ICSE

### A …… lens is used to correct myopia and a ….. lens is used to correct hypermetropia. A Concave, concave B Convex, convex C Convex, concave D Concave, convex

Correct option is (D) Concave, convex Myopia means short-sightedness, in this, a person has a clear vision when looking at objects close to them, but distant objects will appear blurred, to correct...

### Rate of reaction of any substance depends on (A) active mass (B) molecular weight (C) atomic weight (D) equivalent weight

Correct option is (A) active mass The rate at which a substance reacts depends upon its active mass as the rate of reaction is directly proportional to concentration of each reactant and product.

### Formula of oleum is (A) (B) (C) (D)

Correct option is (A) $\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}$ Oleum, or fuming sulfuric acid, is a solution of various compositions of sulfur trioxide in sulfuric acid, or sometimes more...

### A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

A line passes through the point $(2,1,-3)$ and is parallel to the vector $(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$. Find the equations of the line in vector and Cartesian forms....

### Two cells of emf and , internal resistance and , connected in parallel. The equivalent emf of the combination is- (A) (B) (C) (D)

Answer: Option (B)$\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}$

### Prove that:

Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...

### Find the equation of a circle passing through the origin and intercepting lengths a and b on the axes.

Answer:           AD = b units and AE = a units. D(0, b), E(a, 0) and A(0, 0) lies on the circle. C is the centre. The general equation of a circle: (x - h)2 + (y - k)2 = r2...

### Using elementary row transformations, find the inverse of each of the following matrices:

Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...

### The fourth term of a G.P. is the square of its second term and the first term is     . Find its 7th term.

From the question it is given that, The fourth term of a G.P. is the square of its second term $=\text{ }{{a}_{4}}~=\text{}{{({{a}_{2}})}^{2}}$ The first term ${{a}_{1}}~=\text{ }\text{ }3$ We...

### Find the geometric progression whose 4th term is     and 7th term is     .

From the question it is given that, The geometric progression whose 4th term ${{a}_{4}}~=\text{ }54$ The geometric progression whose 7th term ${{a}_{7}}~=\text{ }1458$ We know that, an = arn – 1...

### The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.

From the question it is given that, The angles of a quadrilateral are in A.P. Greatest angle is double of the smallest angle Let us assume the greatest angle of the quadrilateral is a + 3d, Then,...

### The sum of three numbers in A.P. is     and the product is 8. Find the numbers.

From the question it is given that, The sum of three numbers in A.P. = $-3$ The product of three numbers in A.P. = $8$ Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d Now...

### How many three digit numbers are divisible by     ?

The three digits numbers which are divisible by $9$ are $108,\text{ }117,\text{ }126,\text{ }\ldots ,\text{ }999$ Then, first term a = $108$ Common difference = $9$ Last term = $999$ We...

### Which term of the list of numbers     ?

From the question it is given that, First term a = $5$ nth term = $-55$ Common difference d =  $2\text{ }\text{ }5\text{ }=\text{ }\text{ }3$ We know that, an = a + (n – 1)d...

### (i) How many terms of the G.P.     are needed to give the sum     ? (ii) How many terms of the G.P.     must be taken to have their sum equal to     ?

From the question it is given that, Terms of the G.P. $\mathbf{3},\text{ }{{\mathbf{3}}^{\mathbf{2}}},\text{ }{{\mathbf{3}}^{\mathbf{3}}},\text{ }\ldots$ Sum of the terms = $120$ The first term...

### If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

From the question it is given that, a4 = x a7 = y a10 = z   Now we have to prove that, x, y, z are in G.P. Then, by the formula ${{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}$...

### (i) Find the sum of all two digit natural numbers which are divisible by     . (ii) Find the sum of all natural numbers between     and     which are divisible by     .

(i) The two-digit natural numbers which are divisible by $4$ are: $4,\text{ }8,\text{ }12,\text{ }16,\text{ }\ldots ..$ This form an A.P. The last term in this series is found out by dividing...

### (i) Find the sum of first     positive integers. (ii) Find the sum of first     multiples of     .

(i) First $1000$ positive integers are: $1,\text{ }2,\text{ }3,\text{ }4,\text{ }\ldots \ldots ..,\text{ }1000$ This is an A.P with first term a = $1$  and common difference d = $1$ We know...

### In an A.P., the fourth and sixth terms are     and     , respectively. Find the: (i) first term (ii) common difference

From the question it is given that, ${{T}_{4}}~=\text{ }8\text{ }and\text{ }{{T}_{6}}~=\text{ }14$ ⇒ $a\text{ }+\text{ }3d\text{ }=\text{ }8$… (i) ⇒ $a\text{ }+\text{ }5d\text{ }=\text{ }14$…...

### Find the sums given below : (i)     (ii)

From the question, First term a = $34$, Difference d = $32\text{ }\text{ }34\text{ }=\text{ }-2$ So, common difference d = $-2$ Last term Tn = 10 We know that, ${{T}_{n}}$ = a + (n – 1)d...

### The sum of three numbers in A.P. is     and the ratio of first number to the third number is     . Find the numbers.

From the question it is given that, sum of three numbers in A.P. = $30$ The ratio of first number to the third number is $3:7$ Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d...

### The sum of three numbers in A.P. is     and their product is     . Find the numbers.

From the question it is given that, The sum of three numbers in A.P. = $3$ Given, Their product = $-35$ Let us assume the $3$ numbers which are in A.P. are, a – d, a, a + d Now adding $3$...

### If the numbers     are in A.P., find the value of n.

From the question it is given that, $\mathbf{n}\text{ }\text{ }\mathbf{2},\text{ }\mathbf{4n}\text{ }\text{ }\mathbf{1}\text{ }\mathbf{and}\text{ }\mathbf{5n}\text{ }+\text{ }\mathbf{2}$ are in...

### (i) How many two digit numbers are divisible by     ? (ii) Find the number of natural numbers between     and     which are divisible by both     and     .

The two digits numbers divisible by $3$ are, $12,\text{ }15,\text{ }18,\text{ }21,\text{ }24,\ldots ..,99$. The above numbers are A.P. So, first number a = $12$ Common difference d =...

### Find the     term of an A.P. whose     term is     and     term is     .

From the question it is given that, $\begin{array}{*{35}{l}} {{T}_{11}}~=\text{ }38 \\ {{T}_{6}}~=\text{ }73 \\ \end{array}$ Let us assume ‘a’ be the first term and ‘d’ be the common difference,...

### Find the     term of the A.P. whose     term is     less than the     term, first term being     .

From the question it is given that, First term a = $12$ ${{\mathbf{7}}^{\mathbf{th}}}$ term is 24 less than the ${{11}^{th}}$ term = ${{T}_{11}}~\text{ }{{T}_{7}}~=\text{ }24$...

### Why does the acetylation of —NH2 group of aniline reduce its activating effect?

Solution: The acetylation of —NH2 group of aniline reduces its activating effect because the lone pair of electrons on the nitrogen of acetanilide interacts with oxygen atom due to resonance.

### What is Hinsberg reagent?

Solution: Hinsberg's reagent is benzene sulphonyl chloride, also known as $C_6H_5SOCl$. To distinguish between primary, secondary, and tertiary amines, Hinsberg's reagent is used.

### What is the product when C6H5CH2NH2 reacts with HNO2?

Solution: The main product is $C_6H_5CH_2OH$ When $C_6H_5CH_2NH_2$ reacts with HNO2, it produces an unstable diazonium salt, which is then converted to alcohol. When $C_6H_5CH_2NH_2$ reacts with...

### What is the role of HNO3 in the nitrating mixture used for nitration of benzene?

Solution: The nitration of organic compounds is done with a nitration mixture, which is a 1:1 solution of HNO3 and H2SO4. In the nitration of benzene, it acts as a base and provides electrophile.

### The median of the following numbers, arranged in ascending order is 25. Find x. 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46

Solution: Here n = 10, which is even Median = 25 So Median = ½ ( n/2 th term + ((n/2)+1)th term) 25 = ½ (( 10/2 )th term + (10/2)+1)th term) 25 = ½ (( 5 )th term + (6)th term) 25 = ½ (x+2 + x+4) 25...

### The measures of the diameter of the heads of 150 screw is given in the following table. If the mean diameter of the heads of the screws is 51.2 mm, find the values of p and q .

Diameter in mm 32-36 37-41 42-46 47-51 52-56 57-61 62-66 No. of screws 15 17 p 25 q 20 30 Solution: Given mean = 51.2 mm The given distribution is not continuous. Adjustment factor = (37-36)/2 = ½ =...

### The daily expenditure of 100 families are given below. Calculate f1, and f2, if the mean daily expenditure is Rs 188.

Expenditure in Rs 140-160 160-180 180-200 200-220 220-240 No. of families 5 25 f1 f2 5 Solution: Given mean = 188 Class Frequency fi Class mark xi fixi 140-160 5 150 750 160-180 25 170 4250 180-200...

### The mean of the following frequency distribution is 62.8. Find the value of p.

Class 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 5 8 p 12 7 8 Solution: Class Frequency fi Class mark xi fixi 0-20 5 10 50 20-40 8 30 240 40-60 p 50 50p 60-80 12 70 840 80-100 7 90 630 100-120...

### Calculate the Arithmetic mean, correct to one decimal place, for the following frequency

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Students 2 4 5 16 20 10 6 8 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Marks Students fi Class mark xi fixi...

### Find the mean age in years from the frequency distribution given below:

Age in years 25-29 30-34 35-39 40-44 45-49 50-54 55-59 No. of persons 4 14 22 16 6 5 3 Solution: The given distribution is not continuous. Adjustment factor = (30-29)/2 = ½ = 0.5 We subtract 0.5...

### Find the value of p if the mean of the following distribution is 18.

Variate (xi) 13 15 17 19 20+p 23 Frequency (fi) 8 2 3 4 5p 6 Solution: Variate (xi) Frequency (fi) fi xi 13 8 104 15 2 30 17 3 51 19 4 76 20+p 5p 5p2+100p 23 6 138 Total Ʃfi = 23+5p Ʃfi xi =...

### Find the value of p for the following distribution whose mean is 20.6.

Variate (xi) 10 15 20 25 35 Frequency (fi) 3 10 p 7 5 Solution: Variate (xi) Frequency (fi) fx 10 3 30 15 10 150 20 p 20p 25 7 175 35 5 175 Total Ʃfi = 25+p Ʃfi xi = 530+20p Mean = Ʃfx/Ʃf 20.6 =...

### The heights of 50 children were measured (correct to the nearest cm) giving the following results

Height (in cm) 65 66 67 68 69 70 71 72 73 No. of children 1 4 5 7 11 10 6 4 2 Calculate the mean height for this distribution correct to one place of decimal.  Solution: Height x No. of children f...

### The contents of 50 boxes of matches were counted giving the following results.

No. of matches 41 42 43 44 45 46 No. of boxes 5 8 13 12 7 5 Calculate the mean number of matches per box. Solution: No. o f matches x No. of boxes f fx 41 5 205 42 8 336 43 13 559 44 12 528 45 7 315...

### There are 50 students in a class of which 40 are boys and the rest girls. The average weight of the students in the class is 44 kg and average weight of the girls is 40 kg. Find the average weight of boys.

Solution: Total number of students = 50 No. of boys = 40 No. of girls = 50-40 = 10 Average weight of 50 students = 44 kg So sum of weight = 44×50 = 2200 kg Average weight of girls = 40 kg So sum of...

### The average height of 30 students is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for computation of mean. Find the correct mean.

Solution: Average height of 30 students = 150 cm So sum of height = 150×30 = 4500 Difference between correct value and wrong value = 165-135 = 30 So actual sum = 4500+30 = 4530 So actual mean =...

### The mean of 20 numbers is 18. If 3 is added to each of the first ten numbers, find the mean of new set of 20 numbers.

Solution: Given the mean of 20 numbers = 18 Sum of numbers = 18×20 = 360 If 3 is added to each of first 10 numbers, then new sum = (3×10)+360 = 30+360 = 390 New mean = 390/20 = 19.5 Hence the mean...

### Arun scored 36 marks in English, 44 marks in Civics, 75 marks in Mathematics and x marks in Science. If he has scored an average of 50 marks, find x.

Solution: Marks scored in English = 36 Marks scored in Civics = 44 Marks scored in Mathematics = 75 Marks scored in Science = x No. of subjects = 4 Average marks = sum of marks / No. of subjects =...

### 100 pupils in a school have heights as tabulated below

Height in cm 121-130 131-140 141-150 151-160 161-170 171-180 No. of pupils 12 16 30 20 14 8 Draw the ogive for the above data and from it determine the median (use graph paper). Solution: We write...

### The following distribution represents the height of 160 students of a school.

Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180 No. of students 12 20 30 38 24 16 12 8 Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2...

### The marks obtained by 120 students in a Mathematics test are-given below

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 9 16 22 26 18 11 6 4 3 Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for ogive...

### The marks obtained by 100 students in a Mathematics test are given below

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 3 7 12 17 23 14 9 6 5 4 Draw an ogive on a graph sheet and from it determine the :  (i) median  (ii) lower quartile ...

### Using a graph paper, draw an ogive for the following distribution which shows a record of the weight in kilograms of 200 students

Weight 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 5 17 22 45 51 31 20 9 Use your ogive to estimate the following:  (i) The percentage of students weighing 55 kg or more.  (ii) The...

### The monthly income of a group of 320 employees in a company is given below

Monthly income No. of employees 6000-7000 20 7000-8000 45 8000-9000 65 9000-10000 95 10000-11000 60 11000-12000 30 12000-13000 5 Draw an ogive of the given distribution on a graph sheet taking 2 cm...

### Marks obtained by 200 students in an examination are given below

marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of students 5 11 10 20 28 37 40 29 14 6 Draw an ogive for the given distribution taking 2 cm = 10 marks on one axis and 2 cm =...

### The daily wages of 80 workers in a project are given below

Wages in Rs 400-450 450-500 500-550 550-600 600-650 650-700 700-750 No. of workers 2 6 12 18 24 13 5 Use a graph paper to draw an ogive for the above distribution. ( a scale of 2 cm = Rs 50 on x-...

### The table shows the distribution of scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the x-axis and 2 cm = 20 shooters on the y-axis)

Scores 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 No. of shooters 9 13 20 26 30 22 15 10 8 7 Use your graph to estimate the following:  (i) The median.  (ii) The interquartile...

### The weight of 50 workers is given below:

Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120 No. of workers 4 7 11 14 6 5 3 Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5...

### Attempt this question on graph paper.

Age( yrs) 5-15 15-25 25-35 35-45 45-55 55-65 65-75 No. of casualities due to accidents 6 10 15 13 24 8 7 (i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10...

### Use graph paper for this question. The following table shows the weights in gm of a sample of 100 potatoes taken from a large consignment:

Weight (gm) 50-60 60-70 70-80 80-90 90-100 100-110 110-120 120-130 Frequency 8 10 12 16 18 14 12 10 (i) Calculate the cumulative frequencies.  (ii) Draw the cumulative frequency curve and from it...

### Using the data given below construct the cumulative frequency table and draw the-Ogive. From the ogive determine the median.

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 No. of students 3 8 12 14 10 6 5 2 Solution: We write the given data in cumulative frequency table. Marks No of students f Cumulative frequency...

### The following table shows the distribution of the heights of a group of a factory workers.

Height ( in cm ) 150-155 155-160 160-165 165-170 170-175 175-180 180-185 No. of workers 6 12 18 20 13 8 6 (i) Determine the cumulative frequencies. (ii) Draw the cumulative frequency curve on a...

### 33:

Marks obtained 24-29 29-34 34-39 39-44 44-49 49-54 54-59 No. of students 1 2 5 6 4 3 2 Solution: We write the given data in cumulative frequency table. Marks obtained No of students Cumulative...

### Draw an ogive for the following data:

Class intervals 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 3 5 8 7 6 2 Solution: The given distribution is not continuous. Adjustment factor = (11-10)/2 = ½ = 0.5 We subtract 0.5 from lower limit...

### Draw an ogive for the following frequency distribution:

Height ( in cm ) 150-160 160-170 170-180 180-190 190-200 No. of students 8 3 4 10 2 Solution: We write the given data in cumulative frequency table. Height in cm No of students Cumulative frequency...

### Find the mode of the following distribution by drawing a histogram

Mid value 12 18 24 30 36 42 48 Frequency 20 12 8 24 16 8 12 Also state the modal class. Solution: Mid value Frequency 12 20 18 12 24 8 30 24 36 16 42 8 48 12 Here mid value and frequency is given....

### Draw a histogram for the following distribution :

Wt. in kg 40-44 45-49 50-54 55-59 60-64 65-69 No. of students 2 8 12 10 6 4 Hence estimate the modal weight. Solution: The given distribution is not continuous. Adjustment factor = (45-44)/2 = ½ =...

### Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below:

Pocket expenses (in Rs) Number of students (Frequency ) 0-5 10 5-10 14 10-15 28 15-20 42 20-25 50 25-30 30 30-35 14 35-40 12 Draw a histogram representing the above distribution and estimate the...

### IQ of 50 students was recorded as follows.

IQ score 80-90 90-100 100-110 110-120 120-130 130-140 No. of students 6 9 16 13 4 2 Draw a histogram for the above data and estimate the mode. Solution: Construct histogram using given data....

### Draw a histogram and estimate the mode for the following frequency distribution :

Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Solution: Construct histogram using given data. Classes 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 2 8 10 5 4 3 Represent classes...

### A Mathematics aptitude test of 50 students was recorded as follows :

Marks 50-60 60-70 70-80 80-90 90-100 No. of students 4 8 14 19 5 Draw a histogram for the above data using a graph paper and locate the mode. (2011) Solution: Construct histogram using given data....

### Find the modal height of the following distribution by drawing a histogram :

Height (in cm) 140-150 150-160 160-170 170-180 180-190 No. of students 7 6 4 10 2 Solution: Construct histogram using given data. Height (in cm) 140-150 150-160 160-170 170-180 180-190 No. of...

### Draw a histogram for the following frequency distribution and find the mode from the graph :

Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Solution: Construct histogram using given data. Class 0-5 5-10 10-15 15-20 20-25 25-30 Frequency 2 5 18 14 8 5 Represent class on...

### The following table gives the weekly wages (in Rs.) of workers in a factory :

Weekly wages (in Rs) 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 No. of workers 5 20 10 10 9 6 12 8 Calculate: (i) The mean. (ii) the modal class (iii) the number of workers getting weekly wages...

### (i) Using step-deviation method, calculate the mean marks of the following distribution. (ii) State the modal class.

Class interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90 Frequency 5 20 10 10 9 6 12 8 Solution: (i) Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 67.5 Class size (h)...

### At a shooting competition, the scores of a competitor were as given below :

Score 0 1 2 3 4 5 No. of shots 0 3 6 4 7 5 (i) What was his modal score ? (ii) What was his median score ? (iii) What was his total score ? (iv) What was his mean ? Solution: We write the marks in...

### The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained 5 6 7 8 9 10 No. of students 3 9 6 4 2 1 Solution: We write the marks in cumulative frequency table. Marks x Frequency (f) fx Cumulative frequency 5 3 15 3 6 9 54 12 7 6 42 18 8 4 32...

### The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks 0 1 2 3 4 5 No. of students 1 3 6 10 5 5 Calculate the mean, median and mode of the above distribution. Solution: We write the data in cumulative frequency table. Marks x Frequency (f)...

### Find the mode and median of the following frequency distribution :

x 10 11 12 13 14 15 f 1 4 7 5 9 3 Solution: We write the data in cumulative frequency table. x Frequency (f) Cumulative frequency 10 1 1 11 4 5 12 7 12 13 5 17 14 9 26 15 3 29 Here number of...

### Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks : 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8

Solution: Given data is 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 Number of observations, n = 16 Mean = Ʃxi/n = (0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8)/16 = 64/16 = 4 Hence the mean is 4. Here n = 16...

### A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16 (i) What are his modal marks ? (ii) What are his median marks ? (iii) What are his mean marks ?

Solution: (i)We arrange given marks in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19 16 appears maximum number of times. Hence his modal mark is 16. (ii)Here number of observations, n =...

### The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017)

Solution: Given data in ascending order: 13, 35, 43, 46, x, x +4, 55, 61,71, 80 Given median = 48 Number of observations, n = 10 which is even. median = ½ ( n/2 th term + ((n/2)+1)th term) 48 = ½...

### Calculate the mean, the median and the mode of the following numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Solution: We arrange given data in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7 Mean = Ʃxi/n = (1+2+3+3+3+4+5+5+6+7)/10 = 39/10 = 3.9 Here number of observations, n = 10 which is even. So median = ½...

### Find the mean, median and mode of the following distribution : 8, 10, 7, 6, 10, 11, 6, 13, 10

Solution: We arrange given data in ascending order 6, 6, 7, 8, 10, 10, 10, 11, 13 Mean = Ʃxi/n = (6+6+7+8+10+10+10+11+13)/9 = 81/9 = 9 Hence the mean is 9. Here number of observations, n = 9 which...

### Calculate the mean, the median and the mode of the numbers : 3, 2, 6, 3, 3, 1, 1, 2

Solution: We arrange given data in ascending order 1, 1, 2, 2, 3, 3, 3, 6 Mean = Ʃxi/n = (1+1+2+2+3+3+3+6)/8 = 21/8 = 2.625 Hence the mean is 2.625. Here number of observations, n = 8 which is even....

### Find the mode of the following sets of numbers ; (i) 3, 2, 0, 1, 2, 3, 5, 3 (ii) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8 (iii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

Solution: Mode is the number which appears most often in a set of numbers. (i)Given set is 3, 2, 0, 1, 2, 3, 5, 3. In this set, 3 occurs maximum number of times. Hence the mode is 3. (ii) Given set...

### For the following frequency distribution, find : (i) the median (ii) lower quartile (iii) upper quartile

Variate 25 31 34 40 45 48 50 60 Frequency 3 8 10 15 10 9 6 2 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 25 3 3 31 8 11 34 10 21 40 15...

### From the following frequency distribution, find : (i) the median (ii) lower quartile (iii) upper quartile (iv) inter quartile range

Variate 15 18 20 22 25 27 30 Frequency 4 6 8 9 7 8 6 Solution: We write the variates in cumulative frequency table. Variate Frequency (f) Cumulative frequency 15 4 4 18 6 10 20 8 18 22 9 27 25 7 34...

### The daily wages in (rupees of) 19 workers are 41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35. find : (i) the median (ii) lower quartile (iii) upper quartile (iv) inter quartile range

Solution: Arranging the observations in ascending order 21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53 Here n = 19 which is odd. (i)Median = ((n+1)/2)th term = (19+1)/2 =...

### Calculate the mean and the median for the following distribution :

Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: We write the numbers in cumulative frequency table. Marks (x) No. of students (f) Cumulative frequency fx 5 1 1 5 10 2 3 20 15 5 8 75 20...

### Marks obtained by 70 students are given below :

Marks 20 70 50 60 75 90 40 No. of students 8 12 18 6 9 5 12 Calculate the median marks. Solution: We write the marks in ascending order in cumulative frequency table. Marks No. of students (f)...

### Find the median for the following distribution.

Marks 35 45 50 64 70 72 No. of students 3 5 8 10 5 5 Solution: We write the distribution in cumulative frequency table. Marks No. of students (f) Cumulative frequency 35 3 3 45 5 8 50 8 16 64 10 26...

### Find the median for the following distribution:

Wages per day in Rs. 38 45 48 55 62 65 No. of workers 14 8 7 10 6 2 Solution: We write the distribution in cumulative frequency table. Wages per day in Rs. No. of workers (f) Cumulative frequency 38...