SRCC GBO 2010 Question Paper

Since x and y are inversely proportional to each other then xy=k

Now x becomes $$x\left(1+\frac{p}{100}\right)$$

Let's say y decreases by "m" percent. $$y\left(1-\frac{m}{100}\right)$$

Hence $$x\left(1+\frac{p}{100}\right)y\left(1-\frac{m}{100}\right)=k$$

Now xy=k which cancels "k" from RHS$$\left(1+\frac{p}{100}\right)\left(1-\frac{m}{100}\right)=1$$

$$1-\frac{m}{100}=\frac{100}{100+p}$$

$$1-\frac{100}{100+p}=\frac{m}{100\ }$$

$$\frac{100p}{100+p}=m$$

hence option 4 is the correct answer

Tip- For interests that are easier to solve (such as 10%), we shall directly add the compound interest to the amount 'x', and it becomes 1.1x or 1.1*1.1x and so on. 

Assuming the initial deposit to be x three years back. 

After one year, it became 1.1x, from which 1,25,000 was withdrawn. Hence, the remaining amount was (1.1x - 1,25,000).

After the second year, the amount would have become (1.1x - 1,25,000)*1.1, from which 65,000 was withdrawn. Hence, the remaining amount was (1.21x- 2,02,000).

After the third year, the amount would have become (1.21x- 2,02,000)*1.1, from which 1,10,000 was withdrawn and the account was closed. Hence, the remaining amount is 0. 

1.331x-3,32,200=0

x=2,50,000

Let us consider the radius of the biggest circle as 5r.

This makes the radius of the two smallest circles as r, and the middle sized circle as 2r.

Now, we shall calculate the total area of the bigger circle. It shall be $$\pi$$ $$(5r)^{2}$$, which is 25$$\pi$$ $$r^{2}$$. Let us say the area is 25x units.

For the two smaller circles, the total area is 2$$\pi$$ $$r^{2}$$.

And the middle sezied circle's area is $$\pi$$ $$(2r)^{2}$$, which is 4$$\pi$$ $$r^{2}$$.

Now, we shall add the area of two smallest circles with the middle sized circle and subtract it from the area of the bigger circle to find the area which is shaded.

The sum of shaded region is 6$$\pi$$ $$r^{2}$$. Let us say that the sum is 6x units. 

Hence, the part of bigger circle which is not shaded becomes 19x units. 

Now, the ratio of area shaded to the ratio of area of the bigger circle is 19:25.

Let us consider B's efficiency to be x units per hour. This makes A's efficiency to be 3x units per hour. Hence, their combined efficiency is 4x units per hour.

In four hours, A and B together will complete 16x units of work, which is half of the total work. By this, the total work is 32x units. 

Now, B left and C joined. A and C together completed the remaining work (which is 16x units) in 1 hour. In this, A must have done 3x units, and C would have done 13x units of work in one hour. Hence, C's efficiency is 13x units per hour.

To complete the total work, C will take $$\frac{32}{13}$$ hours.

Volume of water flowing per minute $$\pi\ \left(0.5\right)\left(0.5\right)1000$$     Volume of the conical vessel $$\frac{1}{3}\pi\ \left(20\right)\left(20\right)\left(21\right)$$

Total time required $$\frac{1}{3}\pi\ \frac{\left(20\right)\left(20\right)\left(21\right)}{\pi\ \left(0.5\right)\left(0.5\right)\left(1000\right)}$$

This on solving comes out to be 11.2 minutes

Two persons working 2 hours a day in 2 days assemble 2 machines.

Two persons working 2 hours a day in 6 days assemble 6 machines

Two persons working 6 hours a day in 6 days assemble 18 machines

6 persons working 6 hours a day in 6 days assemble 54 machines

It is clear that the first person (say A) takes 3 hours to cover the distance between P and Q, while the second person (say B) takes 5 hours to cover the same distance. Hence, we can assume their speeds to be 5x kmph and 3x kmph (the speeds are in the reverse ratio of time taken when the distance travelled is same).

Hence, the distance is 15x km (speed$$\times$$time).

Now, until B did not start (4 a.m.), A would have covered 10x km in 2 hrs (from 2 a.m.). The distance left is 5x km between A and B. 

Now, B will start moving and we shall use the concept of relative speed on the remaining 5x km. Since A and B are moving towards each other, their speed will be added. Hence, the resultant speed is 8x kmph. 

They will meet at t= 5x/8x, which is 5/8 hours. This is 37 minutes and 30 seconds, which will be counted from when B started (4 a.m.). Hence, they will meet at 4:37:30 a.m.

$$average\ speed\ =\frac{Total\ dis\tan ce\ travelled}{Total\ time\ taken}=\frac{mw+hk}{h+k}$$

Assume the time spent on each commutation mode is t, 16t, and 2t. 

Similarly, the average speed of each of the commutation modes is 20s, s, and 3s.

Now, the distance travelled on each of the commutation modes is 20st, 16st, and 6st. The sum is 42st, which shall be equal to 3990 km. 

Hence, st= $$\frac{3990}{42}$$, which is 95.

We are given that the overall average speed is 42 kmph. Average speed is the ratio of total distance (3990 km) and total time (19t).

So, 42= $$\frac{3990}{19t}$$, which means t= 5. Hence, from the value of 'st' (95), we can say that s=19.

The distance by air is 20st, which is 20$$\times$$19$$\times$$5, which is 1900 km.

According to the equation we can say that $$P\left(1+\frac{r}{100}\right)^2-P=102.50$$

Now r=5% So $$P\left(1.05\right)^2-P=102.50$$ or P(0.1025)=102.50  or P=1000

So new interest will be $$1000\left(1.04\right)^2-1000=81.6$$