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Let the speed of B be x kmph ⇒ Time taken by B to cover 40 km = 40/x ⇒ Time taken by A = (40/x) + 2 So, the speed of A = (20x)/(20 + x) ----(1) Distance = 80 km ⇒ Time taken by B to cover 80 km = 80/x ----(2) ⇒ Speed of A when he doubles his speed = (40x)/(20 + x) ⇒ Time taken by A to cover 80 km = (40 + 2x)/(x) ----(3) A.T.Q., From (2) and (3) (80/x) + (3/2) = (40 + 2x)/x ⇒ (40 + 2x)/2x - 80/x = 3/2 ⇒ 2 (40 + 2x - 80) = 3x ⇒ 2 (2x - 40) = 3x ⇒ 4x - 80 = 3x ⇒ 4x - 3x = 80 ⇒ x = 80 ⇒ Speed of B = 80 kmph Distance = 90 km Time taken by B = 90/80 = 1(1/8) hrs
120×27÷4-10% Of 370 = ?
If (3/7) of 'x' is greater than (2/11) of 'x' by 38, then determine the value of 'x'.
A person borrows ₹5000 at 12% per annum simple interest. He repays ₹3200 at the end of 2 years. How much does he still owe after 2 years?
Evaluate: 1 + tan²(cos⁻¹x)