Club house is 8m north of Swimming Pool, which is 6m to the west of Block-A. Block –B is 13m to the south-west of Mall, which is 12m to the east of Block-A. Parking is 10m to north of Block- B, which is neither in the south east or south west of Block ‘A’. As Block ‘B’ is towards the south-west of Mall, thus it could be towards south, south-east or south-west of Block ‘A’, but with the help of third hint we can say that Block ‘B’ is in south of Block ‘A’, Thus using Pythagoras theorem, we can identify the distance between Block ‘B’ and Block ‘A’ as 5m. Parking is 15m to the west of Park. Entry gate is 9m to the south of Park Playground is situated exactly in the middle of Parking and School, which is neither to the north-east of Swimming Pool nor to the south-east of Swimming Pool. The shortest distance between Club house and School is 5m. As School is not situated in the north-east of Swimming Pool which means it can’t be towards the north of Parking. Also School is not situated towards the south-east of Swimming Pool, which means it can’t be towards south of Parking. We also know that the shortest distance between Club house and school is 5m, this clearly means that School is not in south or north of Swimming Pool. Thus the only left direction in which the distance criteria is also met is North-west. Using Pythagoras theorem the shortest distance between School and Club house is calculated as 5m and thus School is towards the north-west of Swimming Pool.
I. x2 + 13x + 42 = 0
II. y² + 13y + 40 = 0
What is the speed of the stream if a ship takes 15 hours to travel 240 km in calm waters, given that the ratio of its speed against the current to its s...
I. 3x2 – 16x + 21 = 0
II. y2 – 13y + 42 = 0
I. x3 = 1728
II. y2 – 15y + 56 = 0
I. 2x² - 7x + 3 = 0
II. 8y² - 14y + 5 = 0
I. 2y2 + 31y + 99 = 0
II. 4x2 + 8x – 45 = 0
How many values of x and y satisfy the equation 2x + 4y = 8 & 3x + 6y = 10.
I. 3p² - 17p + 22 = 0
II. 5q² - 21q + 22 = 0
If 4x = 40, 3y = 33, what is the value of 6x + 4y?
I. (x13/5 ÷7) = 5488 ÷ x7/5
II. (y2/3 × y2/3 ) ÷ √4 = (343y)1/3...