KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê, d È 2018 S. S. L. C. EXAMINATION, JUNE, 2018
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1 CCE RR REVISED & UN-REVISED O %lo ÆË v ÃO y Æ fio» flms ÿ,» fl Ê«fiÀ M, ÊMV fl KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê, d È 08 S. S. L. C. EXAMINATION, JUNE, 08» D} V fl MODEL ANSWERS MO : ] MOÊfi} MSÊ : 8-E Date : ] CODE NO. : 8-E Œ æ fl : V {} Subject : MATHEMATICS ( Ê Æ p O» fl / New Syllabus ) ( Æ» ~%} % / Regular Repeater ) (BMW«ŒÈ Œ M} / English Version ) B [ V Œ V fl : 80 [ Max. : 80 Ans. Key I.. A and B are two sets, such that n ( A ) 7, n ( B ) 6 and n ( A U B ) 5 ; then n ( A I B ) is (A) (B) 6 (C) 4 (D) 5 (A)
2 8-E CCE RR Ans. Key. Geometric mean between and 8 is (A) 6 (B) 6 (C) 4 (D) 4 (C) 4. HCF of any two prime numbers is (A) a prime number (B) a composite number (C) an odd number (D) an even number (C) an odd number 4. If f ( x ) x + x x + 6 then the value of f ( ) is (A) 0 (B) 0 (C) 8 (D) 8 5. (D) 8. In ABC, ABC 90, BD AC if BD 8 cm and AD 4 cm then the length of CD is (A) 6 cm (B) 4 cm (C) 64 cm (D) cm (A) 6 cm
3 CCE RR 8-E Ans. Key sin ( 90 θ ) 6. cos ( 90 θ ) where θ is acute, is equal to (A) sec θ (B) cot θ (C) tan θ (D) cosec θ (B) cot θ 7. The co-ordinates of the mid-point of the line segment joining the points (, ) and ( 4, 7 ) are (A) (, 5 ) (B) (, ) (C) (, 5 ) (D) ( 6, 0 ) (C) (, 5 ) 8. Formula used to find the surface area of a sphere whose radius r units is (A) πr (B) πr (C) πr (D) 4πr (D) 4π r.
4 8-E 4 CCE RR II. Answer the following : 6 6 ( Question Numbers 9 to 4, give full marks to direct answers ) 9. A boy has pants and 4 shirts. How many different pairs of a pant and a shirt can he dress up with? Number of ways of pairing a pant and a shirt Write sample space for the random experiment tossing two fair coins simultaneously once. S { HH, TT, HT, TH }. The given pie chart shows the annual agricultural yield of different crops in a certain place. If the total production is 600 tons, what is the yield of Ragi in tons? Yield of Ragi tons
5 CCE RR 5 8-E. If ( x + ) is one of the factor of f ( x ) x + 5x + 6, find the other factor. Method : Factor method x + 5x + 6 x + x + x + 6 x ( x + ) + ( x + ) ( x + ) ( x + ) The other factor is ( x + ) Method : Division method x + ) x + 5x + 6 ( x + ( ) x + x x + 6 x + 6 ( ) ( ) 0 The other factor is ( x + ). What are concentric circles? Circles having the same centre but different radii are called concentric circles. 4. Two straight lines are perpendicular to each other. If the slope of one line is, find the slope of the other line. m m m m Slope of the other line.
6 8-E 6 CCE RR III. 5. If A {,, } and B {,, 4, 5 } are the subsets of U {,,, 4, 5, 6, 7, 8 }, verify ( A I B ) l A l U B l. A I B {, } ( A I B ) l U (A I B ) {, 4, 5, 6, 7, 8 }... i) A l { 4, 5, 6, 7, 8 } B l {, 6, 7, 8 } A l U B l {, 4, 5, 6, 7, 8 }... ii) From (i) and (ii) ( A I B ) l A l U B l 6. Find the sum of infinite terms of the geometric series a, r, S? S a r S
7 CCE RR 7 8-E 7. Prove that + is an irrational number. Let us assume + is a rational number. p + where p, q z, q 0 q p q q is a rational number p q Q is rational. q But is not a rational number. This leads to a contradiction. Our assumption that + is a rational number is wrong. + is an irrational number. 8. Find the number of diagonals that can be drawn in an octagon. An octagon has 8 vertices n 8 8 Total number of sides and diagonals C 4 n n ( n ) 8 8 ( 8 ) C C lines includes 8 sides. Number of diagonals 8 8 0
8 8-E 8 CCE RR Alternate method : Number of diagonals in a polygon of n sides n ( n ) In an octagon n 8 Number of diagonals 4 8 ( 8 ) Any other correct alternate method may be given marks. 9. Find the sum of all two digit natural numbers which are divisible by 5. Two-digit numbers which are divisible by 5 0, 5, 0, Sum of all two-digit numbers a 0, d 5, T 95 n T a + ( n ) d n ( n ) 5 85 ( n ) 5 ( n ) 7 n 8 Method : Sum of n natural numbers n S n [ a + ( n ) d ] S 8 8 [ 0 + ( 8 ) 5 ] 9 ( ) 9 05 S OR
9 CCE RR 9 8-E n 8, a 0, l 95 n ( a + l ) S n 9 8 ( ) S Alternate method : ( ) 5 ( 9 ) 5 ( 90 ) Find how many 4 digit numbers can be formed by using the digits,,, 4, 5 without repetition? How many of these are less than 000? If OR ( n P ) + 50 n P, find the value of n. Number of 4-digit numbers formed 5 P digit numbers which are less than Thousand s Hundred s Ten s place Unit place place place P 4 P P P 4 4 numbers. OR
10 8-E 0 CCE RR 4-digit numbers which are less than P 4 4 OR ( n P ) + 50 n P n ( n ) + 50 n ( n ) n n n n 4n n 50 n 50 n 5 n ± 5 n 5. Two unbiased dice whose faces are numbered to 6 are rolled once. Find the probability of getting a sum equal to 7 on their top faces. Total number of possible outcomes Event of getting a sum equal to 7 n ( s ) 6 (, 6 ) A ( 4, ) (, 5 ) ( 5, ) (, 4 ) ( 6, ) n ( A ) 6 n ( A ) Probability of getting the event A P ( A ) n ( S ) 6 6 6
11 CCE RR 8-E. Rationalise the denominator and simplify :. 5 Rationalising factor of 5 is ( ( 5 ) 5 + ) ( ) ( ) ( 0 + ) Simplify ( ) ( 0 + ). ( ) ( 0 + ) ( ) ( ) ( 5 5 ) ( 5 + ) 5 ( 5 + ) 5 ( 5 + ) ( ) 6 ( 5 )
12 8-E CCE RR 4. Find the quotient and remainder by using synthetic division : ( x x + 7x 5 ) ( x ) OR Verify whether ( x ) is a factor of f ( x ) x x + 6x 0 by using factor theorem Quotient x + 7x + 8 Remainder 79. OR Let f ( x ) x x + 6x 0 If ( x ) is a factor of f ( x ), then f ( ) 0 Now f ( x ) x x + 6x 0 f ( ) ( ) + 6 ( ) f ( ) 0 x is not a factor of x x + 6x 0.
13 CCE RR 8-E 5. In Δ ABC, DE BC, if AD cm, DB 5 cm and AE 4 cm, find AC. In Δ ABC, DE BC AD AE DB EC BPT 4 5 EC EC cm AC AE + EC cm. Alternate method : In Δ ABC, DE BC AD AE Cor. BPT AB AC AC AC cm.
14 8-E 4 CCE RR 6. Draw a circle of radius 4 5 cm and a chord PQ of length 7 cm in it. Construct a tangent at P. r 4 5 cm Chord PQ 7 cm Circle Chord Tangent 7. Find the distance between the co-ordinates of the points (, 4 ) and ( 8, ) by using distance formula. Coordinates of ( x y ) Point A (, 4 ) ( x y ) Point B ( 8, )
15 CCE RR 5 8-E Distance between the points d ( x x ) + ( y y ) ( 8 ) + ( 4 ) In a hockey match team A scored one goal less than twice the number of goals scored by team B. If the product of the number of goals scored by both the teams is 5, find the number of goals scored by each team. Let the goals scored by team A be x and goals scored by team B be y. x ( y ) Product of the goals scored by both teams 5 xy 5 ( y ) y 5 y y 5 0 y 6y + 5y 5 0 y ( y ) + 5 ( y ) 0 ( y ) ( y + 5 ) 0 y If y, then x 6 5 Goals scored by team A 5 Goals scored by team B.
16 8-E 6 CCE RR 9. In the given ABC, θ is acute. Write the values of the following trigonometric ratios related to θ : (a) sin θ (b) cos θ (c) cosec θ (d) sec θ. a) sin θ Opp Hyp BC AC b) cos θ Adj Hyp AB AC c) cosec θ sin θ d) sec θ cos θ. Direct answers may be given marks.
17 CCE RR 7 8-E 0. Draw a plan by using the information given below : ( Scale 0 metres cm ) 0 0 m cm 0 Metre to C to D 90 0 to E to B From A m cm m 4 5 cm 0 40 m 40 7 cm m cm m 4 cm m 5 cm 0
18 8-E 8 CCE RR IV.. In a harmonic progression 5th term is and th term is 5. Find its 5th term. OR If the third term of a geometric progression is and its sixth term is 96, find the sum of first 9 terms. T 5 and T 5 Reciprocals of HP are in AP. a + 4d... (i) a + 0d 5... (ii) By solving (i) and (ii) a + 0d 5 a + 4d ( ) ( ) 6d d 6 If d, then a + 4 ( ) a + a a 0 If a 0 and d then T n a + ( n ) d
19 CCE RR 9 8-E T ( 5 ) T 5 Alternate method : The corresponding T and 5 T of AP are T and 5 T 5 d T p T q p q T 5 T If d then a + 4 ( ) a + a 0 If a 0 and d T n a + ( n ) d T ( 5 )
20 8-E 0 CCE RR T 5 OR T ar... (i) T 96 6 ar (ii) ar 5 ar 96 8 OR ar ( r r r 8 ) r 8 r If r then a ( ) 4a a If a and r, n 9 then a ( r n ) S n r ( 9 ) S 9 ( 5 ) 5 S 5 9
21 CCE RR 8-E. Calculate the variance of the following data : Class-interval Frequency ( f ) 5 4 i) Step deviation method : A i 5 C.I. f x d x A i d f d f d Variance N 5 fd 6 f d f d fd σ i N N ( )
22 8-E CCE RR Direct method : C.I. f x fx x f x Variance N 5 fx 0 f f x f x σ N N x Assumed mean method : Assumed mean A C.I. f x d x A fd d f d N 5 fd 0 f d 550
23 CCE RR 8-E Variance f d f d σ N N Actual mean method : C.I. f x fx d x x d f d Mean N 5 fx 0 f x f x N d 490 Variance f d σ N
24 8-E 4 CCE RR. Solve ( x + ) ( x ) + 0 by using formula. OR If one root of the equation x + px + q 0 is four times the other, prove that 4p 5q 0. ( x + ) ( x ) + 0 x ( x ) + ( x ) + 0 6x 4x + 9x x + 5x 4 0 where a 6, b 5. c 4 x b ± b 4 ac a 5 ± ( 4 ) 6 5 ± ± 5 ± 5 + or 5 6 or 6 x or 4. OR
25 CCE RR 5 8-E x + px + q 0 where a, b p, c q If m and n are the roots then m 4n Sum of the roots m + n 4n + n b a p 5n p n p 5... (i) Product of the roots mn a c 4n n q 4 n q... (ii) Substituting (i) in (ii) Then p 4 q 5 4p 5 q 4p 5q 4p 5q 0
26 8-E 6 CCE RR 4. Prove that The tangents drawn from an external point to a circle are equal. Data : A is the centre of the circle. To prove : BP BQ B is an external point. BP and BQ are the tangents. Construction : AP, AQ and AB are joined. Proof : In APB and AQB, A P B AQ B Radius drawn at the point of contact is perpendicular to the tangent. hyp. AB hyp AB AP AQ Common side Radii of the same circle. APB AQB RHS theorem. BP BQ CPCT.
27 CCE RR 7 8-E 5. In Δ ABC, AB AC. P is a point on BC such that PN AC and PM AB as shown in the figure. Prove that MB. CP NC. BP. OR In Δ ABC, DE BC. If DE BC and the area of Δ ABC is 8 cm, show that the area of Δ ADE is 6 cm. In Δ ABC, AB AC B C angles opposite to equal sides In BMP and CNP B M P M B P P N C right angles N C P equal angles MBP ~ NCP equiangular triangles MB NC BP MP AA - criteria CP NP MB. CP BP. NC. OR
28 8-E 8 CCE RR Given DE BC DE BC In Δ ADE and Δ ABC, A D E D A E A B C Corresponding angles B AC Common angle Δ ADE ~ Δ ABC Equiangular triangles Area of Area of Δ ADE Δ ABC Area of Δ ADE 8 Area of Δ ADE DE BC cm. 6. Prove that ( + cot A cosec A ) ( + tan A + sec A ). OR From the top of a building 0 m high, the angle of elevation of the top of a vertical pole is 0 and the angle of depression of the foot of the same pole is 60. Find the height of the pole. cos A sin A sin A sin A cos A cos A sin A + cos A sin A cos A + sin A + cos A ( sin A + cos A ) sin A cos A sin A + cos A + sin A cos A sin A cos A
29 CCE RR 9 8-E but sin A + cos A + sin A cos A sin A cos A sin A cos A sin A cos A OR In BED, D B E 0 DE tan 0 BE x 0 BE BE ( x 0 ) In ABC, AC B 60 AB tan 60 AC 0 ( x 0 )
30 8-E 0 CCE RR ( x 0 ) 0 x 60 0 x 80 x m. Height of the pole 6 6 m ( approximate ). V. 7. Solve the equation x + x 6 0 graphically. x + x 6 0 y x + x 6 x 0 4 y Alternate method : Table Drawing parabola Identifying roots 4 x + x 6 0 y x, y 6 x y x x 0 y y 6 x x 0 y Table Drawing parabola Identifying roots 4
31 CCE RR 8-E
32 8-E CCE RR Alternate method :
33 CCE RR 8-E 8. Construct a direct common tangent to two circles of radii 4 cm and cm whose centres are 9 cm apart. Measure and write the length of the direct common tangent. R 4 cm, r cm R r 4 cm d 9 cm Length of the tangent EF 8 8 cm
34 8-E 4 CCE RR Drawing AB and marking mid-point Drawing C, C, C Joining DB, EF Measuring and writing the length of the tangent 4 9. Prove that In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other two sides. Figure Data To prove Construction Data : In ABC, A B C 90 To prove : Construction : AC AB + BC BD AC is drawn. Proof : Comparing ABC and ABD A B C B AC Statement A D B B A D Right angles common angle Reason BAC ~ DAB Equiangular triangles BA AC AA criteria DA AB AB AC. AD... (i)
35 CCE RR 5 8-E Comparing ABC and BDC A B C B D C Right angles AC B B C D common angle BCA ~ DCB Equiangular triangles BC AC AA criteria DC BC BC AC. DC... (ii) By adding (i) and (ii) AB + BC AC AD + AC DC AC ( AD + DC ) Q AD + DC AC AC AC AB + BC AC A solid is in the shape of a cylinder with a cone attached at one end and a hemisphere attached to the other end as shown in the figure. All of them are of the same radius 7 cm. If the total length of the solid is 6 cm and height of the cylinder is 0 cm, calculate the cost of painting the outer surface of the solid at the rate of Rs. 0 per 00 cm. OR
36 8-E 6 CCE RR A solid metallic cylinder of diameter cm and height 5 cm is melted and recast into toys in the shape of right circular cone mounted on a hemisphere as shown in the figure. If radii of the cone and hemisphere are each equal to cm and the height of the toy is 7 cm, calculate the number of such toys that can be formed. Height of the cone Total height of the solid ( height of the cylinder + radius of the hemisphere ) 6 ( ) But 7, 4, 5 are Pythagorian triplets cm. Slant height of the cone l 5 cm. TSA of the solid LSA of the cone + LSA of the cylinder + LSA of the hemisphere π rl + πrh + πr πr ( l + h + r ) 7 ( ) sq.cm sq.cm. Cost of painting at the rate of Rs. 0 per 00 cm Rs
37 CCE RR 7 8-E Alternate method : Height of the cone h 4 cm Slant height of the cone l 5 cm. LSA of the cone πrl π 7 5 sq.cm 75 π sq.cm. LSA of the cylinder πrh π 7 0 sq.cm 40 π sq.cm. LSA of the hemisphere π r π 7 98π sq.cm. TSA of the solid LSA of the cone + LSA of the cylinder + LSA of the hemisphere ( 75 π + 40 π + 98 π ) sq.cm sq.cm. Cost of painting Rs OR
38 8-E 8 CCE RR Volume of the metal cylinder π r h cubic units r 6 cm π 6 5 c.c. h 5 cm Volume of the toy Volume of the cone r cm + h 7 4 cm Volume of the hemisphere π r h + πr πr ( h + r ) π ( 4 6 ) + 0 π Number of toys Volume of the cylinder Volume of the toy π 0 π 8 4 Alternate method : Cylinder Cone Hemisphere r 6 cm r cm r cm h 5 cm Number of toys h 4 cm Volume of the metal cylinder Volume of the toy
39 CCE RR 9 8-E πr h πr h + πr π ( 6 5 ) π ( )
KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE G È.G È.G È.. Æ fioê, d È 2018 S. S. L. C. EXAMINATION, JUNE, 2018
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