CBSE CCE SAMPLE QUESTION PAPER

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CBSE CCE SAMPLE QUESTION PAPER FIRST TERM (SAI) MATHElMATICS (Wzth Solutions) CLASS X Ttme Allowed : 3 to 3% Hours lmaximu,n Marks : 80 General Instructions : (i) All questions are compulsory.

Section A comprises of 0 questions of mark each, Section B comprises of 8 questions of marks each, Sectioi C comprises of 0 questions of 3 marks each and Section D comprises of 6 questions of 4 marks

1 CBSE CCE SAMPLE QUESTION PAPER FIRST TERM (SAI) MATHElMATICS (Wzth Solutions) CLASS X Ttme Allowed : 3 to 3% Hours lmaximu,n Marks : 80 General Instructions : (i) All questions are compulsory. (ii) The question paper consists of 34 questions divided into four sections A, B, C and D. Section A comprises of 0 questions of mark each, Section B comprises of 8 questions of marks each, Sectioi C comprises of 0 questions of 3 marks each and Section D comprises of 6 questions of 4 marks each. (iii) Question numbers to 0 in Section A are multiple choice questions where you are to select one correct option out of the given four: (iv) There is no overall choice. However, internal choice has been provided in question of two marks, 3 questions of three marks each and questions of four marks each. You have to attempt only one of the alternatives in all such questions. (u) Use of calculators is not permitted. Question numbers to 0 are of oiie mark each.. Euclid's Division Lemma states that for any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy. (a) l<r<b (b) O<r<b (c) O<r<b (d) O<r<b Solution. Choice (c) is correct.. In figure, the graph of a polynomial p(x) is shown. The number of zeroes of p(x) is Y (a) 4 Y' (b) (c) (d) 3 Solution. Choice (b) is correct The number of zeroes is as the graph intersects the zaxis at one point in the given figure.

2 3. In figure, if DE BC, then x equals (a) 6 cm (b) 8 cm (c) 0 cm. (d).5 cm Solution. Choice (c) is correct. In figure, since DE BC, therefore LD=LB [Corresponding Lsl In AADE and AABC, we have LD=LB proved above. LA=A [Common] So, by A criterion of similarity, we have AADE AABC * = AD DE [Corresponding sides of similar triangles are proportional] AB BC 4 * = [.: AB =AD +QB = ( + 3) cm = 5 c ~ I 5 x. * x = 0 cm. 4. f sin 30 = cos (0 69, where (30) and (9 6") are both acute angles, then the value of 0 is (a) 8" (b) 4" (c) 36' (d) 30" Solution. Choice (b) is correct. Given, sin 30 = cos (0 6") cos (90" 30) = cos (0 6") ==) 90"30=06" a 96" = 40 =$ 0=96'+4 * 0 = 4" 5. Given that tan 0 = cosec 0 sec 0 is cosec 0 + sec 0 (a) (b) [.; cos (90" A) = sin A Solution. Choice (c) is correct. Given, tan 0 =, then J5

cosecz e see" cosecz e + seez e = (9 )3 (6+9+ )3 83 63. 8 6 6. In figure, AD 4 cm, BD 36m and CB cm, then cot I3 equals 'ADm '.

3 cosecz e see" cosecz e + seez e = (9 )3 (6+9+ ) In figure, AD 4 cm, BD 36m and CB cm, then cot I3 equals 'ADm '. c B. 3 5 (a) (b) 4 4 (c). (d) 7 3 Solution. Choice (d) is correct. In right AADB, ABZ AD^ t BD' a AB~ = (4Y t (3 ==+ ~~=6+9=5=(5)~

4 CB In MCB, cot 8 = AB 7. The decimal expansion of 4' will terminate after how many places. of 0 decimal? (a) (b) (c) 3 Solution. Choice (c) is correct. (d) wilhnot terminate 47 It shows that arer 3 places of decimal, decimal expansion will terminate The pair of linear equations 3 + y = 5; 3y = 7 have (a) one solution (b) two solutions (c) many solutions (d) no solution Solution. Choice (a) is. correct. Here, al=3,bl=,cl=5 az=,bz=3,cz=7 Clearly, al h Gf% The given pair of lines have one solution. 9. If sec A = cosec B, then A + B is equal to (a) Zero.(b) 90" (c) < 90' (d) z 90" Solution. Choice (b) is correct. Given, sec A = cbsec B * sec A = cosec B cosec (90"A) = cosec B 3 90"A= B 3 A+B=90 [.; cosec (90" 0) = sec 8

0. For a given data with 70 observations the 'less than ogive' and the 'more than ogive' intersect at (0.5,35). The median of the data is (a) 0 (b) 35 (c) 70 (dl 0.5 Solution. Choice (d) is correct.

5 0. For a given data with 70 observations the 'less than ogive' and the 'more than ogive' intersect at (0.5,35). The median of the data is (a) 0 (b) 35 (c) 70 (dl 0.5 Solution. Choice (d) is correct. The xcoordinate of the point of intersection of the 'less than ogive' and the 'more than ogive' is the median. Since the given intersection point is (0.5, 35, therefore the median is 0.5. Question numbers to 8 carry marks each.. Is 7 x 5 x 3 x + 3 a composite number? Justify your answer. Solution. We have 7x5x3x+3 =7x5xx3+x3 [Writing 3 as x 3 =(7~5xZ+l)x3 [Taking 3 as common =(70+)x3 =7x3 = 3 Fundamental Theorem of Arithmetic states that "Every composite number can be expressed (factorised) as a product of primes and their factorisation is unique, apart from the order in which the prime.:x factors occur". So, 7 x 5 x 3 x + 3 = 7 x 3 is a composite number.. Can (x ) be the remainder on division of a polynomial p(x) by ( + 3)? Justify your answer. Solution. No, (x ) cannot be the remainder on division of given polynomidp(x) by (% + 3) as the degree of remainder pblynomid and quotient polynomial is same. 3. In figure, AECD is a rectangle. Find the values of x and y. Dx+YC., T AB Solution. Since ABCD is a rectangle, therefore AB =DC and AD = BC (opposite sides of a rectangle) (opposite sides of a rectangle) = =x+y... () and xy =8.. $) Solving these equations by adding and subtracting the given equations, i.e., we have (x+y)+(xy)=+8 (x+y)(xy).=8 = 0 y =4

6 4. If 7 sinz cos 0 = 4, show that tan 8 =. J5 Solution. Given : 7~in~e+3~0~~es4 [Dividing both sides by cos 0 sin 0 c0s 8 4 =, 7 +3T=cos 0 cos 8 cos 0 a 7 tag = 4 sec 0 j 7tane+3=4(i+tane) a 7tan8+3=4+4tan0 * 7tanz04tan0=43 * 3tan0=l [.; 8 lies in st quadrant] 5 ( + sin 0)( sin 0) If cot 8 =, evaluate 8 (I+ cos 8)( cos 0) ' Solution. We have ( + sin 0)( sin 8) ( + cos 0)( cos 8) [(+ sin 8)( sin 0) ( + cos 0)[( cos 0 ( sin 0) ( cos 0) I [cot 8 5. = ("ven)

7 FE BF EC BE', 5. In figure, DE AC and DF AE. Prove that = ' Solution. In ABCA, DE AC BE BD.. = EC DA In ABEA, DF AE BF BD.. FE DA From () and (), we conclude that...(l)'by Thales Theorem]...() m y Thales Theoreml 6. In figure, AD I BC and BD CD. Prove that 3 BD = CD 3 =j 3BD = CD a 3BD+BD=CD+BD =+ 4BD = BC In right AAE3D;rightangled at D, we have AB~ AD^ + BD~ MB = AD + ~BD' In right AACD, rightangled at D, we have CA~ = + CD~ ==j = AD + ~CD' [Adding BD to both sides [Pythagoras Theoreml... () [Pythagoras Theoreml... ().

8 Subtracting () from (, we get ~CA' AB = (0' + ~CD') (.4~~ + BD) 3 CA AB = [9B~~ BD] a CA AB = (8BD) CA ~ A = B 6BD ~ a CA AB = ( ~BD)~ CA AB = BC [.; BC = 4BD (proved above)] a ~ C = A AB ~ + BC'. 7. The following distribution gives the dailv income of 50 workers of a factory : Daily income (in 7) Number of Workers 8. Find the mode of the following distribution of marks obtained by 80 students : Marks obtained 00 / Number of students I 6 Solution. Since the maximum number of students is 3, therefore, the modal class is Thus, the lower limit (I) of the modal class = fi = 3, fo =, f = 0, h = 0. Using the formula : 4 Write the above distribution as less than type cumulative frequency distribution. Solution. We calculate 'less than type" cumulative frequency distribution by adding frequencies from top to bottom. Less Than TyGe Cumulative Frequency Distribution Daily income (in 7) Number of Workers, Daily income Less than Cumulative Frequency

=.30 + 5 4 = 30 + 6.5 = 36.5 So, the maximum number of students have 36.5 marks.. ~uestio numbers 9 to 8 carry 3 marks each. 9. Show that any positive odd integer is of the form 49 + or 4q + 3, where q is a positive integer.

9 = = = 36.5 So, the maximum number of students have 36.5 marks.. ~uestio numbers 9 to 8 carry 3 marks each. 9. Show that any positive odd integer is of the form 49 + or 4q + 3, where q is a positive integer. Solution. Let a be anypositive integer, when a is divided by 4. Then, by Euclid's division lemna, we get a = 4q + r, 0 5 r c 4, where q and r are whole numbers. So, r = O,,,3. When r = 0 or, then a becomes an even integer, i.e., a = 49 or 4q +. When r or 3, then a becomes positive odd integer, i.e., a = 4q + or 4q + 3. Hence any positive odd integer is of the form 4i~ + or 4q + 3, where q is a positive integer... & 0. Prove that is irrational. 5 h. Solution. Let us assume, to the contrary, that is rational. 5 h, wherep and q are coprime, i.e., their ~ Cis F 5 q => &=5p q Since is a rational number, therefore, 5 ~. s also rational number. q q => h is rational number But this contradicts the fact that h is irrational. h. This contradiction has arisen because of our incorrect assumption that is rationd. 5 So, we conclude that h is irrational. 5 or Prove that (5 &) is irrational. Solution. Let us assume, to the contrary, that (5 A) is rational, i.e., we can find copde a and b (b * 0) such that 5 A = E b a b * 5 = A

10 Since a.and b are integers, we get 5 a is rational.. b + is rational But this contradicts the fact that & is irrational. This contradiction has arisen because of our incorrect assumption that 5 fi is rational. So, we conclude that (5 A) is irrational.. Aperson rowing a boat at the rate of 5 kmhour in still water, takes thrice as much time in going 40 inn upstream as in going 40 km downstream. Find the speed of the stream. Solution. Let the speed of the stream be x kmh and the speed of the boat in still water be 5 kmh. Relative speed of the boat downstream = (5 + x) kmlh and relative speed of the boat upstream = (5 x) km/h. Time taken by the boat downstream = 40 h 5 + x and time taken by the boat upstream =.% h 5x According to the given condition, a boat takes thrice as much time in going 40 krn upstream as in going 40 km downstream, we have. 40 = 3[40) 5x 5+x ==.. 3 5x 5+x ==. 5+x=53x + 3x+x=55 ==t. 4 = 0 x =.5 Hence the speed of the stream is.5 km/h. Or In a competitive examination, one mark is awarded for each correct answer while mark is deducted for each wrong answer. Jayanti answered 0 questions and got 90 marks. How many questions did she answer correctly? Solution. Let the number of correctly answered questions be x and the number of wrongly answered questions bey. It is given that "the total number questions answered is 0"... x+y=0...() Also, it is given that "one mark is awarded for each correct answer and mark is deducted for each wrong answer and Jayanti has scored 90 marks".

11 ...x.y=90 + y=80...() Adding () and (, we get (x+y)+(&y)=0+ 80 a 3 = 300 * x = 00 Substituting x = 00 in (, we get : y 0 x = 0 00 = 0. Hence, the number of correctly answered questions by Jayanti are 00. Alternative Method : Total number of questions = 0. Let the number of correctly answered questions be x, then the number of wrongly answered questions = 0 Lx. It is given that "one mark is awarded for each correct answer and mark is deducted for, each wrong answer". Thus, marks obtained for correctly answered question =.x = x. and, marks obtained for wrongly answered question = 40 x).. Total marks obtained = x 40 x) But the total marks obtained by Jayanti = 90 (given) Hence, the number of correctly answered questions by Jayanti are 00.. If a, 8 are zeroes of the polynomial % 5, then form a quadratic polynomial whose zeroes are (a) and (8). Solution. Since a and P are zeroes of the polynomial, x 5, therefore Let S and P denote respectively the sum and product of zeroes of the required polynomial, then S = a + 8 = (a + $ = () = 4

and P = (n)(p) = 4(ap) = 4( 5) = 60 Hence, the required polynomial p(x) is given by p(x) = (x SX + P) * p(x)=.%46~.., 3. Prove that (cosec 0 sin g)(sed 0 cos 0). Solution. We have L.H.S. = (cosec 8 sin 0)(sec cos 8) tan0 +cote' =.

12 and P = (n)(p) = 4(ap) = 4( 5) = 60 Hence, the required polynomial p(x) is given by p(x) = (x SX + P) * p(x)=.%46~.., 3. Prove that (cosec 0 sin g)(sed 0 cos 0). Solution. We have L.H.S. = (cosec 8 sin 0)(sec cos 8) tan0 +cote' =. cosz 8 sin 8 sin 8 cos 0 sin e cos 0 sin 0 cos 0 sin 0 + c0sz0 sin 8 cos 0lsin 0 cos.0 = [Dividing numerator and denominator by.sin 0 cos 8 sin e cos 0 + sinecose S~~OCOSO sin 0 cos 0. +cos 0 sin 0 tane+cote = R.H.S. 4. If cos 0 + sin 0 & cos 0, show that cos 0 sin 0 = & sin 0. Solution. Given, cos 8 + sin 8 = ficos 8 a sin 8 = & cos 0 cos e =) sin e = (& ) cos o sin 0 = cos e.l * (JZ.+ ) sine, = cos 0 (JZ I)(& + )

* sin 0 + sin 0 = cos 0 a cosesine= sin0 5. In figure, AB I BC, FG I BC and DE I AC. Prove that AADE AGCF. A Solution.. Since AB I BC and GF I BC, thereforem.. L= L3 [Corresponding Lsl.

13 * sin 0 + sin 0 = cos 0 a cosesine= sin0 5. In figure, AB I BC, FG I BC and DE I AC. Prove that AADE AGCF. A Solution.. Since AB I BC and GF I BC, thereforem.. L= L3 [Corresponding Lsl...() In ME, we have L+ L = 90 L3+L=90 GF.... () [using (I)] In AGCF, we have L3 + L4 = 90'... (3) From () and (3), we have A In M E and AGCF, we have L= L3 and ~ = L4 So, by AA smilarity of criterion, we have AADE AGCF Alternative Method : Since AB I BC and GF I BC, therefore AB GF [Proved in ( [Proved in (4.. L= L3... () [Corresponding Lsl In M E and AGCF, we have L= L3 Proved in ()l and LAED = LGFC [Each = 907 So, by AA similarity criterion, we have AADE 0 AGCF 6. AABC and hdbc are on the same base BC and on opposite sides of BC and 0 is the point of intersection of AD and BC. area(aabc) A0 Prove that = area (ADBC) DO '

14 Solution. In the figure AABC and ADBC are on the same base BC, AD and BC intersect at 0. Draw AE I BC and DF I BC. In h4eo and ADFO, we have LAEO = LDFO [Each= 90 B C and LAOE=LDOF [Vertically opposite Lsl So, by AA similarity criterion, we have AAEO ADFO AE A0 =sides are proportional] DF DO... () [In similar triangles corresponding % xbcxae Now, area (AABC) area (ADBC) BC DF [using ( Hence ~roved. A 7. Find mean of the following frequency distribution, using stepdeviation method : Classes Frequency Solution. Let the assumed mean be a = 5 and h = 0. Calculation of Mean Frequency Classmark x, 5 Classes d, = x, 5 U, = cf,) (r,) 0 f, u' * Total n =Xf, = 50 X&uz = 0 From the table, n = Zfi = 50, Zfiui = 0. Using the formula : '

Or The mean of the following fiequency 'distribution is 5. Find the value ofp. Classes 00 00 030 3040.4050 Frequency Solution.

15 Or The mean of the following fiequency 'distribution is 5. Find the value ofp. Classes Frequency Solution. 3 Calculation of Mean 5 3 P I Total 5 3 P n=z&=3+p' From the table, n = Zfi = 3 +p,.zfixi = p Using the formula : Zfifixi Mean = Zfi p (given) 5 = 3 + p * 5(3 +p) = p =? p = p a 45p 5p = * 0p = 40 * P 8: Find the median of the following data : Classes 00 ~00~030)3040)4050/5060~6070~7080)8090/9000 Frequency Solution. Calculation of Median Classes Frequency (f) ~ Zf?, = p Cumulative Freqeuncy (cf) 5 8 5

n 53 Here, n = 53 and = = 6.5,. n NOW, 66 70 is the class whose cumulative frequency is 9 is greater than = 6.5. :. 60 70 is the median class.

16 n 53 Here, n = 53 and = = 6.5,. n NOW, is the class whose cumulative frequency is 9 is greater than = 6.5. : is the median class. From the table, f = 7, cf =, h = 0 Using the formula : Question numbers 9 to 34 carry 4 marks each. 9. Find other zeroes of the polynomialp(z) " s + 30 if two of its zeroes are & and &. Solution. Since two zeroes are fi and fi, therefore (x &)(x + &) = (x ) is a factor of the given polynomial. Now, we divide the given polynomial by ( ). x + 7x 5 x x4+7x39x4x+30, ~~ , First term of quotient is x x 7x3 5x 4x x3 4x + Second term of quotient is '" = 5x + 30 x 7. 5x Third term of quotient is = 0 x so, x4 + 7x3 9x 4x + 30 = (x )(x + 7 5) =(x fi)(x+ h)[x + lox3x5

3 So, the zeroes of x + 7 5 = (x + 5)(x 3) are.given by 5 and. Hence all the zeroes of the given polynomial are fi, fi, 5 and. 3 30.

ar(aabc) AB BC' CA' To prove : = =ar (APQR) PQ~ QR~ RP' Construction : Draw AD I BC and PS I QR. P I ar (.AABC) xbcxad Proof :, ar (PQR) QR,. i: BS [Area of A = (base) x height] ar (AABC) BC x AD %.

17 3 So, the zeroes of x = (x + 5)(x 3) are.given by 5 and. Hence all the zeroes of the given polynomial are fi, fi, 5 and Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Solution. Given : AABC and APQR such that AABC APQR. ar(aabc) AB BC' CA' To prove : = =ar (APQR) PQ~ QR~ RP' Construction : Draw AD I BC and PS I QR. P I ar (.AABC) xbcxad Proof :, ar (PQR) QR,. i: BS [Area of A = (base) x height] ar (AABC) BC x AD %. ar(apqr) QRx PS Now, in AADB we have LB=LQ [AS AABC APQR] LADB = LPSQ [Each = 9Oo 3rd LBAD = 3rd LQPS Thus, AADB and PSQ are equiangular and hence, they are similar. Consequently AD AB = PS PQ [If A's are similar, the ratio of their corresponding sides is same] But [.: AABC APQR] Now, from () and (3, we get...(3) [using (

18 As M C APQR, therefore ar(h4bc) AB' BC~ CA' Hence, == [From (4) and (5 ~~(APQR) PQ'. QR' w Or Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle. Solution. Given : A triangle ABC such that : AC = AB + BC~ To prove : AABC is a rightangled at B, i.e., L B = 90". Construction : Construct a APQR such that LQ = 90' and PQ = AB and QR = BC. Proof : In APQR, as LQ = 90, we have PR~ = PQ~ + QR~ [By Pythagoras Theorem PR~ = AB~ + I)c () [AsPQ=ABandQR=BCI Ac = AB + BC~...() From () and (), we have PR~ = AC~ =+ PR = AC...(3) Now in AABC and APQR, we have AB=PQ BC = QR and AC = PR [using (3.. AABC s APQR [SSS congruency] *. LB=LQ=90 [CPCTI Hence, L B 90".. sece+tane cose 3. Wove that tane:sece+l sin8 Solution. We have secd+tane L.H.S. = tanesece+

cos 8 sine = R.H.S. sec 0 cosec (90" 8) tan 0 cot (90" 0) + sin 55" + sin 35" Evaluate : tan 0" tan 0" tan 60" tan 70" tan 80" Solution.

19 sec8+tan0. tan 8 sec 8 + (sec 0 tan 8) seco+tano (tan 8 sec 8) + (sec 8 tan 8)(sec 8 + tan 8) sec8+tan8 (set 8 tan 0) + (sec 8 tan 0)(sec 8 + tan 8) sec8+tan8 (set 0 tan 0 '+ sec 0 + tan 0 sin 8 cos 8 cos! cos 8 sine = R.H.S. sec 0 cosec (90" 8) tan 0 cot (90" 0) + sin 55" + sin 35" Evaluate : tan 0" tan 0" tan 60" tan 70" tan 80" Solution. We have Or sec 0 cosec (90" 0) tan 8 cot (90" 8) + sin 55" + sin 35" tan 0" tan 0" tan 60" tan 70" tan 80" see 8 sec 8 tan 8 tan 8 + sin (90" 35") + sin 35" tan 0" tab 0" tan 60" tan (90" 0") tan (90" 0") = see 0 tan 0 + ~os"5~ + sin 35". tan 0" tan 0" (&I cot 0" cot 0" + (tan 0". cot lo0)(tan 0". cot 0 )(&) ['.' cosec (90" 8) = sec 0, cot (90" 0) =tan 8 p,. sin (900 8) = cos 0, tan (90" 0) = cot 0 [.: sec 0 tad 8 =, sin 0 + cos 8 =, tan (90" >0) = cot 0 [:. tan 0.,cot9 =

20 p 3. If see 0 + tan 0 = p, prove that sin 0 = p+' Solution. We have p (see 0 +tan 0)' = p + (sec e + tan el + (sec 0 ) + tan 0 + sec 0 tan 0. sec 0 + (I+ tan 0) + sec 0 tan 0 tan 0+tane+sec0tan0 [.: sec 0 = tan 0 and + tanz 0 = sec 0 sec 0 + set" +' sec 0 tan 0 tan 0+sec0tan0 sec 0 + sec 0 tan sec 0) sec 0(sec 0 + tan 0) tan 0 set 0 sin 0 cos 0.sec 0 = sig Draw th6 graphs of following equations : y=l,x+y3 (i) Find the solution of the equation from the graph. ' (ii) Shade the triangular region formed by the lines and the yaxis. Splution. We have &ly= and x +y = 3...() * y=x * 3 x y= Table of y = c 3 x Table of y = Plot the pointsa(0, l), B, 0, C 0,, D(3,O) and E(3,5) on graph paper and join the [: j ( 3 points to form the lines and m as shown in figure.

21 (i) From the graph, clearly lines () and () intersect each other at the point E(3,5). So, x 3, y = 5 is the required solution of the given equations. (ii) Shaded triangdar region ACE is shown in the figure. 34. The following table gives the production yield per hectare of wheat of 00 farms of a village : in kglhectare Number of farms , Change the above distribution to more than type distribution and draw its ogive. Solution. First we change the distribution to a more than type distribution and draw its Here 50, 55, 60, 65, 70, 75 are the lower limits of the respective class intervals more than 5055,5560,6065,6570,7075and7580. To represent the data in the table graphically,,we mark the lower limits of class intervals on thehorizontal axis (xaxis) and their coirespondig cumulative frequencies on the vertical axis Cyaxis), choosing convenient scale. Now plot the point corresponding to the orderedpairs given by (lowerlimit, corresponding cumulative frequency) wive (of more than type) (see figure).

Classes 50 55 55 60 60 65 65 70 70 75 75 80 Production field (in kglhectare) more than or equal to 50 more than or equal to 55 more than or equal to 60 more than or equal to

22 Classes Production field (in kglhectare) more than or equal to 50 more than or equal to 55 more than or equal to 60 more than or equal to 65 more than or equal to 70 more than or equal to 75 Number of farms (Frequency) Cumulative Frequency (more than type) Lower limit Yield in kghectare

Paper: 02 Class-X-Math: Summative Assessment - I

1 P a g e Paper: 02 Class-X-Math: Summative Assessment - I Total marks of the paper: 90 Total time of the paper: 3.5 hrs Questions: 1] The relation connecting the measures of central tendencies is [Marks:1]