Sponsor Area
L.H.S. = (x + 1)2 = x2 + 2x + 1
R.H.S. = 2(x – 3) = 2x –6
∴ 6x2 + 2x + 1 = 2x - 6
⇒ x2 + 2x + 1–2x + 6 = 0
⇒ x2+ 7 = 0
∵ It is of the form ax2 + bx + c = 0, where a = 1,b = 0, c = 7.
Hence, the given equation is a quadratic equation.
R.H.S. = (–2) (3 – x) = – 6 + 2x
∴ x2 – 2x = – 6 + 2x
⇒ x2 – 2x + 6 – 2x = 0
⇒ x2 – 4x + 6 = 0
∵ It is of the form ax2 + bx + c = 0 where a = 1,b = –4, c = 6.
Hence, the given equation is a quadratic equation.
L.H.S. = (x–2) (x + 1)
= x2 – x – 2
R.H.S. = (x – 1) (x + 3)
= x2 + 2x – 3
∴ x2 – x – 2 = x2 + 2x – 3
⇒ – x –2 – 2x + 3 = 0
⇒ – 3x + 1 = 0
∵ It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
L.H.S. = (x – 1) (2x + 1)
= 2x2 + x – 2x – 1
= 2x2 – x – 1
R.H.S. = x(x + 5) = x2 + 5x
∴ 2x2 – x – 1 = x2 + 5x
⇒ 2x2 – x2 – x – 5x – 1 = 0
⇒ x2 – 6x – 1 = 0
∵ It is of the form ax2 + bx + c = 0 where a = 1, b = –b, c = –1.
Hence, the given equation is a quadratic equation.
L.H.S. = (2x – 1) (x – 3)
= 2x2 – 6x– + 3
= 2x2 – 7x + 3
R.H.S. = (x + 5) (a – 1)
= x2 + 5x – x – 5
= x2 + 4x – 5
∴2x2 – 7x + 3 = x2 + 4x – 5
⇒ 2x2 –x2-7x–4x+ 3 + 5 = 0
⇒ x2 – 11x + 8 = 0
∵ It is of the form ax2 + bx + c = 0 where a = 1, b = –11, c = 8.
Hence, the given equation is a quadratic equation.
R.H.S. = (x – 2)2 = x2 – 4x + 4
∴ x2 + 3x + 1 = x2 –4x + 4
⇒ 3x + 4x + 1 – 4 = 0
⇒ 7x – 3 = 0
∵ It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
L.H.S. = (x + 2)3
= x3 + 3x2.2 + 3x.(2)2 +8
= x3 + 6x2 + 12x + 8
R.H.S. = 2x(x2 – 1) = 2x3– 2x
∴ x3 + 6x2 + 12x + 8 = 2x3 - 2x
⇒ x3 – 2x3 + 6x2 + 12x + 2x + 8 = 0
⇒ –x3 + 6x2 + 14x + 8 = 0
∵ It is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
R.H.S. = (x – 2)3
= a3 – 23 – 3x.2 (a – 2)
= x3 – 8 – 6x2 + 12x
= a3 – 6x2 + 12x – 8
∴ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
⇒ –4x2 + 6x2 – x – 12x + 1 + 8 = 0
⇒ 2x2– 13x + 9 = 0
∵ It is of the form ax2 + bx + c = 0 where a = 2, b = –13, c = 9.
Hence, the given equation is a quadratic equation.
Let two consecutive positive integers are x and (x + 1). Then
x(x + 1) = 306
⇒ x2 + x – 306 = 0
which is a quadratic equation in x.
Solving this equation by factorisation method, we get
x2 + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x + 18) (x – 17) = 0
⇒ x + 18 = 0 or x – 17 = 0
⇒ x = –18 or x = 17
But integers are +ve
∴ x = 17
Hence, two consecutive positive integers are 17 and 18.
Let Rohan's present age = x years.
Then, Rohan’s mother age = (x + 26) years
Rohan’s age after 3 years = (x + 3) years and Rohan’s mother age after 3 yrs.
= (x + 29) years.
According to the given condition, (x + 3) (x + 29) = 360
⇒ x2 + 32x – 360 =0
⇒ x2 + 32x – 273 = 0
⇒ x2 + 39x – 7x – 273 = 0
⇒ x(x + 39) – 7(x + 39) = 0
⇒ (x + 39) (x – 7) = 0
⇒ x = –39 or x – 7 = 0 But age cannot be –ve.
∴ x = 17
Hence, Rohan's present age = 7 years.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
x2 – 3x – 10 = 0
⇒ x2 – 5x + 2x– 10 = 0
⇒ x(x – 5) + 2(x – 5) =0
⇒ (x – 5) (x + 2) = 0
⇒ x–5 = 0 or x + 2 = 0
x = 5 or x = –2
Hence, the roots of the given quadratic equation are 5 and (-2).
2x2 + x – 6 = 0
Here, a = 2, b = 1, c = –6
ac = –12 = 4 x (–3)
b = 1= 4 – 3
∴ 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x - 3 = 0
⇒ x = –2 or x = 3/2
Hence, the roots of the given quadratic equation are (-2) and 3/2.
So, this root is repeated twice.
Hence, both the roots of the given quadratic equation are .
Sponsor Area
Let the number of toys produced on that day = x
And, cost of production = (55 – x)
According to the given question,
x (55 – x) = 750
⇒ 55x – x2 – 750 = 0
x2 – 55x + 750 = 0
⇒ x2 – 30x – 25x + 750 = 0
⇒ x(x – 30) – 25(x – 30) = 0
⇒ (x – 30) (x – 25) = 0
⇒ x – 30 = 0 or x – 25 = 0
⇒ x = 30 or x = 25.
Hence, the number of toys produced on that day are 30 or 25.
Let the base (BC) of the right triangle = x cm
So, altitude (AB) = (x – 7) cm
And hypotenuse (AC) = 13 cm
According to Pythagoras theorem,
AC2 = AB2 + BC2 ⇒ (13)2 = (x – 7)2 + (x)2
⇒ 169 = x2 + 49 – 14x + x2
⇒ 169 = 2x2 – 14x + 49
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x - 12) + 5(x – 12) = 0
⇒ (x + 5) (x – 12) = 0
⇒ x + 5 = 0
or x- 12 = 0
⇒ x = –5
or x = 12
Since side of triangle is never negative So,
x = 12
Now required sides of the triangle are
AB = x – 7
= 12 – 7 = 5 cm
BC = x = 12 cm.
Let the number of articles produced be x
Therefore, cost of production of each article (in Rs.) on that particular day = Rs. (2x + 3)
So, total cost of production that day = Rs. x(2x + 3)
According to question,
x(2x+3) = 90
or
or
Since is not possible
So, x = 6
Hence, number of articles produced be x = 6 And cost of production of each article = Rs. (2x + 3)= Rs. (2 x 6 + 3) = Rs. (15).
(i) We have,
2x2 – 1x + 3 = 0
Here, a = 2, b = –7 and c = 3
Now, b2 – = (–7)2 – 4(2) (3)
= 49 – 24 = 25 Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 – 1x + 3 = 0 has distinct roots i.e. α and β.
Now,
So,
Therefore, the required roots are 3 and
(ii) We have,
2x2 + x – 4 = 0
Here, a = 2, b = 1 and c = –4
Now, b2 – 4ac = (1)2 –4 (2) (–4)
= 1 + 32 = 33
Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 + x – 4 = 0 has distinct roots i.e. α and 946;.
Now,
So,
(iii) We have,
Here,
Now,
Since,
Therefore, the quadratic equation has equal roots i.e.
Now,
Therefore, the requried roots are
(iv) We have,
2x2 + x + 4 = 0
Here, a = 2, b = 1 and c = 4
Now, b2 – 4ac = (1)2 – 4(2) (4)
= 1 – 32 = – 31
Since, b2 – 4ac = 0
Therefore, the quadratic equation 2x2 + x + 4 = 0 has no roots.
We have,
Here,
Now,
Therefore, quadratic equation x2 – 3x – 1 =0 has distinct roots, i.e. α and β.
Now,
So,
and
Therefore, the required roots are
Let marks got by Shefali in Mathematics = x
Then, marks got by her in English = (30 – x)
Now, according to the given problem, we have
(x + 2) (30 – x – 3) = 210 (x + 2) (27 – x) = 210
⇒ 27x – x2 + 54 – 2x – 210 = 0
⇒ –x2 + 25x – 156 = 0
⇒ x2 – 25x + 156 = 0
x2 – 13x – 12x + 156 = 0
⇒ x(x – 13) – 12(x – 13) = 0
⇒ (x – 13) (x – 12) = 0
⇒ x – 13 = 0 or x –12 = 0
x = 13 or x = 12
Hence, if marks got by Shefali in Mathematics = 13 then marks got by her in English 30 – 13 = 17 Ans.
and if marks got by Shefali in Mathematics = 12 then marks got by her in English = 30 – 12 = 18 Ans.
Let the smaller number = x
and the larger number = y
Then, ...(i)
and ...(ii)
From (i) and (ii), we get
which is a quadratic equation in y. We can solve it by factorisation method.
which is a quadratic equation in y. We can solve it by factorisation method.
When y = -10, then
∴
When y = 18, then
If smaller number = 12
Then, larger number =
If smaller number = -12
Then, larger number =
Hence, the required numbers are 18, 12 or 18, –12. Ans.
Let the side of first square = x m and the side of second square = y m
Then, the area of first square = x2 m2 and the area of second square = y2 m2 Perimeter of the first square = 4x m and perimeter of the second square = 4y m
Now, according to the given problem, we have x2 + y2 = 468 ...(i)
and 4x – 4y = 24
⇒ x – y = 6 ...(ii)
From equation (ii), we get
x – 6 + y = 0 ...(iii)
Putting this value of x in (i), we get
(6 + y)2 + y2 = 468 ⇒ 36+12y + f + y2 = 468 ⇒ 2y2 + 12y + 36 – 468 =0
⇒ 2y2 + 12y – 432 = 0
⇒ y2 + 6y – 216 = 0 which is a quadratic equation in y.
We can solve this equation by quadratic formula.
Here, a = 1, b = 6, c = -216
∴
Now,
But the side of a square cannot be -ve. y = 12 Putting this value of y in (iii), we get
x = 6 + 12 = 18 Hence, the side of the first square = 18 m and the side of the second square = 12 m. Ans.
We have,
Here,
Now,
Since, D < 0
Therefore, the given quadratic equation has no real roots.
We have,
Here,
Now,
Since, D = 0
Therefore, the given quadratic equation has equal roots.
Now,
Hence, roots are
Sponsor Area
We have,
Here,
Now,
For equal roots,
D = 0
We have,
Kx (x – 2) + 6 = 0
⇒ Kx2 – 2Kx + 6 = 0
Here, a = K, b = –2K, c = 6
Now, D = b2 – 4ac
⇒ D = (–2K)2 – 4(K) (6)
D = 4K2 – 24K
For equal roots, D = 0
⇒ 4K2 – 24K = 0
⇒ 4 K (K – 6) = 0
⇒ 4K = 0 or K – 6 = 0
⇒ K = 0 or K = 6.
∵ is not possible
So,
Hence, breadth of the rectangular mango grove is 20 m and length is 40 m.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Let the present age of first friend be x years. So, present age of other be (20 - x).
4 years ago,
Age of first friend = (x – 4) yrs. and age of second friend = (20 – x – 4) yrs = (16 – x) yrs.
According to question,
(x – 4) (16 – x) = 48
⇒ 16 x – x2 – 64 + 4x = 48
⇒ –x2 + 20x – 112 = 0
⇒ x2 – 20x + 112 = 0
Here, a = 1, b = –20, c = 112
Now, D = b2 – 4 ac
= (–20)2 – 4(1) (112) = 400 – 448 = –48
Since, D < 0
So, the given quadratic equation has no real roots and hence it is not possible to calculate the present ages.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Let length of the rectangular park be x m and breadth be y m.
According to question,
perimeter = 80 m ⇒ 2(l + b) = 80 m
⇒ 2 (x + y) = 80
⇒ x + y = 40
⇒ y = 40 – x
i.e., breadth = (40 – x)m
Now, area = 400
⇒ l x b = 400
⇒ x(40 – x) = 400
40x – x2 = 400
⇒ x2 – 40x + 400 = 0
Here, a = 1, b = –40, c = 400
Now, D = b2 – 4ac
= (–40)2 – 4(1) (400) = 1600 – 1600 = 0
Since D = 0
So, the given quadratic equation has equal roots and hence it is possible to design a rectangular park.
Now solving the quadratic equation x2 – 40x + 400 = 0 by quadratic formula, we get
Now,
Hence, length of the rectangular park (x) = 20 m and breadth = (40 – x) m = 40 – 20 = 20 m.
(x –2)2 (x – 1) = (x – 4) (x + 3)
⇒ (x2 + 4 – 4x) (x – 1) = (x – 4) (x + 3)
⇒ (x3 + 4x – 4x2 – x2 – 4 + 4x)
= (x2 + 3x – 4x – 12)
x3 – 5x2 + 8x – 4 = x2 – x – 12
x3 – 5x2 – x2 + 8x + x – 4 + 12 = 0
⇒ x3 – 6x2 + 9x + 8 = 0
Since, x3 – 6x2 + 9x + 8 = 0 is a polynomial of degree 3, hence this is not a quadratic equation.
We have,
(x–4) (x–3) = (x–3) (x2 – 4)
⇒ (x – 4) = x2 – 4
⇒ x2 – x – 4 + 4 = 0
⇒ x2 – x = 0
Since, x2 – x = 0 is a polynomial of degree 2, hence this is a quadratic equation.
We have,
∵ D > 0
Hence, roots are real and distinct.
2x2 + 5x = 7 ⇒ 2x2 + 5x – 1 = 0
Since 2x2 + 5x – 7 is a quadratic polynomial
Hence, 2x2 + 5x – 7 = 0 is a quadratic equation.
Let the first number be x
∴ Other number = (60 – x)
According to question,
x (60 – x) = 800
⇒ 60x – x2 = 800
⇒ x2 – 60x + 800 = 0
Therefore, the required equation in the form of quadratic equation is
x2 – 60x + 800 = 0
Let the breadth of the rectangular hall be ‘x’ m.
So, length be (x + 3) m
According to question,
Area = 180
i.e., Length x breadth = 108
⇒ x(x + 3) = 108
⇒ x2 + 3x = 108
⇒ x2 + 3x – 108 = 0
Therefore, the required representation in the form of quadratic equation be x2 + 3x – 108 = 0.
Let the present age of Rohan be x years.
So, present age of Rohan's mother be (x + 26) years.
After 3 years
Rohan’s age = (x + 3) years
Rohan’s mother age = (x + 26 + 3) years
= (x + 29) years
According to question,
(x + 3) (x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Therefore, the required representation in the form of quadratic equation
x2 + 32x – 273 = 0
Let one side (AB) = (x – 4)m
Other side (BC) = x m
and hypotenuse = 20 m
We know,
AC2 = AB2 + BC2
⇒ (20)2 = (x – 4)2 + (x)2
⇒ 400 = x2 + 16 – 8x + x2
⇒ 400 = 2x2 – 8x + 16
⇒ 2x2 – 8x – 384 = 0
⇒ 2(x2 – 4x –192) = 0
⇒ x2– 4x – 192= 0
Therefore, the required representation in the form of quadratic equation be x2– 4x –192 = 0.
Let the present age of son be x years. So, present age of father be (35 - x) years.
According to question,
x(35 - x) = 150
⇒ 35x – x2 = 150
⇒ x2 – 35x +150 = 0
Therefore, the required representation in the form of quadratic equation be x2 – 35x + 150 = 0
Let the present age of Aditya’s be x years.
Aditya’s age five years ago = (x – 5) years
Aditya’s age nine years later = (x + 9) years
According to question,
(x – 5) (x + 9) = 15
x2 + 9x – 5x – 45 = 15
x2 + 4x – 60 = 0
Therefore, the required representation in the form of quadratic equation be x2 + 4x – 60 = 0.
We have,
2x2 – 4x + 3 = 0
Here, a = 2, b = – 4, c = 3
Now, D = b2 – 4ac
= (–4)2 – 4(2) (3) = 16 – 24 = –8
Since, D < 0
Therefore, the given quadratic equation has no real roots.
We have,
Here,
Now,
Since, D = 0
Therefore, the given quadratic equation has equal roots.
2x2 – 6x + 3 = 0
Here, a = 2, b = –6, c = 3
Now, D = b2 – 4ac
= (–6)2 – 4(2) (3) = 36 – 24 = 12
Since, D > 0
Therefore, given quadratic equation has real roots.
We have,
Here,
Now,
Since,
Therefore, given quadratic equation has real roots.
–x2 + 2x + 3 = 0
Here, a = 1, b = 2, c = 3
Now, D = b2 – 4ac
= (2)2 –4(–1)(3) = 4 + 12 = 16
Since, D > 0
Therefore, given quadratic equation has real roots.
x2 + x + 1 = 0
Here, a = 1, b = 1, c = 1
Now, D = b2 – 4ac
= (1)2 – 4(1) (1) = 1 – 4 = –3
Since, D < 0
Therefore, given quadratic equation has no real roots.
The given quadratic equation is
3x2 + 2kx + 27 = 0
Here, a = 3, b = 2k and c = 27
D = b2 – 4ac = (2k2) – 4 x 3 x 27 = 4k2 – 324
The given equation will have real and equal roots, if
Here, a = k – 12, b = 2 (k – 12), c = 2
∴ D = b2 – 4ac = [2(k– 12)]2
– 4(k – 12) x 2
= 4 (k – 12)2 – 8 (k– 12)
Roots are equal, if D = 0
⇒ 4 (k – 12) 2 – 8 (k – 12) = 0
⇒ 4(k – 12)(k – 12 – 2) = 0
⇒ (k – 12) (k – 14) = 0
⇒ k = 12 or 14
But k = 12 does not satisfy the eqn.
k = 14 Ans.
Here a = 9, b = 3k, c = 4
∴ D = b2 – 4ac = (3k)2 – 4 x 9 x 4 = 9k2 – 144
Roots are equal, if D = 0
⇒ 9k2 – 144 = 0
⇒ 9k2 = 144
⇒ 9k2 – 144 = 0
⇒ k2 – 16 = 0
⇒ (k)2 – (4)2 = 0
⇒ (k + 4) (k – 4) = 0
⇒ k + 4 = 0 or, k – 4 = 0
⇒ k = 4
∵ (–5) is a root of the quadratic equation
∴ 2x2 + px – 15 = 0
2(–5)2 + p. (–5) – 15 = 0
⇒ 50 – 5p – 15 = 0
⇒ –5p + 35= 0
⇒ –5p = –35
⇒ P = 7
Now. the quadratic equation 7(x2 + x) + k = 0
i.e. 7x2 + 7x + k = 0 has equal roots.
∴ Its discriminant
Hence,
Let the roots of the quadratic equation be α and β.
We have,
2x2 + px – 4 = 0, then
Product of the roots =
But
Sum of the roots =
The given quadratic equation is
Here,
Discriminant =
=
= 100 - 36
= 64 Ans.
Sponsor Area
The given quadratic equation is
x2 + 6x + 9 = 0 Putting a = – 3, we get
L.H.S. = (– 3)2 + 6. (– 3) + 9
= 9 – 18 + 9
= 19 – 18 = 0 = R.H.S.
Hence, x = – 3 is a solution of x2 + 6x + 9 = 0 Proved.
The given quadratic equation is
2x2 + 5x – 3 = 0 Putting a = – 3, we get
L.H.S. = 2x2 + 5x – 3
= 2 (–3)2 + 5 (– 3) – 3
= 2 x 9 – 15 – 3
= 18 – 15 – 3 = 15 – 15 = 0 = R.H.S.
Hence, x = – 3 is a solution of 2x2 + 5x – 3 = 0 Proved.
The given quadratic equation is
3x2 + 13x + 14 = 0
Putting a = – 2, we get
L.H.S. = 3.(–2)2 + 13.(–2) + 14
= 3 x 4 – 26 + 14
= 12 – 26 + 14
= 26 – 26 = 0 = R.H.S.
Hence, x = – 2 is a solution of 3x2 + 13x + 14 = 0 Proved.
We have,
a = 2, b = – 3 and c = 1
Now, D = b2 – 4ac
= (– 3)2 – 4(2) (1) = 9 – 8 = 1
Hence, Descriminant (D) = 1.
We have,
Here, a = K, b = -6 and c = -2
∴
For real roots:
We have,
Here, a = 1; b = 5K and c = 16
∴
The given equation will have no real roots, if
Here, a = 1,
b = k and c = 9
For equal roots D = 0
⇒ b2 – 4ac = 0
⇒ (–k)2 – 4(1) (9) = 0
⇒ k2 – 36 = 0
⇒ k = ± 6.
Here, a = 4, b = –12 and c = –9
D = b2 – 4ac
= (–12)2 – 4 (4) (–9)
= 144 + 144 = 288
Since D > 0, therefore, roots of the given equation are real and unequal. Ans.
Let p(–2) = x2 – 2x + 8
p(–2) = (–2)2 –2 (–2) + 8
= 4 + 4 + 8
= 8 + 8 = 16
∴ x = –2 is not a solution of the given equation
We have
Now
Hence, roots are real and equal.
Solution not provided.
Ans. 3, 15 or 15, 3
Solution not provided.
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Ans. Yes; 8, 9
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Ans. Yes; 6 yrs
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Ans. Yes; 30, 40
Solution not provided.
Ans. Yes, 5, 10 or 10, 5
Solution not provided.
Ans. Yes, 10 and 6
Solution not provided.
Ans. Yes, 12, 13.
Solution not provided.
Ans. Yes, 10 and 5.
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Ans. a - b, a + b
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Ans. 2a - b, 2a + b
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Sponsor Area
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Ans. x = 4, 4
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Ans. x = a + b, a - b
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Ans. y = 0.6
Solution not provided.
Ans. y = -3, -3
Solution not provided.
Ans. x = -1
Solution not provided.
Ans. y = 2, -5
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Ans. x = 4, 100
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Ans. x = 5, 10
Solution not provided.
Ans. y = 5, 7
Solution not provided.
Ans. x = 750, -1000
Solution not provided.
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Ans. x = 12, -13
Solution not provided.
Ans. x = 16, 25
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans.
Let side of bigger square is x m. And, side of smaller square is y m.
According to given condition,
x2 + y2 = 260 ...(i)
And 4x – 4y = 24
⇒ 4(x – y) = 24
⇒ x – y = 6
⇒ x = 6 + y ...(ii)
Putting the value of (ii) in (i), we get
(6 + y) 2 + y2 = 260
⇒ 36 + y2 + 12y + y2 = 260
⇒ 2y2 + 12y – 224 = 0
⇒ y2 + 6y – 112 = 0
⇒ y2 + 14y – 5y – 112 = 0
⇒ y(y + 14) –8(y + 14) = 0
⇒ (y – 8) (y + 14) = 0
⇒ y–8 = 0, y + 14 = 0
⇒ y = 8, y = – 14
Putting the value of y in (ii), we get
x = 6 + y
= 6 + 8 = 14.
Hence, sides of squares are 14 cm and 8 cm.
Let the digits at tens and units place of the number be x and y respectively. Then, Number = 10x + 7
It is given that,
⇒ 10x + y = 4(x + y)
and 10x + y = 3xy
⇒ 6x – 3y = 0 and 10x + y = 3y
⇒ y – 2x and 10x + y = 3xy
⇒ 10x + 2x = 3x X 2x
⇒ 6x2 – 12x = 0
⇒ 6x (x – 2) = 0
⇒ x = 0 or x = 2
Since the given number is a two-digit number. So, its tens digit cannot be zero.
∴ x = 2
⇒ y = 2 x 2 = 4 [∵ = 2x]
Hence,
required number = 10x+y = 10 x 2 + 4 = 24.1
Let the speed of the stream be x km/hr.
Then, speed downstream = (5 + x) km/hr.
And, speed upstream = (5 – x) km/hr.
Now, Time taken by the boat to cover 12 km upstream =
And, Time taken by the boat to cover 12 km downstream =
According to given condition:
Hence, speed of stream = 1 km/hr.
Distance travelled by the first train in 2 hours
= OA = Speed x Time
= 2 (x + 5) km
Distance travelled by the second train in 2 hours
= OB = Speed x Time
= 2x km
By using pythagoras theorem, we have,
AB2 = OA2 + OB2
⇒ 502 = {2(x + 5)}2+ {2x}2
⇒ 2500 = 4(x + 5)2 + 4x2
⇒ 8x2 + 40x – 2400 = 0
⇒ x2 + 5x – 300 = 0
⇒ x2 + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x + 20) (x – 15) = 0
⇒ x = –20 or x = 15
⇒ x = 15
[∵ cannot be negative]
Hence, the speed of the second train is 15 km/hr and the speed of the first train is 20 km/hr.
Let the speed of the fast train = x km/hr
⇒ The speed of slow train = (x – 10) km/hr
Time taken by the fast train to cover 600 km and Time taken by the slow train to cover 600 km
According to given condition,
Neglecting x = -40, we get: x = 50 and x - 10 = 50 - 10 = 40 Speed of fast train = 50 km/hr and speed of slow train = 40 km/hr.
Seven years ago, let Swati’s age be x years.
Then, seven years ago Varun’s age was 5x2 years.
∴ Swati’s present age = (x + 7) years
Varun’s present age = (5x2 + 7) years
Three years hence,
Swati’s age = (x + 7 + 3) years
= (x + 10) years
and Varun’s age = (5x2 + 7 + 3) years
= (5x2 + 10) years
According to given condition,
∴
Hence, Swati's present age = (2 + 7) years = 9 years
Varun's present age =
Let the present age of father be x years. Then,
Son’s present age = (45 – x) years
Five years ago,
Age of Father = (x – 5) years
and Age of Son = (45 – x – 5) years = (40 – x) years
According to given condition,
(x – 5) (40 – x) = 124
⇒ 40x – x2 – 200 + 5x = 124
⇒ x2 – 45x + 324 = 0
⇒ x2 – 36x – 9x + 324 = 0
⇒ x (x – 36) – 9 (x – 36) – 0
⇒ (x - 9) (x - 36) = 0
⇒x = 9 or x = 36
When x = 36, we have
Father’s present age = 36 years
Son’s present age = 9 years
When x = 9, we have
Father’s present age = 9 years
Son’s present age = 36 years
Clearly, this is not possible
Hence, Father’s present age = 36 years and Son’s present age = 9 years.
Let present age of the son be x years
∴ Present age of the father = x2 years
one year ago,
Age of the son = (x – 1) years
and Age of the father = (x2 – 1) years
According to the given condition,
x2 – 1 = 8(x – 1)
⇒ x2 – 1 – 8x + 8 = 0
x2 – 8x + 7 = 0
⇒ (x – 1) (x– 7) = 0
⇒ x – 1 = 0, x – 7 = 0
i.e., x = 1, x = 7
When x = 1, age of son = age of father, which is impossible
∴ x = 1 is rejected.
∴ x = 7
Thus present age of son = 7 years
and present age of father = 49 years.
Let present age of son = x years
Two years ago, Age of son = (x – 2) years
Age of man = 3(x – 2)2
∴ Man’s present age = 3(x – 2)2 + 2
= 3(x2 – 4x + 4) + 2
= 3x2 – 12x + 14
Age of man = 3(x – 2)2
∴ Man’s present age = 3(x – 2)2 + 2
= 3(x2 – 4x + 4) + 2
= 3x2 – 12x + 14
After 3 year
The age of son = (x + 3) years
Age of man
According to given condition,
Since,
∴ The present age of man
Let the present ages of son be x years.
Then, Age of father = 2x2
After 8 years Age of son = (x + 8) yrs.
And Age of father = (2x2+8) yrs.
According to given condition
Hence, present age of son = 4 years
And, present age of father = 2(4)2 = 32 yrs.
Let the shortest side (BC) = x m
Then, hypotenuse (AC) = (2x – 1) m
and third side (AB) = (x + 1) m
Now in right ΔABC, We have
AC2 = AB2 + BC2 [By Pythagoras Theorem)
⇒ (2x – 1)2 = (x + 1)2 + (x)2
⇒ 4x2 + 1 – 4x = x2 + 1 + 2x + x2
⇒ 4x2 + 1 – 4x = 2x2 + 2x + 1
⇒ 2x2– 6x = 0
⇒ 2x(x – 3) = 0
⇒ 2x = 0
or x – 3 = 0
x = 0
or x = 3
Since x = 0 is not possible.
So, x = 3
Hence, required sides of the triangle are
AB = x + 1 = 4 m.
BC = x = 3 m
and AC = 2x – 1 = 5 m.
In right triangle ABC, We have
AC2 = AB2 + BC2 [By Pythagoras theorem)
⇒ (25)2 = (x)2 + (x – 5)2
⇒ 625 = x2 + x2 + 25 – 10x
⇒ 625 = 2x2 – 10x + 25
⇒ 2x2 – 10x – 600 = 0
⇒ x2 – 5x – 300 = 0
⇒ x2 – 20x + 15x – 300 = 0
⇒ x(x – 20) + 15(x – 20) = 0
⇒ (x + 15) (x – 20) = 0
⇒ x + 15 = 0
or x – 20 = 0
⇒ x = – 15
or x = 20
Since x = –15 is not possible
Therefore, x = 20
Hence, required sides of the triangle are
BC = x – 5 = 20 – 5 = 15 cm
and AB = x = 20 cm.
Let the base (BC) of a right triangle be x cm and altitude (AB) be y cm.
Then, according to question, Hypotenuse (AC) = (x + 2) cm ...(i)
and hypotenuse (AC) = (2y + 1) cm ...(ii)
Comparing (i) and (ii), we get
x + 2 = 2y + 1
⇒ 2y = x + 2 – 1
⇒ 2y = x + 1
Now,
Now, in ΔABC, we have
[Using Pythagoras theorem]
x = -1 is not possible
Therefore, x = 15
Now, requried sides are
BC = x = 15 cm
AC = x + 2 = 17 cm
and AB =
Let the base (BC) of the right triangle = x cm
So, altitude (AB) = (x – 7) cm
And hypotenuse (AC) = 13 cm
According to Pythagoras theorem,
AC2 = AB2 + BC2
⇒ (13)2 = (x – 7)2 + (x)2
⇒ 169 = x2 + 49 – 14x + x2
⇒ 169 = 2x2 –14x + 49
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x + 5) (x – 12) = 0
⇒ x + 5 = 0
or x – 12 = 0
⇒ x = – 5
or x = 12
Since side of triangle is never negative
So, x = 12
Now required sides of the triangle are
AB = x – 7
= 12 – 7 = 5 cm and BC = x = 12 cm.
If the smaller side is tripled and the larger side be doubled, the new hypotenuse is 15 cm. Then,
(3x)2 + (2y)2 = 152
⇒ 9x2 + 4y2 = 225 ...(ii)
From equation (i), we get y2 = 45 – x2
Putting y2 = 45 – x2 in equation (ii), we get
9x2 + 4 (45 – x2) = 225
⇒ 5x2 +180 = 225
⇒ 5x2 = 45
⇒ x2 = 9
⇒ x = ±3
But, length of a side cannot be negative.
Therefore, x = 3.
Putting x = 3 in (i), we get
9 + y2 = 45 ⇒ y2 = 36 ⇒ y = 6
Hence, the length of the smaller side is 3 cm and the length of the larger side is 6 cm.
⇒ xy = 100
⇒ x (30 – 2x) = 100
⇒ 30x – 2x2 = 100
⇒ 15x – x2 = 50
⇒ x2 – 15x + 50 = 0
⇒ x2 – 10x – 5x + 50 = 0
⇒ (x – 10) (x – 5) = 0
⇒ x = 5, 10
When x = 5, we have
y = 30 – 2 x 5 = 20
When x = 10, we have
y = 30 – 2 x 10= 10
Hence, the dimensions of the vegetable garden are
5 m x 20 m or 10 m x 10 m.
Let the width of the gravel = x metres. Then,
Each side of the square flower bed
= (44 – 2x) metres.
Now, Area of the square field
= 44 x 44 = 1936 m2
and Area of the flower bed
= (44 – 2x)2 m2
∴ Area of the gravel path
= Area of the field – Area of the flower bed
= 1936 – (44 – 2x)2
= 1936 – (1936 – 176x +4x2)
= (176x – 4x2) m2
Cost of laying the flower bed
= (Area of the flower bed) (Rate per sq. m)
Cost of gravelling the path
= (Area of the path) x (Rate per sq. m)
=
It is given that the total cost of laying the flower bed and gravelling the path is Rs. 4904.
∴ 11 (22 – x)2 + 6 (44x – x2) = 4904
⇒ 11(484 – 44x + x)2 + (264x – 6x2) = 4904
⇒ 5x2 – 220x + 5324 = 4908
5x2 – 220x + 420 = 0
⇒ x2 – 44x + 84 = 0
⇒ x2 – 42x – 2x + 84 = 0
⇒ x (x – 42) – 2 (x – 42) = 0
⇒ (x – 2) (x – 42) = 0
⇒ x = 2 or x = 42
But, x ≠ 42, as the side of the square is 44 m.
Therefore, x = 2.
Hence, the width of the gravel path is 2 metres.
Suppose the time taken by the faster pipe is x minutes.
∴ Time taken by the slower pipe = (x + 3) minutes
Now,
Portion of cistern filled by the faster pipe and portion of cistern filled by the slower pipe =
and portion of the cistern filled by the both in 1 minute
It is given that,
Portion of the cistern filled by both 1 minute ...(ii)
Comparing (i) and (ii), we get
(which is rejected) Hence, time taken by the faster pipe = x = 5 minutes and time taken by the slower pipe = (x + 3) = 5 + 3 = 8 minutes.
Let the number of books bought be x. Then,
Cost of x books = Rs. 80 Cost of one book = Rs.
If the number of books bought is x + 4, then
Cost of one book = Rs.
According to given condition,
∴
or
[∵ x cannot be negative]
Hence, the number of books is 16.
Suppose B alone takes x days to finish the work. Then, A alone can finish it in (x - 6) days.
Now, (A’s one day’s work) + (B’s one day’s work)
and
∴
But, x cannot be less than 6.
So, x = 12.
Hence, B alone can finish the work in 12 days.
Let marks got by P in Mathematics = x
Then, marks got by P in Science = (28 – x)
Now, according to the given condition, we have
(x + 3) (28 – x – 4) = 180
⇒ (x + 3) (24 – x) = 180
⇒ 24x – x2 + 72 – 3x = 180
⇒ –x2 + 21x + 72 – 180 = 0
⇒ –x2 + 21x – 108 = 0
⇒ x2 – 21x + 108 = 0
⇒ x2 – 12x – 9x + 108 = 0 ⇒ x(x – 12)– 9 (x – 12) = 0
⇒ (x – 12) (x – 9) = 0
⇒ x = 12
or x = 9
Hence if marks got by P in mathematics 12 then marks got by him in science = 28 – 12 =16 Ans.
And if marks got by P in mathematics 9 then
Marks got by him in science = 28 – 9 = 19 Ans.
No. of articles = x,
Cost of each article = y
According to question.
Ist condition: xy = 900,
2nd condition:
2x2 + 890x – 4500 = 980x
2x2 – 90x – 4500 = 0
x2 – 45x – 2250 = 0
x2 – 75x + 30x – 2250 = 0
x(x –75) + 30(x – 75) = 0
(x + 30) (x – 75) = 0
x = – 30 (reject),
x = 75
No. of articles = 75.
Solution not provided.
Ans. 92
Solution not provided.
Ans. 84
Solution not provided.
Ans. 57
Solution not provided.
Ans. 8, 9, 10
Solution not provided.
Ans. 6, 9, 15
Solution not provided.
Ans. 4, 6
Solution not provided.
Ans. 8, 12
Solution not provided.
Ans. 18, 12 or -18, -12
Solution not provided.
Ans.
Solution not provided.
Ans. 48 km/hr, 32 km/hr.
Solution not provided.
Ans. 3 km/hr
Solution not provided.
Ans. 45 km/hr, 60 km/hr
Solution not provided.
Ans. 45 km/hr
Solution not provided.
Ans. 40 km/hr
Solution not provided.
Ans. 5 km/hr
Solution not provided.
Ans. 10 km/hr
Solution not provided.
Ans. 6 km/hr
Solution not provided.
Ans. 5 km/hr
Solution not provided.
Ans. 3 yrs.
Solution not provided.
Ans. 36 yrs, 6 yrs
Solution not provided.
Ans. 7 yrs.
Solution not provided.
Ans. 5 yrs, 25 yrs
Solution not provided.
Ans. 3 yrs, 24 yrs
Solution not provided.
Ans. 2 m
Solution not provided.
Ans. 17 cm, 15 cm, 8cm
In a right angled triangle, the largest side is 1 cm more than two times its shortest sides; whereas the other side is 1 cm less than two times the shortest side. Taking the shortest side to be x cm; find (in terms of x) the lengths of other sides of the triangle. Now, find:
(i) the value of x (ii) lengths of the sides of the triangle (iii) area of the triangle.
Solution not provided.
Ans. (2x+1) cm, (2x - 1) cm (i) 8, (ii) 8 cm, 15 cm, 17 cm, (iii) 60 cm2
Solution not provided.
Ans. 24 cm
Solution not provided.
Ans. 10 minutes, 15 minutes
Solution not provided.
Ans. 5 minutes, 6 minutes
Solution not provided.
Ans. 15 hrs, 10 hrs, 6 hrs
Solution not provided.
Ans. 30 days
Solution not provided.
Ans. Maths = 12, English = 18 or Maths = 13, English = 17
Solution not provided.
Ans. Rs 8
Solution not provided.
Ans. 2.72 m
Solution not provided.
Ans. 12 m.
Let the speed of the stream be x km/hr.
∴ Speed of the boat upstream = (18 – x) km/hr.
Speed of the boat downstream = (18 + x) km/hr.
Time taken for going 24 km upstream =
Time taken for going 24 km downstream =
According to given condition,
∴
Using the quadratic formula, we get
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = –54. Therfore, x = 6 gives the speed of the stream as 6km/h. is rejected.
Hence, the speed of the stream is 2 km/hr.
Solution not provided.
Solution not provided.
Solution not provided.
Solution not provided.
Ans. p = 0, 3
Solution not provided.
Solution not provided.
Ans. K = 5, -3
Solution not provided.
Ans. n = 23
Solution not provided.
Solution not provided.
Ans. Rs. 50
Solution not provided.
Ans.
Solution not provided.
Ans. K = 4
Solution not provided.
Ans. length = 12 m, breadth = 9 m
Solution not provided.
Ans. 16 m
Solution not provided.
Ans. Rs 20
Solution not provided.
Ans.
Solution not provided.
Ans. length = 20 met., Rate = 2.10 met.
Solution not provided.
Ans. 2yrs., 22 yrs.
Solution not provided.
Ans. equal and real roots.
Solution not provided.
Ans.
Solution not provided.
Ans. 8, 9, 10
Solution not provided.
Ans. 300 km/hr.
Solution not provided.
Ans. 30 k.m./hr
Solution not provided.
Ans. 30 k.m./hr.
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans. 10m and 10m or 20m. and 5 m.
Solution not provided.
Ans.
Solution not provided.
Ans. 2cm, 13cm, 5cm.
Solution not provided.
Ans. 5cm, 8 cm.
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans. x = 50
Solution not provided.
Ans. 12 and 4 or 4 and 12
Solution not provided.
Ans. 6, 12 or 12, 6
Solution not provided.
Ans. 35
Solution not provided.
Ans.
Solution not provided.
Ans. 27 yrs
Solution not provided.
Ans. 45 k.m./hr.
Solution not provided.
Ans. -5, -10 or 10, 5
Solution not provided.
Ans. Rs. 30
Solution not provided.
Ans. 26 cm., 24 cm.
Solution not provided.
Ans. 25 k.m./hr
Solution not provided.
Ans. 9, 6 or 9, -6
Solution not provided.
Ans. 9, 19 or 12, 16
Solution not provided.
Ans. 6 km/hr.
Solution not provided.
Ans. 25 hrs, 15 hrs.
Solution not provided.
Ans. 12 m
Solution not provided.
Ans. 13, 15 or -15, +3
Solution not provided.
Ans.
Solution not provided.
Ans.
Solution not provided.
Ans. k = 7/4, p = 7
Solution not provided.
Ans.
Solution not provided.
Ans. 1 km/hr.
Solution not provided.
Ans. 8 m, 14 m
Solution not provided.
Ans. 32 yrs., 4 years
Solution not provided.
Ans. Real and equal.
Solution not provided.
Ans. 29
D.
both (a) and (b)If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.
We Given -5 is a root of the quadratic equation 2x2+px-15 =0
, -5 satisfies the given equation.
∴ 2 (-5)2 +p(-5)-15=0
50-5p-15 =0
35-5p=0
5p=35 ⇒ p=7
Substituting p=7 in (x2+x)+k =0, we get
7(x2+x)+k =0
7x2+7x+k=0
The roots of the equation are equal
∴ Discriminant =b2-4ac =0
Here, a= 7, b=7,c=k
b2-4ac=0
∴(7)2-4(7)(k) =0
49-28k =0
28k=49
k= 49/28
=7/4
If the quadratic equation has two roots, then find the value of p.
The given quadratic equation has two equal roots.
therefore,
by using the discriminant method
⇒ 20p2 - 60p = 0
⇒ 20p (p - 3) = 0
⇒ p = 0 or p - 3 = 0
⇒ p = 0 or p = 3
p cannot be zero.
Hence, the value of p is 3.
Solve the following quadratic equation for x,
4x2 + 4bx - (a2 - b2) = 0
Given equation,
4x2 + 4bx - (a2 - b2) = 0
⇒ 4x2 +4bx - ( a + b) (a - b) = 0
⇒ 4x2 + 2[(a + b) - (a - b)] x - (a + b)(a-b) = 0
⇒ 4x2 +2 (a + b)x - 2(a - b)x - (a + b ) (a - b) = 0
⇒ 2x [2x + (a + b)] [(a - b)[2x + (a + b)] = 0
⇒ 2x + (a + b) = 0 or 2x - (a - b) = 0
⇒ x = - (a + b)/2 OR x = (a - b)/2
Find the value of k for which the equation x2 + k(2x + k − 1) + 2 = 0 has real and equal roots.
The given equation is x2+k(2x+k−1)+2=0.
⇒x2+2kx+k(k−1)+2=0
So, a = 1, b = 2k, c = k(k − 1) + 2
We know D=b2−4ac
⇒D=(2k)2−4×1×[k(k−1)+2]
⇒D=4k2−4[k2−k+2]
⇒D=4k2−4k2+4k−8
⇒D=4k−8=4(k−2)
For equal roots, D = 0
Thus, 4(k − 2) = 0
So, k = 2.
If the equation (1 + m2) x2 + 2mcx + c2 – a2 = 0 has equal roots then show that c2 = a2 (1 + m2).
Given quadratic equation: (1 + m2)x2 + 2mcx + (c2 – a2) = 0
The given equation has equal roots, therefore
D = b2 − 4ac = 0 ..... (1)
From the above equation, we have
a = (1 + m2)
b = 2mc
and c = (c2 – a2)
Putting the values of a, b and c in (1), we get
D = (2mc)2 − 4(1 + m2) (c2 – a2) = 0
⇒ 4m2c2 − 4 (c2 + c2m2 − a2 − a2m2) = 0
⇒ 4m2c2 − 4c2 − 4c2m2 + 4a2 + 4a2m2 = 0
⇒ −4c2 + 4a2 + 4a2m2 = 0
⇒ 4c2 = 4a2 + 4a2m2
⇒ c2 = a2 + a2m2
⇒ c2 = a2(1 + m2)
Hence, proved.
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Let the speed of the stream be x km/h.
It is given that the speed of a boat in still water is 15 km/h.
Now,
Speed of the boat upstream = Speed of the boat in still water − Speed of the stream = (15 − x) km/h
Speed of the boat downstream = Speed of the boat in still water + Speed of the stream = (15 + x) km/h
We know that
Time=Distance/Speed
Time is taken for upstream journey + Time taken for the downstream journey = 4 h 30 min
Since speed can not be negative, therefore, x = 5.
Thus, the speed of the stream is 5 km/h.
If x = 3 is one root of the quadratic equation x2-2kx -6 =0, the find the value of k.
x = 3 is one of the root of x2-2kx -6 =0
(3)2 -2k(3) - 6 =0
9-6k - 6 = 0
3 - 6k = 0
3 = 6k
k = 3/6 = 1/2
A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Let the usual speed of the plane be x km/hr
Time taken to cover 1500 km with usual speed, t1 =
Time taken to cover 1500 km with speed of (x +100) km/hr, t2
Difference between t1 - t2 =30 min = 1/2
But speed can't be negative
Hence, usual speed 500 km/hr.
A motor boat whose speed is 18 km/hr in still water 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Let the speed of stream be x km/hr
Now, for upstream,speed = (18-x) km/hr
Now, for downstream;speed = (18 +x) km/hr
Given that,
Speed can never be negative, -54 km/hr
hence speed of the stream 6 km/hr
A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. It takes 3 hours to complete the total journey, what is the original average speed ?
Let x be the original average speed of the train for 63 km.
Then, (x + 6) will be the new average speed for the remaining 72 km.
Total time taken to complete the journey is 3 hrs.
Since speed can not be negative.
Therefore x = 42 km/hr.
Find the values of k for which the quadratic equation
9x2 - 3kx + k = 0 has equal roots.
Given Quadratic equation 9x2 - 3kx + k = 0 has equal roots.
Let
A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
let speed of stream = x kmh
Speed of boat in steel water = 18 km
speed of boat in upstream = (18 - x ) km
speed of boat in downstream = (18 + x ) km
Distance = 24 km
As per question,
If the roots of the equation (a2 + b2) x2 + 2(ac + bd) x + (c2 + d2 ) = 0 are
equal, prove that .
We have,
(a2 + b2) x2 + 2(ac + bd) x + (c2 + d2 ) = 0
The discriminant of the given equation is given by
A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?
Suppose B alone takes x days to to finish the work.
Then, A alone can finish it in (x-6) days.
(A's one days work) + (B's one day work) =
And, ( A+B)'s one day's work =
But, x cannot be less than 6.
So, x = 12
Hence, B alone can finish the work in 12 days.
If 1 is a root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0, then ab equals
3
6
-3
A.
3
It is given that 1 is the root of the equations
Therefore, y = 1 will satisfy both the equations.
Find the values (s) of k so that the quadratic equation 3x2 - 2kx + 12 = 0 has equal roots.
Given quadratic equation is 3x2 - 2kx + 12 = 0
Here a = 3, b = -2k, c = 12
The quadratic equation will have equal roots if = 0.
b2 - 4ac = 0
Putting the values of a, b and c we get
= ( -2k )2 - 4 (3) (12) = 0
4k 2 - 144 = 0
4k 2 = 144
Therefore, the required values of k are 6 and -6.
A shopkeeper buys some books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. Find the number of books he bought.
OR
The sum of two number is 9 and the sum of their reciprocals is . Find the numbers.
Total cost of books = Rs.80
Let the number of books = x
So the cost of each book = Rs
Cost of each book if he buy 4 more books = Rs
As per given in equations:
Since number of books cannot be negative,
so the number of books he brought is 16.
OR
Let the frist number be x then the second number be 9 - x as the sum of both number is 9.
Now the sum of their reciprocal is , therefor
If x = 6 then other number is 3.
And, if x = 3 then other number is 6.
Hence numbers are 3 and 6.
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