SOLUTIONS 0th Mathematics Solution Sample paper -0 Sample Question Paper 6 SECTION A. The smallest prime number and smallest composite number is. Required HCF (, ).. y...(i) and + y...(ii) Adding both the equations, 6 Put the value of in the eqn. (ii), we get + y y Hence, a, b.. For similar triangles, area of triangle F area of triangle perimeter of triangle I HG perimeter of triangle K J F I HG 5 K J 6 65. From the table it is clear that the frequency is maimum for the class 0 0, so modal class is 0 0. 5. According to question, Sum of zeroes SECTION B Product of zeroes 5 The required quadratic polynomial (sum of zeroes) + product of zeroes () 5 5 5 (5 5 ) The quadratic polynomial is 5 (5 5 ). 6. From fig., + y...(i) y 6...(ii) Adding (i) and (ii), we get 8 or 9 Put the value of in equation (i), we get 9 + y or y 9 Hence, 9 and y.
C.B.S.E. (CCE) Term-I, Mathematics, Class - X 7. In ABD, from Pythagoras theorem, AB AD + BD A BC AD + BD, as AB BC CA (given) (BD) AD + BD, ( is the median in an equi. ) BD AD. B C D 8. C 90º (Angle in a semi-circle) AB ( ) + ( ) 9 + (By pythagoras theorem) tan A BC AC tan B AC BC O cm tan Atan B.. 9. cos 5 + sec 0 sec 60 + + 6 + 6 + 0. Class Frequency Cumulative Frequency 5 0 0 5 7 5 50 8 9 50 55 0 55 60 6 6 60 65 5 5 SECTION C. Let is rational, p, where p and q are co-prime integers and q 0 q p q p q
p is divisible by p is divisible by Let p r for some positive integer r p r q r q r Sample Question Paper-6 Solutions...(i) q is divisible by q is divisible by From (i) and (ii), p and q are divisible by. Which contradicts the fact that p and q are co-primes. Hence, our assumption is false. is irrational.. By Euclid s algorithm 50 65 + 65...(ii) 65 65 8 + 5 65 5 + 0 5 0 + 5 Now, 05 5 87 + 0 0 5 + 0 HCF of 50 and 65 is 5. HCF (50, 65, 05) 5.. Let α and β are the zeroes of the quadratic polynomial, then as per question β 7α F 8I α + 7α 8α HG K J α and α 7α k + 7α k + I 7 HG K J k + 7 9 k + 7 k. Let y a, the given equations become F k. + 6a 5 6 8a Multiply eqn. (i) by and eqn. (ii) by and adding 6 + a 60 8 a On adding, 0...(i)...(ii) 0
C.B.S.E. (CCE) Term-I, Mathematics, Class - X Put the value of in eqn. (), we get () + 6a 5 6a 5 a 6 a y y Hence and y Again y p p () p p 5. In ABC, PQ AB Again in BCD, BP PC AQ QC PR BD BP PC DR RC (By BPT)...(i) (By BPT)...(ii) AQ From (i) and (ii), QC DR RC B P C QR AD (By converse of BPT) 6. In ADB, AB AD + BD (Pythagoras Theorem)...(i) (given AD BC) In ADC, AC AD + CD (Pythagoras Theorem)...(ii) Subtracting eqn. (ii) from eqn. (i), we get AB AC BD CD A F I HG K J F H G I K J BC BC BC (AB AC ) BC (AB) AC + BC. 7. Let sec θ + tan θ λ B D C...(i) We know that sec θ tan θ (sec θ + tan θ) (sec θ tan θ) λ(sec θ tan θ) A Q D R Adding eqns. (i) and (ii), we get sec θ tan θ λ...(ii) F HG sec θ λ + λ + I K J λ + λ + λ + λ Comparing both sides, we get λ or λ sec θ + tan θ or
Sample Question Paper-6 Solutions 5 8. cos (5º+ q)+cos (5º q) + cosec (75º + θ) sec (5º θ) tan (60º+ q)tan (0º q) cos (5º+ q)+sin (90º 5º+ q) tan (60º+ q)cot(90º 0º+ q ) + cosec (75º + θ) cosec (90º 5º + θ) cos (5º+ q)+sin (5º+ q) tan (60º+ q).cot (60º+ q ) + cosec (75º + θ) cosec (75º + θ). 9. i f i f i Class (Class marks) i 0 0 0 0 0 0 0 5 50 0 60 50 600 60 80 70 p 70p 80 00 90 70 Total Sf i 7 + p Sf i i 0 + 70p We know that Mean, Σ fii Σfi 5 0 + 70p 7 + p 0 + 70p 5 (7 + p) 0 + 70p 86 + 5p 70p 5p 86 0 7p 76 p 76 7 8. 0. Modal class : 5 7 Here l 5, f 80, f 0 5, h, f 55 Mode f f0 l + f f0 f h 80 5 5 5+ 5+ 60 5 55 60 6 7. SECTION D. Let be any positive integer, then it is of the form q or q + or q +. where q is a natural number. Squaring, we get (q) 9q q m, m q
6 C.B.S.E. (CCE) Term-I, Mathematics, Class - X (q + ) 9q + 6q + (q + q) + m +, m q + q (q + ) 9q + q + 9q + q + + (q + q + ) + m +, m q + q + Square of any positive integer is of the form m or m + for some integer m.. p() + 5 + k coeff. of Sum of zeroes coeff. of α + β 5 Product of zeroes constant coeff. of αβ k According to question, α + β + αβ (α + β) αβ + αβ F 5I HG K J 5 k k [ (α + β) α + β + αβ] k k Hence, k. Let the cost of one pencil be ` and the cost of one chocolate be ` y. According to question, + y...(i) Now, + y y + y 7...(ii) and y 5 + y 7 7 y y 0 7 5
Sample Question Paper-6 Solutions 7 y 8 7 (, 5) 6 5 + y 7 (, ) ' 6 5 0 5 6 7 8 + y (, ) (5, ) (7, 0) 5 6 7 y' Plotting the above points, we get the graph of the above equations. Clearly the, two lines intersect at point (, ). Solution of eqns. (i) and (ii) is and y Cost of one pencil ` and cost of one chocolate `.. Let the speed of bus be km/hr and the speed of the train be y km/hr. According to question, and Let a, y Apply [(i) (ii)], we get 0 0 + y 8 0 0 + y 7 b, then 0a + 0b 8 0a + 0b 7 80a + 0b 6 0a + 0b 7...(i)...(ii)...(iii) On subtracting, 60a 9 a 9 60 0 Putting this value of a in eqn.(i), we get b 60 b 60 y y 60 a 0 0 Hence, speed of bus 0 km/hr and speed of train 60 km/hr.
8 C.B.S.E. (CCE) Term-I, Mathematics, Class - X 5. Given : In figure, D and E trisect BC. To Prove : 8AE AC + 5 AD Proof : Let and BD DE EC (given D and E trisect + BC) A BE BC AE AB + BE AB +...(i) AC AB + BC AB + 9 AD AB + BD AB + B D E C Now, 8AE 8AB + [Multiply eqn. () by 8]...(ii) and AC + 5AD (AB + 9 ) + 5(AB + ) AB + 7 + 5AB + 5 8AB +...(iii) AC + 5AD 8AE. [ From eqn. (ii) & (iii)] Proved. 6. (i) Let AB and CD be the two trees of heights a and b metres such that the trees are p metres apart i.e. AC p. Let the lines AD and BC meet at O such that OL h m. D b C h O L y B a A Let In ABC and LOC, we have In ALO and ACD, Adding eqns. (i) and (ii), we get CL and LA y, then + y p CAB CLO C C CAB ~ CLO CA CL p AB LO a h ph a ALO ACD Α A ALO ~ ACD AL AC OL DC y p h b y ph b + y ph a p ph + a b F HG...(i)...(ii) ph + b I K J
h + a b Sample Question Paper-6 Solutions 9 h ab a + b (ii) Similarily of triangles. (iii) Trees are our life line. They should be saved at any cost. 7. sin θ cos θ 7 and tan θ 7 cosec q cot sec q q + cotq 7 + cos q + 7 + tan q 7 + tanq 7 7 + 7 7 + 7 7 7 + 7 7 7 7 7 7 8. We have sin (A + B C ) sin 0º A + B C 0º...(i) and cos ( B + C A ) cos 5º B + C A 5º...(ii) Adding eqns. (i) and (ii), we get B 75º B 7 5º Now subtracting eqn. (ii) from eqn. (i), we get (A C) 5º A C 7 5º...(iii) We know that, A + B + C 80º A + C 5º...(iv) Adding eqns. (iii) and (iv), we get A 5º A 67 5º C 75º Hence, A 67 5º, B 7 5º C 75º. 9. cosec (90º θ) sec θ sec θ tan θ cos 0º + cos 50º cos (90º 50º) + cos 50º sin 50º + cos 50º
0 C.B.S.E. (CCE) Term-I, Mathematics, Class - X tan 0º F HG I KJ sec 5º sin 8º sec 5º.sin (90º 5º) sec 5º.cos 5º cosec 70º tan 0º cosec (90º 0º) tan 0º sec 0º tan 0º Given epression ( ) 9 8 9 6 6 0. Classses i (Class marks) f i f i i 0 00 50 600 00 00 50 6 00 00 00 50 6 500 00 00 50 7 50 00 500 50 9 050 Total Sf i 50 Sf i i 000 Mean Σ i fi Σfi 000 50 0 00 Average daily income ` 0 00.. i f i f i i Classes (Class marks) 0 0 5 80 0 60 5 95 60 90 75 75 90 0 05 5 560 0 50 5 y 5y 50 80 65 85 Total Sf i 50 Sf i i 800 + 75 + 5y 96 + + y 50...(i) + y 5 Σ fii Σfi 800 + 75 + 5y 9 50 650 800 + 75 + 5y 75 + 5y 550 5 + 9y 50...(ii) Solving eqns. (i) and (ii), we get and y 0.