Speed Time And Distance - Study Mode

[#96] A candle of 6 cm long burns at the rate of 5 cm in 5 hour and an another candle 8 cm long burns at the rate of 6 cm in 4h. What is the time required to each candle to remain of equal lengths after burning for some hours, when they starts to burn simultaneously with uniform rate of burning?
Correct Answer

1 hours

Explanation

Solution: (6 - x) = (8 - 1.5x) Or, x = 4 cm. So, It will take 4 hours to burn it in such a way that they will remain equal in lengths.

[#97] A man walking at the speed of 4 km/hr,cross a square field diagonally in 3 minutes.The area of the field is?
Correct Answer

(A) 20000 sqm

Explanation

Solution: Speed of man, = 4 kmph = $$frac{{4 imes 5}}{{18}}$$ m/sec In 3 min (180 sec) man will go = $$frac{{20 imes 180}}{8}$$ = 200 m. That means the diagonal of the square field = 200 m. Diagonal of square, = Side of Square × $$sqrt 2 $$xa0 = 200 m. → Side of Square = $$frac{{200}}{{sqrt 2 }}$$ Area of Square = Side 2 Area = $$frac{{200 imes 200}}{2}$$ xa0 = 20000 sqm.

[#98] A train approaches a tunnel AB. Inside the tunnel a cat located at a point i.e. $$frac{5}{{12}}$$ of the distance AB measured from the entrance A. When the train whistles the Cat runs. If the cat moves to the exit B, the train catches the cat exactly the exit. The speed of the train is greater than the speed of the cat by what order ?
Correct Answer

(C) 6 : 1

Explanation

Solution: Train(T)__________ A_____5k____CAT__________B T<-------x--------------><----12k--------------------------> Let the speed of train be u and the speed of Cat be v and train whistles at a point T, X km away from A. Let AB = 12k and Cat was 5k distance away from A. Time was constant for both, then $$eqalign{
& Rightarrow frac{v}{u} = frac{x}{{5k}} = frac{{ {x + 12k} }}{{7k}} cr
& Rightarrow 7x = 5left( {x + 12k}
ight) cr
& Rightarrow frac{x}{k} = frac{{30}}{1} cr
& { ext{Thus}}, cr
& Rightarrow frac{u}{v} = frac{{30}}{5} = frac{6}{1} cr} $$ $${ ext{or,}},,6:1$$

[#99] A person travels equal distance with speeds of 3 km/hr, 4 km/hr and 5 km/hr and taken a total time of 47 minutes. The total distance (in km) is :
Correct Answer

(B) 3 km

Explanation

Solution: Let the total distance be 3x km Then, $$eqalign{
& Leftrightarrow frac{x}{3} + frac{x}{4} + frac{x}{5} = frac{{47}}{{60}} cr
& Leftrightarrow frac{{47x}}{{60}} = frac{{47}}{{60}} cr
& Leftrightarrow x = 1 cr} $$ ∴ Total distance = (3 × 1) km = 3 km

[#100] Mary jogs 9 km at a speed of 6 km per hour. At what speed would she need to jog during the next 1.5 hours to have an average of 9 km per hour for the entire jogging session ?
Correct Answer

(C) 12 kmph

Explanation

Solution: Let speed of jogging be x km/hr Total time taken : $$eqalign{
& = left( {frac{9}{6}{ ext{hrs}} + 1.5{ ext{ hrs}}}
ight) cr
& = 3{ ext{ hrs}} cr} $$ Total distance covered = $$left( {9 + 1.5x}
ight){ ext{ km}}$$ $$eqalign{
& herefore frac{{9 + 1.5x}}{3} = 9 cr
& Leftrightarrow 9 + 1.5x = 27 cr
& Leftrightarrow frac{3}{2}x = 18 cr
& Leftrightarrow x = left( {18 imes frac{2}{3}}
ight) cr
& Leftrightarrow x = 12{ ext{ kmph}} cr} $$