time and distance Model Questions & Answers, Practice Test for ssc cgl tier 1 2024
ssc cgl tier 1 2024 SYLLABUS WISE SUBJECTS MCQs
Number System
Ratio & Proportion
Percentage
Time & Work
Time & Distance
Trigonometric Ratios & Identities
In a race A, B and C take part. A beats B by 30 m, B beats C by 20 m and A beats C by 48 m. Which of the following is/ are correct ?
l. The length of the race is 300 m.
2. The speeds of A, B and C are in the ratio 50 :45 : 42.
Select the correct answer using the code given below :
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.
A train 100 m long passes a platform 100 m long in 10 seconds. The speed of the train is
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
A person travels a certain distance at 3 km/h and reaches 15 min late. If he travels at 4 km/h, he reaches 15 min earlier. The distance he has to travel is
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.
With a uniform speed a car covers a distance in 8 hours. Had the speed been increased by 4 km/hr, the same distance could have been covered in $7{1/2}$ hours. The distance covered is
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
A boy goes to his school from his house at a speed of 3 kmph and returns at a speed of 2 kmph. If he takes 5 hours in going and coming, then the distance between his house and school is
Answer: (d)
Let total distance = x km
∴ $x/3+x/2=5$
x = 6 km
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