CBSE Sample Papers for Class 10 Maths<\/a>. Here we have given CBSE Sample Papers for Class 10 Maths Paper 5<\/p>\nTime Allowed : 3 hours<\/strong>
\nMaximum Marks : 80<\/strong><\/p>\nGeneral Instructions<\/strong><\/span><\/p>\n\n- All questions are compulsory.<\/li>\n
- The question paper consists of 30 questions divided into four sections – A, B, C and D.<\/li>\n
- Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.<\/li>\n
- There is no overall choice. However, an internal choice has been provided in 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.<\/li>\n
- Use of calculator is not permitted.<\/li>\n<\/ul>\n
SECTION-A<\/strong><\/p>\nQuestion 1.<\/strong><\/span>
\nAOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). Find the length of its diagonal CO.<\/p>\nQuestion 2.<\/strong><\/span>
\nIn the given figure, if \u2220AOB = 125\u00b0, then find the measure of \u2220DOC.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1952\/44477489474_2ee1995605_o.png?resize=212%2C151&ssl=1\")
<\/p>\nQuestion 3.<\/strong><\/span>
\nA vessel has 136 lit. of oil. Another vessel has 92 lit. of oil. What is the capacity of the largest possible ladle which can measure out all the oil in each vessel in exact number of times ?<\/p>\nQuestion 4.<\/strong><\/span>
\nFor what value of k, the system : kx + 3y = 11; 2x + 5y = 3, has unique solution.<\/p>\nQuestion 5.<\/strong><\/span>
\nIf one root of 2x\u00b2 + kx – 6 = 0 is 2, find the value of k.<\/p>\nQuestion 6.<\/strong><\/span>
\nEvaluate : sin 30\u00b0 cos 60\u00b0 + sin 60\u00b0 cos 30\u00b0.<\/p>\nSECTION-B<\/strong><\/p>\nQuestion 7.<\/strong><\/span>
\nIf the HCF of 85 and 153 is expressible in the form 85m – 153, find the value of m.<\/p>\nQuestion 8.<\/strong><\/span>
\nCheck whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle or not.<\/p>\nQuestion 9.<\/strong><\/span>
\nA bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is :
\n(i) Black ball
\n(ii) Red ball
\n(iii) Not a green ball
\n(iv) Green or white ball<\/p>\nQuestion 10.<\/strong><\/span>
\nA two digit number is selected at random. What is the chance that will be :
\n(i) A composite number which is odd
\n(ii) Successor of a prime number<\/p>\nQuestion 11.<\/strong><\/span>
\nCan x – 1 be the remainder on division of a polynomial p(x) by 2x + 3 ? Justify your answer.<\/p>\nQuestion 12.<\/strong><\/span>
\nFind k if the sum of roots of equation x\u00b2 – x + k(2x – 1) is Zero.<\/p>\nSECTION-C<\/strong><\/p>\nQuestion 13.<\/strong><\/span>
\nShow that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.<\/p>\nQuestion 14.<\/strong><\/span>
\nIf P(9a – 2, – b) divides line segment joining A(3a + 1,- 3) and B(8a, 5) in the ratio 3 : 1, find the value of a and b.<\/p>\nQuestion 15.<\/strong><\/span>
\nIn the given figure, D and E trisect BC. Prove that 8AE\u00b2 = 3AC\u00b2 + 5AD\u00b2
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1962\/44477489374_62933b8cdd_o.png?resize=171%2C149&ssl=1\")
<\/p>\nQuestion 16.<\/strong><\/span>
\nIn the given figure, AB is a chord of the circle and AOC is its diameter such that \u2220ACB = 50\u00b0. If AT is the tangent to the circle at the point A, then find the measure of \u2220BAT.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1927\/43386263510_3803ff8daa_o.png?resize=189%2C182&ssl=1\")
\nA triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1907\/44477489284_5cf3437cc5_o.png?resize=221%2C218&ssl=1\")
<\/p>\nQuestion 17.<\/strong><\/span>
\nA path of 4 m width runs round a semi-circular grassy plot whose circumference is \\(163 \\frac { 3 }{ 7 } \\) m.
\nFind :
\n(i) The area of the path.
\n(ii) The cost of gravelling the path at the rate of Rs 1.50 per m\u00b2.
\n(iii) The cost of turfing the plot at the rate of 45 paise per m\u00b2.<\/p>\nOR<\/strong><\/p>\nAll the vertices of a square lie on a circle. Find the area of the square, if area of the circle is 1256 cm\u00b2. (Use \u03c0 = 3.14)<\/p>\n
Question 18.<\/strong><\/span>
\n500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3 ?<\/p>\nOR<\/strong><\/p>\nHow many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm ?<\/p>\n
Question 19.<\/strong><\/span>
\nShow that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.<\/p>\nOR<\/strong><\/p>\nIf sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.<\/p>\n
Question 20.<\/strong><\/span>
\nSolve for x :\\(2\\left( { x }^{ 2 }+\\frac { 1 }{ { x }^{ 2 } } \\right) -9\\left( x+\\frac { 1 }{ x } \\right) +14=0\\)<\/p>\nQuestion 21.<\/strong><\/span>
\nIf tan (A + B) = \u221a3 and tan (A – B) = \\(\\frac { 1 }{ \\sqrt { 3 } } \\);0\u00b0 < A + B \u2264 90\u00b0; A > B, find sin 2A and cos 6B.<\/p>\nQuestion 22.<\/strong><\/span>
\nFind the mean of the following data :
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1941\/44477489224_d50f2e6938_o.png?resize=381%2C165&ssl=1\")
<\/p>\nSECTION-D<\/strong><\/p>\nQuestion 23.<\/strong><\/span>
\nThe angle of elevation of a stationary cloud from a point 2500 m above a lake is 15\u00b0 and the angle of depression of its reflection in the lake is 45\u00b0. What is the height of the cloud above the lake level ?<\/p>\nOR<\/strong><\/p>\nThe angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60\u00b0. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45\u00b0. Calculate the height of the tower.<\/p>\n
Question 24.<\/strong><\/span>
\nConstruct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5\/3 times the corresponding sides of the given triangle.<\/p>\nQuestion 25.<\/strong><\/span>
\nIn a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects separately.<\/p>\nQuestion 26.<\/strong><\/span>
\nA metallic right circular cone 20 cm high and whose vertical angle is 60\u00b0 is cut into two parts at the middle of its height by a plane parallel of its base. If the frustum so obtained be drawn into a wire of diameter 1\/16 cm, find the length of the wire.<\/p>\nOR<\/strong><\/p>\nA pen stand made of wood is in the shape of a cuboid with four conical depressions and a cuboidal depression to hold the pens and pins, respectively. The dimensions of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.<\/p>\n
Question 27.<\/strong><\/span>
\nSolve graphically x – y = 0, 2x + 3y – 30 = 0. Also find coordinates of points where 2x + 3y – 30 = 0 meets axis of X and Y.<\/p>\nOR<\/strong><\/p>\nStudents of a class are made to stand in rows. If 4 students are extra in each row, there would be two rows less. If 4 students are less in each row, there would be 4 more rows. Find the number of students in the class.<\/p>\n
Question 28.<\/strong><\/span>
\nTwo pipes running together can fill a tank in \\(5 \\frac { 5 }{ 21 } \\) minutes. If one pipe takes 1 minute more than the other, then find the time in which each pipe would individually fill the tank.<\/p>\nQuestion 29.<\/strong><\/span>
\nIf sin \u03b1 = a sin \u03b2 and tan \u03b1 = b tan \u03b2, then prove that cos\u00b2 \u03b1 = \\(\\frac { { a }^{ 2 }-1 }{ { b }^{ 2 }-1 } \\)<\/p>\nQuestion 30.<\/strong><\/span>
\nDraw more than ogive’ and ‘less than ogive’ for the following distribution on the same graph. Also find the median from the graph.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1921\/45199812201_5f00078fcb_o.png?resize=258%2C171&ssl=1\")
<\/p>\nSOLUTIONS<\/strong><\/span><\/p>\nSECTION-A<\/strong><\/p>\nSolution 1:<\/strong><\/span>
\nWe know, diagonals of rectangle are equal.
\nOC = AB
\n=> \\(OC=\\left| \\sqrt { { (5-0) }^{ 2 }+{ (0-3) }^{ 2 } } \\right| \\)
\n=> \\(OC=\\left| \\sqrt { 25+9 } \\right| =\\left| \\sqrt { 34 } \\right| \\)
\n=> \\(OC=\\sqrt { 34 } \\) units
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1978\/31325902158_a9067a615a_o.png?resize=232%2C176&ssl=1\")
<\/p>\nSolution 2:<\/strong><\/span>
\nGiven
\n\u2220AOB = 125\u00b0
\nWe know, opposite sides of a quadrilateral circumscribing a circle subtend supplements angles at the centre of a circle.
\n\u2234\u2220AOB +\u2220DOC = 180\u00b0
\n=> 125\u00b0 + \u2220DOC = 180\u00b0
\n=> \u2220DOC = 180\u00b0 – 125\u00b0
\n= 55\u00b0 Ans.<\/p>\nSolution 3:<\/strong><\/span>
\nLargest capacity of the ladle which can measure out all the oil
\n= H.C.F. of 136 and 92
\nNow, 136 = 2 x 2 x 2 x 17
\nand 92 = 2 x 2 x 23
\nH.C.F. (136, 92) = 4
\n\u2234 4 litre capacity ladle can measure out all the oil in each vessel in exact number of times. Ans.<\/p>\nSolution 4:<\/strong><\/span>
\nGiven equations are,
\nkx + 3y – 11 = 0
\nand 2x + 5y – 3 =0
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1949\/45200030111_2f9ca37eea_o.png?resize=767%2C100&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1908\/45150269572_06699ca271_o.png?resize=385%2C164&ssl=1\")
<\/p>\nSolution 5:<\/strong><\/span>
\nGiven equation is,
\n2x\u00b2 + kx – 6 = 0
\nOne root of the equation is 2.
\nx = 2
\n2(2)\u00b2 + k(2) – 6 = 0
\n8 + 2k – 6 = 0
\n2k = – 2
\nk = – 1.<\/p>\nSolution 6:<\/strong><\/span>
\nWe have,
\nsin 30\u00b0 cos 60\u00b0 + sin 60\u00b0 cos 30\u00b0
\n=> \\(\\frac { 1 }{ 2 } .\\frac { 1 }{ 2 } +\\frac { \\sqrt { 3 } }{ 2 } .\\frac { \\sqrt { 3 } }{ 2 } \\)
\n=> \\(\\frac { 1 }{ 4 } +\\frac { 3 }{ 4 } \\)
\n=> 1<\/p>\nSECTION-B<\/strong><\/p>\nSolution 7:<\/strong><\/span>
\n85 = 5 x 17
\n153 = 3 x 3 x 17
\n\u2234H.C.F. (85, 153) = 17
\nBut
\n\u2234H.C.F. = 85m – 153
\n17 = 85m – 153
\n17 + 153 = 85m
\n\\(m= \\frac { 170 }{ 85 } \\)
\n= 2<\/p>\nSolution 8:<\/strong><\/span>
\nLet the given points be A(5, – 2), B(6, 4), C(7, – 2).
\nNow,
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1905\/44288727895_04d1ed1472_o.png?resize=442%2C238&ssl=1\")
\nHere, we observe that
\nAB = BC
\nand AB+BC\u2260AC
\n\u2234Points A,B,C forms an isosceles triangle with AB = BC<\/p>\nSolution 9:<\/strong><\/span>
\nWe have,
\nNumber of red balls = 5
\nNumber of white balls = 8
\nNumber of green balls = 4
\nNumber of black balls = 7
\nTotal balls = 24
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1936\/44478013324_a7480dfd98_o.png?resize=656%2C345&ssl=1\")
<\/p>\nSolution 10:<\/strong><\/span>
\nTotal two-digit numbers 10 to 99 = 90
\n\u2234n(S) = 90
\n(i) Composite numbers which is odd = {15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}
\n\u2234 n(E) = 24
\nProbability of odd composite number
\n= \\(\\frac { n(E) }{ n(S) } =\\frac { 24 }{ 90 } =\\frac { 4 }{ 15 } \\)
\n(ii) Successor of a prime number = {12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98}
\n\u2234 n(E) = 21
\nProbability of a successor of a prime number
\n= \\(\\frac { n(E) }{ n(S) } =\\frac { 21 }{ 90 } =\\frac { 7 }{ 30 } \\)<\/p>\nSolution 11:<\/strong><\/span>
\nNo, degree of the remainder is always less than degree of the divisor.<\/p>\nSolution 12:<\/strong><\/span>
\nGiven equation is,
\nx\u00b2 – x + k(2x – 1) = 0
\n=> x\u00b2 – x + 2kx – k = 0
\n=> x\u00b2 + (2k – 1)x – k = 0
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1926\/30261113107_506a755162_o.png?resize=784%2C249&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1944\/44288727655_671db385bf_o.png?resize=197%2C106&ssl=1\")
<\/p>\nSECTION-C<\/strong><\/p>\nSolution 13:<\/strong><\/span>
\nLet q be the quotient and r be the remainder when n is divisible by 3.
\nTherefore, n = 3q + r, where r = 0, 1, 2
\n=> n = 3q or n = 3q + 1 or n = 3q + 2
\nCase I: If n = 3q, then n is divisible by 3 but n + 2 = 3q + 2 and n + 4 = 3a + 4 are not divisible by 3.
\nCase II: If n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and n + 4 = 3q + 5, which is not divisible by 3.
\nSo, only (n + 2) is divisible by 3.
\nCase III: If n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and n + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
\nSo, only (n + 4) is divisible by 3.
\nHence, one and only one out of n, (n + 2), (n + 4) is divisible by 3.<\/p>\nSolution 14:<\/strong><\/span>
\nBy section formula,
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1942\/44288944405_d9ba8c7a1b_o.png?resize=520%2C521&ssl=1\")
<\/p>\nSolution 15:<\/strong><\/span>
\nHere, D and E trisect BC.
\n=>BD = DE = EC = \\(\\\\ \\frac { 1 }{ 3 } \\) BC
\nor BE = 2 BD,
\nBC = 3BD
\nIn right \u2206ABD,
\nwe have
\nAD\u00b2 = AB\u00b2 + BD\u00b2….(i)
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1938\/45150816142_e12a65e342_o.png?resize=780%2C360&ssl=1\")
<\/p>\nSolution 16:<\/strong><\/span>
\nIn the given circle,
\n\u2220ABC = 90\u00b0
\n\u2220ACB + \u2220CAB = 90\u00b0
\n50\u00b0 + \u2220CAB = 90\u00b0
\n\u2220CAB = 90\u00b0 – 50\u00b0
\n\u2220CAB = 40\u00b0
\nNow, OA is radius of the circle and AT is the tangent of circle
\n\u2220OAT = 90\u00b0
\n\u2220OAB + \u2220BAT = 90\u00b0
\n40\u00b0 + \u2220BAT = 90\u00b0
\n\u2220BAT = 90\u00b0 – 40\u00b0
\n\u2220BAT = 50\u00b0
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1922\/45150816312_81697656e0_o.png?resize=267%2C232&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1939\/45150816792_da9e54bb03_o.png?resize=791%2C497&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1936\/45150817122_0729ba5e25_o.png?resize=645%2C682&ssl=1\")
<\/p>\nSolution 17:<\/strong><\/span>
\nWe have
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1969\/45150815752_5762f69629_o.png?resize=711%2C468&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1948\/45150817462_84932802f4_o.png?resize=803%2C593&ssl=1\")
<\/p>\nSolution 18:<\/strong><\/span>
\nGiven
\nlength of pond, l = 80m
\nbreadth of pond, b = 50m
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1915\/45150814962_a8ca4744a9_o.png?resize=501%2C546&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1969\/45150815182_4789bb15c4_o.png?resize=461%2C291&ssl=1\")
<\/p>\nSolution 19:<\/strong><\/span>
\nOdd integers between 1 and 1000 divisible by 3 are 3, 9,15, 21, … 999
\nThe above series form an A.P.
\nwhere a = 3, d = 9 – 3 = 6 and an = 999
\nWe know
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1960\/45150813802_17d85e0c48_o.png?resize=500%2C383&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1962\/45150814702_242d037d7c_o.png?resize=796%2C405&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1909\/45150813982_e82674608d_o.png?resize=510%2C332&ssl=1\")
<\/p>\nSolution 20:<\/strong><\/span>
\nWe have
\n\\(2\\left( { x }^{ 2 }+\\frac { 1 }{ { x }^{ 2 } } \\right) -9\\left( x+\\frac { 1 }{ x } \\right) +14=0\\)
\n\\(2\\left( { x }^{ 2 }+\\frac { 1 }{ { x }^{ 2 } } +2 \\right) -9\\left( x+\\frac { 1 }{ x } \\right) +14-4=0\\)
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1949\/45150813402_3dc7ce139f_o.png?resize=442%2C717&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1978\/44288943595_4d12883657_o.png?resize=378%2C220&ssl=1\")
<\/p>\nSolution 21:<\/strong><\/span>
\nWe have
\ntan(A+B) = \u221a3
\n=> tan(A+B) = tan 60\u00b0
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1918\/45150813392_8e85fdcfa5_o.png?resize=765%2C450&ssl=1\")
<\/p>\nSolution 22:<\/strong><\/span>
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1973\/45150812952_dbe0596217_o.png?resize=647%2C359&ssl=1\")
<\/p>\nSECTION-D<\/strong><\/p>\nSolution 23:<\/strong><\/span>
\nLet B be the point of oservation, P be the cloud and P’ be its reflection in the take.
\nAB = 2500 m
\n\u2220PBC = 15\u00b0
\nand \u2220CBP = 45\u00b0
\nLet the height of cloud from lake be h m.
\nOP = OP\u2019= h m
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1961\/44478714894_f6d1ecdbfa_o.png?resize=255%2C228&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1943\/44478715074_02a600e0be_o.png?resize=480%2C590&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1931\/44478715174_49786959cd_o.png?resize=732%2C373&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1959\/44478715444_46a2eb6f47_o.png?resize=774%2C387&ssl=1\")
<\/p>\nSolution 24:<\/strong><\/span>
\nSteps, of construction :
\n(1) Draw a line segment AB = 4 cm.
\n(2) Construct \u2220XAB = 90\u00b0.
\n(3) With A as centre, draw an arc of radius 3 cm intersecting AX at C.
\n(4) Join BC. Thus, \u2206ABC is obtained.
\n(5) Draw an acute angle \u2220BAY below of AB.
\n(6) Mark points A1, A2, A3, A4, A5 on AY such that
\nAA1 = A1A2 = A2A3 = A3A4 = A4A5.
\n(7) Join A3B.
\n(8) Draw a line through A5, parallel to A3B intersecting extended line segment AB at B’.
\n(9) Through B’, draw a line parallel to BC intersecting AX at C’.
\n(10) Thus, \u2206AB’C’ is obtained whose sides are \\(\\\\ \\frac { 5 }{ 3 } \\) times the corresponding sides of \u2206ABC.
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1941\/44478715614_63e58bcd0f_o.png?resize=216%2C367&ssl=1\")
<\/p>\nSolution 25:<\/strong><\/span>
\nLet the marks obtained in Maths be x, then marks obtained in English = 40 – x
\nAccording to the question,
\n(x + 3) (40 – x – 4) = 360
\n=> (x + 3) (36 – x) = 360
\n=> – x\u00b2 + 36x + 108 – 3x = 360
\n=> – x\u00b2 + 33x – 252 = 0
\n=> x\u00b2 – 21x – 12x + 252 = 0
\n=> x(x – 21) – 12(x – 21) = 0
\n(x – 21) (x – 12) = 0
\nEither x – 21 = 0 or x – 12 = 0
\n=> x = 21 or x = 12
\nIf x = 21, then marks obtained in Maths = 21
\nand marks obtained in English = 40 – 21 = 19
\nIf x = 12, then marks obtained in Maths = 12
\nand marks obtained in English = 40 – 12 = 28<\/p>\nSolution 26:<\/strong><\/span>
\nGiven
\nHeight of the cone = 20 cm
\nPQ (h) = 10 cm
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1975\/44478715904_2edcec6dae_o.png?resize=757%2C743&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1968\/44478716044_03b58069d6_o.png?resize=471%2C357&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1940\/44478716384_0095caa0f5_o.png?resize=700%2C130&ssl=1\")
<\/p>\nSolution 27:<\/strong><\/span>
\nGiven equations are
\nx – y = 0
\ny = x
\n=> x – y = 0
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1935\/44478714834_508232242b_o.png?resize=571%2C386&ssl=1\")
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1966\/44478717824_cfebf2840c_o.png?resize=635%2C573&ssl=1\")
\n![\"CBSE](\"https:\/\/i0.wp.com\/farm2.staticflickr.com\/1972\/44478718194_dbd64e7d1d_o.png?resize=781%2C557&ssl=1\")
<\/p>\nSolution 28:<\/strong><\/span>
\nLet one pipe fill the tank in x minute
\nthen other will fill the tank in (x + 1) minutes.
\nAccording to the question,
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1922\/44478714154_307bdcbfb3_o.png?resize=431%2C457&ssl=1\")
<\/p>\nSolution 29:<\/strong><\/span>
\nGiven : sin \u03b1 = a sin \u03b2
\nsin\u00b2 \u03b1 = > a\u00b2 sin\u00b2 \u03b2
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1966\/44478714024_dc66ccb823_o.png?resize=789%2C421&ssl=1\")
\n![\"CBSE](\"https:\/\/i2.wp.com\/farm2.staticflickr.com\/1944\/43387441650_5364f3cd37_o.png?resize=546%2C743&ssl=1\")
<\/p>\nSolution 30:<\/strong><\/span>
\n![\"CBSE](\"https:\/\/i1.wp.com\/farm2.staticflickr.com\/1973\/44478713784_512855816a_o.png?resize=492%2C626&ssl=1\")
<\/p>\nWe hope the CBSE Sample Papers for Class 10 Maths Paper 5 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 5, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"
CBSE Sample Papers for Class 10 Maths Paper 5 These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 5 Time Allowed : 3 hours Maximum Marks : 80 General Instructions All questions are compulsory. The question paper consists of …<\/p>\n
CBSE Sample Papers for Class 10 Maths Paper 5<\/span> Read More »<\/a><\/p>\n","protected":false},"author":5,"featured_media":7756,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nCBSE Sample Papers for Class 10 Maths Paper 5 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/cbse-sample-papers-for-class-10-maths-paper-5\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CBSE Sample Papers for Class 10 Maths Paper 5 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"CBSE Sample Papers for Class 10 Maths Paper 5 These Sample Papers are part of CBSE Sample Papers for Class 10 Maths. 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