( )( ) PR PQ = QR. Mathematics Class X TOPPER SAMPLE PAPER-1 SOLUTIONS. HCF x LCM = Product of the 2 numbers 126 x LCM = 252 x 378
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1 Mathematics Class X TOPPER SAMPLE PAPER- SOLUTIONS Ans HCF x LCM Product of the numbers 6 x LCM 5 x 378 LCM 756 ( Mark) Ans The zeroes are, 4 p( x) x + x 4 x 3x 4 ( Mark) Ans3 For intersecting lines: a b 3 3 a b k 46 k 6 ( Mark) Ans4 Since PR PQ QR PR QR + PQ RQP 90 o (Converse of Pythagoras Theorem) Therefore, In PQM Since QPM 30 o and QMP 90 o So MQP 60 o Hence, MQR 30 o ( Mark)
2 Ans5 M N Q P O OM MQ + QO QP + QN [Since Tangents from external point are equal] PN 9cm ( Mark) Ans6 Ans7 The two curves namely less than and more than ogives intersect at the median so the point of intersection is (45.5, 75) ( Mark) Total outcomes HH, TT, HT, TH Favourable outcomes HH P( E : Both Heads ) ( Mark) 4 Ans8 Let a 3 and a 4 be the third and fourth term of the AP According to given Condition 3. a 4. a 3 4 ( a d ) ( a d ) a 6d a + 6d 0 a 0 7 ( Mark)
3 Ans9 sinα + cosα sinα cosα sinα cosα sinα cotα ( Mark) Ans0 A 6cm B 8cm C + AC AB BC U sin g Pythagoras Theorem cm Circumference of semi circle Perimeter cm π r cm ( Mark)
4 SECTION B Ans Since, AP 5 AB So AP: PB : 3 P divides AB in :3 ratios P, 5 5 P 6, 5 5 Ans Area ( ACB) AC. CB. a b Also, area ( ACB). AB. CD cp
5 ab cp ab cp Now b + a + a b a b c a b c a b ( By Pythagoras theorem) c ( Since, ab cp) c p p Hence Proved ( Mark) Ans3 3 x + y 7xy 6x + 3y 7 xy () 3 x + 3y xy 3x + 9 y xy () Eq () gives : 6x + 8y xy (3) When x 0 and y 0 eq() eq (3) gives 5y 5 xy x y 3 Also x 0, y 0 is a solution.
6 Ans4 August has 3 days 4 weeks and 3 days. So 4 weeks means 4 Wednesdays Now remaining 3 days can be S M T T W Th Th F Sa Sa. S M M T W W Th F F Sa S Favorable outcomes are M T W P (3 Wednesdays) 3 7 T W Th W Th F ( Mark) Ans5 sin (A + B) Since o sin 90 A + B 90 o () cos ( A B) since cos30 o 3 3 A B 30 o () Solving () and () A 60 o B 30 o OR
7 7 tan A 4 So the ratio of adjacent and opposite side of the triangle is in the ratio 7:4 Let the common ratio term be k Using Pythagoras Theorem AC 5 k. Consider cos A + cos A ( Mark) 49 7 SECTION C Ans6 Let us assume 5 is rational. 5 p q Where p and q are co prime integers and q 0 5q p 5q p 5divides p 5divides p ()
8 So p 5a for some integer a Substituting p 5a in 5q p 5q 5a q 5a 5divides q 5divides q () From () & () 5 is a common factor to p and q which contradicts the fact that P and q are co prime Our assumption is wrong and hence 5 is irrational. mark Ans7 Let (, ) A x y be the required point which is at a distance of 5 units from the point P(0,5) and 3 units from Q(0,) So AP 5 and AQ 3 ( x ) ( y ) ( x ) ( y ) x + y y 0 0 () ( x ) ( y ) x ( y ) + 9 x + y y 8 0 () Equation () Equation () gives: 8y y
9 Substituting y in () x x 9 0 ± 3 The required points are (3, ) and (-3, ) Ans8 ( sin A+ coseca) + ( cos A + sec A ) sin A cosec A sin A coseca cos A + sec A + cos Asec A sin A + cos A cosec A+ sec (Since sin A.cosec A and cosa.seca ) (Mark) A A cot + + tan ( Mark) (Since, cosec A + cot A and sec A + tan A ) 7 + cot A + tan RHS A OR ( + cotθ - cosecθ)( + tanθ + secθ ) cosθ sinθ sinθ sinθ cosθ cosθ sinθ + cosθ - cosθ + sinθ + sinθ cosθ ( sinθ + cosθ) - () sinθ cosθ sin θ + cosθ + sinθ cosθ - sinθ cosθ + sin θcosθ - siin cosθ
10 sinθ cosθ sinθ cosθ Ans9 Let a, b x + y y x 0a + 4b () 5a 7b 0 () () 3 and () gives 30 a + b 6 30 a 4b 0 6b 6 b Substituting b - in (): 0A 4 0a a 5 x + y 5 x + y y 4 y x 3 OR For real and distinct roots: D > 0 Discriminant D b -4ac ( m) ( m)( m) > 0 ( m) ( m)( m) > 0 + 4m + 4m 6m 4m > 0
11 m > 0 > m > m m < ( mark ) Ans0 We know that area of triangle formed by joining the midpoint of sides of a triangle is th 4 the area of the triangle. ar( PQR) (ar ABC) 4 ar( PQR) 0 ( 6 5 ) + 6 ( 5 5 ) + 8 ( 5 6 ) [ 0 8 ] sq unit So ar ( ABC) 4 sq unit
12 Ans Ans 3x x α + β, αβ 3 3 α + β α + β αβ We know that tangents drawn from an external point are equal. Let AP AQ a BP BS b CS CR C DQ DR d Since ABCD is a parallelogram, opposite sides are equal. a + b c + d a + d c + b on subtracting, we get
13 b d d b b d b d AB a + b a + d AD Since adjacent sides are equal ABCD is a rhombus. OR We know that the two tangents drawn from an external point are equally inclined to the line joining the point and centre Let OAP OAS a OCQ OCR c OBP OBQ b ODR ODS d In AOB : a + b + x 80 In COD : c + d + y On adding a + b + c + d + x + y x + y 360 (Using angle sum property of quadrilateral a+b+c+d360)
14 So x + y 80 o Hence proved. Ans3 Construction of circle and radii OA,OB at an angle of 60 o Construction of tangents through the points on the circle ( marks) Ans4 Let the angles of triangle be x, y, z. Area grazed by the three horses x y z r r r π + π + π π r ( x y z)
15 π r m Ans5 a 46 5 a + 45d 5 9 S9 [ a + 90 d ] (mark) 9( a + 45d ) Section D Ans6 Given: ABC DEF To Prove: ar ABC AB BC AC ar DEF DE EF DF Construction: Draw AM BC and DN EF Proof: In ABCand DEF (mark) ar ABC ar DEF BC AM BC AM....(i) EF DN EF DN
16 Area of base corresponding altitude ABC DEF...(Given) AB BC...(Sidesareproportional)...(ii) DE EF B E... ABC DEF ( Q ) o M N...(each 90 ) ABM DEN...(AASimilarity) AB AM DE DN...( iii)[sides are proportional] From (ii) and (iii), we have BC AM DE DN From (i) and (iv), we have ar ABC BC BC BC. ar DEF EF EF EF Similarly, we can prove that ar ABC AB AC ar DEF DE DF ar ABC AB BC AC ar DEF DE EF DF ( marks) BCQ and ACPare equilateral triangles and therefore similar.
17 AC AB + BC BC (By Pythagoras theorem) Using the above theorem area ACP AC BC area BCQ BC BC OR Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. ( Mark) Given: In ABC,DE BC To prove: AD DB AE EC Construction: Draw EM AD and DN AE. Join B to E and C to D Proof: In ADE and BDE ar ADE ar BDE AD EM AD...( i ) DB EM DB [Area of base corresponding altitude]
18 In ADEand CDE ar ADE ar CDE AE DN AE...( ii ) EC DN EC Q DE BC...(Given) ar BDE ar CDE...( iii) ( Q s on the same base and between the same parallel sides are equal in area) From (i), (ii) and (iii) AD DB AE EC ( marks) Since PQ BC APQ ABC (By AA condition)
19 AP PQ AB BC 3 PQ PQ 3.8cm 9 Ans7 Let length of rectangle x m Breadth y m Area xy m. ( 7)( 3) x + y xy 3x + 7 y 0 () ( 7)( 5) x y + xy 5x 7 y 35 0 () () + () gives: x 56 0 x 8m On substituting x 8 m in equation (), we get y 5 m The length is 8 m and the breadth is 5m. Therefore, area is 40m
20 Ans8 A and R are the positions of the two boys. P is the point where the two kites meet. In ABP o sin 30 PB AP PB 50 PB 75m and QB 5m mark PQ 50m ( mark ) In PQR o sin 45 PQ PR 50 PR mark 50 PR The boy should have a string of length 70.7m OR
21 D is the initial point of observation and C is the next point of observation. AB is the tower of height h. Let BC x AB tanα BD 5 h x + 9 h 5x () tan 3 4 p h x AB BC 3x 4 h () From (): h 9x 9x 5x 960 4x 960 x 40 and substituting in (): ( mark ) 3 40 h from () 4 80 The height of the tower is 80 m.
22 Ans9 Height of cone.8 ( ) Slant height l ( 3.5) + (.8).8 c π π π TSA r + rh + rl ( ) π r r + h + l mark 7 ( ) The surface area of the solid is 69.8 cm mark
23 Ans30 C I f C. f Less than ( marks) ( marks) n n Median class n cf Median l + h f
24 ( marks)
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