SRCC GBO 2011

ABC is sector that is cut from a circle of diameter 12 cm. 

So, AB = AC = $$\frac{12}{2} = 6 cm$$

Let AO = R = Radius of circle that circumscribed the sector. 

OM is perpendicular to AM. AM = $$\frac{AB}{2} = \frac{6}{2} = 3 cm$$

Since, $$\angle{BAC} = \theta, then \angle{BAO} = \frac{\theta}{2}$$

Now, 

In $$\triangle{AMO}:$$

$$sec\frac{\theta}{2} = \frac{AO}{AM}$$

$$sec\frac{\theta}{2} = \frac{R}{3}$$

$$R = 3sec\frac{\theta}{2}$$

River is flowing from A to C, that means A to C is downstream and C to A is upstream. 

Since, Time taken to go from A to C in downstream = 7 hours

Time taken to go from A to B in downstream = $$\frac{7}{2} = 3.5 hours$$

Since, time taken to go from A to B and come back to the same point = 5 hours 15 minutes = $$5 + \frac{15}{60} = 5.25 hours$$

Now, time taken to go from B to A = 5.25 - 3.5 = 1.75 hours

Hence, time taken to go from C to A = 1.75 * 2 = 3.5 hours = $$3\frac{1}{2} hours$$

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

Hence, D has the largest number of students.

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

The number of students scored less than or equal to 60% from A is 1740.

The number of students scored less than or equal to 60% from B is 2730.

The number of students scored less than or equal to 60% from C is 1170.

The number of students scored less than or equal to 60% from D is 6800.

The number of students scored less than or equal to 60% from A is 2400.

The sum of these numbers is 14840.

The average will be 14840/5=2986.

So, option B is the answer.

In A, 40% of students scored above 60% marks, the number of them being equal to 1160. Hence, the total number of students is 2900.

In B, 30% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 3900.

In C, 50% of students scored above 60% marks, the number of them being equal to 1170. Hence, the total number of students is 2340.

In D, 15% of students scored above 60% marks, the number of them being equal to 1200. Hence, the total number of students is 8000.

In E, 40% of students scored above 60% marks, the number of them being equal to 1600. Hence, the total number of students is 4000.

The number of students of schools A and E together = 2900+4000=6900.

The number of students of school D is more than the number of students of schools A and E together = 8000-6900=1100.

The percentage more is by $$\frac{1100}{6900}\cdot100=15.9$$

So, option A is the answer.

Number of students of school B scoring marks less than or equal to 60% = $$1170 * \frac{70}{30} = 2730$$

Number of students of school C scoring marks less than or equal to 60% = $$1170 * \frac{50}{50} = 1170$$

Number of students of schools A and D together scoring marks above 60% = 1160 + 1200 = 2360

Required ratio = (2730 + 1170): 2360 = 3900: 2360 = 

Production of scooters in 2002 = 140

Production of scooters in 2003 = 80

Drop in number of production of scooters in 2003 from 2002 = 140 - 80 = 60

Required percent = (60/140) * 100 = 42.9% (Approx)

Production of motorbikes in 2003 = 130

Pproduction of scooters in 2003 = 80

Required percent = (130/80) * 100 = 162.5%

Average production of motorbikes per year during the period 2000 to 2004 = $$\frac{140 + 120 + 100 + 130 + 110}{5} = 120$$

Hence, in the year 2001 production of motorbikes was equal to the average production of motorbikes per year during the period 2000 to 2004.

Number of scooters produced during the period 2000 to 2003 = 120 + 100 + 140 + 80 = 440

Number of motorbikes produced during the period 2000 to 2003 = 140 + 120 + 100 + 130 = 490

Required ratio = 440: 490 = 44: 49