प्रस्तुत पाठ ‘शुकसप्ततिः’ नामक प्रसिद्ध कथाग्रन्थ से सम्पादित किया गया है। ‘शुकसप्ततिः’ के लेखक और काल के विषय में आज भी भ्रान्ति बनी हुई है। शुकसप्तति अत्यन्त सरल और मनोरंजक कथासंग्रह है। प्रस्तुत कहानी में अपने दो पुत्रों के साथ जंगल के रास्ते से पिता के घर जा रही बुद्धिमती नामक नारी के बुद्धिकौशल को दिखाया गया है जो अपनी चातुर्य से सामने आए बाघ को भी डरा कर भगा देती है।
बुद्धिमती बाघ को समक्ष देख अपने पुत्रों को डाँटने का नाटक करती हुई कहती है कि झगड़ा मत करो। आज एक ही बाघ को बाँटकर खा लो फिर दूसरा देखते हैं। यह सुन ‘यह व्याघ्रमारी है’ ऐसा मानकर बाघ डरकर भांग जाता है। भयभीत बाघ को देखकर शृगाल बाघ का उपहास उड़ाता हुआ कहता है कि मुझे अपने गले में बाँधकर चलो जहाँ वह धूर्ता है। शृगाल के साथ पुनः आते बाघ को देखकर बुद्धिमती अपनी प्रत्युत्पन्नमति से शृगाल को ही आक्षेप लगाती हुई कहती है कि तुमने तीन बाघ देने के लिए कहा था। आज एक ही बाघ क्यों लाए हो? यह सुनते ही शृगाल सहित बाघ भाग जाता है। इस प्रकार अपने बुद्धिबल से वह अपनी और अपने पुत्रों की प्राणरक्षा करती है।
प्रस्तुत पाठ आधुनिक संस्कृत कवि हरिदत्त शर्मा के रचना संग्रह ‘लसल्लतिका’ से संकलित है। इनमें कवि ने महानगरों में बढ़ते प्रदूषण पर चिन्ता व्यक्त करते हुए और मानव कल्याण के लिए पर्यावरण शुद्धि व स्वच्छता का संदेश देकर शुद्ध पर्यावरण की ओर जनजागरण करने का प्रयास किया है।
महानगरों में दिन-रात दौड़ते हुए यातायात साधनों से मनुष्य का शारीरिक व मानसिक ह्रास हो गया है। इस कारण वहाँ रहना मनुष्यों के लिए दुष्कर हो गया है। काले धुएँ को छोड़ती असंख्य गाड़ियों के कोलाहलपूर्ण मार्गों पर चलना भी कठिन है। प्रदूषित वातावरण के कारण वायुमण्डल, जल, भोज्य पदार्थ और समस्त धरातल सब कुछ अति दूषित हो गया है। अतः प्रदूषण को दूर कर पर्यावरण को शुद्ध करना चाहिए। प्रदूषित पर्यावरण से दु:खी कवि की इच्छा है कि कुछ समय के लिए महानगर से दूर प्रकृति की गोद में, एकान्त वन में वास करें, जहाँ पवित्र व शुद्ध जलपूर्ण नदियाँ, झरने और जलाशय हैं। जहाँ पक्षियों के मधुर कलरव से वन प्रदेश गुञ्जित हैं। चकाचौंध पूर्ण इस वातावरण से दूर प्राकृतिक दृश्यों को निहारते हुए जीवन को रसमय करना चाहता है।
कवि की आशंका है कि इस प्रदूषित पर्यावरण में हमारी संस्कृति, प्राकृतिक छटा, पाषाणी सभ्यता लुप्त न हो जाए। सुखमय मानव जीवन की कामना करता हुआ कवि पर्यावरण को शुद्ध व सुरक्षित करने की प्रार्थना करता है।
Q1: Find the centroid of the triangle XYZ whose vertices are X (3, – 5) Y (- 3, 4) and Z (9, – 2). (a) (0, 0) (b) (3, 1) (c) (2, 3) (d) (3,-1) Ans: (d) Centroid of the triangle is given by
Q2: If the mid-point of the line segment joining the points A (3, 4) and B (a, 4) is P (x, y) and x + y – 20 = 0, then find the value of a (a) 0 (b) 1 (c) 40 (d) 45 Ans: (d) mid point (3+a)/2, 4 Now (3+a)/2 -4 -20 = 0 3 + a = 48 a = 45
Q3: Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7). (a) 2:9 (b) 1:9 (c)1:2 (d) 2:3 Ans: (a)
Short Answer type Q4: The coordinates of one end point of a diameter of a circle are (4, -1) and the co-ordinates of the centre of the circle are (1, -3). Find the co- ordinates of the other end of the diameter. Ans:
Q5: If the mid- point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of k. Ans:
Long Answer Type Q6: Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3) Ans:
Q7: Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus. Also, find its area. Ans: The given points are P(3, 0), Q(4, 5), R(-1, 4) and S(-2, -1).We have
Therefore Sides are equal PQ = QR = RS = SP = √26 and Diagonal are not equal PR ≠ QS Hence PQRS is a rhombus but not a square. Now Area of rhombus PQRS is given by A = 1/2 D1 × D2 Where D1 and D2 are diagonals So A = 1/2 4√2 × 6√2 = 24sq.units
Q8: Two opposite vertices of a square are (-1, 2) and (3, 2). Find the co-ordinates of other two vertices. Ans: Let ABCD is the square and A (-1,2) and C(3,2). Lets us assume the coordinate of B(x,y) Now Sides are equal in square, Therefore AB = BC
Putting the value of x from (1), we get [(1 + 1)2 + (y − 2)2] = 8 Solving it ,we get y2 − 4 y = 0 Solving this quadratic equation, we get y = 0 or 4 So coordinates of the other vertices are (1, 0) and (1, 4)
Q9: Find a point on y – axis which is equidistant from the points (5, -2) and (-3, 2). Ans: Let the point on y axis is P(0,y) which is equidistant from the points A(5,-2) and B(-3,2). Then PA = PB
Q10: If the coordinates of the mid points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of its vertices. Ans:
Q1: Which of the following triangles have the same side lengths? (a) Scalene (b) Isosceles (c) Equilateral (d) None of these Ans: (c) Sol: Equilateral triangles have all sides and all angles equal.
Q2: Area of an equilateral triangle with side length a is equal to: (a) (√3/2)a (b) (√3/2)a2 (c) (√3/4) a2 (d) (√3/4) a Ans: (c) Sol: Area of an equilateral triangle with side length a = √3/4 a2
Q3: D and E are the midpoints of side AB and AC of a triangle ABC, respectively, and BC = 6 cm. If DE || BC, then the length (in cm) of DE is: (a) 2.5 (b) 3 (c) 5 (d) 6 Ans: (b) Sol: By midpoint theorem, DE = ½ BC; DE = ½ of 6; DE = 3 cm
Q4: The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of the rhombus in length is: (a) 20 cm (b) 8 cm (c) 10 cm (d) 9 cm Ans: (c) Sol: Here, half of the diagonals of a rhombus are the sides of the triangle, and the side of the rhombus is the hypotenuse. By Pythagoras theorem, (16/2)2 + (12/2)2 = side2 82 + 62 = side2 side = 10 cm
Q5: Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the small triangle is 48 sq.cm, then the area of the large triangle is: (a) 230 sq.cm. (b) 106 sq.cm (c) 107 sq.cm. (d) 108 sq.cm Ans: (d) Sol: Let A1 and A2 be the areas of the small and large triangles. Then, A2/A1 = (side of large triangle/side of small triangle); A2/48 = (3/2)2 A2 = 108 sq.cm.
Q6: If the perimeter of a triangle is 100 cm and the length of two sides are 30 cm and 40 cm, the length of the third side will be: (a) 30 cm (b) 40 cm (c) 50 cm (d) 60 cm Ans: (a) Sol: Perimeter of the triangle = sum of all its sides; P = 30 + 40 + x 100 = 70 + x x = 30 cm
Q7: If triangles ABC and DEF are similar and AB = 4 cm, DE = 6 cm, EF = 9 cm, and FD = 12 cm, the perimeter of triangle ABC is: (a) 22 cm (b) 20 cm (c) 21 cm (d) 18 cm Ans: (d) 18 cm Sol: ABC ~ DEF AB = 4 cm, DE = 6 cm, EF = 9 cm, and FD = 12 cm; AB/DE = BC/EF = AC/DF BC = (4.9)/6 = 6 cm AC = (12.4)/6 = 8 cm Perimeter of triangle ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm
Q8: The height of an equilateral triangle of side 5 cm is: (a) 4.33 cm (b) 3.9 cm (c) 5 cm (d) 4 cm Ans: (a) Sol: The height of the equilateral triangle ABC divides the base into two equal parts at point D. Therefore, BD = DC = 2.5 cm. In triangle ABD, using Pythagoras theorem, AB2 = AD2 + BD2 52 = AD2 + 2.52 AD2 = 25 – 6.25 AD2 = 18.75 AD = 4.33 cm
Q9: If ABC and DEF are two triangles and AB/DE = BC/FD, then the two triangles are similar if (a) ∠A = ∠F (b) ∠B = ∠D (c) ∠A = ∠D (d) ∠B = ∠E Ans: (b) Sol: If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if ∠B=∠D.
Q10: Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio (a) 2: 3 (b) 4: 9 (c) 81: 16 (d) 16: 81 Ans: (d) Sol: Let ABC and DEF be two similar triangles, such that, ΔABC ~ ΔDEF and AB/DE = AC/DF = BC/EF = 4/9. As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides, ∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2 ∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81
Solve the following Questions
Q1: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. Ans: We know that
QR = 6cm
Q2: Determine whether the triangle having sides (b − 1) cm, 2√b cm and (b + 1) cm is a right angled triangle. Ans: These are sides of the Right angle triangle
Q3: Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm,4 cm,5 cm (iii) 40 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm Ans: i. 252 = 72 + 242 Hence right angle triangle with 25 cm as hypotenuse ii. 52 = 32 + 42 Hence right angle triangle with 5 cm as hypotenuse iii. Longest side = 100 cm Check: 100² = 10000 40² + 80² = 1600 + 6400 = 8000 No, it is not a right triangle. iv. 132 = 122 + 52 Hence right angle triangle with 13 cm as hypotenuse
Q4: In ΔABC, AD is perpendicular to BC. Prove that: a. AB2 + CD2 = AC2 + BD2 b. AB2 − BD2 = AC2 − CD2 Ans: In ΔABC, AD is perpendicular to BC
Now ΔABD is an right -angle triangle So from Pythagoras theorem
Similarly ΔADC is an right -angle triangle
So from Pythagoras theorem From (1) and (2) AB2 – BD2 = AC2 – CD2 Which proved part (b) Now rearranging,
Which proved part (a).
Q5: Triangle ABC is right- angled at B and D is the mid – point of BC. Prove that: AC2 = 4AD2 − 3AB2 Ans:
In right angle triangle ABC AC2 = AB2 − BC2___(1) In right triangle ABD
Substituting this in equation (1) AC2 = 4AD2 − 3AB2
Q6: ABC is an isosceles triangle, right -angled at C. Prove that AB2 = 2BC2 . Ans:
Now Since it is an isosceles triangle AC=BC -(1) Now from Pythagorean theorem
Q7: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. Ans:
AB = 6m BC= 4 m Let Height of tower (DE)=h m EF = 28 m
h = 42m
Q8: The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach? Ans:
First case is depicted in the figure (a) Now from Pythagoras theorem
This is the length of the ladder. Now Second case is depicted in figure (b) AC=10 cm BC=8 cm Now from Pythagoras theorem
Q9: The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. Ans: We know that
QR = 8.8 cm
Q10: DEF is an equilateral triangle of side 2b. Find each of its altitudes. Ans:
Multiple Choice Questions Q1: For what value of a, are (2a – 1), 7 and 3a three consecutive terms of an A.P? (a) 6 (b) 3 (c) 2 (d) 8 Ans: (b) 7 − (2a − 1 ) = 3a − 7 15 = 5 a α = 3
Q2: Find the value of p, so that (3p + 7), (2p + 5), (2p + 7) are in A.P (a) 5 (b) -5 (c) 4 (d) -4 Ans: (d) (2p + 5)−(3p + 7) =(2p + 7)−(2p + 5) −p − 2 = 2 p = −4
Q3: Solve the equation: 1 + 4 + 7 + 10 + . . . + x = 287 (a) 40 (b) 41 (c) 59 (d) 54 Ans: (a) Given, the sum of the terms up to x is 287. We have to solve the equation. The sum of the first n terms of an AP is given by Sₙ = n/2[2a + (n-1)d] Here, first term, a = 1 Common difference, d = 4 – 1 = 3 So, 287 = n/2[2(1) + (n – 1)3] 287 = n/2[2 + 3n – 3] 287 = n/2[3n – 1] 574 = n(3n – 1) 3n² – n = 574 3n² – n – 574 = 0 Using the quadratic formula, n = -b±√(b²-4ac)/2a Here, a = 3, b = -1 and c = -574 n = -(-1)±√((-1)²-4(3)(-574))/2(3) = 1±√(1+6888)/6 = 1±√6889/6 n = (1±83)/6 Now, n = (1+83)/6 = 84/6 = 14 n = (1-83)/6 = -82/6 = -41/3 Since a negative term is not possible, n = -41/3 is neglected. So, n = 14 The nth term of the series in AP is given by aₙ = a + (n – 1)d So, x = 1 + (14 – 1)(3) = 1 + 13(3) = 1 + 39 = 40 Therefore, the value of x is 40.
Q4: Find the sum ..upto 11 terms (a) (b) (c) (d)
Ans: (d)
Q5: How many multiples of 4 lie between 10 and 250? (a) 66 (b) 65 (c) 60 (d) 64 Ans: (c) The AP will be 12, 16 … 248 Now 248 = 12 + (n−1)4 n = 60
Short Answer Questions Q6: Find the value of the middle most term (s) of the AP -11, -7, -3,…, 49 Ans: Here a = – 11, d = 4 Now 49 = − 11 + (n − 1)4 n = 16 So middle most terms will 8 and 9 T8 = − 11 + ( 8 − 1 ) 4 = 17 T9 = − 11 + ( 9 − 1 ) 4 = 21
Q7: The 4th term of an A.P is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the A.P Ans: Let 1st term be a and common difference = d According to the question, T4 = 3T1 α + 3d = 3α 3d = 2a -(1) Also, T7 = 2T3 + 1 α + 6d = 2 [a + (2 d] + 1 2d = a + 1 -(2) Solving equation (1) and (2) a = 3, d = 2 So, the AP formed is 3, 5, 7, 9, ………
Q8: The angles of a triangle are in A.P, the least being half the greatest. Find the angles Ans: Let the angles be, (a-d), (a), (a+d) Sum of the angles of a triangle is 180. So, (a-d)+(a)+(a+d) = 180 ______(1) 3a = 180 a = 60 Also, from the given condition, 60-d = 1/2 (60+d) 2(60-d) = 60+d 120-2d = 60+d 120-60 = d+2d 60 = 3d d = 20 Thus, the required angles are 40, 60, 80.
Long Answer Question Q9: Find the sum of all three digit numbers which leave the remainder 3 when divided by 5. Ans: The smallest 3 digit no. = 100 and greatest 3 digit number is 999 Since 100 is divisible by 5,adding 3 on 100 will provide the number which leave the remainder 3 when divided by 5 So the smallest three digit number which is divisible by 5 and gives reminder 3 = 103 The largest 3 digit no , 999/5 gives reminder 4 ,So subtracting 1 will the number which leave the remainder 3 when divided by 5 So the Largest three digit number which is divisible by 5 and gives reminder 3 = 998 Similarly we can find other numbers, the number will be given as 103, 108, 111,…998 Now S = 103 + 108 + 111…+ 998 This is a AP with first term =103 and common difference = 5,last term = 998 Number of term can found from nth term formula αn = α1 + (n−1) × d 998 = 103 + (n−1) × 5 n = 180 Therefore S = 103 + 108 + 111…+ 998 =
Q10: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th terms is 44. Find the first three terms of the AP Ans: The formula for nth term of an AP is aₙ = a + (n – 1) d Here, aₙ is the nth term, a is the first term, d is the common difference and n is the number of terms. Given, a₄ + a₈ = 24 (a + 3d) + (a + 7d) = 24 ⇒ 2a + 10d = 24 ⇒ a + 5d = 12 ….. Equation(1) Also, a₆ + a₁₀ = 44 (a + 5d ) + (a + 9d) = 44 ⇒ 2a + 14d = 44 ⇒ a + 7d = 22 …. Equation(2) On subtracting equation (1) from (2), we obtain (a + 7d ) – (a + 5d) = 22 – 12 a + 7d – a – 5d = 10 2d = 10 d = 5 By substituting the value of d = 5 in equation (1), we obtain a + 5d = 12 a + 5 × 5 = 12 a + 25 = 12 a = – 13 The first three terms are a , (a + d) and (a + 2d) Substituting the values of a and d , we get – 13, (- 13 + 5) and (- 13 + 2 × 5) The first three terms of this A.P. are – 13, – 8, and – 3.
and a1/a2 ≠ b1/b2 Hence , have a unique solution. So, The correct option is Option c
Q2: If a system of a pair of linear equations in two unknowns is consistent, then the lines representing the system will be (a) parallel (b) always coincident (c) always intersecting (d) intersecting or coincident Ans: (d) intersecting or coincident
Sol: A consistent system of linear equations means that there is at least one solution. This can occur in two scenarios:
Intersecting Lines: The lines intersect at exactly one point, meaning there is one unique solution.
Coincident Lines: The lines lie on top of each other, meaning there are infinitely many solutions.
Q3: The pair of equations x = 0 and y = 0 has (a) one solution (b) two solutions (c) infinitely many solutions (d) no solution Ans: (a) one solution
Sol: The equations x=0 and y=0 represent the x-axis and y-axis respectively. The only point where these two lines intersect is at the origin, which is the point (0,0). Therefore, there is exactly one solution to this system of equations
Q4: A pair of system of equations x = 2, y = -2; x = 3, y = – 3 when represented graphically enclose (a) Square (b) Trapezium (c) Rectangle (d) Triangle Ans: (c) Rectangle
Sol: The equations x=2 andx=3 represent vertical lines, while y=−2 and y=−3 represent horizontal lines. The intersection points of these lines are (2,−2), (2,−3), (3,−2), and (3,−3). These points form the vertices of a rectangle, making the shape enclosed by these lines a rectangle on the graph.
Q5: If two lines are parallel to each other, then the system of equations is (a) consistent (b) inconsistent (c) consistent dependent (d) (a) and (c) both Ans: inconsistent
Sol: If two lines are parallel, they never intersect, meaning there is no solution to the system of equations. Thus, the system is inconsistent.
B. Fill in the blanks
Q6:If in a system of equations corresponding to coefficients of member, equations are proportional then the system has ______________ solution (s). Ans:infinitely many
Q7: A pair of linear equations is said to be inconsistent if its graph lines are ____________. Ans: parallel Q8:A pair of linear equations is said to be ____________ if its graph lines intersect or coincide. Ans:Consistent Q9: A consistent system of equations where straight lines fall on each other is also called _____________ system of equations. Ans:Dependent Q10:Solution of linear equations representing 2x – y = 0, 8x + y = 25 is ____________ . Ans: x = 2.5, y = 5
Sol:
1. From 2x – y = 0, we can express y as:
y = 2x
2. Substitute y into the second equation 8x + y = 25:
8x + 2x = 25
10x = 25 → x = 25/10 = 2.5
3. Substitute x = 2.5 back into y = 2x:
y = 2(2.5) = 5
Thus, the solution is x = 2.5 and y = 5,
C. Very Short Answer Questions
Q11:In Fig., ABCD is a rectangle. Find the values of x and y.
Sol: Since ABCD is a rectangle ⇒ AB = CD and BC = AD x + y = 30 …………….. (i) x – y = 14 ……………. (ii) (i) + (ii) ⇒ 2x = 44 ⇒ x = 22 Plug in x = 22 in (i) ⇒ 22 + y = 30 ⇒ y = 8
Q12: Name the geometrical figure enclosed by graph of the equations x + 7 = 0, y – 2 = 0 and x – 2 = 0, y + 7 = 0.
Sol: Clearly, a square of side 9 units is enclosed by lines.
Q13: If 51x + 23y = 116 and 23x + 51y = 106, then find the value of (x – y).
Sol: Given equations are: 51x + 23y = 116 ……….. (i) 23x + 51y = 106 ………… (ii) Subtracting (ii) from (i) 28x – 28y = 10 28(x – y) = 10 ⇒ (x – y) = =
Q14: For what value of V does the point (3, a) lie on the line represented by 2x – 3y = 5?
Sol: Since (3, a) lies on the equation 2x – 3y = 5. ∴ (3, a) must satisfy this equation ⇒ 2(3) – 3(a) = 5 ⇒ 6 – 3a = 5 ⇒ – 3a = 5 – 6 = – 1 ∴ a = 1/3
Q15: Determine whether the following system of linear equations is inconsistent or not. 3x – 5y = 20 6x – 10y = -40
Sol: Given 3x – 5y = 20 ……… (i) 6x – 10y = – 40 ………… (ii) Hence given pair of linear equations are parallel.∴ It is inconsistent.
D. Short Answer Type Questions
Q16: The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. Find the present ages of the son and the father.
Sol:
Let the father’s age be y and the son’s age be x. First condition: y = 6x Second condition: y + 4 = 4 (x + 4) So, we have 6x + 4 = 4 x + 16 x = 6 and y = 36
Q17: If the lines x + 2y + 7 = 0 and 2x + ky + 18 = 0 intersect at a point, then find the value of k.
Sol: For a unique solution Substituting the values
Q18: Find the value of k for which the system of equations x + 2y -3 = 0 and ky + 5x + 7 = 0 has a unique solution.
Sol: For a unique solution
Substituting the values
Q19: Find the values of a and b for which the following system of linear equations has an infinite number of solutions: 2x + 3 y = 7 2ax + (a + b) y = 28
Sol: We have
Solving this , we get a = 4 and b = 8
Q20: In a cyclic quadrilateral ABCD, Find the four angles. a. ∠A = (2 x + 4), ∠B = (y + 3), ∠C = (2y + 10) , ∠D = (4x − 5) . b. ∠A = (2 x − 1) , ∠ B = (y + 5) , ∠C = (2 y + 15) and ∠D = (4 x − 7)
Sol: a. ∠ A = (2 x + 4) , ∠B = (y + 3), ∠C = (2 y + 10), ∠D = (4 x − 5) In a cyclic quadrilateral, Opposite angles are supplementary. ∠A + ∠C = 180° and ∠ B + ∠ D = 180° So, 2x + 4 + 2y + 10 = 180 or x + y = 83 – (1) y + 3 + 4 x − 5 = 180 or y + 4 x = 182 -(2) Solving the equations (1) and (2)by Substitution method x = 33 and y = 50 So Angles are 70°, 53°,110°,127°
b. ∠A = (2 x − 1) , ∠B = (y + 5) , ∠C = (2 y + 15) and ∠D = (4 x − 7) Solving similarly, we get 65°, 55°, 115°, 125°
E. Long Answer Type Questions
Q21:Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and the x-axis. Find the area of the shaded region.
Sol: The given system of equations is 2x + y – 6 = 0 ………. (i) 2x – y + 2 = 0 ……… (ii) Let us write three solutions for each equation of the system in a table. (i) ⇒ y = 6 – 2x Table of solutions for 2x + y – 6 = 0
Similarly (ii) ⇒ y = 2x + 2 Table of solutions for 2x – y + 2 = 0
Plotting these points of each table of solutions on the same graph paper and joining them with a ruler, we obtain the graph of two lines represented by equations (i) and (ii) respectively as shown in the graph below. Since, the two lines intersect at point P(1, 4). Thus x = 1, y = 4 is the solution of the given system of equations. In the graph, the area is bounded by the lines and the x-axis is ∆PAB which is shaded.Draw PM ⊥ x-axis Clearly, PM = y-coordinate of P(1, 4) = 4 units Also, AB = 1 + 3 = 4 units ∴ Area of the shaded region = Area of ∆PAB = × AB × PM= × 4 × 4 = 8 sq. units.
Q22: Draw the graphs of the following equations: 2x – y = 1, x + 2y = 13 (i) Find the solution of the equations from the graph. (ii) Shade the triangular region formed by lines and the y-axis.
Sol: 2x – y = 1 …………….. (i) x + 2y = 13 ………………. (ii) Let us draw a table of values for (i) and (ii) Plotting these points on the graph paper, we see that the two lines representing equations (i) and (ii) intersect at points (3, 5). (i) Therefore, (3, 5) is the solution of a given system, (ii) Also, two lines enclose a triangular region (∆ABC) with a y-axis shaded in the graph.
Q23: A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Sol: Let the speed of the train= x km/ hour Speed of the car = y km/hour Case I: Total distance travelled = 370 km Distance traveled by train = 250 km Distance travelled by car = (370 – 250) km = 120 km
or 250y + 120x = 4xy or 60x + 125y = 2xy
Case II: Total distance covered = 370 km Distance covered by train = 130 km Distance covered by car = (370 -130) km = 240 Time is taken by train = hour Time is taken by car = hour According to 2nd condition, = 4 hours 18 minutes 130x+240y=4310 1300y + 2400x = 43xy or 2400x + 1300y = 43xy Multiplying equation (i) by 40, we get
or x = 100 From (i) and (iii), we get 60(100) + 125y = 2(100)y 6000 + 125y = 200y or (200 – 125) y = 6000 or 75y = 6000 or y = = 80 Hence, the speeds of the train and car is 100 km/hour and 80 km/hour respectively
Q24: The taxi charges in a city comprise a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is ₹75 and for a journey of 15 km, the charge paid is ₹110. What will a person have to pay for traveling a distance of 25 km?
Sol: Let fixed charges for taxi be ₹x and charges for covering distance be ₹y per km. Then, according to the question, we have x + 10y = 75 ……… (i) and x + 15y = 110 Subtracting (i) from (ii), we get 5y = 35 ⇒ y = 35 ÷ 5 = 7 Putting y = 7 in (i), we get x + 10 (7) = 75 ⇒ x = 75 – 70 = 5 ∴ The person will have to pay for travelling a distance of 25 km = x + 25y = 5 + 25(7) = ₹180.
Q25: Solve the following system by drawing their graph: (3/2)x – (5/4)y = 6, 6x – 6y = 20. Determine whether these are consistent, inconsistent, or dependent.
Sol:
Plotting the points and joining by a ruler in each case. Here, we see that the graph of given equations are parallel lines. The two lines have no point in common. The given system of equations has no solution and is, therefore, inconsistent.
Q1: If a and b are roots of the equation x2 + 7 x + 7 . Find the value of a−1 + b−1 − 2αb ? Ans: for f ( x ) = x2 + 7 x + 7 we get α + b = − 7 αb = 7 Now a−1 + b−1 − 2αb
Q2: If the zeroes of the quadratic polynomial x2 + (α + 1 ) x + b are 2 and -3, then find the value of a and b. Ans: Let f (x) = x2 + (a + 1) x + b Then 2 − 3 = − ( a + 1 ) or α = 0 − 6 = b So a = 0 and b = -6
Q.3. If a and b are zeroes of the polynomial f (x) = 2x2 − 7x + 3, find the value of α2 + b2.
Ans: f (x) = 2x2 − 7 x + 3 = 2x2 − x − 6 x + 3 = x(2x − 1) − 3(2 x − 1) = (x − 3) (2 x − 1) So zeroes are 3 and 1/2 Now α2 + b2
Q.4. Find the zeroes of the quadratic polynomial x2 + x − 12 and verify the relationship between the zeroes and the coefficients. Ans: x2 + x − 12 = x2 + 4x − 3x − 12 = x (x + 4) − 3 (x + 4) = (x − 3) (x + 4) So zeroes are 3 and -4 as we know sum of roots = -b/a = -(1)/1 = -1 ie 3+(-4) = -1 product of roots = c/a = -12/1 = -12 ie 3x-4 = -12
Q5: If p and q are zeroes of f (x) = x2 − 5x + k, such that p − q = 1 , find the value of k. Ans: for f ( x ) = x2 − 5 x + k we get p + q = 5 pq = k Now p − q = 1 (p − q)2 = 1 (p + q)2 − 4pq = 1 25 − 4k = 1 k = 6
Q6: Given that two of the zeroes of the cubic polynomial αx3 + bx2 + cx + d are 0, then find the third zero. Ans: Two zeroes = 0, 0 Let the third zero be k. The, using relation between zeroes and coefficient of polynomial, we have: k + 0 + 0 = − b a Third zero = k = -b/a
Q.7. If one of the zeroes of the cubic polynomial x3 + αx2 + bx + c is -1, then find the product of the other two zeroes. Ans:
Q8: If a-b, a , a+b , are zeroes of x3 − 6x2 + 8x , then find the value of b Ans: Let f ( x ) = x3 − 6x2 + 8 x Method -1 = x (x2 − 6 x + 8) = x(x − 2) (x − 4) So 0,2,4 are zeroes of the polynomial. or a=2 and b=2 or -2 Method -2
Q9: Quadratic polynomial 4x2 + 12x + 9 has zeroes as p and q . Now form a quadratic polynomial whose zeroes are p − 1 and q − 1 Ans: 4x2 + 12 x + 9 = 4x2 + 6x + 6x + 9 = 2x(2x + 3) + 3(2x + 3) = (2 x + 3)2 So p = -3/2 and q = -3/2 So, p − 1 = − 5/2 and q − 1 = − 5/2 So quadratic polynomial will be (x + 5/2)2 or 4x2 + 20 x + 25
Long Answer Type Questions
Q10: p and q are zeroes of the quadratic polynomial x2 − (k + 6 ) x + 2(2 k − 1) . Find the value of k if 2(p + q) = p q Ans: for f (x) = x2 − (k + 6) x + 2 (2 k − 1) We get, p + q = k + 6 p q = 2(2 k − 1) Now 2 (p + q ) = pq Therefore, 2 (k + 6) = 2(2k − 1) or k + 6 = 2 k − 1 or k = 7
Q11: Given that the zeroes of the cubic polynomial x3 − 6 x2 + 3 x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial. Ans: k1 + k2 + k3 = a + a + b + a + 2b = 6 a + b = 2 a = 2 − b Now
b=-3 or b=3 So a= 5 or -1 The zeroes with a = 5, b= -3 can be expressed as 5, 2, -1 The zeroes with a = -1, b = 3 can be expressed as -1, 2, 5
Q12: If one zero of the polynomial 2x2−5x−(2k + 1) is twice the other, find both the zeroes of the polynomial and the value of k.
Ans: Let a be one zero ,then another will be 2a
Now α + 2α = 5/2 or a= 5/6 Also
k= -17/9
Q.13: Using division show that 3y2 + 5 is a factor of 6y5 + 15y4 + 16y3 + 4y2 + 10y − 35 . Ans:
Q14: If (x – 2) and [x – 1/2 ] are the factors of the polynomials qx2 + 5x + r prove that q = r. Ans: 4q + 10 + r = 0 -(1) q/4 +5/2 + r = 0 or q + 10 + 4r = 0 -(2) Subtracting 1 from 2 3q-3r = 0 q = r
Q15: Find k so that the polynomial x2 + 2x + k is a factor of polynomial 2x4 + x3 – 14x2 + 5x + 6. Also, find all the zeroes of the two polynomials. Ans: For x2 + 2x + k is a factor of the polynomial 2x4 + x3 – 14x2 + 5x + 6, it should be able to divide the polynomial without any remainder
Comparing the coefficient of x, we get. 21 + 7k = 0 k = -3 So x2 + 2x + k becomes x2 + 2x -3 = (x-1)(x+3) Now 2x4 + x3 – 14x2 + 5x + 6= (x2 + 2x -3)(2x2-3x-8+2k) =(x2 + 2x – 3)(2x2-3x – 2) =(x – 1)(x + 3)(x – 2)(2x + 1) or x = 1, -3, 2,= -1/2
Q1: State whether the given statement is true or false : (i) The product of two rationals is always rational Ans: True
(ii) The product of two irrationals is an irrational Ans: False
(iii) The product of a rational and an irrational is irrational Ans: True
(iv) The sum of two rationals is always rational Ans: True
(v) The sum of two irrationals is an irrational. Ans: False
Very Short Questions
Q2: Classify the following numbers as rational or irrational: (i) 3.1416 Ans: Rational
(ii) 3.142857 Ans: Rational
(iii) 2.040040004…… Ans: Irrational
(iv) 3.121221222… Ans: Irrational
(v) 3√3 Ans: Irrational
Q.3. Find the prime factorization of 1152 Ans: 1152 = 27 × 32
Short Questions
Q4: Show that the product of two numbers 60 and 84 is equal to the product of their HCF and LCM Ans: Prime factorisation of 60 = 2 × 2 × 3 × 5 Prime factorisation of 84 = 2 × 2 × 3 × 7 Hence, LCM of 60 , 84 = 2 × 2 × 3 × 5 × 7 = 420 And HCF of 60 , 84 = 2 × 2 × 3 = 12 Now, LCM × HCF = 420 × 12 = 5040 Also,60 × 84 = 5040 i.e., HCF × LCM = Product of the two numbers
Q5: If p and q are two prime number, then what is their LCM? Ans: The Least Common Multiple (LCM) of two prime numbers p and q is their product, p×q. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves (e.g., 2, 3, 5, 7). Since two distinct prime numbers share no common factors other than 1, their LCM is calculated by multiplying them. For example, if p=2 and q=3, the LCM is 2×3=6. Thus, the HCF of p and q is p × q
Q6: Given that HCF (306, 657) = 9, find LCM (306, 657).. Ans: HCF of two numbers 306 and 657 is 9 To find: LCM of number We know that, LCM × HCF = first number x second number LCM × 9 = 306 × 657 LCM=306 × 657/9 = 22338
Q7: The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers. Ans: Since, the product of two numbers = Their H.C.F. × Their L.C.M. ⇒ One no. × other no. = H.C.F. × L.C.M. ⇒ Other no. =12×240 / 48 = 60.
Long Questions
Q8: In a school, the duration of a period in junior section is 40 minutes and in senior section is 1 hour: If the first bell for each section ring at 9:00 a.m., when will the two bells ring together again? Ans: 1 hour = 60 minutes 40 = 2 × 2 × 2 × 5 = 23 × 5 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5 ∴ LCM (40, 60) = 23 × 3 × 5 = 120 120 minutes = 2 hours Hence, the two bells will ring together again at 9:00 + 2:00 = 11:00 a.m.
Q9: The HCF of 408 and 1032 is expressible in the form 1032 m – 2040. Find the value of m. Also, find the LCM of 408 and 1032. Ans: Let us find HCF of 408 and 1032. Here, 1032 > 408 ∴ 1032 = 2 × 408 + 216 408 = 1 × 216 + 192 216 = 1 × 192 + 24 192 = 8 × 24 + 0 Thus, HCF of 408 and 1032 is 24. Now, HCF (408, 1032) i.e., 24 = 1032 × m – 2040 ⇒ 1032 × m = 24 + 2040 ⇒ 1032 × m = 2064 ⇒ m = 20641032 = 2 408 = 23 × 3 × 17 1032 = 23 × 3 × 43 , ∴ LCM of 408 and 1032 = 23 × 3 × 17 × 43 = 17544.
Q10: Prove that (i) √2 is irrational number (ii) √3 is irrational number Similarly √5, √7, √11…… are irrational numbers. Ans:(i) Let us assume, to the contrary, that 2 is rational. So, we can find integers r and s (≠ 0) such that .√2= r/s Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get ,√2=a/b where a and b are coprime. So, b √2= a. Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. Now, by Theorem it following that 2 divides a. So, we can write a = 2c for some integer c. Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2. This means that 2 divides b2, and so 2 divides b (again using Theorem with p = 2). Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factors other than 1. This contradiction has arisen because of our incorrect assumption that √2 is rational. So, we conclude that √2 is irrational.
(ii) Let us assume, to contrary, that √3 is rational. That is, we can find integers a and b (≠ 0) such that √3=a/b Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b √3= a . Squaring on both sides, and rearranging, we get 3b2 = a2. Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3. So, we can write a = 3c for some integer c. Substituting for a, we get 3b2 = 9c2, that is,b2 = 3c2. This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3). Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √3 is rational. So, we conclude that √3 is irrational.
Multiple Choice Questions Q1: The probability of an impossible event is (a) 0.01 (b) 100 (c) zero (d) 1
Q2: The probability that a leap year will have 53 Sundays or 53 Mondays is (a) 4/7 (b) 2/7 (c) 1/7 (c) 3/7
Q3: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and these values are equally likely outcomes. The probability that it will point at a number greater than 5 is (a) 1/2 (b) 1/4 (c) 1/5 (d) 1/3
Q4: The probability of an impossible event is (a) 0.01 (b) 100 (c) zero (d) 1
Q5: Cards marked with numbers 1, 2, 3, ______, 25 are placed in a box and mixed thoroughly and one card is drawn at random from the box. The probability that the number on the card is a multiple of 3 and 5 is (a) 12/25 (b) 4/25 (c) 1/25 (d) 8/25
Q6: A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and these values are equally likely outcomes. The probability that it will point at a number greater than 5 is (a) 1/2 (b) 1/4 (c) 1/5 (d) 1/3
Very Short Answer Questions Q7: A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears a two digit number
Q8: A card is drawn from a pack of 52 cards. Find the probability of getting a king of red colour
Q9: A bag contains 40 balls out of which some are red, some are blue and remaining are black. If the probability of drawing a red ball is 11/20 and that of blue ball is 1/5 , then what is the no. of black balls?
Short Answer Questions Q10: A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the drawn card is neither a king nor a queen.
Q11: Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.
Q12: Three unbiased coins are tossed together. Find the probability of getting at least two heads?
Q13: Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the taken out card is (1) a prime number less than 10 (2) a number which is a perfect square.
Q14: Why is tossing a coin considered as the way of deciding which team should get the ball at the beginning of a football match?
Q15: Two coins are tossed together. Find the probability of getting both heads or both tails.
Q16: A bag contains, white, black and red balls only. A ball is drawn at random from the bag. If the probability of getting a white ball is 3/10 and that of a black ball is , 2/5 then find the probability of getting a red ball. If the bag contains 20 black balls, then find the total number of balls in the bag.
Q17: A box contains 90 discs which are numbered 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (1) a two digit number, (2) number divisible by 5.
Long Questions Answer Q18: A box contains cards bearing numbers 6 to 70. If one cards is drawn at random from the box, find the probability that it bears (1) a one digit number. (2) a number divisible by 5, (3) an odd number less than 30,
Q19: All red face cards are removed from a pack of playing cards. The remaining cards are well-shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (1) a red card, (2) a face card, (3) a card of clubs.
Q20: Cards numbered 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is (i) an odd number, (ii) a perfect square number, (iii) divisible by 5, (iv) a prime number less than 20.
You can access the solutions to this worksheet here.
1. If x1, x2, x3,….., xn are the observations of a given data. Then the mean of the observations will be: (a) Sum of observations/Total number of observations (b) Total number of observations/Sum of observations (c) Sum of observations +Total number of observations (d) None of the above
Q2: If the mean of frequency distribution is 7.5 and ∑fi xi = 120 + 3k, ∑fi = 30, then k is equal to: (a) 40 (b) 35 (c) 50 (d) 45
Q3: The mode and mean is given by 7 and 8, respectively. Then the median is: (a) 1/13 (b) 13/3 (c) 23/3 (d) 33
Q4: The mean of the data: 4, 10, 5, 9, 12 is; (a) 8 (b) 10 (c) 9 (d) 15
Q5: The median of the data 13, 15, 16, 17, 19, 20 is: (a) 30/2 (b) 31/2 (c) 33/2 (d) 35/2 Q6: If the mean of first n natural numbers is 3n/5, then the value of n is: (a) 3 (b) 4 (c) 5 (d) 6
Q7: If AM of a, a+3, a+6, a+9 and a+12 is 10, then a is equal to; (a) 1 (b) 2 (c) 3 (d) 4
Q8: The class interval of a given observation is 10 to 15, then the class mark for this interval will be: (a) 11.5 (b) 12.5 (c) 12 (d) 14
Q9: If the sum of frequencies is 24, then the value of x in the observation: x, 5,6,1,2, will be; (a) 4 (b) 6 (c) 8 (d) 10
Q10: The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its (a) mean (b) median (c) mode (d) all the three above
Solve the following Questions
Q1: The class marks of a distribution are 13, 17,21, 25 and 29. Find the true class limits.
Q2: Convert the given simple frequency series into a: (i) Less than cumulative frequency series. (ii) More than cumulative frequency series. Q3: Drawn ogive for the following frequency distribution by less than method. Q4: The temperature of a patient, admitted in a hospital with typhoid fever, taken at different times of the day are given below. Draw the temperature-time graph to reprents the data: Q5: The class marks of a distribution are 82, 88, 94, 100, 106, 112 and 118. Determine the class size and the classes.
Q6: Convert the following more than cumulative frequency series into simple frequency series. Q7: Draw a cumulative frequency curve for the following frequency distribution by less than method.
You can access the solutions to this worksheet here.
