Chapter 4 Quardratic Equations

8/8/2019 Chapter 4 Quardratic Equations

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Mathematics For Class X

Quadritic equationhttp://www.extramarks.com/class_chapter_ques_browse.php?

class=10&subject=Mathematics&chapter=Quadritic+equation&bro=yes

(Q.1) Quadric equation has _________ solutions. ( 1 mark )View Answer

(Ans) Two solutions

(Q.2) X = 1 is a solution of equation( 1 mark )

View Answer

(Ans) on putting x = 1 in equation we got

x = 1 is the solution of given equation.

(Q.3) Product of two consecutive positive integers is 240. What are the integers? ( 1 mark )View Answer

(Ans)

(Q.4) The perimeter of a rectangular room is 34 m and the length

of a diagonal is 13 m. The dimensions of the room are( 1 mark )

(a) 9 m & 2m(b) 15 m & 2m(c) 12 m & 5m(d) 17 m & 7mView Answer

(Ans) (c) 12 m & 5m

Explanation :Let the length be a m and breadth be bm.

Perimeter = 2(a + b) = 34 m
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2(a + b) = 34

a + b = 17 . (1)

Diagonal = 13 m.

a2 + b2 = 169

(a + b)2 2ab = 169

(17)2 2ab = 169

ab = 60

=

a b = 7 . (2)

Solving (1) & (2) we get

a = 12 m and b = 5 m.

(Q.5) Find the value of k, if the expression x2 + kx + 1 isfactorizable into two linear factors.

( 1 mark )

(a)

(b)

(c) Either or

(d) Neither orView Answer

(Ans) (c) Either or

Explanation :We have

b2 4ac 0

k2 4(1)(1) 0

k2 4 0

(k + 2)(k 2) 0

So, k 2 or k 2.


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(Q.6) For a quadratic equation, is one root. The other rootmust be

( 1 mark )

(a)

(b)(c) 3

(d) 5

View Answer

(Ans) (b)

Explanation :

The other root must be conjugate of , that is

(Q.7) If the sum of the roots of the quadratic equation 3x2 + (2k

+1)x (k + 5) = 0 is equal to the product of the roots, then thevalue of k is

( 1 mark )

(a) 4(b) -4(c) 1(d) -1View Answer

(Ans) (a) 4

Explanation :

3x2 + (2k + 1)x (k + 5) = 0

sum of the roots = product of the roots

2k + 1 = k + 5

k = 4

(Q.8)

If one of roots of x

2

+ ax + 4 = 0 is twice the other root,then the value of a is ( 1 mark )

(a)

(b) 2

(c)

(d)View Answer

(Ans) (d)

Explanation :Let the second root be = 2


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x2 + ax + 4 = 0

Product of the roots = + 2n = a

3n = a

(Q.9) Ifa andb are the roots of the quadratic equation x2 + px +12

= 0 with the conditiona b = 1, then the value of p is( 1 mark )

(a) 1(b) 7 or 1(c) 7(d) 7 or 7View Answer

(Ans) (d) 7 or 7

Explanation :x2 + px + 12 = 0

a + b = p, ab = 12, a b = 1

We have,

(a + b)2 = (a b)2 + 4ab

(p)2 = (1)2 + 4(12)

p2 = 49

p = 7 (or) 7

(Q.10) The equation has( 1 mark )

(a)no roots(b)two real roots(c)Infinitely many roots(d)three roots

View Answer

(Ans) (a)no roots

Explanation :

By observation x = 3 is a solution. But if we substitute x = 3 the term 7/X-3 is meaningless. Hence there is noroot.


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(Q.11) The number of roots satisfying the equationis/are

( 1 mark )

(a) 1(b) 2(c) 3

(d)4

View Answer

(Ans) (c) 3

Explanation :

Squaring on both sides

5 x = x2(5 x)

(5 x)(1 x2) = 0

x = 5, 1, 1.

(Q.12) If then 4x equals ( 1 mark )

(a) 0

(b)1

(c)(d) 5

View Answer

(Ans) (d) 5

Explanation :


x 1 = x + 1 + 1

3 =


9 = 4(x + 1)

(Q.13) The sum of the reciprocals of the roots of the equation x2 + ( 1 mark )


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px + q = 0 is

(a)

(b)

(c)

(d)

View Answer

(Ans) (b)

Explanation :

Let , are the roots.

We have,

+ = p, = q

(Q.14) The discriminant of with a, c and realconstants is zero. The roots must be ( 1 mark )

(a)equal and integral(b)rational and equal(c)real and equal(d)imaginaryView Answer

(Ans) (c)real and equal

Explanation :

4ac = 8

ac = 2

Let, , be the roots.


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=

So,

Hence the roots are real and equal.

(Q.15) The roots of the equation ax2 + bx + c = 0 will be reciprocal if ( 1 mark )(a) a = b(b) a = b = c(c) b = c(d) c = a

View Answer

(Ans) (d) c = a

Explanation :

Since roots are reciprocal, product of the roots = 1

(Q.16) The roots of the equation are ( 1 mark )

(a)real and equal(b) rational and equal(c) rational and unequal(d) ImaginaryView Answer

(Ans) (a)real and equal

Explanation :b2 4ac = So, the roots are real and equal.


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(Q.17) If one root of the equation a(b c)x2 + b(c a)x + c(a b)= 0 is 1, then the other root is

( 1 mark )

(a)

(b)

(c)

(d)View Answer

(Ans) (d)

Explanation :

Product of roots =

(Q.18) In the equation the roots are equal when( 1 mark )

(a)

(b)

(c)m = 1(d) m = 3

View Answer

(Ans) (b)

Explanation :

mx2 mx m2 m = mx2 mx x2 + x

x2 x m(m + 1) = 0

a = 1, b = 1, c = m(m + 1).

We have,

b2 4ac = 0


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1 + 4m(m + 1) = 0

4m2 + 4m + 1 = 0

(2m + 1)2 = 0

m = -1/2

(Q.19) The equation has ( 1 mark )

(a) two real roots and one imaginary root(b) one real and one(c) two imaginary roots(d) one real rootView Answer

(Ans) (d) one real root

Explanation :

Squaring on the both sides

x 2 = 16 + x2 8x

x2 9x + 18 = 0

(x 6)(x 3) = 0

x = 6 or 3

But by checking, only x = 3 satisfies the equation.

(Q.20) The least value of ax2 + bx + c (a > 0) is ( 1 mark )

(a)

(b)

(c)(d) cannot be determinedView Answer

(Ans) (b)

Explanation :Remember this important point,


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If a > 0 then, ax2 + bx + c has a minimum value at and is equal to .

If a < 0, then ax2

+ bx + c has a maximum value at and is equal to

(Q.21) The product of two consecutive odd numbers is 143. Thenumbers are

( 1 mark )

(a) 11 & 13(b) 13 & 15(c) 11 & 13(d) 11 & 15View Answer

(Ans) (a) 11 & 13

Explanation :Let the numbers be (2n 1) and (2n 3)

We have,

(2n 1)(2n 3) = 143

4n2 8n + 3 = 143

4n2 8n 140 = 0

n2 2n 35 = 0

(n 7)(n + 5) = 0

n = 7 or 5

Here n = 7 only.

So the required numbers are 11 and 13

(Q.22) If a = b = c, then the roots of the equation (x a)(x b) + (x b)(x c) + (x c)(x a) = 0 are

( 1 mark )

(a) real and unequal(b) imaginary(c) real and equal(d) None of theseView Answer

(Ans) (c) real and equal

Explanation :(x a)(x b) + (x b)(x c) + (x c)(x a) = 0

3x2 2x(a + b + c) + (ab + bc + ca) = 0

D = b2 4ac

= [2(a + b + c)]2 4(3)(ab + bc + ca)


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= 4(a2 + b2 + c2 ab bc ca)

= 2(2a2 + 2b2 + 2c2 2ab 2bc 2ca)

= 2[(a b)2 + (b c)2 + (c a)2]

= 0 (since a = b = c)

So, the roots are real and equal.

(Q.23) If the roots of the equation (a b)2 + (b c)x + (c a) = 0are equal. Then

( 1 mark )

(a) 2b = a + c(b) 2a = b + c(c) 2c = a + b(d) a = b + cView Answer

(Ans) (b) 2a = b + c

Explanation :Since roots are equal, D = 0

b2 4ac = 0

(b c)2 4(a b)(c a) = 0

(b2 + c2 2bc) 4(ac a2 bc + ab) = 0

4a2 + b2 + c2 4ab + 2bc 4ac = 0

(2a + b + c)2 = 0

2a + b + c = 0

2a = b + c.

(Q.24) The coefficient of x in the quadratic equation x2 + px + q = 0 was taken as 17in the place of 13 and its roots were found to be 2 and 15. The roots of the originalequation are

( 1 mark )

(a) 3 or 10

(b) 3 or 10(c) 3 or 10(d) None of theseView Answer

(Ans) (b) 3 or 10

Explanation :x2 + px + q = 0

If p = 17, then the roots are 2 and 15

So, product of the roots = (2)(15) = q

q = 30


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Now, the original equation is,

x2 + 13x + 30 = 0

(x + 3)(x + 10) = 0

x = 3 or 10.

(Q.25) If a & b are the roots of x2 px + q = 0, then a2 + b2 = is ( 1 mark )(a) p2 + q2

(b) p2 + 2q(c) p2 q2

(d) p2 2qView Answer

(Ans) (d) p2 2q

Explanation :

We have,

a + b = p

ab = q

a2 + b2 = (a + b)2 2ab = p2 2q

(Q.26) The values of k for which the equation 2x2 kx + x + 8 = 0 will have real andequal roots are

( 1 mark )

(a) 9 and 7(b) only 9(c) only 7(d) 9 and 7View Answer

(Ans) (a) 9 and 7

Explanation :2x2 kx + x + 8 = 0

2x2 (k 1)x + 8 = 0

a = 2, b = 1 k, x = 8

We have,

b2 4ac = 0


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(1 k)2 4(2)(8) = 0

k2 2k 63 = 0

(k 9)(k + 7) = 0

k = 9 (or) 7

(Q.27) The roots of (x2 3x + 2)(x)(x 4) = 0 are ( 1 mark )(a) 1, 3 and 4(b) 0 and 4(c) 0, 1, 2 and 4(d) 1, 2 and 4View Answer

(Ans) (c) 0, 1, 2 and 4

Explanation :

x2 3x + 2 = 0

(x 2)(x 1) = 1

x = 2, x = 1

x = 0

x 4 = 0

x = 4

So x = 0, 1, 2 and 4

(Q.28) The difference of the roots of x2 7x 9 = 0 is ( 1 mark )(a) 7(b) 9

(c)

(d)

View Answer

(Ans) (c)

Explanation :x2 7x 9 = 0

Difference =


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(Q.29) If then is called ( 1 mark )

View Answer

(Ans) is called discriminate and represented by D(Q.30) The product of two successive multiple of 5 is 300. Determine the multiples. ( 1 mark )View Answer

(Ans)

if x = -4

5x = -20

5(x+1) = -15

if x = 3

5x = 15

5(x+1) = 20

(Q.31) General form of quadric equation is _______________ ( 1 mark )View Answer

(Ans) General Form of quadric equation is

(Q.32) If then roots of equation will be _____ and _____( 1 mark )

View Answer

(Ans) Real and equal

(Q.33) Solve the equation ( 1 mark )

View Answer

(Ans)

(Q.34) In a quadratic equation , D < 0. What will be

the conditions of its roots?( 1 mark )

View Answer

(Ans) Roots are negative and not real

(Q.35) ( 1 mark )


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toy. Find the cost price of the toy.

(a) Rs 10(b) Rs 15(c) Rs 20(d) Rs 25View Answer

(Ans) (c) Rs 20

Explanation :

Let the cost price of the toy be Rs x. Then,

Gain =

S.P = C.P + Gain =

But, S.P. = Rs24.

= 24

100x + x2 = 2400

x2 + 100x 2400 = 0

x2 + 120x -20x 2400 = 0

x(x + 120) 20(x + 120) = 0

(x + 120) (x 20) = 0

x = 20, -120

x = 20

Hence, the cost price of the toy is Rs 20.

(Q.37) The number of roots satisfying the equation is/are ( 1 mark )

(a) 1(b) 2(c) 3(d) UnlimitedView Answer

(Ans) (c) 3

Explanation :

Squaring on both sides,


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2 x = x2(2 x)On solving the equation,

(2 x)(1 x2) = 0

x = 2, 1, 1.

(Q.38) The product of two consecutive positive integers is 240. Formulate thequadratic equation whose roots are these integers.

( 1 mark )

(a) x2 + x 120(b) x2 + x 240(c) x2 x 240(d) x2

View Answer

(Ans) (b) x2 + x 240

Explanation :Let two consecutive positive integers be x and x + 1. Then, their product is x (x+ 1).It is given that the product is 240.x (x + 1) = 240

x2 + x 240 = 0This is the required quadratic equation

(Q.39) Find the zeroes of the polynomials x2- 3. ( 1 mark )

(a)

(b)

(c)

(d)View Answer

(Ans) (a)

Explanation :

=

Value of is zero


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when or

i.e. or

zeroes are

(Q.40) Solve the equation :( 1 mark )

(a)

(b)(c) -2,-3

(d) -4,-3

View Answer

(Ans) (b)

Explanation :

x = -3

( x + 2) = 0

x = -

Hence, x = -3 and x = - are two roots of the given equation.

(Q.41) The sum of two numbers is 15. If the sum of their reciprocal is , find the

two numbers.

( 1 mark )

(a) 1, 5(b) 8,7


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(c) 15, 10(d) 10, 5View Answer

(Ans) (d) 10, 5

Explanation :

Sum of two numbers = 15Therefore let one number be x, then the other number will be 15 x.

Sum of reciprocal of the two numbers =

.

x = 5 or 10

(Q.42) If x = 2 and x = 3 are roots of equation 3x2 2kx + 2m = 0, find the value of kand m.

( 1 mark )

(a)

(b)(c) 14, 3

(d)View Answer

(Ans) (d)

Explanation :

Since x = 2 and x = 3 are roots of equation 3x2 2kx + 2m = 03 x 22 2k x 2 + 2m = 0 and 3 x 32 2k x 3 + 2m = 0

12 4k + 2m = 0 and 27 6k + 2m = 0


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12 = 4k 2m and 27 = 6k 2m

Solving these two equations,

and m = 9

(Q.43) Solve the quadratic equation by factorization method:x2 9 = 0

( 1 mark )

(a) +3(b) 3,3(c) 3,0(d) 0,3View Answer

(Ans) (a) +3

Explanation :x2 9 = 0(x 3) (x + 3) = 0x 3 = 0 or, x + 3 = 0x = 3 or, x = -3x = +3

Thus, x = 3 and x = -3 are roots of the given solution

(Q.44) Solve 20071204453718001196751909 _image022.wmz" o:title=""/>( 1 mark )

(a) 1,0(b) 0,1(c) 1,-1(d) -1,-1View Answer

(Ans) (c) 1,-1

Explanation :

Putting 5x

= y,

5y2 +5 = 26y

5y2 26y + 5 = 0

5y2 25y y + 5 = 0

5y(y 5) 1(y 5) = 0

(y 5) (5y 1) = 0


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y 5 = 0 or 5y 1 = 0

y = 5 or y =

5x = 51 or 5x = 5-1

x = 1 or x = -1

Hence, the roots of the given equation are 1, -1.

(Q.45) Solve the quadratic equation : 15x2 28 = x ( 1 mark )(a) 1, -2(b) -2, 5

(c)

(d)View Answer

(Ans) (d)

Explanation :

15x2 28 = x 15x2 - x -28 = 0Here, a = 15, b = -1, c = -28

D = b2 4ac = (-1)2 4 x 15 x (-28) = 1 + 1680 = 1681 >0

So, the given equation has real roots,

The roots of the given equation are


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(Q.46) Find the discriminant of the quadratic equation:3x2 5x 2 = 0

( 1 mark )

(a) 49(b) 50(c) 51(d) 52

View Answer

(Ans) (a) 49

Explanation :The given quadratic equation is 3x2 5x 2 = 0Here, a = 3, b = -5, and c = -2D = b2 4ac = (-5)2 4 x 3 x (-2) = 25 + 24 = 49

(Q.47) For what value of p, the given equation has real roots?2x2 + 2x + p = 0

( 1 mark )

(a)

(b) 2 p

(c)

(d) 1 pView Answer

(Ans) (c)

Explanation :2x2 + 2x + p = 0Here, a = 2, b = 2, c = pFor real roots, Discriminant = b2 4ac 0 (2)2 4(2) p 0

4 8p 0

1 2p 0(Q.48) Find the value of k so that the equation 9x2 kx + 81 = 0 has equal roots ( 1 mark )(a) +50, -50(b) +52, -52(c) +54,-54(d) +56, -56View Answer

(Ans) (c) +54,-54

Explanation :

Here a = 9, b = -k, c = 81

D = b2 4ac


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= (-k)2 4(9)(81)

= k2 2916

For equal roots, D = 0

k2 2916 = 0k2 = 2916

(Q.49) If -4 is a root of the quadratic equation x2 + px 4 = 0, find the value of p. ( 1 mark )(a) 3(b) 4(c) 5(d) 6View Answer

(Ans) (a) 3

Explanation :

If -4 is a root of the quadratic equation x2 + px 4 = 0,

Then (-4)2 4p -4 = 0

(Q.50) What is the nature of the roots of the quadratic equation: 3x2 5x + 2 = 0 ( 1 mark )(a) Real(b) Unequal(c) Real and unequal(d) Equal

View AnswerAns) (c) Real and unequal

Explanation :Here a = 3, b = -5, c = 2D = b2 4ac= (-5)2 4 x 3 x 2= 25 - 24= 1 > 0

The roots are real and unequal.(Q.51) Find the sum and product of zeroes of the polynomials: x2 + 7x + 10. ( 1 mark )(a) 7, 10

(b) -7, 10(c) -7,-10(d) -10,-12


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View Answer

(Ans) (b) -7, 10

Explanation :On comparing quadratic polynomial

a = 1, b = 7, c = 10Sum of zeroes is given as,

Product of zeroes is given as:

(Q.52) Find the sum and product of zeroes of cubic polynomialsp(x) = 3x3 5x2 11x 3

( 1 mark )

(a)

(b)

(c)

(d)View Answer

(Ans) (d)

Explanation :

Let be the roots of the given polynomials,

On comparing with ax3 + bx2 + cx + d,

a = 3, b = -5, c = -11, d = -3

(Q.53) The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is ( 1 mark )


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13 cm, find the other two sides.

(a) -12, -5(b) -12, 5(c) 12, -5(d) 8, 1View Answer

(Ans) (c) 12, -5

Explanation :

We have, hypotenuse of right triangle = 13cm.

Let the base of the right triangle = x

Then altitude = x - 7.

By Pythagoras theorem,

AB2 = BC2 + AC2

If x 12 = 0 x = 12

If x + 5 = 0 x = -5

(Q.54) Rohans mother is 26 years older than he. The product of their ages, 3 yearsfrom now will be 360.

( 1 mark )

(a) x2+ 32x - 273 = 0(b) x2 + 30x - 275 = 0(c) x2 + 33x - 270 = 0(d) x2 + 32x - 263 = 0View Answer

(Ans) (a) x2+ 32x - 273 = 0

Explanation :

Let the age of Rohan be x years.


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Then his mothers age is (x + 26) years

After three years :

Rohans age = (x + 3) years

Rohans mother age = (x + 26) + 3 years

= (x + 29) years

According to the question,

(x + 3) (x + 29) = 360

x2 + 29x + 3x + 87 = 360

x2 + 32x 273 = 0

(Q.55) Using quadratic formula, solve the following quadratic equation for x :( 1 mark )

(a) a + b, a b(b) a +2b, a 2b(c) 2a + b, 2a b(d) 2a+2b ,2a-2b

View Answer

(Ans) (c) 2a + b, 2a b

Explanation :

Comparing with ax2 + bx + c = 0a = 1 , b = -4a, c = 4a2 b2

D = b2 4ac

= (-4a)2 4 x 1 x (4a2 b2)

= 16a2 16a2 + 4b2

=4b2

x =


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x = 2a + b or x = 2a b

(Q.56) The equation has ( 1 mark )

(a) Two real roots and one imaginary root(b) One real and one imaginary root(c) Two imaginary roots(d) One real rootView Answer

(Ans) (d) One real root

Explanation :

Squaring on the both sides

x 3 = 81 + 4x2 36x

4x2 37x + 84 = 0

(x 4)(x 21) = 0

x = 4 or 21

But by checking, only x = 4 satisfies the equation.

(Q.57) The difference of the roots of x2 5x 7 = 0 is ( 1 mark )

(a)

(b)(c) 5

(d) -7

View Answer


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(Ans) (b)

Explanation :

For the given equation, x2 5x 7 = 0

Difference =

(Q.58) If one root of the equationp(q r)x2 + q(r p)x + r(p q) = 0 is 1, then the other root is

( 1 mark )

(a)

(b)

(c)

(d)View Answer

(Ans) (d)

Explanation :Let another root = .

Product of roots =

=

=

(Q.59) Find the values ofk for which the given equation has equal

roots. ( 2 Marks )

View Answer

(Ans)

Here, the given equation is


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Here

We know

The given equation has equal roots if D=0

(Q.60) Using quadratic formula solve the equation for( 2 Marks )

View Answer

(Ans) Here


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(Q.61) Solve by factorization method

( 2 Marks )

View Answer

(Ans) Here

(Q.62)

Solve the quadric equations by factorization method.( 2 Marks )

View Answer

(Ans) Here,

(Q.63) If one root of the quadratic equation is 2, find

the value of .Also find the other root.( 2 Marks )

View Answer

(Ans)


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Since is a root of the equation

Putting in the equation we get

Hence the other root is

(Q.64) In the sum of n successive odd natural numbers startingfrom 3 is 48. find the value of n

( 3 Marks )

View Answer

(Ans) Here 3+5+7+9+n terms=48

(Q.65) Divide 16 into two parts such that twice the square of thelarger part exceeds the square of the smaller part by 164.

( 3 Marks )

View Answer

(Ans) Let the larger part be then the smaller part

By hypothesis


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Hence the required parts are 10 and 6.

(Q.66) If 4 is a root of the quadratic equation and

the quadratic equation has equal roots, find the

value of

( 3 Marks )

View Answer

(Ans) Since 4 is a root of the

The equation has equal roots

Discriminate = 0

(Q.67) Prove that the equation:

has real root if

( 3 Marks )

View Answer

(Ans) The discriminate of the given equation is given by


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We have

Hence, the given equation has no real roots

(Q.68) A two digit numbers is such that the product of its digit is

18. When 63 is subtracted from the number, the digitsinterchange their places. Find the numbers.

( 3 Marks )

View Answer

(Ans)

Let the tens digit be then the units digits

Numbers

And Number obtained by inter changing the digits

But a digit can never be negative

So


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Hence, the required number

(Q.69) The area of a right angled triangle is . If the base of

the triangle exceeds the altitude by 10cm find the dimensions ofthe triangle.

( 5 Marks )

View Answer

(Ans)Let the altitude AB of right angled triangle ABC be then

Base =

Area

Hence Base =(30+10)cm =40 cm altitude = 30cm


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(Q.70) A person on tour has Rs 360 for his expenses. If heextends his tour for 4 days he has to cut down his daily expenseby Rs 3. Find the original duration of the tour

( 5 Marks )

View Answer

(Ans)Let the original duration days

Total expenditure on tour =Rs 360

Expenditure per day= Rs

Duration of the extended tourExpenditure per day according new

Schedule = RsIt is given that the daily expenses are cut down Rs 3

Hence the original duration of the tour = 20 days

(Q.71) ( 5 Marks )


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Two pipes running together can fill a cistern in minutes. Ifone pipe

takes 3 minutes more than the other to fill it, find the time in

which each pipe would fill the cistern?View Answer

(Ans)Suppose faster pipe take time = min

Slower pipe take time

Portion of the cistern filled by the faster pipe in one minute

Cistern filled by the faster pipe in min

Similarly cistern isfilled by slower pipe in min=

Cistern filled in min

Hence faster pipe takes time = 5 minSlower pipe takes time =8min

(Q.72) Seven years ago Varuns age was five times the square ofSwatis age. Three years hence swati'S age will be two fifth ofvaruns age. Find then present age.

( 5 Marks )

View Answer

(Ans)Seven years agoLet swati age be years

Varuns age years

Swati present age = years

Varuns present age= years


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It is given that three years hence swati age will be of varuns age

Hence swati's present age = (2+7) years = 9 years

Varuns present age= years = 27 years

(Q.73) A plane left 30 minutes later than the schedule time and inorder to reach its destination 1500km away in time, it has toincrease its speed by 250km/hr from its usual speed. Find itsusual speed.

( 5 Marks )

View Answer

(Ans)

Let the usual speed of the place be .Then

Time taken to cover 1500km with the usual speed,t

Time taken to cover 1500km with the speed of

The usual speed of the plane is


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Chapter 4 Quardratic Equations

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