Question
If Ankit drives a certain distance at 18 km/h, he gets
delayed by 3 hours. If he drives at 30 km/h, he reaches 2 hours before time. Find 40% of the distance travelled by Ankit.Solution
ATQ:
Let the distance travelled by Ankit = ‘d’ km
Let the usual time taken by Ankit to travel this distance = ‘t’ hours We have,
d ÷ 18 = t + 3…………………..(i)
Also, d ÷ 30 = t – 2…………………(ii) Equation (i) – Equation (ii), we get
(d/18) – (d/30) = t + 3 – (t – 2) = 5
Or, (5d – 3d) ÷ 90 = 5
So, d = (90 × 5) ÷ 2 = 225 So, 40% of the distance travelled by Ankit = 225 × 0.4 = 90 km
I. 6x² - 23x + 7 = 0
II. 6y² - 29y + 9 = 0
I. 3x6- 19x3+16=0
II. 9y4- 27y2+20=0
- If the quadratic equation x² + 18x + n = 0 has real and equal roots, what is the value of n?
I. 6x² - 49x + 99 = 0
II. 5y² + 17y + 14 = 0
I. x2 – 18x + 81 = 0
II. y2 – 3y - 28 = 0
I. 7p + 8q = 80
II. 9p – 5q = 57
I. 8x² - 74x + 165 = 0
II. 15y² - 38y + 24 = 0
I). 5p2 Â - p - 4 = 0
II). q2 - 12q + 27 = 0
I. 6x2 – 7x - 20 = 0
II. 3y2Â - y - 14 = 0
I. 5q = 7p + 21
II. 11q + 4p + 109 = 0
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