41194.Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even ?
3/4
1/4
7/4
1/2
Explanation:
Total number of cases = 6*6 = 36a
Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27
So Probability = 27/36 = ¾
Total number of cases = 6*6 = 36a
Favourable cases = [(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)] = 27
So Probability = 27/36 = ¾
41195.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither blue nor green?
2/3
8/21
3/7
9/22
Explanation:
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
Total number of balls = (8 + 7 + 6) = 21
Let E = event that the ball drawn is neither blue nor green =e vent that the ball drawn is red.
Therefore, n(E) = 8.
P(E) = 8/21.
41196.A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is
1/13
2/13
1/26
1/52
Explanation:
Total number of cases = 52
Favourable cases = 2
Probability = 2/56 = 1/26
Total number of cases = 52
Favourable cases = 2
Probability = 2/56 = 1/26
41197.A speaks truth in 75% of cases and B in 80% of cases. In what percentage of cases are they likely to contradict each other, narrating the same incident
30%
35%
40%
45%
Explanation:
Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = 1-3/4 = 1/4
P(B-lie) = 1-4/5 = 1/5
Now
A and B contradict each other =
[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
[Please note that we are adding at the place of OR]
= (3/5*1/5) + (1/4*4/5) = 7/20
= (7/20 * 100) % = 35%
Let A = Event that A speaks the truth
B = Event that B speaks the truth
Then P(A) = 75/100 = 3/4
P(B) = 80/100 = 4/5
P(A-lie) = 1-3/4 = 1/4
P(B-lie) = 1-4/5 = 1/5
Now
A and B contradict each other =
[A lies and B true] or [B true and B lies]
= P(A).P(B-lie) + P(A-lie).P(B)
[Please note that we are adding at the place of OR]
= (3/5*1/5) + (1/4*4/5) = 7/20
= (7/20 * 100) % = 35%
41198.From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings
2/121
2/221
1/221
1/13
Explanation:
Total cases =52C2=52∗51/2∗1=1326
Total King cases =4C2=4∗3/2∗1=6
Probability ==6/1326=1/221
Total cases =52C2=52∗51/2∗1=1326
Total King cases =4C2=4∗3/2∗1=6
Probability ==6/1326=1/221
41199.Bag contain 10 back and 20 white balls, One ball is drawn at random. What is the probability that ball is white
1
2/3
1/3
4/3
Explanation:
Total cases = 10 + 20 = 30
Favourable cases = 20
So probability = 20/30 = ⅔
Total cases = 10 + 20 = 30
Favourable cases = 20
So probability = 20/30 = ⅔
41200.There is a pack of 52 cards and Rohan draws two cards together, what is the probability that one is spade and one is heart ?
11/102
13/102
11/104
11/102
Explanation:
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
Let sample space be S
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:
Let sample space be S
41201.AMCAT Quantitative Ability Previous Papers-2
In how many ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.
In how many ways can the letters of the word "CORPORATION" be arranged so that vowels always come together.
5760
50400
2880
None of above
Explanation:
Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
41202.In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there
109
128
138
209
Explanation:
In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have
(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)
This combination question can be solved as
In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
So we can have
(four boys) or (three boys and one girl) or (two boys and two girls) or (one boy and three gils)
This combination question can be solved as
41203.From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done
456
556
656
756
Explanation:
From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)
From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
So we can have
(5 men) or (4 men and 1 woman) or (3 men and 2 woman)
41204.A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw
64
128
132
222
Explanation:
From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.
Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]
=1+[3×6]+[3×((6×5)/(2×1))]=1+18+45=64
From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.
Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]
=1+[3×6]+[3×((6×5)/(2×1))]=1+18+45=64
41205.Ques. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours
12
24
48
168
Explanation:
This question seems to be a bit typical, isnt, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
This question seems to be a bit typical, isnt, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways
Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24
Please note that we have multiplied the combination results, we use to add when their is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.
41206.Three unbiased coins are tossed, what is the probability of getting at least 2 tails ?
1/3
1/6
1/2
1/8
Explanation:
Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]
Favoured cases are = [TTH, THT, HTT, TTT] = 4
So required probability = 4/8 = ½
Total cases are = 2*2*2 = 8, which are as follows
[TTT, HHH, TTH, THT, HTT, THH, HTH, HHT]
Favoured cases are = [TTH, THT, HTT, TTT] = 4
So required probability = 4/8 = ½
41207.Ques. In a throw of dice what is the probability of getting number greater than 5
1/2
1/3
1/5
1/6
Explanation:
Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]
So probability = ⅙
Number greater than 5 is 6, so only 1 number
Total cases of dice = [1,2,3,4,5,6]
So probability = ⅙
41208.A man can row at 5 kmph in still water. If the velocity of the current is 1 kmph and it takes him 1 hour to row to a place and come back. how far is that place.
.4 km
1.4 km
2.4 km
3.4 km
Explanation:
Let the distance is x km
Rate downstream = 5 + 1 = 6 kmph
Rate upstream = 5 - 1 = 4 kmph
then
x/6 + x/4 = 1 [because distance/speed = time]
=> 2x + 3x = 12
=> x = 12/5 = 2.4 km
Let the distance is x km
Rate downstream = 5 + 1 = 6 kmph
Rate upstream = 5 - 1 = 4 kmph
then
x/6 + x/4 = 1 [because distance/speed = time]
=> 2x + 3x = 12
=> x = 12/5 = 2.4 km
41209.The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes is
1.6 km
2 km
3.6 km
4 km
Explanation:
Speed downstreams =(15 + 3)kmph
= 18 kmph.
Distance travelled = (18 x 12/60)km
= 3.6km
Speed downstreams =(15 + 3)kmph
= 18 kmph.
Distance travelled = (18 x 12/60)km
= 3.6km
41210.Sahil can row 3 km against the stream in 20 minutes and he can return in 18 minutes. What is rate of current ?
1/2 km/hr
1/3 km/hr
2 km/hr
4 km/hr
Explanation:
Speed Upstream=3/(20/60)=9km/hrSpeed
Downstream=3(18/60)=10km/hr
Rate of current will be (10−9)/2=1/2km/hr
Speed Upstream=3/(20/60)=9km/hrSpeed
Downstream=3(18/60)=10km/hr
Rate of current will be (10−9)/2=1/2km/hr
41211.In how many words can be formed by using all letters of the word BHOPAL
420
520
620
720
Explanation:
Required number
=6!=6∗5∗4∗3∗2∗1=720
Required number
=6!=6∗5∗4∗3∗2∗1=720
41212.In how many way the letter of the word "APPLE" can be arranged
20
40
60
80
Explanation:
Easy way to solve this type of permutation
tion is as,
So word APPLE contains 1A, 2P, 1L and 1E
Required number =
=5!/(1!∗2!∗1!∗1!)
=(5∗4∗3∗2!)2!
=60
41213.In how many way the letter of the word "RUMOUR" can be arranged
2520
480
360
180
Explanation:
In above word, there are 2 "R" and 2 "U"
So Required number will be
=6!/(2!∗2!)
=6∗5∗4∗3∗2∗1/4
=180
In above word, there are 2 "R" and 2 "U"
So Required number will be
=6!/(2!∗2!)
=6∗5∗4∗3∗2∗1/4
=180
41214.In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still water
6
7
8
9
Explanation:
We know we can calculate it by 1/2(a+b)
=> 1/2(11+5) = 1/2(16) = 8 km/hr
We know we can calculate it by 1/2(a+b)
=> 1/2(11+5) = 1/2(16) = 8 km/hr
41215.If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream is
5 km/hr
4 km/hr
2 km/hr
1 km/hr
Explanation:
Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph
Rate upstream = (15/3) kmph
Rate downstream (21/3) kmph = 7 kmph.
Speed of stream (1/2)(7 - 5)kmph = 1 kmph
41216.A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water is
4 kmph
5 kmph
6 kmph
7 kmph
Explanation:
Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph
Rate upstream = (750/675) = 10/9 m/sec
Rate downstream (750/450) m/sec = 5/3 m/sec
Rate in still water = (1/2)*[(10/9) + (5/3)] m/sec.
= 25/18 m/sec
= (25/18)*(18/5) kmph
= 5 kmph
41217.If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of the boat in still water is
12 kmph
13 kmph
14 kmph
15 kmph
Explanation:
Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 3 kmph.
Let speed in sttil water is x km/hr
Then, speed upstream = (x —3) km/hr.
x-3 = 10 or x = 13 kmph
Rate upstream = (7/42)*60 kmh = 10 kmph.
Speed of stream = 3 kmph.
Let speed in sttil water is x km/hr
Then, speed upstream = (x —3) km/hr.
x-3 = 10 or x = 13 kmph
41218.A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat in still water and stream is
3:1
1:3
2:4
4:2
Explanation:
Let speed downstream = x kmph
Then Speed upstream = 2x kmph
So ratio will be,
(2x+x)/2 : (2x-x)/2
=> 3x/2 : x/2 => 3:1
Let speed downstream = x kmph
Then Speed upstream = 2x kmph
So ratio will be,
(2x+x)/2 : (2x-x)/2
=> 3x/2 : x/2 => 3:1
41219.A mans speed with the current is 20 kmph and speed of the current is 3 kmph. The Mans speed against the current will be
11 kmph
12 kmph
14 kmph
17 kmph
Explanation:
If you solved this question yourself, then trust me you have a all very clear with the basics of this chapter.
If not then lets solve this together.
Speed with current is 20,
speed of the man + It is speed of the current
Speed in still water = 20 - 3 = 17
Now speed against the current will be
speed of the man - speed of the current
= 17 - 3 = 14 kmph
If you solved this question yourself, then trust me you have a all very clear with the basics of this chapter.
If not then lets solve this together.
Speed with current is 20,
speed of the man + It is speed of the current
Speed in still water = 20 - 3 = 17
Now speed against the current will be
speed of the man - speed of the current
= 17 - 3 = 14 kmph
41220.A man in a train notices that he can count 41 telephone posts in one minute. If they are known to be 50 metres apart, then at what speed is the train travelling?
60 km/hr
100 km/hr
110 km/hr
120 km/hr
Explanation:
Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour
Number of gaps between 41 poles = 40
So total distance between 41 poles = 40*50
= 2000 meter = 2 km
In 1 minute train is moving 2 km/minute.
Speed in hour = 2*60 = 120 km/hour
41221.A person travels equal distances with speed of 3 km/hr, 4 km/hr and 5 km/hr and takes a total of 47 minutes. Find the total distane
3 km
4 km
6 km
9 km
Explanation:
Let the distance be 3x km,
then,
x/3+x/4+x/5=47/60
47x/60=47/60
x=1
So total distance = 3*1 = 3 Km
Let the distance be 3x km,
then,
x/3+x/4+x/5=47/60
47x/60=47/60
x=1
So total distance = 3*1 = 3 Km
41222.A walks around a circular field at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start at same point at 7:30 am. They shall first cross each other at ?
7:15 am
7:30 am
7: 42 am
7:50 am
Explanation:
Relative speed between two = 6-1 = 5 round per hour
They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.
So 7:30 + 12 mins = 7:42
Relative speed between two = 6-1 = 5 round per hour
They will cross when one round will complete with relative speed,
which is 1/5 hour = 12 mins.
So 7:30 + 12 mins = 7:42
41223.The ratio between the speeds of two trains is 7: 8. If the second train runs 400 kms in 4 hours, then the speed of the first train is ?
83.5 km/hr
84.5 km/hr
86.5 km/hr
87.5 km/hr
Explanation:
Let the speeds of two trains be 7X and 8X km/hr.
8X=400/4
=>X=12.5Km/hr
So speed of first train is 12.5*7 = 87.5 km/hr
Let the speeds of two trains be 7X and 8X km/hr.
8X=400/4
=>X=12.5Km/hr
So speed of first train is 12.5*7 = 87.5 km/hr
41224.A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream.
0,5
5,5
15,5
10,5
Explanation:
Please remember,
If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)
=> Rate in still water = 1/2(20+10) = 15 kmph
=> Rate of current = 1/2(20-10) = 5 kmph
Please remember,
If a is rate downstream and b is rate upstream
Rate in still water = 1/2(a+b)
Rate of current = 1/2(a-b)
=> Rate in still water = 1/2(20+10) = 15 kmph
=> Rate of current = 1/2(20-10) = 5 kmph
41225.Ques. A Man travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly on bicycle at 9 km/hr. What is the distance travelled on foot?
16 km
14 km
12 km
10 km
Explanation:
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr
Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4
So distance traveled on foot = 4(4) = 16 km
Let the time in which he travelled on foot = x hour
Time for travelling on bicycle = (9 - x) hr
Distance = Speed * Time, and Total distance = 61 km
So,
4x + 9(9-x) = 61
=> 5x = 20
=> x = 4
So distance traveled on foot = 4(4) = 16 km
41226.A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. Find the average speed for first 320 km of tour.
70.11 km/hr
71.11 km/hr
72.11 km/hr
73.11 km/hr
Explanation:
We know Time = Distance/speed
So total time taken =(160/64+160/80)=92hours
Time taken for 320 Km = 320∗2/9
=71.11km/hr
We know Time = Distance/speed
So total time taken =(160/64+160/80)=92hours
Time taken for 320 Km = 320∗2/9
=71.11km/hr
41227.Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
9 km/hour
10 km/hour
11 km/hour
12 km/hour
Explanation:
We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,
Let the distance travelled by x km
Time = Distance/Speed
x/10−x/15=2
[because, 2 pm - 12 noon = 2 hours]
3x−2x=60
x=60
Time=Distance/Speed
Time@10km/hr=60/10=6hours
So 2 P.M. - 6 = 8 A.M
Robert starts at 8 A.M.
He have to reach at 1 P.M. i.e, in 5 hours
So, Speed = 60/5 = 12 km/hr
We need to calculate the distance, then we can calculate the time and finally our answer.
Lets solve this,
Let the distance travelled by x km
Time = Distance/Speed
x/10−x/15=2
[because, 2 pm - 12 noon = 2 hours]
3x−2x=60
x=60
Time=Distance/Speed
Time@10km/hr=60/10=6hours
So 2 P.M. - 6 = 8 A.M
Robert starts at 8 A.M.
He have to reach at 1 P.M. i.e, in 5 hours
So, Speed = 60/5 = 12 km/hr
41228.A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
44 km/hour
46 km/hour
48 km/hour
50 km/hour
Explanation:
Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr
Average speed =2xy/(x+y)
=2∗40∗60/(40+60)
=4800/100
=48Km/hr
Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr
Average speed =2xy/(x+y)
=2∗40∗60/(40+60)
=4800/100
=48Km/hr
41229.A sum of money invested at compound interest to Rs. 800 in 3 years and to Rs 840 in 4 years. The rate on interest per annum is.
4%
5%
6%
7%
Explanation:
S.I. on Rs 800 for 1 year = 40
Rate = (100*40)/(800*1) = 5%
S.I. on Rs 800 for 1 year = 40
Rate = (100*40)/(800*1) = 5%
41230.A train covers a distance in 50 minutes, if it runs at a speed of 48kmph on an average. Find the speed at which the train must run to reduce the time of journey to 40 minutes.
50 km/hr
60 km/hr
65 km/hr
70 km/hr
Explanation:
We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.
Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km
New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph
We are having time and speed given, so first we will calculate the distance. Then we can get new speed for given time and distance.
Lets solve it.
Time = 50/60 hr = 5/6 hr
Speed = 48 mph
Distance = S*T = 48 * 5/6 = 40 km
New time will be 40 minutes so,
Time = 40/60 hr = 2/3 hr
Now we know,
Speed = Distance/Time
New speed = 40*3/2 kmph = 60kmph
41231.Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively. What time will they take to be 8.5km apart, if they walk in the same direction?
15 hours
16 hours
17 hours
18 hours
Explanation:
In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs
In this type of questions we need to get the relative speed between them,
The relative speed of the boys = 5.5kmph – 5kmph
= 0.5 kmph
Distance between them is 8.5 km
Time = Distance/Speed
Time= 8.5km / 0.5 kmph = 17 hrs
41232.Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour ?
8 minutes
10 mintues
12 minutes
14 minutes
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = (9/54) hour
= (1/6)*60 minutes
= 10 minutes
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = (9/54) hour
= (1/6)*60 minutes
= 10 minutes
41233.2 trains starting at the same time from 2 stations 200 km apart and going in opposite direction cross each other at a distance of 110 km from one of the stations. What is the ratio of their speeds ?
11:9
13:9
17:9
21:9
Explanation:
We know total distance is 200 Km
If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km]
So ratio of their speed = 110:90 = 11:9
We know total distance is 200 Km
If both trains crossed each other at a distance of 110 km then one train covered 110 km and other 90 km [110+90=200km]
So ratio of their speed = 110:90 = 11:9
41234.A man saves Rs 200 at the end of each year and lends the money at 5% compound interest. How much will it become at the end of 3 years.
Rs 662
Rs 662.01
Rs 662.02
Rs 662.03
Explanation:
=[200(21/20×21/20×21/20)+200(21/20×21/20)+200(21/20)]
=662.02
=[200(21/20×21/20×21/20)+200(21/20×21/20)+200(21/20)]
=662.02
41235.Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually
Rs 312
Rs 412
Rs 512
Rs 612
Explanation:
Please apply the formula
Amount=P(1+R/100)^n
C.I. = Amount - P
Please apply the formula
Amount=P(1+R/100)^n
C.I. = Amount - P
41236.The present worth of Rs.169 due in 2 years at 4% per annum compound interest is
Rs 155.25
Rs 156.25
Rs 157.25
Rs 158.25
Explanation:
In this type of question we apply formula
Amount=P/$(1+R/100)^n$
Amount=169$(1+4/100)^2$
Amount=169∗25∗25/26∗26
Amount=156.25
In this type of question we apply formula
Amount=P/$(1+R/100)^n$
Amount=169$(1+4/100)^2$
Amount=169∗25∗25/26∗26
Amount=156.25
41237.At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years
3%
4%
5%
6%
Explanation:
Let Rate will be R%
$1200(1+R/100)^2$=134832/100
$(1+R/100)^2$=134832/120000
$(1+R/100)^2$=11236/10000
(1+R100)=106100
=>R=6%
Let Rate will be R%
$1200(1+R/100)^2$=134832/100
$(1+R/100)^2$=134832/120000
$(1+R/100)^2$=11236/10000
(1+R100)=106100
=>R=6%
41238.The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is
4 years
5 years
6 years
7 years
Explanation:
As per question we need something like following
$P(1+R/100)^n > 2P$
$(1+20/100)^n > 2$
$(6/5)^n > 2 $
6/5×6/5×6/5×6/5 > 2
So answer is 4 years
As per question we need something like following
$P(1+R/100)^n > 2P$
$(1+20/100)^n > 2$
$(6/5)^n > 2 $
6/5×6/5×6/5×6/5 > 2
So answer is 4 years
41239.In what time will Rs.1000 become Rs.1331 at 10% per annum compounded annually
2 Years
3 Years
4 Years
5 Years
Explanation:
Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs.10%p.a.
Let the time be n years then,
1000(1+10/100)^n=1331
(11/10)^n=1331/1000
$(1110)^3$=1331/1000
So answer is 3 years
Principal = Rs.1000;
Amount = Rs.1331;
Rate = Rs.10%p.a.
Let the time be n years then,
1000(1+10/100)^n=1331
(11/10)^n=1331/1000
$(1110)^3$=1331/1000
So answer is 3 years
41240.If the simple interest on a sum of money for 2 years at 5% per annum is Rs.50, what will be the compound interest on same values
Rs.51.75
Rs 51.50
Rs 51.25
Rs 51
Explanation:
S.I.=P∗R∗T/100
P=(50∗100)/5∗2=500
Amount=500(1+5/100)^2
500(21/20∗21/20)=551.25
C.I.=551.25−500
=51.25
S.I.=P∗R∗T/100
P=(50∗100)/5∗2=500
Amount=500(1+5/100)^2
500(21/20∗21/20)=551.25
C.I.=551.25−500
=51.25
41241.The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Rs 1. Find the sum
Rs 600
Rs 625
Rs 650
Rs 675
Explanation:
Let the Sum be P
S.I. = P∗4∗2/100=2P/25
C.I. = P(1+4/100)^2−P
=(676P/625)−P
=51P/625
As, C.I. - S.I = 1
=>51P/625−2P/25=1
=>(51P−50P)625=1
P=625
Let the Sum be P
S.I. = P∗4∗2/100=2P/25
C.I. = P(1+4/100)^2−P
=(676P/625)−P
=51P/625
As, C.I. - S.I = 1
=>51P/625−2P/25=1
=>(51P−50P)625=1
P=625
41242.A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount
1000
1500
2000
2500
Explanation:
SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000
SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000
41243.In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%
8
9
10
11
Explanation:
Clue:
Firstly we need to calculate the SI with prinical 800,Time 3 years and Rate 9/2%, it will be Rs. 108
Then we can get the Time as
Time = (100*108)/(150*8) = 9
Clue:
Firstly we need to calculate the SI with prinical 800,Time 3 years and Rate 9/2%, it will be Rs. 108
Then we can get the Time as
Time = (100*108)/(150*8) = 9