SRCC GBO 2009 Question Paper

Initial selling price = Rs. $$x$$

Cost price after 15% loss = $$\frac{x}{100-15}\times100$$ --------------(i)

New selling price = Rs. $$y$$

Cost price after 15% gain = $$\frac{y}{100+15}\times100$$ -------------(ii)

Comparing equations (i) and (ii),

=> $$\frac{x}{85}=\frac{y}{115}$$

=> $$\frac{x}{y}=\frac{115}{85}=\frac{23}{17}$$

Using componendo and dividendo

=> $$\frac{y-x}{y+x}=\frac{17-23}{17+23}=\frac{6}{40}$$

=> $$\frac{y-x}{y+x}=3:20$$

=> Ans - (C)

Expression : $$\sqrt{x+\frac{x}{y}}=x\sqrt{\frac{x}{y}}$$

=> $$x+\frac{x}{y}=x^2\times\frac{x}{y}$$

=> $$1+\frac{1}{y}=\frac{x^2}{y}$$

=> $$\frac{1}{y}(x^2-1)=1$$

=> $$y=x^2-1$$

=> Ans - (B)

A number when divided by 195 leaves a remainder 47

Let the number be $$N=195q+47$$

Now, if N is divided by 15, the remainder is = $$(195q+47)\%15$$

Since, 195 is completely divided by 15, thus remainder depends on 47, => $$47=3\times15+2$$

=> Remainder = 2

=> Ans - (C)

Given = a : b = 3 : 4; b : c = 4 : 7

=> $$a:b:c=3:4:7$$

Let $$a=3,b=4,c=7$$

$$\therefore$$ $$\frac{a+b+c}{c}$$

= $$\frac{3+4+7}{7}=\frac{14}{7}=2$$

=> Ans - (B)

The speeds of the two trains in m/s would be $$80\times\ \frac{5}{18}=22.22$$ m/s and $$55\times\ \frac{5}{18}=15.277$$ m/s
Giving their relative speed to be 37.5 m/s

The total distance both trains essentially have to cover would be the distance between them + length of train 1 + length of train 2
Distance to be travelled = 90+120+90 = 300m

The time taken would then be $$\frac{300}{37.5}=8$$ second

Therefore, Option C is the correct answer. 

Average speed is the harmonic mean of both speeds = $$2\div(\frac{1}{60}+\frac{1}{40})$$

= $$2\div(\frac{2+3}{120})$$

= $$2\times\frac{120}{5}=48$$ km/hr

=> Ans - (B)

Let total work to be done = 96 units

Thus, 12 men complete 12 units of work each day, thus 1 man's efficiency = 1 unit/day

Work done in first 3 days = 36 units

Now, remaining work, i.e. 60 units must be completed in 4 days

=> Number of men required = $$\frac{60}{4}=15$$ 

Thus, 3 more men are required.

=> Ans - (A)

Let total work = 18 units

Let B's efficiency = $$x$$ units/day and A's efficiency = $$2x$$ units/day

According to ques, working together, they finish work in = $$(x+2x)\times18=18$$

=> $$3x=1$$

=> $$x=\frac{1}{3}$$

$$\therefore$$ Time taken by A alone to finish the work = $$\frac{18}{\frac{1}{3}}=54$$ days

=> Ans - (D)

Two pipes A and B can fill a tank in 24 hours and 30 hours respectively.

Time taken by both pipes together = $$1\div(\frac{1}{24}+\frac{1}{30})$$

= $$1\div(\frac{5+4}{120})$$

= $$\frac{120}{9}=\frac{40}{3}$$ hours

=> Ans - (B)

Let quantity of bucket = $$9$$ units

Here, 4 units is filled in 1 minute

=> Remaining, 5 units will be filled in = $$\frac{5}{4}$$ min

=> Ans - (C)