# CBSE [ Delhi ]_X_Mathematics_2004_Set I

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• Q1

In Fig. 1, D is a point on the side BC of ABC, such that ADC = BAC

Prove that

Marks:2

In triangle BAC and triangle AD,

D = A           (Given)

C = C           (common angles)

By AA,

Proved.

• Q2

Solve the following system of linear equations:

Marks:3

• Q3

Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.

Marks:3

Let the present age of the father be x year and present age of son be y year

Two year ago the age of father  = x-2 year

Two year ago the age of son = y-2 year

A.T.Q

case I

x-2 = 5(y-2)

x-2 = 5y-10

x-5y = -8

5y-x = 8

5y-8 = x

case II

x+2 =3(y+2) +8

putting the value of y

5y-8+2=3y+6+8

2y = 20

y = 10 years

Putting the value of y in equation 1

X = 5(10)-8 = 42years
Present age of father =42 years
Son's present age = 10 years

• Q4

Solve for x: 4x2 - 2(a2 + b2)x + a2b2 = 0

Marks:3

• Q5

The HCF and LCM of two polynomials P(x) and Q(x) are (2x - 1) and (6x3 + 25x2 - 24x + 5) respectively, if P(x) = 2x2 + 9x - 5, determine Q(x).

Marks:3

Out of syllabus

• Q6

If (x - 2)(x + 3) is the GCD of the polynomials , find the value of a and b.

Marks:3

It is out of syllabus

• Q7

If as a rational expression in lowest terms.

Marks:3

It is not in syllabus.

• Q8

The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Marks:3

Let the first term of A.P be a and the common difference be d

7th term = a + 6d = 32                               …(1)

13th term = a + 12d = 62                           …(2)

Solving equation (1) and (2), we get

-6d = -30

d = 5

putting the value of d in equation (1)

a + 6 x 5 = 32

a = 32 - 30

a = 2

A.P series = 2, 7, 12, 17, ...,

• Q9

Find the sum of the first 25 terms of an A.P. whose nth term is given by tn = 2 - 3n.

Marks:3

Let n = 1

t1 = 2 - 3 = -1

t2 = 2 - 6 = -4

t3 = 2 - 9 = -7

first term = -1 ,  common difference = -4 - (-1) = -3

t25 = 2 - 3 25 = -73

S25 = (n/2)(a + l)

S25 = (n/2)(-1 - 73)

= (25/2)(-74)

= 25 -37 = -925