time and distance Model Questions & Answers, Practice Test for ssc cgl tier 1 2024

Answer: (a)

Let the length of race be 'x' meters and velocity of A, B, C be respectively $v_A , v_B and v_C$.

According to question

⇒$x/{v_A} = {x - 30}/{v_B}$

⇒${v_A}/{v_B} = {x - 30}/x$ ---(i)

⇒$x/{v_B} = {x - 20}/{v_C}$

⇒${v_C}/{v_B} = {x - 20}/x$ ---(ii)

⇒$x/{v_A} = {x - 48}/{v_C}$

⇒${v_C}/{v_A} = {x - 48}/x$ ---(iii)

Multiplying eq (i) & (ii)-

${v_B}/{v_A} × {v_C}/{v_B} = ({x - 30}/x)({x - 20}/x)$

${v_C}/{v_A} = ({x - 30}/x)({x - 20}/x)$ ---(iv)

From eq (iii) & eq (iv)-

$({x - 48}/x) = ({x - 30}/x)({x - 20}/x)$

⇒$x^2 – 48x = x^2$ + 600 – 50x

⇒2x = 600

⇒x = 300

Putting x = 300 in equation (i) & eq (ii)

⇒${v_B}/{v_A} = {300 - 30}/{300} = {270}/{300} = 9/{10}$

⇒$v_B = 9/{10} v_A$ ---(v)

⇒${v_C}/{v_B} = {300 - 20}/{300} = {280}/{300} = {28}/{30}$

⇒$v_c = {28}/{30} v_B = {28}/{30} × 9/{10} v_A = {84}/{100} v_A$ ---(vi)

From eq (v) 7 eq (vi)-

$v_A : v_B : v_C = v_A : 9/{10} v_A : {84}/{100}v_A$

= 100 : 90 : 84 = 50 : 45 : 42

so, option (c) is correct.

Answer: (d)

Speed of the train (m/sec)

= ${\text"Length of (train plateform)"}/{\text"Time to cross the plateform"}$

= ${(100 + 100)}/{10} = {200}/{10}$ = 20 m/sec

Speed (in Kmph) = 20 × ${18}/5$ = 72 Kmph

Answer: (b)

According to question,

$t_1 -t_2$ = 15 – (–15)⇒15 + 15

⇒30 min ⇒$x/{v_1} - x/{v_2} = {30}/{60}$h

Here, $v_1$ = 3 km/h, $v_2$ = 4 km/h

⇒$x/3 - x/4 = 1/2⇒{4x - 3x}/{12} = 1/2⇒x/{12} = 1/2$

∴ x = ${12}/2$ = 6 km.

Answer: (b)

Let the distance be x km.

According to question

$ x/{7{1/2}}-x/8=4$

⇒${2x}/15-x/8=4$

⇒${16x-15x}/120=4$

⇒ x = 480 km

Answer: (d)

Let total distance = x km

∴ $x/3+x/2=5$

x = 6 km