SCRA Whole test paper

SCRA  Special Class Railway Apprentices    Solved Paper 2009  Mathematics questions with answers,SCRA model question papers,SCRA previous years solved question papers with detailed explanations

1 What is the coefficient of x⁵ in the expansion of (1 + x)²¹ + (1 + x)²² + (1 + x)²³ + ...... + (1 + x)³⁰ ?

(a) C(51, 5)

(b) C(31, 5) - C(21, 5)

(c) C(31, 6) - C(21, 6)  (Ans)

(d) C(31, 6) + C(21, 6)

Solution :  (1 + x)²¹  + (1 + x)²² + (1 + x)²³ + ...... + (1 + x)³⁰

  ⇒   (²¹C₀ + ²¹C₁x + ²¹C₂x² + ...... + ²¹C₂₁x²¹)+ (²²C₀ + ²²C₁x + ²²C₂x² + ...... + ²²C₂₂x²²) + ...... +  (³⁰C₀ + ³⁰C₁x + ³⁰C₂x² + ...... + ³⁰C₃₀x³⁰)

Now, equating the coefficient of x⁵

= (²¹C₅ + ²²C₅ + ²³C₅ + ...... + ³⁰C₅)

= ⁽³⁰⁺¹⁾C₍₅₊₁₎ - {⁵C₅ + ⁶C₅ + ...... + ²⁰C₅}

{∵  ^rC_r +  ^r+1C_r + ...... +ⁿC_r =  ⁿ⁺¹C_r+1}

= ³¹C₆ - ⁽²⁰⁺¹⁾C₍₅₊₁₎

= ³¹C₆ - ²¹C₆

=C(31,6) - C(21,6)

So, option (c) is correct.

2. 

For any θ ∈ R, denote the matrix

Consider the following statements.

1.  Adj (A_θ) = - A_θ for some  θ ∈ R then  θ  =  (2r + 1)p  for some integer r.
                                                                    2

3.  Adj (A_θ) = A_θ^Tholds for only finitely many θ ∈ R.

Which of the statements given above is/are correct ?                                                                  

(a) 1 only

(b) 2 only  (Ans)

(c) Both 1 and 2

(d) Neither 1 nor 2

Solution : 

Cofactors ofA_θ

a₁₁ = cos θ, a₁₂ = sin θ, a₂₁ = -sin θ, a₂₂ = cos θ

  ⇒   Adj  A_θ = A_θ^T; θ ∈ R

But  Adj  A_θ ≠ - A_θ

So, option (b) is correct.

4.  Two events A and B have probabilities 0.25 and 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then, the probability that neither A nor B occurs is

(a) 0.39   (Ans)

(b) 0.25

(c) 0.11

(d) None of these

Solution :  Given,

            P (A) = 0.25,  P (B) = 0.50,  P (A Ç B) = 0.14

 ∴   P (A È B) = P (A) + P (B) - P (A Ç B)

 ⇒   P (A È B) = 0.25 + 0.50 - 0.14

      P (A È B) = 0.61

         _____    
 ∴   P (A È B) = 1 -  P (A È B) = 1 - 0.61 = 0.39

5. If cos (θ -  a) = a and cos (θ - β) = b, then what is the value of sin² (a - β) + 2ab cos (a - β)?

(a) a²+ b²  (Ans)

(b) a²- b²

(c) - a²- b²

(d) - a²+ b²

Solution :     cos (θ -  a) = a ⇒  θ -  a = cos^-1a 

and cos (θ - β) = b ⇒  θ - β = cos^-1 b

⇒  (θ - β) - (θ -  a) = cos^-1b - cos^-1 a 

(a - β) =  cos^-1 {ab + √1 - a²  √1 - b²}

cos (a - β) = ab + √1 - a² . √1 - b²

cos² (a - β) = a²b² + (1 - a²)(1 - b²) + 2ab √1 - a² √1 - b²

cos² (a - β) =  1 - a² - b² + 2a²b² + 2ab √1 - a² √1 - b²

sin² (a - β) =  a² + b² - 2a²b² - 2ab √1 - a² √1 - b²

Now,  sin ² (a - β) + 2ab cos(a - β)

= a² + b² - 2a²b² - 2ab √1 - a² √1 - b² + 2ab {ab + √1 - a²√1 - b²}

= a² + b² - 2a²b² - 2ab √1 - a² √1 - b² + 2a²b² + 2ab √1 - a² . √1 - b²

= (a² + b²)

6. Consider the set A of all determinants of order 3 with entries 0 or 1 only.  Let B be the subset of A consisting of all determinants with value 1. Let C be the subset of A consisting of all determinants with value - 1. Then,

(a) C is empty

(b) B has as many elements as C   (Ans)

(c) A = B È C

(d) B has twice as many elements as C

Solution :   Since, A is the determinant of order 3 with entries 0 or 1 only

Also, B is the subset of A consisting of all determinants with value 1.

(We know by interchanging any two rows or columns among them self sign changes)

Given that, C is the subset having determinant with value - 1.

∴     B has as many elements as C.

7.  If sin x + sin y = 3(cos y - cos x),then what is the value of sin 3x cosec 3y ?

(a) -1  (Ans)

(b) 0

(c) 1

(d) None of these

Solution :   sin x + sin y = 3(cos y - cos x)

sin x + 3 cos x = 3 cos y - sin y

r cos (x - a) = r cos (y + a)

where  r = √10, tan  a = 1/3

⇒  x - a = ± (y + a)

⇒  x = - y    or   x - y = 2a

Thus, (x = - y) satisfies relation

∵    sin 3x . cosec 3y

            =    sin 3x    =     sin 3x    =  -1
                  sin 3y          -sin 3x

8.  What is the image of [0,  √3 - 1/2√2] under the inverse sine function ?

(a) [0,  p/10]

(b) [0,  p/8]

(c) [0,  p/15]

(d) [0,  p/12]  (Ans)

Solution :   Given, interval  [0,  √3 - 1/2√2]

⇒     [0, (1/√2 . √3/2 - 1/2 . 1/√2)]

⇒     [0, (sin p/3 . cos p/4 - cos p/3. sin p/4)]

⇒     [0, sin (p/3 - p/4)]

        =  [0, sin  p/12]                    ............(i)

Now, the image of Eq. (i)

Under the inverse sine function is

= [sin^-1 (0),sin^-1 (sin p/12)]

= [sin^-1 (sin 0),sin^-1 (sin p/12)]

= [0, p/12]

9. The real value of x such that e^{sin x} - e^{-sin x} = 4

(a) is 3p/4

(b) is p/3

(c) is p/6

(d) Does not exist  (Ans)

Solution :   e^{sin x} - e^{-sin x} = 4

e^{sin x} - 1/e^{sin x} = 4

(e^{sin x})² - 4 . (e^{sin x}) - 1 = 0

Let  (y = e^{sin x})

⇒    y² - 4y - 1 = 0

⇒    y =  4 ± 2√5  
                  2

⇒    (y = 2 ± √5)

⇒    e^{sin x}  =  2 ± √5 =

SCRA

∵    2 < e^{sin x} < 3

ie, no solution (does not exist).

10  AB is a vertical pole. The end A is on the level of ground. C is the middle point of AB. P is a point on the level of ground. The portion BC subtends an angle β at P. If AP = n (AB),then what is tan β is equal to ?

(a) n/2n² + 1 (Ans)

(b) n/n² - 1

(c) n/n² + 1

(d) n/2n² - 1

Solution :  

Given condition; AP = nAB

In ∆APB,

            tan (a + β) = AB/AP

            tan (a + β) = 1/n            .........(i)

In ∆APC,

            tan a = AC/AP = AB/2AP         (tan a = 1/2n)

From Eq. (i)

              tan a + tan β     = 1/n
            1 - tan a . tan β

               1/2n + tan β      = 1/n
               1 -  tan β/2n

              (1 + 2n tan β)     = 1/n
               (2n  -  tan β)

⇒   n + 2n² tan β = 2n - tan β

     (2n² + 1) tan β = n;

      tan β =        n       
                   2n² + 1

11.  The points (-a, -b),(0, 0),(a, b) and are

(a) collinear  (Ans)

(b) vertices of a rectangle

(c) vertices of a parallelogram

(d) None of the above

Solution :  The point O (0, 0) is the mid point of A (-a, -b) and B (a, b).

Therefore, A, O, B are collinear and equation of line AOB is

y = b/a x

Since, the fourth point D (a², ab) satisfies the above equation.

Hence, the four points are collinear.

12. What does the equation,

x² - 5x + 6 = 0

represent in three-dimensional space ?

(a) Skew lines

(b) A pair of non-parallel planes

(c) A pair of parallel planes

(d) None of the above  (Ans)

Solution :  Given equation,

x² - 5x + 6 = 0

⇒  (x - 3)(x - 2) = 0

⇒  x - 3 = 0 and  x - 2 = 0

Which represent parallel lines perpendicular to x-axis in the first quadrant these lines cannot be skew lines, because skew lines is not parallel to each other.

Also, the both equation will not represents a equation of plane because the lines represents a two-dimensional space. 

13.  The equation

 x²/1 - r  - y²/1 + r = 1, |r| < 1 represents

(a) an ellipse

(b) a hyperbola  (Ans)  

(c) a circle

(d) None of these

Solution :  Given equation is

x²/1 - r  - y²/1 + r = 1, where |r| < 1

⇒  1 - r is positive and 1 + r is positive.

∴    Given  equation is of the form x²/a²- y²/b²= 1

Hence, it represents a hyperbola when |r| < 1.

14. If tan θ = - 4/3, then sin θ is

(a) - 4/5 but not 4/5

(b) - 4/5 or 4/5  (Ans)

(c) 4/5 but not - 4/5

(d) None of these

Solution :   Since, tan θ < 0.

∴    Angle θ is either in the second or fourth quadrant.

Then,     sin θ > 0 or  < 0

∴    sin θ  may be 4/5 or - 4/5

15. A symmetric coin is tossed until the first head is observed. What is the probability that more than seven tosses will be required ?

(a) 1/7

(b) 1/49

(c) 1/64

(d) 1/128  (Ans)

Solution :  The probability that more than seven tosses will be required

= (1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 )

= (1/2)⁷ = (1/128)

16.  The equation 2 cos² (x/2) sin² x = x² + x^-2 , x ≤ p/9 has

(a) no real solution   (Ans)

(b) one real solution

(c) more than one real solution

(d) None of the above

Solution :  Given equation is

 2 cos² (x/2) sin² x = x² + x^-2 , x ≤ p/9

LHS = 2 cos² (x/2) sin² x < 2

and RHS =  x² + 1/x²  ≥ 2

∴    The equation has no real solution.

17.  A number is selected at random from a set of first 100 natural numbers. What is the probability that it will be divisible neither by 5 nor by 6 ?

(a) 0.67   (Ans)

(b) 0.54

(c) 0.33

(d) 0.16

Solution :  Sample events = {1, 2, 3, ...., 100}

Unlikely events

= {5, 10, 15, ...., 100, 6, 12, 18, 24, 36, ...., 84, 96}

Total numbers of 5 multiples = 20

Total number of 6 multiples

= (16 - 3) = 13

So, total number of unlikely events = (20 + 13) = 33

That is like events = 100 - 33 = 67

So, probability =               like events        
                               total sample events

= 67/100 = 0.67

18.  The value of k such that  x - 4  =  y - 2   =  z - k   lies in the plane 2x - 4y + z = 7, is
                                            1           1            2

(a) 7   (Ans)

(b) -7

(c) Nor real value

(d) 4

Solution :  Given equation of straight line

x - 4  =  y - 2   =  z - k
  1           1            2

Since, the line lies in the plane 2x - 4y + z = 7

∴    Point (4, 2, k) must satisfy the plane.

⇒     8 - 8 + k = 7

⇒            k = 7

19. Two persons A and B alternately toss a die, A starting the process. He who throws a six score first wins the game. What is the probability that A wins the game ?

(a) 1/2   (Ans)

(b) 2/3

(c) 4/7

(d) 6/11

Solution :  Sample event for a game =  {win, loss}

Favorable event = {win}

So, probability for A wins = 1/2

20.  Consider any set of 201 observations x₁, x₂, ....., x₂₀₀, x₂₀₁. It is given that  x₁ < x₂< .... < x₂₀₀, x₂₀₁. Then the mean deviation of this set of observations about a point k is minimum when k equals

(a) (x₁ + x₂ + ... + x₂₀₀ + x₂₀₁) / 201

(b) x₁

(c) x₁₀₁  (Ans)

(d) x₂₀₁

Solution :  Given that,  x₁ < x₂ < x₃ < ... <x₂₀₁

∴    Median of the given observation

  = (201 + 1/2)th item

  = x₁₀₁

Now, deviations will be minimum, if we taken from the median.

∴    Mean deviation will be minimum, if k = x₁₀₁.

21. Consider the following statements.

1. Three planes may intersect in a point.

2. There planes may intersect in a line.

3. Three planes may be such that two of them are parallel and the third one may intersect them in parallel lines.

4. Three planes may not intersect.

How many possibilities are correct ?

(a) One

(b) Two   (Ans)

(c) Three

(d) Four

Solution :   In a plane three planes may intersect in a line and three planes may be such that two of them are parallel and the third one may intersect them in parallel lines.

SCRA

3 planes intersect in a line

2  planes parallel and 1 plane intersect both planes intersecting lines are parallel.

22. The area bounded by the curves y = f (x),the x-axis and the ordinates x = 1 and x = b is (b - 1) sin (3b + 4). Then, f (x) is

(a) (x - 1) cos (3x + 4)

(b) 8 sin (3x + 4)

(c) sin (3x + 4) + 3(x - 1) cos (3x + 4)    (Ans)

(d) None of the above

Solution :   Since, ∫^b₁  f (x) dx = (b - 1) sin (3b + 4)

On differentiating both side w.r.t. b, we get

f (b) = 3(b - 1) . cos (3b + 4) + sin (3b + 4)

∴    f (x) = sin (3x + 4) + 3(x - 1) cos (3x + 4)

23.  A point moves along the curve y = (sin 2x) + 1, -2p ≤ x ≤ 2p.  How many times does the point attain the maximum height from the x-axis ?

(a) 8

(b) 4  (Ans)

(c) 2

(d) 1

Solution :  Curve ; y = sin 2x + 1

So, from figure we observe that the curve 4 times does the point attains the maximum height from the x-axis.

So, option (b) is correct.

24.  If  2f (x) - f (1/x) = x², x ≠ 0, then what is f (2) is equal to ?

(a) 3/4

(b) - 3/4

(c) 5/4

(d) - 7/4  (Ans)

Solution :  2 f (x) - 3 f (1/x) = x², x ≠ 0        ........(i)

Put x = 2 in Eq. (i),

                2 f (2)  -  3 f (1/2) = 4               ........(ii)

Again, put x = 1/2 in Eq. (i),

                2 f (1/2)  -  3 f (2)  = 1/4          ........(iii)

Eq. (i) * 2 + Eq. (iii) * ,

                4 f (2)  -  6 f (1/2) = 8           

⇒             6 f (1/2)  - 9 f (2)  = 3/4

On adding,   - 5 f (2) = 35/4

⇒                    f (2) =  - 7/4

25.  Two circles  x² + y² = 6 and   x² + y²  - 6x + 8 = 0 are given. Then the equation of the circle through their points of intersection and the point (1, 1) is

(a) x² + y² - 6x + 4 = 0

(b) x² + y² - 3x + 1 = 0  (Ans)

(c) x² + y² - 4y + 2 = 0

(d) None of the above

Solution :  The required equation of circle is, S₁ + λ (S₂ - S₁) = 0.

∴    (x² + y² - 6) + λ (- 6x + 14) = 0

Also, passing through (1, 1).

⇒              - 4 + λ (8) = 0

⇒                           λ =  1/2

∴    Required equation of circle is,

        x² + y² - 6 - 3x + 7 = 0

or      x² + y² - 3x + 1 = 0

26. The number of values of c such that the straight line y = 4x + c touches the curve x²/4  + y² = 1 is

(a) 0

(b) 2 (Ans)

(c) 1

(d) ∞

Solution :    For ellipse, condition of tangency is c² = a² m² + b²

Given line is y = 4x + c and curve x²/4  + y² = 1

⇒             c² = 4* 4² + 1 = 65

⇒              c = ± √65