INTRODUCTION:
Percent means out of 100.
We can find any of the asking value by comparing on this 100 with the given value.
N:B: IF we can remember the percent value in terms of fractional value and vice versa, we can find the answer easily & quickly.
CONVERT OF PERCENT To FRACTIONAL VALUE :
10%=1/10
20%=1/5
25%=1/4
40%=2/5
50%=1/2
60%=3/5
75%=3/4
80%=4/5
Similarly we can convert fraction in to percentage.
The price of a Maruti car rises by 30% while the sales of the car comes down by 20%.What is the percentage change in the total revenue?
(a) -4%
(b) -2%
(c) +4%
(d) +2%
. A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fare and buys 20 mangoes and the balance, he purchases oranges. Number of oranges he can purchase is:
(a) 36
(b) 40
(c) 15
(d) 20
I bought 5 pens, 7 pencils and 4 erasers. Rajesh bought 6 pens, 8 erasers and 14 pencils for an amount which has half more what I had paid. What percent of the total amount paid by me was paid for the pens?
(a) 37.5%
(b) 62.5%
(c) 50%
(d) None of these
One bacteria splits in to eight bacteria of the next generation. But due to environment, only 50% of one generation can produce the next generation. If the seventh generation number is 4096 million, what is the number in the first generation?
(a) 1 million
(b) 2 million
(c) 4 million
(d) 8 million
. 2/5 of the voters promise to vote for P and the rest promised to vote for Q. Of these, on the last day 15% of the voters went back of their promise to vote for P and 25% of voters went back of their promise to vote for Q, and P lost by 2 votes. Then, the total number of voters is:
(a) 100
(b) 110
(c) 90
(d) 95
A report consists of 20 sheets each of 55 lines and each such line consists of 65 characters. This report is reduced on to sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in the number of sheets closet to:
(a) 20
(b) 5
(c) 30
(d) 35
The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994 is Rs20 per kg. The inflation rates of the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be:
(a)Rs23.60
(b) Rs24.00
(c) Rs24.20
(d) Rs24.60
A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction?
33.33%
75%
300%
150%
A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss?
Rs. 404.80
Rs. 440
Rs. 536.80
Rs. 160
A trader professes to sell his goods at a loss of 8% but weights 900 grams in place of a kg weight. Find his real loss or gain per cent.
2% loss
2.22% gain
2% gain
None of these
Rajiv sold an article for Rs.56 which cost him Rs.x. If he had gained x% on his outlay, what was his cost?
Rs. 40
Rs. 45
Rs. 36
Rs. 28
A trader buys goods at a 19% Amount on the label price. If he wants to make a profit of 20% after allowing a Amount of 10%, by what % should his marked price be greater than the original label price?
+8%
-3.8%
+33.33%
None of these
If apples are bought at the rate of 30 for a rupee. How many apples must be sold for a rupee so as to gain 20%?
28
25
20
22
One year payment to the servant is Rs. 200 plus one shirt. The servant leaves after 9 months and receives Rs. 120 and a shirt. Then find the price of the shirt.
Rs. 80
Rs. 100
Rs. 120
Cannot be determined
Two merchants sell, each an article for Rs.1000. If Merchant A computes his profit on cost price, while Merchant B computes his profit on selling price, they end up making profits of 25% respectively. By how much is the profit made by Merchant B greater than that of Merchant A?
Rs.66.67
Rs. 50
Rs.125
Rs.200
A merchant marks his goods in such a way that the profit on sale of 50 articles is equal to the selling price of 25 articles. What is his profit margin?
25%
50%
100%
66.67%
A merchant marks his goods up by 75% above his cost price. What is the maximum % Amount that he can offer so that he ends up selling at no profit or loss?
75%
46.67%
300%
42.85%
Showing posts with label aptitude written test question and answers. Show all posts
Showing posts with label aptitude written test question and answers. Show all posts
Quantitative analysis aptitude test on trains
INTRODUCTION:
The problems on Trains are same to Time & Distance. But incase of train, when it passes a stationary man or stationary pole , it passes the distance which is its own length.
And when it passes another train or plat form or a bridge, it passes the length of its own with the length of other train or the length of platform or bridge.
QUESTION 1:
How many seconds will a train200metres long running at the rate of 36 km an hour take to pass a certain telegraph post?
A.10 seconds.
B. 15 seconds.
C.20 seconds.
D. 25 seconds.
.
Sol: The speed of train is 36 km/h =36*5/18=20 m/sec.
Required time to pass the post =200/10=20 seconds
QUESTION 2:
How long does a train 240metres long running @ 72km/h take to pass a bridge 120metres in length?
A.15 seconds.
B. 18 seconds.
C.20seconds.
D. 25seconds.
Sol: Total distance to be cover is 240 +120 =360 metres and the speed of train is 72 km/h=72*5/18=20 m/sec.
So, required time is 360/20=18 seconds.
QUESTION 3:
Two trains 360metres &240metres in length are running in opposite directions, one @50km/h and the other @22km/h. In what time will they be completely clear each other from the moment they meet?
A.20 seconds.
B. 25 seconds.
C.30 seconds.
D.40 seconds.
Sol: Total distance to be covered is 360+240=600metres & the relative speed is 50+22=72km/h or 20 m/sec.
So, required time =600/20=30 seconds
QUESTION 4:
Two trains start their journey from Delhi & Hyderabad and proceed towards each other @ 80 & 95 km/h respectively. When they meet, it is found that one train has covered 180 km more than the other. Find the distance between Delhi & Hyderabad.
A.2000km.
B.2100km.
C.2200km.
D. 2400km.
Sol: One train covers 15 km more than other in 1 hour in a distance of 80+95=175 km.
So, One train covers 180km more than other in 12 hours in a distance of 2100km.
QUESTION 5:
A train over takes two persons who are walking in the same directions as the train is moving, @ 2km/h and 4km/h and passes them completely in 9 and 10 seconds respectively. Find the speed & the length of the train.
A.20 km/h,50m.
B. 22 km/h,50m.
C.20 km/h,60m.
D.30 km/h,50m.
Sol: The ratio between time taken by train to pass the persons is 9:10.
So, the ratio between the relative speed is 10:9.
1 part of difference is due to 4-2=2km/h.So in first case 10 parts is 20km/h.So speed of train=20+2=22km/h.
And the length of train =Relative speed with first man *time =20*5/18*9=50metres
QUESTION 6: Two trains can run @ 54km/h & 36km/h respectively on parallel tracks. When they are running in opposite directions they pass each other in 10 sec. When they are running in the same direction, a person sitting in the faster train observes that he passes the other train in 30 seconds. Find the length of the slower train.
A.100metres.
B.150metres.
C.200metres.
D.300metres.
QUESTION 7:
A train after traveling 50 km meets with an accident and then proceeds at ¾ of it’s former speed and arrives at it’s destination35 min late. Had the accident occurred 24 km further, it would have reached the destination only 25min late. Find the speed of the train.
A.48 km/h.
B.36 km/h.
C.50 km/h. D.72 km/h.
QUESTION 8:
A train covers a distance between stations A & B in 45 min. If the speed is reduced by 5 km/h, it will cover the same distance in 48 min. Find the distance between the two stations A & B , also the speed of the train.
A.80km/h,60km. B.60km/h,80km.
C.80km/h,80km. D.60km/h,80km
QUESTION 9:
Two trains A & B start from Delhi & Patna towards Patna & Delhi respectively. After passing each other they take 4 hours 48 min and 3 hours 20 min to reach Patna & Delhi respectively. If the train from Delhi is moving at 45 km/h then find the speed of the other train.
A.45km/h. B.54km/h.
C.60km/h. D.64km/h.
The problems on Trains are same to Time & Distance. But incase of train, when it passes a stationary man or stationary pole , it passes the distance which is its own length.
And when it passes another train or plat form or a bridge, it passes the length of its own with the length of other train or the length of platform or bridge.
QUESTION 1:
How many seconds will a train200metres long running at the rate of 36 km an hour take to pass a certain telegraph post?
A.10 seconds.
B. 15 seconds.
C.20 seconds.
D. 25 seconds.
.
Sol: The speed of train is 36 km/h =36*5/18=20 m/sec.
Required time to pass the post =200/10=20 seconds
QUESTION 2:
How long does a train 240metres long running @ 72km/h take to pass a bridge 120metres in length?
A.15 seconds.
B. 18 seconds.
C.20seconds.
D. 25seconds.
Sol: Total distance to be cover is 240 +120 =360 metres and the speed of train is 72 km/h=72*5/18=20 m/sec.
So, required time is 360/20=18 seconds.
QUESTION 3:
Two trains 360metres &240metres in length are running in opposite directions, one @50km/h and the other @22km/h. In what time will they be completely clear each other from the moment they meet?
A.20 seconds.
B. 25 seconds.
C.30 seconds.
D.40 seconds.
Sol: Total distance to be covered is 360+240=600metres & the relative speed is 50+22=72km/h or 20 m/sec.
So, required time =600/20=30 seconds
QUESTION 4:
Two trains start their journey from Delhi & Hyderabad and proceed towards each other @ 80 & 95 km/h respectively. When they meet, it is found that one train has covered 180 km more than the other. Find the distance between Delhi & Hyderabad.
A.2000km.
B.2100km.
C.2200km.
D. 2400km.
Sol: One train covers 15 km more than other in 1 hour in a distance of 80+95=175 km.
So, One train covers 180km more than other in 12 hours in a distance of 2100km.
QUESTION 5:
A train over takes two persons who are walking in the same directions as the train is moving, @ 2km/h and 4km/h and passes them completely in 9 and 10 seconds respectively. Find the speed & the length of the train.
A.20 km/h,50m.
B. 22 km/h,50m.
C.20 km/h,60m.
D.30 km/h,50m.
Sol: The ratio between time taken by train to pass the persons is 9:10.
So, the ratio between the relative speed is 10:9.
1 part of difference is due to 4-2=2km/h.So in first case 10 parts is 20km/h.So speed of train=20+2=22km/h.
And the length of train =Relative speed with first man *time =20*5/18*9=50metres
QUESTION 6: Two trains can run @ 54km/h & 36km/h respectively on parallel tracks. When they are running in opposite directions they pass each other in 10 sec. When they are running in the same direction, a person sitting in the faster train observes that he passes the other train in 30 seconds. Find the length of the slower train.
A.100metres.
B.150metres.
C.200metres.
D.300metres.
QUESTION 7:
A train after traveling 50 km meets with an accident and then proceeds at ¾ of it’s former speed and arrives at it’s destination35 min late. Had the accident occurred 24 km further, it would have reached the destination only 25min late. Find the speed of the train.
A.48 km/h.
B.36 km/h.
C.50 km/h. D.72 km/h.
QUESTION 8:
A train covers a distance between stations A & B in 45 min. If the speed is reduced by 5 km/h, it will cover the same distance in 48 min. Find the distance between the two stations A & B , also the speed of the train.
A.80km/h,60km. B.60km/h,80km.
C.80km/h,80km. D.60km/h,80km
QUESTION 9:
Two trains A & B start from Delhi & Patna towards Patna & Delhi respectively. After passing each other they take 4 hours 48 min and 3 hours 20 min to reach Patna & Delhi respectively. If the train from Delhi is moving at 45 km/h then find the speed of the other train.
A.45km/h. B.54km/h.
C.60km/h. D.64km/h.
Quantitative analysis Time and Distance
INTRODUCTION:
The problems based on time & distances are based on speed, time & distance. The general process to find out distance is as follows:
DISTANCE= SPEED*TIME.
Where S is the speed of the moving object, T is the time period & D is the distance covered.
Process to convert km/h in to m/s or vice versa is as follows:
1km/ h = 5/18 m/s
1 m/s = 18/5 km/h
Relative speed: The different between the speeds of two moving object with respect to their direction is known as relative speed. The relative speed according to the two different directions is as follows:
Case I: In case of same direction it is the difference between the speeds of the two moving objects.
Case II: In case of opposite direction it is the sum of the speeds of the two moving objects.
Example: Two objects are moving @ 30 km/h & 50 km/h. then the relative speed between them in two different cases are as follows:
Case I: In the same direction it is (50-30) i.e. 20 km/h
Case II: In case of opposite direction it is (50+30) i.e.80 km/h.
QUESTION:
A boy goes to school @ 10 km/h & returns to his home @ 12 km/h respectively. If he takes 5 ½ hrs in all, find the distance between school & his home.
A.20km.
B.25 km.
C.30 km .
D.30 km.
Sol:
The L.C.M of two different speeds of 10 &12 is 60.
If 60 km is the distance, then at the rate of two different speeds the times are 6 & 5 hrs respectively.
When the total time is 6+5 i.e. 11hrs, the distance between the two points is 60 km.
So when the total time is 5 ½ hrs, the distance between school & home is 30 km.
QUESTION:
A person covers half of his journey @ 21 km/h, & the rest half @ 24 km/h respectively. If he takes 15 hrs in all, find the distance covered by the person.
A.168 km.
B.224 km.
C.336 km.
D.280 km.
Sol:
The L.C.M of 21 &24 i.e. 168.
If 168 km is the distance, then the person covers ½ of it i.e. 84 km @ 21 km/h in 4 hrs and the rest half @ 24 km/h in 3 ½ hrs.
So when the total time is 7 ½ hrs, the distance is 168 km.
And if the total time is 15 hrs, the distance is 336 km.
QUESTION:
A person goes a certain distance @ 48 km/h & returns to the starting point @ 96 km/h respectively. Find his average speed during the whole journey.
A.60 km/h.
B.64 km/h
C.68 km/h
D.72 km/h
Sol:
The L.C.M of 48 @ 96 is 96.
If the distance is 96 km, then @ two different speeds the required time are 2 &1 hr respectively.
Average speed= Total distance/Total time
So the average speed i.e.(96+96)/2+1=64 km/h.
QUESTION:
A person covers a certain distance between his house & office on scooter. Having an average speed of 45 km/h, he is late by 15 minutes. How ever, with a speed of 60 km/h, he reaches his office 7 ½ min earlier. Find the distance between his house & office.
A.67.5 km.
B.60 km.
C.75 km.
D.120 km.
Sol: Let the L.C.M of the two different speeds i.e. the L.C.M of 45 & 60 =180 is the distance.
So @ 45km/h, it will take him 4 hrs & @ 60 km/h, it will take him 3 hours.
In this case the difference between the two times is 1hr for a distance of 180 km.
So when difference between the two time i.e. [15-(-7 ½)], the distance is 67.5 km.
QUESTION:
A boy walking at a speed of 15 km/h reaches his school 20 minutes late. Next time he increases his speed 5km/h, but still he is late by 10 minutes. Find the distance of his school from his house.
A.10 km.
B.12 km.
C.15 km.
D.20 km.
Sol: In this case the two different speeds are 15 & 20 km/h of which the L.C.M is 60.
If the distance is 60 km, then @ 15km/h it will take him 4 hours where as @ 20 km/h it will take him 3 hours. So the difference between two times is 1 hr.
When the difference is 1 hr, the distance is 60 km.
So when the difference is (20-10) i.e. 10 minutes or 1/6 hr, the distance is 10 km.
QUESTION:
: The distance between two stations is 900 km. A train starts from A and moves towards B at an average speed of 30 km/h. Another train starts from B, 20 minutes earlier than the train A, and moves towards A at an average speed of 40 km/h. How far from A will the two trains meet?
Sol: If the train B starts 20minutes earlier than the train A, it must have covered (1/3*40) i.e.40/3 km. So at the time of the starting of A, the difference between the distances between them is (900-40/3) i.e.2660/3 km.
The ratio between the speeds of the two trains is 30: 40 i.e.3:4.So at the same time the difference between the two train is 3+4 i.e. 7 parts which is 2660/3
The distance from A at which they will meet i.e. 3 parts =380 km.
QUESTION:
Walking ¾ of his usual speed, a person is 15 minutes late to reach his office. Find his usual time to cover the distance.
A.30 min.
B.45 min.
C.60 min.
D.75 min.
Sol:
If the usual speed is 1 part, then the new speed is ¾.
So the ratio between the speed of usual & new is 4:3.
There fore the ratio between usual time & new time is 3:4. (As speed α 1/Time)
Due to 1 part more time in case of new time it takes 15 minutes more.
In this case the usual time is 3 parts i.e. 45 minutes.
QUESTION:
Walking 5/3 of his usual speed, a person is 20 minutes earlier to reach his office. Find his usual time to cover the distance.
A.25 min.
B.50 min.
C.60 min.
D.100min.
Sol:
The ratio between the usual & new speed is 3:5, so that the ratio between the time between usual & new is 5:3.
New time is less than the usual time by 2 parts, so it takes 20 minutes earlier.
Usual time is 5 parts i.e. 50 minutes.
QUESTION:
A train leaves from kolkata at 7.30 am and travels @ 40 km/h; another train leaves kolkata at noon and travels @ 64 km/h, when and where the second train overtakes the first?
A.400 km.
B.460 km.
C.480 km.
D.500 km.
Sol:
The first train starts its journey 4 hrs 30 minutes before the second train in which it must have covered 180 km.
When the second train starts its journey, the difference between the distances from the first train is 180 km.
So the second train meets the first train after (180/24) i.e.7 hrs 30 minutes. (As required time to meet= difference between the distance/ relative speed)
And the distance after which the second train meets i.e. 64*7 ½ =480 km.
QUESTION:
One man takes 150 steps a minute, each 4 decimeters long, another walks 4 km/h, if they start together, how soon will one of them be 60 meters ahead to the other?
A.3 min.
B.6 min.
C.9 min.
D.10min.
Sol:
The speed of the first man is 150* 4/10 m i.e.60 meters/min & the second man it is 200/3 meters/min.
The difference between them is 20/3 meter in 1minute.
So one will be ahead of 60 meters to the other in 9 minutes.
QUESTION:
Two men A & B start from a place P walking @ 5.5 & 6.5 km/h respectively. How many km will they be apart at the end of 5 hrs, if they walk in the opposite direction?
A.20km.
B.30 km.
C.40km.
D.60 km.
Sol: In 1 hr, they are 12 km apart.
So in 5 hrs, they are 60 km apart.
QUESTION:
Two men A & B start their journey from P to Q, a distance of 36 km, at 4 & 5 km/h respectively. B reaches Q and returns immediately and meets A at the point R.
(i) Find the distance from P to R.
(ii) And also find the distance covered by B.
(iii) the difference between the distances covered by B & A.
Sol: The ratio between the speeds of A & B is 4:5, so that the ratio between the distance covered by A & B is also 4: 5. (As speed α Distance)
So that the total distance covered by A & B by both i.e. 9 parts. And also the total distance covered by A & B by both i.e. twice of 36km
So 9parts = 72 km.
The distance from P to R is 4 parts = 32 km.
Similarly, the distance traveled by B is 5 parts i.e.40 km.
And the difference between the distances covered by B & A is 1 part i.e. 8 km.
QUESTION:
A man sets out to cycle from Mumbai to kolkata, and at the same time another man starts from Kolkata to Mumbai. After passing each other they complete their journey in 4 & 9 hrs respectively. At what rate does the second man cycle if the first cycle at 9 km/h?
A.4 km/h.
B.5 km/h.
C.6 km/h.
D.8 km/h.
Sol: According to the method
A’s speed : B’s speed=√B’s time :√A’s time.
Speed of second man= 6 km/h.
QUESTION:
Two bullets were fired at a place at an interval of 38 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 36 minutes. If the speed of the sound is 330 m/sec, what is the speed of the car?
A.60km/h.
B. 66km/h.
C.72km/h.
D.76km/h.
Sol: The speed of car in 36 min could be traveled by sound in (38-36) i.e. in 2 min.
The speed of the car in 36 min=330*120 m/sec.
So the speed of car per hour =66 km/h.
QUESTION:
Two persons do the same journey by traveling respectively @ 9 & 10 km/h. Find the length of the journey when one takes 32 minutes longer than the other.
A.36km.
B.40km.
C.48km.
D.50km.
Sol: If the total distance is the L.C.M of 9 & 10i.e.90 km, then @ 9 km/h, it will take 10 hrs & @ 10 km/h, it will take 9 hrs.
When the difference between two timing is 1 hr, the distance is 90km.
So when the difference between two timing is 32min, the distance is 48km.
QUESTION:
A carriage driving in a fog passed a man who was walking @ 5km/h in the same direction. He could see the carriage for 6 minutes and it was visible to him up to a distance of 120metres. What was the speed of the carriage?
A.5km/h.
B.6 km/h.
C.6.2km/h.
D.7.2 km/h.
Sol: The speed of carriage in 6 min = The speed of man in 6 min @ 5km/h + 120metres.
So, the speed of carriage per hour=6.2 km/h.
QUESTION:
A man takes 6 hrs 30 min in walking to a certain place and riding back. He would have gained 2hrs 10 min by riding both ways. How long would he take to walk both ways? And also find how long would he take to ride both ways?
Sol: Required time to walk both ways= Required time to walk + riding back +Time of Gain.
=6 hrs 30min+2 hrs 10 min=8 hrs 40 min.
Required time to ride both ways= Required time to walk + riding back -Time of Gain.
=6 hrs 30min-2 hrs 10 min=4 hrs 20 min.
QUESTION:
A man leaves a point P and reaches the point Q in 4 hrs. Another man leaves the point Q, 2 hrs later and reaches the point P in 4 hours. Find the time in which first man meets to the second man.
QUESTION:
A person covers a certain distance in 24 minutes if he runs at a speed of 27 km/h on an average. Find the speed at which he must run to reduce the time of journey to 18minutes.
A.36 km/h.
B.40.5km/h.
C.44km/h.
D.48km/h.
Sol: The ratio between two time =24:18 i.e.4:3.
The ratio between two speeds =3:4.
( As time α1/speed)
So if 3 parts of speed = 27 km/h.
Then 4 parts of speed = 36 km/h.
QUESTION:
Without any stoppage a person travels a certain distance at an average speed of 15 km/h, and with stoppage he covers the same distance at an average speed of 12 km/h. How many minutes per hour does he stop?
A.10min.
B.12min.
C.15min.
D.20min.
Sol: The ratio between speed without stoppage & with stoppage =15:12 i.e.5:4.
So, the ratio between time without stoppage & with stoppage =4:5.
Therefore, in 5 parts there is a rest of 1 part .
So, in 60 parts or 60 min, there is a rest of 12 part or12min..
QUESTION:
A person has to cover a distance of 100km in 10 hrs. If he covers half of the journey in 3/5 of the time, what should be his speed to cover the remaining distance in the time left?
QUESTION:
A man travels 480 km in 6 hrs, partly by air and partly by train. If he had traveled all the way by air, he would have saved3/4 of the time he was in train and would have arrived at his destination 3hrs early .Find the distance traveled by the train.
QUESTION:
One aeroplane started 1 hr later than the scheduled time from a place 3000 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by500 km/h. What was the speed of the aeroplane per hour during the journey?
QUESTION:
A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km/h. At the next station 300 km away, the train reached on time .Find the original speed of the train.
QUESTION:
When a person travels equal distance at speeds V1 and V2 km/h, his average speed is 8 km/h. But when he travels at these speeds for equal times his average speed is 9 km/h. find the difference of the two speeds.
QUESTION:
A person covers a certain distance on scooter. Had he moved 6 km/h faster, he would have taken 30 minutes less. If he had moved 4 km/hr slower, he would have taken 1 hr 30 min more.Find the original speed.
QUESTION:
A train does a journey without stopping in 9 hrs. If it had traveled 8 km/h faster, it would have done the journey in 6 hours. What is its original speed?
QUESTION:
A car travels a distance of 80 km in 3 hrs partly at a speed of 45 km/h & partly at 20 km/h. Find the distance traveled at a speed of 45 km/h.
.
QUESTION:
A distance is covered by a person in 6 hrs. He covers ¾ of it at 12 km/h and the remaining at 16 km/h. Find the total distance
QUESTION:
The ratio between the speeds of Ajay & Bijay is 6:7. If Ajay takes 30 minutes more than Bijay to cover a distance, then find the actual time taken by Ajay and Bijay.
QUESTION:
A person covers 2/3 rd of his journey at 30 km/h and remaining journey at 60 km/h. If the total journey is of 270 km, what is his average speed for the whole journey?
QUESTION:
A hare sees a dog 200 meters away from her and scuds off in the opposite direction at a speed of 24 km/h. Two minutes later the dog perceives her and gives chase at a speed of 32 km/h. How soon will the dog overtake the hare, and at what distance from the spot whence the hare took flight?
QUESTION:
Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260kms?
A.60 km/hr
B.90 km/hr
C.120 km/hr
D.135 km/hr
QUESTION:
A man moves from A to B at the rate of 4 km/hr. Had he moved at the rate of 3.67 km/hr, he would have taken 3 hours more to reach the destination. What is the distance between A and B?
33 kms
132 kms
36 kms
144 kms
The problems based on time & distances are based on speed, time & distance. The general process to find out distance is as follows:
DISTANCE= SPEED*TIME.
Where S is the speed of the moving object, T is the time period & D is the distance covered.
Process to convert km/h in to m/s or vice versa is as follows:
1km/ h = 5/18 m/s
1 m/s = 18/5 km/h
Relative speed: The different between the speeds of two moving object with respect to their direction is known as relative speed. The relative speed according to the two different directions is as follows:
Case I: In case of same direction it is the difference between the speeds of the two moving objects.
Case II: In case of opposite direction it is the sum of the speeds of the two moving objects.
Example: Two objects are moving @ 30 km/h & 50 km/h. then the relative speed between them in two different cases are as follows:
Case I: In the same direction it is (50-30) i.e. 20 km/h
Case II: In case of opposite direction it is (50+30) i.e.80 km/h.
QUESTION:
A boy goes to school @ 10 km/h & returns to his home @ 12 km/h respectively. If he takes 5 ½ hrs in all, find the distance between school & his home.
A.20km.
B.25 km.
C.30 km .
D.30 km.
Sol:
The L.C.M of two different speeds of 10 &12 is 60.
If 60 km is the distance, then at the rate of two different speeds the times are 6 & 5 hrs respectively.
When the total time is 6+5 i.e. 11hrs, the distance between the two points is 60 km.
So when the total time is 5 ½ hrs, the distance between school & home is 30 km.
QUESTION:
A person covers half of his journey @ 21 km/h, & the rest half @ 24 km/h respectively. If he takes 15 hrs in all, find the distance covered by the person.
A.168 km.
B.224 km.
C.336 km.
D.280 km.
Sol:
The L.C.M of 21 &24 i.e. 168.
If 168 km is the distance, then the person covers ½ of it i.e. 84 km @ 21 km/h in 4 hrs and the rest half @ 24 km/h in 3 ½ hrs.
So when the total time is 7 ½ hrs, the distance is 168 km.
And if the total time is 15 hrs, the distance is 336 km.
QUESTION:
A person goes a certain distance @ 48 km/h & returns to the starting point @ 96 km/h respectively. Find his average speed during the whole journey.
A.60 km/h.
B.64 km/h
C.68 km/h
D.72 km/h
Sol:
The L.C.M of 48 @ 96 is 96.
If the distance is 96 km, then @ two different speeds the required time are 2 &1 hr respectively.
Average speed= Total distance/Total time
So the average speed i.e.(96+96)/2+1=64 km/h.
QUESTION:
A person covers a certain distance between his house & office on scooter. Having an average speed of 45 km/h, he is late by 15 minutes. How ever, with a speed of 60 km/h, he reaches his office 7 ½ min earlier. Find the distance between his house & office.
A.67.5 km.
B.60 km.
C.75 km.
D.120 km.
Sol: Let the L.C.M of the two different speeds i.e. the L.C.M of 45 & 60 =180 is the distance.
So @ 45km/h, it will take him 4 hrs & @ 60 km/h, it will take him 3 hours.
In this case the difference between the two times is 1hr for a distance of 180 km.
So when difference between the two time i.e. [15-(-7 ½)], the distance is 67.5 km.
QUESTION:
A boy walking at a speed of 15 km/h reaches his school 20 minutes late. Next time he increases his speed 5km/h, but still he is late by 10 minutes. Find the distance of his school from his house.
A.10 km.
B.12 km.
C.15 km.
D.20 km.
Sol: In this case the two different speeds are 15 & 20 km/h of which the L.C.M is 60.
If the distance is 60 km, then @ 15km/h it will take him 4 hours where as @ 20 km/h it will take him 3 hours. So the difference between two times is 1 hr.
When the difference is 1 hr, the distance is 60 km.
So when the difference is (20-10) i.e. 10 minutes or 1/6 hr, the distance is 10 km.
QUESTION:
: The distance between two stations is 900 km. A train starts from A and moves towards B at an average speed of 30 km/h. Another train starts from B, 20 minutes earlier than the train A, and moves towards A at an average speed of 40 km/h. How far from A will the two trains meet?
Sol: If the train B starts 20minutes earlier than the train A, it must have covered (1/3*40) i.e.40/3 km. So at the time of the starting of A, the difference between the distances between them is (900-40/3) i.e.2660/3 km.
The ratio between the speeds of the two trains is 30: 40 i.e.3:4.So at the same time the difference between the two train is 3+4 i.e. 7 parts which is 2660/3
The distance from A at which they will meet i.e. 3 parts =380 km.
QUESTION:
Walking ¾ of his usual speed, a person is 15 minutes late to reach his office. Find his usual time to cover the distance.
A.30 min.
B.45 min.
C.60 min.
D.75 min.
Sol:
If the usual speed is 1 part, then the new speed is ¾.
So the ratio between the speed of usual & new is 4:3.
There fore the ratio between usual time & new time is 3:4. (As speed α 1/Time)
Due to 1 part more time in case of new time it takes 15 minutes more.
In this case the usual time is 3 parts i.e. 45 minutes.
QUESTION:
Walking 5/3 of his usual speed, a person is 20 minutes earlier to reach his office. Find his usual time to cover the distance.
A.25 min.
B.50 min.
C.60 min.
D.100min.
Sol:
The ratio between the usual & new speed is 3:5, so that the ratio between the time between usual & new is 5:3.
New time is less than the usual time by 2 parts, so it takes 20 minutes earlier.
Usual time is 5 parts i.e. 50 minutes.
QUESTION:
A train leaves from kolkata at 7.30 am and travels @ 40 km/h; another train leaves kolkata at noon and travels @ 64 km/h, when and where the second train overtakes the first?
A.400 km.
B.460 km.
C.480 km.
D.500 km.
Sol:
The first train starts its journey 4 hrs 30 minutes before the second train in which it must have covered 180 km.
When the second train starts its journey, the difference between the distances from the first train is 180 km.
So the second train meets the first train after (180/24) i.e.7 hrs 30 minutes. (As required time to meet= difference between the distance/ relative speed)
And the distance after which the second train meets i.e. 64*7 ½ =480 km.
QUESTION:
One man takes 150 steps a minute, each 4 decimeters long, another walks 4 km/h, if they start together, how soon will one of them be 60 meters ahead to the other?
A.3 min.
B.6 min.
C.9 min.
D.10min.
Sol:
The speed of the first man is 150* 4/10 m i.e.60 meters/min & the second man it is 200/3 meters/min.
The difference between them is 20/3 meter in 1minute.
So one will be ahead of 60 meters to the other in 9 minutes.
QUESTION:
Two men A & B start from a place P walking @ 5.5 & 6.5 km/h respectively. How many km will they be apart at the end of 5 hrs, if they walk in the opposite direction?
A.20km.
B.30 km.
C.40km.
D.60 km.
Sol: In 1 hr, they are 12 km apart.
So in 5 hrs, they are 60 km apart.
QUESTION:
Two men A & B start their journey from P to Q, a distance of 36 km, at 4 & 5 km/h respectively. B reaches Q and returns immediately and meets A at the point R.
(i) Find the distance from P to R.
(ii) And also find the distance covered by B.
(iii) the difference between the distances covered by B & A.
Sol: The ratio between the speeds of A & B is 4:5, so that the ratio between the distance covered by A & B is also 4: 5. (As speed α Distance)
So that the total distance covered by A & B by both i.e. 9 parts. And also the total distance covered by A & B by both i.e. twice of 36km
So 9parts = 72 km.
The distance from P to R is 4 parts = 32 km.
Similarly, the distance traveled by B is 5 parts i.e.40 km.
And the difference between the distances covered by B & A is 1 part i.e. 8 km.
QUESTION:
A man sets out to cycle from Mumbai to kolkata, and at the same time another man starts from Kolkata to Mumbai. After passing each other they complete their journey in 4 & 9 hrs respectively. At what rate does the second man cycle if the first cycle at 9 km/h?
A.4 km/h.
B.5 km/h.
C.6 km/h.
D.8 km/h.
Sol: According to the method
A’s speed : B’s speed=√B’s time :√A’s time.
Speed of second man= 6 km/h.
QUESTION:
Two bullets were fired at a place at an interval of 38 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 36 minutes. If the speed of the sound is 330 m/sec, what is the speed of the car?
A.60km/h.
B. 66km/h.
C.72km/h.
D.76km/h.
Sol: The speed of car in 36 min could be traveled by sound in (38-36) i.e. in 2 min.
The speed of the car in 36 min=330*120 m/sec.
So the speed of car per hour =66 km/h.
QUESTION:
Two persons do the same journey by traveling respectively @ 9 & 10 km/h. Find the length of the journey when one takes 32 minutes longer than the other.
A.36km.
B.40km.
C.48km.
D.50km.
Sol: If the total distance is the L.C.M of 9 & 10i.e.90 km, then @ 9 km/h, it will take 10 hrs & @ 10 km/h, it will take 9 hrs.
When the difference between two timing is 1 hr, the distance is 90km.
So when the difference between two timing is 32min, the distance is 48km.
QUESTION:
A carriage driving in a fog passed a man who was walking @ 5km/h in the same direction. He could see the carriage for 6 minutes and it was visible to him up to a distance of 120metres. What was the speed of the carriage?
A.5km/h.
B.6 km/h.
C.6.2km/h.
D.7.2 km/h.
Sol: The speed of carriage in 6 min = The speed of man in 6 min @ 5km/h + 120metres.
So, the speed of carriage per hour=6.2 km/h.
QUESTION:
A man takes 6 hrs 30 min in walking to a certain place and riding back. He would have gained 2hrs 10 min by riding both ways. How long would he take to walk both ways? And also find how long would he take to ride both ways?
Sol: Required time to walk both ways= Required time to walk + riding back +Time of Gain.
=6 hrs 30min+2 hrs 10 min=8 hrs 40 min.
Required time to ride both ways= Required time to walk + riding back -Time of Gain.
=6 hrs 30min-2 hrs 10 min=4 hrs 20 min.
QUESTION:
A man leaves a point P and reaches the point Q in 4 hrs. Another man leaves the point Q, 2 hrs later and reaches the point P in 4 hours. Find the time in which first man meets to the second man.
QUESTION:
A person covers a certain distance in 24 minutes if he runs at a speed of 27 km/h on an average. Find the speed at which he must run to reduce the time of journey to 18minutes.
A.36 km/h.
B.40.5km/h.
C.44km/h.
D.48km/h.
Sol: The ratio between two time =24:18 i.e.4:3.
The ratio between two speeds =3:4.
( As time α1/speed)
So if 3 parts of speed = 27 km/h.
Then 4 parts of speed = 36 km/h.
QUESTION:
Without any stoppage a person travels a certain distance at an average speed of 15 km/h, and with stoppage he covers the same distance at an average speed of 12 km/h. How many minutes per hour does he stop?
A.10min.
B.12min.
C.15min.
D.20min.
Sol: The ratio between speed without stoppage & with stoppage =15:12 i.e.5:4.
So, the ratio between time without stoppage & with stoppage =4:5.
Therefore, in 5 parts there is a rest of 1 part .
So, in 60 parts or 60 min, there is a rest of 12 part or12min..
QUESTION:
A person has to cover a distance of 100km in 10 hrs. If he covers half of the journey in 3/5 of the time, what should be his speed to cover the remaining distance in the time left?
QUESTION:
A man travels 480 km in 6 hrs, partly by air and partly by train. If he had traveled all the way by air, he would have saved3/4 of the time he was in train and would have arrived at his destination 3hrs early .Find the distance traveled by the train.
QUESTION:
One aeroplane started 1 hr later than the scheduled time from a place 3000 km away from its destination. To reach the destination at the scheduled time the pilot had to increase the speed by500 km/h. What was the speed of the aeroplane per hour during the journey?
QUESTION:
A train leaves the station 1 hour before the scheduled time. The driver decreases its speed by 50 km/h. At the next station 300 km away, the train reached on time .Find the original speed of the train.
QUESTION:
When a person travels equal distance at speeds V1 and V2 km/h, his average speed is 8 km/h. But when he travels at these speeds for equal times his average speed is 9 km/h. find the difference of the two speeds.
QUESTION:
A person covers a certain distance on scooter. Had he moved 6 km/h faster, he would have taken 30 minutes less. If he had moved 4 km/hr slower, he would have taken 1 hr 30 min more.Find the original speed.
QUESTION:
A train does a journey without stopping in 9 hrs. If it had traveled 8 km/h faster, it would have done the journey in 6 hours. What is its original speed?
QUESTION:
A car travels a distance of 80 km in 3 hrs partly at a speed of 45 km/h & partly at 20 km/h. Find the distance traveled at a speed of 45 km/h.
.
QUESTION:
A distance is covered by a person in 6 hrs. He covers ¾ of it at 12 km/h and the remaining at 16 km/h. Find the total distance
QUESTION:
The ratio between the speeds of Ajay & Bijay is 6:7. If Ajay takes 30 minutes more than Bijay to cover a distance, then find the actual time taken by Ajay and Bijay.
QUESTION:
A person covers 2/3 rd of his journey at 30 km/h and remaining journey at 60 km/h. If the total journey is of 270 km, what is his average speed for the whole journey?
QUESTION:
A hare sees a dog 200 meters away from her and scuds off in the opposite direction at a speed of 24 km/h. Two minutes later the dog perceives her and gives chase at a speed of 32 km/h. How soon will the dog overtake the hare, and at what distance from the spot whence the hare took flight?
QUESTION:
Train A traveling at 60 km/hr leaves Mumbai for Delhi at 6 P.M. Train B traveling at 90 km/hr also leaves Mumbai for Delhi at 9 P.M. Train C leaves Delhi for Mumbai at 9 P.M. If all three trains meet at the same time between Mumbai and Delhi, what is the speed of Train C if the distance between Delhi and Mumbai is 1260kms?
A.60 km/hr
B.90 km/hr
C.120 km/hr
D.135 km/hr
QUESTION:
A man moves from A to B at the rate of 4 km/hr. Had he moved at the rate of 3.67 km/hr, he would have taken 3 hours more to reach the destination. What is the distance between A and B?
33 kms
132 kms
36 kms
144 kms
Aptitude questions on Height and Distance-Interview written test questions
INTRODUCTION:
The questions based on height & distance are based on the following factors:
The angle of elevation or depression.
One of the side out of height and distance.
Generally the angle of elevation or depression which are coming in different examinations are based on 300, 450, & 600.
When the angle of elevation or depression is 300 or 600:
Take the side infront of 300 as 1 part.
Take the side infront of 600 as P 3 parts.
Take the side infront of 900 as 2 parts.
By taking the ratio we can find the asking one
When the angle of elevation or depression is 450 :
Take the side infront of 450 as 1/P 2 part.
Take the side infront of 900 as 1part.
By taking the ratio we can find the asking one
QUESTION:
A person standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is 60º.When he retires 100 m from the bank he finds the tree to be 30º. Find the height of the tree and the breadth of the river.
SOLUTION:
In D ABC, the side opposite to 30º which is the breadth of the river is 1 part
Let i.e. x meter
Then the side opposite to 60ºwhich is the height of tree is Ö 3 part = Ö 3x
Again in case of D ABD the side opposite to 30º which is the height of the tree as 1 part=Ö 3x And the side opposite tom 60 which is Ö 3parts=3x
3x=x+100
Therefore x i.e. breadth of the river= 50 m
And the height of the tree which is Ö 3x=50Ö 3m
QUESTION:
An observer on the top of a cliff 200 m above the sea level observes the angles of depression of the two ships on opposite sides of the cliff to be 45 and 30 respectively. Find the distance between the ships if the line joining them passes through the base of the cliff.
Solution:
In D ABD AD = BD
(as the corresponding angle 45)= 200 m
So AD = DC
Again in ADC the side opposite to 300 which is 1 part i.e. 200m
Therefore the side opposite to60 which is CD i.e. Ö 3 parts = 200Ö 3 m
The distance between the two
ships= (200 + 200Ö 3) m
= 200(1+Ö 3) m
= 200(1+1.732) m
= 5464 m
QUESTION:
A lizard is moving up a wall. From the foot of the wall at a certain distance the angle of elevation of dog’s eye with the lizard is 30.When the lizard reaches the top of the wall, the dog is 15 m away from the wall and at this point the angle of elevation of the dog’s eye with the lizard is 60.During this time if the lizard has covered 5Ö 3 m of height, the dog has covered how many meters of distance?
Solution:
Let initially the lizard is at a point E and latter it reaches at the point A
initially the dog is at a point C and latter it reaches at the point D
In D ABD, BD is opposite to angle 300 which is 1 part is 15 m
So the side opposite to 60 which is AB i.e. Ö 3 parts is 15 Ö 3m
Therefore BE = 10Ö 3 m
Then in D BCE, BE which is opposite to 30 i.e. 1 part is 10Ö 3 m
The distance BC which is opposite to 60 i.e. Ö 3 parts is 30 m.
The distance covered by the dog is CD = (30 -15) m
= 15 m
QUESTION:
Two unequal poles are erected on either side of the road of width 60 m .A man is standing at the exact middle position at which the angle of elevations are 30º and 60º respectively. Then find the
height of the shorter pole
height of the longer pole
Distance between the topmost points of the two poles
Solution:
(i) In case of D ABC the side BC is opposite to the angle
60º which is Ö 3 parts = 20 m
Therefore the height of the shorter pole i.e. AB opposite to the angle 30º
which is 1 part = 30/Ö 3 = 10Ö 3 m
(ii) In case of D CDE, CD is opposite to the angle 30º which is 1 part = 30 m
The side opposite to the angle 30º which is 1 part = 30 m
Therefore the side i.e. DE the height of the longer pole is opposite to 60º is Ö 3 parts= 30Ö 3m
(iii) In case of D ACE,Ð ACE = 90º so it is a right angled triangle.
In this case CE = opposite of angle of 900
From D CDE i.e. 2 parts = 60 m
Similarly AC is opposite side of the angle of 900 from D ABC i.e. 2 parts= 20 Ö 3 m
So the ratio between the side AC & CE = 1:Ö 3 and in this case 1 part = 20Ö 3 m
Therefore the distance between the topmost points of two poles AE which is opposite to the angle of 90º is 2 parts = 40Ö 3 m.
QUESTION:
If the shadow of a tower is 30 m when the sun’s altitude is 30º, what is the length of the shadow when the sun’s altitude is 60º?
Solution:
In D ABC, BC is opposite to the angle of 600 i.e. Ö 3 parts = 30 m
The height of the tower AB is opposite to the angle of 30º i.e.
1 part = 30/Ö 3 = 10Ö 3m
In case of D ABD,AB is opposite of 60º i.e. Ö 3 parts= 10Ö 3 m
So the length of shadow BD which is opposite to 300 is 1 part = 10m
QUESTION:
A man on the top of a vertical light house observes a boat is coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30º to 60º how soon will it reach the light house?
Solution:
In D ABC let the length of AB = x meter
AB is opposite to 30º i.e. 1 part = x meters
So BC is opposite to 60º
i.e. Ö 3 parts = Ö 3 x meters
Again in D ABD AB opposite of 60º
which is Ö 3 parts= x meters
BD which one is opposite to 30º
i.e. 1 part = x/Ö 3 m
CD= (Ö 3 x-x/Ö 3)m = 2 x/Ö 3 m
The ship covers 2 x/Ö 3 m in 10 minutes
QUESTION:
A man in a boat being rowed away from a cliff 120(Ö 3+1)m high takes two minutes to change the angle of elevation of the top of the cliff from 45º to 30º.Find the speed of the boat in km/hr.
Solution:
In D ABD AB = BD =120(Ö 3+1)m
Again in D ABC AB is opposite
to 30º is 1 part = 120(Ö 3+1)
BC which is opposite to 60º is
Ö 3 parts= Ö 3 [120(Ö 3+1)]
SO that DC=Ö 3 [120(Ö 3+1)]-120(Ö 3+1)
=240m
There fore speed of boat is 240/120=2m/s
Which one is 7.2km/h
QUESTION:
From the top of a cliff150 m high, the angles of depression of two boats which are due north of the observer are 600 and 300.Find the distance between them.
Solution:
In the D ABC, the side opposite to 600 is Ö 3 parts = 150 m
The side opposite to 300 which is 1 part = 150/Ö 3 = 50Ö 3 m
Again in the D ABD the side opposite to 300 is 1 part=150 m
So the side opposite to 600 i.e. BD = Ö 3 parts which is 150Ö 3 m
DB=150Ö 3 m and BC = 50Ö 3 m
Therefore the distance between the two ships is
BD – BC=100Ö 3 m
The questions based on height & distance are based on the following factors:
The angle of elevation or depression.
One of the side out of height and distance.
Generally the angle of elevation or depression which are coming in different examinations are based on 300, 450, & 600.
When the angle of elevation or depression is 300 or 600:
Take the side infront of 300 as 1 part.
Take the side infront of 600 as P 3 parts.
Take the side infront of 900 as 2 parts.
By taking the ratio we can find the asking one
When the angle of elevation or depression is 450 :
Take the side infront of 450 as 1/P 2 part.
Take the side infront of 900 as 1part.
By taking the ratio we can find the asking one
QUESTION:
A person standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is 60º.When he retires 100 m from the bank he finds the tree to be 30º. Find the height of the tree and the breadth of the river.
SOLUTION:
In D ABC, the side opposite to 30º which is the breadth of the river is 1 part
Let i.e. x meter
Then the side opposite to 60ºwhich is the height of tree is Ö 3 part = Ö 3x
Again in case of D ABD the side opposite to 30º which is the height of the tree as 1 part=Ö 3x And the side opposite tom 60 which is Ö 3parts=3x
3x=x+100
Therefore x i.e. breadth of the river= 50 m
And the height of the tree which is Ö 3x=50Ö 3m
QUESTION:
An observer on the top of a cliff 200 m above the sea level observes the angles of depression of the two ships on opposite sides of the cliff to be 45 and 30 respectively. Find the distance between the ships if the line joining them passes through the base of the cliff.
Solution:
In D ABD AD = BD
(as the corresponding angle 45)= 200 m
So AD = DC
Again in ADC the side opposite to 300 which is 1 part i.e. 200m
Therefore the side opposite to60 which is CD i.e. Ö 3 parts = 200Ö 3 m
The distance between the two
ships= (200 + 200Ö 3) m
= 200(1+Ö 3) m
= 200(1+1.732) m
= 5464 m
QUESTION:
A lizard is moving up a wall. From the foot of the wall at a certain distance the angle of elevation of dog’s eye with the lizard is 30.When the lizard reaches the top of the wall, the dog is 15 m away from the wall and at this point the angle of elevation of the dog’s eye with the lizard is 60.During this time if the lizard has covered 5Ö 3 m of height, the dog has covered how many meters of distance?
Solution:
Let initially the lizard is at a point E and latter it reaches at the point A
initially the dog is at a point C and latter it reaches at the point D
In D ABD, BD is opposite to angle 300 which is 1 part is 15 m
So the side opposite to 60 which is AB i.e. Ö 3 parts is 15 Ö 3m
Therefore BE = 10Ö 3 m
Then in D BCE, BE which is opposite to 30 i.e. 1 part is 10Ö 3 m
The distance BC which is opposite to 60 i.e. Ö 3 parts is 30 m.
The distance covered by the dog is CD = (30 -15) m
= 15 m
QUESTION:
Two unequal poles are erected on either side of the road of width 60 m .A man is standing at the exact middle position at which the angle of elevations are 30º and 60º respectively. Then find the
height of the shorter pole
height of the longer pole
Distance between the topmost points of the two poles
Solution:
(i) In case of D ABC the side BC is opposite to the angle
60º which is Ö 3 parts = 20 m
Therefore the height of the shorter pole i.e. AB opposite to the angle 30º
which is 1 part = 30/Ö 3 = 10Ö 3 m
(ii) In case of D CDE, CD is opposite to the angle 30º which is 1 part = 30 m
The side opposite to the angle 30º which is 1 part = 30 m
Therefore the side i.e. DE the height of the longer pole is opposite to 60º is Ö 3 parts= 30Ö 3m
(iii) In case of D ACE,Ð ACE = 90º so it is a right angled triangle.
In this case CE = opposite of angle of 900
From D CDE i.e. 2 parts = 60 m
Similarly AC is opposite side of the angle of 900 from D ABC i.e. 2 parts= 20 Ö 3 m
So the ratio between the side AC & CE = 1:Ö 3 and in this case 1 part = 20Ö 3 m
Therefore the distance between the topmost points of two poles AE which is opposite to the angle of 90º is 2 parts = 40Ö 3 m.
QUESTION:
If the shadow of a tower is 30 m when the sun’s altitude is 30º, what is the length of the shadow when the sun’s altitude is 60º?
Solution:
In D ABC, BC is opposite to the angle of 600 i.e. Ö 3 parts = 30 m
The height of the tower AB is opposite to the angle of 30º i.e.
1 part = 30/Ö 3 = 10Ö 3m
In case of D ABD,AB is opposite of 60º i.e. Ö 3 parts= 10Ö 3 m
So the length of shadow BD which is opposite to 300 is 1 part = 10m
QUESTION:
A man on the top of a vertical light house observes a boat is coming directly towards it. If it takes 10 minutes for the angle of depression to change from 30º to 60º how soon will it reach the light house?
Solution:
In D ABC let the length of AB = x meter
AB is opposite to 30º i.e. 1 part = x meters
So BC is opposite to 60º
i.e. Ö 3 parts = Ö 3 x meters
Again in D ABD AB opposite of 60º
which is Ö 3 parts= x meters
BD which one is opposite to 30º
i.e. 1 part = x/Ö 3 m
CD= (Ö 3 x-x/Ö 3)m = 2 x/Ö 3 m
The ship covers 2 x/Ö 3 m in 10 minutes
QUESTION:
A man in a boat being rowed away from a cliff 120(Ö 3+1)m high takes two minutes to change the angle of elevation of the top of the cliff from 45º to 30º.Find the speed of the boat in km/hr.
Solution:
In D ABD AB = BD =120(Ö 3+1)m
Again in D ABC AB is opposite
to 30º is 1 part = 120(Ö 3+1)
BC which is opposite to 60º is
Ö 3 parts= Ö 3 [120(Ö 3+1)]
SO that DC=Ö 3 [120(Ö 3+1)]-120(Ö 3+1)
=240m
There fore speed of boat is 240/120=2m/s
Which one is 7.2km/h
QUESTION:
From the top of a cliff150 m high, the angles of depression of two boats which are due north of the observer are 600 and 300.Find the distance between them.
Solution:
In the D ABC, the side opposite to 600 is Ö 3 parts = 150 m
The side opposite to 300 which is 1 part = 150/Ö 3 = 50Ö 3 m
Again in the D ABD the side opposite to 300 is 1 part=150 m
So the side opposite to 600 i.e. BD = Ö 3 parts which is 150Ö 3 m
DB=150Ö 3 m and BC = 50Ö 3 m
Therefore the distance between the two ships is
BD – BC=100Ö 3 m
Aptitude test on Partnership -Interview written test aptitude questions
INTRODUCTION:
When two or more persons they put their money in order to carry their business that deal is known as partnership.
According to the deal, the amount of the profit is divided according to the product of the ratio of their money and duration of investment respectively.
TYPES OF PARTNERSHIP:
There are two types of partnership. Such as:
1)Simple partnership
2)Compound partnership
Simple partnership:
In case of simple partnership the duration of investment is constant. So the amounts of profit is divided according to the ratio among their capital respectively.
If the ratio of their amount is a: b: c, then the profit will be divided in the ratio a: b: c respectively
Compound partnership:
In case of compound partnership the duration of investment is not constant. So the amounts of profit is divided according to the product ratio among their capital and duration of investment respectively.
If the ratio of their amount is a: b: c, and duration of investment is x: y: z, then the profit will be divided in the ratio ax: by: cz respectively
TYPES OF PARTNER:
There are two types of partners in partnership business. Such as :
(i) Working partner.
The partner who works for the business is known as working partner.
(ii) Sleeping partner.
The partner who does not works for the business but puts his money, is known as working partner.
Below are Questions to solve
QUESTION:
Three partners A, B and C invests Rs16000, Rs18000 and Rs23000 respectively in a business. How should they divide a profit of Rs19380?
QUESTION:
A, B and C enter in to partnership. A advances Rs12000 for 4 months, B Rs14000 for 8 months, and C Rs10000 for 10 months. They gain Rs5850 altogether. Find the share of each.
QUESTION:
A starts a business with Rs20000. B joins him after 3 months with Rs40000. C puts a sum of Rs10000 in the business for 2 months only. At the end of the year the business gave a profit of Rs5000. How should the profit be divided among them?
QUESTION:
A and B entered in to a partnership investing Rs16000 and Rs12000 respectively. After 3 months, A withdrew Rs5000 while B invested Rs5000 more. After 3 months more C joins the business with a capital of Rs21000. The share of B exceeds that of C, out of a total profit of Rs26400 after one year, by:
QUESTION:
Manoj got Rs6000 as his share out of the total profit of Rs9000 which he and Ramesh earned at the end of one year. If Manoj investedRs20000 for 6 months, where as Ramesh invested his amount for the whole year, the amount invested by Ramesh was:
QUESTION:
A and B enter in to a partnership with their capitals in the ratio 7: 9. At the end of 8 months, A withdraws his capital. If they receive the profits in the ratio 8:9, find how long B’s capital was used?
QUESTION:
A, B and C invested capitals in the ratio 2:3:5; the timing of their investments being in the ratio 4:5:6. In what ratio would their profit be distributed?
QUESTION:
A, B and C are partners. A receives 2/5 of profit and B and C share the remaining profit equally. A’s income is increased by Rs220 when the profit rises from 8% to 10%. Find the capitals invested by A, B and C.
QUESTION:
Two partners invest Rs125, 000 and Rs85, 000 respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be treated as interest on capital. If one partner gets Rs300 more than the other, find the total profit made in the business.
When two or more persons they put their money in order to carry their business that deal is known as partnership.
According to the deal, the amount of the profit is divided according to the product of the ratio of their money and duration of investment respectively.
TYPES OF PARTNERSHIP:
There are two types of partnership. Such as:
1)Simple partnership
2)Compound partnership
Simple partnership:
In case of simple partnership the duration of investment is constant. So the amounts of profit is divided according to the ratio among their capital respectively.
If the ratio of their amount is a: b: c, then the profit will be divided in the ratio a: b: c respectively
Compound partnership:
In case of compound partnership the duration of investment is not constant. So the amounts of profit is divided according to the product ratio among their capital and duration of investment respectively.
If the ratio of their amount is a: b: c, and duration of investment is x: y: z, then the profit will be divided in the ratio ax: by: cz respectively
TYPES OF PARTNER:
There are two types of partners in partnership business. Such as :
(i) Working partner.
The partner who works for the business is known as working partner.
(ii) Sleeping partner.
The partner who does not works for the business but puts his money, is known as working partner.
Below are Questions to solve
QUESTION:
Three partners A, B and C invests Rs16000, Rs18000 and Rs23000 respectively in a business. How should they divide a profit of Rs19380?
QUESTION:
A, B and C enter in to partnership. A advances Rs12000 for 4 months, B Rs14000 for 8 months, and C Rs10000 for 10 months. They gain Rs5850 altogether. Find the share of each.
QUESTION:
A starts a business with Rs20000. B joins him after 3 months with Rs40000. C puts a sum of Rs10000 in the business for 2 months only. At the end of the year the business gave a profit of Rs5000. How should the profit be divided among them?
QUESTION:
A and B entered in to a partnership investing Rs16000 and Rs12000 respectively. After 3 months, A withdrew Rs5000 while B invested Rs5000 more. After 3 months more C joins the business with a capital of Rs21000. The share of B exceeds that of C, out of a total profit of Rs26400 after one year, by:
QUESTION:
Manoj got Rs6000 as his share out of the total profit of Rs9000 which he and Ramesh earned at the end of one year. If Manoj investedRs20000 for 6 months, where as Ramesh invested his amount for the whole year, the amount invested by Ramesh was:
QUESTION:
A and B enter in to a partnership with their capitals in the ratio 7: 9. At the end of 8 months, A withdraws his capital. If they receive the profits in the ratio 8:9, find how long B’s capital was used?
QUESTION:
A, B and C invested capitals in the ratio 2:3:5; the timing of their investments being in the ratio 4:5:6. In what ratio would their profit be distributed?
QUESTION:
A, B and C are partners. A receives 2/5 of profit and B and C share the remaining profit equally. A’s income is increased by Rs220 when the profit rises from 8% to 10%. Find the capitals invested by A, B and C.
QUESTION:
Two partners invest Rs125, 000 and Rs85, 000 respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be treated as interest on capital. If one partner gets Rs300 more than the other, find the total profit made in the business.
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