Race MCQ Quiz in मल्याळम - Objective Question with Answer for Race - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

P and Q take part in 400 m race. P runs at 12 km/hr. P gives Q a start of 20 m and still beats him by 13 seconds.

Formula used:

Speed = Distance / Time

Calculations:

Speed of P = 12 km/hr = 12 × (5/18) m/s = 10/3 m/s

Distance covered by P = 400 m

Time taken by P = Distance / Speed = 400 / (10/3) = 120 seconds

Since P beats Q by 13 seconds, time taken by Q = 120 + 13 = 133 seconds

Distance covered by Q = 400 m - 20 m = 380 m

⇒ Speed of Q = Distance / Time = 380 / 133 m/s

⇒ Convert speed to km/hr: 380 / 133 × (18/5) = 10.29 km/hr

∴ The correct answer is option (2).

Shortcut Trick 

Given:

Distance of the race = 5 km

A beats B by 750 metres

A beats C by 1260 metres

Formula Used:

B's distance when A finishes = Total distance - Distance A beats B

C's distance when A finishes = Total distance - Distance A beats C

Distance B beats C = C's distance when A finishes - B's distance when A finishes

Calculation:

B's distance when A finishes = 5000 - 750 = 4250 metres

C's distance when A finishes = 5000 - 1260 = 3740 metres

Let time taken by A to complete the race = x

Time taken by B to cover 4250 m = x

Speed of B = 4250/x

Time taken by B to cover 5000 m = 5000/(4250/x) = 1.1764x

Distance covered by C in x time = 3740 m

Distance covered by C in 1.1764x time = 3740 m × 1.1764 = 4400

Distance B beats C = 5000 - 4400 = 600m

The correct answer is option 2.

Let track length be equal to T.

Time taken to meet for the first time = \( \frac{T}{6-s} \) or \( \frac{T}{s-6} \)

Time taken for a lap for Ram = \( \frac{T}{6} \)

Time taken for a lap for Radha = \( \frac{T}{s} \)

So, time taken to meet for the first time at the starting point = \( \text{LCM}\left( \frac{T}{6}, \frac{T}{s} \right) = \frac{T}{\text{HCF}(6, s)} \)

Number of meeting points on the track = Time taken to meet at starting point / Time taken for first meeting = Relative speed / \( \text{HCF}(6, s) \).

So, in essence, we have to find values for s such that \( \frac{6-s}{\text{HCF}(6, s)} = 2 \) or \( \frac{s-6}{\text{HCF}(6, s)} = 2 \).

The question is "If two people cross each other at exactly two points on the circular track and s is a natural number less than 30, how many values can s take?"

s = 2, 10, 18 satisfy this equation. So, there are three different values that s can take.