Total number of balls = 12 Total number of black balls = x P (getting a black ball) = x/12 If 6 more black balls are put in the box, then Total number of balls = 12 + 6 = 18 Total number of black balls = x + 6 P (getting a black ball now) = (x + 6)/18 According to the condition given in the questionRead more
Total number of balls = 12
Total number of black balls = x
P (getting a black ball) = x/12
If 6 more black balls are put in the box, then
Total number of balls = 12 + 6 = 18
Total number of black balls = x + 6
P (getting a black ball now) = (x + 6)/18
According to the condition given in the question,
2(x/12) = (x + 6)/18
⇒ 3x = x + 6
⇒ 2x = 6
⇒ x = 3
Let the number of blue balls be x. Number of red balls = 5 Total number of balls = x + 5 P (getting a red all) = 5/(5 + x) P (getting a blue ball) = x/(5 + x) Given that, 2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x - 50 = 0 ⇒ x² - 10x + 5x - 50 = 0 ⇒ x(x - 10) + 5(x - 10) = 0 ⇒ (x - 10)(x + 5)Read more
Let the number of blue balls be x.
Number of red balls = 5
Total number of balls = x + 5
P (getting a red all) = 5/(5 + x)
P (getting a blue ball) = x/(5 + x)
Given that,
2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x – 50 = 0
⇒ x² – 10x + 5x – 50 = 0 ⇒ x(x – 10) + 5(x – 10) = 0
⇒ (x – 10)(x + 5) = 0
⇒ Either x – 10 = 0 or x + 5 = 0
⇒ x – 10 or x = -5
However, the number of balls cannot be negative.
Hence, number of blue balls = 10
QR is diameter of circle. Therefore, ∠RPQ = 90° [Angle in semicircle is right angle] In APQR, by Pythagoras theorem, RP² + PQ² = RQ² (7)² + (24)² = RQ² ⇒ RQ² = 576 + 49 = 625 ⇒ RQ = √625 = 25 Therefore, the radius of circle = RQ/2 = 25/2 cm Area of shaded region = Area of semicircle - Area of APQR =Read more
QR is diameter of circle.
Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]
In APQR, by Pythagoras theorem,
RP² + PQ² = RQ²
(7)² + (24)² = RQ²
⇒ RQ² = 576 + 49 = 625
⇒ RQ = √625 = 25
Therefore, the radius of circle = RQ/2 = 25/2 cm
Area of shaded region = Area of semicircle – Area of APQR
= 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12
= 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = (-1)/4 = (-b)/a αβ = 1/4 = c/a On comparing, a = 4, b = 1 and c = 1 Hence, the required quadratic polynomial is 4x² + x + 1. For Video Solution See Here 👀
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = (-1)/4 = (-b)/a
αβ = 1/4 = c/a
On comparing,
a = 4, b = 1 and c = 1
Hence, the required quadratic polynomial is 4x² + x + 1.
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 4 = 4/1 = (-b)/a αβ = 1 = 1/1 = c/a On comparing, a = 1, b= - 4 and c = 1 Hence, the required quadratic polynomial is x² - 4x + 1. See Here 😃✌
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = 4 = 4/1 = (-b)/a
αβ = 1 = 1/1 = c/a
On comparing,
a = 1, b= – 4 and c = 1
Hence, the required quadratic polynomial is x² – 4x + 1.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Total number of balls = 12 Total number of black balls = x P (getting a black ball) = x/12 If 6 more black balls are put in the box, then Total number of balls = 12 + 6 = 18 Total number of black balls = x + 6 P (getting a black ball now) = (x + 6)/18 According to the condition given in the questionRead more
Total number of balls = 12
Total number of black balls = x
P (getting a black ball) = x/12
If 6 more black balls are put in the box, then
Total number of balls = 12 + 6 = 18
Total number of black balls = x + 6
P (getting a black ball now) = (x + 6)/18
According to the condition given in the question,
2(x/12) = (x + 6)/18
⇒ 3x = x + 6
⇒ 2x = 6
⇒ x = 3
See here for video explanation🙌
See lessA bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Let the number of blue balls be x. Number of red balls = 5 Total number of balls = x + 5 P (getting a red all) = 5/(5 + x) P (getting a blue ball) = x/(5 + x) Given that, 2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x - 50 = 0 ⇒ x² - 10x + 5x - 50 = 0 ⇒ x(x - 10) + 5(x - 10) = 0 ⇒ (x - 10)(x + 5)Read more
Let the number of blue balls be x.
See lessNumber of red balls = 5
Total number of balls = x + 5
P (getting a red all) = 5/(5 + x)
P (getting a blue ball) = x/(5 + x)
Given that,
2(5/(5 + x)) = (x/(5 + x)) ⇒ 10(x + 5) = x² + 5x – 50 = 0
⇒ x² – 10x + 5x – 50 = 0 ⇒ x(x – 10) + 5(x – 10) = 0
⇒ (x – 10)(x + 5) = 0
⇒ Either x – 10 = 0 or x + 5 = 0
⇒ x – 10 or x = -5
However, the number of balls cannot be negative.
Hence, number of blue balls = 10
Find the area of the shaded region in Figure.
QR is diameter of circle. Therefore, ∠RPQ = 90° [Angle in semicircle is right angle] In APQR, by Pythagoras theorem, RP² + PQ² = RQ² (7)² + (24)² = RQ² ⇒ RQ² = 576 + 49 = 625 ⇒ RQ = √625 = 25 Therefore, the radius of circle = RQ/2 = 25/2 cm Area of shaded region = Area of semicircle - Area of APQR =Read more
QR is diameter of circle.
Therefore, ∠RPQ = 90° [Angle in semicircle is right angle]
In APQR, by Pythagoras theorem,
RP² + PQ² = RQ²
(7)² + (24)² = RQ²
⇒ RQ² = 576 + 49 = 625
⇒ RQ = √625 = 25
Therefore, the radius of circle = RQ/2 = 25/2 cm
Area of shaded region = Area of semicircle – Area of APQR
= 1/2 × πr² – 1/2 PR × PQ = 1/2 × π(25/2)² – 1/2 × 7 × 24 = 1/2 × 22/7 × 25/2 × 25/2 – 7 × 12
= 6875/28 – 84 = (6875 – 2352)/28 = 4523/28 cm²
Here is the video explanation ✌😇
See lessFind a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. -1/4,1/4
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = (-1)/4 = (-b)/a αβ = 1/4 = c/a On comparing, a = 4, b = 1 and c = 1 Hence, the required quadratic polynomial is 4x² + x + 1. For Video Solution See Here 👀
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = (-1)/4 = (-b)/a
αβ = 1/4 = c/a
On comparing,
a = 4, b = 1 and c = 1
Hence, the required quadratic polynomial is 4x² + x + 1.
For Video Solution See Here 👀
See lessFind a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 4,1
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have α + β = 4 = 4/1 = (-b)/a αβ = 1 = 1/1 = c/a On comparing, a = 1, b= - 4 and c = 1 Hence, the required quadratic polynomial is x² - 4x + 1. See Here 😃✌
Let α and β are the zeroes of the polynomial ax² + bx + c, then we have
α + β = 4 = 4/1 = (-b)/a
αβ = 1 = 1/1 = c/a
On comparing,
a = 1, b= – 4 and c = 1
Hence, the required quadratic polynomial is x² – 4x + 1.
See Here 😃✌
See less