SRCC GBO 2012

Since, each has the same remainder, we need to find the H.C.F. [(1417-1059),(2312-1417),(2312-1059)]

= H.C.F. (358,895,1253) = 179

Now, when we divide 1059, 1417 and 2312 by 179, we get remainder as 164

$$\therefore$$ $$r+k=164+179=343$$

=> Ans - (C)

Let length of each candle be $$l$$ cm

Time taken to burn each candle is 3 hours and 4 hours respectively.

=> Candle burn in 1 minute = $$\frac{l}{180}$$ cm and $$\frac{l}{240}$$ cm

Let required time be $$t$$ minutes.

Thus, candle burnt in $$t$$ minutes = $$l-\frac{tl}{180}$$ cm and $$l-\frac{tl}{240}$$ cm

According to ques,

=> $$(l-\frac{tl}{180}):(l-\frac{tl}{240})=1:2$$

=> $$2l-\frac{tl}{90}=l-\frac{tl}{240}$$

=> $$l=\frac{tl}{30}(\frac{1}{3}-\frac{1}{8})$$

=> $$\frac{30}{t}=\frac{5}{24}$$

=> $$t=30\times\frac{24}{5}=144$$ minutes

$$\therefore$$ Candles should be lit 144 minutes prior to 5 pm, i.e. at 2:36 pm

=> Ans - (B)

Using, $$\frac{M_1D_1H_1}{W_1}$$ $$=\frac{M_2D_2H_2}{W_2}$$

Let number of articles produced be $$x$$

=> $$\frac{2\times2\times2}{2}=\frac{4\times4\times4}{x}$$

=> $$x=\frac{64}{4}=16$$

=> Ans - (C)

Let speed of A, B and C be $$a,b,c$$ m/s respectively.

Time taken by A to cover $$d$$ metres = Time taken by B to cover $$(d-20)$$ metres

=> $$\frac{s_a}{s_b}=\frac{d}{d-20}$$ ---------------(i)  [Since, speed is directly proportional to distance]

Similarly, Time taken by A to cover $$d$$ metres = Time taken by C to cover $$(d-28)$$ metres

=> $$\frac{s_a}{s_c}=\frac{d}{d-28}$$ --------------(ii)

and Time taken by B to cover $$d$$ metres = Time taken by C to cover $$(d-10)$$ metres

=> $$\frac{s_b}{s_c}=\frac{d}{d-10}$$ -------------(iii)

Equating equations (i) and (ii), => $$\frac{s_b}{s_c}=\frac{d-20}{d-28}$$

Comparing it with equation (iv), => $$\frac{d}{d-10}=\frac{d-20}{d-28}$$

Solving above equation, we get : $$d=100$$ m

=> Ans - (C)

Time taken by the machines working together = $$x$$ hours

According to ques,

=> $$\frac{1}{x+6}+\frac{1}{x+1}+\frac{1}{x+x}=\frac{1}{x}$$

=> $$\frac{(x+6)+(x+1)}{(x+6)(x+1)}=\frac{1}{x}-\frac{1}{2x}$$

=> $$\frac{2x+7}{(x+6)(x+1)}=\frac{1}{2x}$$

=> $$4x^2+14x=x^2+7x+6$$

=> $$3x^2+7x-6=0$$

=> $$3x^2+9x-2x-6=0$$

=> $$(3x-2)(x+3)=0$$

=> $$x=\frac{2}{3},-3$$

$$\because$$ $$x$$ cannot be negative, hence $$x=\frac{2}{3}$$ hours

=> Ans - (A)

Out of first 50 balls, number of blue balls = 49

From now, for each $$x$$ balls, number of blue balls = $$\frac{7x}{8}$$

Total number of balls = $$m=(x+50)$$

According to ques,

=> $$\frac{\frac{7x}{8}+49}{x+50}\geq \frac{9}{10}$$

=> $$\frac{70x}{8}+490\geq9x+450$$

=> $$40\geq\frac{x}{4}$$

=> $$x\leq160$$

$$\therefore$$ To maximize $$m$$, we need to maximize $$x$$, which has a maximum value of 160

=> Maximum number of balls = $$160+50=210$$

=> Ans - (C)

Shortcut : SP = Rs. 636 and CP = Rs. 600

=> Profit % = $$\frac{636-600}{600}\times100$$

= $$\frac{36}{6}=6\%$$

=> Ans - (A)

Let the numbers be $$a$$ and $$b$$

According to ques, => $$ab:(a+b):(a-b)=24:7:1$$

Thus, $$\frac{a+b}{a-b}=\frac{7}{1}$$

=> $$a+b=7a-7b$$

=> $$6a=8b$$

=> $$\frac{a}{b}=\frac{4}{3}$$

Let $$a=4y$$ and $$b=3y$$

Also, $$\frac{ab}{a-b}=\frac{24}{1}$$

=> $$\frac{4y\times3y}{4y-3y}=24$$

=> $$12y^2=24y$$

=> $$y=2$$

$$\therefore$$ Numbers are $$a=8$$ and $$b=6$$

=> Product = $$ab=8\times6=48$$

=> Ans - (B))

Let price paid be each individual be Rs. $$a,b,c,d$$ respecitvely.

=> $$(a+b+c+d)=6000$$ -------------(i)

Also, $$2a=b+c+d$$ --------------(ii)

Solving above equations, => $$a=2000$$ and $$b+c+d=4000$$

Similarly, $$3b=a+c+d$$, => $$b+1500$$ and $$c+d=2500$$

and $$4c=a+b+d$$, => $$c=1200$$

$$\therefore$$ Amount paid by 4th person = Rs. $$1300$$

=> Ans - (D)

Since, the jars are identical (have same capacity). Let quantity of each jar be L.C.M. (2,3) = 6 units

=> Acid in jar one  = 3 units and acid of water in jar 1 = 3 untis

Similarly, Acid in jar two = 4 units and water in jar two = 2 units

When mixed together, ratio of water to acid = $$(3+2):(3+4)=5:7$$

=> Ans - (A)