# Elementary Statistics Questions for RRB Group-D PDF Set-2

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## Elementary Statistics Questions for RRB Group-D PDF

Download Top-15 RRB Group-D Elementary Statistics Questions set-2 PDF. RRB GROUP-D Maths  questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: In a triangle ABC, the median from A passes through a point on the side BC, find that point if A(2,4) B(1,-3) C(5, 1) ?

a) (3,-1)

b) (3/2, 1/2)

c) (7/2,5/2)

d) none of these

Question 2: Find the point of intersection of medians of a triangle having vertices A(1,2), (2,3), (3,1) ?

a) (1,1)

b) (2,1)

c) (1,2)

d) (2,2)

Instructions

In archery trials for selection into the national team, each participant is given 15 arrows and he has to shoot the target. A person can either hit the target or miss it. The table given below gives the number of targets hit by different number of participants. For example, there were 11 who missed the target in each of their 15 attempts. There were 6 people who hit the target only once and so on. Some of the data in the table has been intentionally left blank.

It is known that the number of participants who hit 3 or more targets, hit 7 targets on an average.
It is also known that the number of participants who hit 12 or fewer targets, hit 6 targets on an average.

Question 3: The participants who hit fewer than 4 targets are eliminated from the second round of the event and the number of people who hit more than 12 targets have directly qualified for the third round. What is the median score among the people who participated in the first round but not in second round?

a) 13

b) 12

c) 3

d) 2

Question 4: Find the length of the median BE of a triangle ABC, with vertices A(1,2) , B(2,3) and C(3,6) ?

a) 1 unit

b) 2 units

c) 3 units

d) 4 units

Question 5: The mean of six positive integers is 15. The median is 18, and the only mode of the integers is less than 18. The maximum possible value of the largest of the six integers is

a) 26

b) 28

c) 30

d) 32

e) 34

Question 6: The point where the 3 medians of a triangle meet is called

a) centroid

b) Incentre

c) Circumcentre

d) orthocentre

Question 7: Three medians AD, BE and CF of ∆ABC intersect at G; Area of ∆ABC is 36 sq cm. Then the area of ∆CGE is

a) 12 sq cm

b) 6 sq cm

c) 9 sq cm

d) 18 sq cm

Question 8: In Δ ABC, the median AD is 6 cm and CB is 12 cm, measure of angle CAB is

a) 90°

b) 30°

c) 60°

d) 120°

Question 9: The shortest median of a right angled triangle is 15 units. If the area of the triangle is 216 square units then what is the length of the longest median of the triangle?

a) 5$\sqrt{45}$ units

b) 9$\sqrt{5}$ units

c) 3$\sqrt{45}$ units

d) 3$\sqrt{73}$ units

Question 10: In a set of 5 distinct positive integers, 3, 5, 9, 15 and x, x < 20. For how many values of x will the mean of the set greater than or equal to the median of the set?

Question 11: Ramesh analysed the monthly salary figures of five vice presidents of his company. All the salary figures are integers. The mean and the median salary figures are 5 lakh, and the only mode is 8 lakh. Which of the options below is the sum (in lakh) of the highest and the lowest salaries?

a) 9

b) 10

c) 11

d) 12

e) None of the above.

Question 12: The median of 11 different positive integers is 15 and seven of those 11 integers are 8, 12, 20, 6, 14, 22, and 13.

Statement I: The difference between the averages of four largest integers and four smallest integers is 13.25.
Statement II: The average of all the 11 integers is 16.

Which of the following statements would be sufficient to find the largest possible integer of these numbers?

a) Statement I only.

b) Statement II only.

c) Both Statement I and Statement II are required.

d) Neither Statement I nor Statement II is sufficient.

e) Either Statement I or Statement II is sufficient.

Question 13: In a triangle ABC, median is AD and centroid is O. AO = 10 cm. The length of OD (in cm) is

a) 6

b) 4

c) 5

d) 3.3

Question 14: If the median drawn on the base of a triangle is half its base, the triangle will be:

a) right-angled

b) acute-angled

c) obtuse-angled

d) equilateral

Question 15: The scores of a group of students are 23, 45, 32, 41, 48, 20, 36 and 32. What is the median of their scores?

a) 32

b) 33

c) 34

d) 35.5

Median from A passes through the midpoint of BC, B(1,-3) C(5, 1)

Mid point of BC = ($\frac{1+5}{2} , \frac{-3+1}{2}$) = $(3 , -1)$

So the answer is option A.

point of intersection of medians of a triangle is centroid.

Centroid = $(\frac{x_1+x_2+x_3}{3} , \frac{y_1+y_2+y_3}{3}) = (\frac{1+2+3}{3} , \frac{2+3+1}{3}) = (2 , 2)$

So the answer is option A.

The number of people who participated in first round but not in second round = 11 + 6 + 4 + 18 + 4 + 3 + 1 = 47
Hence, the median score will be the score of the 24th person. Thus, the median score will be 3.

E = mid point of AC = $(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$ = $(\frac{1+3}{2} , \frac{2+6}{2})$ = $(2, 4)$

B = (2 , 3) & E = ( 2 , 4)

Since x values are same,distance = $y_{2}-y_{1} = 4-3 = 1$

So the answer is option A.

Mean of the six numbers = 15
So, the sum of the numbers = 15 * 6 = 90
As the median is 18, the mean of middle two numbers must be 18 and thus, their sum must be 36.
Also, the mode is a number less than 18. So, the mode must be appearing as the first and the second number of the six given integers, when arranged in ascending order. To maximize the largest integers, the mode must be equal to 1.
Therefore, out of the six integers, two are 1 and 1.
For the middle two numbers whose sum is 36, we cannot have 18 and 18 because then we will have two modes which is inappropriate as per the question.
So, the middle numbers must be 17 and 19.
The fifth integer can be 20.
Maximum possible value of the largest integer = 90 – (1 + 1 + 17 + 19 + 20) = 32
Hence, option D is the correct answer.

The point where the 3 medians of a triangle meet is called the centroid of the triangle.

=> Ans – (A)

The medians of a triangle divide into 6 triangles of equal areas.

=> $ar(\triangle AFG)$ = $ar(\triangle BFG)$ = $ar(\triangle BDG)$ = $ar(\triangle CGD)$ = $ar(\triangle CGE)$ = $ar(\triangle AGE)$

Thus, $ar(\triangle ABC)$ = $6 \times ar(\triangle CGE)$

=> $ar(\triangle CGE)$ = $\frac{1}{6} \times 36=6$ $cm^2$

=> Ans – (B)

AD = 6 cm and BC = 12 cm

Since, AD is the median, => BD = DC = 6 cm

In $\triangle$ ADC, => AD = CD = 6 cm

=> $\angle$ DCA = $\angle$ DAC = 45°

Similarly, in $\triangle$ ADB => $\angle$ DAB = 45°

$\therefore$ $\angle$ CAB = $\angle$ DAC + $\angle$ DAB = 45° + 45° = 90°

=> Ans – (A)

In a right angled triangle, the shortest median is the one which is drawn on the hypotenuse. Let the triangle be ABC and let AC be the hypotenuse of this triangle. Length of hypotenuse = 2*15 = 30 cm (median is equal to the circumradius. Hypotenuse is twice the circumference.)
Area = $\frac{1}{2}*AB*BC = 216$
AB*BC = 432
=> $AB^2 + BC^2 = 900$
Solving these two equations, we get
AB = 18, BC = 24.
Hence, the longest median will be
$\sqrt{9^2 + 24^2} = \sqrt{657}$
Hence, the length of the longest median = 3$\sqrt{73}$ units.

Mean of the given numbers = (3 + 5 + 9 + 15 + x)/5 = $\frac{32 + x}{5}$
For, x = 1, 2 the mean will be 6.xx. In both these cases, the median will be 5. Hence, they will satisfy the given conditions.
Similarly x = 4 will also satisfy the given condition. If n = 6, 7 or 8. It will be the median. In this also the mean will be greater than it. If x = 10, then mean will be 8.xx but median will be 9. For all values of x greater than or equal to 13, the given conditions will be satisfied. Hence, the required values of x can be
1, 2, 4, 6, 7, 8, 13, 14, 16, 17, 18 and 19
Thus, 12 values of x satisfy the given conditions.

Median = 5 , Mode = 8

Mean = 5, => Sum of 5 salaries = $25$

As mode is 8, it will occur 2 times but not 3 ($\because$ sum is 25)

Also, median is 5, the third salary is 5 and the first two are less than 5

=> Sum of third , fourth and fifth salary = 5 + 8 + 8 = 21

Sum of first two = 25 – 21 = 4

First and second salaries cannot be same ($\because$ mode is 8)

=> First and second salary = 1 and 3

$\therefore$ Sum (in lakh) of the highest and the lowest salaries = 1 + 8 = 9

Median of 11 integers is 15, => In ascending order 6th integer = 15

=> Numbers = 6,8,12,13,14,15,20,22

Statement I : Average of four smallest = 6 + 8 + 12 + 13

= $\frac{39}{4} = 9.75$

It is given that, avg of 4 largest – avg of 4 smallest = 13.25

=> Average of 4 largest = 13.25 + 9.75 = 23

=> Sum of 4 largest numbers = 23 * 4 = 92

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement I is sufficient.

Statement II : Sum of 11 integers = 11 * 16 = 176

Sum of given 8 integers = 6+8+12+13+14+15+20+22 = 110

Sum of remaining numbers = 176 – 110 = 66

So, we can easily allocate other three numbers different minimum values but more than 15 and maximize the remaining one value

Thus, statement II is sufficient.

$\therefore$ Either statement I or II is sufficient.

A centroid divides a median in the ratio 2 : 1

=> $\frac{AO}{OD} = \frac{2}{1}$

=> $\frac{10}{OD} = \frac{2}{1}$

=> OD = 5 cm

The median AD drawn on the base BC of $\triangle$ABC is equal to half of BC

=> AD = BD = BC

Here, we can consider D as centre of circle with A,B and C lying on the circumference of circle

=> BC is diameter which subtends an angle of 90.

=> ABC is right angled triangle.