X Maths Sqp1 Sol

An

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Mock Test Paper – I Mathematics Class X

m o .r c u o H g n i n r a e .L w w w Solution

Section A

1.

17 has none of the 2, 3, 5 as its factors so given rational no. is in simplest form Also 90 = 2 × 32 × 5  2m ×5n. 

2. 3.

17 is a non-terminating repeating. 90

The graph intersects the x-axis at one point the number of zero is one. 3x + y – 1 = 0

... (1)

2(3x + y – 1) = 0

… (2)



a1 b c  1  1 a2 b2 c2

The pair of linear equation is consistent with infinite solutions.

4.

Put x  3 in the equation

( 3) 2  3 3  3  6  0

3 – 9 + 6  0  0 = 0, which is true

 x  3 is soln of the equation.

5.

tan  = 1, AB2 = AC2 + BC2  AB = sin =

2

1 1 , cos = 2 2

 2sin  cos = 2 

1 1   1 ; proved. 2 2

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A

6.

5 .6

cm

m o .r c u o H g n i n r a e .L w w w B

3.2 cm

C

6 cm

In  ABC, AD is the bisector of A



BD AB 3.2 5.6    DC AC 6  3.2 AC

 AC

7.

2.8  5.6  4.9 cm 3.2

 POQ + PTQ = 180°

 PTQ = 180° – 110° = 70°  x =70°

8.

2r – 2r = 105 2r (– 1) = 105

22  105  7  1  105  r   24.5 cm 2  15  7 

 2r 

9.

P (E) + P ( not E ) = 1

= P(E) = 1 – P (not E) = 1 

10.

1 5  6 6

a 

5 6

x 10   xi 10  5  50

new x 

new x 

x

i

5

  xi  50  3  5  35

3 5 7 5

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Solution B 11.

Sum of zeros =

(coefficient of x ) (coefficient of x 2 )

c  ab  c  ab 1

m o .r c u o H g n i n r a e .L w w w 

Product of zeroes =

 (constant term) coeff .of x 2

=

 abc  – abc 1

Or

3x  8 2x  5 6 x2  31x  47  6 x2 15x

–

+

16x  47 16x  40

+

12.

– 7

 Quotient = 3x – 8 Remainder = 7

sin 5 A = cos 4A

sin 5A = sin (90° – 4A) 5 A = 90 – 4 A

5 A + 4A = 90  A =

13.

90 = 10° 9

In  ABP and  AQD, we have, BAQ = AQD (alt S) B = D

ABP ~  QDA (AA ~ Rule)



AB BP AB × AD = BP × DQ  DQ AD

Hence AB × BC = BP × DQ = [ AD = BC]

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14.

Let r be the radius of circle area of circle = r2

m o .r c u o H g n i n r a e .L w w w  r2 2

Area of semicircle =

2

 r  45  r  360 8

Area of sector ROQ =

 Area of sector POR 

2

 r 2  r 2 3 r 2   2 8 8

 P(E) of the spinner to land in region II =

15.

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3 r 2 / 8 3 = 8 r2

AB = BC

2

2



1  6   3  1



 25  16  =



41 = K2 – 2K + 26



K2 –2K – 15 = 0



(K – 5) (K + 3) = 0



K = 5 or – 3



2

2

 K  1  8  3 

K 2  2 K  1  25

Section C

16.

Let the length of altitude be xcm length of hypotenuse = 2x + 1

length of base = 2x + 1 –2 = 2x – 1 Using pythogoras thm, we get x2 + (2x –1)2 = (2x + 1)2  x2 – 8x = 0

 x(x – 8) = 0  0 or x = 8 rejecting x = 0 length of sides of  are 8 cm, 17cm and 15 cm.

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Or

 a 2  b2    0 17. x² + bx –  4  

m o .r c u o H g n i n r a e .L w w w b a2  b2    x² + 2 2 x =   4 b x + 2  2x +   2

2

a2  b2  b  b   = +   4 2 2

2

2

b a2  x    =  2 4 

 x

18.

b a = 2 2

 x

b a  2 2

 x

b  a b  a , 2 2

tn = 5 – 6 n

t1 = 5 – 6×1 = –1

 a = –1 and l = 5 – 6n  Sum of n terms, Sn 



n a  l  n  1  5  6n  2 2 n 4  6n  n 2  3n 2

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19.

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Let a be any positive integer of the form 5q + 1 a = 5q + 1 a2 = (5q + 1)2

m o .r c u o H g n i n r a e .L w w w  a2 = 25q2 + 10q + 1 = 5 (5q2 + 2q) + 1 =5m+1

where m = 5q2 + 2q Hence proved.

20.

To prove;

cos A sin A   sinA  cosA 1  tanA 1  cotA

L.H.S,

cos A sin A  1  sinA/cosA 1  cotA/sinA

=

cos 2 A sin 2 A  cosA–sinA sinA–cosA

cosA  sinA cosA  sinA  cos 2 A sin 2 A  = cosA–sinA cosA–sinA  cosA  sinA  = cos A + sin A = RHS Hence proved.

Or

4  cos70  cosec 90   70   3cos 55   7sin 90   55   7  tan5  tan85   tan25  tan65   tan45 

=

3cos 55 4 cos70 sec70  7 cos55 7 tan5 tan 90 –5  tan 25 tan 90  25  1

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

22.

3 4  7 7  tan 5 cot 5 tan 25 cot 25 

m o .r c u o H g n i n r a e .L w w w =

21.

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3 4 3  4 1    7 7 7 7

Solution for Q. 21 to be done by students.

Let A (4,6) , B (0, 4) andC(6,2) be the vertices of the triangle ABC. Let P(x, y) be the circumcentre of ABC. Then, PA= PB = PC

 PA² = PB² = PC²

Now, PA² = PB²

 (x – 4)2 + (y – 6)2 = (x – 0)2 + (y – 4)2  2x + y = 9

… (i)

Also PB2 = PC2

 (x – 0)2 + (y – 4)2 = (x – 6)2 + (y – 2)2

 12x – 4y = 24  3x – y = 6 Adding (i) & (ii), we get

… (ii)

5x = 15  x = 3

23.



y = 9 – 2x (from i)



y=9–6=3



x = 3, y = 3

The coordinates of any point on y-axis are of type (O, y). Let point be P(O, y) Let A and B denote given points (–5, –2) and (3, 2) respectively

 AP = BP 

2

2

0  5    y  2 



2

2

0  3    y  2 

 25  y 2  4  4 y  9  y 2  4  4 y

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 16   8 y  y  2

m o .r c u o H g n i n r a e .L w w w  The coordinates are (0, –2)

24.

Since the tangents from an external point to a circle are equal in length

A

B

P

C

and

 BP = BQ

… (i)

CP = CR

… (ii)

AQ = AR

… (iii)

R

Q

Now, perimeter of ABC =

AB + BC + AC

=

AB + (BP + PC ) + AC

=

AB + BQ + AC + CR (from (i) & (ii))

=

AQ + AR

=

2 AQ [ AQ = AR]

Or

In ABC, B=90°

 CA2 = AB2 + BC2 [by pythagoras theorem]

(Now AD2 = AB2 + BC2 + CA2 – given) = CA2 + CD2 [ CA = CD]

Now , in ACD, we have AD2 = CA2 + CD

 ACD = 90°

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25.

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Area of region ABCD = area of sector AOC – area of sector BOD

40 22  40 22    14  14    7  7  cm 2 360 7  360 7 

m o .r c u o H g n i n r a e .L w w w =

22 154 cm² = (28.7)cm2 = 9 3

Or

Area of field 

Total cost of ploughing Rate per m 2

 5775  2  m  3850 m²  1.5 

=

Let r be the radius of field, area of circle = r2  r2 = 3850  r  3850

7 = 22

Circumference of field = r = 2  Cost of fencing the field = 220 

35  35  35

22  35 = 220 m 7

17  Rs.1870 2

Section D

26.

If a line is drawn parallel to one side of a triangle intersecting the other two sides then it divides the two sides in the same ratio. A

F

D

Given : A  ABC in which DE | | BC and intersects AB in D and AC in E

G

E

To prove :

B

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AD AE  DB EC

C

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Construction : Join BE, CD and draw EF  BA and DG  CA. Proof : Since EF  AG.EF the height of s ADE and DBE.

m o .r c u o H g n i n r a e .L w w w Now, Area (ADE)=

1 (AD.EF) 2

1 Area ( DBE) = DB . EF 2 

Area  ADE  Area  DBE 



½ AD.EF  ½ DB.EF 



AD DB

… (i)

Similarly

Area  ADE  Area  DEC 



AE EC

… (ii)

But DBE and  DEC are on the same base and between the same parallel DE and BC. ar (DBE) = ar (DEC)

… (iii)

from (i), (ii) and (iii) ar ADE  ar DEC 





ar ADE  ar  DBE 

AE AD = EC DB

Now, LM | | AB in ABC

AL BM = LC MC



AL BM  AC  AL BC  BM



x3 x2  2 x  ( x  3) (2 x  3)  ( x  2)



(x  3) ( x  5)  (x  2)( x  3)



x ²  2 x  15  x ²  x  6



x=9

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27.

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For the upper portion of the shuttle cock r1 = 1 cm, r2 = 5/2 cm, h = (7 – 1) cm = 6

m o .r c u o H g n i n r a e .L w w w 2

l  h2  r2  r1  

36  2.25 cm =

38.25 cm 6.2(approx)

C.S.A. of frustum part = l (r1 + r2) =  × 6.2 ×(2.5 + 1) cm² = 21.7  cm² C.S.A of hemisphere part with radius 1cm = 2 r12 = 2 (1)2 = 2 cm²

Total S.A. = (21.7+2) cm² = 23.7 

22  74.49 cm²  74.5 cm² 7 Or

Inner radius of the glass (r) =

7 cm Height (h) = 12 cm 2

Apparent capacity of the glass = r2h

22 7 7    12  462 cm³ 7 2 2

volume of inverted hemispherical bottom =

2 22 7 7 7 539 2     cm³ r³ =  3 7 2 2 2 6 3

 Actual capacity of glass = 462 

28.

539 2233 1  cm³  372 cm³ 6 6 6

Let the original no. of student in room Aand room B be x and y resp. 5 students shifted from : x – 5 = y + 5 A to B

5 students are shifted : x + 5 = 2(y – 5) from B to A

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 The pair of linear equation is

x  y  10

… (i)

x  2 y  15

… (ii)

m o .r c u o H g n i n r a e .L w w w x

10

5

x

5

–5

y

0

–5

y

10

5

From the graph (to be drawn) we see that the lines intersect at P (35, 25) No. of students in room A = 35



No. of students in room B = 25

29.

Let B be the window of a house AB and let CD be the other house. Then AB = h m Let CD = H metres, CE = AB = h m

D

ED = (H – h) m

(H-h)m

From rt BED

BE BE  cot    cot  ED (H  h)

 BE = cot  (H – h)

… (i)



E

In rt d  BEC tan  =

Hm



CE  BE  h cot  BE

… (ii)

hm

From (i) and (ii)



(H – h) cot  = h cot 

H = h

30.

(cot   cot )  h(1 + tan  cot ) cot 

Class

Frequency

Med pt. ni

f in i

0-20

17

10

170

20-40

f1

30

30f1

40-60

32

50

1600

60-80

f2

70

70f2

80-100

19

90

1710

Total

120

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f

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17  f1  32  f 2  19  120



68 + f1 + f2 = 120



f1 + f2 = 120 – 68



f1 + f2 = 52

m o .r c u o H g n i n r a e .L w w w  f x  17  30 f i i

Mean =

1

 1600  70 f 2  1710 = 3480 + 30f1 + 70f2

3480  30 f1  70 f 2  50 120

 30 f1 + 70 f2 = 2520

…(ii)

By solving (i) and (ii) we get f1 =

112  28 4

f2 = 52 – 28 = 24

Or

The classes are discontinuous form New Class Intervals. Height (in cm)

No. of student

159.5-162.5

15

162.5-165.5

118

165.5-168.5

142

168.5-171.5

127

171.5-174.5

18

1 = 165.5, h = 3, f1 = 142, fo = 118 f2 = 127



f1  f o    h 2 f  f  f 2   1 0

Mo = l + 

142  118   24   3 = 165.5 + 3 39  2  142  118  127 

= 165.5 + 

= 165.5 +

24  165.5  1.85 = 167.35. 13

........End of Solution........

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