{"id":6040,"date":"2023-11-03T13:54:53","date_gmt":"2023-11-03T08:24:53","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=6040"},"modified":"2023-11-06T11:40:22","modified_gmt":"2023-11-06T06:10:22","slug":"karnataka-sslc-maths-model-question-paper-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/karnataka-sslc-maths-model-question-paper-4\/","title":{"rendered":"Karnataka SSLC Maths Model Question Paper 4 with Answers"},"content":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 4 with Answers, 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 4 with Answers<\/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

Question 1.
\nIf the H.C.F of 65 and 117 is expressible in the form of 65m – 117, then the value of m is
\na) 4
\nb) 3
\nc) 11
\nd) 2
\nAnswer:
\nSolution:
\nThe H.C.F. of 65 and 117 = 65m- 117
\nNow 13 = 65m- 117
\n65m 117+13
\n65rn = 130
\nm = 130\/65 = 2
\n\u2234 The value of m = 2
\n\u2234 HCF(65, 117)
\n= 13
\n\"Karnataka<\/p>\n

\"Karnataka<\/p>\n

Question 2.
\nIf sinx = sin 60\u00b0 cos 30\u00b0 – cos 60 sin 30!l, then the value of x is
\na) 0\u00b0
\nb) 30\u00b0
\nc) 45\u00b0
\nd) 60\u00b0
\nAnswer:
\nb) 30\u00b0
\nSolution:
\nSinx = sin 60\u00b0cos30 – cos 60 sin 30
\n\"Karnataka<\/p>\n

Question 3.
\nThe angle between the radius of a circle and the tangent drawn at the point of contact is
\na) 0\u00b0
\nb) 60\u00b0
\nc) 90\u00b0
\nd) 30\u00b0
\nAnswer:
\nc) 90\u00b0<\/p>\n

Question 4.
\nThe T.S.A. of a cuboid of dimension, l = 30cm, b = 20cm, c = 10cm, is ______
\na) 600cm2<\/sup>
\nb) 60cm2<\/sup>
\nc) 6000cm2<\/sup>
\nd) 2200 cm2<\/sup>
\nAnswer:<\/p>\n

Question 5.
\nWhich of the following is a polynomial
\na) x2<\/sup> – 5x + 3\u221ax
\nb) x1\/2<\/sup> + x1\/2<\/sup> – x +1
\nc) \\(\\sqrt{x}-\\frac{1}{\\sqrt{x}}\\)
\nd) x2<\/sup> – 4x + \u221a2
\nAnswer:
\nx2<\/sup> – 4x + \u221a2
\nSolution:
\nx2<\/sup> – 4x + \u221a2 because variable has no negative or decimal powers.<\/p>\n

Question 6.
\nThe value of p if x, 2x + p and 3x + 6 are in A.P.
\na) p = 3
\nb) p = 2
\nc) p = 1
\nd) p = 0
\nAnswer:
\nSolution:
\nx, 2x + p, 3x + 6 are inAP a, A, b are A.P.
\n\"Karnataka
\n2(2x + p) = 2(2x + 3)
\n2x + p = 2x + 3
\np = 2x + 3 – 2k
\nP = 3<\/p>\n

Question 7.
\nIn triangle PQR, The value of y is
\na) 4\u221a3
\nb) 6\u221a3
\nc) 5\u221a3
\nd) \u221a3
\nAnswer:
\nb) 6\u221a3
\nSolution:
\n\"Karnataka
\nQS2<\/sup> = PS x SR
\nQS2<\/sup> = \\(\\sqrt{9 \\times 12}\\)
\nQS = \\(\\sqrt{9 \\times 4 \\times 3}\\)
\nQS = 3 x 2\u221a3
\nQS = 6\u221a3<\/p>\n

Question 8.
\nWhen 2 unbiased coins are tossed at ‘ a time, the probability of getting 2 heads is _____
\na) 1\/4
\nb) 1\/2
\nc) 1
\nd) 0
\nAnswer:
\na) 1\/4
\nSolution:
\nS = {(H, H) (H, T) (T, H) (T, T)}
\nn(s) = 4
\nAn event of getting 2 heads = A={(H, H)}
\n\u2234 P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{1}{4}\\)<\/p>\n

\"Karnataka<\/p>\n

II. Answer the following Questions : ( 1 x 8 = 8 )<\/span><\/p>\n

Question 9.
\nIf the product of zeros of polynomial f(y) = ay3<\/sup> – 6y2<\/sup> + 11y – 6 is 4 then find the value of \u2018a\u2019.
\nAnswer:
\n\u03b1 \u03b2 \u03b3 = \\(-\\frac{\\mathrm{d}}{\\mathrm{a}}\\)
\n4 = \\(\\frac{-(-6)}{a}\\)
\n4a = 6
\na = \\(\\frac{6}{4}\\)
\na = \\(\\frac{3}{2}\\)<\/p>\n

Question 10.
\nWhat is the value of C, if ax2<\/sup> + bx + c = 0 has equal roots?
\nAnswer:
\nIf ax2<\/sup> + bx + c = 0 has equal roots then,
\nb2<\/sup> – 4ac = 0
\nb2<\/sup> = 4ac
\n4ac = b2<\/sup>
\nc = \\(\\frac{b^{2}}{4 a}\\)<\/p>\n

Question 11.
\nFind the second term if sum of the \u2018n\u2019 term of an AP is 2n2<\/sup> + 1.
\nAnswer:
\nSn = 2n2<\/sup> + 1
\nS1<\/sub> = 2(1)2<\/sup> + l= 2xl + l= 2+ l= 3
\n\u2234 S1<\/sub> = T1<\/sub> = a = 3
\nS2<\/sub> = 2(2)2<\/sup> + 1 =8+ 1 =9
\na + T2<\/sub> = 9
\nT2<\/sub> = 9 – 3 = 6<\/p>\n

Question 12.
\nState converse of Pythagoras Theorem.
\nAnswer:
\n\u201cThe square on the greatest side of a triangle is equal to the sum of the squares on the other two sides, then the other two sides contian the right angle\u201d.<\/p>\n

Question 13.
\nWhat is the form of \\(\\frac{p}{q}\\) (p,q \u2208 z q \u2260 0) form of 0.57 ?
\nAnswer:
\n\\(0.5 \\overline{7}=\\frac{57-5}{90}=\\frac{52}{90}=\\frac{26}{45}\\)<\/p>\n

Question 14.
\nIf sin\u03b8 = \\(\\frac{1}{3}\\) then find the value of (cot2<\/sup>\u03b8 + 2)
\nAnswer:
\nAC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\nBC2<\/sup> = AC2<\/sup> – AB2<\/sup>
\n= 32<\/sup> – 12<\/sup>
\n= 9 – 1 = 8
\nBC = \u221a8 = 2\u221a2
\ncot\u03b8 = \\(\\frac{2 \\sqrt{2}}{1}\\) = 2\u221a2
\n2cot2<\/sup>\u03b8 + 2 = 2(2\u221a2)2<\/sup> + 2
\n= 2 x 8 + 2
\n= 16 + 2
\n= 18
\n\"Karnataka<\/p>\n

Question 15.
\nIn sin(A + B) = \\(\\frac{\\sqrt{3}}{2}\\) and cos (A – B) = 1, 0< A + B < 90\u00b0, A \u2265 B
\nAnswer:
\nsin(A + B) = \\(\\frac{\\sqrt{3}}{2}\\) = sin60\u00b0
\nA + B = 60\u00b0 …. (1)
\ncos (A – B) = 1 = cos0\u00b0
\nA – B = 0 ….(2)
\nSolve (1) and (2)
\nA + B = 60
\nA – B = 0
\n2A = 60 or A = 30\u00b0
\nA + B = 60
\n30 + B = 60
\nB = 60 – 30\u00b0 = 30\u00b0
\n\u2234 A = 30\u00b0, B = 30\u00b0<\/p>\n

\"Karnataka<\/p>\n

Question 16.
\nThe surface area of a sphere is same as the C.S.A. of a right circular cylinder whose height and diameter are 4 cm each. Find the radius of the sphere.
\nSoln:
\n4\u03c0r\u00b2 = 2\u03c0rh
\n4R\u00b2 = dh .
\n4R\u00b2= 4 x 4
\nR\u00b2 = \\(\\frac{4 \\times 4}{4}\\) = 4
\nR = \u221a4 = 2cm
\nA = 30\u00b0, B = 30\u00b0
\n\u2234 Radius of the sphere = 2cm.<\/p>\n

III. Answer the following : ( 2 x 8 = 16 )<\/span><\/p>\n

Question 17.
\nBy Euclid\u2019s division lemma, show that thte square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
\nAnswer:
\nLet x be any positive integer. Then it is of the form 3q, 3q+1 or 3q + 2.
\nNow, (3q)2<\/sup> = 9q2<\/sup> = 3(3q2<\/sup>) = 3m where m = 3q2<\/sup>
\n(3q + 1)2<\/sup> = 9q2<\/sup> + 6q+ 1 = 3(3q2<\/sup> + 2q) + 1
\n= 3m + 1, where m = 3q2<\/sup> + 2q and (3q+2)2<\/sup> = 9q2<\/sup> + 12q + 4
\n= 3(3q2<\/sup> + 4q + 1) +1
\n=3m +1, where m = 3q2<\/sup> + 4q + 1
\nHence, it is proved.<\/p>\n

Question 18.
\nSolve : \\(\\frac{x+y}{x y}\\) = 2 and \\(\\frac{x-y}{x y}\\) = 6
\nAnswer:
\nx + y = 2xy
\nx – y = 6xy
\n2x = 8xy
\n2 = 8y
\n\"Karnataka<\/p>\n

Question 19.
\nSolve : y2<\/sup> – (\u221a3 + 1)y + \u221a3 = 0
\nAnswer:
\ny2<\/sup> – \u221a3y – y + \u221a3 = 0
\ny(y – \u221a3) – 1(y – \u221a3) = 0
\n(y – \u221a3)(y – 1) = 0
\ny – \u221a3 = 0, y – 1 = 0
\ny = \u221a3 and y = 1<\/p>\n

\"Karnataka<\/p>\n

Question 20.
\nShow that the points (3, 2) (-2, -3) and (2, 3) are collinear Or non – collinear.
\nAnswer:
\nLet A= (3, 2) B = (-2, -3) C = (2, 3)
\n\"Karnataka
\nAC \u2260 AB + BC
\n\u2234 They are non – collinear.<\/p>\n

Question 21.
\nIn the given fig \u2206DGH ~ \u2206DEF, DH = 8cm, DF = 12cm, DG = (3x – l)cm and DE = (4x +. 2)cm, Find the lengths of DG and DE.
\nAnswer:
\n\u2206DGH ~ \u2206DEF
\n\\(\\frac{\\mathrm{DG}}{\\mathrm{DE}}=\\frac{\\mathrm{DH}}{\\mathrm{DF}}\\)
\n\\(\\frac{3 x-1}{4 x+2}=\\frac{8}{12}\\)
\n\"Karnataka
\n12 (3x – 1) = 8(4x + 2)
\n36x – 12 = 32x + 16
\n36x – 32x = 16 + 12
\n4x = 28
\nx = 28\/4
\nx = 7
\n\u2234 DG = 3x – 1 = 3 x 7 – 1 = 21 – 1 = 20
\nDE = 4x + 2 = 4 x 7 + 2 = 28 + 2 = 30<\/p>\n

OR<\/p>\n

D is a point on the side BC of A ABC such that \u2220ADC = \u2220BAC. Prove that
\n\\(\\frac{\\mathbf{C A}}{\\mathbf{C D}}=\\frac{\\mathbf{C B}}{\\mathbf{C A}}\\)
\nAnswer:
\nIn \u2206 ABC and \u2206 DAC,
\n\u2220ADC = \u2220BAC and
\n\u2220C = \u2220C
\n\u2206ABC ~ \u2206DAC
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 22.
\nA card is drawn at random from a box containing 21 cards numbered 1 to 21. Find the probability that the card drawn is
\na) Prime number b) Divisible by 3.
\nAnswer:
\nn(s) = 21
\nA= {2,3,5,7,11,13,17,19}
\nn(A) = 8
\nP(A) = \\(\\frac{n(A)}{n(S)}=\\frac{8}{21}\\)<\/p>\n

b) n(S) = 21
\nB= {3,6, 9,12,15,18}
\nn(B) = 6
\n\u2234 P(B) = \\(\\frac{n(B)}{n(S)}=\\frac{6}{21}=\\frac{2}{7}\\)<\/p>\n

Question 23.
\nDraw a circle of radius 3cms. Construct a pair of tangents to it, from a point 5cm away from the circle.
\nAnswer:
\nr = 3cm d = 3 + 5 = 8cm
\n\"Karnataka<\/p>\n

Question 24.
\nExpress sinA and sec A in terms of cot A.
\nAnswer:
\nW.K.T cosec2<\/sup>A = 1 + cot2<\/sup>A
\n\"Karnataka<\/p>\n

OR<\/p>\n

If tan\u03b8 + sin\u03b8 = m and tan \u03b8 – sin \u03b8= n, S.T m2<\/sup> – n2<\/sup> = 4\\(\\sqrt{\\mathbf{m} \\mathbf{n}}\\)
\nAnswer:
\nL.H.S = m2<\/sup> – n2<\/sup> = (m +n) (m – n)
\n= {(tan \u03b8 + sin \u03b8 ) + (tan \u03b8 – sin \u03b8 )x(tan \u03b8 +sin \u03b8) – (tan \u03b8 – sin \u03b8)}
\n= {(2tan\u03b8) x (2sin\u03b8) = 4tan\u03b8sin\u03b8
\n\"Karnataka
\n= 4\\(\\sqrt{\\mathbf{m} \\mathbf{n}}\\)
\n= RHS.<\/p>\n

IV. Answer the following : ( 3 x 9 = 27 )<\/span><\/p>\n

Question 25.
\nThe sum of the numerator and denominator of a fraction is 24. If 4 is subtracted from the numerator and 5 from its denominator, then it reduces to 1\/4. Find the fraction.
\nAnswer:
\nLet the fraction be \\(\\frac{\\mathrm{x}}{\\mathrm{y}}\\)
\nsum of its numerator and denominator = 24.
\nx + y = 24 ……..(1)
\nIf 4 is subtracted from the numerator and
\n5 from its denominator we get = 1\/4
\n\u21d2 \\(\\frac{x-4}{y-5}=\\frac{1}{4}\\)
\n4(x – 4) = 1(y – 5)
\n4x – 16 = y – 5
\n4x – y = -5 + 16
\n4x – y = 11 ……..(2)
\nFrom (1) and (2)
\n\"Karnataka
\nConsider,
\nx + y = 24
\n7 + y = 24
\ny = 24 – 7
\ny = 17
\n\u2234 The fraction is \\(\\frac{x}{y}=\\frac{7}{17}\\)<\/p>\n

OR<\/p>\n

Two women and five men can together finish an embroidary work in 4 days. While three women and 6 men can finish in 3 days. Find the time taken by one women alone and also that taken by one man alone.
\nAnswer:
\nLet 1 woman can finish the embroidary in x days and 1 man can finish the embroidary in y days.
\n1 woman\u2019s 1 day\u2019s work = \\(\\frac{1}{x}\\)
\n1 man\u2019s 1 day\u2019s work = \\(\\frac{1}{y}\\)
\n\"Karnataka
\nLet \\(\\frac{1}{x}\\) = u, \\(\\frac{1}{y}\\) = v
\n2u + 5v = \\(\\frac{1}{4}\\)
\n3u + 6v = \\(\\frac{1}{3}\\)
\nsolve (1) and (2)
\n(8u + 20v = 1) x 9
\n(9u+ 18v= 1) x 8
\n\"Karnataka
\nv = \\(\\frac{1}{36}\\)
\nPut v = \\(\\frac{1}{36}\\)…….(1)
\n2u + 5v = \\(\\frac{1}{4}\\)
\n2u + 5\\(\\frac{1}{36}\\) = \\(\\frac{1}{4}\\)
\n\"Karnataka
\nThus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.<\/p>\n

\"Karnataka<\/p>\n

Question 26.
\nIf two zeroes of the polynomial p(x) = x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35 are 2 – \u221a3 and 2 + \u221a3. Find the other two zeroes.
\nAnswer:
\nThe two zeroes of p(x) are,
\n(2 – \u221a3) and (2 + \u221a3)
\n\u21d2 [x – (2 – \u221a3] and [x – [2 + \u221a3] are the factors p(x).
\n\u21d2 [ x – 2 + \u221a3] [x – 2 – \u221a3]
\n(x – 2)2<\/sup> – (\u221a3)2<\/sup>
\nx2<\/sup> – 4x + 4 – 3
\n\u21d2 x2<\/sup> – 4x + 1 is a factor of p(x)
\nDivide p(x) by x2<\/sup> – 4x + 1 to get quotient
\n\"Karnataka
\nConsider the quotient, x2<\/sup> – 2x – 35 = 0
\nx2<\/sup> – 7x + 5x – 35 = 0
\nx(x – 7) + 5(x – 7) = 0
\n(x – 7) (x + 5) = 0
\nx-7 = 0 orx + 5 = 0
\nx = 7 or x = -5
\n\u2234 The other two zeroes are 7, -5.<\/p>\n

Question 27.
\nA two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchange their places. Find the number.
\nAnswer:
\nLet the two digits be x and y.
\n\u21d2 The two digit number is = 10x + y
\nThe product of three digits = xy = 18
\nWhen 63 is subtracted from the number, the digits interchange their places.
\n10x + y – 63 = 10y + x
\n10x – x – 10y + y = 63
\n9x – 9y = 63
\n9(x – y) = 63
\nx – y = 63\/9 = 7
\nx – y = 7
\nx = 7 + y
\nConsider xy = 18 and substitute x = (y + 7) in it.
\ny(y + 7) = is . .
\ny2<\/sup> + 7y – 18 = 0
\ny2<\/sup> + 9y – 2y – 18 = 0
\ny (y + 9) – 2(y + 9) = 0 –
\n(y + 9) (y – 2) = 0 .
\ny + 9 = 0, y -2 = 0
\ny = -9 or y = 2
\nxy= 18
\nx(2) = 18
\nx = 18\/2 = 9
\nx = 2
\nBut negative digit is not considered.
\n\u2234 The two digit number is
\n= 10x + y
\n= 10(9) + 2 = 92<\/p>\n

OR<\/p>\n

A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km \/ hr from its usual speed. Find its usual speed.
\nAnswer:
\nLet the usual speed of the plane be x km \/ hr. Time taken to cover 1500 km with the usual speed.
\n\\(\\frac{1500}{x}\\) hrs
\nTime taken to cover 1500 km with the speed of (x + 250) km \/ hr = \\(\\frac{1500}{x+250}\\)
\n\"Karnataka
\n2(1500 x 250) = x2 + 250x
\nx2<\/sup> + 250x-750000 = 0
\nx2<\/sup> + 1000x – 750x – 750000 = 0
\nx(x + 1000) – 750(x + 1000) = 0
\n(x+ 1000)(x-750) = 0
\nx + 1000 = 0 or x – 750 = 0
\nx = -1000 or x= 750 ,
\nx = 750
\nHence, the usual speed of the plane = 750 km\/hr.<\/p>\n

\"Karnataka<\/p>\n

Question 28.
\nIf the co – ordinates of the mid points of \u2206 ABC are D(1, 2) E(0, -1) and (2, -1). Find the respective co – ordinates of \u2206 ABC.
\nAnswer:
\n\"Karnataka
\nD is the midpoint of BC.
\n\"Karnataka
\nx1<\/sub> + x2<\/sub> = 4 and y1<\/sub> + y2<\/sub> = -2 ………(1)
\nE is the mid point of BC
\n\"Karnataka
\nx2<\/sub> + x3<\/sub> = 0 and y2<\/sub> + y3<\/sub> = -2 …….(2)
\nF is the mid point of BC
\n\"Karnataka
\nx1<\/sub> + x3<\/sub> = 4 and y2<\/sub> + y3<\/sub> = -2 ….(3)
\nSolve (1) and (3)
\n\"Karnataka
\nx2<\/sub> = 0
\nx2<\/sub> = 0
\n0 + x3<\/sub> = 0
\nx3<\/sub> = 0
\nx1<\/sub> + x3<\/sub> = 4
\nx1<\/sub> + 0 = 4
\nx1<\/sub> = 4 – 0 = 4
\nx1<\/sub> = 4<\/p>\n

y2<\/sub> = -i
\ny2<\/sub> + y3<\/sub> = -2
\n-1 + y3<\/sub> = -2
\ny3<\/sub> = -2 +1
\ny3<\/sub> = -1
\ny1<\/sub> + y3<\/sub> = -2
\ny1<\/sub> + (-1) = -2
\ny1<\/sub> -1 = -2
\ny1<\/sub> = -2 + 1 = -1
\ny1<\/sub> = -1
\nA(x1<\/sub>,y1<\/sub>) = A(4,-1)
\nB (x2<\/sub> y2<\/sub>) = B(0, -1)
\nC(x3<\/sub>, y3<\/sub>) = C(0,-1)<\/p>\n

OR<\/p>\n

Find the length of the median through the vertex A(5,1) drawn to the triangle ABC where other two vertices are B(1, 5) andiC(-3, -1)
\nAnswer:
\n\"Karnataka<\/p>\n

Question 29.
\nProve that the tangents drawn from an external point are equal.
\nAnswer:
\n\"Karnataka
\nData : \u2018O is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
\nTo prove that : \u2220APB + \u2220AOB = 180\u00b0
\n\u2220OAP + \u2220OBP = 90\u00b0
\n(\u2235 Angle between the radius and tangent at the point of contact is 90\u00b0)
\nOP = OP (Common side)
\nOA = OB (Radii of the same circle)
\nAccording to RHS postulate
\n\u2206APO = \u2206BPO
\n\u2220 OAP = \u2220OBP = 90\u00b0
\n\u2220OAP + \u2220OBP = 90\u00b0 +90\u00b0 =180\u00b0
\n\u21d2 Opposite angles of OAPB quadrilateral are supplementary.
\nOAPB is a cyclic quadrilateral
\n\u21d2 \u2220 APB + \u2220 AOB = 180\u00b0<\/p>\n

\"Karnataka<\/p>\n

Question 30.
\nIf a chord of circle of radius 10cm subtend an angle of 60\u00b0 at the centre of the circle. Find the area of the corresponding segment of the circle.
\n(Take \u03c0 =3.14, \u221a3 = 1.7)
\nAnswer:
\nO is the centre of the circle having radius 10 cm.
\n\u2220 AOB = 60\u00b0 is angle subtended at the centre O of the circle by a chord AB.
\nDraw BL \u22a5 OA
\nIn right angled \u2206 OLB,
\n\"Karnataka
\n\"Karnataka
\nArea of segment \u2206 APB
\n= The area of sector OAPB – Area of \u2206 OAB
\n\"Karnataka<\/p>\n

OR<\/p>\n

Find the area of the shaded region where PQRS is a square of side lOcms and semicircles are drawn with each side of square as diameter.
\nAnswer:
\nThe four shaded regions are marked as I, II, III, IV.
\nAll the four regions have equal areas due to \u00a7 symmetry.
\nArea of I + Area of III
\n= Area of square ABCD – Sum of the areas of two semicircles of each of radius 5cm.
\n\"Karnataka
\n{10 x 10 – 2 x \\(\\frac{1}{2}\\) \u03c0 x 52<\/sup>}cm2<\/sup>
\n= {100 – 25 x 3.14} cm2<\/sup>
\n= {100 – 78.5} cm2<\/sup>
\n= 21.5 cm2<\/sup>
\nSimilarly Area of II + Area of IV = 21.5 cm2. Total Area of shaded region = 21.5 + 21.5
\n= 43 cm2<\/sup><\/p>\n

Question 31.
\nThe following table gives the production yield per hectare of paddy of 50 farms of a village.
\n\"Karnataka
\nAnswer:
\n\"Karnataka<\/p>\n

Question 32.
\nFind the mean of the following frequency distribution.
\n\"Karnataka
\nAnswer:
\n\"Karnataka
\nA = 50, h = 20, N = 100 and \u03a3 fi<\/sub>ui<\/sub> = 15
\nMean = A + h\\(\\left\\{\\frac{1}{\\mathrm{N}} \\sum \\mathrm{f}_{i} \\mathrm{u}_{\\mathrm{i}}\\right\\}\\)
\nMean = 50 + 20 x \\(\\frac{15}{100}\\)
\nMean = 50 + 3 = 53<\/p>\n

\"Karnataka<\/p>\n

Question 33.
\nConstruct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose Sides are 2\/3 of the corresponding sides of it.
\nAnswer:
\n\"Karnataka<\/p>\n

V. Answer the following : ( 4 x 4 = 16 )<\/span><\/p>\n

Question 34.
\nSolve the pair of equations graphically.
\n4x – 3y + 4 = 0
\n4x + 3y – 20 = 0
\nAnswer:
\n4x – 3y + 4 = 0 ….. (1)
\n4x + 4 = 3y
\ny = \\(\\frac{4 x+4}{3}\\)
\n\"Karnataka
\n4x + 3y – 20 = 0
\n3y = -4x + 20
\ny = \\(\\frac{-4 x+20}{3}\\)
\n\"Karnataka
\n\"Karnataka<\/p>\n

Question 35.
\nHow many terms of the series 93 + 90 + 87 + ……….. amounts to 975. Find also the last term.
\nAnswer:
\n93 + 90 + 87 + …………
\na = 93, d = 90 – 93 = -3, Sn<\/sub> = 975, n = ?,
\nT = ?
\nS = \\(\\frac{n}{2}\\)[2a + (n – 1)d]
\n975 = \\(\\frac{n}{2}\\)[2(93) + (n-1)(-3)]
\n975 x 2 = n[186 – 3n + 3]
\n1950 = n [189 – 3n]
\n1950 = n[189 – 3n]
\n1950 = 189n – 3n2
\n(3n2<\/sup>-189n+ 1950-0) H-3
\nn2<\/sup> – 63n + 650 = 0
\nn2<\/sup> – 50n-13n +650 = 0
\nn(n – 50) – 13(n – 50) = 0
\n(n – 50) (n -13) = 0
\nn – 50 = 0 or n – 13 = 0
\nn= 50 or n= 13
\nWhen n = 50, the last term
\nTn<\/sub> = a + (n – 1)d
\n= 93 + (49) (-3)
\n= 93 – 147
\nT50<\/sub> =-54
\nWhen n = 13, the last term
\nTn<\/sub> = a + (n – 1)d \u201d
\n=93 + (13-1) (-3)
\n= 93 – 36
\nT13<\/sub> = 57<\/p>\n

OR<\/p>\n

If m times the mth term of an A.P is equal to n times its n,h term, show that (m + n)th term is zero.
\nSoln: Let T1<\/sub> = a and c.d = d
\nGiven : m times m,h term = n times nthterm
\n=> mTm<\/sub> = nTn<\/sub>
\nm[a + (m – 1)d] = n[a + (n – 1)d]
\nm[a +(m – l)d] – n[a+(n -l)d] = 0
\nm[a + md – d] – n[a + nd – d] = 0
\nma + m2<\/sup>d – md – na – n2<\/sup>d + nd = 0
\na(m – n) + d(m – n) (m + n) – d(m – n) = 0
\n(m – n) {a + d(m +n) – d} = 0 . (m – n) {a + (m + n – l)d} = 0
\nm-n = 0, a + (m + n-l)d = 0
\nm= n Tm+n<\/sub> , = 0
\nBut m \u2260 n<\/p>\n

Question 36.
\nA tower is 50m high. Its shadow is x mtrs shorter when the suns altitude is 45\u00b0 than when it is 30\u00b0. Find the value of x.
\nAnswer:
\n\"Karnataka
\nLet PQ be the tower and let PA and PB be its shadows when the altitudes of the sum are 45\u00b0 and 30\u00b0 respectively.
\n\u2220 PAQ = 45\u00b0, \u2220 PBQ = 30\u00b0
\n\u2220 BPQ = 90\u00b0 and PQ = 50m ,
\nLet AB = x metres
\nFrom right angled \u2206 APQ, we have
\n\\(\\frac{\\mathrm{AP}}{PQ}\\) = cot 45\u00b0=1
\n\\(\\frac{\\mathrm{AP}}{50}\\) = 1 => AP = 50m
\nFrom right angled \u2206 BPQ,
\n\\(\\frac{\\mathrm{BP}}{\\mathrm{PQ}}\\) = cot30\u00b0 = \u221a3
\n\u21d2 \\(\\frac{x+50}{50}\\) = \u221a3
\nx = 50(\u221a3 – 1)m
\n= 50 (1.73 – l)m
\n= 50 x 0.732
\n= 36.6 m<\/p>\n

Question 37.
\nProve that \u201cIn a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides.\u201d
\nAnswer:
\n\"Karnataka
\nVerify
\nBC = 8cm
\nBC’= \\(\\frac{5}{3} \\times 8=\\frac{40}{3}\\)
\nAB =6cms
\nA’B= \\(\\frac{5}{3} \\times 6\\)
\n=10cms.<\/p>\n

\"Karnataka<\/p>\n

VI. Answer the following : ( 5 x 1 = 5 )<\/span><\/p>\n

Question 38.
\nThe height of cone is 20m. A small cone is cut off from it at its top by the plane parallel to the base. If the volume of small cone 1\/1000 th of the volume of given cone, at what height above the base the section is made.
\nAnswer:
\nIn A OPB and A OQC A OPB ~aOQC OP _ PB OQ ~ QC
\n\"Karnataka
\n\"Karnataka<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Karnataka SSLC Maths Model Question Paper 4 with Answers, 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 4 with Answers Time: 3 Hours Max Marks: 80 I. …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6040"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=6040"}],"version-history":[{"count":1,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6040\/revisions"}],"predecessor-version":[{"id":90890,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/6040\/revisions\/90890"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=6040"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=6040"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=6040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}