{"id":23295,"date":"2020-02-21T13:01:50","date_gmt":"2020-02-21T07:31:50","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=23295"},"modified":"2021-07-02T15:24:43","modified_gmt":"2021-07-02T09:54:43","slug":"karnataka-sslc-maths-model-question-paper-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-6\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern)"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern), Karnataka SSLC Maths Model Question Papers with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/span><\/p>\n

Karnataka State Syllabus SSLC Maths Model Question Paper 1 (Old Pattern)<\/h2>\n

Time: 3 Hours
\nMax Marks: 80<\/p>\n

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 \u00d7 8 = 8 )<\/span><\/p>\n

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.<\/strong><\/p>\n

(8 \u00d7 1 = 8)<\/strong>
\nQuestion 1.
\nThe distance between two points p(x1<\/sub>, y1<\/sub>) and q (x2<\/sub>, y2<\/sub>) is given by
\n\"Karnataka<\/p>\n

Question 2.
\nThe degree of polynomial p(x) = x2<\/sup> – 3x + 4x3<\/sup> – 6 is
\n(A) 2
\n(B) 1
\n(C) 3
\n(D) 6<\/p>\n

Question 3.
\nWhich one of the following cannot be the probability of an event?
\n(A) \\(\\frac { 2 }{ 3 }\\)
\n(B) -1.5
\n(C) 15%
\n(D) 0.7<\/p>\n

\"Karnataka<\/p>\n

Question 4.
\nThe curved surface area of frustum of a cone is given by
\n(A) \u03a0 (r1<\/sub> + r2<\/sub>) l
\n(B) \u03a0 (r1<\/sub> + r2<\/sub>) h
\n(C) \u03a0 (r1<\/sub> – r2<\/sub>) l
\n(D) \u03a0 (r1<\/sub> – r2<\/sub>) h<\/p>\n

Question 5.
\nThe solutions for the equations x + y = 10 and x – y = 2 are
\n(A) x = 6, y = 4
\n(B) x = 4, y = 6
\n(C) x = 7, y = 3
\n(D) x = 8, y = 2<\/p>\n

Question 6.
\nIn the adjoining figure, TP and TQ are the tangents to the circle with centre O. The measure of \u2220PTQ is
\n\"Karnataka
\n(A) 90\u00b0
\n(C) 70\u00b0
\n(B) 110\u00b0
\n(D) 40\u00b0<\/p>\n

Question 7.
\nThe coordinates of origin are
\n(A) (1, 1)
\n(B) (2, 2)
\n(C) (0, 0)
\n(D) (3, 3)<\/p>\n

Question 8.
\nIf the discriminant of quadratic equation b2<\/sup> – 4ac = 0 then the roots are
\n(A) Real and distinct
\n(B) Roots are equal
\n(C) No Real Roots
\n(D) Roots are unequal and irrational<\/p>\n

(6 \u00d7 1 = 6)<\/strong>
\nQuestion 9.
\nState \u201cBasic proportionality theorem\u201d.<\/p>\n

Question 10.
\nIdentify the tangent to the circle in the adjoining figure and write its name.
\n\"Karnataka<\/p>\n

Question 11.
\nState Euclid\u2019s division lemma.<\/p>\n

Question 12.
\nFind the number of zeroes of a polynomial p(x) from the graph given
\n\"Karnataka<\/p>\n

Question 13.
\nFind the distance of the point p(3, 4) from the origin.<\/p>\n

\"Karnataka<\/p>\n

Question 14.
\nExpress 140 as a product of prime factors.<\/p>\n

(16 \u00d7 2 = 32)<\/strong>
\nQuestion 15.
\nHow many two – digit numbers are divisible by 3?<\/p>\n

Question 16.
\n\u2206ABC ~ \u2206DEF , Area of \u2206ABC = 64cm2<\/sup> and area of \u2206DEF = 121cm2<\/sup>. If EF = 15.4 cm, Find BC.<\/p>\n

Question 17.
\nSolve for x and y : 2x + y = 6 and 2x – y = 2.<\/p>\n

Question 18.
\nFive years ago, Gouri was thrice as old as Ganesh. Ten years later Gouri will be twice as old as Ganesh. How old are Gouri and Ganesh?<\/p>\n

Question 19.
\nFind the area of the shaded region in the figure, where ABCD is a square of side 14 cm.
\n\"Karnataka<\/p>\n

Question 20.
\nConstruct a pair of tangents to a circle of radius 5 cm Which are inclined to each other at an angle of 60\u00b0.<\/p>\n

Question 21.
\nFind the value of k, if the points A (2, 3), B(4, k) and C (6, -3) are collinear.<\/p>\n

Question 22.
\nProve 3 + \u221a5 is irrational.<\/p>\n

Question 23.
\nFind the zeroes of polynomial p(x) = 6x2<\/sup> – 3 – 7x.<\/p>\n

Question 24.
\nFind the quadratic polynomial whose sum and product of zeroes are \\(\\frac { 1 }{ 4 }\\) and -1 respectively.<\/p>\n

Question 25.
\nSolve the equation 3x2<\/sup> – 5x + 2 = 0 by using the formula.<\/p>\n

Question 26.
\nEvaluate : 2 tan2<\/sup> 45\u00b0 + cos2<\/sup> 30\u00b0 = sin2<\/sup> 60\u00b0<\/p>\n

Question 27.
\nThe angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30\u00b0, Find the height of the tower.<\/p>\n

Question 28.
\nAs observed from the top of a 100 m high lighthouse from the sea level the angle of depression of two ships are 30\u00b0 and 45\u00b0. If one of the ships is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (\u221a3 = 1.73)<\/p>\n

Question 29.
\nA die is thrown once. Find the probability of getting a number between 2 and 6.<\/p>\n

Question 30.
\nThe volume of a cube is 64 cm2. Find the total surface area of the cube.<\/p>\n

(6 \u00d7 3 = 18)<\/strong>
\nQuestion 31.
\nProve that \u201cThe tangent at any point of a circle is perpendicular to the radius through the point of contact\u201d.
\nOR
\nProve that \u201cThe lengths of tangents drawn from an external point to a circle are equal\u201d.<\/p>\n

Question 32.
\nConstruct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it, whose sides are \\(\\frac { 2 }{ 3 }\\) of the corresponding sides of the first triangle.<\/p>\n

Question 33.
\nA two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
\nOR
\nThe numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \\(\\frac { 29 }{ 20 }\\) Find the original fraction.<\/p>\n

Question 34.
\nIf 4tan\u03b8 = 3, Evaluate \\(\\frac { 4sin\\theta -cos\\theta +1 }{ 4sin\\theta +cos\\theta -1 }\\)
\nOR
\nIf tan 2A = cot (A – 18\u00b0) where 2A is an acute angle. Find the value of A.<\/p>\n

Question 35.
\nCalculate the median for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Class interval<\/strong><\/td>\nFrequency (F)<\/strong><\/td>\n<\/tr>\n
0-20<\/td>\n6<\/td>\n<\/tr>\n
20-40<\/td>\n8<\/td>\n<\/tr>\n
40-60<\/td>\n10<\/td>\n<\/tr>\n
60-80<\/td>\n12<\/td>\n<\/tr>\n
80-100<\/td>\n6<\/td>\n<\/tr>\n
100-120<\/td>\n5<\/td>\n<\/tr>\n
120-140<\/td>\n3<\/td>\n<\/tr>\n
<\/td>\nn = 50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

OR
\nCalculate the mode for the following frequency distribution table<\/p>\n\n\n\n\n\n\n\n\n\n\n
Class Interval<\/strong><\/td>\nFrequency (F)<\/strong><\/td>\n<\/tr>\n
5-15<\/td>\n6<\/td>\n<\/tr>\n
15-25<\/td>\n11<\/td>\n<\/tr>\n
25-35<\/td>\n21<\/td>\n<\/tr>\n
35-45<\/td>\n23<\/td>\n<\/tr>\n
45-55<\/td>\n14<\/td>\n<\/tr>\n
55-65<\/td>\n5<\/td>\n<\/tr>\n
<\/td>\nn = 80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 36.
\nConstruct \u2018ogive\u2019 for the following construction.<\/p>\n\n\n\n\n
C31<\/td>\n0-3<\/td>\n3-6<\/td>\n6-9<\/td>\n9-12<\/td>\n12-15<\/td>\n<\/tr>\n
F<\/td>\n9<\/td>\n3<\/td>\n5<\/td>\n3<\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

(4 \u00d7 4 = 16)<\/strong>
\nQuestion 37.
\nThe sum of four consecutive terms which are in an arithmetic progression is 32 and the ratio of the product of the first and the last term in the product of two middle terms is 7.15. Find the number.
\nOR
\nIn an arithmetic progression of 50 terms, the sum of the first ten terms is 210 and the sum of the last fifteen terms is 2565. Find the arithmetic progression.<\/p>\n

Question 38.
\nProve that \u201cIn a right-angled triangle the square on the hypotenuse is equal to the sum of the square on the other two-siders\u201d.<\/p>\n

Question 39.
\nSolve the equations graphically.
\n2x – y = 2; 4x – y = 4<\/p>\n

Question 40.
\nA wooden article was made by scooping out a hemisphere from one end of a cylinder and a cone from other ends as shown in the figure. If the height of the cylinder is 40 cm, a radius is 7 cm and height of the cone is 24 cm, find the volume of the wooden article.
\n\"Karnataka<\/p>\n

Solutions<\/strong><\/p>\n

Solution 1.
\n(B) or (D)
\n\"Karnataka<\/p>\n

Solution 2.
\n(C) 3<\/p>\n

Solution 3.
\n(B) -1, 5<\/p>\n

Solution 4.
\n(A) \u03a0 (r1<\/sub> + r2<\/sub>) l<\/p>\n

Solution 5.
\n(A) x = 6, y = 4<\/p>\n

Solution 6.
\n(C) 70\u00b0<\/p>\n

Solution 7.
\n(C) (0, 0)<\/p>\n

Solution 8.
\n(B) Roots are equal<\/p>\n

Solution 9.
\nIf a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.<\/p>\n

Solution 10.
\nRS<\/p>\n

\"Karnataka<\/p>\n

Solution 11.
\nLet a and b be any two positive integers.
\nThen there exist two unique whole nos. q and r such that a = bq + r, (0 \u2264 r < b)<\/p>\n

Solution 12.
\n4<\/p>\n

Solution 13.
\n\"Karnataka<\/p>\n

Solution 14.
\n\"Karnataka<\/p>\n

Solution 15.
\nNos. divisible by 3 = 12,15,18, 21, …….., 99
\na = 12, d = 3, T = 99, n = ?
\nTn<\/sub> = a + (n – 1)d
\n\u21d2 99 = 12 + (n – 1) 3
\n\u21d2 99 – 12 = 3n – 3
\n\u21d2 87 + 3 = 3n
\n\u21d2 90 = 3n
\n\u21d2 n = 30
\n30 two digits are divisible by 3.<\/p>\n

Solution 16.
\n\"Karnataka
\n\"Karnataka<\/p>\n

Solution 17.
\n\"Karnataka
\nSubstitute the value of x in Eqn 1
\n2x + y = 6
\n\u21d2 2(2) + y = 6
\n\u21d2 4 + y = 6
\n\u21d2 y = 6 – 4 = 2
\n\u2234 x = 2, y = 2<\/p>\n

Solution 18.
\nLet the age of Gauri be \u2018x\u2019 years.
\nLet the age of Ganesh be \u2018y\u2019 years.
\n5 years ago
\n(x – 5) = 3(y – 5)
\n\u21d2 x – 5 = 3y – 15
\n\u21d2 x – 3y = -15 + 5
\n\u21d2 x – 3y = -10 …..(1)
\n10 years later
\n(x + 10) = 2(y + 10)
\n\u21d2 x + 10 = 2y + 20
\n\u21d2 x – 2y = 20 – 10
\n\u21d2 x – 2y = 10 ……(2)
\n\"Karnataka
\nSubstitute the value of y in equation 1
\nx – 3y = -10
\n\u21d2 x – 3(20) = -10
\n\u21d2 x – 60 = -10
\n\u21d2 x = -10 + 60
\n\u21d2 x = 50
\nGouri\u2019s Age = 50 yrs, Ganesha\u2019s Age = 20 yrs.<\/p>\n

\"Karnataka<\/p>\n

Solution 19.
\n\"Karnataka<\/p>\n

Solution 20.
\n\"Karnataka<\/p>\n

Solution 21.
\n\"Karnataka<\/p>\n

Solution 22.
\nLet us assume that (3 + \u221a5) is rational
\n\"Karnataka
\nBut \u221a5 is irrational
\nIt contradicts the fact that \u221a5 is irrational
\nour assumption is wrong.
\n3 + \u221a5 is irrational.<\/p>\n

Solution 23.
\np(x) = 6x2<\/sup> – 7x – 3
\nBy splitting the middle term we get
\n6x2<\/sup> – 9x + 2x – 3
\n3x(2x – 3) + 1 (2x – 3)
\n(2x – 3)(3x + 1)
\nfor zero\u2019s of p(x) put 2x – 3 = 0 & 3x + 1 = 0
\n1. 2x – 3=0
\n2x = 3
\nx = \\(\\frac { 3 }{ 2 }\\)
\n2. 3x + 1 = 0
\n3x = -1
\nx = \\(\\frac { -1 }{ 3 }\\)
\nZeros of p(x) are \\(\\frac { 3 }{ 2 }\\) and \\(\\frac { -1 }{ 3 }\\)<\/p>\n

Question 24.
\nThe quadratic polynomial is given by
\n\"Karnataka<\/p>\n

Question 25.
\n3x2<\/sup> – 5x + 2 = 0
\nCompare it with the standard form
\n\"Karnataka<\/p>\n

Solution 26.
\n\"Karnataka<\/p>\n

Solution 27.
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Solution 28.
\n\"Karnataka
\n\"Karnataka<\/p>\n

Solution 29.
\nNo. of all possible outcome = {1, 2, 3, 4, 5, 6}
\nn(S) = 6
\nLet E be the event of getting a no. between 2 and 6.
\nNo. of favourable outcomes n(E) = {3, 4, 5} = 3
\nP(E) = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 3 }{ 6 }\\)<\/p>\n

Solution 30.
\nVolume of the cube = l3<\/sup>
\n64 = l3<\/sup>
\nl = 4
\nT.S.A. of the cube = 6l2<\/sup> = 6 \u00d7 42<\/sup> = 6 \u00d7 16 = 96 Sq.cm.<\/p>\n

Solution 31.
\nTheorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
\n\"Karnataka
\nProof: We are given a circle with centre O a tangent XY to the circle at a point R We need to prove that OP is perpendicular to XY.
\nTake a point Q on XY other the P and join OQ see in below fig.
\nThe point Q must lie outside the circle. (Why? Note that if Q lies inside the circle XY will become a secant and not a tangent to the circle). Therefore OQ is longer than the radius OP of the circle. That is,
\nOQ > OR
\nSince this happens for every point on the line XY except the point R OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. (as shown in theorem).
\nOR
\nTheorem: The lengths of tangents drawn from an external point to a circle are equal.
\n\"Karnataka
\nProof: We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see in fig.). We are required to prove that PQ = PR,
\nFor this, we join OR OQ and OR. Then \u2220OQP and \u2220ORP are right angles because these are angles between the radii and tangents, and according to the theorem, they are right angles.
\nNow in right triangles OQP and ORP
\nOQ = OR (Radii of the same circle)
\nOP = OP (Common)
\nTherefore, \u0394OQP = \u0394ORP (RHS)
\nThis gives PQ = PR<\/p>\n

Solution 32.
\n\"Karnataka<\/p>\n

Steps of construction:<\/strong><\/p>\n

    \n
  1. Draw BC = 6 cm.<\/li>\n
  2. With B as the centre & radius 4 cms. draw an arc.<\/li>\n
  3. With C as centre and radius, 5 cm cut the first arc at A<\/li>\n
  4. Join AB and AC to get \u0394ABC<\/li>\n
  5. Draw any ray BX below BC making an acute angle with BC.<\/li>\n
  6. Locate 3 points B1<\/sub>, B2<\/sub>, B3<\/sub> as BX such that BB1<\/sub> = B1<\/sub>B2<\/sub> = B2<\/sub>B3<\/sub><\/li>\n
  7. Join B3<\/sub>C<\/li>\n
  8. Draw a line through B2<\/sub> parallel to B3<\/sub>C intersecting BC at C’<\/li>\n
  9. Draw a line through C’ parallel to AC to intersect BA at A’<\/li>\n
  10. Now A’BC’ is the required \u0394 each of whose sides is \\(\\frac { 2 }{ 3 }\\) of the corresponding sides of \u0394ABC<\/li>\n<\/ol>\n

    Solution 33.
    \nSet the no. be xy.
    \n10x + y = 4(x + y)
    \n10x + y = 4x + 4y
    \n10x – 4x = 4y – y
    \n2x = y
    \n10x + y = 3xy
    \n10x + 2x = 3x(2x)
    \n12x = 6×2
    \n12 = 6x
    \nx = 2
    \ny = 2 \u00d7 2 = 4
    \nNo. = 24
    \nOR
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 34.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 35.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 36.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 37.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 38.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 39.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 40.
    \n\"Karnataka<\/p>\n

    Karnataka State Syllabus SSLC Maths Model Question Paper 2 (Old Pattern)<\/h2>\n

    Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.<\/strong><\/p>\n

    (8 \u00d7 1 = 8)<\/strong>
    \nQuestion 1.
    \nThe co-ordinates of the origin are
    \n(a) (0, 0)
    \nb) (1, 0)
    \n(c) (0, 1)
    \n(d) (1, 1)<\/p>\n

    Question 2.
    \nIf p(x) = x3<\/sup> – 1 then p(-1) = ……..
    \n(a) 2
    \n(b) -2
    \n(c) 1
    \n(d) 0<\/p>\n

    Question 3.
    \nThe total number of all possible events when a dice is thrown once is.
    \n(a) 4
    \n(b) 5
    \n(c) 6
    \n(d) 8<\/p>\n

    Question 4.
    \nThe area of a square is given by
    \n(a) l \u00d7 b
    \n(b) l \u00d7 b \u00d7 h
    \n(c) \u03c0r2<\/sup>
    \n(d) l2<\/sup><\/p>\n

    Question 5.
    \nIf y = 2x – 1 and x + y = 5, then the value of x is
    \n(a) 2
    \n(b) -2
    \n(c) 1
    \n(d) -1<\/p>\n

    Question 6.
    \nIn the adjoining fig AB and AC are tangents from A. If \u2220BAC = 80\u00b0 then BQC
    \n\"Karnataka
    \n(a) 110\u00b0
    \n(b) 100\u00b0
    \n(c) 80\u00b0
    \n(d) 60\u00b0<\/p>\n

    Question 7.
    \nThe graph of an equation cuts x-axis at two points. Then the no. of solutions of that equation is
    \n(a) 0
    \n(b) 1
    \n(c) 2
    \n(d) 3<\/p>\n

    Question 8.
    \nIf the discriminant of a quadratic equation b2<\/sup> – 4ac > 0 then the roots are
    \n(a) equal
    \n(b) imaginary
    \n(c) unequal & irrational
    \n(d) Real & District<\/p>\n

    (1 \u00d7 6 = 6)<\/strong>
    \nQuestion 9.
    \nState the converse of \u201cBasic Proportionality theorem\u201d<\/p>\n

    Question 10.
    \nHow are the radius and diameter of a circle related?<\/p>\n

    Question 11.
    \nFind the HCF of 24 and 36 by factorisation.<\/p>\n

    Question 12.
    \nIf p(x) = x2<\/sup> – x – 2 find p(o)<\/p>\n

    Question 13.
    \nFind the distance between the point p (4, 5) and Q (1, 2)<\/p>\n

    Question 14.
    \nIf a no. is divided by 5. which are the remainders?<\/p>\n

    (2 \u00d7 16 = 32)<\/strong>
    \nQuestion 15.
    \nHow many two digits no. are divisible by 4?<\/p>\n

    Question 16.
    \n\u2206ABC ~ \u2206DEF, Area of \u2206DEF = 25 Sqcm. and Area of \u2206ABC = 100Sqcm., If DE = 16 cms find AB.<\/p>\n

    Question 17.
    \nSolve: 3x + 2y = 8, x + 3y = 5<\/p>\n

    Question 18.
    \nFive years ago A was 7 times as old as B. 20 yrs. later A will be twice as old as B. What are their present ages?<\/p>\n

    Question 19.
    \nFind the are of the shaded region in the figure given.
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Question 20.
    \nConstruct a tangent to a circle of radius 4 cms. at a point on the circumference.<\/p>\n

    Question 21.
    \nIf the coordinates of the vertices of a \u0394 are (1, 2), (3, 4) and (5, 6). find the perimeter of the triangle.<\/p>\n

    Question 22.
    \nProve that 2 + \u221a3 ls irrational.<\/p>\n

    Question 23.
    \nFind the zeros of the polynomial p(x) = 2x2<\/sup> – 3x – 5.<\/p>\n

    Question 24.
    \nFind the quadratic polynomial whose sum & product of zeros are \\(\\frac { 1 }{ 3 }\\) and 2 respectively.<\/p>\n

    Question 25.
    \nSolve by using formula 5x2<\/sup> – 9x + 3 = 0<\/p>\n

    Question 26.
    \nEvaluate : tan2<\/sup> 45\u00b0 + 2 cos 30\u00b0 – sin 60\u00b0<\/p>\n

    Question 27.
    \nA tower stands vertically on the ground. From a point on the ground. 15 mts. away from the foot of the tower, the angle of elevation of the top of the tower is 60\u00b0. Find the height of the tower.<\/p>\n

    Question 28.
    \nAn observer 1.5 mts tall is 28.5 mts. away from a chimney. The angle of elevation by the top of the chimney from her eyes is 45\u00b0, What is the height of the chimney?<\/p>\n

    Question 29.
    \nA dice is thrown once. Find the probability of getting a number lying between 3 & 6.<\/p>\n

    (6 \u00d7 3 = 18)<\/strong>
    \nQuestion 30.
    \nThe T.S.A. of a cube is 96 Sqcms. Find the volume of the cube.<\/p>\n

    Question 31.
    \nProve that the tangents drawn to a circle from an external point are equal.
    \nOR
    \nProve that tangents drawn to a circle from an external point subtend equal angles at the centre.<\/p>\n

    Question 32.
    \nConstruct a \u0394 of sides 6 cms, 9 cms and 12 cm and then another \u0394 similar to it, whose sides are \\(\\frac { 2 }{ 3 }\\) of the corresponding sides of the first triangle.<\/p>\n

    Question 33.
    \nSolve the following: 8x + 5y = 9, 3x + 2y = 4
    \nOR
    \nA fraction becames \\(\\frac { 1 }{ 3 }\\) when 1 is subtracted from the numerator & it becomes \\(\\frac { 1 }{ 4 }\\) when 8 is added to its denominator. Find the fraction.<\/p>\n

    Question 34.
    \nIf 4 tan\u03b8 = 3, Evaluate \\(\\left[ \\frac { 5sin\\theta -cos\\theta +2 }{ 5cos\\theta +sin\\theta -2 } \\right]\\)<\/p>\n

    Question 35.
    \nCalculate the median for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    20-40<\/td>\n4<\/td>\n<\/tr>\n
    40-60<\/td>\n5<\/td>\n<\/tr>\n
    60-80<\/td>\n6<\/td>\n<\/tr>\n
    80-100<\/td>\n7<\/td>\n<\/tr>\n
    100-120<\/td>\n8<\/td>\n<\/tr>\n
    <\/td>\nN=30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    OR
    \nCalculate the mode for the following frequency distribution.<\/p>\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    15-25<\/td>\n1<\/td>\n<\/tr>\n
    25-35<\/td>\n6<\/td>\n<\/tr>\n
    35-45<\/td>\n15<\/td>\n<\/tr>\n
    45-55<\/td>\n18<\/td>\n<\/tr>\n
    55-65<\/td>\n9<\/td>\n<\/tr>\n
    <\/td>\nN=50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Question 36.
    \nConstant OGIVE for the following distribution<\/p>\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    3-6<\/td>\n3<\/td>\n<\/tr>\n
    6-9<\/td>\n6<\/td>\n<\/tr>\n
    9-12<\/td>\n4<\/td>\n<\/tr>\n
    12-15<\/td>\n2<\/td>\n<\/tr>\n
    15-18<\/td>\n5<\/td>\n<\/tr>\n
    <\/td>\nN=20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    (4 \u00d7 4 = 16)<\/strong>
    \nQuestion 37.
    \nHow many two digit numbers are divisible by 3?
    \nOR
    \nFind the 31st term of an AR whose 11th term is 38 and the 16th term is 73.<\/p>\n

    Question 38.
    \nThe ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides Prove this.<\/p>\n

    \"Karnataka<\/p>\n

    Question 39.
    \nSolve graphically : 3x – y = 5, 2x + y = 5<\/p>\n

    Question 40.
    \nA metallic sphere of radius 4.2 cms is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.<\/p>\n

    Solutions<\/strong><\/p>\n

    Solution 1.
    \n(a) (0, 0)<\/p>\n

    Solution 2.
    \n(b) -2<\/p>\n

    Solution 3.
    \n(c) 6<\/p>\n

    Solution 4.
    \n(d) l2<\/sup><\/p>\n

    Solution 5.
    \n(a) 2<\/p>\n

    Solution 6.
    \n(b) 100\u00b0<\/p>\n

    Solution 7.
    \n(c) 2<\/p>\n

    Solution 8.
    \n(d) Real & distinct.<\/p>\n

    Solution 9.
    \nIt states that \u201cIf a line divides the two sides of a triangle in proportion then that line is parallel to the third side<\/p>\n

    Solution 10.
    \ndiameter = 2 \u00d7 radius<\/p>\n

    Solution 11.
    \n\"Karnataka
    \n24 = 2 \u00d7 2 \u00d7 2 \u00d7 3
    \n36 = 2 \u00d7 2 \u00d7 3 \u00d7 3
    \nHCF = 2 \u00d7 2 \u00d7 3 = 12<\/p>\n

    Solution 12.
    \np(x) = x2<\/sup> – x – 2
    \n\u21d2 p(0) = 02<\/sup> – 0 – 2
    \n\u21d2 p(0) = -2<\/p>\n

    Solution 13.
    \nDistance
    \n\"Karnataka<\/p>\n

    Solution 14.
    \nThe Remainders are {0, 1, 2, 3, 4}<\/p>\n

    Solution 15.
    \nThe nos. are 12,16, 20, 24, 28, ……. ,96
    \na = 12, d = 4, Tn<\/sub> = 96, n = ?
    \nTn<\/sub> = a + (n – 1) d
    \n\u21d2 96 = 12 + (n – 1) 4
    \n\u21d2 96 – 12 = 4n – 4
    \n\u21d2 84 + 4 = 4n
    \n\u21d2 4n = 88
    \n\u21d2 n = 22
    \nThere are 22 two digit nos. divisible by 4.<\/p>\n

    Solution 16.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 17.
    \n\"Karnataka<\/p>\n

    Solution 18.
    \nLet the present ages of A = x years, B = y years.
    \n\"Karnataka
    \n\"Karnataka
    \nAge of A = 40 years
    \nAge of B = 10 years<\/p>\n

    Solution 19.
    \n\"Karnataka<\/p>\n

    Solution 20.
    \n\"Karnataka<\/p>\n

    Solution 21.
    \n\"Karnataka<\/p>\n

    Solution 22.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 23.
    \n2x2<\/sup> – 3x – 5 = 0
    \n2x2<\/sup> – 5x + 2x – 5 = 0
    \nx[2x – 5] + 1[2x – 5] = 0
    \n(2x – 5)(x + 1) = 0
    \nif 2x – 5 = 0 or if x + 1 = o
    \n2x = 5 or x = 1
    \nx = \\(\\frac { 5 }{ 2 }\\)
    \nThe zeroes are 1 and \\(\\frac { 5 }{ 2 }\\)<\/p>\n

    Solution 24.
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 25.
    \nIt is in the form ax2<\/sup> + bx + c = 0
    \na = 5, b = -a, c = 3
    \n\"Karnataka<\/p>\n

    Solution 26.
    \n\"Karnataka<\/p>\n

    Solution 27.
    \n\"Karnataka
    \nAB = Tower
    \nCB = distance of the point C from the foot of the tower.
    \nACB is a Right Angle Triangle
    \ntan60\u00b0 = \\(\\frac { AB }{ CB }\\)
    \n\u221a3 = \\(\\frac { AB }{ 15 }\\)
    \nAB = 15\u221a3 mtrs
    \nHeight of the tower = 15\u221a3 Mtrs<\/p>\n

    Solution 28.
    \n\"Karnataka
    \nAB = Chimney
    \nCD = Observer
    \nADE = angle of elevation
    \nADE is the Right angled \u2206
    \nAB = AE + EB = AE + 1.5
    \nDE = CB = 28.5
    \ntan45\u00b0 = \\(\\frac { AE }{ DE }\\)
    \n1 = \\(\\frac { AE }{ 28.5 }\\)
    \nAE = 28.5
    \nHeight of the Chimney = AB = AE + EB = 28.5 + 1.5 = 30 mts.<\/p>\n

    Solution 29.
    \nNo. of all Possible outcomes n(s) = 6 {1, 2, 3, 4, 5, 6}
    \nLet E be the vent of getting a no. between 3 and 6.
    \nn(E) = {4, 5} = 2
    \np(E) = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 2 }{ 6 }\\)<\/p>\n

    Solution 30.
    \nT.S.A. of a cube = 6l2<\/sup>
    \n96 = 6l2<\/sup>
    \n\u21d2 l2<\/sup> = 16
    \n\u21d2 l = \u221a16 = 4
    \nVolume of the cube = l3<\/sup> = 43<\/sup> = 64 cc.<\/p>\n

    Solution 31.
    \nData: O is the centre of the circle. AB and AC are tangents from A
    \n\"Karnataka
    \nTo prove :
    \nAB = AC
    \nConstruction:
    \nJoin AO, OB and OC.
    \nProof:
    \nIn \u0394les AOB and AOC
    \nOA is common
    \nOBA = OCA = 90\u00b0 (angle between radius and tangent)
    \nOB = OC (radii)
    \nBy RHS Postulate
    \n\u0394AOB = \u0394AOC
    \nAB = AC
    \nOR
    \nData: O is the centre of the circle AB and AC are tangents from A. Join OA, OB & OC.
    \n\"Karnataka
    \nTo Prove : BQA = QOA
    \nProof:
    \nIn \u0394les AOB and AOC
    \nOA is common
    \n\u2220OBA = \u2220OCA = 90\u00b0 (angle between radius & tangent)
    \nOB = OC (radii)
    \nBy RHS postulate
    \n\u0394AOB = \u0394AOC
    \nBOA = COA<\/p>\n

    Solution 32.
    \nAXY is the required \u0394
    \n\"Karnataka<\/p>\n

    Solution 33.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 34.
    \n\"Karnataka<\/p>\n

    Solution 35.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 36.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 37.
    \nList of two-digit nos.
    \ndivisible by 3 = 12, 15, 18, ……. , 99.
    \nThis is an AP
    \na = 12, d = 3, Tn<\/sub> = 99, n = ?
    \nTn<\/sub> = a + (n – 1 ) d
    \n\u21d2 99 = 12 + (n – 1)3
    \n\u21d2 99 – 12 = 3n – 3
    \n\u21d2 87 + 3 = 3n
    \n\u21d2 90 = 3n
    \n\u21d2 n = 30
    \n30 two digit nos. are divisible by 3.
    \nOR<\/strong>
    \n\"Karnataka
    \nSubstitute the value of d in equation 1
    \na + 10d = 38
    \n\u21d2 a + 10(7) = 38
    \n\u21d2 a + 70 = 38
    \n\u21d2 a = 38 – 70
    \n\u21d2 a = -32
    \n\u21d2 T31<\/sub> = a + 30d
    \n\u21d2 T31<\/sub> = -32 + 30(7)
    \n\u21d2 T31<\/sub> = -32 + 210
    \n\u21d2 T31<\/sub> = 178<\/p>\n

    \"Karnataka<\/p>\n

    Solution 38.
    \ndata: \u0394ABC || \u0394PQR
    \nTo Prove that:
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 39.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka
    \nQ = (2, 1) intersecting of two lines.
    \nThe two lines intersecting at the point (2, 1), so x = 2, y = 1 is the required solution of the pair of linear equations.<\/p>\n

    Solution 40.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Karnataka State Syllabus SSLC Maths Model Question Paper 3 (Old Pattern)<\/h2>\n

    Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.<\/strong><\/p>\n

    (8 x 1 = 8)<\/strong>
    \nQuestion 1.
    \nThe coordinates of the mid-point P of the join of the points A(x1<\/sub>, y1<\/sub>) and B(x2<\/sub>, y2<\/sub>) is
    \n\"Karnataka<\/p>\n

    Question 2.
    \nThe quadratic polynomial among the following is
    \n(a) p(x) = x2<\/sup> + 4x + 3
    \n(b) q(x) = ax + b
    \n(c) g(x) = 4x – 3
    \n(d) f(x) = x3<\/sup> + 4x2<\/sup> – 5x + 2<\/p>\n

    Question 3.
    \nWhen a dice is thrown twice n(s) = ………
    \n(a) 6
    \n(b) 12
    \n(c) 24
    \n(d) 36<\/p>\n

    Question 4.
    \nThe volume of a sphere is given by
    \n(a) \\(\\frac { 2 }{ 3 }\\) \u03c0r3<\/sup>
    \n(b) \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/sup>
    \n(c) 4\u03c0r3<\/sup>
    \n(d) 2\u03c0r3<\/sup><\/p>\n

    Question 5.
    \nThe values of x and y which satisfy the given equations are 2x – 3y = 2, 3x – 2y = 8
    \n(a) x = 2, y = 2
    \n(b) x = 3, y = 1
    \n(c) y = 2 , y = 2
    \n(d) x = 2, y = 4<\/p>\n

    Question 6.
    \nIn the given fig AB and AC are tangents from A and BAC = 80\u00b0 value of BOA = …….
    \n\"Karnataka
    \n(a) 50\u00b0
    \n(b) 60\u00b0
    \n(c) 70\u00b0
    \n(d) 90\u00b0<\/p>\n

    Question 7.
    \nIf A(3, 4) and B(5, 6) are two points, then the coordinates of the mid-point of AB are
    \n(a) (4, 5)
    \n(b) (3, 4)
    \n(c) (5, 6)
    \n(d) (6, 7)<\/p>\n

    Question 8.
    \nThe discriminant of the equation ax2<\/sup> + bx + c = 0
    \n(a) c2<\/sup> – 4ab
    \n(b) a2<\/sup> – 4ab
    \n(c) b2<\/sup> – 4ac
    \n(d) a2<\/sup> – 2ac<\/p>\n

    \"Karnataka<\/p>\n

    (1 \u00d7 6 = 6)<\/strong>
    \nQuestion 9.
    \nState Pythagoras Theorem<\/p>\n

    Question 10.
    \nWhat is the length of the longest chord in a circle of radius 5 cms?<\/p>\n

    Question 11.
    \nFind the HCF of 135 and 225.<\/p>\n

    Question 12.
    \nFrom the graph, find the number of zeroes of the polynomial p(x).
    \n\"Karnataka<\/p>\n

    Question 13.
    \nFind the distance between A and B if the coordinates of A and B are (2, 3) and (5, 6) respectively.<\/p>\n

    Question 14.
    \nExpress 226 as a product of Prime facotrs.<\/p>\n

    (2 x 16 = 32)<\/strong>
    \nQuestion 15.
    \nHow many two digit nos. are divisible by 5?<\/p>\n

    Question 16.
    \nIf a pair of corresponding sides of two triangles are 5 cms and 7 cms then find the ratio of the areas of triangles<\/p>\n

    Question 17.
    \nSolve:
    \n5x + 4y = 14 \u2192 (1)
    \n4x + 2y = 10 \u2192 (2)<\/p>\n

    Question 18.
    \n5 Years ago A was thrice as old as his son 20 years, later A will be twice as old as his son. Find their present ages.<\/p>\n

    Question 19.
    \nFind the area of the sector OAPB
    \n\"Karnataka<\/p>\n

    Question 20.
    \nDraw a circle of radius 4 cms. construct a tangent to the circle from a point 10 cms away from the centre.<\/p>\n

    Question 21.
    \nFind the value of K. If the points A (4, 6), B(8, 2k) and C(12, -6) are collinear.<\/p>\n

    Question 22.
    \nProve that 5 + \u221a7 is irrational<\/p>\n

    Question 23.
    \nFind the zeros of the polynomial p(x) = 6x2<\/sup> + 7x – 3<\/p>\n

    Question 24.
    \nFind the quadratic polynomial, whose sum & Product of the zeros are \\(\\frac { 1 }{ 2 }\\) & 2 respectively.<\/p>\n

    Question 25.
    \nSolve by using formula 2x2<\/sup> – 5x + 3 = 0<\/p>\n

    Question 26.
    \nEvaluate : 2 sin2<\/sup> 60\u00b0 + 3 tan2<\/sup> 45\u00b0 – cos2<\/sup> 30\u00b0<\/p>\n

    Question 27.
    \nThe angle of elevation of the top of a tower from a point on the ground, which is 60 M away from the foot of the tower is 30\u00b0. Find the height of the tower.<\/p>\n

    Question 28.
    \nAs observed from the top of a 200M height tower from the sea level the angle of depression of two boats are 30\u00b0 and 45\u00b0, if one of the boats is exactly behind the other on the same side of the tower, find the distance between two boats (\u221a3 = 1.73).<\/p>\n

    Question 29.
    \nA dice is thrown once. Find the probability of getting a perfect square number.<\/p>\n

    Question 30.
    \nA copper rod of diameter 1 cm and length 8 cms. is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.<\/p>\n

    (16 x 2 = 32)<\/strong>
    \nQuestion 31.
    \nProve that in two concentric circles the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.
    \nOR
    \nProve that the tangent at any point of a circle is perpendicular to the radius through the point of contact.<\/p>\n

    Question 32.
    \nConstruct a triangle of side 6 cm., 9 cm and 1 cm. and then a triangle similar to it whose side, are \\(\\frac { 2 }{ 3 }\\) of the corresponding sides of the first triangle.<\/p>\n

    Question 33.
    \nSolve the following equations by reducing them to a pair of linear equation.
    \n\"Karnataka
    \nOR
    \n5 pencils and 7 books together cost Rs. 50. whereas 7 pencils & 5 books together cost Rs. 46. Find the cost of 1 pencil and 1 book.<\/p>\n

    Question 34.
    \nIf sin 3A = cos (A – 26\u00b0). Where 3A is an acute angle, find the value of A.
    \nOR
    \n\"Karnataka<\/p>\n

    Question 35.
    \nCalculate the median for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    0 – 20<\/td>\n4<\/td>\n<\/tr>\n
    20 – 40<\/td>\n6<\/td>\n<\/tr>\n
    40 – 60<\/td>\n8<\/td>\n<\/tr>\n
    60 – 80<\/td>\n10<\/td>\n<\/tr>\n
    80 – 100<\/td>\n4<\/td>\n<\/tr>\n
    100 – 120<\/td>\n5<\/td>\n<\/tr>\n
    120 – 140<\/td>\n3<\/td>\n<\/tr>\n
    <\/td>\nn = 40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    OR
    \nCalculate the mode for the following frequency distribution<\/p>\n\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    5 – 15<\/td>\n3<\/td>\n<\/tr>\n
    15 – 25<\/td>\n8<\/td>\n<\/tr>\n
    25 – 35<\/td>\n17<\/td>\n<\/tr>\n
    35 – 45<\/td>\n20<\/td>\n<\/tr>\n
    45 – 55<\/td>\n11<\/td>\n<\/tr>\n
    55 – 65<\/td>\n1<\/td>\n<\/tr>\n
    <\/td>\nn = 60<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Question 36.
    \nConstruct ‘OGIVE’ for the following:<\/p>\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    0 – 3<\/td>\n8<\/td>\n<\/tr>\n
    3 – 6<\/td>\n2<\/td>\n<\/tr>\n
    6 – 9<\/td>\n4<\/td>\n<\/tr>\n
    9 – 12<\/td>\n2<\/td>\n<\/tr>\n
    12 – 15<\/td>\n1<\/td>\n<\/tr>\n
    <\/td>\nn = 17<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    (4 x 4 = 16)<\/strong>
    \nQuestion 37.
    \nFind the 11th term from the last term (towards the first-term) of the AR 10, 7, 4,…….., -62.
    \nOR
    \nIn a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third & so on. There are 5 rose-plants in the last row. How many rows are there in the flower bed ?<\/p>\n

    Question 38.
    \nIn a triangle if square as one side is equal to the sum of the squares on the other two sides, then the angle opposite to the first side is a right angle, prove this.<\/p>\n

    Question 39.
    \nSolve Graphically :
    \n2x – y = 5
    \n4x – y = 13<\/p>\n

    Question 40.
    \nRam made a bird bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cms. Find the T.S.A. of the bird bath.<\/p>\n

    Solutions<\/strong><\/p>\n

    Solution 1.
    \n(b)
    \n\"Karnataka<\/p>\n

    Solution 2.
    \n(a) p(x) = x2<\/sup> + 4x + 3<\/p>\n

    Solution 3.
    \n(d) 36<\/p>\n

    Solution 4.
    \n(b) \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/p>\n

    Solution 5.
    \n(c) x = 4, y = 2<\/p>\n

    Solution 6.
    \n(a) 50\u00b0<\/p>\n

    Solution 7.
    \n(a) (4, 5)<\/p>\n

    Solution 8.
    \n(c) b2<\/sup> – 4ac<\/p>\n

    Solution 9.
    \n\u201cIn a right-angled triangle the square on the hypotenuse is equal to the sum of the squares as the other two sides\u201d.<\/p>\n

    Solution 10.
    \n10 cm.<\/p>\n

    Solution 11.
    \n\"Karnataka
    \nBy division lemma
    \n225 = 135 x 1 + 90
    \n135 = 90 x 1 + 45
    \n90 = 45 x 2 + 0
    \nThe last divisor is 45
    \nHCF = 45<\/p>\n

    Solution 12.
    \n4<\/p>\n

    Solution 13.
    \n\"Karnataka<\/p>\n

    Solution 14.
    \n\"Karnataka<\/p>\n

    Solution 15.
    \nNos. divisible by 5 are 10, 15, 20, ………, 95
    \nThey are in AP
    \na = 10, d = 5, Tn<\/sub> = 95, n = ?
    \nTn<\/sub> = a + (n – 1)d
    \n\u21d2 95 = 10 + (n – 1)5
    \n\u21d2 95 – 10 = 5n – 5
    \n\u21d2 85 + 5 = 5n
    \n\u21d2 90 = 5n
    \n\u21d2 n = 18
    \n18 two digit nos. are divisible by 5.<\/p>\n

    Solution 16.
    \nLet ABC and PQR be the triangles.
    \n\"Karnataka<\/p>\n

    Solution 17.
    \n\"Karnataka
    \nSubstitute the value of x in eqn. 1.
    \n5x + 4y = 14
    \n\u21d2 5(2) + 4y = 14
    \n\u21d2 10 + 4y = 14
    \n\u21d2 4y = 14 – 10
    \n\u21d2 4y = 4
    \n\u21d2 y = 1
    \n\u2234 x = 2, y = 1<\/p>\n

    Solution 18.
    \nLet the age of A = x years
    \nLet the age of son = y years
    \n5 years ago
    \n(x – 5) = 3(y – 5)
    \nx – 5 = 3y – 15
    \nx – 3y = -15 + 5
    \nx – 3y = -10 …..(1)
    \n20 years later
    \n(x + 20) = 2(y + 20)
    \nx + 20 = 2y + 40
    \nx – 2y = 40 – 20
    \nx – 2y = 20 …… (2)
    \nFrom Eqn (1) & (2)
    \n\"Karnataka
    \nPresent age of the father = 80 years
    \nAge of the son = 30 years.<\/p>\n

    Solution 19.
    \n\"Karnataka<\/p>\n

    Solution 20.
    \n\"Karnataka<\/p>\n

    Solution 21.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 22.
    \nLet us assume that 5 + \u221a7 is rational.
    \n\"Karnataka
    \nbut \u221a7 is irrational.
    \nIf contradicts the fact that \u221a7 is irrational
    \nour assumption is wrong.
    \n(5 + \u221a7) is irrational.<\/p>\n

    Solution 23.
    \np(x) = 6x2<\/sup> + 7x – 3
    \nBy splitting the middle term
    \n6x2<\/sup> + 9x – 2x – 3
    \n\u21d2 3x(2x + 3) – 1(2x + 3)
    \n\u21d2 (2x + 3) (3x – 1)
    \nfor zeroes of p(x) put
    \n2x + 3 = 0 and 3x – 1 = 0
    \nx = \\(\\frac { -3 }{ 2 }\\) and x = \\(\\frac { 1 }{ 3 }\\)
    \nzeroes of p(x) are \\(\\frac { -3 }{ 2 }\\) and x = \\(\\frac { 1 }{ 3 }\\)<\/p>\n

    Solution 24.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 25.
    \nCompare this with ax2<\/sup> + bx + c = 0
    \na = 2, b = -5, c = 3
    \n\"Karnataka<\/p>\n

    Solution 26.
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 27.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 28.
    \n\"Karnataka<\/p>\n

    Solution 29.
    \nNo. of all possible outcomes n(S) = 6
    \nLet \u2018E\u2019 be the event of getting a perfect square no.
    \nn(E) = 2 {1, 4}
    \np(E) = \\(\\frac { n(E) }{ n(S) }\\) = \\(\\frac { 2 }{ 6 }\\)<\/p>\n

    Solution 30.
    \n\"Karnataka<\/p>\n

    Solution 31.
    \n\"Karnataka
    \nData: Two concentric circles C1<\/sub> and C2<\/sub> have the same centre O.
    \nAB is the chord of the larger circle C2<\/sub>.
    \nIt touches C1<\/sub> at P
    \nTo prove AP = PB
    \nConstruction: Join OP
    \nProof: AB is the tangent to C1<\/sub> at P and OP is the radius.
    \nOP \u22a5 AB
    \nAB is the chord for C2<\/sub> & OP \u22a5 AB
    \nOP is the bisector of AB, as the perpendicular from the centre bisects the chord.
    \ni.e., AP = BR
    \nOR<\/strong>
    \n\"Karnataka
    \nData: \u2018O\u2019 is the centre of the circle XY is a tangent at P.
    \nTo prove: QP \u22a5 XY
    \nConstruction: Take a point Q as XY other than P and join OQ.
    \nProof: The point P lies outside the circle (if it lies inside XY becomes a secant).
    \nOQ is longer than OP (radius)
    \nOQ > OP
    \nSince this happens for every point on XY except the point P
    \nOP is the shortest of all the distances from the point O to the points on XY.
    \nSo, OP is the perpendicular to XY.<\/p>\n

    Solution 32.
    \n\"Karnataka
    \n\u0394ABC is required \u0394
    \n\u0394A’BC’ is required \u0394<\/p>\n

    Solution 33.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 34.
    \nsin3A = cos(A – 26\u00b0)
    \n\u21d2 sin 3A = cos (90 – 3A)
    \n\u21d2 cos (90 – 3A) = cos (A – 26)
    \n\u21d2 90 – 3A = A – 26
    \n\u21d2 90 + 26 = A + 3A
    \n\u21d2 116 = 4A
    \n\u21d2 A = 29\u00b0
    \nOR<\/strong>
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 35.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 36.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 37.
    \nIf we write the given AP in the reverse order, then a = -62 and d = 3.
    \nNow find the 11th term with these a and d.
    \nso, a11<\/sub> = -62 + (11 – 1) x 3
    \na11<\/sub> = -62 + 30
    \nan<\/sub> = -32
    \nOR<\/strong>
    \nThe number of rose plants in 1st, 2nd, 3rd, ……… rows are
    \n23, 21, 19, ………. 5
    \nIt forma an AP
    \nLet the number of rows = n
    \na = 23, d = -2, an<\/sub> = 5
    \nan<\/sub> = a + (n – 1)d
    \n\u21d2 5 = 23 + (n – 1)(-2)
    \n\u21d2 5 – 23 = -2n + 2
    \n\u21d2 -18 – 2 = -2n
    \n\u21d2 -20 = -2n
    \n\u21d2 n = 10
    \nNo. of rows = 10<\/p>\n

    Solution 38.
    \n\"Karnataka
    \nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
    \nwe need to prove \u2220B = 90\u00b0
    \nconstruct \u0394PQR , right angled at Q such that
    \nPQ = AB & QR = BC
    \nNow from \u0394PQR , we have
    \nPR2<\/sup> = PQ2<\/sup> + QR2<\/sup> (by pythogora\u2019s theorem)
    \nPR2<\/sup> = AB2<\/sup> + BC2<\/sup> (by construction)
    \nAC2<\/sup> = AB2<\/sup> + BC2<\/sup> (data)
    \nAC = PR
    \nIn \u0394ABC and \u0394PQR
    \nAB = PQ (by construction)
    \nBC = QR (by construction)
    \nAC = PR (Proved)
    \n\u0394ABC = \u0394PQR
    \n\u2220B = \u2220Q (CPCT)
    \n\u2220Q = 90\u00b0 (by Construction)
    \n\u2220B = 90\u00b0<\/p>\n

    Solution 39.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 40.
    \n\"Karnataka
    \nLet h be the height of the cylinder and r the common radius of cylinder
    \nhemi-sphere T.S.A. of bird-bath = CSA of cylinder + CSA of the hemisphere.
    \n= 2\u03c0rh + 2\u03c0r2<\/sup>
    \n= 2\u03c0r (h + r)
    \n= 2 x \\(\\frac { 22 }{ 7}\\) x 30 (145 + 30)
    \n= 2 x \\(\\frac { 22 }{ 7}\\) x 30 x 175
    \n= 33000 cm2<\/sup><\/p>\n

    \"Karnataka<\/p>\n

    Karnataka State Syllabus SSLC Maths Model Question Paper 4 (Old Pattern)<\/h2>\n

    Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.<\/strong><\/p>\n

    (1 \u00d7 8 = 8)<\/strong>
    \nQuestion 1.
    \nThe distance between two points A (1, 7) and B (4, 2) is
    \n(a) \u221a34
    \n(b) \u221a43
    \n(c) \u221a68
    \n(d) \u221a86<\/p>\n

    Question 2.
    \nIf p(x) = x2<\/sup> + x + 1 then the value of p(-1) is
    \n(a) -1
    \n(b) 1
    \n(c) 2
    \n(d) -2<\/p>\n

    Question 3.
    \nWhich is correct for the event E?
    \n(a) P(\\(\\bar { E }\\)) = 1 + P(E)
    \n(b) P(\\(\\bar { E }\\)) = P(E)
    \n(c) P(\\(\\bar { E }\\)) = 1 – P(E)
    \n(d) P(E) = 1 + P(\\(\\bar { E }\\))<\/p>\n

    Question 4.
    \nThe volume of a cylinder is 154 cc. and the radius is 7 cms. then the height is
    \n(a) 2 cms
    \n(b) 3 cms
    \n(c) 4 cms
    \n(d) 1 cm<\/p>\n

    Question 5.
    \nThe value of x and y for the following equations are x + y = 15, x – 7 = 1
    \n(a) (8, 7)
    \n(b) (7, 8)
    \n(c) (10, 5)
    \n(d) (5, 10)<\/p>\n

    Question 6.
    \nABCD is a cyclic quadrilateral. A = 70\u00b0, then C = ?
    \n(a) 100\u00b0
    \n(b) 110\u00b0
    \n(c) 120\u00b0
    \n(d) 70\u00b0<\/p>\n

    Question 7.
    \nOA is a radius in the circle. If coordinates of A are (2, 3) then OA =…
    \n\"Karnataka
    \na) \u221a9
    \nb) \u221a4
    \nc) \u221a13
    \nd) \u221a5<\/p>\n

    Question 8.
    \nIf the discriminant of a quadratic equation b2<\/sup> – 4ac < 0 then the roots are
    \n(a) Real and distinct
    \n(b) equal
    \n(c) not real
    \n(d) unequal&rational<\/p>\n

    (1 \u00d7 6 = 6)<\/strong>
    \nQuestion 9.
    \nIn \u0394ABC, XY || BC, If AX = 2 cm, AB = 5 cms and AY = 4 cms. then AC = ………
    \n\"Karnataka<\/p>\n

    Question 10.
    \nIdentify the largest chord in the given figure.
    \n\"Karnataka<\/p>\n

    Question 11.
    \nFind the HCF of 25 and 15<\/p>\n

    Question 12.
    \nFrom the graph find the number of zeros of the polynomial p(x)
    \n\"Karnataka<\/p>\n

    Question 13.
    \nWhich are the coordinates of the origin?<\/p>\n

    Question 14.
    \nExpress 200 as a product of prime factors.<\/p>\n

    (16 \u00d7 2 = 32)<\/strong>
    \nQuestion 15.
    \nHow many 2 digit numbers are divisible by 9.<\/p>\n

    Question 16.
    \nDiagonals of a trapezium ABCD with AB || DC intersect at O. If AB = 2CD. Find the ratio of the areas of triangles AOB and COD.<\/p>\n

    Question 17.
    \nSolve for x and y
    \n2x + y = 4
    \n3x + 4y = 6<\/p>\n

    Question 18.
    \nHalf the perimeter of a rectangle whose length is 4 m. more than its width is 36 m. Find the dimensions of the garden.<\/p>\n

    Question 19.
    \nFind the area of the shaded region in the fig given. If each side of the square is 14 cm.
    \n\"Karnataka<\/p>\n

    Question 20.
    \nDraw a circle of radius 4 cms. Draw two radii in it such that the angle between them is 130\u00b0. Construct two tangents at the ends of radii.<\/p>\n

    Question 21.
    \nThe coordinates of the vertices of a triangle are (2, 3), (4, 5) and (6, 9). Find its area.<\/p>\n

    Question 22.
    \nProve that (5 + \u221a3) is irrational.<\/p>\n

    Question 23.
    \nFind the zeros of the polynomial p(x) = 5x2<\/sup> – 2x – 3<\/p>\n

    Question 25.
    \nSolve the given equation using formula x2<\/sup> – 6x – 4 = 0<\/p>\n

    Question 26.
    \nIf A = \u221a2 – 1 show that \\(\\frac { tanA }{ 1+{ tan }^{ 2 }A } =\\frac { \\surd 2 }{ 4 }\\)<\/p>\n

    Question 27.
    \nFrom a point an the ground, the angles of elevation of the bottom and top of a tower fixed at the top of 20 m high building is 45\u00b0 & 60\u00b0 respectively. Find the height of the tower.<\/p>\n

    Question 28.
    \nThe angle of elevation of an aeroplane from a point on the ground is 60\u00b0. After 15 seconds flight the elevation changes to 30\u00b0. If the aeroplane is flying at a height of 1500\u221a3 m, find the speed of the plane in km\/hr.<\/p>\n

    Question 29.
    \n12 defective pens are mixed with 132 good ones one pen is taken out at random from this lot. Find the probability that the pen is taken out is a good one.<\/p>\n

    (6 \u00d7 3 = 18)<\/strong>
    \nQuestion 30.
    \nTwo cubes each of volume 64 cm3<\/sup> are joined end to end. Find the surface area of the resulting cuboid.<\/p>\n

    Question 31.
    \nProve that in two concentric circles the chord of the longer circle, which touches the smaller circle is bisected at the point of contact.
    \nOR<\/strong>
    \nProve that the tangents drawn to a circle from an external point are equal.<\/p>\n

    Question 32.
    \nConstruct a \u0394 of sides, 8 cms., 10 cms & 12 cms. and then A similar to it, whose sides are \\(\\frac { 3 }{ 4 }\\) the corresponding sides of the first \u0394<\/p>\n

    Question 33.
    \n90% and 97% acid solutions are mixed to get 21 Its. of 95% pure acid solution. Find the amount of each type of acid to be mixed to form the mixture.
    \nOR<\/strong>
    \nOne says \u201cgive me a hundred friend I shall then become twice as rich as you\u201d. The other says \u201cIf you give me 10,1 shall be 6 times as rich as you\u201d. Find the amount of their capital.<\/p>\n

    Question 34.
    \nIf A, B, C are interior angles of a triangle ABC show that
    \n\"Karnataka<\/p>\n

    Question 35.
    \nCalculate the median for the following data<\/p>\n\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    65 – 85<\/td>\n4<\/td>\n<\/tr>\n
    85 – 105<\/td>\n5<\/td>\n<\/tr>\n
    105 – 125<\/td>\n13<\/td>\n<\/tr>\n
    125 – 145<\/td>\n20<\/td>\n<\/tr>\n
    145 – 165<\/td>\n14<\/td>\n<\/tr>\n
    165 – 185<\/td>\n8<\/td>\n<\/tr>\n
    185 – 205<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    OR<\/strong>
    \nCalculate the mode for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    1000 – 1500<\/td>\n24<\/td>\n<\/tr>\n
    1500 – 2000<\/td>\n40<\/td>\n<\/tr>\n
    2000 – 2500<\/td>\n33<\/td>\n<\/tr>\n
    2500 – 3000<\/td>\n28<\/td>\n<\/tr>\n
    3000 – 3500<\/td>\n30<\/td>\n<\/tr>\n
    3500 – 4000<\/td>\n22<\/td>\n<\/tr>\n
    4000 – 4500<\/td>\n16<\/td>\n<\/tr>\n
    4500 – 5000<\/td>\n07<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Question 36.
    \nDraw the \u2018less than type OGIVE for the following data.<\/p>\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    50 – 60<\/td>\n6<\/td>\n<\/tr>\n
    60 – 70<\/td>\n5<\/td>\n<\/tr>\n
    70 – 80<\/td>\n9<\/td>\n<\/tr>\n
    80 – 90<\/td>\n12<\/td>\n<\/tr>\n
    90 – 100<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    (4 \u00d7 4 = 16)<\/strong>
    \nQuestion 37.
    \nWhich term of the A.P 45, 41, 37, 33, ………… is the first negative term?
    \nOR<\/strong>
    \n\"Karnataka<\/p>\n

    Question 38.
    \nProve that in a \u0394, if square as one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.<\/p>\n

    Question 39.
    \nSolve Graphically
    \nx + 3y = 6
    \n2x – 3y = 12<\/p>\n

    Question 40.
    \nThe base radius and height of a right circular solid cone are 2 cms and 8 cms respectively. It is melted & recast into spheres of diameter 2 cms. each. Find the number of spheres formed.<\/p>\n

    Solutions<\/p>\n

    Solution 1.
    \n(a) \u221a34<\/p>\n

    Solution 2.
    \n(b) 1<\/p>\n

    Solution 3.
    \n(c) P(\\(\\bar { E }\\)) = 1 – P(E)<\/p>\n

    Solution 4.
    \n(d) 1 cm<\/p>\n

    Solution 5.
    \n(a) (8, 7)<\/p>\n

    Solution 6.
    \n(b) 110\u00b0<\/p>\n

    Solution 7.
    \n(c) \u221a13<\/p>\n

    Solution 8.
    \n(c) not real<\/p>\n

    Solution 9.
    \n\"Karnataka<\/p>\n

    Solution 10.
    \nXY<\/p>\n

    Solution 11.
    \n25 = 5 \u00d7 5
    \n15 = 5 \u00d7 3<\/p>\n

    Solution 12.
    \n2<\/p>\n

    Solution 13.
    \n(0, 0)<\/p>\n

    Solution 14.
    \n\"Karnataka
    \n200 = 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5 = 23<\/sup> \u00d7 52<\/sup><\/p>\n

    Solution 15.
    \nTwo digit Nos. are 18, 27, 36, ……. 99
    \nIt is in AP
    \na = 18, d = 9, an<\/sub> = 99, n = ?
    \nan<\/sub> = a + (n – 1) d
    \n\u21d2 99 = 18 + (n – 1)9
    \n\u21d2 99 – 18 = 9n – 9
    \n\u21d2 81 + 9 = 9n
    \n\u21d2 90 = 9n
    \n\u21d2 n = 10
    \nThere are 10 two digit nos.<\/p>\n

    Solution 16.
    \nArea of \u2206AOB and \u2206COD
    \nA\u00d4B = COD (V.O.A.)
    \nCOO = OA (alternate angles)
    \nBy AA Similarity criterion
    \n\u2206AOB ~ \u2206DOC
    \n\"Karnataka<\/p>\n

    Solution 17.
    \n\"Karnataka
    \nSubstitute the value of x in equation 1
    \n2x + y = 4
    \n\u21d2 2(2) + y = 4
    \n\u21d2 4 + y = 4
    \n\u21d2 y = 4 – 4 = 0
    \nx = 2, y = 0<\/p>\n

    Solution 18.
    \n\"Karnataka
    \nl + b = 36
    \n\u21d2 l = 36 – b
    \nl = b + 4
    \nb + 4 = 36 – b
    \n\u21d2 b + b = 36 – 4
    \n\u21d2 2b = 32
    \n\u21d2 b = 16
    \nl = b + 4
    \n\u21d2 l = 16 + 4
    \n\u21d2 l = 20
    \nLength = 20 m
    \nWidth = 16 m<\/p>\n

    Solution 19.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 20.
    \nPA and PB are tangents to the circle at the ends of radii
    \n\"Karnataka<\/p>\n

    Solution 21.
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 22.
    \nLet us assume that (5 + \u221a3)
    \n\"Karnataka
    \nRHS is rational
    \n\u21d2 \u221a3 is rational
    \nBut \u221a3 is irrational
    \nIt is contradictory to the fact that \u221a3 is irrational
    \nHence our assumption is wrong.
    \n(5 + \u221a3) is irrational.<\/p>\n

    Solution 23.
    \nTo get the zeros of the, polynomial
    \nput p(x) = 0
    \n\u21d2 5x2<\/sup>\u00a0– 2x – 3 = 0
    \n\u21d2 5x2<\/sup>\u00a0– 5x + 3x – 3 = 0
    \n\u21d2 5x(x – 1) + 3(x – 1) = 0
    \n\u21d2 (x – 1)(5x + 3) = 0
    \nx – 1 = 0 or 5x + 3 = 0
    \nx = 1 or x = \\(\\frac { -3 }{ 5 }\\)
    \nZeros of the polynomial are 1 and \\(\\frac { -3 }{ 5 }\\)<\/p>\n

    Solution 24.
    \nHere a = a, b = -5, c = c
    \nLet the zeros be \u03b1 and \u03b2 then
    \n\"Karnataka<\/p>\n

    Solution 25.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 26.
    \n\"Karnataka<\/p>\n

    Solution 27.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 28.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 29.
    \nNo. of defective pens = 12
    \nNo. of good pens = 132
    \nTotal no. of pens = 144
    \nNo. of all possible outcomes = 144
    \nLet E be the event that the pen taken out is a good one
    \nNo. of outcomes favourable to E = 132
    \n\"Karnataka<\/p>\n

    Solution 30.
    \n\"Karnataka
    \nLet each edge of the cube be a
    \nVolume of each cube = a3<\/sup>
    \n64 = a3<\/sup>
    \n\u21d2 a = 4
    \nFor the resulting cuboid l = 4 + 4 = 8 cm
    \nbreadth = 4 cm
    \nHeight = 4 cm.
    \nSurface area of the resulting cuboid = 2(lb + bh + hl)
    \n= 2[(8 \u00d7 4) + (4 \u00d7 4) + (4 \u00d7 8)]
    \n= 2[32 + 16 + 32]
    \n= 2 [80]
    \n= 160 cm2<\/sup><\/p>\n

    Solution 31.
    \nC1<\/sub> and C2<\/sub> are the concentric circle with centre O A chord AB of the larger circle C touches the smaller circle C2<\/sub> at P
    \n\"Karnataka
    \nTo prove: AP = BP
    \nConstruction: Join OP
    \nProof: AB is a tangent to the circle C2<\/sub> at P and OP is the radius.
    \nOP \u22a5 AB
    \nAB is the chord of circle Q & OP \u22a5 AB
    \nOP is the bisector of chord AB
    \nAP = PB
    \nOR
    \nData: O is the centre of the circle. AB and AC are tangents from A
    \n\"Karnataka
    \nTo prove AB = AC
    \nConstruction: Join OA, OB and OC
    \nProof: In \u0394AOB and \u0394AOC
    \nABO = ACO = 90\u00b0 [Angle between radius & tangent]
    \nOB = OC (Radii of the circle)
    \nOA is common
    \nBy RHS criterian
    \n\u0394AOB = \u0394AOC
    \nAB = AC<\/p>\n

    Solution 32.
    \nA’CC’ is the required triangle
    \n\"Karnataka<\/p>\n

    Solution 33.
    \nLet x Lts of 90% pure acid solution & Y Lts of 97% pure acide solution be mixed,
    \ntotal volume of the mixture = (x + y) Lts
    \nx + y = 21 ……… (1)
    \n90% of x + 97% of y = 95% of 21
    \n\"Karnataka
    \nSubstituting the value of y in eqn 1
    \nx + y = 21
    \n\u21d2 x + 15 = 21
    \n\u21d2 x = 21 – 15 = 6
    \n\u21d2 x = 6
    \nQuantity of 90% pure acid solution = 6 Lts.
    \nQuantity of 97% pure acid solution = 15 Lts.
    \nOR<\/strong>
    \nLet the amounts with them be respectively.
    \nx and y
    \nx + 100 = 2(y – 100)
    \n\u21d2 x + 100 = 2y – 200
    \n\u21d2 x – 2y = -200 – 100
    \n\u21d2 x – 2y = -300 ……. (1)
    \ny + 10 = 6(x – 10)
    \n\u21d2 y + 10 = 6x – 60
    \n\u21d2 y – 6x = -60 – 10
    \n\u21d2 -6x + y = -70
    \n\u21d2 6x – y = 70 ……… (2)
    \nFrom Equation 1
    \nx = 2y – 300
    \nSubstituting the value of x in equation 2
    \n6x – y = 70
    \n\u21d2 6(2y – 300) – y = 70
    \n\u21d2 12y – 1800 – y = 70
    \n\u21d2 11y = 70 + 1800
    \n\u21d2 11y = 1870
    \n\u21d2 y = 170
    \nSubstitute the value Y in (1)
    \nx = 2y – 300
    \n\u21d2 x = 2(170) – 300
    \n\u21d2 x = 340 – 300
    \n\u21d2 x = 40
    \nTheir capitals are 40 Rs. & 170 Rs.<\/p>\n

    \"Karnataka<\/p>\n

    Solution 34.
    \nBecause A, B, C are the angles of a \u2206ABC
    \nA + B + C = 180\u00b0
    \n\u21d2 B + C = (180 – A)
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 35.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 36.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 37.
    \na = 45
    \nd = a2<\/sub> – a1<\/sub> = 41 – 45 = -4
    \nLet nth term of the AP be the first negative term then an<\/sub> < 0
    \na + (n – 1)d < 0
    \n\u21d2 45 + (n – 1) (-2) < 0
    \n\u21d2 45 – 4n + 4 < 0
    \n\u21d2 49 – 4n < 0
    \n\u21d2 49 < 4n
    \n\u21d2 n > \\(\\frac { 49 }{ 4 }\\)
    \n\u21d2 n > 12\\(\\frac { 1 }{ 4 }\\)
    \nleast positive integral value of n = 13
    \n13th term is the first negative term.
    \nOR<\/strong>
    \nLet \u2018a\u2019 be the first term and d the common difference
    \na4<\/sub> = a + (4 – 1) d = a + 3d
    \na6<\/sub> = a + 5d
    \na7<\/sub> = a + 6d
    \na8<\/sub> = a + 7d
    \n\"Karnataka<\/p>\n

    Solution 38.
    \n\"Karnataka
    \ndata : ABC is a \u2206 in which AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
    \nTo prove: \u2220B = 90\u00b0
    \nConstruction: Construct \u2206PQR right angled at Q such that
    \nPQ = AB & RQ = CB
    \nProof: PR2<\/sup> = PQ2<\/sup> + RQ2<\/sup> (by Pythagoras theorem)
    \nPR2<\/sup> = AB2<\/sup> + BC2<\/sup> (by construction)
    \nBut AC2<\/sup> = AB2<\/sup> + BC2<\/sup> (data)
    \nAC = PR
    \nIn \u2206ABC and \u2206PQR
    \nAB = PQ
    \nBC = QR (by construction)
    \nAC = PR (proved)
    \n\u2206ABC = \u2206PQR
    \n\u2220B = \u2220Q
    \nbut \u2220Q = 90\u00b0 (by construction)
    \n\u2220B = 90\u00b0<\/p>\n

    Solution 39.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 40.
    \n\"Karnataka<\/p>\n

    Karnataka State Syllabus SSLC Maths Model Question Paper 5 (Old Pattern)<\/h2>\n

    Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.<\/strong><\/p>\n

    (1 \u00d7 8 = 8)<\/strong>
    \nQuestion 1.
    \nThe distance between the points (2, 3) and (4, 1) is
    \n(a) 3\u221a2
    \n(b) 2\u221a3
    \n(c) \u221a2
    \n(d) 2\u221a2<\/p>\n

    Question 2.
    \nThe cubic polynomial among the following is
    \n(a) p(x) = x2<\/sup> – x – 2
    \n(b) q(x) = 2x2<\/sup> + x – 7
    \n(c) r(x) = x3<\/sup> + x2<\/sup> – 1
    \n(d) s(x) = x4<\/sup> – x3<\/sup> + x2<\/sup> – 2<\/p>\n

    Question 3.
    \nThe probability of a certain event is
    \n(a) 0
    \n(b) 1
    \n(c) \\(\\frac { 1 }{ 2 }\\)
    \n(d) \\(\\frac { 1 }{ 3 }\\)<\/p>\n

    Question 4.
    \nThe volume of a sphere of radius 21 cm is
    \n(a) 38808 cm3<\/sup>
    \n(b) 80838 cm3<\/sup>
    \n(c) 83808 cm3<\/sup>
    \n(d) 88380 cm3<\/sup><\/p>\n

    Question 5.
    \nIf y = x – 2 and x + y = 8, then the values of x and y are respectively,
    \n(a) -5, 3
    \n(b) -3, 5
    \n(c) 3, 5
    \n(d) 5, 3<\/p>\n

    Question 6.
    \nIn the adjoining figure, AB and AC are tangents from A. Then ABC =?
    \n\"Karnataka
    \n(a) 30\u00b0
    \n(b) 40\u00b0
    \n(c) 50\u00b0
    \n(d) 60\u00b0<\/p>\n

    Question 7.
    \nThe distance between points (a, b) and (-a, -b) is
    \n(a) \\(2\\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\\)
    \n(b) \\(\\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\\)
    \n(c) 2\\(\\sqrt { { a }+{ b } }\\)
    \n(d) \\(\\sqrt { { a }+{ b } }\\)<\/p>\n

    Question 8.
    \nThe roots of the equation 2x2<\/sup> – 200 = 0 are
    \n(a) \u00b120
    \n(b) \u00b110
    \n(c) 20
    \n(d) 10<\/p>\n

    (6 \u00d7 1 = 6)<\/strong>
    \nQuestion 9.
    \nState the converse of Basic proportionality theorem.<\/p>\n

    Question 10.
    \nAB is a chord in a circle of radius 5 cms. OP \u22a5 AB. If OP = 4 cms then AB = ……..
    \n\"Karnataka
    \n(a) 8 cm
    \n(b) 10 cm
    \n(c) 12 cm
    \n(d) 6 cm<\/p>\n

    Question 11.
    \nThe product of HCF and LCM of two numbers is 75. Then the product of the numbers is
    \n(a) 125
    \n(b) 25
    \n(c) 75
    \n(d) 57<\/p>\n

    Question 12.
    \nThe zeroes of the Polynomial p(x) = x2<\/sup> + 4x + 3 are
    \n(a) -3, -1
    \n(b) 1, 3
    \n(c) -1, 3
    \n(d) -3, 1<\/p>\n

    \"Karnataka<\/p>\n

    Question 13.
    \nIf a point P (x, y) divides the line segment joining, A (x1<\/sub>, y1<\/sub>) and B (x2<\/sub>, y2<\/sub>) in the ratio m1<\/sub> and m2<\/sub>, then the coordinates of P are
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Question 14.
    \nExpress 120 as a product of prime factors.<\/p>\n

    (16 \u00d7 2 = 32)<\/strong>
    \nQuestion 15.
    \nWrite the first term and common difference for the following A.P.
    \n\\(\\frac { 1 }{ 3 }\\) , \\(\\frac { 5 }{ 3 }\\) , \\(\\frac { 9 }{ 3 }\\) , \\(\\frac { 13 }{ 3 }\\) , ……..<\/p>\n

    Question 16.
    \nIn the figure of \u2206ABE = \u2206ACD show that \u2206ADE ~ \u2206ABC
    \n\"Karnataka<\/p>\n

    Question 17.
    \nSolve for x and y
    \n3x + 2y = 3
    \n2x + 3y = 2<\/p>\n

    Question 18.
    \nIf twice the son\u2019s age is added to the age of his father the sum is 90. If twice the father\u2019s age is added to the age of the son the sum is 120. Find their ages.<\/p>\n

    Question 19.
    \nIn the figure given O is the centre of the bigger circle and AC is its diameter. Another circle with BA as the diameter is drawn.
    \nIf AC = 54 cms & BC = 10 cms. Find the area of the shaded region.
    \n\"Karnataka<\/p>\n

    Question 20.
    \nConstruct a tangent of length 4 cm. from an external point to a circle of radius 3 cms.<\/p>\n

    Question 21.
    \nFind the relation between x and y, such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).<\/p>\n

    Question 22.
    \nProve that 7 + \u221a5 is irrational.<\/p>\n

    Question 23.
    \nFind the quadratic polynomial, whose zeroes are (2 + \u221a3) and (2 – \u221a3)<\/p>\n

    Question 24.
    \nFind a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time & product of its zeroes as 2, -7, -14 respectively.<\/p>\n

    Question 25.
    \nSolve the equation 2x2<\/sup> – 5x + 3 = 0 using formula.<\/p>\n

    Question 26.
    \nEvaluate:
    \n\"Karnataka<\/p>\n

    Question 27.
    \nFrom the top of a 7 m high building, the angle of elevation of the top of a tower is 60\u00b0 & the angle of depression of its foot is 45\u00b0. Find the height of the tower.<\/p>\n

    Question 28.
    \nFrom the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be 30\u00b0 and 60\u00b0. If the height of the lighthouse is h mts & the line joining the ships passes through the foot of the lighthouse find the distance between the ships.<\/p>\n

    Question 29.
    \nA lot of 20 bulbs contains 4 defective ones, one bulb is drawn at random from the lot. What is the probability that the bulb is defective?<\/p>\n

    Question 30.
    \nThe sum of the radius of base and height of the solid right circular cylinder is 37 cms. If the total surface area of the solid cylinder is 1628 cm2. Find the volume of the cylinder.<\/p>\n

    (6 \u00d7 3 = 18)<\/strong>
    \nQuestion 31.
    \nA line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. Prove this.<\/p>\n

    Question 32.
    \nDraw a line segment of length 4.5 cms and divide it is the ratio 3 : 6 & measure each part.<\/p>\n

    Question 33.
    \nPocket money of A and B are in the ratio 6 : 5 & the ratio of their expenditure are in the ratio 4 : 3. If each of them saves Rs. 50 at the end of the month find their pocket money.
    \nOR<\/strong>
    \nThe sum of a two-digit number & the no. obtained by reversing the digits is 99. If the no. obtained by reversing the digits is 9 more than the original no. find the no.<\/p>\n

    Question 34.
    \nIf tan2A = cot (A – 18\u00b0) where 2A is an acute angle, find A.
    \nOR<\/strong>
    \nShow that tan 48\u00b0. tan 23\u00b0. tan 42 \u00b0. tan 67\u00b0 = 1<\/p>\n

    Question 35.
    \nCalculate the median for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    1500 – 2000<\/td>\n14<\/td>\n<\/tr>\n
    2000 – 2500<\/td>\n50<\/td>\n<\/tr>\n
    2500 – 3000<\/td>\n60<\/td>\n<\/tr>\n
    3000 – 3500<\/td>\n86<\/td>\n<\/tr>\n
    3500 – 4000<\/td>\n74<\/td>\n<\/tr>\n
    4000 – 4500<\/td>\n62<\/td>\n<\/tr>\n
    4500 – 5000<\/td>\n48<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    OR<\/strong>
    \nCalculate the mode for the following data.<\/p>\n\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    150 – 155<\/td>\n12<\/td>\n<\/tr>\n
    155 – 160<\/td>\n13<\/td>\n<\/tr>\n
    160 – 165<\/td>\n10<\/td>\n<\/tr>\n
    165 – 170<\/td>\n8<\/td>\n<\/tr>\n
    170 – 175<\/td>\n5<\/td>\n<\/tr>\n
    175 – 180<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Question 36.
    \nContant OGIVE for the following distribution.<\/p>\n\n\n\n\n\n\n\n\n
    C.I.<\/strong><\/td>\nf<\/strong><\/td>\n<\/tr>\n
    50 – 60<\/td>\n6<\/td>\n<\/tr>\n
    60 – 70<\/td>\n5<\/td>\n<\/tr>\n
    70 – 80<\/td>\n9<\/td>\n<\/tr>\n
    80 – 90<\/td>\n12<\/td>\n<\/tr>\n
    90 – 100<\/td>\n6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    (4 \u00d7 4 = 16)<\/strong>
    \nQuestion 37.
    \nIf the 3rd and 9th terms of an A.P. are 4 and -8 respectively, which term of this AP is zero?
    \nOR<\/strong>
    \nIf (2n + 3) is the nth term of an A.P. find (i) first term, (ii) common difference, (iii) 15th term.<\/p>\n

    Question 38.
    \nProve that in a right-angled triangle, the square as the hypotenuse is equal to the sum of the squares on the other two sides?<\/p>\n

    Question 39.
    \nSolve the following in graphically:
    \n3x – y = 3
    \n2x + y = 2<\/p>\n

    Question 40.
    \nThe cost of painting the total outer surface of a closed cylindrical tank at 60 per square cm. is Rs. 237.60. The height of the tank is 6 times the radius of the base. Find the height & radius of the tank.<\/p>\n

    Solutions<\/strong><\/p>\n

    Solution 1.
    \n(d) 2\u221a2<\/p>\n

    Solution 2.
    \n(c) r(x) = x3<\/sup> + x2<\/sup> – 1<\/p>\n

    Solution 3.
    \n(b) 1<\/p>\n

    Solution 4.
    \n(a) 38808 cm3<\/sup><\/p>\n

    Solution 5.
    \n(d) (5, 3)<\/p>\n

    Solution 6.
    \n(c) 50\u00b0<\/p>\n

    Solution 7.
    \n(a) \\(2\\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\\)<\/p>\n

    Solution 8
    \n(b) \u00b110<\/p>\n

    Solution 9.
    \nIf a line divides the two sides of a triangle proportionally then that line is parallel to the third side.<\/p>\n

    Solution 10.
    \n(d) 6 cms<\/p>\n

    Solution 11.
    \n(c) 75<\/p>\n

    Solution 12.
    \n(a) (-3, -1)<\/p>\n

    \"Karnataka<\/p>\n

    Solution 13.
    \n\"Karnataka<\/p>\n

    Solution 14.
    \n\"Karnataka<\/p>\n

    Solution 15.
    \n\"Karnataka<\/p>\n

    Solution 16.
    \n\u0394ABE = \u0394ACD (data)
    \nAB = AC …… (1) (CPCT)
    \nAE = AD (CPCT)
    \n\u21d2 AD = AE ….. (2) (CPCT)
    \nDividing (1) & (2)
    \n\\(\\frac { AB }{ AD }\\) = \\(\\frac { AC }{ AE }\\) …..(3)
    \nalso DAE = BAC ….. (4) (Common angle)
    \nFrom (3) & (4)
    \n\u0394ADE ~ \u0394ABC (SAS Similarity criterian)<\/p>\n

    Solution 17.
    \n\"Karnataka
    \nSubstitute the value of y in equation (1)
    \n3x + 2y = 3
    \n\u21d2 3x + 2(0) = 3
    \n\u21d2 3x + 0 = 3
    \n\u21d2 3x = 3
    \n\u21d2 x = 1
    \n\u2234 x = 1, y = 0<\/p>\n

    Solution 18.
    \nLet the age of father = x years
    \nLet the age of the son = y years
    \n\"Karnataka
    \nSubstitute the value of y in equation 1
    \nx + 2y = 90
    \n\u21d2 x + 2(20) = 90
    \n\u21d2 x + 40 = 90
    \n\u21d2 x = 90 – 40 = 50
    \nAge of the father = 50 years
    \nAge of the son = 20 years<\/p>\n

    Solution 19.
    \nArea of the shaded region = Area of the circle with AC as diameter – Area of the circle with AB as the diameter
    \nArea of the shaded region
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 20.
    \nDraw the rough figure & calculate OA using Pythagora\u2019s theorem. Then construct the tangent.
    \n\"Karnataka<\/p>\n

    Solution 21.
    \nLet P \u2192 (x, y)
    \nA \u2192 (3, 6) & B \u2192 (-3, 4)
    \nPA = PB (data)
    \nPA2<\/sup> = PB2<\/sup>
    \n(3 – x)2<\/sup> + (6 – y)2<\/sup> = (-3 – x)2<\/sup> + (4 – y)2<\/sup>
    \n\u21d2 9 + x2<\/sup> – 6x + 36 + y2<\/sup> – 12y = 9 + x2<\/sup> + 6x +16 + y2<\/sup> – 8y
    \n\u21d2 -6x + 36 – 12y = 6x + 16 – 8y
    \n\u21d2 -12x – 4y + 20 = 0 (\u00f74)
    \n\u21d2 3x + y – 5 = 0
    \nThis is required relation.<\/p>\n

    Solution 22.
    \nLet us assume that 7 + \u221a5 is rational
    \n\"Karnataka
    \nRHS is rational
    \n\u21d2 LHS is also rational but \u221a5 is irrational.
    \nIt is contradictory to the fact that \u221a5 is irrational.
    \nour assumption is wrong
    \n7 + \u221a5 is irrational<\/p>\n

    Solution 23.
    \nS = Sum of the zeros = 2 + \u221a3 + 2 – \u221a3 = 4
    \nP = Product of the zeroes = (2 + \u221a3)(2 – \u221a3) = 4 – 3 = 1
    \nThe required quadratic polynomial is
    \nk[x2<\/sup> – Sx + P] where \u2260 0 is real
    \n\u21d2 k[x2<\/sup> – 4x + 1]<\/p>\n

    Solution 24.
    \nLet \u03b1, \u03b2 & \u03b3 be the zeroes of the required cubic polynomial
    \n\"Karnataka<\/p>\n

    Solution 25.
    \n2x2<\/sup> – 5x + 3 = 0
    \nIt is in the form ax2<\/sup> + bx + c = 0
    \na = 2, b = -5, c = 3
    \n\"Karnataka<\/p>\n

    Solution 26.
    \n\"Karnataka<\/p>\n

    Solution 27.
    \n\"Karnataka
    \nIn the Rt. angled \u0394ABD
    \ntan 45\u00b0 = \\(\\frac { AB }{ BD }\\)
    \n1 = \\(\\frac { AB }{ BD }\\)
    \nBD – AB = 7m
    \nAE = 7m
    \nIn the Rt. angled \u0394AEC
    \ntan60\u00b0 = \\(\\frac { CE }{ AE }\\)
    \n\u221a3 = \\(\\frac { CE }{ 7 }\\)
    \nCE = 7\u221a3
    \nHeight of the tower = CD
    \n= DE + CE
    \n= AB + CE
    \n= 7 + 7\u221a3
    \n= 7(\u221a3 + 1)<\/p>\n

    \"Karnataka<\/p>\n

    Solution 28.
    \n\"Karnataka<\/p>\n

    Solution 29.
    \nTotal no. of bulbs = 20
    \nNo. of all possible outcomes = 20
    \nLet E1<\/sub> be the event that the bulb-drawn at random from the lot is defective
    \nNo. of outcomes favourable to E1<\/sub> is 4 Since there are 4 defective bulbs.
    \n\"Karnataka<\/p>\n

    Solution 30.
    \nr + h = 37
    \nT.S.A. of the cylinder = 2\u03c0rh + 2\u03c0r2<\/sup> = 2\u03c0r (h + r)
    \nBut 2\u03c0r (h + r) = 1628
    \n\"Karnataka<\/p>\n

    Solution 31.
    \n\"Karnataka
    \nData: A radius OP of a circle is drawn. Through the circle is drawn. Through the end point P of this radius a line AB is drawn the perpendicular to radius OR
    \nTo prove: AB is a tangent to the circle at P
    \nProof: Let Q be any point different from P on this line.
    \nNow OP \u22a5AB (data)
    \nOP is the shortest of all the distances from the point O to the line APB.
    \nOP < OQ \u21d2 OQ > Radius OP
    \nQ is an exterior point of the circle i.e. Q lies outside the circle, for all positions of Q different from P
    \nLine AB meets the circle only at the point P Line AB is a tangent to the circle at P<\/p>\n

    Solution 32.
    \nAC : CB = 3 : 6
    \nAB = 4.5 cms.
    \nAC = 1.5 cms
    \nCB = 3 cms.
    \n\"Karnataka<\/p>\n

    Solution 33.
    \nLet the pocket money of A and B respectively = 6x and 5x
    \nLet their expenditure be 4y and 3y respectively.
    \nMonthly savings of A and B = (6x – 4y) & (5x – 3y)
    \n\"Karnataka
    \nThe pocket money of A = 6x = 6 \u00d7 25 = Rs. 150
    \nThe pocket money of B = 6x = 5 \u00d7 25 = Rs. 125
    \nOR<\/strong>
    \nLet the two digit no. be yx = 10y + x
    \nas reversing the digits it becomes 10x + y
    \n10y + x + 10x + y = 99
    \n\u21d2 11x + 11y = 99
    \n\u21d2 x + y = 9 …….. (1)
    \nAccording to the second condition
    \n10x + y = 10y + x + 9
    \n\u21d2 9x – 9y = 9
    \n\u21d2 x – y = 1 ……… (2)
    \n\"Karnataka
    \nSubstitute the value of x in equation (2)
    \nx – y = 1
    \n\u21d2 5 – y = 1
    \n\u21d2 y = 4
    \nThe required no. is yx = 45<\/p>\n

    Solution 34.
    \ntan 2A = cot (A – 18\u00b0)
    \ncot(90 – 2A) = cot( A – 18) [\u2234 tan \u03b8 = cot (90 – \u03b8)]
    \n90 – 2A = A – 18
    \n90 + 18 = A + 2A
    \n108 = 3A
    \nA = 36\u00b0
    \nOR<\/strong>
    \nLHS = tan 45\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0
    \n= tan(90 – 42) tan 23\u00b0 tan 42\u00b0 tan(90 – 23\u00b0)
    \n= cot42\u00b0 tan 23\u00b0 tan 42\u00b0 cot 23\u00b0
    \n\"Karnataka<\/p>\n

    Solution 35.
    \n\"Karnataka
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 36.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 37.
    \nLet the first term and common difference of the AP be a and d respectively.
    \nT3<\/sub> = 4 \u21d2 a + 2d = 4 …… (1)
    \nT9<\/sub> = -8 \u21d2 a + 8d = -8 ……. (2)
    \nSolving Equation 1 & 2
    \nd = -2
    \nSubstitute the value of d in equation 1
    \na + 2d = 4
    \n\u21d2 a + 2(-2) = 4
    \n\u21d2 a – 4 = 4
    \n\u21d2 a = 8
    \nLet the nth term be zero
    \nTn<\/sub> = 0
    \na + (n – 1)d = 0
    \n\u21d2 8 + (n – 1)(-2) = 0
    \n\u21d2 8 – 2n + 2 = 0
    \n\u21d2 10 – 2n = 0
    \n\u21d2 10 = 2n
    \n\u21d2 n = 5
    \n5th term of the AP = zero
    \nOR<\/strong>
    \na. an<\/sub> = 2n + 3
    \nput n = 1
    \na1<\/sub> = 2(1) + 3
    \n\u21d2 a1<\/sub> = 5
    \nfirst term = 5
    \nPut n = 2
    \nb. a2<\/sub> = 2(2) + 3
    \n\u21d2 a2<\/sub> =4 + 3
    \n\u21d2 a2<\/sub> = 7
    \nd = a2<\/sub> – a1<\/sub>
    \n\u21d2 d = 7 – 5
    \n\u21d2 d = 2
    \nCommon difference = 2
    \nc. Tn<\/sub> = (2n + 3)
    \nPut n = 15
    \nT15<\/sub> = 2(15) + 3 = 30 + 3 = 33
    \nFifteenth term = 33<\/p>\n

    Solution 38.
    \ndata: ABC is a \u2206 in which B = 90\u00b0
    \nTo Prove: AC2<\/sup> = AB2<\/sup> + BC2<\/sup>
    \nConstruction: Draw BD \u22a5 AC
    \nProof: In \u2206ABC and \u2206ABD
    \nA is common
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Solution 39.
    \n\"Karnataka<\/p>\n

    \"Karnataka<\/p>\n

    Solution 40.
    \n\"Karnataka
    \n\"Karnataka<\/p>\n

    Karnataka SSLC Maths Model Question Papers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

    Students can Download Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern), Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus SSLC Maths Model Question Paper 1 (Old Pattern) Time: 3 Hours Max Marks: …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23295"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=23295"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/23295\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=23295"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=23295"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=23295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}