ML Aggarwal

### 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 = ${{T}_{11}}~\text{ }{{T}_{7}}~=\text{ }24$...

### Draw a cumulative frequency curve for the following data :

Marks obtained 0-10 10-20 20-30 30-40 40-50 No. of students 8 10 22 40 20 Hence determine:  (i) the median  (ii) the pass marks if 85% of the students pass.  (iii) the marks which 45% of the...

### Using the data given below, construct the cumulative frequency table and draw the ogive. From the ogive, estimate : (i) the median (ii) the inter quartile range.

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 3 8 12 14 10 6 5 2 Also state the median class Solution: Arranging the data in cumulative frequency table. Marks Frequency Cumulative...

### The daily wages of 30 employees in an establishment are distributed as follows:

Daily wages in Rs 0-10 10-20 20-30 30-40 40-50 50-60 No. of employees 1 8 10 5 4 2 Estimate the modal daily wages for this distribution by a graphical method. Solution: Daily wages in Rs. No. of...

### Calculate the mean, the median and the mode of the following distribution.

Age in years 12 13 14 15 16 17 18 No. of students 2 3 5 6 4 3 2 Solution: Age in years xi No. of students fi Cumulative frequency fixi 12 2 2 24 13 3 5 39 14 5 10 70 15 6 16 90 16 4 20 64 17 3 23 51...

### Find the median and mode for the set of numbers : 2, 2, 3, 5, 5, 5, 6, 8, 9

Solution: Here n = 9 which is odd. Median = ((n+1)/2)th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5 th term Median = 5 Mode is the number which appears most often in a set of...

### The marks scored by 16 students in a class test are : 3, 6, 8, 13, 15, 5, 21, 23, 17, 10, 9, 1, 20, 21, 18, 12 Find (i) the median (ii) lower quartile (iii) upper quartile

Solution: Arranging data in ascending order 1,3,5,6,8,9,10,12,13,15,17,18,20,21,21,23 Here n = 16 which is even (i) So median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (16/2 th term +...

### Find the median of: 17, 26, 60, 45, 33, 32, 29, 34, 56 . If 26 is replaced by 62, find the new median.

Solution: Arranging numbers in ascending order 17,26,29,32,33,34,45,56,60 Here n = 9 which is odd. So Median =( (n+1)/2) th term Median = ((9+1)/2)th term Median = (10/2)th term Median = 5th term...

### If the median of 5, 9, 11, 3, 4, x, 8 is 6, find the value of x.

Solution: Arranging numbers in ascending order 3,4,5,x,8,9,11 Here n = 7 which is odd Given median = 6 So Median =( (n+1)/2) th term 6 = ((7+1)/2)th term 6 = ((8/2)th term 6 = 4th term 6 = x Hence...

### 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...

### The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m-1 and median q. Find (i) p (ii) q (iii) the mean of p and q.

Solution: (i) Mean of 1, 7, 5, 3, 4, 4 is m. Here n = 6 Mean, m = (1+7+5+3+4+4)/6 m = 24/6 m = 4 Given the numbers 3, 2, 4, 2, 3, 3, p have mean m-1. So m-1 = (3+2+4+2+3+3+p)/7 4-1 = (17+p)/7 3 =...

### The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.

Solution: Observation are as follows : 11, 12, 14, (x-2), (x+4), (x+9), 32, 38, 47 n = 9 Here n is odd. So median = ((n+1)/2)th term = (9+1)/2 )th term = 5th term = x+4 Given median = 24 x+4 = 24 x...

### Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5

Solution: Arranging the numbers in ascending order : 0, 1, 1, 2, 2, 3, 3, 3, 4, 5 Here, n = 10 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (10/2 th term + ((10/2)+1)th term) = ½...

### (a) Find the median of the following set of numbers : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7 (1990) (b)For the following set of numbers, find the median: 10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15.

Solution: (a) Arranging the numbers in ascending order : 0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9 Here, n = 12 which is even Median = ½ ( n/2 th term + ((n/2)+1)th term) = ½ (12/2 th term +...

### A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.

Solution: Arranging the data in the ascending order 1,2,2,3,4,5,5,6,7,7,8 Here number of terms, n = 11 Here n is odd. So median = [(n+1)/2 ]th observation = (11+1)/2 = 12/2 = 6th observation Here...

### Using the information given in the adjoining histogram, calculate the mean correct to one decimal place.

Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45 Class interval frequency (fi) Class mark (xi) di = xi – A fidi 20-30 3 25 -20 -60 30-40 5 35 -10 -50 40-50...

### The following table gives the life time in days of 100 electricity tubes of a certain make :

Life time in days No. of tubes Less than 50 8 Less than 100 23 Less than 150 55 Less than 200 81 Less than 250 93 Less than 300 100 Find the mean life time of electricity tubes. Solution: Class mark...

### The mean of the following frequency distribution is 57.6 and the sum of all the frequencies is 50. Find the values of p and q.

Class intervals 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 7 P 12 q 8 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20...

### The mean of the following distribution is 50 and the sum of all the frequencies is 120. Find the values of p and q.

Class intervals 0-20 20-40 40-60 60-80 80-100 Frequency 17 P 32 q 19 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-20 17 10...

### The following distribution shows the daily pocket allowance for children of a locality. The mean pocket allowance is Rs. 18. Find the value of f.

Daily pocket allowance in Rs. 11-13 13-15 15-17 17-19 19-21 21-23 23-25 No. of children 3 6 9 13 f 5 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Daily pocket allowance in...

### The mean of the following distribution is 23.4. Find the value of p.

Class intervals 0-8 8-16 16-24 24-32 32-40 40-48 Frequency 5 3 10 P 4 2 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class intervals Frequency fi Class mark xi fixi 0-8 5 4...

### A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

No. of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 No. of students 11 10 7 4 4 3 1 Solution: Class mark, xi = (upper class limit + lower class limit)/2 No. of days Frequency fi Class mark xi fixi...

### Calculate the mean of the distribution given below using the short cut method.

Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80 No. of students 2 6 10 12 9 7 4 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Assumed mean, A = 45.5 Marks No. of students (fi)...

### The following table gives the daily wages of workers in a factory:

Wages in Rs. 45-50 50-55 55-60 60-65 65-70 70-75 75-80 No. of workers 5 8 30 25 14 12 6 Calculate their mean by short cut method. Solution: Class mark, xi = (upper class limit + lower class limit)/2...

### Find the mean of the following frequency distribution:

Class intervals 0-50 50-100 100-150 150-200 200-250 250-300 frequency 4 8 16 13 6 3 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi...

### Calculate the mean of the following distribution using step deviation method:

Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 10 9 25 30 16 10 Solution: Class mark (xi) = (upper limit + lower limit)/2 Let assumed mean (A) = 25 Class size (h) = 10 Class Interval No....

### Calculate the mean of the following distribution:

Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 8 5 12 35 24 16 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 8...

### Find the mean of the following distribution.

Class interval 0-10 10-20 20-30 30-40 40-50 Frequency 10 6 8 12 5 Solution: Class mark, xi = (upper class limit + lower class limit)/2 Class interval Frequency fi Class mark xi fixi 0-10 10 5 50...

### Marks obtained by 40 students in a short assessment are given below, where a and b are two missing data.

Marks 5 6 7 8 9 No. of students 6 a 16 13 b If the mean of the distribution is 7.2, find a and b. Solution: Marks (x) No. of students (f) fx 5 6 30 6 a 6a 7 16 112 8 13 104 9 b 9b Total Ʃf = 35+a+b...

### Find the value of the missing variate for the following distribution whose mean is 10

Variate (xi) 5 7 9 11 _ 15 20 Frequency (fi) 4 4 4 7 3 2 1 Solution: Let the missing variate be x. Variate (xi) Frequency (fi) fixi 5 4 20 7 4 28 9 4 36 11 7 77 x 3 3x 15 2 30 20 1 20 Total Ʃfi =25...

### If the mean of the following distribution is 7.5, find the missing frequency ” f “.

Variate 5 6 7 8 9 10 11 12 Frequency 20 17 f 10 8 6 7 6 Solution: Variate (x) Frequency (f) fx 5 20 100 6 17 102 7 f 7f 8 10 80 9 8 72 10 6 60 11 7 77 12 6 72 Total Ʃf = 74+f Ʃfx = 563+7f Given mean...

### (i) Calculate the mean wage correct to the nearest rupee (1995) (ii) If the number of workers in each category is doubled, what would be the new mean

Category A B C D E F G Wages (in Rs) per day 50 60 70 80 90 100 110 No. of workers 2 4 8 12 10 6 8 Category Wages in Rs. (x) No. of workers f fx A 50 2 100 B 60 4 240 C 70 8 560 D 80 12 960 E 90 10...

### Find the mean for the following distribution.

Numbers 60 61 62 63 64 65 66 Cumulative frequency 8 18 33 40 49 55 60 Solution: Numbers (x) Cumulative frequency Frequency (f) fx 60 8 8 60×8 = 480 61 18 18-8 = 10 61×10 = 610 62 33 33-18 = 15 62×15...

### Six coins were tossed 1000 times, and at each toss the number of heads were counted and the results were recorded as under

No. of heads 6 5 4 3 2 1 0 No. of tosses 20 25 160 283 338 140 34 Calculate the mean for this distribution. Solution: No. of heads (x) No. of tosses (f) fx 6 20 6×20 = 120 5 25 5×25 = 125 4 160...

### Calculate the mean for the following distribution :

Pocket money (in Rs) 60 70 80 90 100 110 120 No. of students 2 6 13 22 24 10 3 Solution: Pocket money in Rs (x) Number of students (f) fx 60 2 60×2 = 120 70 6 70×6 = 420 80 13 80×13 = 1040 90 22...

### The contents of 100 match boxes were checked to determine the number of matches they contained

No. of matches 35 36 37 38 39 40 41 No. of boxes 6 10 18 25 21 12 8 (i) Calculate, correct to one decimal place, the mean number of matches per box. (ii) Determine how many extra matches would have...

### Find the mean of the following distribution:

Number 5 10 15 20 25 30 35 Frequency 1 2 5 6 3 2 1 Solution: Number (x) Frequency (f) fx 5 1 5×1 = 5 10 2 10×2 = 20 15 5 15×5 = 75 20 6 20×6 = 120 25 3 25×3 = 75 30 2 30×2 = 60 35 1 35×1 = 35 Total...

### Find the mean of 25 given numbers when the mean of 10 of them is 13 and the mean of the remaining numbers is 18.

Solution: Mean of 10 numbers = 13 Sum of numbers = 13×10 = 130 Mean of remaining 15 numbers = 18 Sum of numbers = 15×18 = 270 Sum of all numbers = 130+270 = 400 Mean = sum of numbers/25 = 400/25 =...

### (a) The mean age of 33 students of a class is 13 years. If one girl leaves the class, the mean becomes years. What is the age of the girl ? (b) In a class test, the mean of marks scored by a class of 40 students was calculated as 18.2. Later on, it was detected that marks of one student was wrongly copied as 21 instead of 29. Find the correct mean.

Solution: (a)Given mean age = 13 Number of students = 33 Sum of ages = mean ×number of students = 13×33 = 429 After a girl leaves, the mean of 32 students becomes = 207/16 Now sum of ages =...

### (a) The mean of the numbers 6, y, 7, x, 14 is 8. Express y in terms of x. (b) The mean of 9 variates is 11. If eight of them are 7, 12, 9, 14, 21, 3, 8 and 15 find the 9th variate.

Solution: (a)Given observations are 6, y, 7, x, 14. Mean = 8 Number of observations = 5 Mean = Sum of observations/number of observations 8 = (6+y+7+x+14)/5 40 = 27+x+y 40-27 = x+y 13 = x+y y = 13-x...