Model Paper - 1

Answers

1. 3x+2y=6 ; (k+1)x +4y = (2k+2).
   Here, a1 = 3, b1= 2, c1 = 6.
         a2 =k+1, b2=4, c2 = 2k+2.
   For a system of linear equations to have infinite solutions,
   We must have a1 = b1 = c1           Þ    3  =  2 =  6  
                         a2   b2     c2                  k+1    4   2k+2 
   \  Taking  3   =  2                        & Taking  2  =   6   
                  k+1    4                                       4     2k+2 
   \  12 = 2k +2                               \ 4k + 4 = 24 
   \  2k = 10                                    \ 4k = 20 
   \  k  = 5                                       \ k = 5 
   \ Ans is k = 5.

2. The Prime numbers from 20-50 are 23,29,31,37,41,43,47
   As the number of terms is seven (odd)
   \ The median = (7+1)th term  = 8th term 
                               2                 2
    =  4th term 
   median = 37 Ans.

3. Roots of the equation are3/2 & -5
   \   sum of the roots  =  3 + (-5)
                                      2
                                     = 3 - 5 
                                        2
                                     = 3 - 10
                                           2 
                                     =  - 7 
                                           2
   Product of the roots   = ( 3 )* (-5) = -15
                                        2                2
   Equation x2 -(sum of the roots) x + product of the roots = 0
    \ x2 + 7 x - 15 = 0
               2       2
   
    \ 2x2+ 7x - 15 = 0 Ans.

4. L.H.S. = sec2A - tan 2A
                 tan A 
              = sec2A - tan 2A 
                           tan A
              =    1       [ since sec2 A- tan2A = 1]
                 tan A 
              = cot A    [ since 1/tan A = cot A]
              =  R.H.S. 

5.     
   
           
    
             
   
               
   Age 
                 Group
             
             
   
               
   Population
             
             
   
               
   Number 
                 of Deaths
             
           
           
    
             
   
               
   0 -10
             
             
   
               
   35650
             
             
   
               
   310
             
           
           
    
             
   
               
   10 - 
                 20
             
             
   
               
   37750
             
             
   
               
   250
             
           
           
    
             
   
               
   20 - 
                 35
             
             
   
               
   -
             
             
   
               
   110
             
           
           
    
             
   
               
   35 - 
                 50
             
             
   
               
   30450
             
             
   
               
   -
             
           
           
    
             
   
               
   above 
                 50
             
             
   
               
   20150
             
             
   
               
   390
             
           
           
    
             
   
               
   1,40,000
             
             
   
               
   1200 
                 
             
           
         
   Population of age group 20-35 = 1,40,000 - (35650+37750+30450+20150)
                                                    = 26000
   No. of deaths of age group 35-50 = 1200 - (310+250+110+390)
                                                        = 140
   C.D.R. =  No of deaths       * 1000  
                  Total Population
               =  1200          * 1000
                   1,40,000
               = 8.57 per thousand. Ans.

6. x2 - 4x - 12 = (x-6) (x+2)
   x3 + 8 = (x+2) (x2-2x+4) 
   \ G.C.D. =  x+2. 

7.  sin44 ° + sin 46° = sin (90 - 46°)+sin (90 - 44°)
    cos46°    cos44°         cos46°          cos44°
     = cos46°/cos46° + cos44°/cos44° = 1 + 1 = 2. 

8. 
   
   
   Given  Þ ÐQ = 90° in D PQS
   Prove that  Þ QS2 + PR2 = PS2 +QR2
   Proof  Þ In D PQR
   PR2 = PQ2 + QR2 ....................... (By Pythegorus theorem) (1)
   \ QR.
    In D PQS.
    PS2 = PQ2 +QS2 ........................ (By pythegorus theorem)
    \ QS2 = (- PQ2 + PS2 )  .............. (2)
    \ (1) + (2)
    PR2 + QS2 = PQ2 + QR2 - PQ2 + PS2 
    PR2  + QS2 = QR2 + PS2.     ........ Hence Proved. 

9. 
   
   
   Given  Þ (1)  PQ || YZ
                (2) XP = 4 
                    PY    3
                (3) XQ = 6.6 cms.
   Find  Þ  l (QZ) 
   Proof  Þ since PQ || YZ. Then,
                  XP = XQ
                  PY    QZ
                \   4 =  6.6
                    3     QZ
                \ QZ = 6.6 * 3
                               4
                \ QZ = 19.8
                              4
                l (QZ) = 4.95 cms.

10. x3 +y3 = 91               By componendo & dividendo
    x3- y3     37
    x3+y3 + x3-y3  = 91+37       \ 2x3 = 128
    x3+y3 - x3+y3     91-37           2y3     54
    \x3 = 128                            \ x3 =64\x= 4 (Taking cube root on both sides)
     y3     54                                y3    27  y  3
    \ x : y=4:3 Ans proved.

11. Sales tax (difference) = 2700-2500
                                       = Rs. 200/-
    \ Rate of sales tax     = Sales tax   * 100
                                       Cost price 
                                       = 200   * 100 
                                          2500 
                                       = 8% 

12. 
    
    Given  Þ (1) ABCD is a square 
                 (2) l(AC) = 12 CM.
    Find  Þ Side of square
    Proof  Þ because  ABCD is a square
                 AB = BC = CD = AD .       ..........(1)
                 ÐA = ÐB = ÐC = ÐD = 90° .......(2) 
                 \ By Pythagorus theorem
                 AC2 =AD2 + DC2
                (12)2 = AD2 + AD2 ............ from (1)
                144 = 2AD2
                \AD2 =72
                \AD =6Ö2 cms. Ans 
          

13. Given 1) ABCD is a cyclic quadrilateral 
              2) ÐA = 2ÐC
    Find  Þ ÐA
    Proof  ÞÐA+ÐB+ÐC+ÐD=3600 
            ÐA+ÐC=1800 given (1)
            2 ÐC+ ÐC = 1800 given (2)
            3 ÐC = 1800
            ÐC = 600
            \ÐA = 2ÐC
            =2 * 600
            = 1200 Ans.

14.  
    
            
     
              
     
                
    x
              
              
     
                
    f
              
              
     
                
    fx
              
            
            
     
              
     
                
    5
              
              
     
                
    3
              
              
     
                
    15
              
            
            
     
              
     
                
    10
              
              
     
                
    6
              
              
     
                
    60
              
            
            
     
              
     
                
    15
              
              
     
                
    4
              
              
     
                
    60
              
            
            
     
              
     
                
    20
              
              
     
                
    k
              
              
     
                
    20k
              
            
            
     
              
     
                
    255
              
              
     
                
    3
              
              
     
                
    75
              
            
            
     
              
     
                
    Total
              
              
     
                
    16 + k
              
              
     
                
    210 + 20k
              
            
          
                Mean = 14.5
    
          Mean = åf(x) .x.
            	        åf
          14.5 = 260+20k
                        16+k
          14.5k + 232 = 210 + 20k
          \ 5.5k =22
          \k = 4 Ans 

15. 
    
    Given 1) AB --- diameter 2)P-centre
             3)BC =10
             4)AC=24               5) ÐC =900
    Find   Þ AP
    Proof  Þ In DABC ÐC=900     given 5)
                   \ AB2  =  AC2+BC2  .......... Pythagorous theoram
                   \ AB2  = (24)2+(10)2
                                 =576+100
                       AB2  = 676
                   \ AB = 26
                   
                   AP= ½ AB ---- given 1) & 2)
                   =1/2 * 26
                  AP = 13 units Ans.

16.  
         
    
            
     
              
     
                
    Commodity 
                  
              
              
     
                
    Quantity(q0)
              
              
     
                
    Price 
                  in 1999(p0)
              
              
     
                
    Price 
                  of 2000 (p1)
              
              
     
                
    p0q0
              
              
     
                
    p1q0
              
            
            
     
              
     
                
    P
              
              
     
                
    5
              
              
     
                
    4.20
              
              
     
                
    6.00
              
              
     
                
    21
              
              
     
                
    30
              
            
            
     
              
     
                
    R
              
              
     
                
    4
              
              
     
                
    5.00
              
              
     
                
    8.50
              
              
     
                
    20
              
              
     
                
    34
              
            
            
     
              
     
                
    A
              
              
     
                
    3
              
              
     
                
    8.00
              
              
     
                
    10.50
              
              
     
                
    24
              
              
     
                
    31.50
              
            
            
     
              
     
                
    G
              
              
     
                
    7
              
              
     
                
    9.00
              
              
     
                
    11.50
              
              
     
                
    63
              
              
     
                
    80.50
              
            
            
     
              
     
                
    S
              
              
     
                
    8
              
              
     
                
    12.50
              
              
     
                
    16.00
              
              
     
                
    100
              
              
     
                
    128
              
            
            
     
              
     
              
     
              
     
              
     
                
    Total :
              
              
     
                
    228.00
              
              
     
                
    304.00
              
            
          
          Cost of living index = åp1q0 * 100
                                         åp0q0   
                    = 304  * 100
                                       228
                                    = 133.33 Ans.

17. Cos2 45 + Sin 30 = 2Cos2A
    \( 1)2 + 1
       (Ö2)     2 = 2Cos2A 
    \ 1 + 1
        2     2 = 2Cos2A
    \ 2cos2A = 1
    \Cos2A = 1 
                       2
    because Cos A = 1
                             Ö2
    because Cos 450= 1
                               Ö2
     becauseA=450
    \ Sec2A=Sec2 450 =(Ö2)2
     = 2 Ans.

18. 3x-y = -6                                                      2x+y =4
    
    
            
    
              
    x
              
    0
              
    2
              
    x
              
    0
              
    2
            
            
    
              
    y
              
    -6
              
    0
              
    y
              
    4
              
    0
            
            
    
              
    (x,y)
              
    (0,-6)
              
    (2,0)
              
    (x,y)
              
    (0,4)
              
    (2,0)
            
          
    

19. LHS = a2 (b+c)+b2(c+a)+c2(a+b)+3abc
            = a2b+a2c+b2c+ab2+ac2+bc2+3abc
            = a2b+ab2+a2c+ac2+b2c+bc2+3abc
            = a2b+ab2+abc+a2c+ac2+abc+b2c+bc2+abc
            = ab(a+b+c)+ac(a+b+c)+bc(b+c+a)
            = (ab+ac+bc) (a+b+c)
            = R.H.S.

20. 
    
    Given  Þ in DPQR 
    1) O is midpoint of seg QR
    2) M is midpoint of Seg OP
    3) Seg QS meets side PR at S
    4) QS ïïOT
    Prove  Þ PS =1/3PR
    Proof  Þ In D POT
    MS ïïOT ..................Given 4)
    M is the midpoint of seg OP ...............given 2)
    \ S is the midpoint of seg PT -(A line is drawn through the midpoint of 
    one side and parallel to the second side bisects the third side)
    \PS = ST ............I
    In D SQR    OT ïï QS .............given 4)
                       O is the midpoint of QR
                       \ T is the midpoint of SR
                       \ ST = TR ..............II
    \ from I & II        PS=ST=TR
    But PS+ST+TR=PR
    \PS+PS+PS=PR
    \3PS=PR        \PS =1/3 PR is proved.

21. Let radius of smaller circle be x
     \radius of larger circle = 2x-1
     \Area of smaller circle = px2
     Area of larger circle = p(2x-1)2
     \As per conditions given
     px2+p(2x-1)2 =34 p
     p[x2+4x2-4x+1]=34p
     \5x2-4x+1=34
     \5x2-4x-33=0
     \5x2-15x+11x-33=0
     \5x(x-3)+11 (x-3)=0
     \(x-3) (5x+11) =0
     \x-3=0 & 5x+11=0
     \x=3 & x = -11/5
     because radius can not be -ve
     \ x  = 3  Ans

23.  For the right circular cyliner.
      X =10cm & h=30 cm
      \volume of the right circular cylinder = pr2
      =p * 102 * 30 =p x100 * 30cm3
      For the cone, radius (R) of the base = diameter = 2 =1cm
                                                                2           2 
     height (H) =10cm
     \volume of each cone =1/3pR2H=1/3p*12*10 =10p cm3
                                                                         3 
     Number of cones prepared from the cylinder
     = volume of the cylinder = px100*30
        volume of each cone        10p/3
     =p *100 * 30* 3/10p = 900 Ans
    

24.       
    
            
     
              
     
                
    Class
              
              
     
                
    Frequency(f)
              
              
     
                
    Mid - Interval (a)
              
              
     
                
    f(a)
              
            
            
     
              
     
                
    00 
                  - 10
              
              
     
                
    3
              
              
     
                
    5
              
              
     
                
    15
              
            
            
     
              
     
                
    10 
                  - 20
              
              
     
                
    9
              
              
     
                
    15
              
              
     
                
    135
              
            
            
     
              
     
                
    20 
                  - 30
              
              
     
                
    15
              
              
     
                
    25
              
              
     
                
    375
              
            
            
     
              
     
                
    30 
                  - 40
              
              
     
                
    8
              
              
     
                
    35
              
              
     
                
    280
              
            
            
     
              
     
                
    40 
                  - 50
              
              
     
                
    5
              
              
     
                
    45
              
              
     
                
    225
              
            
            
     
              
     
                
    Total
              
              
     
                
    40
              
              
     
                
    125
              
              
     
                
    1030
              
            
          
    
          Mean = åf(x) =1030 = 25.75 Ans
                         åf      40 

25. a , b are roots of the equation
    \ a+b =5/4 & ab =4/4 =1
    for new equation the roots are a2,b2.
    \sum of the roots  Þ a2+b2 =(a+b)2-2ab
    = (5/4)2 -2(1)
           
    = 25 -2
       16
    = 25-32
         16
    = -7
       16
    Product of the root  Þa2b2 = (ab)2
    =(1)2
    = 1
    \The required equation is x2-(sum of roots)x + Product of roots =0
    \x2+7x/16+1 =0
    \16x2+7x+16=0 Ans

26.     
    
          
     
            
    Total Annual Income
            
    Rs.74,600/-
          
          
     
            
    Standard deduction 1/3of Rs.74,600 or Rs. 10,000 whichever is less
            
    (- Rs. 15,000)
          
          
     
            
    Income after Std. Deduction
            
    Rs.59,600/-
          
          
     
            
    Income tax on Rs.35,000/- 
            
    NIL
          
          
     
            
    Income tax on 24,600/- @ 20%
            
    Rs.4920/-
          
          
     
            
    Total income tax 
            
    Rs.4920/-
          
          
     
            
    Savings for tax rebate 
            
    
          
          
     
            
    P.F.600x12 = Rs.7200/-L.I.C. Rs.2600/-
            
    
          
          
     
            
    Total Savings Rs.9800/- 
            
    
          
          
     
            
    Rebate in tax @ 20%
            
    (-) Rs.1960/-
          
          
     
            
    Income tax payable                                                            Ans.
            
    Rs.2960/-
          
        
     
        

27. 
    
    

    Let QR = Tower, SR = H, RS=Flagstaff 
    P is any point on the plane
    ÐRPQ = A0 & ÐSPQ = B0
    From D ABC
    Tan A = QR
                   PQ
    In D ABD
    Tan B = QS = RQ+h
                 PQ      PQ 
    Tan B  =   RQ+h
    Tan A         QR
    \ QR . Tan B = (RQ+h) tan A
    \ QR. tan B = RQ. tan A +h . tan A
    \ h. tan A = QR tan A-QR tan B
    \ h. tan A = QR(tan A-tan B)
    \ QR  =    h.  tan A   
                    tan A-tan B 

28.       a (x-a)    
    å (a-b) (a-c)
    =   a (x-a)    +       b(x-b)    +     c (x-c)  
       (a-b) (a-c)     (b-c) (b-a)     (c-a) (c-b)
    =  -a (x-a)(b-c) -b (x-b) (c-a) -c (x-c) (a-b)
                      (a-b) (b-c) (c-a)
    =  -a ( bx - cx - ab + ac) -b ( cx -ax- bc + ab ) -c (ax - bx -ac +bc)
                                       (a-b) (b-c) (c-a)
    =  a2b- a2c+b2c-ab2+ac2-bc2
                 (a-b) (b-c) (c-a)
    =  a2(b-c) +bc (b-c)-a2(b2-c2) 
                 (a-b) (b-c) (c-a)
    = (b-c) [a2+bc-a (b+c)]
            (a-b) (b-c) (c-a)
    = a2+bc-ab-ac 
         (a-b) (c-a)
    = a(a-c)-b(a-c)
      (a-b) (c-a)
    = (a-b) (a-c)
       (a-b) (c-a)
    = -(c-a)
        (c-a)
    = -1  Ans.

29. Given 1) ÐQ=90 0; 2) A & B are midpoints of PQ & QR respectively.
                   PB2 =PQ2+QB2
    Prove  Þ 4 (RA2+PB2) =5(PR)2   
    Proof  Þ In D AQR 
          AR2=AQ2+QR2 ..................1)
    
    In D PQB 
          PB2=PQ2+QB2 ...................2)
    by adding 1 & 2
    AR2+PB2=AQ2+QR2+PQ2+QB2 ........... 3)
    
    In D PQR & D AQB 
          PR2=PQ2+QR2 & AB2 = AQ2+QB2........4)
          \ AR2+PB2=PR2+AB2      By substituting 3  Þ 4
         A & B are midpoints of PQ & QR respectively.
          \ AB = ½ PR2
    
          \ AR2+PB2 = PR2+(1/2PR) ............... Substituting in 4
          \ AR2+PB2 = PR2+1/4PR2
          \ 4(AR2+PB2) = 4PR2+PR2
             4 (AR2+PB2) = 5PR2 Ans.

30. 
    

     
    In right D APC OP = 5-y
    
    y2 = 62-x2
    y2 = 36-x2
    x2 = 36-y2 .............(I)    
    In right D OPC,
    x2 + (5-y)2 = 52
    x2 + 25 - 10y + y2 =25
    x2 - 10y - y2 ...................(II)
    
    From (I) & (II)
    10y - y2 = 36 - y2
    10y = 36
    y =  3.6
    Substituting y = 3.6 in (I) we get
    x2 = 36 - (3.6)2 
    x2 = 36 - 12.96
    x2 = 23.04
    x = Ö23.04
    x = 4.8
    BC = 2x = 2*4.8 = 9.6 cm Ans.