{"id":40199,"date":"2020-01-28T16:14:16","date_gmt":"2020-01-28T10:44:16","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40199"},"modified":"2020-01-28T17:31:31","modified_gmt":"2020-01-28T12:01:31","slug":"time-and-distance-questions-for-ssc-chsl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/time-and-distance-questions-for-ssc-chsl-pdf\/","title":{"rendered":"Time and Distance Questions for SSC-CHSL PDF"},"content":{"rendered":"

Time and Distance Questions for SSC-CHSL PDF<\/strong><\/span><\/h1>\n

Download SSC CHSL Time and Distance Questions with answers\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams.<\/p>\n

Download Time and Distance Questions PDF<\/a><\/p>\n

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Question 1:\u00a0<\/b>Buses start from a bus terminal with a speed of 20 km\/hr at intervals of 10 minutes. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of 8 minutes ?<\/p>\n

a)\u00a03 km\/hr<\/p>\n

b)\u00a04 km\/hr<\/p>\n

c)\u00a05 km\/hr<\/p>\n

d)\u00a07 km\/hr<\/p>\n

Question 2:\u00a0<\/b>By walking at 3\/4 of his usual speed, a man reaches his office 20 minutes later than his usual time. The usual time taken by him to reach his office is<\/p>\n

a)\u00a075 minutes<\/p>\n

b)\u00a060 minutes<\/p>\n

c)\u00a040 minutes<\/p>\n

d)\u00a030 minutes<\/p>\n

Question 3:\u00a0<\/b>In a 100m race, Kamal defeats Bimal by 5 seconds. If the speed of Kamal is 18 Kmph, then the speed of Bimal is<\/p>\n

a)\u00a015.4 kmph<\/p>\n

b)\u00a014.5,kmph<\/p>\n

c)\u00a014.4 kmph<\/p>\n

d)\u00a014 kmph<\/p>\n

Question 4:\u00a0<\/b>A train, 240 in long crosses a man walking along the line in opposite direction at the rat4 of 3 kmph in 10 seconds. The speed of the train is<\/p>\n

a)\u00a063 kmph<\/p>\n

b)\u00a075 km ph<\/p>\n

c)\u00a083.4 kmph<\/p>\n

d)\u00a086.4 kmph<\/p>\n

Question 5:\u00a0<\/b>The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is<\/p>\n

a)\u00a0500<\/p>\n

b)\u00a0600<\/p>\n

c)\u00a0700<\/p>\n

d)\u00a0800<\/p>\n

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Question 6:\u00a0<\/b>Walking at 3\/4th of his usual speed , aman is $1\\frac{1}{2}$ hours late. his usual time to cover the same distance, in hours, is?<\/p>\n

a)\u00a0$4\\frac{1}{2}$<\/p>\n

b)\u00a0$4$<\/p>\n

c)\u00a0$5\\frac{1}{2}$<\/p>\n

d)\u00a0$5$<\/p>\n

Question 7:\u00a0<\/b>Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete the round. After how much time to they meet at the starting point for the first time?<\/p>\n

a)\u00a01800 seconds<\/p>\n

b)\u00a03600 seconds<\/p>\n

c)\u00a02400 seconds<\/p>\n

d)\u00a04800 seconds<\/p>\n

Question 8:\u00a0<\/b>Walking at 6\/7th of this usual speed a man is 25 minutes too late. His usual time to cover this distance is<\/p>\n

a)\u00a02 hours 30 minutes<\/p>\n

b)\u00a02 hours 15 minutes<\/p>\n

c)\u00a02 hours 25 minutes<\/p>\n

d)\u00a02 hours 10 minutes<\/p>\n

Question 9:\u00a0<\/b>Walking at 5 km\/hr a student reaches his school from his house 15 minutes early and walking at 3 km\/hr he is late by 9 minutes. What is the distance between his school and his house?<\/p>\n

a)\u00a05 km<\/p>\n

b)\u00a08 km<\/p>\n

c)\u00a03 km<\/p>\n

d)\u00a02 km<\/p>\n

Question 10:\u00a0<\/b>With average speed of 40 km\/hour, a train reaches its destination in time. If it goes with an average speed of 35 km hour, it is late by 15 minutes. The total journey is<\/p>\n

a)\u00a030 km<\/p>\n

b)\u00a040 km<\/p>\n

c)\u00a070 km<\/p>\n

d)\u00a080 km<\/p>\n

Question 11:\u00a0<\/b>A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25$^{\\circ}$ in a distance of 40 metres?<\/p>\n

a)\u00a091.64 metres<\/p>\n

b)\u00a090.46 metres<\/p>\n

c)\u00a089.64 metres<\/p>\n

d)\u00a093.64 metres<\/p>\n

Question 12:\u00a0<\/b>Two cars are moving with speeds $v1$ and $v2$ towards a crossing along two roads. If their distance from the crossing be 40 m and 50 m at an instant of time, then they do not collide, if their speeds are such that<\/p>\n

a)\u00a0$ v_1:v_2 \\neq 4:5 $<\/p>\n

b)\u00a0$ v_1:v_2 \\neq 5:4 $<\/p>\n

c)\u00a0$ v_1:v_2=16:25 $<\/p>\n

d)\u00a0$ v_1:v_2 = 25:16 $<\/p>\n

Question 13:\u00a0<\/b>A man rides at the rate of 18 km\/hr, but stops for 6 mins. to change horses at the end of every 7th km. The time that he will take to cover a distance of 90 km is<\/p>\n

a)\u00a06 hrs.<\/p>\n

b)\u00a06 hrs 12 min.<\/p>\n

c)\u00a06 hrs 18 min.<\/p>\n

d)\u00a06 hrs 24 min.<\/p>\n

Question 14:\u00a0<\/b>A farmer travelled a distance of 61 km in 9 hrs. He travelled partly on foot at the rate of 4 km\/hr and partly on bicycle at the rate of 9 km\/hr. The distance travelled on foot is<\/p>\n

a)\u00a014 km<\/p>\n

b)\u00a0171cm<\/p>\n

c)\u00a016 km<\/p>\n

d)\u00a015 km<\/p>\n

Question 15:\u00a0<\/b>Raj and Prem walk in opposite directions at the rate of 3 km and 2 km per hour respectively. How far will they be from each other after 2 hours ?<\/p>\n

a)\u00a010 km<\/p>\n

b)\u00a08 km<\/p>\n

c)\u00a06 km<\/p>\n

d)\u00a02 km<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Distance between buses will $20\\times\\frac{10}{60}$ = $\\frac{10}{3}$ km.
\nNow man is travelling this distance in 8 min. with the relative speed of (20+$x$) (let’s assume speed of man is $x$ km\/hr )
\nhence\u00a0(20+$x$) = $\\frac{\\frac{10}{3}}{\\frac{8}{60}}$
\n$x$= 5<\/p>\n

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

As distance is constant and we know s =\u00a0\\frac{d}{t} (where s is distance and t is time)
\nhence st = constant
\nor $s_{1}t_{1}$= $s_{2}t_{2}$
\nor $s_{1}t_{1}$=$ \\frac{3}{4}s_{1}(t_{1}+ 20)$
\nhence $t_{1}$ = 60<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

As we know distance is constant for vimal and kamal and that is equal to 100 m.
\nhence $V_{bimal}\\times t_{bimal}$ = 100
\nor $V_{bimal}\\times (t_{kamal} + 5)$ = 100
\nwhere $t_{kamal}$ will be $\\frac{Distance}{V_{kamal}}$ i.e. $\\frac{100 m}{18 kmph}$
\nso $V_{bimal}\\times (\\frac{100 m}{18 kmph} + 5)$ = 100
\nAfter \u00a0solving above equation we will get $V_{bimal}$ = 4 $\\frac{m}{s}$ or 14.4 kmph<\/p>\n

4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

As man and train are coming to each other
\nand we assume speed of train = $x$
\nhence its relative speed will be ($x$ + 3)
\nrelative distance travelled = 240m = .24 km
\ntime taken = 10 sec. = $\\frac{10}{3600}$ hr<\/p>\n

so\u00a0($x$ + 3) =\u00a0$\\frac{.24}{\\frac{10}{3600}}$<\/p>\n

solving the eq. we will $x$ equals to 83.4 kmph<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

In 1 revolution wheel will complete a distance of $2\\pi r = 2 \\times \\pi \\times \\frac{98}{2}$ = 308 cm.
\nHence to cover 1540 m. , revolutions will be = $\\frac{1540\\times100}{308} = 500$<\/p>\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

As distance is constant
\nHence $v_1 \\times t_1 = \\frac{3v_1}{4} \\times (t_1 + \\frac{3}{2})$ (Where is $v_1$ is speed, t_1 is time taken to travel)
\nOn solving above equation, we will get $t_1 = \\frac{9}{2}$<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Meeting at the first time will be L.C.M. of time taken by individuals to complete
\ni.e. L.C.M. of 200,300,360 and 450 will be equal to 1800 sec.<\/p>\n

8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the initial speed and time be s,t\u00a0 respectively,
\nthen speed and time in the next case are 6s\/7 and (t+25)
\nAs distance = speed * time, and distance travelled in both cases is the same,
\n(6s\/7)*(t+25) = s*t
\nSolving the\u00a0 above equation results in\u00a0 t=150min<\/p>\n

9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the time and distance be t mins and d respectively,
\nIn the first case:
\nTotal time taken = (t – 15) mins = (t-15)\/60 hrs.
\nDistance travelled\u00a0 =\u00a0 5*(t-15)\/60
\nIn the second case:
\nTotal time taken = (t + 9) mins = (t+9)\/60 hrs.
\nDistance travelled\u00a0 = 3*(t+9)\/60<\/p>\n

So, 5*(t-15)\/60 = 3*(t+9)\/60,
\nSolving the above equation we get, t=51
\nSo, d=3*(51+9)\/60
\n=3 KMs
\n10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the time and distance be t mins and d km respectively,
\nIf it goes with an average speed of 40 km\/hour, a train reaches its destination in time.
\nSo, distance = (40*t)\/60
\nIf it goes with an average speed of 35 km hour, it is late by 15 minutes.
\nSo, distance = 35*(t+15)\/60
\nIn both the cases, distance is same,
\nSo,\u00a0 40*t = 35*(t+15)
\nSolving the above equation gives t = 105
\nand d = (40*105)\/60 = 70<\/p>\n

11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

With 25 degree angle length of arc will be ($r \\times$ ($\\theta$ in radian) = 40m (where $r$ is radius of arc and $\\theta$ will be angle made by it)
\nNow solving above equation, we will get $r$ = 91.64 m.<\/p>\n

12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

they will collide if time to reach the crossing is same.<\/p>\n

it is the case when<\/p>\n

time taken by 1st car = time taken by 2nd car.<\/p>\n

40\/v1 = 50\/v2<\/p>\n

v1\/v2 = 4\/5<\/p>\n

but condition not to collide is v1\/v2 $\\neq$ 4\/5.<\/p>\n

so the answer is option A.<\/p>\n

13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Speed of man = 18 km\/h<\/p>\n

Total distance = 90 km<\/p>\n

As he stops after 7th km, => $90=(12\\times7)+6$<\/p>\n

=> He stops 12 times in the journey.<\/p>\n

Total stoppage time = $12\\times6=72$ min<\/p>\n

Real time = $\\frac{90}{18}=5$ hours<\/p>\n

$\\therefore$ Required time taken = 5hr + 72 min<\/p>\n

= 6 hrs 12 min<\/p>\n

=> Ans – (B)<\/p>\n

14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let the distance travelled on foot be $x km$<\/p>\n

=> Distance travelled on bicycle = $(61-x) km$<\/p>\n

Time taken to travel on foot = $\\frac{x}{4}$ hrs<\/p>\n

Time taken to travel on bicycle = $\\frac{61-x}{9}$ hrs<\/p>\n

=> Total time = 9 = $\\frac{x}{4} + \\frac{61-x}{9}$<\/p>\n

=> $9x + 244 – 4x = 324$<\/p>\n

=> $5x = 324-244$<\/p>\n

=> $x = \\frac{80}{5} = 16$ km<\/p>\n

15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Since Raj and Prem walks in opposite directions, their relative speed will be sum of their individual speeds i.e.<\/p>\n

Speed, $s$ = 3+2 = 5 kmph<\/p>\n

Now, distance covered in 2 hours = time * speed<\/p>\n

= 5*2 = 10 km<\/p>\n

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We hope this Time and Distance Questions for SSC CHSL Exam preparation is so helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Time and Distance Questions for SSC-CHSL PDF Download SSC CHSL Time and Distance Questions with answers\u00a0 PDF based on previous papers very useful for SSC CHSL Exams. Top-15 Very Important Questions for SSC Exams. Take a free mock test for SSC CHSL Download SSC CHSL Previous Papers More SSC CHSL Important Questions and Answers PDF […]<\/p>\n","protected":false},"author":49,"featured_media":40205,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1441],"tags":[358,3397],"class_list":{"0":"post-40199","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-stenographer","17":"tag-ssc-chsl","18":"tag-time-and-distance-questions"},"better_featured_image":{"id":40205,"alt_text":"Time and Distance Questions for SSC CHSL PDF","caption":"Time and Distance Questions 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