Model 5 Problems with Races Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

2 hours 45 minutes

= $(2 + 45/60)$ hours

= $(2 + 3/4)$ hours = $11/4$ hours

Distance = Speed × Time

= 4 × $11/4$ = 11 km.

Time taken in covering 11 km at 16.5 kmph

= $11/{16.5}$ hour

= $({11 × 10 × 60}/165)$ minutes

= 40 minutes

Answer: (c)

According to the question,

Since, When B runs 200 m metres, A runs 190 metres

When B runs 180 metres, A runs

= $190/200 × 180$ = 171 metres

When C runs 200m, B runs 180 metres.

Hence, C will give a start to A by

= 200 - 171 = 29 metres

Answer: (d)

According to the question,

When A runs 800 metres, B runs 760 metres

When A runs 200 metres, B runs

= $760/800 × 200$ = 190 metres

Again, when B runs 500 metres, C runs 495 metres.

When B runs 190 metres, C runs

= $495/500 × 190$ = 188.1 metres

∴ Hence, A will beat C by

200 - 188.1 = 11.9 metres in a race of 200 metres.

Answer: (a)

Let the time taken to complete the race by A,B, and C be x minutes.

Speed of A = $1000/x$,

B = ${1000 - 50}/x = 950/x$

C = ${1000 - 69}/x = 931/x$

Now, time taken to complete the race by

B = $1000/{950/x} = {1000 × x}/950$

and distance travelled by C in

${1000x}/950$ min = ${1000x}/950 × 931/x$ = 980 km.

B can allow C

= 1000 - 980 = 20 m

Answer: (a)

Let A take x seconds in covering 1000m and b takes y seconds

According to the question,

x + 20 = $900/1000$ y

x + 20 = ${9y}/10$ ...(i)

and, $950/1000$ x + 25 = y ...(ii)

From equation (i),

${10x}/9 + 200/9 = y$

${10x}/9 + 200/9 = {950x}/1000 + 25$

${10x}/9 + 200/9 = {19x}/20 + 25$

${10x}/9 - {19x}/20 = 25 - 200/9$

${200x - 171x}/180 = {225 - 200}/9$

${29x}/180 = 25/9$

$x = 25/9 × 180/29 = 500/29$ seconds.