{"id":25949,"date":"2019-03-12T12:46:35","date_gmt":"2019-03-12T07:16:35","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=25949"},"modified":"2019-03-12T12:46:35","modified_gmt":"2019-03-12T07:16:35","slug":"height-and-distance-questions-for-ssc-cgl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/height-and-distance-questions-for-ssc-cgl-pdf\/","title":{"rendered":"Height and Distance Questions for SSC CGL PDF"},"content":{"rendered":"

Height and Distance Questions for SSC CGL PDF:<\/strong><\/span><\/h2>\n

Download SSC CGL Height and Distance questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Height and Distance objective questions for SSC exams.<\/p>\n

Download SSC CGL Height and Distance Questions PDF<\/a><\/p>\n

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 <\/p>\n

Question 1:\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.
\nb)\u00a06 hrs 12 min.
\nc)\u00a06 hrs 18 min.
\nd)\u00a06 hrs 24 min.<\/p>\n

Question 2:\u00a0<\/b>In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. If both the chords are on the same side of the centre, then the distance between the chord is<\/p>\n

a)\u00a09 cm
\nb)\u00a07 cm
\nc)\u00a023 cm
\nd)\u00a011 cm<\/p>\n

Question 3:\u00a0<\/b>N is the foot of the perpendicular from a point P of a circle with radius 7 cm, on a diameter AB of the circle. If the length of the chord PB is 12 cm, the distance of the point N from the point B is<\/p>\n

a)\u00a0$12\\frac{2}{7}$
\nb)\u00a0$3\\frac{5}{7}$
\nc)\u00a0$10\\frac{2}{7}$
\nd)\u00a0$6\\frac{5}{7}$<\/p>\n

Question 4:\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 $
\nb)\u00a0$ v_1:v_2 \\neq 5:4 $
\nc)\u00a0$ v_1:v_2=16:25 $
\nd)\u00a0$ v_1:v_2 = 25:16 $<\/p>\n

Question 5:\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
\nb)\u00a090.46 metres
\nc)\u00a089.64 metres
\nd)\u00a093.64 metres<\/p>\n

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Question 6:\u00a0<\/b>If the distance between two point (0, -5) and (x, 0) is 13 unit, then x =<\/p>\n

a)\u00a010
\nb)\u00a0\u00b1 10
\nc)\u00a012
\nd)\u00a0\u00b1 12<\/p>\n

Question 7:\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
\nb)\u00a08 km
\nc)\u00a03 km
\nd)\u00a02 km<\/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
\nb)\u00a02 hours 15 minutes
\nc)\u00a02 hours 25 minutes
\nd)\u00a02 hours 10 minutes<\/p>\n

Question 9:\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$4\\frac{1}{2}$
\nb)\u00a0$4$
\nc)\u00a0$5\\frac{1}{2}$
\nd)\u00a0$5$<\/p>\n

Question 10:\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
\nb)\u00a0600
\nc)\u00a0700
\nd)\u00a0800<\/p>\n

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Question 11:\u00a0<\/b>Jogging at a speed of 6.5 km\/hr a man can cover a distance in 18 hours. Running at a speed of 12 km\/hr the man covers the same distance in<\/p>\n

a)\u00a07.85 hours
\nb)\u00a010.5 hours
\nc)\u00a08.36 hours
\nd)\u00a09.75 hours<\/p>\n

Question 12:\u00a0<\/b>Raju travels half of the distance to his school on foot at 10 km\/hr and remaining half of the distance by bus at 30 km\/hr. If the distance between school and his starting point is 60 km, find the time taken by Raju to reach his school?<\/p>\n

a)\u00a05 hours
\nb)\u00a04.8 hours
\nc)\u00a03.6 hours
\nd)\u00a04 hours<\/p>\n

Question 13:\u00a0<\/b>A boat in a still water moves with a speed of $9$ $\\text{km\/h}$ and the speed of the stream is $1.5$ $\\text{km\/h}$. A man rows for a distance of $52.5$ $\\text{km}$ and returns back to the starting point. The total time taken( in hours ) is:<\/p>\n

a)\u00a0$7$
\nb)\u00a0$5$
\nc)\u00a0$12$
\nd)\u00a0$6$<\/p>\n

Question 14:\u00a0<\/b>A person walks with $\\frac{3}{5}$ times the original speed and reaches the destination 1 hour late. The time taken to travel the distance at the original speed would be<\/p>\n

a)\u00a045 Minutes
\nb)\u00a060 Minutes
\nc)\u00a075 Minutes
\nd)\u00a090 Minutes<\/p>\n

Question 15:\u00a0<\/b>The floor of a room is of size $4m \\times 3m$ and its height is 3m. The walls and ceiling of the room require painting. The area to be painted is<\/p>\n

a)\u00a066 m2
\n<\/sup>b)\u00a054 m2
\n<\/sup>c)\u00a043 m2
\n<\/sup>d)\u00a033 m2<\/sup><\/p>\n

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Question 16:\u00a0<\/b>The tops of two poles of height 24 m and 36 m are connected by a wire. If the wire makes an angle of 60 $^{\\circ}$ with the horizontal, then the length of the wire is<\/p>\n

a)\u00a0$8\\sqrt{3}$
\nb)\u00a0$8$
\nc)\u00a0$6\\sqrt{3}$
\nd)\u00a0$6$<\/p>\n

Question 17:\u00a0<\/b>Two poles of height 7 metre and 12 metre stand on a plane ground. If the distance between their feet is 12 metre, the distance between their top will be<\/p>\n

a)\u00a015 metre
\nb)\u00a013 metre
\nc)\u00a019 metre
\nd)\u00a017 metre<\/p>\n

Question 18:\u00a0<\/b>A boat is sailing towards a lighthouse of height 20\u221a3 m at a certain speed. The angle of elevation of the top of the lighthouse changes from 30\u00b0 to 60\u00b0 in 10 seconds. What is the time taken (in seconds) by the boat to reach the lighthouse from its initial position?<\/p>\n

a)\u00a010
\nb)\u00a015
\nc)\u00a020
\nd)\u00a060<\/p>\n

Question 19:\u00a0<\/b>The length of two parallel sides of a trapezium are 30 cm and 40 cm. If the area of the trapezium is 350 cm2, then what is the value (in cm) of its height?<\/p>\n

a)\u00a08
\nb)\u00a010
\nc)\u00a015
\nd)\u00a012<\/p>\n

Question 20:\u00a0<\/b>The length of two parallel sides of a trapezium are 15 cm and 20 cm If its area is 175 sq.cm, then its height is :<\/p>\n

a)\u00a010 cm
\nb)\u00a015 cm
\nc)\u00a025 cm
\nd)\u00a020 cm<\/p>\n

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Question 21:\u00a0<\/b>The average height of a class comprising of 40 students in 155 cm. When two of the students leave the class, the average height of the class drops down to 153 cm. What is the average height of the two students who left the class?<\/p>\n

a)\u00a0193 cm
\nb)\u00a0178 cm
\nc)\u00a0185 cm
\nd)\u00a0166 cm<\/p>\n

Question 22:\u00a0<\/b>The average height of a class is 160 cm. If the average height of boys is 172cm and that of girls is 136cm, what is the ratio of number of boys to that of number of girls in the class?<\/p>\n

a)\u00a01:2
\nb)\u00a03:2
\nc)\u00a02:1
\nd)\u00a02:3<\/p>\n

Question 23:\u00a0<\/b>The average height of 30 boys out of a class of 50 is 160 cm. If the average height of the remaining boys is 165 cm, the average height of the whole class (in cm) is:<\/p>\n

a)\u00a0161
\nb)\u00a0162
\nc)\u00a0163
\nd)\u00a0164<\/p>\n

Question 24:\u00a0<\/b>The average height of 8 students is 152 cm. Two more students of heights 144 cm and 155 cm join the group. What is the new average height ?<\/p>\n

a)\u00a0151.5 cm
\nb)\u00a0152.5 cm
\nc)\u00a0151 cm
\nd)\u00a0150.5 cm<\/p>\n

Question 25:\u00a0<\/b>Out of 20 boys, 6 are each of 1 m 15 cm height, 8 are of 1 m 10 cm and rest of 1m 12cm. The average height of all of them is<\/p>\n

a)\u00a01m 12cm
\nb)\u00a01m 121cm
\nc)\u00a01m 211cm
\nd)\u00a01m 21cm<\/p>\n

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

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

Speed of man = 18 km\/h
\nTotal distance = 90 km
\nAs he stops after 7th km, => $90=(12\\times7)+6$
\n=> He stops 12 times in the journey.
\nTotal stoppage time = $12\\times6=72$ min
\nReal time = $\\frac{90}{18}=5$ hours
\n$\\therefore$ Required time taken = 5hr + 72 min
\n= 6 hrs 12 min
\n=> Ans – (B)<\/p>\n

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

<\/figure>\n

Given : CD = 30 cm and AB = 16 cm
\nTo find : MN = ?
\nSolution : Perpendicular distance from the centre bisects the chord.
\n=> CM = $\\frac{30}{2}=15$ cm and AN = $8$ cm
\nNow, in $\\triangle$ OCM,
\n=> $(OM)^2=(OC)^2-(CM)^2$
\n=> $(OM)^2=(17)^2-(15)^2$
\n=> $(OM)^2=289-225=64$
\n=> $OM=\\sqrt{64}=8$ cm
\nSimilarly, in $\\triangle$ OAN,
\n=> $(ON)^2=(OA)^2-(AN)^2$
\n=> $(ON)^2=(17)^2-(8)^2$
\n=> $(ON)^2=289-64=225$
\n=> $ON=\\sqrt{225}=15$ cm
\n$\\therefore$ MN = ON – OM
\n= $15-8=7$ cm
\n=> Ans – (B)<\/p>\n

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

As we know $cos\\theta$ = $\\frac{a^2 + b^2 – c^2}{2ab}$ (where $\\theta$ is angle between sides a and b, side c is opposite to angle $\\theta$ )
\nHence for triangle POB (where O is centre of circle)
\nOP=OB=radius of circle=7
\nPB=12
\nLet’s say angle between PB and OB is $\\theta$.
\nSo putting up values and solving it, we will get $cos\\theta$ = $\\frac{6}{7}$
\nHence value of BN will be BP $cos\\theta = 12cos\\theta = \\frac{72}{7}$<\/p>\n

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

they will collide if time to reach the crossing is same.
\nit is the case when
\ntime taken by 1st car = time taken by 2nd car.
\n40\/v1 = 50\/v2
\nv1\/v2 = 4\/5
\nbut condition not to collide is v1\/v2 $\\neq$ 4\/5.
\nso the answer is option A.<\/p>\n

5)\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

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

$Distance^{2} = diff.of x cordinates^{2}+diff.of y coordinates^{2}$
\n$13^{2}=x^{2}+5^{2}$
\nHence $x=\\pm12$<\/p>\n

7)\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 = 5*(t-15)\/60
\nIn the second case:
\nTotal time taken = (t + 9) mins = (t+9)\/60 hrs.
\nDistance travelled = 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<\/p>\n

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

Let the initial speed and time be s,t 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 above equation results in t=150min<\/p>\n

9)\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

10)\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

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

Distance he covers = 6.5*18 = 117 km
\nTime taken by the man to cover the distance while running = 117\/12 = 9.75 hours<\/p>\n

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

Total time taken to reach Raju\u2019s school = $\\frac{30}{10} + \\frac{30}{30}$ = 3+1 = 4 hours<\/p>\n

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

Speed of the man in downstream = $( 9 + 1.5 )$ km\/h = $10.5$ km\/h
\nSpeed of the man in upstream = $( 9 – 1.5 )$ km\/h = $7.5$ km\/h
\nTime taken while going upstream = $\\frac{22.5}{7.5}$ hr = $7$ hr
\nTime taken while going upstream = $\\frac{22.5}{10.5}$ hr = $5$ hr
\nTotal Time taken = $(7+5)$ hr = $12$ hr
\nHence , the correct option is C<\/p>\n

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

Since Time and Speed are inversely proportional
\nTherefore , $\\frac{5}{3}$( Usual Time ) – ( Usual Time ) = 1
\n$\\implies$ $\\frac{2}{3}$( Usual Time ) = 1
\nHence , Usual Time = $\\frac{3}{2}$ Hours = 90 Minutes
\nCorrect Option is D<\/p>\n

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

Area of walls and ceiling of room = area of 4 walls + area of 1 ceiling
\n= $(2 \\times (3 \\times 3) + 2\\times (4 \\times 3)) + 4 \\times 3$
\n= 18 + 24 + 12 = 54<\/p>\n

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

Let’s say height of bigger pole is 12+24.
\nWhere length of wire will be hypotenuse, pole length of 12 in bigger pole will be perpendicular and distance between poles will be base.
\nAnd angle between base and hypotenuse will be $\\theta$
\nHence length of wire will be = $12cosec\\theta$ (where $\\theta=60^o$)
\nSo length of wire will be = $12 \\frac{2}{\\sqrt3} = 8\\sqrt3$<\/p>\n

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

<\/p>\n

Height of pole AB = 7 metre = DE and CD = 12 metre
\nBD = 12 metre = AE
\nNow, CE = CD-DE = 12-7 = 5 metre
\nIn $\\triangle$ ACE
\nAC = $\\sqrt{(AE)^2 + (CE)^2}$
\n= $\\sqrt{12^2 + 5^2} = \\sqrt{169}$
\n= 13 metre<\/p>\n

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

 <\/p>\n

<\/figure>\n

Given : CD = $20\\sqrt3$ m and time taken to reach B from A = 10 seconds
\nTo find : Time taken to reach D from A
\nSolution : In $\\triangle$ BCD,
\n=> $tan(60^\\circ)=\\frac{CD}{BD}$
\n=> $\\sqrt3=\\frac{20\\sqrt3}{BD}$
\n=> $BD=20$ m
\nSimilarly, in $\\triangle$ ACD,
\n=> $tan(30^\\circ)=\\frac{CD}{AD}$
\n=> $\\frac{1}{\\sqrt3}=\\frac{20\\sqrt3}{20+AB}$
\n=> $AB+20=60$
\n=> $AB=60-20=40$ m
\n=> Speed of boat (while travelling from A to B) = distance\/time
\n= $\\frac{40}{10}=4$ m\/s
\n=> AD = BD + AB = 20 + 40 = 60 m
\n$\\therefore$ Time taken to reach D from A = $\\frac{60}{4}=15$ seconds
\n=> Ans – (B)<\/p>\n

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

Sum of two parallel sides of a trapezium = $30+40=70$ cm
\nLet height = $h$ cm
\n=> Area of trapezium = $\\frac{1}{2}\\times h\\times$ (sum of parallel sides)
\n=> $\\frac{1}{2}\\times(h)\\times(70)=350$
\n=> $\\frac{h}{2}=\\frac{350}{70}=5$
\n=> $h=5\\times2=10$ cm
\n=> Ans – (B)<\/p>\n

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

Sum of the two parallel sides of the trapezium = $15+20=35$ cm
\nLet its height = $h$ cm
\n=> Area of trapezium = $\\frac{1}{2}\\times$ (sum of parallel sides) $\\times$ height
\n=> $\\frac{1}{2}\\times35\\times h=175$
\n=> $h=\\frac{175}{35}\\times2$
\n=> $h=5\\times2=10$ cm
\n=> Ans – (A)<\/p>\n

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

Let the average height of the two students be x.
\nSo, sum of their ages will be 2x
\nThe sum of the heights of all the students in the class will be 155 X 40
\nThis should be equal to-
\n155 X 40 = 153 X 38 + 2x
\n38(155-153) + 2 X 155 = 2x
\n=> x = 38 + 155 = 193cm<\/p>\n

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

Let the number of boys in the class be \u2018b\u2019 and number of girls in the class be \u2018g\u2019.
\n$b\\times 172 +g \\times 136 = (b+g) \\times 160$
\n$12b = 24g$
\n$b:g = 2:1$
\nHence, option c is the correct answer.<\/p>\n

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

Total boys in class = 50
\nAverage height of first 30 boys = 160 cm
\n=> Total height of 30 boys = $30 \\times 160=4800$ cm
\nSimilarly, total height of remaining 20 boys = $20 \\times 165=3300$ cm
\n$\\therefore$ Average height of the whole class = $\\frac{(4800+3300)}{30+20}$
\n= $\\frac{8100}{50}=162$ cm
\n=> Ans – (B)<\/p>\n

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

Average height of 8 students = 152 cm
\n=> Total height of 8 students = $152\\times8=1216$ cm
\nAfter addition of 2 students, total students of 10 students = $1216+144+155=1515$
\n$\\therefore$ New average height = $\\frac{1515}{10}=151.5$ cm
\n=> Ans – (A)<\/p>\n

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

Total number of boys = 20
\nTotal height of first 6 boys = $6\\times1.15=6.9$ m
\nTotal height of next 8 boys = $8\\times1.10=8.8$ m
\nTotal height of last 6 boys = $6\\times1.12=6.72$ m
\n=> Average height = $\\frac{(6.9+8.8+6.72)}{20}$
\n= $\\frac{22.42}{20}=1.121=$ 1m 121 cm
\n=> Ans – (B)<\/p>\n","protected":false},"excerpt":{"rendered":"

Height and Distance Questions for SSC CGL PDF: Download SSC CGL Height and Distance questions with answers PDF based on previous papers very useful for SSC CGL exams. 25 Very important Height and Distance objective questions for SSC exams.   Question 1:\u00a0A man rides at the rate of 18 km\/hr, but stops for 6 mins. 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