 years

Therefore, the age of the class teacher is 29 years.

Total runs after 25 innings = 25 x 56= 1400

Since, after 26 innings average run is increased by 2, hence new average = 56+2 = 58

Now,

Total runs after 26 innings = 26 x 58 = 1508

Score in 26th inning = 1508 - 1400 = 108 runs.

Class A average is 20. And their range is 18 to 22

Class B average is 25. And their range is 23 to 31

Class A average is 30. And their range is 26 to 33

If 5 students transferred from A to B, As average cannot be determined but Bs average comes down as the highest score of A is less than lowest score of B.

If 5 students transferred from B to C, Cs average cannot be determined the Bs range fo marks and Cs range

of marks are overlapping.

It will definitely decrease since the highest possible transfer is lower than the lowest value in C

So, the correct answer is B. 

If the present age of Barney’s wife is x years. Then according to the question:
(33 X 6 + x + 2) / 8 = 32
x + 2 + 198 = 256 x + 2 = 58
x = 56
x = 56 years 

You can take 53 as the base to reduce your calculations. Otherwise, the question will become highly calculation-intensive. Let the fifth experiment’s measurement be ‘x’ above 53.
Then you get:
0.735 X 7 = 1.005 X 3 + (x + 0.004) + x + 0.995 X 2
-> 5.145 = 3.015 + 2x + 0.004 + 1.99.
On solving this you get x = 0.068. Hence, the weight of the fifth body is 53.068 and the weight of the fourth body is 53.072. Hence, the option (d) is correct. 

Let's solve the problem again step by step, ensuring accuracy:

1. Let the area of the first farm be AAA acres.
2. Therefore, the area of the second farm is A−12A - 12A−12 acres.

Given that:

• The first farm gets an average harvest of 21 tons of wheat per acre.
• The second farm gets an average harvest of 25 tons of wheat per acre.
• The second farm harvested 300 tons more wheat than the first.

Set up the equations:

1. The total harvest of the first farm: 21A21A21A tons
2. The total harvest of the second farm: 25(A−12)25(A - 12)25(A−12) tons
3. The difference in harvest: 25(A−12)−21A=30025(A - 12) - 21A = 30025(A−12)−21A=300

Solve the equation: 25A−300−21A=30025A - 300 - 21A = 30025A−300−21A=300 4A−300=3004A - 300 = 3004A−300=300 4A=6004A = 6004A=600 A=150A = 150A=150

The area of the first farm is 150 acres, and the area of the second farm is 150−12=138150 - 12 = 138150−12=138 acres.

Calculate the total harvest for each farm:

1. The first farm: 21×150=315021 \times 150 = 315021×150=3150 tons
2. The second farm: 25×138=345025 \times 138 = 345025×138=3450 tons

The correct answer is:

1. 3150, 3450

It is given that the date of births of four of them are prime numbers

For the average to be maximum, the dates we need to choose must be closest to the maximum dates

Therefore, the four prime dates are 29th, 23th, 19 and 5th

Sum of the date of birth = 29+23+19+10+5+12+11+10+9+8 =136

Average = 136/5  = 27.2

The maximum average when the birth dates of four of them are prime numbers is 27.2

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x – 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

It is given that the average of the 30 integers = 5
Sum of the 30 integers = 30*5=150
There are exactly 20 integers whose value is less than 5.
To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers
So the sum of 10 integers = 10*6=60
The sum of the 20 integers = 150-60= 90
Average of 20 integers = 90/20 = 4.5

If C increases, then the average of C goes up from 30. For this to happen it is definite that the average of B should drop.

So, the correct option is B.

Let x , y and z be no. of students in class X, Y ,Z respectively.
From 1st condition we have
83*x+76*y = 79*x+79*y which give 4x = 3y.
Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .
Now overall average of all the classes can be given as 
Substitute the relations in above equation we get,
= (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

Ten years ago, the total age of the family is 231 years.

Seven years ago, (Just before the death of the first person), the total age of the family would have been
231+8*3 = 231+24 = 255.
This is because, in 3 years, every person in the family would have aged by 3 years,
Total change in age = 231+24 = 255

After the death of one member, the total age is 255-60 = 195 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8
Four years ago, (i.e. 6 years after start date) one of the member of age 60 dies,
therefore, total age of the family is 195+24-60 = 159 years.
Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8
After 4 more years, the current total age of the family is = 8x4 + 159 = 191 years
The average age is 191/8 = 23.875 years = 24 years (approx)

Alternatively,
Since the number of members is always the same throughout
The 2 older members dropped their age by 60
So, after 10yrs, total age = 231 + 8*10 - 2*60 = 191
Average age = 191/8 = 23.875≌24