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).

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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.

Given:

A can give B a start of 100 m in a 1 km race.

A can give C a start of 200 m in a 1 km race.

B can give C a lead of 20 m in a 200 m race.

Formula used:

Speed is inversely proportional to the time taken when the distances are equal.

Calculation:

Let the speeds of A, B, and C be S_A, S_B, and S_C respectively.

In a 1 km race:

⇒ A runs 1000 m while B runs 900 m.

⇒ S_A / S_B = 1000 / 900

⇒ S_A / S_B = 10 / 9

A runs 1000 m while C runs 800 m:

⇒ S_A / S_C = 1000 / 800

⇒ S_A / S_C = 5 / 4

From the above ratios:

⇒ S_B / S_C = (S_B / S_A) × (S_A / S_C)

⇒ S_B / S_C = (9 / 10) × (5 / 4)

⇒ S_B / S_C = 45 / 40

⇒ S_B / S_C = 9 / 8

In a 200 m race:

B runs 200 m while C runs 180 m:

⇒ Time taken by B = Time taken by C × (S_C / S_B)

⇒ Time taken by B = Time taken by C × (8 / 9)

Let time taken by B to run 200 m be t:

⇒ t = (Distance run by B / Speed of B)

⇒ t = (200 / S_B)

We know S_B / S_C = 9 / 8:

⇒ S_B = 9 / 8 × S_C

Substitute S:

⇒ t = 200 / (9 / 8 × SC)

⇒ t = (200 × 8) / (9 × SC)

⇒ t = 1600 / (9 × SC)

From 200 m race ratio:

⇒ SC = 180 / Time taken by C

⇒ t = 1600 / (9 × (180 / Time taken by C))

⇒ Time taken by B = 90 / 4

∴ The correct answer is option (3).

Given:

When A runs 1200 m, B runs 1200 − 120 = 1080 m

When B runs 800 m, C runs 800 − 40 = 760 m

Formula used:

Speed ∝ Distance (for same time)

If A beats B by x meters in d meters race ⇒ Speed of A : B = d : (d − x)

Calculation:

Speed of A : B = 1200 : 1080 = \(\dfrac{1200}{1080} = \dfrac{10}{9}\)

Speed of B : C = 800 : 760 = \(\dfrac{800}{760} = \dfrac{20}{19}\)

⇒ Speed of A : C = A : B × B : C = \(\dfrac{10}{9} \times \dfrac{20}{19} = \dfrac{200}{171}\)

Now in 400 m race, A runs full 400 m, so C runs:

\(\dfrac{171}{200} \times 400 = 342\) meters

⇒ A beats C by = 400 − 342 = 58 meters

∴ The correct answer is 58 meters.

Given:

Length of the circular track = 800 m

Ram's speed = 25 m/sec

Mohan's speed = 40 m/sec

Direction of running = Same direction

Formula used:

Time = Distance / Speed

To meet at the starting point, the time will be the LCM of the individual times taken to complete one round.

Calculations:

Time taken by Ram to complete one round (T_Ram) = 800 m / 25 m/sec = 32 seconds

Time taken by Mohan to complete one round (T_Mohan) = 800 m / 40 m/sec = 20 seconds

To find when they will meet for the first time at the starting point, we need to find the LCM of their individual times to complete one round.

Time to meet at the starting point = LCM(T_Ram, T_Mohan)

⇒ Time to meet at the starting point = LCM(32, 20)

Prime factorization of 32 = 2⁵

Prime factorization of 20 = 2² × 5

LCM(32, 20) = 2⁵ × 5 = 32 × 5 = 160 seconds

∴ They will meet for the first time at the starting point after 160 seconds.

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.