Hint Let the speed of stream be \(x \mathrm{~km} / \mathrm{h}\).
\(\therefore\) Speed of upstream \(=(18-x) \mathrm{km} / \mathrm{h}\)
and speed of downstream \(=(18+x) \mathrm{km} / \mathrm{h}\)
According to the question, we get
\(\frac{24}{18-x}-\frac{24}{18+x}=1\)
Now, solve it. Ans. Speed of stream \(=6 \mathrm{~km} / \mathrm{h}\)

The helicopter is equally likely to crash anywhere in the region.
Area of the entire region where the helicopter can crash
= (4.5 x 9) km² = 40.5 km²
Area of the lake = (2.5 x 3) km² = 7.5 km²
Therefore, P (helicopter crashed in the lake) = \(\frac{7.5}{40.5}=\frac{75}{405}=\frac{5}{27}\)

Let us first split the middle term - 5x as -2x -3x [because (-2x) x (-3x) = 6x² = (2x² ) x 3].
So, 2x² - 5x + 3 = 2x² - 2x - 3x + 3 = 2x (x - 1) -3(x - 1) = (2x - 3)(x - 1)
Now, 2x² - 5x + 3 = 0 can be rewritten as (2x - 3)(x - 1) = 0.
So, the values of x for which 2x² - 5x + 3 = 0 are the same for which (2x - 3)(x - 1) = 0,
i.e., either 2x - 3 = 0 or x - 1 = 0.
Now, 2x - 3 = 0 gives x = \(\frac{3}{2}\) and x - 1 = 0 gives x = 1.
So, x = \(\frac{3}{2}\) and x = 1 are the solutions of the equation.
In other words, 1 and \(\frac{3}{2}\) are the roots of the equation 2x² - 5x + 3 = 0.
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x² - 5x + 3 = 0 by factorising 2x² - 5x + 3 into two linear factors and equating each factor to zero.

kx(x - 2) + 6 = 0
or kx² - 2kx + 6 = 0
Comparing this equation with ax² + bx + c = 0, we get
a = k, b = - 2k and c = 6
= ( - 2k)² - 4 (k) (6)
= 4k² - 24k
For equal roots,
b² - 4ac = 0
4k² - 24k = 0
4k (k - 6) = 0
Either 4k = 0 or
k = 6 = 0
k = 0 or k = 6
However, if k = 0, then the equation will not have the terms 'x²' and 'x'.
Therefore, if this equation has two equal roots, k should be 6 only.

There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25  
Therefore, P (card with name of a girl) = P(Girl) = \(\frac{25}{40}=\frac{5}{8}\)
(ii) The number of outcomes favourable for a card with the name of a boy = 15 
Therefore, P(card with name of a boy) = P(Boy) \(=\frac{15}{40}=\frac{3}{8}\)
Note : We can also determine P(Boy), by taking
P(Boy) = 1 – P(not Boy) = 1 – P(Girl) \(=1-\frac{5}{8}=\frac{3}{8}\)

Total number of pieces = 8 + 10 = 18
(i) No.of triangles = 8. Hence, P(triangle is lost) = \(\frac{8}{18}=\frac{4}{9}\)
(ii) No.of squares = 10. Hence, P(square is lost) = \(\frac{10}{18}=\frac{5}{9}\)
(iii) No.of squares of blue colour = 6. So, P(square of blue colour is lost) = \(\frac{6}{18}=\frac{1}{3}\)
(iv) No.of triangles of red colour = 8 - 3 = 5. So, P(triangle of red colour is lost) = \(\frac{5}{18}\)

Let the number of marbles John had be x
Then the number of marbles Jivanti had be = 45 – x (Why?).
The number of marbles left with john, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marble = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= – x² + 45x – 200
So, – x² + 45x – 200 = 124 (Given that product = 124)
i.e., – x² + 45x – 324 = 0
i.e., x² – 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x² – 45x + 324 = 0
which is the required representation of the problem mathematically

One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100 equally likely outcomes
(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 (Why?)
Therefore, P (shirt is acceptable to Jimmy) = \(\frac{88}{100}=0.88\)
(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96 (Why?)
So, P (shirt is acceptable to Sujatha) = \(\frac{96}{100}=0.96\)

(i)Total number of days in the year 1990=365
Total number of ways in which two friends Ankita and Nagma may have their birthday = 365 x 365,
the number of ways in which both have same birthday = 365
Probability that both have same birthday
\(={365\over 365\times365}={1\over 365}\)
(ii)Probability that both have different birthday
\(=1-{1\over 365}={364\over 365}\)

\(\left[-{1\over 3} , -1\right]\)

Let the two consecutive integers be x and x+1
ATQ x²+(x+1)2=365
\(\Rightarrow\) x²+x2+2x+1=365 \(\Rightarrow\) 2x²+2x-364=0
\(\Rightarrow\) x²+x-182=0 \(\Rightarrow\) x²+14x-13x-182=0
\(\Rightarrow\) x(x+14)-13(x+14)=0 \(\Rightarrow\) (x-13)(x+14)=0
\(\Rightarrow\) x=13, -14 (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and 13+1=14

Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x² = 750
i.e., – x² + 55x – 750 = 0
i.e., x² – 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x² – 55x + 750 = 0
which is the required representation of the problem mathematically.

Total number of outcomes in a single throw of a six faces die = 6
(i) Let E₁ = Event of getting a letter A
then, number of outcomes favourable to E₁ = 2
Hence, probability of getting a letter A,
\(P\left(E_1\right)=\frac{2}{6}=\frac{1}{3}\)
(ii) Let E₂ = Event of getting a letter D
Then, number of outcomes favourable to E₂ = 1
Hence, probability of getting a letter D,
\(P\left(E_2\right)=\frac{1}{6}\)

(i) The car starts normally but when there is some defect, then car does not start. So, the outcomes are not equally likely.
(ii) The outcomes in this situation are not equally likely because the outcomes depends on many factors such as training ofthe player, quality of basketball, etc.
(iii) The outcomes in trial of true-false question is either true or false. Hence, the two outcomes are equally likely.
(iv) A new baby can be either a boy or a girl, so both the outcomes are equally likely.

Yes x²+2-2 \(\sqrt { 2x } -2x-2=0\)
\(\Rightarrow x^{ 2 }-2(\sqrt { 2 } +1)x=0\)
b²-4ac=4 \(\sqrt { 2 } +1)^{ 2 }\) -4X1X0
4(2+1+\(2\sqrt { 2 } )=12+8\sqrt { 2 } \)
\(\therefore \) D>0 Distinct real roots

Let the breadth of the park be x m.
Given, perimeter of a rectangular park = 80 m
\(\Rightarrow\) 2(Length + Breadth) = 80 m
\(\Rightarrow\) Length + Breadth = 40 m
\(\Rightarrow\) Length = (40 - x)m
Now, area of a rectangular park = Length \(\times\)Breadth
=(40 - x)x m²
But according to the question, area of the rectangular park is 400 m².
\(\therefore\) (40 - x) x = 400
\(\Rightarrow\) x² - 40x + 400 = 0
\(\Rightarrow\) x² - 2x \(\times\) 20 + (20)² = 0
\(\Rightarrow\) (x - 20)² = 0 [\(\because\) a² - 2ab + b² = (a - b)²]
\(\Rightarrow\) x = 20
thus, breadth of the park = 20 m
and length of the park = 40 - 20 = 20 m
Hence, it is possible to design the rectangular park having perimeter 80 m and area 400 m with equal length and breadth, i.e. 20 m each.

Let one number be x,
Then, another number be 27 - x
[\(\because\) sum of two numbers = 27]
According to the question,
Product of these two numbers = 182
\(\therefore\) x(27-x)=182
\(\Rightarrow\) 27x-x² = 182
\(\Rightarrow\) x²-27x+182=0 
\(\Rightarrow\) x²-14x-13x+182=0
[\(\because\) (-14) \(\times\) (-13) =182 and -14-13=-27
\(\Rightarrow\) x(x-14)-13(x-14)=0 \(\Rightarrow\) (x-14)(x-13)=0
\(\Rightarrow\) x-14 = 0 or x-13 = 0
\(\Rightarrow\) x=14 or x=13
If x = 14, then 27-x= 27-14 =13 and if x=13, then 27-x= 27-13 = 14 Hence, in both cases, the numbers are 13 and 14.