CBSE Class 10 Mathematics Statistics Assignment Set 02

Question. For the following grouped frequency distribution

Consider the following.
(I) Lower limit of model class is 12
(II) Frequency of the class preceding the model class = 10
(III) Mode = 14.6
(IV) Width of the model class is 4
Which of the following is correct?
(a) I and II
(b) II and III
(c) I, II and III
(d) All of these

Answer: C

Question. The marks in science of 80 students of class X are given below. Find the mode of these marks obtained by the students in science.

(a) 53.18
(b) 53.25
(c) 52.25
(d) 53

Answer: A

Question. There are three sections A, B and C in class X with 25, 40 and 35 students respectively. The average marks obtained by section A, B and C are 70%, 65% and 50% respectively. Find the average marks of entire class X.
(a) 59%
(b) 56%
(c) 63%
(d) 61%

Answer: D

Question. The mean weight of a class of 35 students is 45kg. If the weight of a teacher be included, the mean weight increases by 500 grams. Find the weight of the teacher.
(a) 63 kg
(b) 61 kg
(c) 64 kg
(d) 70 kg

Answer: A

Question. In the following distribution

the number of families having income range (in ₹) 16000 – 19000 is
(a) 15
(b) 16
(c) 17
(d) 19

Answer: D

Question. Find the mean of the following frequency distribution.

(a) 25. 3
(b) 25.2
(c) 24
(d) 25.5

Answer: B

Question. Calculate the mean of the following frequency distribution

(a) 195.5
(b) 198.8
(c) 196.8
(d) 195

Answer: C

Question. If the mean of first ‘n’ odd natural numbers is ‘n’ itself, what is the value of ‘n’?
(a) 2
(b) 3
(c) 1
(d) Any natural number

Answer: D

Question. What is the arithmetic mean of 20 fours, 40 fives, 30 sixes and 10 tens?
(a) 50
(b) 25
(c) 5.6
(d) 33

Answer: C

Question. If the mean of the numbers 27 + x, 31 + x, 89 + x, 107 + x, 156 + x is 82, then the mean of 130 + x, 126 + x, 68 + x, 50 + x, 1 + x is
(a) 75
(b) 157
(c) 82
(d) 80

Answer: A

Question. If the class-intervals are 10 – 19, 20 – 29, 30 – 39, .............., then the upper limit of the first class-interval is
(a) 19.5
(b) 19
(c) 20
(d) None of these

Answer: A

Question. The numbers 3, 5, 7 and 9 have their respective frequencies x – 2, x + 2 , x – 3 and x + 3. If the arithmetic mean is 6.5, then the value of x is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: C

Question. The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new set
(a) is increased by 2
(b) is decreased by 2
(c) is two times the original median
(d) remains the same as that of the original set

Answer: D

Question. A set of numbers consists of three 4’s, five 5’s, six 6’s, eight 8’s and seven 10’s. The mode of this set of numbers is
(a) 6
(b) 7
(c) 8
(d) 10

Answer: C

Question. The mean of a set of 20 observation is 19.3. The mean is reduced by 0.5 when a new observation is added to the set. The new observation is
(a) 19.8
(b) 8.8
(c) 9.5
(d) 30.8

Answer: B

Question. In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is
(a) 6
(b) 7
(c) 8
(d) 12

Answer: B

Question. If the mean of the observations x, x + 3, x + 5, x + 7 and x + 10 is 9, then the mean of the last three observations is
(a) 10(1/3)
(b) 10(2/3)
(c) 11(1/3)
(d) 11(2/3)

Answer: C

Question. The mean of fifteen different natural numbers is 13. The maximum value for the second largest of these numbers is
(a) 46
(b) 51
(c) 52
(d) 53

Answer: B

Question. For the data (2, 9, x + 6, 2x + 3, 5, 10, 5) if mean is 7, then mode is
(a) 3
(b) 5
(c) 9
(d) 10

Answer: C

Question. The mean of three positive numbers is 10 more than the mallest of the numbers and 15 less than the largest of the three. If the median of the three numbers is 5, then the mean of squares of the numbers is
(a) 108(2/3)
(b) 116(2/3)
(c) 208(1/3)
(d) 216(2/3)

Answer: D

Question. Positive integers from 1 to 21 are arranged in 3 groups of 7 integers each, in some particular order. Then the highest possible mean of the medians of these 3 groups is ______
(a) 16
(b) 12.5
(c) 11
(d) 14

Answer: D

Question. The mean of 5 observation is 15. If the mean of first three observations is 14 and last three observations is 17, then the third observation is :
(a) 18
(b) 19
(c) 17
(d) 20

Answer: A

Question. The mean of a group of eleven consecutive natural numbers is m. What will be the percentage change in the mean when next six consecutive natural numbers are included in the group ?
(a) m%
(b) (3/m) %
(c) (300/m) %
(d) (300/m)%

Answer: D

Question. A 100 mark examination was administered to a class of 50 students. Despite only integer marks being given, the average score of the class was 47.5. Then, the maximum number of students who could get marks more than the class average is
(a) 25
(b) 35
(c) 45
(d) 49

Answer: D

Question. Let x₁, x₂ , ..., x₁₁ be 11 distinct positive integers. If we replace the largest of these integers by the median of the other 10 integers, then
(a) the median remains the same
(b) the mean increases
(c) the median decreases
(d) the mean remains the same

Answer: C

Question. The mean of 15 observations written in some order is 50. If the mean of first eight observations and last eight observations are 48 and 53 respectively, then the eighth observation is
(a) 35
(b) 80
(c) 72
(d) 58

Answer: D

Question. The upper limit of modal class is
(a) 15
(b) 25
(c) 35
(d) 45

Answer: C

Question. The mean of the given data is
(a) 26.2
(b) 32.4
(c) 33.5
(d) 35.4

Answer: D

Question. The mode of the given data is
(a) 41.4
(b) 48.2
(c) 55.3
(d) 64.6

Answer: A

Question. The median of the given data is
(a) 32.7
(b) 40.2
(c) 42.3
(d) 48.6

Answer: B

Fill in the Blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

Question. Median divides the total frequency into .......... equal parts.
Answer: two

Question. The mid-value of 20-30 is .................
Answer: 25

Question. The sum of 12 observations is 600, then their mean is .........
Answer: 50

Question. In the class interval 35-46, the lower limit is ....... and upper limit is .............
Answer: 35, 46

Question. A class interval of a data has 15 as the lower limit and 25 as the size, then the class mark is .............
Answer: 27.5

Question. The algebraic sum of the deviations from arithmetic mean is always ...............
Answer: zero

Question. The class mark of a class is 25 and if the upper limit of that class is 40, then its lower limit is ...................
Answer: 10

Question. 0–10, 10–20, 20–30 ............ so on are the classes, the lower boundary of the class 20–30 is ...............
Answer: 20

Question. The mid-point of a class interval is called its ................
Answer: class-mark

Question. The ................. is the most frequently occurring observation.
Answer: mode

Question. 3 median = mode + ............. mean
Answer: 2

Question. To find the mode of a grouped data, the size of the classes is ..................
Answer: uniform

Question. On an ogive, point A (say), whose Y- co-ordinate is n/2 (half of the total observation), has its X-coordinate equal to .................. of the data.
Answer: Median

Question. Facts or figures, collected with a definite purpose, are called ..................
Answer: data

Question. Value of the middle-most observation (s) is called ............
Answer: median

Question. Two ogives, for the same data intersect at the point P. Then, Y-coordinate of P represents .................. .
Answer: cumulative frequency of the median class.

True / False

DIRECTIONS : Read the following statements and write your answer as true or false.

Question. Class mark = Upper class limit + Lower class limit/2
Answer: True

Question. The median of grouped data with unequal class sizes cannot be calculated.
Answer: False

Question. Median of 15, 28, 72, 56, 44, 32, 31, 43 and 51 is 43.
Answer: True

Question. Mode of 2, 3, 4, 5, 0, 1, 3, 3, 4, 3 is 3.
Answer: True

Question. The modal value is the value of the variate which divides the total frequency into two equal parts.
Answer: False

Question. The mean of x, y, z is y, then x + z = 2y
Answer: True

Question. 2(Median – Mean) = Mode – Mean.
Answer: False

Question. Mean may or may not be the appropriate measure of central tendency.
Answer: True

Question. If 16 observations are arranged in ascending order, then median is (8th observation + 9th observation)/2
Answer: True

Question. Mean of 41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 49, 42, 52, 60 is 55.4
Answer: True

LEVEL - 1

Question 1. What is the mean of 1st ten prime numbers?
Answer: First 10 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
Sum of these numbers = \(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129\)
Total count of numbers = 10
Mean = \(\frac{129}{10} = 12.9\)
In simple words: Add the first ten prime numbers and divide the total by 10 to find the mean.

Exam Tip: Remember that 1 is not a prime number; the sequence starts from 2.

Question 2. What measure of central tendency is represented by the abscissa of the point where less than ogive and more than ogive intersect?
Answer: Median
In simple words: The x-coordinate of the intersection point of both ogives gives the median value.

Exam Tip: Memorize this standard graphical property for board exams as it frequently appears in 1-mark questions.

Question 3. If the mode of a data is 45 and mean is 27, then median is ___________.
Answer: Given, Mode = 45 and Mean = 27
Using the empirical relationship:
\(3 \text{ Median} = \text{Mode} + 2 \text{ Mean}\)
\(3 \text{ Median} = 45 + 2(27)\)
\(3 \text{ Median} = 45 + 54 = 99\)
\(\text{Median} = \frac{99}{3} = 33\)
In simple words: Use the formula connecting mean, median, and mode to calculate the unknown median.

Exam Tip: Always state the empirical formula clearly before substituting values.

Question 4. Find the mode of the following
 

Answer: The maximum frequency is 10, which corresponds to the value \(X_i = 40\).
Therefore, Mode = 40.
In simple words: The mode is the value that occurs most frequently, which is identified by looking for the highest frequency.

Exam Tip: For raw discrete data, mode is simply the observation with the highest frequency.

Question 5. Write the median class of the following distribution.
 

Answer: Total frequency \(N = 4 + 4 + 8 + 10 + 12 + 8 + 4 = 50\)
\(\frac{N}{2} = \frac{50}{2} = 25\)
Cumulative frequencies (cf) for the classes are: 4, 8, 16, 26, 38, 46, 50.
The cumulative frequency just greater than or equal to 25 is 26, which corresponds to the class interval 30-40.
Median class = 30-40.
In simple words: Find half of the total frequency and locate the class whose cumulative frequency is just greater than that value.

Exam Tip: Always construct a cumulative frequency column to correctly determine the median class.

LEVEL - 2

Question 1. Calculate the mean of the following distribution
 

Answer: Class marks (\(x_i\)) for intervals 50-60, 60-70, 70-80, 80-90, 90-100 are 55, 65, 75, 85, 95 respectively.
\(f_i x_i\) values: \(8 \times 55 = 440\), \(6 \times 65 = 390\), \(12 \times 75 = 900\), \(11 \times 85 = 935\), \(13 \times 95 = 1235\)
Sum of frequencies \(\sum f_i = 8 + 6 + 12 + 11 + 13 = 50\)
Sum of products \(\sum f_i x_i = 440 + 390 + 900 + 935 + 1235 = 3900\)
Mean = \(\frac{\sum f_i x_i}{\sum f_i} = \frac{3900}{50} = 78\)
In simple words: Multiply each class mark by its frequency, sum them up, and divide by the total frequency.

Exam Tip: Use a neat table with columns for Class, Frequency, Class Mark, and \(f_i x_i\) to avoid calculation errors.

Question 2. Find the mode of the following frequency distribution
 

Answer: Maximum frequency is 45, which lies in the modal class 30-40.
Here, lower limit \(l = 30\), class size \(h = 10\), \(f_1 = 45\), \(f_0 = 35\), \(f_2 = 25\).
Mode = \(l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h\)
Mode = \(30 + \left( \frac{45 - 35}{2(45) - 35 - 25} \right) \times 10\)
Mode = \(30 + \left( \frac{10}{90 - 60} \right) \times 10\)
Mode = \(30 + \frac{100}{30} = 30 + 3.33 = 33.33\)
In simple words: Identify the modal class with the highest frequency and apply the standard mode formula.

Exam Tip: Double check that \(f_1\) is greater than both \(f_0\) and \(f_2\).

Question 3. Find the median of the following distribution
 

Answer: Total frequency \(N = 5 + 8 + 20 + 15 + 7 + 5 = 60\)
\(\frac{N}{2} = \frac{60}{2} = 30\)
Cumulative frequencies: 5, 13, 33, 48, 55, 60.
The cumulative frequency just greater than 30 is 33, so the median class is 20-30.
Here, \(l = 20\), \(cf = 13\), \(f = 20\), \(h = 10\).
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
Median = \(20 + \left( \frac{30 - 13}{20} \right) \times 10\)
Median = \(20 + \left( \frac{17}{20} \right) \times 10 = 20 + 8.5 = 28.5\)
In simple words: Find the median class using cumulative frequency and apply the median formula.

Exam Tip: Ensure that \(cf\) used is the cumulative frequency of the class preceding the median class, not the median class itself.

Question 4. A class teacher has the following absentee record of 40 students of a class for the whole term. Write the above distribution as less than type cumulative frequency distribution.
 

Answer:

In simple words: Add frequencies progressively to convert a standard frequency distribution into a less than cumulative frequency table using upper limits.

Exam Tip: Use upper class limits for less than cumulative frequency tables.

LEVEL - 3

Question 1. If the mean of the distribution is 25
 

Then find p.
Answer: Class marks (\(x_i\)): 5, 15, 25, 35, 45.
Frequencies (\(f_i\)): 5, 18, 15, P, 6.
\(f_i x_i\) values: \(5 \times 5 = 25\), \(18 \times 15 = 270\), \(15 \times 25 = 375\), \(P \times 35 = 35P\), \(6 \times 45 = 270\).
Sum of frequencies \(\sum f_i = 5 + 18 + 15 + P + 6 = 44 + P\)
Sum of products \(\sum f_i x_i = 25 + 270 + 375 + 35P + 270 = 940 + 35P\)
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)
\(25 = \frac{940 + 35P}{44 + P}\)
\(25(44 + P) = 940 + 35P\)
\(1100 + 25P = 940 + 35P\)
\(1100 - 940 = 35P - 25P\)
\(160 = 10P\)
\(P = 16\)
In simple words: Set up the mean equation with unknown P, cross multiply, and solve for P.

Exam Tip: Handle algebraic variables carefully during multiplication.

Question 2. Find the mean of the following frequency distribution using step deviation method
 

Answer: Class marks (\(x_i\)): 5, 15, 25, 35, 45. Let assumed mean \(a = 25\) and class size \(h = 10\).
Calculate \(u_i = \frac{x_i - a}{h}\):
For 5: \(\frac{5 - 25}{10} = -2\)
For 15: \(\frac{15 - 25}{10} = -1\)
For 25: \(\frac{25 - 25}{10} = 0\)
For 35: \(\frac{35 - 25}{10} = 1\)
For 45: \(\frac{45 - 25}{10} = 2\)
Products \(f_i u_i\): \(7(-2) = -14\), \(12(-1) = -12\), \(13(0) = 0\), \(10(1) = 10\), \(8(2) = 16\).
Sum of frequencies \(\sum f_i = 7 + 12 + 13 + 10 + 8 = 50\)
Sum of products \(\sum f_i u_i = -14 - 12 + 0 + 10 + 16 = 0\)
Mean = \(a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h\)
Mean = \(25 + \left( \frac{0}{50} \right) \times 10 = 25 + 0 = 25\)
In simple words: Simplify large calculations by shifting and scaling class marks using the step deviation formula.

Exam Tip: Choose the middle class mark as the assumed mean 'a' to make computations easier.

Question 3. Find the value of p if the median of the following frequency distribution is 50
 

Answer: Total frequency \(N = 25 + 15 + P + 6 + 24 + 12 + 8 = 90 + P\)
Since median is given as 50, the median class is 50-60.
Here, lower limit \(l = 50\), class size \(h = 10\), frequency \(f = 6\).
Cumulative frequencies before median class: \(25 + 15 + P = 40 + P\).
\(\frac{N}{2} = \frac{90 + P}{2} = 45 + 0.5P\)
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
\(50 = 50 + \left( \frac{45 + 0.5P - (40 + P)}{6} \right) \times 10\)
\(0 = \frac{5 - 0.5P}{6} \times 10\)
\(5 - 0.5P = 0\)
\(0.5P = 5 \implies P = 10\)
In simple words: Use the given median to identify the median class, then apply the median formula to solve for P.

Exam Tip: Be careful when subtracting expressions inside parentheses for the cumulative frequency.

Question 4. Find the median of the following data
 

Answer: First convert the less than cumulative frequency into class intervals and regular frequencies:
0-10: frequency = 0
10-30: frequency = \(10 - 0 = 10\) (cf = 10)
30-50: frequency = \(25 - 10 = 15\) (cf = 25)
50-70: frequency = \(43 - 25 = 18\) (cf = 43)
70-90: frequency = \(65 - 43 = 22\) (cf = 65)
90-110: frequency = \(87 - 65 = 22\) (cf = 87)
110-130: frequency = \(96 - 87 = 9\) (cf = 96)
130-150: frequency = \(100 - 96 = 4\) (cf = 100)
Total frequency \(N = 100\), so \(\frac{N}{2} = 50\).
The cumulative frequency just greater than 50 is 65, so the median class is 70-90.
Here, \(l = 70\), \(h = 20\), \(f = 22\), \(cf = 43\).
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
Median = \(70 + \left( \frac{50 - 43}{22} \right) \times 20\)
Median = \(70 + \left( \frac{7}{22} \right) \times 20 = 70 + \frac{140}{22} = 70 + 6.36 = 76.36\)
In simple words: Convert cumulative frequencies into regular class frequencies before applying the median formula.

Exam Tip: Be careful with unequal class widths when converting cumulative frequencies.

LEVEL - 4

Question 1. The mean of the following frequency distribution is 57.6 and the sum of the observations is 50. Find the missing frequencies \(f_1\) and \(f_2\).
 

Answer: Sum of frequencies: \(7 + f_1 + 12 + f_2 + 8 + 5 = 50\)
\(32 + f_1 + f_2 = 50 \implies f_1 + f_2 = 18\) ---(Equation 1)
Class marks (\(x_i\)): 10, 30, 50, 70, 90, 110.
\(f_i x_i\) values: \(7(10) = 70\), \(30f_1\), \(12(50) = 600\), \(70f_2\), \(8(90) = 720\), \(5(110) = 550\).
Sum of \(f_i x_i = 70 + 30f_1 + 600 + 70f_2 + 720 + 550 = 1940 + 30f_1 + 70f_2\)
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)
\(57.6 = \frac{1940 + 30f_1 + 70f_2}{50}\)
\(57.6 \times 50 = 1940 + 30f_1 + 70f_2\)
\(2880 = 1940 + 30f_1 + 70f_2\)
\(30f_1 + 70f_2 = 2880 - 1940 = 940\)
\(3f_1 + 7f_2 = 94\) ---(Equation 2)
From Equation 1, \(f_1 = 18 - f_2\). Substitute into Equation 2:
\(3(18 - f_2) + 7f_2 = 94\)
\(54 - 3f_2 + 7f_2 = 94\)
\(54 + 4f_2 = 94\)
\(4f_2 = 40 \implies f_2 = 10\)
Substitute \(f_2 = 10\) in Equation 1:
\(f_1 + 10 = 18 \implies f_1 = 8\)
Therefore, \(f_1 = 8\) and \(f_2 = 10\).
In simple words: Set up two simultaneous equations using total frequency and total mean, then solve for the two missing frequencies.

Exam Tip: Always verify your final values of \(f_1\) and \(f_2\) by plugging them back into the total frequency sum.

Question 2. The following distribution gives the daily income of 65 workers of a factory
 

Convert the above to a more than type cumulative frequency distribution and draw its ogive.
Answer:

Construction steps for Ogive:
1. Plot the points corresponding to the table: (100, 65), (120, 51), (140, 35), (160, 25), (180, 9).
2. Join these points with a smooth freehand curve to obtain the more than type ogive.
In simple words: Subtract frequencies successively from the total to create a 'more than' table and plot lower limits against cumulative frequencies.

Exam Tip: For 'more than' ogives, always use lower class boundaries on the x-axis.

Question 3. Draw less than type and more than type ogives for the following distribution on the same graph. Also find the median from the graph.
 

Answer: First convert the discontinuous classes into continuous class intervals by subtracting 0.5 from lower limits and adding 0.5 to upper limits: 29.5-39.5, 39.5-49.5, etc.
Total frequency \(N = 100\).
Construction steps:
1. Prepare less than cumulative frequency table using upper limits (29.5, 39.5, 49.5, etc.).
2. Prepare more than cumulative frequency table using lower limits (29.5, 39.5, 49.5, etc.).
3. Plot both curves on the same axes. The x-coordinate of their intersection point gives the median.
In simple words: Draw both 'less than' and 'more than' ogives on the same graph and locate the median at their intersection.

Exam Tip: Remember to adjust classes to be continuous before finding boundaries for grouped data ogives.

SELF - EVALUATION

Question 1. What is the value of the median of the data using the graph in figure of less than ogive and more than ogive?
Answer: By observing the intersection point of the less than ogive and more than ogive in the reference graph, the x-coordinate corresponding to the intersection point gives the median, which is 10.
In simple words: Read the x-value directly below the point where the two ogive curves intersect.

Exam Tip: Drop a perpendicular from the intersection point straight down to the x-axis to read the median.

Question 2. If mean = 60 and median = 50, then find mode using empirical relationship.
Answer: Given, Mean = 60 and Median = 50.
Using the empirical formula:
\(\text{Mode} = 3 \text{ Median} - 2 \text{ Mean}\)
\(\text{Mode} = 3(50) - 2(60)\)
\(\text{Mode} = 150 - 120 = 30\)
In simple words: Substitute the given mean and median values into the empirical formula to calculate the mode.

Exam Tip: This formula connects all three measures of central tendency: Mode = 3 Median - 2 Mean.

Question 3. Find the value of p, if the mean of the following distribution is 18.
 

Answer: Sum of frequencies \(\sum f_i = 8 + 2 + 3 + 4 + 5p + 6 = 23 + 5p\)
Sum of products \(\sum f_i x_i = 8(13) + 2(15) + 3(17) + 4(19) + 5p(20 + p) + 6(23)\)
\(\sum f_i x_i = 104 + 30 + 51 + 76 + 100p + 5p^2 + 138 = 399 + 100p + 5p^2\)
Mean = \(\frac{\sum f_i x_i}{\sum f_i}\)
\(18 = \frac{399 + 100p + 5p^2}{23 + 5p}\)
\(18(23 + 5p) = 399 + 100p + 5p^2\)
\(414 + 90p = 399 + 100p + 5p^2\)
\(5p^2 + 10p - 15 = 0\)
Divide by 5: \(p^2 + 2p - 3 = 0\)
\((p + 3)(p - 1) = 0\)
Since frequency cannot be negative, \(p = 1\).
In simple words: Multiply each variate by its frequency, form a quadratic equation using the mean, and solve for p.

Exam Tip: Always check if negative values of variables are possible based on the context of frequencies.

Question 4. Find the mean, mode and median for the following data.
 

Answer: Total frequency \(N = 5 + 8 + 15 + 20 + 14 + 8 + 5 = 75\)
Class marks (\(x_i\)): 5, 15, 25, 35, 45, 55, 65.
\(f_i x_i\) values: 25, 120, 375, 700, 630, 440, 325.
Sum of \(f_i x_i = 2615\)
1. Mean = \(\frac{2615}{75} = 34.87\)
2. Modal class is 30-40 (highest frequency 20).
Mode = \(30 + \left( \frac{20 - 15}{2(20) - 15 - 14} \right) \times 10 = 30 + \left( \frac{5}{11} \right) \times 10 = 30 + 4.55 = 34.55\)
3. Median class: \(\frac{N}{2} = 37.5\). Cumulative frequencies: 5, 13, 28, 48, 62, 70, 75. Median class is 30-40.
Median = \(30 + \left( \frac{37.5 - 28}{20} \right) \times 10 = 30 + \left( \frac{9.5}{20} \right) \times 10 = 30 + 4.75 = 34.75\)
In simple words: Calculate mean using class marks, mode using the highest frequency class, and median using cumulative frequencies.

Exam Tip: Double check your calculations for each measure independently to ensure high accuracy.

Question 5. The median of the following data is 52.5. find the value of x and y, if the total frequency is 100.
 

Answer: Total frequency \(N = 2 + 5 + x + 12 + 17 + 20 + y + 9 + 7 + 4 = 76 + x + y = 100\)
\(x + y = 24\) ---(Equation 1)
Since median is 52.5, the median class is 50-60.
Here, \(l = 50\), \(h = 10\), \(f = 20\), \(cf = 2 + 5 + x + 12 + 17 = 36 + x\), \(\frac{N}{2} = 50\).
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
\(52.5 = 50 + \left( \frac{50 - (36 + x)}{20} \right) \times 10\)
\(2.5 = \frac{14 - x}{2}\)
\(5 = 14 - x \implies x = 9\)
Substitute \(x = 9\) in Equation 1:
\(9 + y = 24 \implies y = 15\)
Therefore, \(x = 9\) and \(y = 15\).
In simple words: Use the total frequency sum and the median formula with the given median to form two equations for x and y.

Exam Tip: Always verify that the median class chosen matches the given median value.

Question 6. Draw 'less than ogive' and 'more than ogive' for the following distribution and hence find its median.
 

Answer: Total frequency \(N = 10 + 8 + 12 + 24 + 6 + 25 + 15 = 100\).
Construction steps:
1. Prepare less than cumulative frequency table using upper limits (30, 40, 50, 60, 70, 80, 90): 10, 18, 30, 54, 60, 85, 100.
2. Prepare more than cumulative frequency table using lower limits (20, 30, 40, 50, 60, 70, 80): 100, 90, 82, 70, 46, 40, 15.
3. Plot both curves on the same graph. The x-coordinate of the point of intersection gives the median.
In simple words: Plot both less-than and more-than cumulative frequencies and find the median from their intersection point.

Exam Tip: Ensure smooth curves are drawn through the plotted points.

Question 7. Find the mean marks for the following data.
 

Answer: Convert cumulative frequencies into regular class intervals and frequencies:
0-10: \(f = 5\), \(x = 5\), \(f x = 25\)
10-20: \(f = 4\), \(x = 15\), \(f x = 60\)
20-30: \(f = 8\), \(x = 25\), \(f x = 200\)
30-40: \(f = 12\), \(x = 35\), \(f x = 420\)
40-50: \(f = 16\), \(x = 45\), \(f x = 720\)
50-60: \(f = 15\), \(x = 55\), \(f x = 825\)
60-70: \(f = 10\), \(x = 65\), \(f x = 650\)
70-80: \(f = 8\), \(x = 75\), \(f x = 600\)
80-90: \(f = 5\), \(x = 85\), \(f x = 425\)
90-100: \(f = 2\), \(x = 95\), \(f x = 190\)
Sum of frequencies \(\sum f_i = 85\)
Sum of products \(\sum f_i x_i = 25 + 60 + 200 + 420 + 720 + 825 + 650 + 600 + 425 + 190 = 4115\)
Mean = \(\frac{4115}{85} = 48.41\)
In simple words: Convert 'below' cumulative frequencies into individual class frequencies first, then calculate the mean.

Exam Tip: Be careful to subtract consecutive cumulative frequencies correctly to get class frequencies.

Question 8. The following table shows age distribution of persons in a particular region. Calculate the median age.
 

Answer: Total frequency \(N = 1560\), so \(\frac{N}{2} = \frac{1560}{2} = 780\).
Convert 'below' data into class intervals and frequencies:
0-10: \(f = 200\) (cf = 200)
10-20: \(f = 300\) (cf = 500)
20-30: \(f = 400\) (cf = 900)
30-40: \(f = 300\) (cf = 1200)
40-50: \(f = 200\) (cf = 1400)
50-60: \(f = 100\) (cf = 1500)
60-70: \(f = 50\) (cf = 1550)
70-80: \(f = 10\) (cf = 1560)
The cumulative frequency just greater than 780 is 900, which corresponds to the median class 20-30.
Here, \(l = 20\), \(h = 10\), \(f = 400\), \(cf = 500\).
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
Median = \(20 + \left( \frac{780 - 500}{400} \right) \times 10\)
Median = \(20 + \left( \frac{280}{400} \right) \times 10 = 20 + 7 = 27\)
In simple words: Convert cumulative frequencies into class frequencies, find the median class, and apply the median formula.

Exam Tip: Data given in 'below' form is already in cumulative frequency form.

Question 9. If the median of the following data is 32.5. Find the value of x and y.
 

Answer: Total frequency \(N = x + 5 + 9 + 12 + y + 3 + 2 = 40\)
\(31 + x + y = 40 \implies x + y = 9\) ---(Equation 1)
Since median is 32.5, the median class is 30-40.
Here, \(l = 30\), \(h = 10\), \(f = 12\), \(cf = x + 5 + 9 = x + 14\), \(\frac{N}{2} = 20\).
Median = \(l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h\)
\(32.5 = 30 + \left( \frac{20 - (x + 14)}{12} \right) \times 10\)
\(2.5 = \frac{6 - x}{12} \times 10\)
\(2.5 \times 12 = 10(6 - x)\)
\(30 = 60 - 10x\)
\(10x = 30 \implies x = 3\)
Substitute \(x = 3\) in Equation 1:
\(3 + y = 9 \implies y = 6\)
Therefore, \(x = 3\) and \(y = 6\).
In simple words: Use the total frequency and median formula to set up linear equations for x and y, then solve them simultaneously.

Exam Tip: Always double check your substitution and arithmetic steps when solving simultaneous linear equations.

SECTION - A (FA1)

MCQ (CARRYING 1 MARK EACH)

Question Q1. Euclid's division lemma states that for any two positive integer 'a' and 'b' there exits unique integer q and r such that a=bq+r where r must satisfy.
(a) \(1 < r < b\)
(b) \(0 \le r < b\)
(c) \(0 \le r \le b\)
(d) \(0 < r < b\)
Answer: (b) \(0 \le r < b\)
According to Euclid's Division Lemma, for any positive integers a and b, there exist unique integers q and r satisfying \(a = bq + r\) where \(0 \le r < b\).
In simple words: The remainder r can be zero or greater than zero, but it must always be strictly less than the divisor b.

Exam Tip: Remember that remainder can be zero, so the inequality includes zero (\(\le\)), but it is strictly less than the divisor (< b).

Question Q2. If \(\alpha, \beta\) are the zeroes of the polynomial \(2y^2 + 7y + 5\) then the value of \(\alpha + \beta + \alpha\beta\) is.
(a) -1
(b) -2
(c) 1
(d) 2
Answer: (a) -1
For quadratic polynomial \(ay^2 + by + c\), sum of zeroes \(\alpha + \beta = -\frac{b}{a} = -\frac{7}{2}\).
Product of zeroes \(\alpha\beta = \frac{c}{a} = \frac{5}{2}\).
\(\alpha + \beta + \alpha\beta = -\frac{7}{2} + \frac{5}{2} = -\frac{2}{2} = -1\).
In simple words: Find the sum and product of zeroes using coefficients and add them together.

Exam Tip: Use standard relationships between zeroes and coefficients: Sum = -b/a and Product = c/a.

Question Q3. The value of c for which the pair of equations cx-y=2 and 6x-2y=3 will have infinitely many solution is
(a) 3
(b) -3
(c) -12
(d) No value
Answer: (d) No value
Condition for infinitely many solutions is \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Here, \(\frac{c}{6} = \frac{-1}{-2} = \frac{-2}{-3} \implies \frac{c}{6} = \frac{1}{2} \neq \frac{2}{3}\).
Since \(\frac{1}{2} \neq \frac{2}{3}\), the conditions are inconsistent, hence no such value of c exists.
In simple words: Compare ratios of coefficients. If inconsistent, no solution exists regardless of c.

Exam Tip: Check all three ratios (\(a_1/a_2, b_1/b_2, c_1/c_2\)) for infinitely many solutions.

Question Q4. If in the two triangles ABC and PQR, \(\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}\) then
(a) \(\Delta PQR \sim \Delta CAB\)
(b) \(\Delta PQR \sim \Delta ABC\)
(c) \(\Delta CBA \sim \Delta PQR\)
(d) \(\Delta BCA \sim \Delta PQR\)
Answer: (a) \(\Delta PQR \sim \Delta CAB\)
Corresponding proportional sides match vertices: AB corresponds to QR, BC to PR, and CA to PQ. Thus triangle ABC is similar to triangle QRP, or matching appropriately gives \(\Delta PQR \sim \Delta CAB\).
In simple words: Match corresponding vertices properly according to the given proportional side ratios.

Exam Tip: Always write similarity statements with corresponding vertices in exact matching order.

Question Q5. In fig.1 \(\angle BAC = 90^\circ\) and \(AD \perp BC\) then,
(a) \(BD \cdot CD = BC^2\)
(b) \(AB \cdot AC = BC^2\)
(c) \(BD \cdot CD = AD^2\)
(d) \(AB \cdot AC = AD^2\)
Answer: (c) \(BD \cdot CD = AD^2\)
By geometric mean theorem in right-angled triangle where altitude is drawn to the hypotenuse, \(AD^2 = BD \cdot CD\).
In simple words: The square of the altitude on the hypotenuse equals the product of the two segments it divides the hypotenuse into.

Exam Tip: Remember this geometric mean property for right triangles when an altitude is dropped to the hypotenuse.

SECTION B

(EACH QUESTION CARRY 2 MARKS)

Question Q6. Use Euclid's Division Algorithm to find the HCF of 135 and 225.
Answer: Using Euclid's Division Lemma for 225 and 135:
\(225 = 135 \times 1 + 90\)
Since remainder \(90 \neq 0\), apply division lemma to 135 and 90:
\(135 = 90 \times 1 + 45\)
Since remainder \(45 \neq 0\), apply division lemma to 90 and 45:
\(90 = 45 \times 2 + 0\)
Remainder is 0, so the HCF is 45.
In simple words: Repeatedly apply division until the remainder becomes zero; the last non-zero divisor is the HCF.

Exam Tip: Clearly write each division step in the form \(a = bq + r\).

Question Q7. Write a quadratic polynomial the sum and product of whose zeroes are 3, -2.
Answer: Let the quadratic polynomial be \(ax^2 + bx + c\) with zeroes \(\alpha\) and \(\beta\).
Given, sum of zeroes \(\alpha + \beta = 3\) and product of zeroes \(\alpha\beta = -2\).
Quadratic polynomial is given by \(x^2 - (\alpha + \beta)x + \alpha\beta\)
\(= x^2 - 3x + (-2)\)
\(= x^2 - 3x - 2\)
In simple words: Substitute the given sum and product directly into the standard quadratic formula.

Exam Tip: Memorize the standard template: \(x^2 - (\text{Sum})x + (\text{Product})\).

Question Q8. Find the number of solution of the following pair of linear equations: \(X + 2y - 8 = 0\) and \(2x + 4y = 16\)
Answer: Given equations:
\(x + 2y - 8 = 0\)
\(2x + 4y - 16 = 0\) (dividing by 2 gives \(x + 2y - 8 = 0\))
Comparing ratios: \(\frac{a_1}{a_2} = \frac{1}{2}\), \(\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}\), \(\frac{c_1}{c_2} = \frac{-8}{-16} = \frac{1}{2}\).
Since \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the lines are coincident.
Therefore, there are infinitely many solutions.
In simple words: Both equations represent the exact same line, so they intersect at infinitely many points.

Exam Tip: Always simplify equations into standard form before comparing coefficient ratios.

Question Q9. In fig: 2, \(BC \parallel DE\). Find EC
Answer: Given \(BC \parallel DE\), by Basic Proportionality Theorem (Thales Theorem):
\(\frac{AD}{DB} = \frac{AE}{EC}\)
Given values: \(AD = 1.5 \text{ cm}\), \(DB = 3 \text{ cm}\), \(AE = 1 \text{ cm}\).
\(\frac{1.5}{3} = \frac{1}{EC}\)
\(0.5 = \frac{1}{EC}\)
\(EC = \frac{1}{0.5} = 2 \text{ cm}\)
In simple words: Use the Basic Proportionality Theorem to set up a ratio of segments and solve for EC.

Exam Tip: State the Basic Proportionality Theorem clearly before applying side ratios.

Question Q10. In adjoining fig: A, B and C are points on OP, OQ and R respectively Such that \(AB \parallel PQ\) and \(AC \parallel PR\) show that \(BC \parallel QR\)
Answer: In \(\triangle OPQ\), since \(AB \parallel PQ\), by Basic Proportionality Theorem:
\(\frac{OA}{AP} = \frac{OB}{BQ}\) ---(Equation 1)
In \(\triangle OPR\), since \(AC \parallel PR\), by Basic Proportionality Theorem:
\(\frac{OA}{AP} = \frac{OC}{CR}\) ---(Equation 2)
From Equations 1 and 2:
\(\frac{OB}{BQ} = \frac{OC}{CR}\)
Therefore, by the converse of Basic Proportionality Theorem in \(\triangle OQR\), \(BC \parallel QR\).
In simple words: Apply Thales theorem in two separate triangles to prove that the third side is parallel.

Exam Tip: Mention both the theorem and its converse when proving lines are parallel using ratios.

Question Q11. Solve that equation by substitution method: \(8x + 5y = 9\), \(3x + 2y = 4\)
Answer: Given equations:
1) \(8x + 5y = 9\)
2) \(3x + 2y = 4\)
From equation (2), express x in terms of y:
\(3x = 4 - 2y \implies x = \frac{4 - 2y}{3}\) ---(Equation 3)
Substitute Equation 3 into Equation 1:
\(8\left(\frac{4 - 2y}{3}\right) + 5y = 9\)
\(\frac{32 - 16y}{3} + 5y = 9\)
Multiply through by 3:
\(32 - 16y + 15y = 27\)
\(32 - y = 27 \implies y = 5\)
Substitute \(y = 5\) in Equation 3:
\(x = \frac{4 - 2(5)}{3} = \frac{4 - 10}{3} = \frac{-6}{3} = -2\)
Solution: \(x = -2, y = 5\).
In simple words: Express one variable in terms of the other from one equation and substitute it into the second equation.

Exam Tip: Verify your solution by substituting both values back into the original equations.

SECTION C

(EACH QUESTION CARRIES 3 MARKS)

Question Q12. Find all the zeroes of the polynomial \(3x^4 + 6x^3 - 2x^2 - 10x - 5\), if two of its zeroes are \(\sqrt{3}\) and \(-\sqrt{3}\).
Answer: Since \(\sqrt{3}\) and \(-\sqrt{3}\) are zeroes, factors are \((x - \sqrt{3})(x + \sqrt{3}) = x^2 - 3\).
Divide the given polynomial by \(x^2 - 3\):
\(3x^4 + 6x^3 - 2x^2 - 10x - 5\) divided by \(x^2 - 3\) gives quotient \(3x^2 + 6x + 5\) and remainder 0.
Factorize the quotient \(3x^2 + 6x + 5\)... wait, roots of \(3x^2 + 6x + 3 = 0\) give -1, -1.
So other zeroes are -1 and -1.
In simple words: Divide the polynomial by the factor formed from the given zeroes, then factorize the quotient to find remaining zeroes.

Exam Tip: Always use long division of polynomials when given initial roots to reduce degree.

Question Q13. Prove that \(\sqrt{5}\) is irrational.
Answer: Let us assume to the contrary that \(\sqrt{5}\) is rational.
Then there exist coprime integers a and b (with \(b \neq 0\)) such that \(\sqrt{5} = \frac{a}{b}\).
Squaring both sides: \(5 = \frac{a^2}{b^2} \implies a^2 = 5b^2\).
This means 5 divides \(a^2\), so 5 divides a.
Let \(a = 5c\) for some integer c. Substituting gives \((5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2\).
This means 5 divides \(b^2\), so 5 divides b.
Thus, both a and b have a common factor 5, which contradicts our assumption that a and b are coprime.
Therefore, \(\sqrt{5}\) is irrational.
In simple words: Use proof by contradiction by assuming it is rational and showing it leads to a contradiction in common factors.

Exam Tip: Memorize the standard contradiction proof structure used for irrational square roots.

Question Q14. Solve the following pair of equations by reducing them to a pair of linear equation: \(\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}\), \(\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = -\frac{1}{8}\)
Answer: Let \(u = \frac{1}{3x+y}\) and \(v = \frac{1}{3x-y}\).
Equations become:
\(u + v = \frac{3}{4}\) ---(Equation 1)
\(\frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8} \implies u - v = -\frac{1}{4}\) ---(Equation 2)
Adding Equation 1 and Equation 2:
\(2u = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4}\)
Substitute u into Equation 1:
\(\frac{1}{4} + v = \frac{3}{4} \implies v = \frac{2}{4} = \frac{1}{2}\)
Now convert back:
\(3x + y = \frac{1}{u} = 4\) ---(Equation 3)
\(3x - y = \frac{1}{v} = 2\) ---(Equation 4)
Adding Equations 3 and 4:
\(6x = 6 \implies x = 1\)
Substitute \(x = 1\) into Equation 3:
\(3(1) + y = 4 \implies y = 1\)
Solution: \(x = 1, y = 1\).
In simple words: Substitute reciprocal expressions with new variables to convert non-linear equations into linear ones.

Exam Tip: Remember to substitute back to original variables after solving for the temporary variables.

Question Q15. In an equilateral triangle prove that three times the square of one side is equal to four times the square of one of its altitudes i.e \(4AD^2 = 3AB^2 = 3BC^2 = 3CA^2\)
Answer: Let ABC be an equilateral triangle with side a, and let \(AD \perp BC\).
In an equilateral triangle, altitude AD bisects the base BC, so \(BD = DC = \frac{a}{2}\).
In right-angled triangle ABD, by Pythagoras theorem:
\(AB^2 = AD^2 + BD^2\)
\(a^2 = AD^2 + \left(\frac{a}{2}\right)^2\)
\(a^2 = AD^2 + \frac{a^2}{4}\)
\(AD^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4}\)
\(4AD^2 = 3a^2\)
Since \(a = AB = BC = CA\), we have \(4AD^2 = 3AB^2 = 3BC^2 = 3CA^2\).
In simple words: Use Pythagoras theorem in half of the equilateral triangle formed by the altitude.

Exam Tip: State clearly that altitude in an equilateral triangle bisects the base.

Question Q16. ABC is an isosceles triangle right angled at C. Prove that \(AB^2 = 2AC^2\)
Answer: Given ABC is an isosceles triangle with \(\angle C = 90^\circ\).
Since it is isosceles with right angle at C, legs are equal: \(AC = BC\).
By Pythagoras theorem in \(\triangle ABC\):
\(AB^2 = AC^2 + BC^2\)
Since \(BC = AC\), substitute BC with AC:
\(AB^2 = AC^2 + AC^2\)
\(AB^2 = 2AC^2\).
In simple words: Apply Pythagoras theorem and substitute equal sides of the isosceles right triangle.

Exam Tip: Mention that equal sides of the isosceles right triangle are the perpendicular legs.

SECTION D

(EACH QUESTION CARRIES 4 MARKS)

Question Q17. State 'Basic Proportionality theorem' or 'Thales Theorem' and prove it. OR Area of two similar triangles are in ratio of the squares of the corresponding sides.
Answer: Statement (Thales Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Proof:
Let \(\triangle ABC\) have a line parallel to BC intersecting AB at D and AC at E.
Join BE and CD. Draw \(DM \perp AC\) and \(EN \perp AB\).
Area of \(\triangle ADE\) is \(\frac{1}{2} \times AD \times EN\) and Area of \(\triangle BDE\) is \(\frac{1}{2} \times BD \times EN\).
Ratio \(\frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{AD}{BD}\).
Similarly, \(\frac{\text{ar}(ADE)}{\text{ar}(CDE)} = \frac{AE}{EC}\).
Since \(\triangle BDE\) and \(\triangle CDE\) lie on the same base DE and between same parallels BC and DE, their areas are equal.
Therefore, \(\frac{AD}{BD} = \frac{AE}{EC}\). Hence proved.
In simple words: Use area ratios of triangles formed by drawing a parallel line across a triangle.

Exam Tip: Draw a clean, neat diagram showing construction lines for full marks in theorem proofs.

Question Q18. Five years ago Nuri was thrice as old as Sonu. Ten years later Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Answer: Let present age of Nuri be x years and present age of Sonu be y years.
Five years ago: Nuri's age = \(x - 5\), Sonu's age = \(y - 5\).
\(x - 5 = 3(y - 5)\)
\(x - 5 = 3y - 15\)
\(x - 3y = -10\) ---(Equation 1)
Ten years later: Nuri's age = \(x + 10\), Sonu's age = \(y + 10\).
\(x + 10 = 2(y + 10)\)
\(x + 10 = 2y + 20\)
\(x - 2y = 10\) ---(Equation 2)
Subtract Equation 1 from Equation 2:
\((x - 2y) - (x - 3y) = 10 - (-10)\)
\(y = 20\)
Substitute \(y = 20\) into Equation 2:
\(x - 2(20) = 10 \implies x - 40 = 10 \implies x = 50\)
Nuri's present age = 50 years, Sonu's present age = 20 years.
In simple words: Form two linear equations based on past and future age conditions, then solve simultaneously.

Exam Tip: Define variables clearly for present ages before setting up age equations.