A boat can travel 60 km downstream in the same time that it ta... | Stage I (CBT I) MCQ | Examoogle

Stage I (CBT I)RRB ALP CBT-I 11 Mar, 2026 Shift 1

A boat can travel 60 km downstream in the same time that it takes to travel 36 km upstream. The boat then makes a round trip of 240 km, going 120 km downstream and returning 120 km upstream. The total time taken for this round trip is 1 hour more than the time it would have taken to cover 240 km in still water. Find the speed (in km/h) of the boat in still water.

$16$

$18$

$15$

$17$

Correct Answer

📚 GK • GK Ratio & Speed MethodStage I (CBT I)RRB ALP CBT-I 11 Mar, 2026 Shift 1

Option A

$16$

Quick Summary:

Given: Distance downstream = 60 km, Distance upstream = 36 km in equal time. Round trip = 120 km downstream + 120 km upstream = 240 km. Total time taken for round trip is 1 hr more than 240 km in still water.

📐MAMath SolutionRatio & Speed Method

📋 Given

Distance downstream = 60 km, Distance upstream = 36 km in equal time. Round trip = 120 km downstream + 120 km upstream = 240 km. Total time taken for round trip is 1 hr more than 240 km in still water.

🔢 Formula Used

D=u+v,U=u−v,T=dsD = u + v, \quad U = u - v, \quad T = \frac{d}{s}D=u+v,U=u−v,T=sd​

⚡ Exam Hall Shortcut / Speed Trick

Since time is equal, downstream to upstream speed ratio is $60 : 36 = 5 : 3$ . Thus $u : v = (5+3) : (5-3) = 4 : 1$ . Let $u = 4x$ and $v = x$ . Downstream speed $D = 5x$ , Upstream speed $U = 3x$ . Using round trip time condition: $\frac{120}{5x} + \frac{120}{3x} - \frac{240}{4x} = 1 \implies \frac{24}{x} + \frac{40}{x} - \frac{60}{x} = 1 \implies \frac{4}{x} = 1 \implies x = 4$ . So, $u = 4 \times 4 = 16\text{ km/h}$ .

⚠️ Common Student Trap / Pitfall

Confusing the speed of the boat in still water ( $u$ ) with the speed of stream ( $v$ ) or downstream/upstream speeds ( $D, U$ ), leading to incorrect substitution in the time equation.

📊 Diagram / Illustration

🔢 Step-by-Step Solution

Establish the Speed Ratios from Time Equivalence

Since time is equal for downstream ( $60\text{ km}$ ) and upstream ( $36\text{ km}$ ), the speeds are directly proportional to distance. Let downstream speed be $D$ and upstream speed be $U$ .

DU=6036=53\frac{D}{U} = \frac{60}{36} = \frac{5}{3}UD​=3660​=35​

Express Speeds in Terms of a Common Variable $x$

Let $D = 5x$ and $U = 3x$ . The speed of the boat in still water ( $u$ ) and stream speed ( $v$ ) are given by:

u=D+U2=5x+3x2=4x,v=D−U2=xu = \frac{D + U}{2} = \frac{5x + 3x}{2} = 4x, \quad v = \frac{D - U}{2} = xu=2D+U​=25x+3x​=4x,v=2D−U​=x

Formulate the Total Time Equation for Round Trip

The round trip consists of $120\text{ km}$ downstream, $120\text{ km}$ upstream, and comparison with $240\text{ km}$ in still water.

Tdown=1205x=24x,Tup=1203x=40x,Tstill=2404x=60xT_{\text{down}} = \frac{120}{5x} = \frac{24}{x}, \quad T_{\text{up}} = \frac{120}{3x} = \frac{40}{x}, \quad T_{\text{still}} = \frac{240}{4x} = \frac{60}{x}Tdown​=5x120​=x24​,Tup​=3x120​=x40​,Tstill​=4x240​=x60​

Solve for Variable $x$

Given that the total time taken for the round trip is $1\text{ hour}$ more than the time in still water:

24x+40x−60x=1  ⟹  4x=1  ⟹  x=4\frac{24}{x} + \frac{40}{x} - \frac{60}{x} = 1 \implies \frac{4}{x} = 1 \implies x = 4x24​+x40​−x60​=1⟹x4​=1⟹x=4

Calculate Speed of Boat in Still Water

Substitute $x = 4$ into the expression for $u$ :

u=4x=4×4=16 km/hu = 4x = 4 \times 4 = 16\text{ km/h}u=4x=4×4=16 km/h

✅

A is correct because substituting $x = 4$ into $u = 4x$ gives $16\text{ km/h}$ as the speed of the boat in still water.

Core Concepts Used

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Relative Speed in Downstream & Upstream Ratio Method for Equal Time Time-Speed-Distance Equations

💡 EXAM TIP

Whenever time is constant, speed is directly proportional to distance ( $S \propto D$ ). This ratio principle instantly simplifies complex word problems in Time, Speed & Distance without solving quadratic equations.

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