CCE SAMPLE QUESTION PAPER I FIRST TERM (SA-I) MATHEMATICS (With Solutions) CLASS X lFme Allowed : 3 to 3% Hogs] [Maximum 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 10 questions of 1 mark each, ;Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions of4marks each. (iii) Question numbers I to 10 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 prooided in I question of two marks, 3 questions of three marks each and 2 questions offour marks each. You have to attempt only one of the alternatives in all such questions. ( v ) Use of calculators is not permitted. 'Question numbers 1 to 10 are of one mark each. 23 win tkrminate after 1. The decimal expansion of the rational number - 23~2 ' - - how many places of decimal ? (a) 1 (b) 3 (c) 1 (dl 5 Sylution. Choice ( b )i s correct. 23 will terminate after Thus, shows that decimal expansion of the rational number - ~ 3 5 ~ ' three places of decimal. 2. If HCF (105,120) = 15, then LCM (105,120) is (a) 210 . (b) 420 (c) 840 (d) 1680 Solution. Choice (c)is correct. ' HCF x LCM = Product of two numbers 15 x LCM = 105 x 120 a LCM = 105 x 120 15 + LCM = 105 x 8 = 840.
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FIRST TERM (SA-I) MATHEMATICS CLASS X I...U-Like CCE Sample Question Paper 1 27 Solution. We divide the given polynomial by x2 + 5. x2+2X+3 x2 + 5 x4 + 2x3 + 8x2 + 12% + 18 First term
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CCE SAMPLE QUESTION PAPER I FIRST TERM (SA-I)
MATHEMATICS (With Solutions) CLASS X
I lFme Allowed : 3 to 3% H o g s ] [Maximum 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 10 questions of 1 mark each, ;Section B comprises of 8 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 6 questions o f 4 m a r k s each.
( i i i ) Question numbers I to 10 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 prooided in I question of two marks, 3 questions of three marks each and 2 questions offour marks each. You have to attempt only one of the alternatives i n all such questions.
( v ) Use of calculators is not permitted.
'Question numbers 1 to 10 are of one mark each.
23 win tkrminate after 1. The decimal expansion of the rational number - 2 3 ~ 2 ' - -
how many places of decimal ? ( a ) 1 (b) 3 1 (c) 1 (dl 5 Sylution. Choice (b) i s correct.
23 will terminate after Thus, shows that decimal expansion of the rational number - ~ 3 5 ~ '
three places of decimal. 2. I f HCF (105,120) = 15, then LCM (105,120) is
( a ) 210 . (b) 420 (c) 840 ( d ) 1680
Solution. Choice (c) i s correct. '
HCF x LCM = Product of two numbers 15 x LCM = 105 x 120
a LCM = 105 x 120
15 + LCM = 105 x 8 = 840.
trainer
Rectangle
3. In figure, the graph of a polynomial p(x) is shown. The number of zeroes of p(x) is
Y t
Y" (a) 1 (b) 2 . . (c) 3 (d) 4
Solution. Choice (b) is correct. The number of zeroes ofp(x) is 2 as the graph intersects the x-axis at two points A and B in
figure. 4. The value of k for which 62,- 3y + 10 = 0 and 2r + ky + 9 = 0 has no solution is
(a) - 1 (b) -2 (c) 1 (dl 3
Solution. Choice (a) is correct. Here al = 6, bl = - 3, cl = 10 and az = 2, bz = k, c2 = 9 The given system of equations will have no solution if
- k = - 1 5. Let AABC - APQR, area of AABC - 81 cm2, area of APQR - 144 cm2 and QR =
6 cm, then length of BC is (a) 4 cm (b) 4.5 cm (c) 9 cm (d) 12 cm
Solution. Choice (b) is correct. Since AABC - APQR, therefore
.: area (AABC) = 81 cm2 (given) and area (APQR) = 144 em2 (given) 1
U-Like CCE Sample Question Paper 1 25
+ 3 x 3 BC = - 2
* BC = 4.5 cm cos (90" - 0) COs 0 - is 6. The value of
tan 0 (a) - sin2 0 (c) - cos2 0
Solution. Choice (a) is correct. cos (90" - 0) cos 0 - 1 - s i n e c o s e
t an 0 ' t an 0
- - s i n e c o s e sin 01cos 0
2 = cos 0 - 1 = - (1 -?sZ 0) =-sin 8
[.; QR = 6 cm (given)]
(b) - cosec2 0 (d) - cot 0
1 2 1 7. If tan 8 + - = 2, then the value of tan 0 + 7 is . tan 0 tan 0
(a) 3 (b) 4 (c) 2 (d) - 4
Solution. Choice ( c ) is correct.
Given, t a n 0 + - I - 2 t a n 0
1 + 2 t a n 0 . = 4 * t d 0 + 7 [.; (a + b12 = a2 + b2 + 2abl t an 0 t a n 0
* t a ~ P e + ~ + 2 = 4 ' t a n 0
* 1 t a n 2 0 + T = 4 - 2 t a n 0
* 1 t a n 2 0 c 7 = 2 tan 0
5sina-3cosrr is 8. If 5-tan a = 4, then the value of
. 5sinac2cosa
1 2 (a) - (b) 3
3
Solution. Choice (c) is correct. 4 5 t a n a = 4 * t a n a = - 5
- - (5 sin a - 3 cos a)/cos a [Dividing numerator and denominator by cos a]
(5 sin a + 2 cos a)/cos a
- - 5 tan a - 3 5(4/5) - 3 - - 5 tan a + 2 5(4/5) + 2 . 4 - 3 1 =-=- ~. 4 + 2 ' 6
9. If sec 4A - cosec (A - 209, where 4.4 is an acute angle, then the value of A is (a) 21" (b) 22" (c) 23" (d) 24"
5A = 110" * - A-22"- 10. Which measure of central tendency is given.by the x-coordinate of the point
of intersection of 'more than ogive' and 'less than ogive' ? (a) Mean (b) Mode (c) Median (d) Mean and Median both.
Solution. Choice (c ) is correct. '
The median of a grouped data of central tendency is given by the x-coordinate of the point of intersection of the 'more than ogive' and 'less than ogive'.
-1 . .
Question numbers I1 to I8 carry 2 marks each. 11. Write 21975 as a product of its prime factors. Solution. We have
21975 = 5 x 4395 = 5 x 5 x 8 7 9 = 5 x 5 x 3 x 2 9 3 -3~5~x293. .
12. If the polynomial x4 + b3 + 8x2 + 122 + 18 is divided by another polynomial 2 + 5, the remainder comes out to be px + q. Fiqd the values ofp and q. . .
U-Like CCE Sample Question Paper 1 27
Solution. We divide the given polynomial by x2 + 5.
x2+2X+3
x2 + 5 x4 + 2x3 + 8x2 + 12% + 18 First term of qiotient is = x x4 + 5x2 x - -
2.2 2X3 + 3x2 + + of quotient is = 2x
- 2 ~ 3 + IOX . X . - 3x2 + 2x + 18 Third term ofquotient is = 3
= RH.S. 15. Prove that.in the figure if ABCD is a trapezium with AB 11 DC 11 EF, then
Solution. Given : Atrapezium ABCD in which AB 1 1 DC 11 EF: AE BF To prove : - = - ED FC
Proof: In U C , we have EP 1 1 DC
Also, in W C , we have PFIIAB
[.: EF I / DC (given)]
... (1) [using BPTl
From (1) and (21, we have
Hence, proved.
16. In AABC, LA = 90" and AD I BC. Prove that AB2 + CD2 = BD' +A@. Solution. Let ABC be a triangle such that L A = 90" and AD I BC. We have to prove that
AB2 + C D ~ = B D ~ + A C ~ c In right AABD, we have
AB2 = AD2 + BD2 ... (1) [using Pythagoras Theorem] In right AACD, we have
We cgculate more than cumulative frequencies by adding successive classes from bottom to top.
18. For the following grouped frequency distribution find the mode :
I Class 1 3 - 6 1 6 - 9 1 9-12 112-15 1 15-18 1 18-21 121-24
400 450 500 550 600 650 700 750
Frequency
230 210 175 135 103 79 52 34
Solution. Here the maximum frequency is23 and the class corresponding to this frequency is 12 - 15.
2 5 1 10 23 21 12 3
U-Like CCE Sample Question Paper 1 31
So, the modal class is 12 - 15. Lower limit (1) of the modal class = 12. Class size (h) = 15 - 12 = 3 Frequency (fi) of the modal class = 23 Frequency (fo) of the class preceeding the modal class = 10 Frequency (fi) of the class following the modal class = 21 Using the formula :
Mode =' 1 + fl - fo h 2fl - fo - f 2
Question numbers 19 to 28 carry 3 marks each. 19. Show that any positive odd integer is of the form 69 + 1 or 69 + 3 or 6q + 5,
where q is a positive integer. . Solution. Let a be any positive integer, when a is divided by 6. Then, by Euclid's division
lemna, we get a = 6q + r, 0 a r < 6, where q and r are whole numbers.
So,.r=O,1,2,3,4,5. When r = 0 or 2 or 4, then a becomes an even integer, i.e., a = 6q or 6q + 2 or 6q + 4. When r = 1 or 3 or 5. then a becomes a positive odd integer, i.e., a = 6q + 1 or 6q + 3 or
6q + 5. Hence any positive odd integer is of the form 69 + 1 or 6q + 3 or 69 + 5, where q is a positive
integer. .
20. Prove that & + & is irrational.
Solution. Let us assume, to the contrary, that & + & is a rational number. That is we can find coprimep and q (q 0) such that
=> [Squaring both sides1
2 2 Since, p and q are integers, - 3q is rational, and so & is rational.
2pq
But this contradicts the fact that fi is irrational.
So, we conclude that f i + & is irrational. Or
Prove that - 5fi is irrational. 7
Solution. Let us assume to the contrary, that is a rational number. 7
. . 5f i - , wherep and q are coprime, i.e., their HCF is 1. - 7 q
7 P . Since P is a rational number, therefore, - = - 1s also rational number. 4 5 q
* & is a rational number.
But this contradicts the fact that is irrational.
5& . . This contradiction has arises because of our incorrect assumption that - 1s rational. 7
So, we conclude that - 5fi is irrational. 7
21. A two digit number is four times the sum of its digits and twice the product of the digits. Find the number.
Solution. Let the unit place digit be x and ten's place digit bey. Then, Original number = IOy + x It is given that a two digit number is four times the sum of its digits. . . lOy+x=4Cy+x) =, lOy-4y=42-x
U-Like CCE Sample Question Paper 1 33
* 6y = 3x * x 7 2y ... (1) Further, it is also given that a two digit number is twice the product of the digits.
1oy + x = 2(xy) * 1oy + 2y = 2[(2y)yI =z$ 12y = 4y2 * 3 = y Substituting y = 3 in ( I ) , we get
x = 2(3) = 6 Hence, the original number is 36.
Or Solve for x a i ~ d y :
3 ' - 2 a+b
a x - b y - a 2 - b 2 Solution. The given.system of equations can be written as
Y - 2 j b x = 2 = z $ b x + a y = 2 a b - + - - a b ab
and az -by=a2-b2 Multiplying (1) by b and (2) by a, we have
b2x + aby = 2ab2 and a2x - aby = a3-ab2 . '
Adding (3) and (41, we get b2x + a2x = 2ab2 + a3 - ab2
+ (b2 + a2)x = ab2 + a3 * (b2 + a2)x = a(b2 + a2)
+ x = a Substituting x = a in (2), we get
- by = - b2 + by=a2-a2+b2 * by = b2 + y = b Hence, x - a, y - b is the solution of the given system of equations. 22. I f a and p are the zeroes of the quadratic polynomial f ix) = & - &v + 7, find the
quadratic polynomial whose.zeroes are 2a + 38 and 3 a + 28. Solution. Since a and p are zeroes of the polynomial 2x2 - 5x i 7 , therefore
Let S and P denote respectively the sum and product of zeroes of the required polynomial, then
A A
and P = (2a + 3p)(3a + 2p) * P = 6a2 + 4ap + gap + 6p2 - P = 6a2 + 6p2 + 13up * P = 6(a2 + p2 + Zap) + up * P = 6(a + p12 + ap
=). . 75 7 82 p = - + - = - -41 2 2 2
Hence, the required polynomial p(x) is given by : p(x) = k(x2 - Sx + P)
or
or p(x) = k(@ - 252 + 82), where k is any non-zero real number.
l + c o s A + s i n A . .
23. Prove that : - 2 cosec A sin A 1 + cos A
Solution. We have
L.H.S. = l+cosA sinA
t sinA l+cosA
- - (1 + cos A)' + sin2 A sinA(l+cosA) .
- - ~ + ~ C O S A + C O S ~ A + S ~ ~ ~ A sin A (1 + cos A)
= L.H.S. 25. If ABC is an equilateral triangle with AD BC, then prove that m2 = ~DC'. Solution. Let ABC be a n equilateral triangle and AD I BC. I n AADB and AADC, we ha<e
AB =AC [given] LB = LC [Each = 6Oo1
and LADB = LADC [Each = 907 . . AADB - AADC ==s BD = DC ... (1) . . BC = BD + DC = DC + DC = 2DC ... (2) [using (i)]
In right angled AADC, we have AC2 = AD2 + DC'
j B C ~ = AD^ + DC' [.; AC = BC, sides of an equilateral A1 [using (2)l
* AD2 = 4DC2 - DC' =) AD' = 3Dc2 26. In figure, DEFG is a square and LBAC - 90". Show that DEZ = BD x EC.
C
Solution. In AAGF and ADBG, we have LGAF = LBDG
LAGF = LDBG So, by AA-criterion of similarity of triangles, we have
AAGF - ADBG .In AAGF and AEFC, we have
LFAG = LCEF LAFG = LECF
So, by AA-criterion of similarity of triangles, we have AAGF - AEFC -
From (1) and (2), we obtain ADBG - AEFC
=) BD DG - = - FE ' E C
[.; DEFG is a square.
[Each = 9Oo1 [Corresponding Lsl
[Each = 90°1 [Corresponding Lsl
:. EF = DE, DG = DE]
27. Find the mean of the following distribution, using step-deviation method :
So, the zeroes of x2 + 3x - 18 = (x + 6)(x - 3) are given by x = - 6 and x = 3. Hence, the other zeroes of the given polynomial are - 6 and 3. 30. Prove that in a right triangle, the square of the hypotenuse is equal to the
sum of the squares of the other two sides. Solution. Given :A righ't triangle ABC, right angled a t B. To prove : ( ~ ~ ~ o t e n u s e ) ~ = ( ~ a s e ) ~ + (~erpendiculad~
40 U-Like Mathematics-X
= A B ~ + B C ~ . L.e., Construction : Draw BD I AC proof: AADB - AABC. [If a perpendicular is drawn from the vertex of the .A
right angle of a triangle to the hypotenuse then triangles on both sides of thepeaendicular are similar to the whole A D . C triangle and to each other.]
so, A D A B - = -
AB AC [Sides are proportional]
a AD.AC = AB2 ... (1) Also, M D C 7 AABC [Same reasoning as above]
CD BC - = - .~~
so, BC AC
[Sides are proportional]
CD.AC=BC~ ... (2) Adding (1) and (2), we have ADAC.+ CD.AC = AB? + BC' - (AD + CD)AC = AB2 + B c 2
ACAC = AB2 + BC2 * Hence, AC? = -AB2 + B C ~
Or Prove that the ratio of areas of two similar triangles is equal to the square of
their corresponding sides., Solution. Given : AABC and APQR such that AABC - APQR.
To prove : ar(AABC) AB2 BC' C A ~ =-=-=- ar(APQR) PQ' Q R ~ p2
Construction : Draw AD I BC and PS I QR.
, . ' . 'A.A' B D c Q s R
J X B C X A D Proof: ar(AABC) - - 2
ar (APQR) 2 QR ps 2
a r (IIABC) - BC x AD - ar (APQR) QR x PS
Now, in M B and APSQ, we have LB = LQ
1 [Area of A = -(base) 2 x height]
... (1)
[As AABC - APQRI
U-Like CCE Sample Question Paper 1 41
LADB = LPSQ' [Each = 9Oo1 3rd B A D = 3rd LQPS
Thus, AADB and APSQ are equiangular and hence, they are similar. AD AB Consequently - = - ... (2) PS PQ
[Jf A's are similar, the ratio of their corresponding sides is same]
But AB BC' PQ - QR
Now, from (1) and (31, we get
ar(AABC) BC BC ar (APQR) = R &R - ar (AABC) -- B C ~ - ar(APQR) QR'
As AABC - APQR, therefore
Hence, ar f u c ) . A B ~ B C ~ C A ~ =-=-=- ar(APQR) P Q ~ QR2 Rp2
31. Ifx-tanA+sinAandy=tanA-sinA,showthat
2 - y 2 = 4 f i Solution. Given, tan A + sin A = x and tanA-sinA = y Adding (1) and (2), we get
2 t a n A = x + y Subtracting (2) from (I), we get
2 s i n A = x - y Multiplying (3) and (4), we get
(x + y)(x - y) = (2 tanA)(2 sin A) ==2. ~ ~ - ~ ~ = 4 t a n A s i n A
Now, 4& = 4,/(tan A + sin A)(tan A - sin A)
[..' AABC - APQR]
... (3) [using (211
[From (4) and (511
... (5)
[using (1) and (2)l
sin A.sin A = 4
cos A
= 4 tanA sinA From (5) and (61, we obtain
~ ~ - ~ ~ = 4 t a n A s i n A = 4 & '
* x2 -y2=4&
Or Evaluate :
2 2 5 -cosec2 58" --cot 58" tan 32" - -tan 13" tan 37" tan 45" tan 53" tan 77" , 3 3 3
Solution. We have 2 2 5 -cosec2 58' --cot 58" tan 32" - -tan 13" tan 37" tan 45" t an 53" tan 77" 3 3 3
2 = zcosecZ. 58" --cot 58" tan (90" - 58") 3 3
5 --tan 13" tan 37" tan 45" tan (90" - 37") tan (90" - 13") 3
32. Prove that : sin 8 (1 + tan 8) + cos 8 (1 + cot 8) = sec 8 + cosec 8.
Solution. We have L.H.S. = sin 8 (1 +tan 0) + cos 0 (1 + cot 0)
= (sin 8 + sin 0 tan 0) +'(cos 0 + cos 8 cot 0)
sin 8 +sine- cos 8
= (sin 0 + cos 0) + (sin2 8 cos2 01 (cos8+sine)
= (sin 8 + cos 8) + /sin3 e + cos3 e\ ( sinecose ) (sin 0 + cos 0)(sin2 8 + cos2 8 -sin 8 cos 0)
= (sin e + cos 0) + sin 0 cos 0
[.: a3 +.b3 = (a + b)(a2 + b2 - ab)]
sin2 e+cos2 8-siGecos8 = (sin 0 + cos 8)
sin 8 cos 0 I 1-sinecos0
= (sin 8 + cos 8) sin 8 cos 8 I
sinecose + 1-sinOcos0 = (sin 0 + cos 8)
sin 8 cos 8 I - - sin 6 + cos 8
sin 0 cos 8 (1)
- - sin 8 cos 0 + sin 0 cos 8 sin 8 cos 8
- 1 1 =-+- cos 8 sin 8 sec 0 + cosec 0
= R.H.S. 33. The coach of a cricket team buys 3 bats and 6 balls for f 3900. Later, he buys
another .bat and 3 more balls of the same kind for 4 1300. Represent this situation algebraically and geometrically.
Solution. Let the price of a bat be f x and that of a ball be 7 y . It is given that the coach buys 3 bats and 6 balls for 7 3900. . . 3x + 6y = 3900 Also, it is given that the coach buys another bat and 3 more balls for f 1300.
The algebraic representation of the given conditions is 3x + 6y = 3900 ... (1)
x + 3y = 1300 ... (2) Geometrical representation : Let us draw the graphs of the'ecpations (1) and (2). For
this, we find two solutions of each of the equations which are given in tables.
3x + 6y = 3900 x + 3y = 1300
Plot the points A(O,650), B(1300, O ) , C(100,400) and B(1300,O) on graph paper and join the points to form the lines AB and BC as shown in figure.
The two lines (1) and (2) intersect at the point B(1300,O). So, x = 1300, y = 0 is the required solution of the pair of linear equations.
-200*,
-4OOan
-600
-800
-1OOO'*
$1
v y' ..
34. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Draw a more than type ogive for the given data. Hence, obtain the median length of the leaves from the graph and verify the result by using the formula.
Solution. The given frequency distribution is not continuous. So, we &st make it continu- ous and prepare the cumulative frequency table by more than type given below :
Other than the given class-interval, we assume that the class-interval 180.5 - 189.5 with zero frequency. Now, we mark the lower class limits on x-axis and the cumulative frequencies along y-axis on suitable scales to plot the points (117.5,40), (126.5,37), (135.5,32), (144.5, 231, (153.5, 111, (162.5, 61, (171.5, 2). We join these points with a smooth curve to get the "more than" ogive as shown in the figure.
n 40 Locate - = - = 20 on the y-axis (see figure). 2 2
Number of leaves [Frequency 0 1
3 5 9
12 5 4 2
Length more tlian 117.5 126.5 135.5 144.5 153.5 162.5 171.5
Cumulative frequency (cf)
40 37 32 23 11 6 2 .
. From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determine the median length of the data (see figure), i.e., median length is 146.75 mm.
To calculate the median length, we need the frequency distribution table with the given cumulative frequency.