Mathematics. Sample Question Paper. Class 10th. (Detailed Solutions) Sample Question Paper 13

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1 Sample uestion aper Sample uestion aper (etailed Solutions) Mathematics lass 0th 5. We have, k and 5 k as three consecutive terms of an. 8 Their common difference will be same. 5 i.e. k k k k 5 k 8 k 8 8 k 6 ( k ) k + 9k 6 k 6 k 6. We have, point (, 5) and any coordinate on X-ais will have coordinates as (, 0 ). Using distance formula, istance of from X-ais ( + ) + ( 5 0) [ ( a + b) a + b + ab] Hence, the distance of point (, 5) from X-ais will be Let two circles with same centre at. nd R be the chord which touches the inner circle. lso, R 90 [here, tangent R at point of an innercircle is perpendicular to the radius through the point of contact] R [ perpendicular from the centre bisects the chord] In, ( ) ( ) + ( ) 5 cm cm R [by ythagoras theorem] ( 5) ( ) + ( ) [given, 5 and ] ( ) cm [taking positive square root] R + R R + [from Eq. (i)] cm + cm R 8 cm. Let the length of the shadow be m length of pole 8 m In, tan 0 tanθ erpendicular ase 8 8 tan0 Hence, the length of the shadow is 8 m. 5. Let the radii of two circles be r and r. πr Ratio of their circumferences πr r r πr r Ratio of their areas r πr r r 9. Hence, the ratio of their areas is 9 :. 6. First, we will find the HF of 65 and by using Euclid s division algorithm, then put 65 m HF and simplify. y using Euclid s division algorithm, ( 65 ) + 5 Here, divisor is 65 and remainder is 5. gain, by using Euclid s division algorithm, we get 8 m 65 ( 5 ) + (/) Here, divisor is 5 and remainder is. 0º E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M

2 6 Mathematics lass X gain, by using Euclid s division algorithm, we get Here, remainder 0 5 ( ) + 0 HF of 65 and lso, given that HF of 65 and From Eqs. (i) and (ii), 65m 65m 65m 0 (ii) m (/) lternate method y prime factorisation method, factors of 65 and are given as, 65 5 and HF (65, ) ccording to the question, we get 65m HF (65, ) 65m [from Eq. (i)] 65m m. If points ( 0, 0 ), (, ) and ( a, b) are collinear, then area of 0. ( y y) + ( y y) + ( y y ) 0 ( b) + ( b 0) + a ( 0 ) 0 0 b + a ( ) 0 b a 0 a b 8. Letα and be two zeroes of the given polynomial, which α are reciprocal of each other. Now, product of zeroes, onstant term α α oefficient of 6a a + 9 a + 9 6a a 6a ( a ) 0 [ + y y ( y) ] a 9. We have, tanθ tan 0 θ 0 Now, cosec θ sec θ cosec θ + sec θ cosec cosec 0 sec sec 0 [ θ 0 ] ( ) + ( ) + + cosec 0 and sec rime factors of 66 rime factors of 88 and prime factors of Then, HF of 66, 88 and Required number of rows Given, LHS 6 + [from Eq. (i)] n squaring both sides, we get ( 6 + ) ( ) + ( ) 0 ( )( + ) 0 0 or + 0 or ut is a natural number. [by factorisation] sin θ + cosθ [rationalisation] cosθ cosθ + cosθ sin θ ( + cos θ) sin θ ( + cos θ) cos θ sin θ [ cos θ sin θ] + θ cosθ + cosec θ + cotθ RHS cosθ cotθ Hence proved.. Let be the tower of height H m and be the building of height m, such that E 60 and E 5. m 60º 5º 5º Then, E ( H ) m E m ( H ) m H E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M

3 Sample uestion aper and E 5 [alternates angles] In, tan5 [ tan θ perpendicular ] base [ tan5 ] m E m [ E] E In E, E tan 60 H H [ tan 60 ] H ( + ) m Hence, height of the tower ( + ) (. + ) (. ) 9. m. Let r be the radius and h be the height of the cylinder. Then, h r and volume of the cylinder πr h πr. r [ h r] r cm cm learly, volume of the sphere ( ) cm [ Volume of sphere πr and R cm, given](/) ccording to the question, r (/) r. r. r cm 5. For hollow sphere, Eternal radius, R 8 cm Internal radius, r 6 cm h r Volume of the metal in hollow sphere π ( R r ) π ( 8 6 ) [ R 8, r 6] π 5 6 ( ) π 96 8 π cm For smaller cones, Radius of a cone, r cm, Height of a cone, h 8 cm π π Volume of a cone π ( r ) h 8 ( ). cm learly, required number of cones Volume of metal Volume of one cone 8π / 8 π / 6. Let be the given chord of a circle with centre. Then, r 0 cm and 90 (i) rea of minor segment rea of minor sector M rea of (/) 90 πr cm (ii) rea of the major sector rea of circle rea of minor sector πr θ θ πr πr (/) θ πr [ θ 90 ] cm. Since, is a root of + p 0. n putting, we get ( ) + p ( ) 0 6 p 0 p p Now, on comparing + p + q 0 with a + b + c 0, we get a, b p and c q Since, roots of + p + q 0 are equal. iscriminant, 0 p q 0 [ b ac] n putting p, we get Major segment 0 m 0 m Minor segment M ( ) q 0 q 9 p and q 9 E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M

4 8 Mathematics lass X 8. Let a, b and c be the sides of right angled, such that a, b and c. Let the circle touches the sides c F b,, at, E and F, r r respectively. E r Then, E F and F a and E [tangents drawn from an eternal point are equal in length] In quadrilateral E each angle is 90. So, E is a square therefore, we have E E r F E b r and F a r F + F ( b r) + ( a r) b r + a r c a + b r [ c] r a + b c a + b c r Hence proved. 9. Let (, y)be any point on the perpendicular bisector of Then, ( ) + ( y 6) ( + ) + ( y ) [by distance formula] n squaring both sides, we get ( ) + ( y 6) ( + ) + ( y ) y y y 8y + 6 [ ( a ± b) a + b ± ab] + y y 5 0 [dividing both sides by ] (, y) 0. Here, 55 < 0. So, on applying Euclid s division lemma to 0 and 55, we get 0 ( 55 ) + 5 (/) Here, remainder, 5 0. So, we have to apply the division lemma to new dividend 55 and new divisor 5, we get 55 ( 5 ) + 0 (ii) (/) Here, remainder, 0 0. So, we have to apply the division lemma to new dividend 5 and new divisor 0, we get 5 ( 0 ) + 5 (iii) (/) Here, remainder, 5 0. So, again applying the division lemma to new dividend 0 and new divisor 5, we get Here, remainder 0 (/) Thus, HF of 0 and 55 Last divisor 5 Now, to epress the HF as a linear combination of the two given numbers, start with Eq. (iii) and successively eliminates the previous remainders. From Eq. (iii), 5 5 ( 0 ) (/) 5 5 [( 55 5 ) ] [from Eq. (ii), 55 ( 5 ) 0] [from Eq. (i), 0 ( 55 ) 5] 5 5 ( 0 55 ) a + 55b where, a 5 and b 9 (/). Given: and EF in which X and Y are the medians, such that (, 6) (, ) Hence, the equation of the perpendicular bisector of is + y 5 0. We know that, y-coordinate of every point on X-ais is zero. n putting y 0 in Eq. (i), we get Thus, the perpendicular bisector of cuts X-ais at 5, 0. X G E F To prove: ~ EF E Y F X Y onstruction : roduce X to G such that XG X, join G. lso produce Y to H such that YH Y. Join FH. H E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M

5 Sample uestion aper roof: In X and GX, we have, X X X [ X is the median] XG [by construction] and X XG [vertically opp. s] So, by SS-criterion of congruence, we have X GX G [T] gain, in YE and HYF, we have EY Y YF [ Y is the median] YF [by construction] and YE HYF [vertically opp. s] So, by SS-criterion of congruence, we have Now, G ~ HF Similarly, we have YE HYE E HF [T] (ii) X [given] E F Y G X [from Eqs. (i) and (ii)] HF F Y G HF X F Y G HF G F H [ X G and Y H] [by SSS-criterion of similarity] + + (iii) Thus, in and EF, we have and [from Eq. (iii)] E F [given] So, by SS-criterion of similarity, we have. alculation of Mean aily Savings (in `) ~ EF Number of children ( ) f i lass-mark f ( i ) i i y 0 0y Total n Σf i 5 Σf i i y y From the table y 6 [given, Σf i 6] + y y 9 y 9 Using the formula: Mean Σ f i i Σfi y 8 [given, mean 8] ( 9 ) [Using Eq. (i)] (ii) From Eqs. (i) and (ii), we obtain y Hence, the missing frequencies are 9 and y 0.. Given circle of radius cm and two points and each at a distance of cm from its centre, on one of its etended diameter. Required Tangents to the circle from and. Steps of onstruction. (i) raw a circle with centre and radius cm. (ii) Take two points and on one of its etended diameter, such that cm. (iii) Now, draw the perpendicular bisectors ( p and q) of and. Let they intersect at Mand at N. (iv) Taking M as centre and M as radius, draw a circle which intersects the given circle at and. (v) gain, taking N as centre and N as radius, draw a circle which intersects the given circle at and. (vi) Join,, and. Thus,,, and are the required tangents.. Given equation is p ( + m ) + ( mc ) + ( c a ) 0 n comparing with + + 0, we get ( + m ), mc and ( c a ) Since, the given equation has equal roots. iscriminant, 0 0 cm cm M N (/) ( mc ) ( + m )( c a ) 0 m c ( c a + m c m a ) q 0 9 E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M 5

6 0 Mathematics lass X m c ( c a + m c m a ) 0 [dividing by ] m c c + a m c + m a 0 c + a + m a 0 (/) c + a ( + m ) 0 c a ( + m )c a ( + m ) Hence proved. 5. When two coins are tossed, then possible outcomes are HH, HT, TH, TT Total number of outcomes (i) Let E be the event that Neha is getting first chance. Number of outcomes favourable to E (i.e. HT, TH) Hence, ( E) Number of outcomes favourable Total number of outcomes (ii) Let E be the event that Lata is getting first chance. Number of outcomes favourable to E (i.e. H, H) Hence, ( E) (iii) Let E be the event that Sonia is getting first chance. Number of outcomes favourable to E (i.e. TT ) Hence, ( E) (iv) No, the number of cases favourable to each one of them is not equal. (v) Values: ishonesty, as she kept two cases favourable to her and one each for the other two friends. 6. We may prepare the less tan type and more than type series. aily income (in `) Number of workers (cf) Less than 0 Less than 0 6 Less than 50 Less than 80 0 Less than n the graph paper, we plot the points ( 0, ), ( 0, 6, ) ( 60, ), ( 80, 0 ), E ( 00, 50 ). Join,, and E with a free hand to get a less than ogive aily income (in `) Number of workers (cf) More than More than 0 8 More than 0 More than 60 6 More than 80 0 Number of workers Y n the graph paper, we plot the points ( 00, 50 ), ( 0, 8), R( 0,, ) S( 60, 6, ) T( 80, 0 ). Join, R, RS and ST with a free hand to get more (greater) tan ogive. More than ogive Less than ogive (00, 50) E (00, 50) (0, 8) (80, 0) (0, 6) (60, ) R (0, ) (0, ) aily income (in `). Let the cost of full first class fare be `. Then, the cost of half first class fare be `. lso, let reservation charges be ` y per ticket. ase I The cost of one reserved first class ticket from the stations to `50 () + y 50 ase II The cost of one reserved first class ticket and one reserved first class half ticket from stations to ` 80 + y + + y 80 + y 80 + y 60 (ii) n multiplying Eq. (i) by and then subtracting from Eq. (ii), we get + y 60 + y n putting the value of in Eq. (i), we get y S (60, 6) T (80, 0) y 50 y 0 (/) Hence, full first class fare from stations to is 0 ` 500 and the reservation charges for a ticket is ` 0. (/) X E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M 6

7 Sample uestion aper 8. In R, we have R + R [by ythagoras theorem] ( + ) + [given, R + and R ] ( + ) ( + + )( + ) [ a b ( a + b)( a b)] ( + )( ) ( + ) + (i) We have + cot φ + [taking positive square root] R R. R In R, cot φ + + [using Eq. (i)] (ii) We have, + tanθ + tanθ + + θ + + R R. In R, tanθ. [using Eq. (i)] + (iii) We have, cosθ R Given,α andβ are the zeroes of the quadratic polynomial p( s) s 6s +. oefficient of s Sum of zeroes, ( α + β) oefficient of s ( 6) 6 onstant term and product of zeroes, αβ oefficient of s φ Now, α β αβ β α α β α + β αβ α + β + + αβ αβ ( α + β) αβ α + β + + αβ αβ αβ [ a + b ( a + b) ab] ( ) ( / ) + + / / [α + β and αβ / ] Given, and E are altitudes which intersection each other at the point. (i) In E and, (ii) In E [each 90 ] and E [vertically opposite angles] E ~ [by similarity criterion] and E, E [each 90 ] and E [common angle] ~ E [by similarity criterion] (iii) In E and, E [each 90 ] and E [common angle] E ~ (iv) In and E [by similarity criterion] E [each 90 ] and E [common angle] ~ E [by similarity criterion] E:\ashudtp\0\shwani\I-succeed\Math 0th Eng\Final\nline\Sample aper.vp Wednesday, September 06, 0 :: M

C.B.S.E Class X

C.B.S.E Class X SOLVE PPER with SE Marking Scheme..S.E. 08 lass X elhi & Outside elhi Set Mathematics Time : Hours Ma. Marks : 80 General Instructions : (i) ll questions in both the sections are compulsory. (ii) This