Mathematics. Mock Paper. With. Blue Print of Original Paper. on Latest Pattern. Solution Visits:

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1 10 th CBSE{SA I} Mathematics Mock Paper With Blue Print of Original Paper on Latest Pattern Solution Visits: S.C.O. - 36, Sector 40 D, Chd. Phone: ,

2 10 th CBSE First Term {SA- I} Blue Print Topic/Unit MCQs SA(1) SA(II) LA Total Number System () 1() (6) 5(10) Algebra () (4) (6) (8) 8(0) Geometry 1(1) (4) (6) 1(4) 6(15) Trigonometry 4(4) 1() (6) (8) 9(0) Statistics 1(1) (4) (6) 1(4) 6(15) Total 10(10) 8(16) 10(30) 6(4) 34(80) S.C.O. - 36, Sector 40 D, Chd. Phone: ,

3 General instructions: Time: 3hrs. M: M: 90 (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into sections A, B, C and D. Section-A comprises of 8 questions of 1 mark each, section-b comprises of 6 questions of two marks each, section-c comprises of 10 questions of three marks each and section-d comprises of 10 questions of four marks each. (iii) Question numbers 1 to 8 in section-a are multiple choice questions where you are required to select one option out of the given four. (iv) There is no overall choice. However, internal choice has been provided in 1 question of two marks, 3 questions of three marks each and two questions of four marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculator is not permitted. S.C.O. - 36, Sector 40 D, Chd. Phone: ,

4 Section-A Questions number 1 to 8 carry one marks each 1. The pair of equation x and x 4 has : (a) infinitely many solutions (b) no solution (c) two solutions (d) one solution (b). If sides of two similar triangles are in ratio 4 : 9, then area of these triangles are in the ratio (a) : 3 (b) 4 : 9 (c) 81 : 16 (d) 16 : 81 (d) Ratio of Areas The x-coordinate of the point of intersection of more than and less then ogive is : (a) mode (b) mean (c) median (d) Variance (c) 4. If LCM (54, 336) 304, then HCF (54, 336) is: (a) 54 (b) 6 (c) 336 (d) 36 (b) L.C.M H.C.F Product of two numbers H.C.F H.C.F 6 S.C.O. - 36, Sector 40 D, Chd. Phone: ,

5 5. Value of 5 tan A 5sec A is : (a) 1 (b) 0 (c) 5 (d) 5 (c) 5 tan A 5 (1 + tan A) 5 (tan A 1 tan A) 5 6. If sina 3, then A equal to : (a) 90 o (b) 60 o (c) 45 o (d) 30 o (d) sin A 3 SinA Sin60 0 A If sec A q p, then value of 1 p cos A is: (a) 1 p (b) 1 q (c) 1 pq (d) p q (b) 1 1 p 1 cos A p p q q 8. If a positive integer n is divided by, then the remainder can be : (a) 1 or (b) 1,, or 3 (c) 0 or 1 (d) (c) S.C.O. - 36, Sector 40 D, Chd. Phone: ,

6 Section-B Questions number 9 to 14 carry two marks each 9. In the given figures, find measure of X. In PQR and zyx PQ 4. 1 ZY 8.4 PR ZX 6 3 QR 7 1 YX 14 PQ PR QR ZY ZX YX PQR ZYX (By SSS similarity criteria) PRQ ZXY..(1) (By CPST) In PQR PQR + PRQ + RPQ (Angle sum property of ) PRQ PRQ 50..() From (1) & () 0 ZXY X Is a composite number? Justify your answer S.C.O. - 36, Sector 40 D, Chd. Phone: ,

7 13( ) 13(77 + 1) 13(78) As, the no is having more than prime factors, The given no. is a composite no. 11. Solve: and 19 x y x y 3 0 x y 5 19 x y Put 1 a & 1 x y b..(1)..() in (1) & () 3a b 0..(3) a + 5b 19..(4) Multiply (3) & (4) by 5 & respectively, (3a b 0) 5 15a 10b 0..(5) & (a + 5b 19) 4a + 10b 38 (6) Adding (5) & (6) 15a 10b 0 4a 10b 38 19a 38 a Put a in () 3() b 0 6 b 0 S.C.O. - 36, Sector 40 D, Chd. Phone: ,

8 b 6 b x a 1 1 & y b 3 1. The following distribution give the daily income of 50 workers of a factory: Daily income in Rs Number of workers Write the above distribution as more than type cumulative frequency distribution. Daily income (in More than type fi Cf Rs) distribution More than 100 More than 10 More than 140 More than 160 More than Total If 1 and are zeroes of polynomial 4x x + (k 4). Find k. If f(x) 4x x + (k 4) where, x, 1 According to relationship between zeroes & coefficients of a polynomial, 1 c a S.C.O. - 36, Sector 40 D, Chd. Phone: ,

9 1 k k 4 k 8 Value of k If tan A cot (A 18 o ), where A is an acute angle, find the value of A. 0 tan A cot A 18 0 cot 90 A cot (A 18) 90 A A A 3A 108 A 36 0 Or If is an acute angle and sin cos, find the value of 3 tan + sin 1. sin cos (1) (Given) 3 tan sin 1 sin 3 sin sin cos cos [ sin cos 1] sin 3 sin sin cos sin cos from(1) cos 3 3 sin sin sin cos Value of 3tan sin S.C.O. - 36, Sector 40 D, Chd. Phone: ,

10 Section-C Questions number 15 to 4 carry three marks each 15. A survey regarding the height (in cm) of 50 girls of class X of a school was conducted and the following data was obtained: Height(in cm) Total Number of girls Find the mode of the data. Model class f1 0 f0 1 f 8 h 10 l 150 f1 f0 Mode l f f f 1 0 h cm 16. Prove that : To prove : L. H. S. cot A cos A coseca 1 cot A cos A coseca 1. cot A cos A cosec A 1 cot A cos A cosec A 1 cot A cos A cot A cos A S.C.O. - 36, Sector 40 D, Chd. Phone: ,

11 cos A cos A sin A cos A cos A sin A cos A cos A sin A cos A sin A sin A cos A sin A cos A sin A cos A cos A sin A cos A cos A 1 sin A cos A 1 sin A 1 1 cosec A 1 1 cosec A cosec A 1 cosec A cosec A 1 cosec A cosec A 1 cosec A 1 Hence proved. R. H. S. 17. In the given figure, o ACB 90 and CD AB. Prove that BC AC BD AD S.C.O. - 36, Sector 40 D, Chd. Phone: ,

12 Sol: ADC ~ ACB AC AD CD AB AC BC AC AD.AB.(1) CDB ~ ACB CD BC BD CA AB BC BC AB.BD Equation ()/(1) we get.() BC AC BC AC AB.BD AD.AB BD AD In the given figure, if AD BC, prove that AB +CD BD +AC. Or S.C.O. - 36, Sector 40 D, Chd. Phone: ,

13 Given : In ABC, AD BC To Prove : AB + CD BD + AC Proof : In ADC, AD CD (Given) AC CD + AD (By Pythagoras theorem) AD AC CD..(1) In ADB, AD BD (Given) AB AD + BD (By Pythagoras theorem) AD AB BD..() From (1) & () AC CD AB BD AC + BD AB + CD AB + CD BD + AC Hence, proved. S.C.O. - 36, Sector 40 D, Chd. Phone: ,

14 18. In the given figure, the line segment XY is parallel to AC of a ABC and it divides the triangle into two parts of equal area. Find AX AB. Let Area of BXY k Area of BAC k In XBY and ABC XBY ABC (Common angle) BXY BAC (Corresponding angles) BXY ~ BAC by AA similarity criteria Area of BXY BX Area of BAC AB K k BX AB BX 1 AB S.C.O. - 36, Sector 40 D, Chd. Phone: ,

15 AB AX 1 AB AB AX AB AX 1 AB 1 AX AB AX AB 19. Prove that: sin cos 3 3 sin cos tan To prove : 3 sin sin 3 cos cos tan Proof : 3 sin sin 3 cos cos sin 1 sin cos cos cos tan tan tan 1 sin 1 sin 1 1 sin sin 1 1 sin 1 sin tan R. H. S. Hence, proved. 0. Find the mean of the following data: S.C.O. - 36, Sector 40 D, Chd. Phone: ,

16 Class- interval Frequency Or Find the missing frequency for the given data if mean of distribution is 5. Wages (in Rs.) No. of workers f 6 13 Class interval Frequency (fi) xi fixi fi 50 fixi 760 Mean xf f i i i S.C.O. - 36, Sector 40 D, Chd. Phone: ,

17 Or Wages in Rs. No. of workers xi di xi 45 fidi f fi 33 f fidi 80 fd i i Mean A + f f f f 80 7f f 7 i 1. Prove that 5 is an irrational number. Let us assume the 5 is a rational number. Then, according to Euclid s division lemma, use can find two co-prime integers, such them, 5 5 a b a 5 b b a b S.C.O. - 36, Sector 40 D, Chd. Phone: ,

18 As, a & b are integers, this implies, (b a)/b is a rational no. 5 is also a rational number. But it contradicts the fact the 5 is irrational. This contradiction has arisen due to our incorrect assumption that number. 5 is a rational 5is an irrational number.. If one zeros of polynomial p(x) 3x 8x + k + 1 is seven times of the other, then find the zeroes and the value of k. p(x) 3x 8x + (k + 1) x, 7 b 7 a 8 x 1 3 x 7 And, k c k 1 a 3..(1)..() (From (1) & ()) 7 6k + 3 6k 4 k 3 Value of k 3 S.C.O. - 36, Sector 40 D, Chd. Phone: ,

19 3. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Find the speed of the stream and that of the boat in still water. Speed of stream y km/hr Speed of boat x km/hr 5 D T Case I x y x y Case II x y x y Let 1 1 a, b x y x y 1 1 a, b 5 11 on solving x 8 km/hr y 3 km/hr Or Solve by cross multiplication method: ax + by a b; bx ay a + b ax by a b bx ay a b S.C.O. - 36, Sector 40 D, Chd. Phone: ,

20 x y ab b a ab ab b a ab 1 a b x y 1 a b a b a b I II III Equating I & III x 1 a b a b x 1 Equating II & III y 1 a b a b y 1 Values of x 1 & y 1 sin A cos(90 A)cos A cos A sin(90 A)sin A 4. Evaluate: sina cosa. sec(90 A) cosec(90 A) sin A cos A sin A sin A cos A cosec A cos A cos A sin A sec A sin A cos A - sin 3 A cos A cos 3 A sin A sin A cos A sin A cos A (sin A + cos A) sin A cos A sin A cos A S.C.O. - 36, Sector 40 D, Chd. Phone: ,

21 Section-D Questions number 5to 34 carry Four marks each 5. Draw more than ogive and less than ogive for the following distribution and hence obtain the median. Class-interval Frequency Sol: Less then type of distribution Class interval Less than type f.i c.f distribution 5 10 Less than Less than Less than Less than Less than Less than Less than We will plot the points (10, ); (15, 14); (0, 16); (5, 0); (30, 3) (35, 7); (40, 30) More than type of distribution Class interval More than type f.i c.f distribution 5 10 More than More than More than More than More than More than S.C.O. - 36, Sector 40 D, Chd. Phone: ,

22 35 40 Less than Total 30 We will plot the points (5, 30), (10, 8), (15, 16), (0, 14), (5, 10), (30, 7), (35, 3) sin sin (90 ) 3cot 30 sin 54 sec Evaluate: o o o o 3 sec 61 cot 9 cosec 65 tan 5 o o o. Or 1 cos A sin A 1 sin A Prove that : sin A cos A 1 cos A Sol: sin cos 3 3 sin 54 3 sec 61 tan 61 cos 36 sec 5 tan cos sin S.C.O. - 36, Sector 40 D, Chd. Phone: ,

23 1 9 cos 36 3 cos To prove : 1 cos A sin A 1 sin A sin A cos A 1 cos A Or L.H.S. 1 sin A cos A 1 sin A cos A sin A cos A sin A cos A 1 1 sin A cos A sin A cos A 1 1 sin sin A cos A sin A cos A sin Acos A 1 sin A sin A sin A 1 1 sin Acos A sin A sin A 1 sin Acos A sin A 1 R.H.S. cos A Hence proved 7. If tan A + sin A m and tan A sin A n, Show that m n 4 mn. Sol: Given : sin A tan A m & tan A sin A n To prove : m n 4 mn Proof: S.C.O. - 36, Sector 40 D, Chd. Phone: ,

24 m n L.H.S. sin A tan A tan A sin A sin A tan A sin A tan A tan A sin A sin A tan A 4sinA tana R.H.S. 4 mn 4 tan A sin A tan A sin A 4 tan A sin A sin A 4 sin A cos A 4 sin A sin Acos A cos A 4 sin A 1 cos A cos A 4 sin A sin A cos A 4 sin A tan A 4sin A tan A L.H.S. Hence proved 8. If the median of the distribution given below is 3.5. find x and y. Class interval Total Frequency x y S.C.O. - 36, Sector 40 D, Chd. Phone: ,

25 Sol: Class interval f.i c.f 0 10 x x x x x y x+ y x+ y x+ y x+ y x + y 9 Median x 3, x + y y 9 y 6 0 x In the given figure, AB PQ CD, AB x units, CD y units and PQ z units, prove that, x y z S.C.O. - 36, Sector 40 D, Chd. Phone: ,

26 Or In an equilateral triangle ABC, D is a point on BC, such that BD 1 BC. Prove that 9 AD 7 3 AB. ABD ~ PQD x z l m m xm z (l + m) x m zl + zm xm zm zl m (x z) zl S.C.O. - 36, Sector 40 D, Chd. Phone: ,

27 m z..(1) l x z BPQ ~ BCP z l y l m z(l + m) yl zl + mz yl mz yl zl mz l(y z) m y z l z..() From (1) & () z y z x z z y z x z z xy yz xz + z xy yz +zx xy yz zx xyz xyz xyz x y S.C.O. - 36, Sector 40 D, Chd. Phone: ,

28 Or Sol: Given : Equilateral ABC, BD 1 BC 3 To prove : 9AD 7AB Construction: Draw AM BC. Proof: BD 1 BC (1)(Given) 3 BM 1 BC () [Altitude of an equilateral is also its median] DM BM BD DM BC BC 3 [From (1) and ()] DM 3BC BC BC 6 6.(5) In ADM,AM DM AD AM + DM (By Pythagoras theorem) AM AD DM..(3) Similarly, in ABM, S.C.O. - 36, Sector 40 D, Chd. Phone: ,

29 AM AB BM (4) From (3) and (4) AD DM AB BM AD AB DM BM AD AB BC BC 6 (from () and (5) AD AB BC BC 36 4 BC 9BC 36 8BC 36 AD AB BC 9 AD BC 9 9AB 9AD AB +9AB 9AD 7AB Hence proved 30. Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides. Sol: S.C.O. - 36, Sector 40 D, Chd. Phone: ,

30 Given : 0 ABC, BAC 90 ; AD BC To prove : BC AB +AC Proof: According to a theorem, if a is drawn from the right of a to its hypotenuse, then the two s formed are similar to each other & to the whole. BAD ACD BCA (1) BAD BCA From (1) AB BD BC AB AB BC. BD (by CPST)..() ACD BCA (from (1)) AC CD BC AC AC BC. CD Adding (1) & () AC +AB BC.BD +BC.CD BC(BD + CD) BC BC BC AC +AB BC.() Hence proved 31. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m +1 or 9m + 8. Sol: Let a be any +ve integer and b 3, the according to Euclid s division lemma, a can be S.C.O. - 36, Sector 40 D, Chd. Phone: ,

31 written as, a 3q +r, whose, 0 r<3. a is of the form 3q, 3q+ 1&3q +. Case I: a 3q cubing both sides a 3 (3q) 3 7 q 3 9(3q 3 ) 9m, where m 9q 3 a 3 is of the of 9m. Case II: a 3q + 1 cubing both sides a 3 (3q +1) 3 7 q q +9q 9(3q q +q) +1 9m +1, where m 3q 3 +3q +q a 3 is of the form 9m +1. Case III: a 3q + cubing both sides a 3 (3q+) 3 7q q + 36q 9(3q 3 +6q +4q) + 8 9m +8, where m 3q 3 + 6q +4q a 3 is of the form 9m Draw the graph of x + y 6 and x y + 0. Shade the region bounded by these lines with x axis. Find the area of the shaded region. Sol: x + y 6 x y S.C.O. - 36, Sector 40 D, Chd. Phone: ,

32 Area of 1 b h cm 33. Find other zeroes of the polynomial p(x) x 4 1x x 10x 0, if two of its zeroes are 5 ± 5. Sol: P(x) x x 10x 0 x 5 5 x ( x 5) 5 0.(1) x 5 5 x (x 5) () Multiplying (1) & () S.C.O. - 36, Sector 40 D, Chd. Phone: ,

33 (x 5) 5 0 x x 5 0 f(x) x 10x +0 0 x x x 10x 0 x 1x 49x 10x x 0x 40x 3 x 9x 10x 3 x 10x 0x x 10x 0 x 10x 0 0 g(x) x x 1 0 x x + x x ( x 1) + 1(x 1) (x + 1) ( x 1) 0 x 1,1& 5 5, Let days taken by 1 women x Let days taken by 1 man y 5 1 x y 4 (1) Ans. & x y 3 Put 1 x a & 1 y () b in (1) & () S.C.O. - 36, Sector 40 D, Chd. Phone: ,

34 a +5b 1 4 (3) 3a + 6b 1 3 a b 1/9 a + 4b /9..(4) Subtracting (3) from (4) we get a + 5b ¼ a + 4b /9 Put b 1/36 in (3) a a a 4 a x 1 a 18 days y 1 b 36 days All The Best For Solution Visits: S.C.O. - 36, Sector 40 D, Chd. Phone: ,

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