General Test - 33

Let the original speed of the car is \(x \) km/hr.

Then, time taken in moving \(840\) km \(=\frac{840}{x}\) hr

If speed is going \(10 \) km/hr more \(= x+10\)

Then, time taken in moving \(840 \) km\(=\frac{840}{x+10}\) hr

\(\therefore \frac{840}{x}-\frac{840}{x+10}=2\)

\(\Rightarrow 840(x+10)-840 x=2 x(x+10)\)

\(\Rightarrow 2 x^{2}+20 x-8400=0\)

\(\Rightarrow x^{2}+70 x-60 x-4200=0\)

\(\Rightarrow(x+70)(x-60)=0\)

\(\Rightarrow x=60\) or \(x=-70\)

If \(x=-70\) is not true because the speed is not negetive.

So, speed is \(60\) km/hr.

Time taken by Vijaya to walk both ways \(=12\) hours \(20\) minutes

\(=(12 \times 60)+20\)

\(=740\) minutes

Time taken by Vijaya to walk one way \(=\frac{740}{2}=370\)

Time taken by Vijaya in walking a distance and running back to same place \(=9\) hours \(50\) minutes

\(=(9 \times 60)+50\)

\(=590\) minutes

Time taken by Vijaya to run one way \(=590-370=220\) minutes

Time taken by her to run back both ways \(=220 \times 2=440\) minutes

\(=(7 \times 60)+20\)

\(=7\) hours \(20\) minutes

Therefore, the time taken by her to run both ways is \(7\) hours \(20\) minutes.

Speed \(=45 \times \frac{5}{18}\) m/sec

\( =\frac{25}{2}\) m/sec

Time \(=30\) sec

Let the length of the bridge be \(x\) m,

Then, \(\frac{130+x}{30}=\frac{25}{2}\)

\(\Rightarrow 2(130+x)=750\)

\(\Rightarrow x=245\) m

Given,

\(x^{2}+8 x+4=0\)

We know that the

Sum of zeros \(=-\frac{\text { coefficient of } x}{\text { coefficientof } x^{2}}\)

\(\Rightarrow a+b=-\frac{8}{1}=-8\)

Also we know that,

Product of the zeros \(=\frac{\text { constant }}{\text { coefficientof } x^{2}}\)

\(\Rightarrow\) product of the zeros \(=\frac{4}{1}\)

\(\Rightarrow a b=4\)

Now, we will be finding value of the following:-

\(\frac{a}{b}+\frac{b}{a}\)

\(=\frac{a^{2}+b^{2}}{a b}\)

\(=\frac{(a+b)^{2}-2 a b}{a b}\)

\(=\frac{64-8}{4}\)

\(=\frac{56}{4}\)

\(=14\)

Let the numbers be x and y. Now, according to the given condition,

x + y = 50……………………(i)

And x - y = 30………………(ii)

We can write x = 50 - y from equation (i).

Therefore, on substituting the value of x in equation (ii), we get,

50 - y - y = 30

⇒ 50 - 2y = 30

⇒ 2y = 50 - 30 = 20

⇒ y = \(\frac {20} {2} \) = 10

And x = 50 - y

⇒50 - 10 =40

Therefore, the two numbers are 40 and 10.

Let alcohol \(=6 x\) and water \(=5 x\)

In 21 litre mixture, alcohol \(=\left(\frac{6}{11}\right) \times 22=12\) litre

And water \(=\left(\frac{5}{11}\right) \times 22=10\) litre

(6x -12) : (5x-10+22)= 9 : 13

\(\Rightarrow 13(6 x-12)=9(5 x+12) \Rightarrow 33 x=264\Rightarrow x=8\)

So, alcohol after replacement \(=6 \times 8-12=36\) litre

Total number of balls\(=(8+7+6)\)\(=21\)

\(\therefore n(S)=21\)

Let \(E=\) event that the ball drawn is neither red nor green that means the ball drawn is blue.

Number of blue balls \(=7\)\(\therefore n(E)=7\)

\(\therefore P(E)=\frac{n(E)}{n(S)}=\frac{7}{21}\)\(=\frac{1}{3}\)

A line segment that joins any vertex of the triangle and the mid-point of its opposite side is called a median.

The area of the triangle is divided in half by a median.

Since AD is the mid-point of \(B C\),

\(\Rightarrow AD\) is the median on the line \(BC\)

Since the median of a triangle divides it into two triangles of equal area. \(\Rightarrow \operatorname{ar}(\triangle ABD )=\frac{1 }{ 2} \times\) area of \(\triangle ABC \quad \cdots(1)\)

Now, BE is the mid-point of \(AD\),

\(\Rightarrow BE\) is the median on the line \(AD\)

\(\Rightarrow \operatorname{ar}(\triangle BED )=\frac{1 }{ 2} \times\) area of \(\triangle ABD\)

\(\Rightarrow \operatorname{ar}(\triangle BED )=\frac{1 }{ 2} \times\frac{1 }{ 2} \times\) area of \(\triangle ABC \quad[\) Using (1) \(]\)

\(\Rightarrow \operatorname{ar}(\triangle B E D)=\frac{1 }{ 4} \times\) area of \(\triangle A B C\)