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**Equations MCQ**

**1.**

(a) 9

(b) ± 9

(c) 81

(d) ± 6.

So, option **(b)** is correct

_{2.}

_{a. }5 & 3_{}

_{b. }

_{c. 6 & 5}

_{d. }

_{}

Or, 15x^{2} – 18x + 3 = 0, Or, 15x^{2} – 15x – 3x + 3 = 0

Or, 15x(x – 1) – 3(x – 1) = 0, Or, (x – 1) (15x – 3) = 0

So, x – 1 = 0, Or, 15x – 3 = 0

So, option **(d)** is correct

**3.If ****(x ****–****7) ^{1/2} **

**+**

**(x**

**–**

**3)**

^{1/2}= 2, what is the value of x ?(a) 81

(b) 49

(c) 36

(d) 9

(x -7)^{1/2} + (x -3)^{1/2} = 2. Or 2x^{1/2}-10=2. 2x^{1/2}=12, or x^{1/2} = 6 or x=36. So, option **(c)** is correct

**4. What is the sum of the roots of the equation 2x ^{2} **

**–**

**11x**

**+**

**5 = 0?**

We know, For Quadratic Equation of Form : ax^{2}+bx + c = 0, sum of the roots= –

Considering the equation 2x^{2} – 11x + 5 = 0, in form of ax^{2}+bx + c = 0, we find a = 2, b = – 11, c = 5

**5. If the roots of the equation px ^{2 }**

**+**

**qx**

**+**

**3 = 0 are reciprocal to each other, then find value of p**

(a) q = 3

(b) p = 3

(c) p = 5

(d) q = 6

We know, in equation of the form: ax^{2} – bx + c = 0, if one root is reciprocal of other, then c = a

In equation, px^{2 }+ qx + 3 = 0, a = p, c = 3, So, p = 3 . So, option **(b)** is correct

**6. The roots of the equation x ^{2 }**

**+**

**6x**–

**5 = 0 are**

(a) Real and equal

(b) Imaginary

(c) Real and unequal

(d) None of these.

In equation of the form: ax^{2} – bx + c = 0, we may get the nature of roots by examining the value of

b^{2} – 4ac. In x^{2 }+ 6x – 5 = 0,** ** a = 1, b = 6, c = – 5 .

So, b^{2} – 4ac = (6)^{2} – 4 .X1 X (- 5), or, b^{2} – 4ac = 36 – (- 20), or, b^{2} – 4ac = 36 + 20, b^{2} – 4ac = 56

This value is greater than zero. So, roots are Real & unequal. So, option **(c)** is correct

**7. The sum of the squares of three numbers is 725. If the ratio in the numbers be 2 : 3 : 4, what are the numbers ?**

(a) 25, 10, 15

(b) 12, 3, 4

(c) 14, 6, 8

(d) 10, 15, 20

Let numbers be 2x, 3x, 4x, So, (2x)^{2} + (3x)^{2} + (4x)^{2} = 725, Or, 4x^{2} + 9x^{2} + 16x^{2} = 725, Or, 29x^{2} = 725

So, number be 2 x 5 = 10, 3 x 5 = 15, 4 x 5 = 20. So, numbers are 10, 15, 20.

(Note : the numbers may also be 2 x -5 = -10, 3 x -5 = -15, 4 x -5 = -20, or -10, -15,- 20). So, option **(d)** is correct.

**8. Sum of the squares of two numbers is 100. If the ratio in the numbers be 3 : 4, what is their sum ?**

(a) 8

(b) 15

(c) 14

(d) 18.

Let numbers be 3x and 4x . So, (3x)^{2} + (4x)^{2} = 100, Or, 9x^{2} + 16x^{2} = 100, Or, 25x^{2} = 100, Or, x^{2} = 4

Or, x ± 2, assuming x = 2 , One number = 3x=3 x 2 = 6. The other number = 4x= 4 x 2 = 8

So, Sum of the numbers = 6 + 8 = 14.

If we consider x=-2, One number = 3x=3 x (-2) = – 6. The other number = 4x= 4 x -2 = -8. The sum is -6 + (-8) = -14. So, So, Sum of the numbers = 14 is the correct answer. So, option **(c)** is correct.

**9. If the sum of two positive numbers is 5 and the sum of their squares is 17, what is the product of the numbers ?**

(a) 20

(b) 18

(c) 4

(d) 10

Let one number x, so, other number is 5 – x, So, x^{2} + (5 – x)^{2} = 17, Or, x^{2} + (25 – 10x + x^{2} )= 17

Or, 2x^{2} – 10x + 8 = 0, Or, 2x^{2} – 2x – 8x + 8 = 0, Or, 2x(x – 1) – 8 (x – 1) = 0, Or, (x – 1) (2x – 8) = 0

2x – 8 = 0, or x -1 = 0,

When 2x – 8 = 0, 2x = 8, Or, x = 4, the Other number is (5 – 1) = 4, The numbers are 4,1. Product of the numbers is (4 x 1) = 4

When x – 1 = 0, x = 1, The Other number is (5 – 1) = 4, The numbers are 1,4. Product of the numbers is (4 x 1) = 4. So, option **(c)** is correct.

**10. The difference between the ages of two men is 10 years. 15 years ago, the age of the older was twice the age of younger. What are their present ages ?**

(a) 20,15

(b) 35,40

(c) 25,35

(d) 45,25

Let one man’s present age is x, then, Other man’s age is x + 10, 15 years ago, Age of the two men was x-15, & (x+10)-15 = x-5. So, x-5=2 X (x-15), Or, x – 5 = 2x – 30, Or, – 5 + 30 = 2x – x = x

Or, x = 25, So, present age of 1st man is 25 years, and present age of 2nd man is 25+10=35 years,

So, the present age of the present are 25, 35. So, option **(c)** is correct.

**11. The solution of the two simultaneous equations 2x ****+**** y = 8 and 3y = 4 ****+**** 4x is**

(a) x = 4, y = 2

(b) x = 1, y = 2

(c) x = 2, y = 4

(d) x = 3, y = 8

2x + y = 8 .. eq. (1), 3y = 4 + 4x, eq. (2), Multiplying eq. (1) by 2, we get 4x + 2y = 16 .. eq (3)

Deducting Eq (2) from Eq (3), we get 5y=20, or y=4. Putting the value of y in eq. 1, we get 2x+4=8,

Or 2x=4, pr x=2. So, the solution is x=2, y=4. So, option **(c)** is correct.

(a) 3 and 5

(b) 2 and 3

(c) 4 and 1

(d) 3 and 1

So, option **(b)** is correct.

**13. The value of the expression 5x ^{2 }**

**+**

**6x**

**+**

**7 for x = 1 is**

(a) 12

(b) 10

(c) 18

(d) 28

When x = 1, 5x^{2 }+ 6x+ 7 = 5(1)^{2 }+ 6.1+ 7 = 5 + 6 + 7 = 18. So, option **(c)** is correct.

**14. If x = 2 is solution of x ^{2 }**

**+**

**kx**

^{ }**+**

**4 = 0, then value of k is**

(a) 12

(b) – 6

(c) 10

(d) – 4

x = 2. So, x^{2 }+ kx^{ }+ 4 = 0, Or, (2)^{2} + (k X 2) + 4 = 0, Or, 4 + 2k + 4 = 0, Or, 2k = – 8, Or, k = – 4

So, option **(d)** is correct.

**15. Quadratic equation whose roots are 3 and 5 is**

(a) x^{2} – 8x + 15 = 0

(b) x^{2} + 8x + 25 = 0

(c) 2x^{2} + 8x – 15 = 0

(d) 3x^{2} – 8x – 25 = 0.

Standard form of quadratic equation is : x^{2} – (sum of roots)x + product of roots = 0.

Here the roots are 3 & 5. So, Hence, x^{2} – (3 + 5)x + (3 x 5) = 0, Or, x^{2} – 8x + 15 = 0.

So, option **(a)** is correct.

(a) p^{2} + 6q

(b) p^{2} – 2q

(c) p(p^{2} – 3q)

(d) None of these.

Comparing the equation **x ^{2} – px + q = 0, **with genera form, ax

^{2}+bx + c = 0, we get, a=1, b= – p, c=q.

So, option

**(b)**is correct.

(a) 2x^{2} + 5x + 6 = 0

(b) 3x^{2} – 5x – 6 = 0

(c) x^{2} + 5x – 6 = 0

(d) x^{2} – 5x + 6 = 0

Comparing the equation x^{2} – 3x + 2 = 0,** ** with general form, ax^{2} + bx + c = 0, we get, a=1, b= – 3, c=2.

So, Product of the roots of the equation :

So, the Required equation

= x^{2} – (sum of roots)x + product of roots = 0

= x^{2} – 5x + 6 = 0. So, option **(d)** is correct.

**18. If quadratic equation 2x ^{2} **

**+**

**3x**

**+**

**p = 0 has equal roots, then the value of p will be**

**(a) **

**(b) **

**(c) **

**(d) **

If roots are equal, then b^{2} – 4ac = 0. Here, b = 3, a = 2, c = p. So, (3)^{2} – 4 X 2 X p = 0, Or, 9 – 8p = 0

Or, 8p = 9., or p= ** . **So, option **(d)** is correct.

**20. If the sum of the roots of the equation qx ^{2} **

**+**

**3x**

**+**

**3q = 0, is equal to their product, then find value of q.**

**(a) **

**(b) **

**(c) 7**

**(d) -8**

So, option **(a)** is correct.

**21. Find roots of 3x ^{2} **

**–**

**14x**

**+**

**8 = 0**

(a) ± 6

(b) ± 3

(c)

(d) ± 7

3x^{2} – 14x + 8 = 0, Or, 3x^{2} – 2x – 12x + 8 = 0, Or, x(3x – 2) – 4(3x – 2) = 0, Or, (3x – 2) (x – 4) = 0

Either x – 4 = 0, Or, x = 4

So, option **(c)** is correct.

**22. If ****a**** & ****b**** ****are the roots of equation x ^{2 }**

**–**

**5x**

**+**

**6 = 0, Find the equation with roots (**

**a**

**+**

**b**

**)**

**and (**

**a**

**–**

**b**

**)**

**is**

(a) x^{2 }– 6x + 5 = 0

(b) 12x^{2 }– 6x + 5 = 0

(c) 12x^{2 }– 5x + 6 = 0

(d) 3x^{2 }– 5x + 6 = 0

Here,** **a + b = – = 5. ab = =6.

We know, (a – b)^{2} = (a + b)^{2} – 4ab = (5)^{2} – (4 X 6) = 25 – 24 = 1, or, a – b = (1)^{1/2} =1

The required equation with roots (a + b) and (a – b) is

x^{2} – (sum of roots)x + product of roots = 0

Or, x^{2} – (a + b + a – b)x + (a + b) (a – b) = 0

Or, x^{2} – {(a + b) + (a – b)}x + (a + b) (a – b) = 0, Or, x^{2} – (5 + 1)x + 5 X 1 = 0

Or, x^{2} – 6x + 5 = 0. So, option **(a)** is correct.

**23. If ****a****b**** ****are the roots of equation x ^{2 }**

**–**

**5x**

**+**

**6 = 0, Find the equation with roots (**

**a**

**b**

**+**

**a**

**+**

**b**

**)**

**and (**

**a**

**b**

**–**

**a**

**–**

**b**

**)**

(a) x^{2 }– 12x + 11 = 0

(b) 2x^{2 }– 6x + 10 = 0

(c) 5x^{2 }– 12x + 12 = 0

(d) 4x^{2} – 7x + 12 = 0

We know, (a – b)^{2} = (a + b)^{2} – 4ab = (5)^{2} – (4 X 6) = 25 – 24 = 1, or, a – b = (1)^{1/2} =1

The required equation with roots (ab+ a + b) and (ab- a – b) is

x^{2} – (sum of roots)x + product of roots = 0

Here, Sum of Roots = (ab + a + b) + (ab – a – b) = (6 + 5) + {(6 – a + b)} = 11 + (6 – 5) = 11 + 1 = 12

and, Product of roots = {ab + (a + b)} X {ab – (a + b)} = (6 + 5) X (6 – 5) = 11 X1 = 11

Hence, the Required equation : x^{2} – 12x + 11 = 0. So, option **(d)** is correct.

**24. Find value of x & y, from the equations (x/y) ^{1/2} + (y/x)^{1/2} – **

**=0 & x**

**+**

**y**

**–**

**5 = 0**

_{,}(a) 1,4

(b) 1,1

(c) 1,7

(d) None of the above

(x/y)^{1/2}+(y/x)^{1/2}– =0, or, (x/y)^{1/2}+(y/x)^{1/2} = , (x+y)/(xy)^{1/2} = or, 5/(xy)^{1/2} = (as x + y-5 = 0_{, or x+y=5).}

Or, 1/(xy)^{1/2} = , or, (xy)^{1/2}=2. Or xy=4. So, (x-y)^{2} = (x+y)^{2} = 4xy. Or, (x-y)^{2} = (5)^{2 }– 4 X 4 = 25 – 16 = 9

So, x-y=±3.

When x-y= 3, (x+y) + (x-y) = 5+3=8. So, 2x=8, or x=4. So, y=5-4=1. So, the roots are 4,1.

When x-y=- 3, (x+y) + (x-y) = 5-3=2, 2x=2, or x=1. So, y=5-1=4. So, the roots are 1,4.

So, option **(a)** is correct

**25.Find value of x & y, from the equations 1/x ^{2} + 1/y^{2} -13 =0 and **

**+**

**– 5=0.**

(a) ,

(b) ,

(c) ,

(d) ,

1/x^{2} + 1/y^{2} -13 =0, or, 1/x^{2} + 1/y^{2} =13, Also, + – 5=0, or, + = 5

Now, 1/x^{2} + 1/y^{2} = ( + ) ^{2} – 2 X ) =13. Or, 5^{2} – 2X() =13. Or 25 – =13, or =12,

Now, ( – ^{2}= ( + ^{2} – 4 X Or, ( – ^{2} = 5^{2} – )= 25- 2X12 = 25-24=1

So, ( – = ±1, we have, + = 5,

Assuming ( – = -1, Adding the 2 equations, we get 2 X = 5-1=4, or 1/y=2, or y=1/2. 1/x = 5-2=3, or x=1/3, so the roots are ,

Assuming ( – = 1, Adding the 2 equations, we get 2 X = 5+1=6, or 1/y=3, or y=1/3. 1/x = 5-3=2, or x=1/2, so the roots are , ,

So, option **(b)** is correct

**26. On solving the equation (****) ^{1/2} + ( **

**)**

^{1/2}= 2**, you get one value of x as**

(a) 4/13

(b) 1/15

(c) 1/17

(d) 6/26

()^{1/2} + ( )^{1/2} = . or, {x+(1-x)}/{x^{1/2}(1-x)^{1/2}}= or, 1/({x^{1/2}(1-x)^{1/2}}=,

Squaring both sides, we get = . Or, 1/(x-x^{2})= . Or, 169x – 169x^{2} = 36

Or, 169x – 169x^{2} – 36 = 0, Or, 169x^{2} – 169x + 36 = 0, Or, 169x^{2} – 117x – 52x + 36 = 0

Or, 13x (13x – 9) – 4 (13x – 9) = 0, Or, (13x – 9) (13x – 4) = 0.

So, 13x-9=0, or x=9/13. Or, 13x-4=0, or x=4/13. So, option **(a)** is correct

**27. Find the positive value of k for which the equations x**^{2}**+kx+64 = 0 and x**^{2}** – 8x + k = 0 will have real equal roots.**

(a) 15

(b) 16

(c) 17

(d) 21

For equation ax^{2} + bx + c = 0, If roots are real and equal then b^{2} = 4ac.

So, For equation x^{2} + kx + 64 = 0, (k)^{ 2} = 4 X 1 X 64, k^{2}= (16)^{2}. So, k=±16

For equation x^{2} – 8x + k = 0, (-8)^{2}= 4 X 1X k, or 64=4k, or k=16. So, for positive value of k, k=16 is the answer. So, option **(b)** is correct.

**Practical Applications of Equations**

**28. A man sells 6 radios and 4 televisions for Rs.18.480. If 14 radios and 2 televisions are sold for the same amount, what is the price of a televisions?**

(a) Rs.2660

(b) Rs.3660

(c) Rs.1780

(d) Rs.3360

Let price of ratio per unit = x, Price of Television per unit = y

So, 6x + 4y = 18480 …….( Eq 1) and 14x + 2y = 18480 ……( Eq 2)

From equation (1) we get, 3x + 2y = 9240 ….. (Eq 3)

Now, Deducting Eq (2) from Eq (3), we get : – 11x = – 9240, or x=840.

Putting the value of y in (Eq 2), we get, 14.840 + 2y = 18480, Or, 2y = 18480 – 11760

Or, 2y = 6720, Or, y = 3360, Price of Television per set Rs.3360, So, option **(d)** is correct.

**29. A man got exchange of Rs.1, 000. In denominations Rs.5 and Rs.10, in 175 notes. Find how many notes of Rs.5 and Rs.10 respectively, did he receive?**

(a) 25 & 150

(b) 100 & 75

(c) 75 & 100

(d) 150 & 25

Let number of 5 rupees note he get be x and number of 10 rupees note he gets is y

So, x + y = 175. .Eq (1), and 5x + 10y = 1000 Eq (2).

Multiply equation (1) by 5, we get 5x + 5y = 875 .. Eq (3). Subtracting Eq (2) from Eq (3),

we get, -5y=-125, or y= – =25, x= 175-25=150. So, he got 150 notes of Rs 5 and 25 notes of Rs 10.

We may cross check, (150+5) + 25×10 = Rs 750+250=1000, and 150+28=175 notes. So, option **(d)** is correct.

**Practical Applications of Equations in Co-ordinate Geometry**

**30. The equation of a line which is perpendicular to 5x – 2y = 7 and passes through the mid-point of the line joining (2, 7) and (– 4, 1) is**

(a) 2x – 5y – 28 = 0 (b) 2x + 5y + 18 = 0 (c) 2x + 5y – 18 = 0 (d) 2x – 3y – 18 = 0 |

We know, Midpoint (x.y) of the line joining the points (x_{1},y_{1}) & (x_{2}, y_{2}) is (x_{1} + x_{2})/2, (y_{1} + y_{2})/2, So, the co-ordinate of midpoint of line joining (2, 7) and (– 4, 1) is

x==-=-1. and y===4, or -1,4. Putting the value of x,y in equation of perpendicular line 5x – 2y = 7passing through the midpoint (-1,4), we get, 5x – 2y = 7, Or, 2y = 5x – 7, Or, y = x – Putting the value of y in general equation of straight line y=mx+c, we get m=5/2. Since this line is perpendicular, we get, m X = -1, so, m= – . So, the equation of the line is y=- +c. Since this line passes through (– 1, 4), So, putting the value of x,y, we get, 4 = {– X (– 1)} + c, or, 4 = () + c or, c = 4 – () = . The required equation is y = – x + . or, 5y = – 2x +18, or, 2x +5y = 18 or, 2x + 5y – 18 = 0. So, option (c) is correct. |

**31.The area of a triangle with vertices (1,3), (5,6) and (– 3, 4) in terms of square units is:**

(a) 14

(b) 3

(c) 18

(d) 16

Area of a triangle passing through points, (x_{1} y_{1}) (x_{2} y_{2}) (x_{3} y_{3}) is x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})

= 1 (6 – 4) + 5 (4 – 3) + (– 3) (3 – 6) = 2 + 5 + 9 = 16. So, option **(d)** is correct.

**32.The centroid of the triangle ABC is at the point (2, 3). A and B are the points (5, 6) and (– 1, 4) respectively. The coordinates of C are:**

(a) (1,– 2 ) (b) (2,– 1 ) (c) (– 1, – 3) (d) (1, 3) |

We know co-ordinate for centroid is |

& . Here, x_{1}=5, y_{1}=6, x_{2}=-1, y_{2}=4.

So, = 2, or 4+x_{3}=6, x_{3}= 6-4=2. and =3, or 10+y_{3}=9, or y_{3}=9-10=-1

So, the Co-ordinate of c is (2, –1). So, option **(b)** is correct.

**33. If area and perimeter of a rectangle is 6000 cm**^{2}** and 340 cm respectively, then the length of rectangle is:**

(a) 110

(b) 120

(c) 125

(d) 210

Let length = x, Width = y. So, xy = 6000, or 2 (x + y) = 340, Or, x + y = 170, Or, x = (170 – y)

So y X (170 – y) = 6000, Or, 170y – y^{2} = 6000, Or, y^{2} – 170y + 6000 = 0, Or, y^{2} – 120y – 50y + 6000 = 0

Or, (y – 120) (y – 50) = 0, either y = 120, or y=50.

If y = 120, x = (170 – 120) = 50, As length is higher than width, length will be 120 cm. So, option **(b)** is correct.