Question
If Sneha cycles a certain distance at 9 km/h, she
arrives 7 hours late. If she cycles at 15 km/h, she reaches 2 hours early. What is 80% of the distance Sneha covered?Solution
ATQ:
Let the distance travelled by Sneha = ‘d’ km
Let the usual time taken by Sneha to travel this distance = ‘t’ hours We have,
d ÷ 9 = t + 7…………………..(i)
Also, d ÷ 15 = t – 2…………………(ii) Equation (i) – Equation (ii), we get
(d/9) – (d/15) = t + 7 – (t – 2) = 9
Or, (5d – 3d) ÷ 45 = 9
So, d = (45 × 9) ÷ 2 = 202.5 So, 80% of the distance travelled by Sneha = 202.5 × 0.8 = 162 km
I. y² + y – 56 = 0
II. 2x² + 11 x – 40 = 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:Â x2Â - 33x + 242 = 0
II:Â y2Â - 4y - 77 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 19x² - 88x + 100 = 0
Equation 2: 17y² - 79y + 90...
I. 27x² + 120x + 77 = 0
II. 56y² + 117y + 36 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 13x² - 60x + 47 = 0
Equation 2: 17y² - 80y + 63 = 0
I. 4x² -  15x + 9 = 0
II. 20y² -  23y + 6 = 0
I. x2 – 36 = 0
II. y2 - 7y + 6 = 0
I. x2 – 13x + 40 = 0
II. 2y2 – 15y + 13 = 0Â
I. 66x² - 49x + 9 = 0
II. 46y² - 37y - 30 = 0
