Equations MCQ -

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Equations MCQ

1. $\displaystyle \mathbf{If}\frac{x}{{27}}=\frac{3}{x},\text{ }\mathbf{find}\text{ }\mathbf{value}\text{ }\mathbf{of}\text{ }\mathbf{x}$

(a) 9

(b) ± 9

(d) ± 6.

$\displaystyle \frac{x}{{27}}=\frac{3}{x},\text{ Or }{{\text{x}}^{2}}=81.\text{ }Or\text{ }x=\pm 9.$
So, option (b) is correct

_2.

_a.5 & 3

_b.

_{c. 6 & 5}

_d.

Or, 15x² – 18x + 3 = 0, Or, 15x² – 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
$\displaystyle So,\text{ }either\text{ }x=1~~\text{ }Or\text{ }\left( {15x} \right)\frac{2}{3}=18,~\text{ }15x=3,\text{ },\text{ }i.e\text{ }~x=\frac{3}{{15}}=\frac{1}{5}$
$\displaystyle So,\text{ }either\text{ }x=1,\text{ }or\text{ }~x=\frac{1}{5}~,$
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

(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 2x² – 11x + 5 = 0?

$\displaystyle (a)\frac{{11}}{2}$
$\displaystyle (b)-\frac{{11}}{2}$
$\displaystyle (c)\frac{5}{{11}}$
$\displaystyle (d)15$

We know, For Quadratic Equation of Form : ax²+bx + c = 0, sum of the roots= –

Considering the equation 2x² – 11x + 5 = 0, in form of ax²+bx + c = 0, we find a = 2, b = – 11, c = 5

$\displaystyle So,\text{ }sum\text{ }of\text{ }the\text{ }roots=~\frac{{-b}}{a}=\frac{{-(-11)}}{2}=\frac{{11}}{2}.$

5. If the roots of the equation px²+ qx + 3 = 0 are reciprocal to each other, then find value of p

(a) q = 3

(b) p = 3

(d) q = 6

We know, in equation of the form: ax² – bx + c = 0, if one root is reciprocal of other, then c = a

In equation, px²+ qx + 3 = 0, a = p, c = 3, So, p = 3 . So, option (b) is correct

6. The roots of the equation x²+ 6x – 5 = 0 are

(a) Real and equal

(b) Imaginary

(d) None of these.

In equation of the form: ax² – bx + c = 0, we may get the nature of roots by examining the value of

b² – 4ac. In x²+ 6x – 5 = 0, a = 1, b = 6, c = – 5 .

So, b² – 4ac = (6)² – 4 .X1 X (- 5), or, b² – 4ac = 36 – (- 20), or, b² – 4ac = 36 + 20, b² – 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

(d) 10, 15, 20

Let numbers be 2x, 3x, 4x, So, (2x)² + (3x)² + (4x)² = 725, Or, 4x² + 9x² + 16x² = 725, Or, 29x² = 725
$\displaystyle Or,\text{ }{{x}^{2}}=\frac{{725}}{{29}}=25.\text{ }So,\text{ }x=\pm 5.$
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

(d) 18.

Let numbers be 3x and 4x . So, (3x)² + (4x)² = 100, Or, 9x² + 16x² = 100, Or, 25x² = 100, Or, x² = 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

(d) 10

Let one number x, so, other number is 5 – x, So, x² + (5 – x)² = 17, Or, x² + (25 – 10x + x² )= 17

Or, 2x² – 10x + 8 = 0, Or, 2x² – 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

(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

(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.

$\displaystyle 12.\mathbf{If}~\frac{2}{x}+\frac{3}{y}=\text{ }\mathbf{2},\text{ }\mathbf{and}~\frac{6}{x}+\frac{{18}}{y}=\text{ }\mathbf{9},\text{ }\mathbf{find}\text{ }\mathbf{the}\text{ }\mathbf{value}\text{ }\mathbf{of}\text{ }\mathbf{x}\text{ }\And \text{ }\mathbf{y}$

(a) 3 and 5

(b) 2 and 3

(d) 3 and 1

$\displaystyle \frac{2}{x}+\frac{3}{y}=\text{ }2~-\text{ }eq\text{ }\left( 1 \right),~\frac{6}{x}+\frac{{18}}{y}=\text{ }9\text{ }..\text{ }eq\text{ }\left( 2 \right).$
$\displaystyle Multiplying\text{ }Eq\text{ }\left( 1 \right)\text{ }by\text{ }3,\text{ }we\text{ }get,~\frac{6}{x}+\frac{9}{y}=\text{ }6,\text{ }eq\text{ }..\text{ }3.$
$\displaystyle Deducting\text{ }eq\text{ }\left( 2 \right)\text{ }from\text{ }eq\text{ }\left( 3 \right),\text{ }we\text{ }get,-~\frac{9}{y}=-3,\text{ }or-3y=-9,\text{ }or~y=3,~$
$\displaystyle Putting\text{ }value\text{ }of\text{ }y\text{ }in\text{ }eq\text{ }1,\text{ }we\text{ }get~\frac{2}{x}+\frac{3}{y}~=2,\text{ }or,~\frac{2}{x}+1=2$
$\displaystyle Or,~\frac{2}{x}=1,or\text{ }x=2,\text{ }So,\text{ }x=2,\text{ }y=3,$
So, option (b) is correct.

13. The value of the expression 5x²+ 6x+ 7 for x = 1 is

(a) 12

(b) 10

(d) 28

When x = 1, 5x²+ 6x+ 7 = 5(1)²+ 6.1+ 7 = 5 + 6 + 7 = 18. So, option (c) is correct.

14. If x = 2 is solution of x²+ kx+ 4 = 0, then value of k is

(a) 12

(b) – 6

(d) – 4

x = 2. So, x²+ 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) x² – 8x + 15 = 0

(b) x² + 8x + 25 = 0

(d) 3x² – 8x – 25 = 0.

Standard form of quadratic equation is : x² – (sum of roots)x + product of roots = 0.

Here the roots are 3 & 5. So, Hence, x² – (3 + 5)x + (3 x 5) = 0, Or, x² – 8x + 15 = 0.

So, option (a) is correct.

$\displaystyle ~\begin{array}{*{20}{l}} {16.\mathbf{If}\alpha ,\beta \mathbf{are}\text{ }\mathbf{the}\text{ }\mathbf{roots}\text{ }\mathbf{of}\text{ }{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{px}+\mathbf{q}\text{ }=\text{ }\mathbf{0},} \ {\mathbf{Find}\text{ }\mathbf{value}\text{ }\mathbf{of}~{{\alpha }^{\mathbf{2}}}+{{\beta }^{\mathbf{2}}}} \end{array}$

(a) p² + 6q

(b) p² – 2q

(d) None of these.

Comparing the equation x² – px + q = 0, with genera form, ax² +bx + c = 0, we get, a=1, b= – p, c=q.
$\displaystyle Here,\alpha +\beta =-\frac{b}{a}=-\frac{p}{1},\alpha \beta =\frac{c}{a}=\frac{q}{1}=q$
$\displaystyle {{\alpha }^{2}}+{{\beta }^{2}}=\text{ }{{(\alpha +\beta )}^{2}}-2\beta \beta ,\text{ }or{{\alpha }^{2}}+{{\beta }^{2}}={{p}^{2}}-2q.$
So, option (b) is correct.

$\displaystyle \begin{array}{*{20}{l}} {17.\mathbf{If}\alpha ,\beta \mathbf{are}\text{ }\mathbf{the}\text{ }\mathbf{roots}\text{ }\mathbf{of}\text{ }{{\mathbf{x}}^{\mathbf{2}}}-\mathbf{3x}+\mathbf{2}\text{ }=\text{ }\mathbf{0},} \ {\mathbf{Find}\text{ }\mathbf{then}\text{ }\mathbf{the}\text{ }\mathbf{equation}\text{ }\mathbf{whose}\text{ }\mathbf{roots}\text{ }\mathbf{are}\text{ }(\alpha +\mathbf{1}),(\beta +\mathbf{1}):} \end{array}$

(a) 2x² + 5x + 6 = 0

(b) 3x² – 5x – 6 = 0

(d) x² – 5x + 6 = 0

Comparing the equation x² – 3x + 2 = 0, with general form, ax² + bx + c = 0, we get, a=1, b= – 3, c=2.
$\displaystyle So,\alpha +\beta =-\frac{b}{a}=-\frac{{-3}}{1}=3,\alpha \beta =\frac{c}{a}=\frac{2}{1}=2$
$\displaystyle So,\text{ }sum\text{ }of\text{ }the\text{ }roots\text{ }of\text{ }the\text{ }equation:\left( {\alpha +1} \right)+\left( {\beta +1} \right)=\left( {\alpha +\beta } \right)+2=3+2=5$
So, Product of the roots of the equation :
$\displaystyle (\alpha +1)\times (\beta +1)=\alpha \beta +\alpha +\beta +1=(\alpha \beta )+(\alpha +\beta )+1$
$\displaystyle Or\text{ }(\alpha +\text{ }1)\times (\beta +1)=2+3+1=6.$
So, the Required equation
= x² – (sum of roots)x + product of roots = 0
= x² – 5x + 6 = 0. So, option (d) is correct.

18. If quadratic equation 2x² + 3x + p = 0 has equal roots, then the value of p will be

(a) $\displaystyle \frac{8}{2}$

(b) $\displaystyle \frac{5}{2}$

(c) $\displaystyle \frac{1}{5}$

(d) $\displaystyle \frac{9}{8}$

If roots are equal, then b² – 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.

$\displaystyle \mathbf{19}.\mathbf{If}\text{ }\mathbf{A}=\frac{{x+1}}{{x-1}}~~\And \text{ }\mathbf{B}=-\frac{{x-1}}{{x+1}}~,\text{ }\mathbf{then}\text{ }\mathbf{find}\text{ }\mathbf{value}\text{ }\mathbf{of}\text{ }\mathbf{A}+\mathbf{B}$

20. If the sum of the roots of the equation qx² + 3x + 3q = 0, is equal to their product, then find value of q.

(a) $\displaystyle -\frac{2}{3}~$

(b) $\displaystyle \frac{3}{2}~$

(c) 7

(d) -8
$\displaystyle Let\text{ }\alpha ~\And ~\beta \text{ }be\text{ }the\text{ }roots\text{ }of\text{ }the\text{ }equation\text{ }q{{x}^{2}}+3x+3q=0.$
$\displaystyle So,\text{ }\alpha +\beta =-\frac{2}{q},\alpha \beta =~\frac{{3q}}{q}=3.~$
$\displaystyle Here,\text{ }\alpha +\beta =\alpha \beta ,\text{ }So,~-\frac{2}{q}=\text{ }3,\text{ }or\text{ }q=-\frac{2}{3}.$
So, option (a) is correct.

21. Find roots of 3x² – 14x + 8 = 0

(a) ± 6

(b) ± 3

(d) ± 7

3x² – 14x + 8 = 0, Or, 3x² – 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
$\displaystyle OR,~\text{ }3x-2=0,\text{ }or\text{ }x=-\frac{2}{3}.$
So, option (c) is correct.

22. If a & b are the roots of equation x²– 5x + 6 = 0, Find the equation with roots (a + b) and (a – b) is

(a) x²– 6x + 5 = 0

(b) 12x²– 6x + 5 = 0

(d) 3x²– 5x + 6 = 0

Here, a + b = – $\frac{{-5}}{1}$ = 5. ab = $\frac{6}{1}$ =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

x² – (sum of roots)x + product of roots = 0

Or, x² – (a + b + a – b)x + (a + b) (a – b) = 0

Or, x² – {(a + b) + (a – b)}x + (a + b) (a – b) = 0, Or, x² – (5 + 1)x + 5 X 1 = 0

Or, x² – 6x + 5 = 0. So, option (a) is correct.

23. If ab are the roots of equation x²– 5x + 6 = 0, Find the equation with roots (ab + a + b) and (ab – a – b)

(a) x²– 12x + 11 = 0

(b) 2x²– 6x + 10 = 0

(d) 4x² – 7x + 12 = 0

$\displaystyle Here,\text{ }\alpha +\beta =-\frac{{-5}}{1}~=\text{ }5.\text{ }\alpha \beta =\frac{6}{1}=\text{ }6.~$

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 (ab+ a + b) and (ab- a – b) is

x² – (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 : x² – 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

(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/x² + 1/y² -13 =0 and + – 5=0.

(a) ,

(b) ,

(d) ,

1/x² + 1/y² -13 =0, or, 1/x² + 1/y² =13, Also, + – 5=0, or, + = 5

Now, 1/x² + 1/y² = ( + ) ² – 2 X ) =13. Or, 5² – 2X() =13. Or 25 – =13, or =12,

Now, ( – ²= ( + ² – 4 X Or, ( – ² = 5² – )= 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

(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-x²)= . Or, 169x – 169x² = 36

Or, 169x – 169x² – 36 = 0, Or, 169x² – 169x + 36 = 0, Or, 169x² – 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 x²+kx+64 = 0 and x² – 8x + k = 0 will have real equal roots.

(a) 15

(b) 16

(d) 21

For equation ax² + bx + c = 0, If roots are real and equal then b² = 4ac.

So, For equation x² + kx + 64 = 0, (k)² = 4 X 1 X 64, k²= (16)². So, k=±16

For equation x² – 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

(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

(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 (x₁,y₁) & (x₂, y₂) is (x₁ + x₂)/2, (y₁ + y₂)/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

(d) 16

Area of a triangle passing through points, (x₁ y₁) (x₂ y₂) (x₃ y₃) is x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)

= 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, x₁=5, y₁=6, x₂=-1, y₂=4.

So, = 2, or 4+x₃=6, x₃= 6-4=2. and =3, or 10+y₃=9, or y₃=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 cm² and 340 cm respectively, then the length of rectangle is:

(a) 110

(b) 120

(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 – y² = 6000, Or, y² – 170y + 6000 = 0, Or, y² – 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.

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