ssc english medium previous maths exam paper 2019

                                                      Section-1

1) If A={x:x is a factor of 24}, then find n(A)

sol)  Given A={x:x is a factor of 24}

x is a factor of 24 , it mean any number "x" which can divide "24"  is a factor

24/1 -= 24

24/2=  12

24/3 = 8

24/ = 6

24/7 = we cannot able to divide "24" with >>7 

24/8= 3

24/ 9 = we cannot

24/10= we cannot

24/11= we cannot

24/12= 2
"
"
"
24/24 = 1


we got = { 1,2,3,4,6,8,12,24} = total eight numbers

.^. n(A) = 8

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2). Find the HCF of 24 and 33 by using division algorithm

sol) Division algorithm : We find the H.C.F of two numbers by dividing the larger number with smaller, the smaller by the remainder, the first remainder by the second remainder, and so on, until exact division is obtained(when the remainder is zero), which is the HCF of these two numbers.

Here the two numbers are = 24, 33

larger number= 33

smaller number = 24

Lets, apply division algorithm , that is {dividing the larger number with smaller, the smaller by the remainder, the first remainder by the second remainder, and so on, until exact division is obtained(when the remainder is zero)



  24 ) 33 ( 1            
       24
                 9 )24 ( 2 <-----Here, dividing 1st remainder(24) with 
            18               2nd remainder(9)
             6 ) 9 ( 1
                  6  
exact divisor>> 3 )6( 2
                      6  
                      0 <-------- remainder


We found that exact divisor is "3"

Hence, the H.C.F of (24,33) = 3




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3). Radha says "1,1,1......are in A.P and also in G.P".DO you agree with Radha? Give reason

sol) A.P :- We say the series are in A.P if common difference is same

i.e, common difference d1=d2

here a1= 1,   a2=1,    a3= 1

d1 =  a2-a1  = 1-1 =0

d2 = a3 - a2 = 1-1=0

Here the difference of any two consecutive terms in each case is "0".

So, the given list is in A.P whose first term a= 1 and common difference d= 0


Geometric progression or G.P:- we say the series are in G.P if each successive term(next term) are obtained by multiplying the preceding(previous) term by a fixed number. 

This fixed number is called the common ration"r"of GP

or common ration are same

Here the series is "1,1,1.......

a1=1,      a2= 1,     a3= 1

r1 =a2/a1 = (1/1) = 1

r2= a3/a2 = (1/1) = 1

Common ration are same "1"

Hence the series"1,1,1.....both are in A.P and in G.P

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4). If p(x) = x^4 +1, then find P(2) - P(-2)

sol) Given equation : P(x) =X^4 + 1

P(2) =  (2)^4 + 1

=> (2*2*2*2)  +  1

=> 16+1= 17


P(-2) = (-2)^4 + 1

=> (-2 * -2 * -2 *-2) +1

=> 16 + 1

=> 17

Now, 

P(2) - P(-2) = 17 - 17 = 0

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5). Find the roots of quadratic equation x^2+2x-3= 0

sol)  Given :- x^2 +2x - 3= 0

Here a = 1,  b=2,  c=-3

select the  two terms(x,y) in such a way that if you multiply (x*y) we should get "c" here c= -3 and

 if we add (x+y) we should get middle term "b" here it's b= 2 

3*-1 = -3  or ( 3x*(-X=-3x) (we got c=-3)

3-1= 2  or  (3x-x = 2x))( we got middle term b= 2)


x^2 + 3x -x -3 = 0

=> x(x+3) -1(x+3) = 0

=> (x-1)(x+3) = 0

=> x-1=0  => x =1

=> x+3=0 => x =-3

.^. The roots of the quadratic equation x^2+2x-3=0 is (1,-3)

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6). Find the centroid of a triangle PQR , whose vertices are P(1,1), Q(2,2), R(-3,-3)

sol) Centroid formulae=(x1+x2+x3),  (y1+y2+y3)
                                                 3                    3

P(1,1) = (x1, y1)

Q(2, 2) = (x2,y2)

R(-3,-3) = (x3,y3)

(x1+x2+x3) = (1+2-3) = 0/3=0 
        3                   3

(y1+y2+y3)=  (1+2-3) = 0/3=0
        3                   3

.^. (x,y) = centroid = (0,0)


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7) For what value of "t" the following pair of linear equations has a no solution 2x+ty=5 and 3x+2y=11

sol) The pair of linear equations has no solution if it satisfy the following condition (a1 = b1 =/= c1 )
                                   a2    b2        c2

Given:- 2x+ty = 5  or  2x+ty-5=0 (a1=2, b1=t, c1=-5)

           3x+2y=11  or  3x+2y-11=0 (a2=3, b1=2, c1=-11)


a1  = b1
a2     b2

2 =  t
3     2

2*2 =  t
  3

4 = t
3

 4 =t
 3

a1 = b1 =/= c1
a2    b2        c2


2 = ( 4/3) =/=  -5  
3        2           -11

2  = ( 4  *  1) =/=  5
3        3      2         11

2 = 2 =/= 5
3    3       11

Hence, it has no solution

          or

(*) Graphical representation :- parallel lines

(*) Algebraic interpretation :- no solution


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                    section-2

8) If µ=(0,1,2,3,4,5,6,7,8,9,10) ,A={2,3,5,8} and B={0,3,5,7,10}.Then represent A ∩ B in the Venn diagram.

sol)  In A ∩ B {3,5} are common in both sets

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9) Akhila says " points A(1,3) B(2,2) C(5,1) are collinear".Do you agree with Akhila?why?

sol) Given =A(1,3)  B(2,2)  C(5,1)

Q) What is collinear points?
A) Three or more points P1,P2,P3......Pn are said to be collinear if they lie on the single straight  line "L".

A(1,   3)  B(2,    2)  C(5,     1)
  (x1,y1)    (x2, y2)     (x3, y3)

=> 1 [ 1(2-1)  + 2(1-3)  + 5(3-2)] = 0
     2

=> 1  [1(1)  + 2(-2)  +5 (1) ] = 0
     2

=>1/2 [1 -4 +5] = 0

=> 1/2  [ 2] = 0

=> 1=/= 0

I dont agree with akhila because left hand "1"side is not equal to right hand side"0"


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10) Write the quadratic equation, whose roots are 2+_/3 and 
       2-_/3

sol) Here 𝛂 = 2+√3  and 𝛃 = 2 - √3

(*) Sum of the roots = 𝛂 + 𝛃

=> (2+√3)  +  (2 - √3)

𝛂 + 𝛃= 4  ------------------------(1)


(*) product of the roots = 𝛂 * 𝛃

=> (2 + √3)  *  (2 - √3)
      (a + b)    *  (a - b)  = a^2 - b^2

=> (2)^2  - (√3)^2

=> 4 - 3

𝛂𝛃 = 1  ---------(2)

We have, 

=> (x)^2 - (𝛂 + 𝛃)x + 𝛂𝛃 =0 -------(3)

substitute (1) and (2) in (3) we get

=> (x)^2 - (4)x - 1 = 0

.^. the quadratic eq ax^2+bx+c = x^2-4x-1=0

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11). Divide x^3 - 4x^2 +5x - 2 by (x-2)

sol) x-2 )    x^3  -  4x^2  +  5x  -   2(x^2 -2x +1
                   x^3  -  2x^2
                          (+)         
                           -    2x^2  +  5x  -  2
                           -    2x^2  +  4x
                         (+)           ( - )     
                                                1x  -  2
                                                1x  -  2
                                                                      (-)      (+)       
                                                                             0        


p (x)  ➗ g (x)  => q (x)

x^3 - 4x^2 +5x - 2  ➗ (x-2) => (x^2-2x + 1)


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12) write the formula of nth term of G.P and explain the terms in it.
                                                           
sol) nth term of G.P is given by = ar^(n-1)

1) a = is the first term in G.P

2) r = common ration of the term

3) n = number of terms in G.P

example :- 2,4,8,16.....

a = 2

r1= a1/a = 4/2 = 2

r2 = a2/a1 = 8/4 = 2

common ration

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13) solve the pair of linear equations 2x + 3y = 8  and x + 2y = 5 by elimination method.

sol) eq(1) : 2x + 3y = 8  
       eq(2) : x   + 2y = 5

elimination method :-

multiply eq(1)  with 1
multiply eq(2)  with 2,  we get

(2x  +  3y  = 8) * 1  => 2x  +3y  = 8
(x    +  2y  = 5) * 2  => 2x  +4y  = 10
                                    (-)   (-)       (-)
                                            -y     = -2  => y =2


substitute "y=2" in eq (2) we get,

=> x + 2(2) = 5

=> x + 4 = 5

=> x = 1

.^. X=1  and Y=2

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                           section-3

14 (a) draw the graph of the polynomial p (x) = x^2-7x+12, then find its zeroes from the graph.

sol) p (x) = x^2 - 7x  +12

=> x^2 - 3x - 4x + 12 = 0

=> x(x-3) -4(x - 3) = 0

=> (x-4) (x-3) = 0

=> x = 4 or x = 3

=> (0)^2 - 7(0) + 12 => 0-0+12=12

 => (1)^2 - 7(1) + 12=>1 -7+12= 6

=> (2)^2 - 7(2) +12 => 4-14+12=2

=> (3)^2 - 7(3) +12=> 9-21+12= 0

=> (4)^2 - 7(4) +12=> 16-28+12=0

=> (5)^2 - 7(5) +12 => 25-35+12= 2

=> (6)^2 - 7(6)+12 => 36-42+12= 6


    x                =         0   1   2   3   4    5  6
y=(x^2- 7x+12)       12  6   2   0    0   2  6

y= x^2 -7x +12 intersect x-axis at x=3, x=4

Hence, x=3  and x=4 are zeroes of a given polynomial

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b) Solve the equations graphically
      3x+ 4y = 10  ------(1)
      4x-  3y = 5    ------(2)

sol)  Multiply eq(1) with"4"and eq(2) with "3"

4*(3x + 4y = 10) => 12x + 16y  = 40

3*(4x - 3y  = 5)  =>  12x  -  9y   = 15
                                (-)     (+)       (-)   
                                          25y    = 25

=> 25y = 25

=> y = 1


substitute "y=1" in any equation lets take eq(2)

=> 4x - 3y = 5

=> 4x - 3(1) = 5

=> 4x = 5+3

=> x = (8/4) = 2

.^. x= 2  and y=1

Graphically:-

(1) 3x + 4y = 10

     4y = 10 - 3x

       y = 10 - 3x
                  4

y(-2) = 10 - 3(-2)  =>  10+6 => 16 =>4
                   4                   4          4 

y(-1) = 10 - 3(-1)  => 10 +3 => 13 => 3.25
                   4                   4          4      

y(0) = 10 - 3(0) => 10+0 => 2.5
               4                  4 

y(1) = 10 - 3(1) => 10-3 => 7 => 1.75
               4                 4         4

y(2) = 10 - 3(2) => 10-6 => 4 => 1
                4                 4        4

------------------------------------------------
x                   =      -2     -1      0     1      2

y = 10 - 3x   =      
           4               4  | 3.25 | 2.5 |  1.75  | 1


(2) 4x - 3y = 5

      4x - 5 = 3y

      y = 4x - 5
                3

y(-2) = 4(-2) - 5 => (-8 -5) => - 13  => - 4.3
                     3              3             3

y(-1) = 4(-1) -5 => (-4-5) => -9 => -3
                   3               3          3
      
y(0) = 4(0) - 5  => -5 => -1.6
                3             3

y(1) = 4(1) - 5 => 4-5 => -1 => -0.3
                  3           3         3 

y(2) = 4(2) - 5 => 8 -5 => 3 => 1
                  3            3        3
           


   x           =   -2       -1       0       1       2

y= 4x-5   = -4.3      -3     -1.6    -0.3    1  
        3                                                    

.^. The point of intersections of two eq(1) & (2) is at (2,1)


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15 (a) Find the ratio in which x-axis divide the line segment joining the points (2, -3) and (5, 6). Then find the intersecting point on X-axis.

sol) Given :- (2,   -3)  & (5,    6)
                      (x1,y1)       (x2, y2)

Given:- x-axis dividing the line segment.

we know that.In x-axis (y=0)

We have section formula:-

P(x,0) =           "

x1 = 2,  x2=5,  y1 = -3,  y2=6

x = m1(5) + m2(2)         (1)
             m1 +m2


0 = m1(-3) + m2(6) --------(2)
              m1 + m2

-3(m1) + 6(m2) = 0 * (m1+m2)

-3m1 + 6m2 = 0

-3m1 = -6m2

m1 = 6 = 2
m2    3    1

m1 : m2 = 2:1  is the ratio in which it divide the line segment--------(3)

substitute (3) in (1) we get,


x = 2(5) + 1(2)
          2  +  1

x = 10 +2
           3

x = 12/3 = 4

=> p(x,0) = P (4,0) is the intersecting points on x-axis


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(b) find the sum of all two digit odd multiples of "3"

sol) we have to find two digit ODD multiples of "3"

we cant take 6,9,  as its one digit  ,

we  cannot take 12 as its 2 digit but  "even" not odd

=> 15, 21,27,33,39,45,51,57,63,69,75,81,87,93,99 =n= total 15

We have,

Sum of number:-

Where, n = No. of terms = 15

d= diff between two consecutive numbers=21-15=6

=> Sn = 15 [2 * 15 +(15-1) * 6]
              2

=>  Sn     = 15 [ 30 +(14*6)]
                    2

=>  Sn       = 15 [ 30 + 84]
                      2

=>  Sn       = 15 [ 114 ]
                      2


=>  Sn        = 15 * 57

.^. sum of all two digit odd multiples of "3"is Sn = 855


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16 (a) If A ={x:2x +1, x Ɛ N, x ≤ 11},

  If B ={x : x is a composite number, x < 12}

then show that (A U B ) - (A n B)= (A-B) U (B -A)

sol) In set A its given x belongs to(Ɛ ) N(natural number= 1,2,3......). and it also less than or equal to 11

x : 2x+ 1:-

(1) : 2(1) +1 = 3

(2) : 2(2) +1 = 5

(3): 2(3) +1 = 7

(4): 2(4) +1 = 9

(5) : 2(5) +1 = 11  (x < 11)

A= {3,5,7,9,11}


(b) In set B, "x" is a composite number and it should be less than or equal to 12

what is composite number:- A whole number that can be formed by multiplying other whole numbers

example :- 2*2 =4,    2*3=6,  2*4=8

but "5...7...11..." which are prime numbers cannot be composite 

Now B={4,6,8,9,10,12} x < 12

Now 

A U B = all elements of A and B

=> {3,4,5,6,7,8,9,10,11,12}


A n B = common elements of A & B

=> {9}

L.H.S :- (A U B) - (A n B) 

=> {3,4,5,6,7,8, 9 ,10,11,12} - {9}

=> {3,4,5,6,7,8,10,11,12}   --------(1)


Now 

A - B = {3,5,7,9,11} - {4,6,8,9,10,12}

         = { 3,5,7,11}

B - A = {4,6,8,9,10,12} - {3,5,7,9,11}

          = {4,6,8,10,12}


R.H.S = (A -B) U (B-A)

           = {3,4,5,6,7,8,10,11,12}------(2)

.^.  from (1) & (2) 

(A U B ) - (A n B)= (A-B) U (B -A)

Hence proved

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(b) Prove that √2 + √7  is an irrational number.

sol) R.T.P :- √2  + √7

Let us assume the opposite , 

i.e, √2 + √7 is rational

Hence it can be written in the form (a/b)

where a and b(b=/=0) are co-prime(no common factor other than 1)

Hence, √2 + √7 = ( a / b)

√7 = (a/b) - (√2)

squaring on both sides

(√7)^2  = ((a/b) - (√2))^2

                  (a-b)^2 = a^2 +b^2 - 2ab


7 = (a/b)^2 + (√2)^2  - 2 (a/b)(√2)


7 = a^2  +  2  - 2√2(a/b)
      b^2


7 - 2 = a^2  -  2√2(a/b)
            b^2


5 - a^2 = 2√2(a/b)
     b^2

5b^2 - a^2  = 2 √2(a)
       b^2                  b


 5b^2 - a^2  =  2√2(a)
           b


5b^2  -  a^2  = √2
        2ab


we can see clearly √2 is an irrational number

5b^2  -  a^2     but left hand side is rational
        2ab

Since rational =/= irrational

This is a contradiction

.^. our assumption is incorrect

Hence,√2 +√7 is irrational.

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17 (a) Sum of the areas of two squares is 850 m2. If the difference of their perimeter is 40 m. Find the sides of the two squares.

sol) Given :- Sum of the areas of 2 squares = 850 m2

we know area of square = s^2

lets

area of 1st square = s1^2

area of 2nd square = s2^2

Now,

1st condition :- sum of squares = 850

s1^2 + s2^2 = 850 -------(1)

2nd condition:- 

=> difference in perimeter of 2 square = 40

we know,

perimeter of square = 4s

now,

perimeter of 1st square = 4S1

perimeter of 2nd square =4S2


2nd condition:-

4S1  -  4S2  = 40

taking"4" common

4(S1 - S2 ) = 40

S1 - S2 = 10

S1 = 10 +S2----------(2)

substitute (2) in (1) we get,

=> (10+S2)^2  +  S2^2 = 850
      (a +b)^2 form

=> S2^2  + 100  + 20S2  + S2^2  = 850

=> 2S2^2 + 20S2 = 750

=> 2S2^2 + 20S2 - 750 = 0

taking "2" common

=> 2(S2^2 + 10S2 - 375) = 0

=> S2^2 + 10S2 - 375 = 0

factorize to get  factors

=> S2^2 + 25S2 - 15S2  - 375 = 0

=> S2(S2 + 25) - 15(S2 +25) = 0

=> (S2-15) = 0  or (S2 +25) = 0

=> S2 = 15  or S2= -25

Since the sides cannot be in negatives we take S2=15

substitute S2=15 in (2) we get

 S1 = S2 + 10

.^. S1 = 15+10 = 25


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(b) Sum of the present ages of two friends are 23 years. five year ago product of their ages was 42. Find their ages 5 year hence.   

sol) Given :- Sum of ages of two friends =23yrs

as we dont know the exact ages of two friends

let "x" be the age of first friend

let "y" be the age of second friend

sum of their ages (x+y) = 23

Now,

let present age of first friend be = x

Then, present age of 2nd friend be y = 23-x

Also,

5 year ago :-

1st friend age = 5 - x

2nd friend age = 23 - x - 5 = 18 - x

5 year ago product of their ages = 42

(x -5) * (18-x) = 42

x(18-x) -5(18-x) = 42

18x - x^2  - 90 + 5x = 42

-x^2  +  23x = 42 + 90

x^2 - 23x = -132

x^2 - 23x + 132 = 0

x^2 - 12x - 11x +132 = 0

x(x-12) -11(x-12)=0

(x-11)(x-12)=0

x=11,12


when x=11 age of other friend y = 23-11= 12

when x = 12 age of other friend y = 23-12 = 11

(*) After 5yrs their ages are:-

x = 11 +5 = 16 yrs

y = 12+ 5 = 17 yrs