Probashi Kallyan Bank Ltd. || EO (General) (23-11-2019) || 2019

Let, the length of train is x meter and the velocity of the train is v km/hr.

According to first condition,$\frac{x}{(v-4.5){\displaystyle \frac{5}{18}}}=8.4.............\left(i\right) [\because \frac{Dis\tan ce}{Velocity}=Time]$

According to second condition,$\frac{x}{(v-5.4){\displaystyle \frac{5}{18}}}=8.5............\left(ii\right)$

From equation (i) $\frac{x}{(v-4.5){\displaystyle \frac{5}{18}}}=8.4$

$$\begin{gathered} =x=(v-4.5)\times \frac{5}{18}\times 8.4 \\ =x=(v-4.5)\times \frac{42}{18} \\ =x=\frac{7(v-4.5)}{3} \\ =3x=7v-314.5 \\ =18x=42v-182............\left(iii\right) \\ From Equation \left(ii\right)\frac{x}{(v-5.4){\displaystyle \frac{5}{18}}}=8.5 \\ =x=8.5\times \frac{5}{18}(v-5.4) \\ =18x=42.5v-229.5..........\left(iv\right) \\ Subtracting equation \left(iii\right) from equation \left(iv\right) \\ \frac{18x=42.5v-229.5 \\ 18x=42v-189}{0=0.5v-40.5} \\ =0.5v=40.5 \\ \therefore v=\frac{40.5}{0.5}=81 \\ \therefore The velocity of the train is 81kmph. \left(answer\right) \end{gathered}$$

Let, the container primarily contains 7x and 5x of mixtures A and B respetively.

Now, after drawing off 9 litres of mixtures

The quantity of A left in the mixture = $7x-\frac{7\times 9}{5+7}=7x-\frac{63}{12}=7x-\frac{21}{4}$ litres.

And the quantity of B left in the mixture = $5x-\frac{5\times 9}{5+7}=5x-\frac{45}{12}-\frac{15}{4} litres.$

According to question, $\frac{7x-{\displaystyle \frac{21}{4}}}{5x-{\displaystyle \frac{15}{4}}+9}=\frac{7}{9}$

$$\begin{gathered} =\frac{{\displaystyle \frac{28x-21}{4}}}{{\displaystyle \frac{20x-15+36}{4}}}=\frac{7}{9} \\ =\frac{28x-21}{4}\times \frac{4}{20x+21}=\frac{7}{9} \\ =\frac{28x-21}{20x+21}=\frac{7}{9} \\ =252x-189=140x+147 \\ =112x=336 \\ \therefore x=\frac{336}{112}=3 \\ \therefore The container contained = 7\times 3=21 litres of A. \end{gathered}$$

In bag 1: There are 5 red balls and 3 green balls.

In bag 2: There are 4 red balls and 6 green balls.

Now,

In case 1, the probability of getting red ball from bag I and green ball from bag

$2=\frac{5}{5+3}\times \frac{6}{4+6}=\frac{5}{6}\times \frac{6}{10}=\frac{30}{60}$

And, In case 2, the probability of getting green ball from bag I and red ball from bag

$2=\frac{3}{8}\times \frac{4}{10}=\frac{12}{80}$

$\therefore$Probability of getting 1 red ball and 1 green ball = $\frac{30}{80}+\frac{12}{80}=\frac{30+12}{80}=\frac{42}{80}=\frac{21}{40}$ (answer).

Let us depict a picture according to question

Now, let one ship is x meter away and another ship is y meter away from the lighthouse.

Here, AD = 100 meter = height of lighthouse

$\angle$ABD = 45$°$ and $\angle$ACD = 30$°$

From $∆$ADC.

tanθ = $\frac{AD}{DC}$

= tan 30$°$ = $\frac{100}{x}$

$$\begin{gathered} =\frac{1}{\sqrt{3}}=\frac{100}{x} \\ \therefore x=100\sqrt{3} \\ Again, from ∆ABD \\ \tan \theta =\frac{AD}{BD} \\ =\tan 45°=\frac{100}{y} \\ =1=\frac{100}{y} \\ \therefore y=100 \\ \because The dis\tan ce between the two ships will be =x+y \\ =100\sqrt{3}+100=100(1.73+1)=100(2.73)=273 \left(answer\right) \end{gathered}$$

Given, $\frac{x^2+xy+y^2}{x^2-xy+y^2}$

$$\begin{gathered} \therefore x+y=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{{(\sqrt{5}+1)}^2+{(\sqrt{5}-1)}^2}{(\sqrt{5}-1) (\sqrt{5}+1)}=\frac{{\left(\sqrt{5}\right)}^2+2\times \sqrt{5}\times 1+1^2+{\left(\sqrt{5}\right)}^2-2\times \sqrt{5}\times 1+1^2}{{\left(\sqrt{5}\right)}^2-\left(1\right)} \\ =\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{5-1}=\frac{12}{4}=3 \\ And, xy =\frac{\sqrt{5}+1}{\sqrt{5}-1}\times \frac{\sqrt{5}-1}{\sqrt{5}+1}=1 \\ Now, x^2+xy+y^2=x^2+2xy+y^2-xy={(x+y)}^2-xy=3^2-1=9-1=8 \\ And, x^2-xy+y^2=x^2+2xy+y^2-3xy={(x+y)}^2-3xy=3^2-(3\times 1)=9-3=6 \\ \therefore \frac{x^2+xy+y^2}{x^2-xy+y^2}=\frac{8}{6}=\frac{4}{3} \end{gathered}$$

Given that, ab+bc+ca=0

$$\begin{gathered} \therefore ab+bc=-ca \therefore ca=-(ab+bc) \\ =ab+ca=-bc \therefore =bc=-(ab+ca) \\ =bc+ca=-ab \therefore =ab=-(bc+ca) \\ Again given that, \frac{1}{a^2-bc}+\frac{1}{b^2-ac}+\frac{1}{c^2-ab} \\ =\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+bc+ca}=\frac{1}{a(a+b+c)}+\frac{1}{b(b+a+c)}+\frac{1}{c(c+b+a)} \\ =\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}=\frac{1}{a(a+b+c)}\times (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{a+b+c}\times \frac{(bc+ca+ab)}{abc} \\ =\frac{ab+bc+ca}{abc(a+b+c)}=\frac{0}{abc(a+b+c)}=0 \end{gathered}$$