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....
, subject to the constraints 
and 
The feasible region determined by $x+3 y \geq 6, x-3 y \leq 3,3 x+4 y \leq 24$ $-3 x+2 y \leq 6,5 x+y \geq 5, x \geq 0$ and $y \geq 0$ is given by The corner points of the feasible region are $A(4 /...
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...
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...
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...
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...
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 +...
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...
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...
Given $x-y \leq 3$ $\Rightarrow-y \leq 3-x$ Multiplying by minus on both the sides, we'll get $\begin{array}{l} y \geq-3+x \\ y \geq x-3 \end{array}$ Consider the equation $y=x-3$. Finding points on...
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 =...
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...
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...
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...
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 =...
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 =...
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...
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...
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...
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 =...
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...
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 =...
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...
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...
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...
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...
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 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 =...
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-...
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...
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...
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...
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...
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...
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...
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...
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...
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...
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....
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 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....
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...
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....
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...
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...
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...
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)...
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...
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)...
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...
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 =...
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...
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...
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...
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....
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...
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...
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...
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 =...
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 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)...
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...
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...
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 =...
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...
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...
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 +...
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...
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...
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...
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...
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...
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...
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...
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...
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)...
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...
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....
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...
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...
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...
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...
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...
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...
Solution: Steps to construct: Step 1: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
Solution: Steps to construct: Step 1: Mark a point O. Step 2: With center O and radius 4cm and 6cm, draw two concentric circles. Step 3: Join OA and mark its mid-point as M. Step 4: With center M...
Solution: (a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure). As the...
Solution: Radius of the circle = 6 cm and length of tangent = 8 cm Let OP be the distance i.e. OA = 6 cm, AP = 8 cm . OA is the radius OA ⊥ AP Now In right ∆OAP, OP2 = OA2 + AP2 (By Pythagoras...
Solution: CT is the radius CP = 13 cm and tangent PT = 12 cm CT is the radius and TP is the tangent CT is perpendicular TP Now in right angled triangle CPT, CP2 = CT2 + PT2 [using Pythagoras axiom]...
(a) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate (i) ∠QBC (ii) ∠BCP Solution: (i) ABCD is a cyclic quadrilateral ∠A + ∠C = 180o 30o + p =...
Solution: (a) Given, ∠AOC = 150° and AD = CD We know that an angle subtends by an arc of a circle at the center is twice the angle subtended by the same arc at any point on the remaining part of the...
Solution (a) (i) ∠PRB = ∠BAP (Angles in the same segment of the circle) ∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
Solution (a) ∠NYB = 50°, ∠YNB = 20°. In ∆YNB, ∠NYB + ∠YNB + ∠YBN = 180o 50o + 20o + ∠YBN = 180o ∠YBN + 70o = 180o ∠YBN = 180o – 70o = 110o But ∠MAN = ∠YBN (Angles in the same segment) ∠MAN = 110o...
(a) ABCD is cyclic Quadrilateral ∠B + ∠D = 1800 Y + 400 + 45o = 180o (y + 85o = 180o) Y = 180o – 85o = 95o ∠ACB = ∠ADB xo = 40 (a) Arc ADC Subtends ∠AOC at the centre and ∠ ABC at the remaining part...
$\begin{array}{l} x^{2}+6 x-\left(a^{2}+2 a-8\right)=0 \\ \Rightarrow x^{2}+6 x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+[(a+4)-(a-2)] x-(a+4)(a-2)=0 \\ \Rightarrow x^{2}+(a+4) x-(a-2) x-(a+4)(a-2)=0 \\...
$\begin{array}{l} 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0 \\ \Rightarrow 4 x^{2}+4 b x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2[(a+b)-(a-b)] x-(a-b)(a+b)=0 \\ \Rightarrow 4 x^{2}+2(a+b) x-2(a-b)...
$\begin{array}{l} \sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x^{2}-3 \sqrt{2} x+\sqrt{2} x-2 \sqrt{3}=0 \\ \Rightarrow \sqrt{3} x(x-\sqrt{6})+\sqrt{2}(x-\sqrt{6})=0 \\...
$\begin{array}{l} 3 x^{2}+5 \sqrt{5} x-10=0 \\ \Rightarrow 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0 \\ \Rightarrow 3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0 \\ \Rightarrow(x+2 \sqrt{5})(3 x-\sqrt{5})=0...
$\begin{array}{l} x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0 \\ \Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0 \\ \Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0 \\ \Rightarrow(x-\sqrt{3})(x-1)=0 \\ \Rightarrow...
for which the quadratic equation 
has equal roots.It is given that the quadratic equation $9 x^{2}-3 k x+k=0$ has equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-3 k)^{2}-4 \times 9 \times k=0 \\ \Rightarrow 9 k^{2}-36 k=0 \\...
are equal, find the value of 
.It is given that the roots of the quadratic equation $p x^{2}-2 p x+6=0$ are equal. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 p)^{2}-4 \times p \times 6=0 \\ \Rightarrow 4 p^{2}-24 p=0 \\...
is 
, write the other zero.Let the other zero of the given polynomial be $\alpha$. Now, Sum of the zeroes of the given polynomial $=\frac{-(-4)}{1}=4$ $\begin{array}{l} \therefore \alpha+(2+\sqrt{3})=4 \\ \Rightarrow...
has two equal roots then find the value of .It is given that the quadratic equation $p x^{2}-2 \sqrt{5} p x+15=0$ has two equal roots. $\begin{array}{l} \therefore D=0 \\ \Rightarrow(-2 \sqrt{5} p)^{2}-4 \times p \times 15=0 \\ \Rightarrow 20...
.The given quadratic equation is $3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0$ $\begin{array}{l} 3 \sqrt{3} x^{2}+10 x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3} x^{2}+9 x+x+\sqrt{3}=0 \\ \Rightarrow 3 \sqrt{3}...
.The given quadratic equation is $2 x^{2}-x-6=0$ $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\ \Rightarrow(x-2)(2 x+3)=0 \\ \Rightarrow x-2=0...
is a solution of 
.The given equation is $3 x^{2}+13 x+14=0$. Putting $x=-2$ in the given equation, we get $L H S-3 \times(-2)^{2}+13 \times(-2)+14=12-26+14=0=R H S$ $\therefore x=-2$ is a solution of the given...
. are
Answer is (b) $2, \frac{-3}{2}$ The given quadratic equation is $2 x^{2}-x-6=0$. $\begin{array}{l} 2 x^{2}-x-6=0 \\ \Rightarrow 2 x^{2}-4 x+3 x-6=0 \\ \Rightarrow 2 x(x-2)+3(x-2)=0 \\...
has no real roots then
Answer is (c) $\frac{-8}{5}<k<\frac{8}{5}$ It is given that the equation $\left(x^{2}+5 k x+16=0\right)$ has no real roots. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)<0 \\...
, it is given that 
Then, the roots of the equation areAnswer is (b) real and unequal We know that when discriminant, $D>0$, the roots of the given quadratic cquation are real and uncqual.
has equal roots then 
?Answer is (a) 1 or 4 It is given that the roots of the equation $\left(x^{2}+2(k+2) x+9 k=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow\{2(k+2)\}^{2}-4...
are equal then 
?
Answer is (d) $\frac{b^{2}}{4 a}$ It is given that the roots of the equation $\left(a x^{2}+b x+c=0\right)$ are equal. $\begin{array}{l} \therefore\left(b^{2}-4 a c\right)=0 \\ \Rightarrow b^{2}=4 a...
and 
are the roots of the equation 
then 
?
Answer is (c) $-4$ It is given that $\alpha$ and $\beta$ are the roots of the equation $3 x^{2}+8 x+2=0$ $\therefore \alpha+\beta=-\frac{8}{3}$ and $\alpha \beta=\frac{2}{3}$...
is equal to their product then the value of 
Answer is (d) $\frac{-2}{3}$ Given: $k x^{2}+2 x+3 k=0$ Sum of the roots $=$ Product of the roots $\begin{array}{l} \Rightarrow \frac{-2}{k}=\frac{3 k}{k} \\ \Rightarrow 3 k=-2 \\ \Rightarrow...
is
Answer is (d) $2: 3$ Given: $7 x^{2}-12 x+18=0$ $\therefore \alpha+\beta=\frac{12}{7}$ and $\beta=\frac{18}{7}$, where $\alpha$ and $\beta$ are the roots of the equation $\therefore$ Ratio of the...
is
Answer is (b) -2 Sum of the roots of the equation $x^{2}-6 x+2=0$ is $\alpha+\beta=\frac{-b}{a}=\frac{-(-6)}{1}=6$, where $\alpha$ and $\beta$ are the roots of the equation.
is a solution of the equation 
, then 
?
Answer is (b) $-11$ It is given that $x=3$ is a solution of $3 x^{2}+(k-1) x+9=0$; therefore, we have: $\begin{array}{l} 3(3)^{2}+(k-1) \times 3+9=0 \\ \Rightarrow 27+3(k-1)+9=0 \\ \Rightarrow...
Answer is (c) $(\sqrt{2} x+3)^{2}=2 x^{2}+6$ $\begin{array}{l} \because(\sqrt{2} x+3)^{2}=2 x^{2}+6 \\ \Rightarrow 2 x^{2}+9+6 \sqrt{2} x=2 x^{2}+6 \end{array}$ $\Rightarrow 6 \sqrt{2} x+3=0$, which...
Let the shortest side be $x \mathrm{~m}$. Therefore, according to the question: Hypotenuse $=(2 x-1) m$ Third side $=(x+1) m$ On applying Pythagoras theorem, we get: $\begin{array}{l} (2...
Let one side of the right-angled triangle be $x \mathrm{~m}$ and the other side be $(x+4) \mathrm{m}$. On applying Pythagoras theorem, we have: $\begin{array}{l} 20^{2}=(x+4)^{2}+x^{2} \\...
Let the altitude of the triangle be $x \mathrm{~m}$. Therefore, the base will be $3 x \mathrm{~m}$. $\begin{array}{l} \text { Area of a triangle }=\frac{1}{2} \times \text { Base } \times \text {...
Let the length and breadth of the rectangular garden be $x$ and $y$ meter, respectively. Given: $x y=180 \mathrm{sq} \mathrm{m}$$\ldots(i)$ and $\begin{array}{l} 2 y+x=39 \\ \Rightarrow x=39-2 y...
. If the difference in their perimeter be 
, find the sides of the two squareLet the length of the side of the first and the second square be $x$ and $y \cdot$ respectively. According to the question: $x^{2}+y^{2}=640$ Also, $\begin{array}{l} 4 x-4 y=64 \\ \Rightarrow x-y=16...
and its area is 288 sq meters. Find the dimension of the plotLet the length and breadth of the rectangular plot be $x$ and $y$ meter, respectively. Therefore, we have: $\begin{array}{l} \text { Perimeter }=2(x+y)=62 \quad \ldots . .(i) \text { and } \\ \text...
Let the breath of the rectangular hall be $x$ meter. Therefore, the length of the rectangular hall will be $(x+3)$ meter. According to the question: $\begin{array}{l} x(x+3)=238 \\ \Rightarrow...
Let the tap of smaller diameter fill the tank in $x$ hours. $\therefore$ Time taken by the tap of larger diameter to fill the tank $=(x-9) h$ Suppose the volume of the tank be $V$. Volume of the...
minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.Let one pipe fills the cistern in $x$ mins. Therefore, the other pipe will fill the cistern in $(x+3)$ mins. Time taken by both, running together, to fill the cistern $=3 \frac{1}{13} \min...
in still water, goes 
downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. $\therefore$ Downstream speed $=(9+x) \mathrm{km} / \mathrm{hr}$ Upstream speed $=(9-x) \mathrm{km} / \mathrm{hr}$ Distance covered...
, takes 1 hour more to go 
upstream than to return to the same spot. Find the speed of the stream.Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. Given: Speed of the boat $=18 \mathrm{~km} / \mathrm{hr}$ $\therefore$ Speed downstream $=(18+x) \mathrm{km} / h r$ Speed upstream...
at a uniform speed. Had the speed been 
more, it would have taken 30 minutes less for the journey. Find the original speed of the train.Let the original speed of the train be $x \mathrm{~km} / \mathrm{hr}$. According to the question: $\frac{90}{x}-\frac{90}{(x+15)}=\frac{1}{2}$ $\begin{array}{l} \Rightarrow \frac{90(x+15)-90...
and then travels a distance of 63 
at an average speed of 
more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?Let the first speed of the train be $x \mathrm{~km} / \mathrm{h}$. Time taken to cover $54 \mathrm{~km}=\frac{54}{x} h .$ New speed of the train $=(x+6) \mathrm{km} / \mathrm{h}$ $\therefore$ Time...
at a uniform speed. If the speed had been 
less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.Let the usual speed of the train be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Reduced speed of the train $=(x-8) \mathrm{km} / \mathrm{h}$ Total distance to be covered $=480 \mathrm{~km}$ Time...
Solution: Set of all the first elements or $x$-coordinates of the ordered pairs is called Domain. Set of all the second elements or $y$-coordinates of the ordered pairs is called Range. Therefore,...
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It is given that a, b, c, d are in proportion Consider a/b = c/d = k a = b, c = dk (iii) \[({{a}^{4}}~+\text{ }{{c}^{4}}):\text{ }({{b}^{4}}~+\text{ }{{d}^{4}})\text{ }=\text{...
is then the other root is
Answer is (d)3 Given: $3 x^{2}-10 x+3=0$ One root of the equation is $\frac{1}{3}$. Let the other root be $\alpha$. Product of the roots $=\frac{c}{a}$ $\begin{array}{l} \Rightarrow \frac{1}{3}...
is then the value of 
is
Answer is (c) 8 It is given that the product of the roots of the equation $x^{2}-3 x+k=10$ is $-2$. The equation can be rewritten as: $x^{2}-3 x+(k-10)=0$ Product of the roots of a quadratic...
is 2 then 
?
Answer is (b) $-7$ It is given that one root of the equation $2 x^{2}+a x+6=0$ is 2 . $\begin{array}{l} \therefore 2 \times 2^{2}+a \times 2+6=0 \\ \Rightarrow 2 a+14=0 \\ \Rightarrow a=-7...
Answer is (c) $(\sqrt{2} x+3)^{2}=2 x^{2}+6$ $\begin{array}{l} \because(\sqrt{2} x+3)^{2}=2 x^{2}+6 \\ \Rightarrow 2 x^{2}+9+6 \sqrt{2} x=2 x^{2}+6 \end{array}$ $\Rightarrow 6 \sqrt{2} x+3=0$, which...
Let the base be $x \mathrm{~m}$. Therefore, the altitude will be $(x+7) m$. Area of a triangle $=\frac{1}{2} \times$ Base $\times$ Altitude $\begin{array}{l} \therefore \frac{1}{2} \times x...
. If the base of the triangle exceeds the altitude by 
, find the dimensions of the triangle.Let the altitude of the triangle be $x \mathrm{~cm}$ Therefore, the base of the triangle will be $(x+10) \mathrm{cm}$ $\begin{array}{l} \text { Area of triangle }=\frac{1}{2} x(x+10)=600 \\...
more than the width of the rectangle. Their areas being equal, find the dimensions.Let the breadth of rectangle be $x \mathrm{~cm}$. According to the question: Side of the square $=(x+4) \mathrm{cm}$ Length of the rectangle $=\{3(x+4)\} \mathrm{cm}$ It is given that the areas of...
long and 
wide. There is a path of uniform width all around it, having an area of 
. Find the width of the pathLet the width of the path be $x \mathrm{~m}$. $\therefore$ Length of the field including the path $=16+x+x=16+2 x$ Breadth of the field including the path $=10+x+x=10+2 x$ Now, (Area of the field...
Let the breath of the rectangular hall be $x$ meter. Therefore, the length of the rectangular hall will be $(x+3)$ meter. According to the question: $\begin{array}{l} x(x+3)=238 \\ \Rightarrow...
. Find the dimension of the rectangle.Let the length and breadth of the rectangle be $2 x \mathrm{~m}$ and $x \mathrm{~m}$, respectively. According to the question: $2 x \times x=288$ $\Rightarrow 2 x^{2}=288$ $\Rightarrow x^{2}=144$...
minutes. If on pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.Let the time taken by one pipe to fill the tank be $x$ minutes. $\therefore$ Time taken by the other pipe to fill the tank $=(x+5) \min$ Suppose the volume of the tank be $V$. Volume of the tank...
to finish a piece of work. If both 
and together can finish the work in 12 days, find the time taken by to finish the work.Let B takes $x$ days to complete the work. Therefore, A will take $(x-10)$ days. $\begin{array}{l} \therefore \frac{1}{x}+\frac{1}{(x-10)}=\frac{1}{12} \\ \Rightarrow...
. It can go upstream and 
downstream is 5 hours. Fid the speed of the streamSpeed of the boat in still water $=8 \mathrm{~km} / \mathrm{hr}$. Let the speed of the stream be $x \mathrm{~km} / \mathrm{hr}$. $\therefore$ Speed upstream $=(8-x) \mathrm{km} / \mathrm{hr}$ Speed...
. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speeds of the two train differ by 
.Let the speed of the Deccan Queen be $x \mathrm{~km} / \mathrm{hr}$. According to the question: Speed of another train $=(x-20) \mathrm{km} / \mathrm{hr}$ $\begin{array}{l} \therefore...
at a uniform speed. Had the speed been more, it would have taken 30 minutes less for the journey. Find the original speed of the train.Let the original speed of the train be $x \mathrm{~km} / \mathrm{hr}$. According to the question: $\frac{90}{x}-\frac{90}{(x+15)}=\frac{1}{2}$ $\begin{array}{l} \Rightarrow \frac{90(x+15)-90...
away, in time, the pilot increased the speed by 
hour. Find the original speed of the plane. Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?Let the original speed of the plane be $x \mathrm{~km} / \mathrm{h}$. $\therefore$ Actual speed of the plane $=(x+100) \mathrm{km} / \mathrm{h}$ Distance of the journey $=1500 \mathrm{~km}$ Time...
Let the present ages of the boy and his brother be $x$ years and $(25-x)$ years. According to the question: $\begin{array}{l} x(25-x)=126 \\ \Rightarrow 25 x-x^{2}=126 \\ \Rightarrow x^{2}-(18-7)...
Let the present age of the son be $x$ years. $\therefore$ Present age of the $\operatorname{man}=x^{2}$ years One year ago, Age of the son $=(x-1)$ years Age of the man $=\left(x^{2}-1\right)$ years...
Let the total number of pens be $x$. According to the question: $\begin{array}{l} \frac{80}{x}-\frac{80}{x+4}=1 \\ \Rightarrow \frac{80(x+4)-80 x}{x(x+4)}=1 \\ \Rightarrow \frac{80+320-80 x}{x^{2}+4...
Let the original duration of the tour be $x$ days. $\therefore \text { Original daily expenses }=\text { γ } \frac{10,800}{x}$ If he extends his tour by 4 days, then his new daily expenses...
. Find the fraction.Let the denominator of the required fraction be $x$. Numerator of the required fraction $=x-3$ $\therefore$ Original fraction $=\frac{x-3}{x}$ If 1 is added to the denominator, then the new fraction...
Let the digits at units and tens places be $x$ and $y$, respectively. $\therefore x y=14$ $\Rightarrow y=\frac{14}{x}$ According to the question: $\begin{array}{l} (10 y+x)+45=10 x+y \\ \Rightarrow...
Let the three consecutive positive integers be $x, x+1$ and $x+2$. According to the given condition, $\begin{array}{l} x^{2}+(x+1)(x+2)=46 \\ \Rightarrow x^{2}+x^{2}+3 x+2=46 \\ \Rightarrow 2...
Let the two natural numbers be $x$ and $y$. According to the question: $\begin{array}{l} x^{2}+y^{2}=25(x+y) \quad \ldots \ldots(i) \\ x^{2}+y^{2}=50(x-y) \end{array}$ From (i) and (ii), we get:...
.Let the two parts be $x$ and $(27-x)$. According to the given condition, $\begin{array}{l} \frac{1}{x}+\frac{1}{27-x}=\frac{3}{20} \\ \Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20} \\ \Rightarrow...
. Find the number.Let the natural number be $x$. According to the given condition, $\begin{array}{l} x+\frac{1}{x}=\frac{65}{8} \\ \Rightarrow \frac{x^{2}+1}{x}=\frac{65}{8} \\ \Rightarrow 8 x^{2}+8=65 x \\...
. Find the numbers.Let the required natural numbers be $x$ and $(x+5)$. Now, $x<x+5$ $\therefore \frac{1}{x}>\frac{1}{x+5}$ According to the given condition, $\begin{array}{l}...
. Find the numbers.Let the required natural numbers be $x$ and $(15-x)$. According to the given condition, $\begin{array}{l} \frac{1}{x}+\frac{1}{15-x}=\frac{3}{10} \\ \Rightarrow \frac{15-x+x}{x(15-x)}=\frac{3}{10}...
Let the two consecutive positive even integers be $x$ and $(x+2)$. According to the given condition, $\begin{array}{l} x(x+2)=288 \\ \Rightarrow x^{2}+2 x-288=0 \\ \Rightarrow x^{2}+18 x-16 x-288=0...
Let the required consecutive multiples of 3 be $3 x$ and $3(x+1)$. According to the given condition, $\begin{array}{l} 3 x \times 3(x+1)=648 \\ \Rightarrow 9\left(x^{2}+x\right)=648 \\ \Rightarrow...
Let the two consecutive positive integers be $x$ and $(x+1)$. According to the given condition, $\begin{array}{l} x(x+1)=306 \\ \Rightarrow x^{2}+x-306=0 \\ \Rightarrow x^{2}+18 x-17 x-306=0 \\...
Let the two consecutive positive odd numbers be $x$ and $(x+2)$. According to the given condition, $\begin{array}{l} x^{2}+(x+2)^{2}=514 \\ \Rightarrow x^{2}+x^{2}+4 x+4=514 \\ \Rightarrow 2 x^{2}+4...
Let the required number be $x$ and $(28-x)$. According to the given condition, $\begin{array}{l} x(28-x)=192 \\ \Rightarrow 28 x-x^{2}=192 \\ \Rightarrow x^{2}-28 x+192=0 \\ \Rightarrow x^{2}-16...
Find the number.Let the required natural number be $x$. According to the given condition, $x+\sqrt{x}=132$ Putting $\sqrt{x}=y$ or $x=y^{2}$, we get $y^{2}+y=132$ $\Rightarrow y^{2}+y-132=0$ $\Rightarrow y^{2}+12...
and 
are simultaneously real then prove that 
It is given that the roots of the equation $a x^{2}+2 b x+c=0$ are real. $\begin{array}{l} \therefore D_{1}=(2 b)^{2}-4 \times a \times c \geq 0 \\ \Rightarrow 4\left(b^{2}-a c\right) \geq 0 \\...
for which the roots of 
are real and equalGiven: $\begin{array}{l} 9 x^{2}+8 k x+16=0 \\ \text { Here, } \\ a=9, b=8 k \text { and } c=16 \end{array}$ It is given that the roots of the equation are real and equal; therefore, we have:...
for which the quadratic equation 
has real roots.$2 x^{2}+p x+8=0$ Here, $a=2, b=p$ and $c=8$ Discriminant $D$ is given by: $\begin{array}{l} D=\left(b^{2}-4 a c\right) \\ =p^{2}-4 \times 2 \times 8 \\ =\left(p^{2}-64\right) \end{array}$ If $D...
., find the value of so that the roots of the equation 
are equal.It is given that 3 is a root of the quadratic equation $x^{2}-x+k=0$. $\begin{array}{l} \therefore(3)^{2}-3+k=0 \\ \Rightarrow k+6=0 \\ \Rightarrow k=-6 \end{array}$ The roots of the equation...
for which the quadratic equation 
. , 
has equal roots. Hence find the roots of the equation.The given equation is $(p+1) x^{2}-6(p+1) x+3(p+9)=0$. This is of the form $a x^{2}+b x+c=0$, where $a=p+1, b=-6(p+1)$ and $c=3(p+9)$. $\begin{array}{l} \therefore D=b^{2}-4 a c \\ =[-6(p+1)]^{2}-4...
for which the quadratic equation 
. has real and equal roots.The given equation is $(3 k+1) x^{2}+2(k+1) x+1=0$. This is of the form $a x^{2}+b x+c=0$, where $a=3 k+1, b=2(k+1)$ and $c=1$. $\begin{array}{l} \therefore D=b^{2}-4 a c \\ =[2(k+1)]^{2}-4 \times(3...
are the roots of the equation 
. real and equal?The given equation is $4 x^{2}+p x+3=0$. This is of the form $a x^{2}+b x+c=0$, where $a=4, b=p$ and $c=3$. $\therefore D=b^{2}-4 a c=p^{2}-4 \times 4 \times 3=p^{2}-48$ The given equation will have...
are the roots of the quadratic equation 
real and equal.The given equation is $\begin{array}{l} k x(x-2 \sqrt{5})+10=0 \\ \Rightarrow k x^{2}-2 \sqrt{5} k x+10=0 \end{array}$ This is of the form $a x^{2}+b x+c=0$, where $a=k, b=-2 \sqrt{5} k$ and $c=10$....
has no real roots.The given equation is $2\left(a^{2}+b^{2}\right) x^{2}+2(a+b) x+1=0$ $\begin{array}{l} \therefore D=[2(a+b)]^{2}-4 \times 2\left(a^{2}+b^{2}\right) \times 1 \\ =4\left(a^{2}+2 a...
(i) The given equation is $5 x^{2}-4 x+1=0$ This is of the form $a x^{2}+b x+c=0$, where $a=5, b=-4$ and $c=1$. $\therefore$ Discriminant, $D=b^{2}-4 a c=(-4)^{2}-4 \times 5 \times 1=16-20=-4<0$...
, where 
and 
$12 a b x^{2}-\left(9 a^{2}-8 b^{2}\right) x-6 a b=0$ On comparing it with $A x^{2}+B x+C=0$, we get: $A=12 a b, B=-\left(9 a^{2}-8 b^{2}\right)$ and $C=-6 a b$ Discriminant $D$ is given by:...
$3 a^{2} x^{2}+8 a b x+4 b^{2}=0$ On comparing it with $A x^{2}+B x+C=0$, we get: $A=3 a^{2}, B=8 a b$ and $C=4 b^{2}$ Discriminant $D$ is given by: $\begin{array}{l} D=\left(B^{2}-4 A C\right) \\...
.The given equation is $x^{2}-(2 b-1) x+\left(b^{2}-b-20\right)=0$ Comparing it with $A x^{2}+B x+C=0$, we get $A=1, B=-(2 b-1)$ and $C=b^{2}-b-20$ $\therefore$ Discriminant, $D=B^{2}-4 A C=[-(2...
and 
respectively. Find the height of the tower and the flagpole mounted on itSolution:- (i) We have to find the ratios AD/AB, DE/BC, From the question it is given that, AD/DB = 3/2 Then, DB/AD = 2/3 Now add 1 for both LHS and RHS we get, (DB/AD) + 1 = (2/3) + 1 (DB + AD)/AD...
Solution:- From the given figure it is given that, CM and RN are respectively the medians of ∆ABC and ∆PQR. (i) We have to prove that, ∆AMC ~ ∆PQR Consider the ∆ABC and ∆PQR As ∆ABC ~ ∆PQR ∠A = ∠P,...
Solution:- (i) From the ΔABC and ΔPQR AB/PQ = 3.2/4 = 32/40 Divide both numerator and denominator by 8 we get, = 4/5 AC/PR = 3.6/4.5 = 36/45 Divide both numerator and denominator by 9 we get, = 4/5...
(iii)Let E be an event of getting a sum divisible by 5. Favourable outcomes = {(1,4),(2,3), (3,2), (4,1),(4,6), (5,5), (6,4)} Number of favourable outcomes = 7 P(E) = 7/36 Probability of getting a...
Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...
Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...
Solution: Total number of cards = 52-4 = 48 [∵4 cards fell down] So number of possible outcomes = 48 (i) Let E be the event of getting black card. There will be 23 black cards since a black jack and...
(v) Let E be the event of getting a spade. There will be 10 spades. Number of favourable outcomes = 10 P(E) = 10/49 Hence the probability of getting a spade is 10/49. (vi) Let E be the event of...
(iii) Let E be the event of getting a king of red colour. There will be 2 cards of king of red colour. Number of favourable outcomes = 2 P(E) = 2/52 = 1/26 Hence the probability of getting a king of...
Solution: Total number of cards = 52. So number of possible outcomes = 52. (i) Let E be the event of getting ‘2’ of spades. There will be only one card of ‘2’ spades. Number of favourable outcomes =...
(v) Let E be the event of getting the number on the card is divisible by 3 or 2 Outcomes favourable to E are {2,3,4,6,8,9,10,12,14,15} Number of favourable outcomes = 10 P(E) = 10/15 = 2/3 Hence the...
Solution: There are 7 ways in which the month of February can occur, each starting with a different day of the week. (i)Only 1 way is possible for 5 Wednesdays to occur in February with 29 days....
Solution: For a leap year there are 366 days. Number of days in January = 31 Total number of January month types = 7 Number of January months with 5 Mondays = 3 (i)Probability that the month of...
(iii) Let E be the event of arrow pointing a number greater than 2. Outcomes favourable to E are {3,4,5,6,7,8} Number of favourable outcomes = 6 P(E) = 6/8 = 3/4 Hence the probability of arrow...
(iii)Let E be the event of getting a number greater than 5. Outcomes favourable to E is 6. Number of favourable outcomes = 1 P(E) = 1/6 Hence the probability of getting a number greater than 5 is...
Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. Number of possible outcomes = 6 (i) Let E be the event of getting an odd number. Outcomes favourable to E are 1,3,5. Number of...
Solution: Number of blue marbles = 7 Number of white marbles = 8 Number of black marbles = 5 Total number of marbles = 7+8+5 = 20 (i) Probability of getting black , = 5/20 = 1/4 Hence the...
Solution: Number of black balls = 5 Number of red balls = 7 Number of white balls = 3 Total number of balls = 5+7+3 = 15 (i)Probability that the ball drawn is red, = 7/15 (ii) Probability of black...
Solution: Total number of alphabets = 26 Number of vowels = 5 Total number of consonants = 26-5 = 21 Probability that the letter chosen is a consonant , = 21/26 Hence the required probability is...