Question
Ajay and Vishal left the place ‘X’ with different
speeds which are in the ratio 7:6 respectively. Ajay left the place 36 seconds after Vishal. If after 72 seconds of the departure of Vishal, Ajay is 100 meters apart from Vishal, then find the speed of Vishal.(in km/hr)Solution
Let Ajay’s speed = 7x, Vishal’s speed = 6x Vishal travels for 72 sec → Distance = 6x×72=432x Ajay starts 36 sec later, so he travels 36 sec → Distance = 7x×36=252x Gap between them = 432x−252x=180x=100 On solving we get 180x = 100 So, x = 5/9 Hence, Vishal speed = 6x = 6 × (5/9) = 10/3 m/s Hence, To convert Vishal's speed from m/s to km/hr, use the conversion factor:
1 m/s = 3.6 km/hr Given speed of Vishal: Vishal’s speed = 10/3 m/s Speed in km/hr = 10/3 × 3.6 = 36/3 = 12 km/hr
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 45x + 450 = 0
Equation 2: y² - 48y + 540 = 0�...
I. 20y² - 13y + 2 = 0
II. 6x² - 25x + 14 = 0
I. 3x2 - 16x - 12 = 0
II. 2y2 + 11y + 9 = 0
I. 6x² - 23x + 7 = 0
II. 6y² - 29y + 9 = 0
I. 35x² - 46x – 16 = 0
II. 35y² - 116y + 96 = 0
Quantity I: The cash price of a notebook is Rs. 100 but is can also be purchased on 11 monthly equal instalments of Rs. 10 each. Find rate of S.I.?
...I. x2 – 19x + 88 = 0
II. (y + 4)2 = 121
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I. 6p2 – 7p = 5p – 7p2 + 25
II. 11q2 – 63q + 90 = 0
I. 2x2 + 3x - 9 = 0
II. 3y2 - y - 10 = 0
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