Let there be x coins of 20 paise and y coins of 50 paise
x+y=90
20x+50y=3690
Solving, you get
x=27 y=63
Option (C)
Let cost price of cow be x
So, that of calf is (1300-x)
Selling price of cow= 1.25x
Selling price of calf= 1.2(1300-x)
Total selling price was 1300+1300*300/1300= 1600
So,
1.25x-1.2x+1560=1600
0.05x=40
x=800
Option (E)
Please post your questions in a single post next time onwards.
with 90 coins value per coin wll be 41 paisa so make combintn of 20 and 50 paisa for 41 with allegatn ratio will cm 9:21 so actual will be 27 and 63......as it is 90.
A naughty bird is sitting on top of a car .It sees another car approaching it at a distance of 12 km .The speed of two car is 60 kmph each .The bird starts flying from the first car and moves towards the second car ,reaches the second car and comes back to the irst car and o on .If the speed at which the bird flies is 120kmph then answer thefollowing ques .Assume that two cars have a crash Q1 the total distance travelled by the birdbefore the crash is a 6 b 12 c 18 d none of these Q2 the total distance travelled by the bird before it reaches the second car for the second time is a 10.55 b 11.55 c 12.33 d none of these Q3 the total number of times that the bird reaches the bonnet of the second car is????? a 12 times b 18 times c infinite times d cannot be determined
A naughty bird is sitting on top of a car .It sees another car approaching it at a distance of 12 km .The speed of two car is 60 kmph each .The bird starts flying from the first car and moves towards the second car ,reaches the second car and comes back to the irst car and o on .If the speed at which the bird flies is 120kmph then answer thefollowing ques .Assume that two cars have a crash Q1 the total distance travelled by the birdbefore the crash is a 6 b 12 c 18 d none of these Q2 the total distance travelled by the third bird before it reaches the second car for the second time is a 10.55 b 11.55 c 12.33 d none of these Q3 the total number of times that the bird reaches the bonnet of the second car is????? a 12 times b 18 times c infinite times d cannot be determined
pla mention the solution
is answer for Q1. (c) ? i.e. 18 ???
for q2. ..the total distance travelled by the third bird ...........
A naughty bird is sitting on top of a car .It sees another car approaching it at a distance of 12 km .The speed of two car is 60 kmph each .The bird starts flying from the first car and moves towards the second car ,reaches the second car and comes back to the irst car and o on .If the speed at which the bird flies is 120kmph then answer thefollowing ques .Assume that two cars have a crash Q1 the total distance travelled by the birdbefore the crash is a 6 b 12 c 18 d none of these Q2 the total distance travelled by the third bird before it reaches the second car for the second time is a 10.55 b 11.55 c 12.33 d none of these Q3 the total number of times that the bird reaches the bonnet of the second car is????? a 12 times b 18 times c infinite times d cannot be determined
pla mention the solution
For our conviniance let us assume its a scooter and a car
distance is 12km... now they both will crash in 12/120=1/10 hour 1) so in 1/10 hour the bird would have travelled d=speed x time=120x1/10=12km
2)now initially the distance is 12 km..so first time add the speed of bird and car and distance remains the same. 1->t=12/180=1/15 hour (time to reach car from scooter for first time)
now from car to scooter in 1/15 of an hour the distance between scooter and car would have reduced by speedxtime=120x1/15=8km... so 4 km is the distance between scooter and car so t=4/180=1/45 hour
now bird reaches car for second time from scooter in 1/45 hour distance would have reduced by =>(1/45)x120=8/3km.. so distance remining in 4-8/3=4/3km. t=(4/3)/180=1/135 hour
total time taken = 1/15+1/45+1/135=13/135hour in 13/135 hour bird travels a total distance of (13/135)x120=11.55km
A dog sees a cat .It estimates that the cat is 25 leaps away .The cat sees the dog and starts running with the dog in hot pursuit. If in every minute , the dog makes 5 leap and the cat makes 6 leaps and one leap of the dog is equal to 2 leaps of the cat .Find the time in which the cat is caught by the dog ???
A dog sees a cat .It estimates that the cat is 25 leaps away .The cat sees the dog and starts running with the dog in hot pursuit. If in every minute , the dog makes 5 leap and the cat makes 6 leaps and one leap of the dog is equal to 2 leaps of the cat .Find the time in which the cat is caught by the dog ???
This one's from PnC : - I am not able to understand the solution - Can someone please explain: The no of circles that can be drawn out of 10 points of which 7 r collinear. ? Answers :1)130 2)85 3)45 4)cannot be determined
This one's from PnC : - I am not able to understand the solution - Can someone please explain: The no of circles that can be drawn out of 10 points of which 7 r collinear. ? Answers :1)130 2)85 3)45 4)cannot be determined
A circle can be dran through any three non-collinear points... So ur task is to select 3 points out of 10 points(ncr rule) - =>10c3..
But 7 points are collinear so u have considered those collinear points as well.. so subtract them 10c3-7c3
or u know to draw a circle u need three non-collinear points.. That is circle passes through 3 non-collinear points..
so, case 1 - from 3 non-collinear points u can draw 3c3=1 circle case2-from 2 collinear points and 1 non-collinear point-7c2x3c1=63 case3-from 1 collinear and 2 non collinear points=7c1x3c2=21
A number is such that when divided by 4,5,6 or 7 it leaves the remainder 2,3,4, or 5 respectively. Which is the largest number below 4000 that satisfies this property?
a)3358 b)3778 c)3898 d)2938
Please provide a stepwise solution, if possible, Thanks
A number is such that when divided by 4,5,6 or 7 it leaves the remainder 2,3,4, or 5 respectively. Which is the largest number below 4000 that satisfies this property?
a)3358 b)3778 c)3898 d)2938
Please provide a stepwise solution, if possible, Thanks
Simple, take the LCM of 4, 5, 6 and 7 and deduct 2 from it (since dividing by 4 leaves 2, by 5 leaves 3 etc. Always a difference of 2) to get the initial number.
So the initial figure is 420 - 2 = 418
To get the next numbers, just keep adding 420. So the highest number less than 4000 would be 3778 = 418 + 420*8
Three sprinters A,Band C has to sprint from points P to Q and back again ( starting in that order).The time interval between between the starting times of the three sprinters A,B and C was 5 sec each .Thus C started 10 seconds after A while B started 5 seconds after A .The three sprinters passed a certain point R , which is somewhere between P and Q , simultaneously (none of them having reached point Q yet ) Having reached Q and reversed the direction , the third sprinter met the second one 9 m short of Q and met the first sprinter 15 m short of Q .Find the speed of the first sprinter if the distance between PQ is equal to 55m
a 4m/s b 3 m/s c 2 m/s d 1 ms e none of these
PS : i m able to get two equations as per the conditions given , but not able to get how to get ans out of that
34Sol: Length of Pool = l Ratio of speeds = (l-18.5)/18.5 Next time meeting point distances = 2l-10.5 and l+10.5 Time taken is same. So, (l-18.5)/18.5=(2l-10.5)/(l+10.5) =>37l-194.25=l^2-8l-194.25 =>l^2-45l=0 =>l=45m!
33rd, I tried but couldn't arrive at a solution :(
The length of the pool would be = 3*18.5-10.5 = 45 m
Three sprinters A,Band C has to sprint from points P to Q and back again ( starting in that order).The time interval between between the starting times of the three sprinters A,B and C was 5 sec each .Thus C started 10 seconds after A while B started 5 seconds after A .The three sprinters passed a certain point R , which is somewhere between P and Q , simultaneously (none of them having reached point Q yet ) Having reached Q and reversed the direction , the third sprinter met the second one 9 m short of Q and met the first sprinter 15 m short of Q .Find the speed of the first sprinter if the distance between PQ is equal to 55m
a 4m/s b 3 m/s c 2 m/s d 1 ms e none of these
PS : i m able to get two equations as per the conditions given , but not able to get how to get ans out of that
let the speed be Va,Vb and Vc
55/Vc=46/Vb + 5 55/Vc =40/Va + 1
lets assume speeds be va,vb,vc here since it was given that vc met b on way back to q from p , the distance it travels is 55+ 9 so (55+9)/vc=46/vb+5 similarly (55+15)/vc=40/va+10 i think there is some critical data missing since i can only get a reln btwn speeds of a and b only.
actually i gat a solution.. answer is 0.5 m/s for A.. 2/3m/s for B ans 1m/s for C
for the above question.... if v assume they meet at 'd' distance from P. Its clear that C is the fastest.. so assuming time taken for A to reach that point to be 't'. then time for B is 't-5' and for c is 't-10'.. now speed of a is (d/t), B is d/(t-5), and c is d/(t-10). and let point at which they met is L. let third sprinter met the second one at 'y' distance from L and first one at 'x' distance from L. x/(d/t)= (x+30)/(d/(t-10)) y/(d/(t-5))= (y+1/(d/(t-10)) by solving above equations.. 5y=6x and d+x+15=55 d+y+9=55 solving gives y= 6+x hence y=36m x=30m t= 20sec d= 10m speed of A=d/t=0.5m/s B=d/(t-5)=0.67m/s C=d/(t-10)=1m/s
for the above question.... if v assume they meet at 'd' distance from P. Its clear that C is the fastest.. so assuming time taken for A to reach that point to be 't'. then time for B is 't-5' and for c is 't-10'.. now speed of a is (d/t), B is d/(t-5), and c is d/(t-10). and let point at which they met is L. let third sprinter met the second one at 'y' distance from L and first one at 'x' distance from L. x/(d/t)= (x+30)/(d/(t-10)) y/(d/(t-5))= (y+1/(d/(t-10)) by solving above equations.. 5y=6x and d+x+15=55 d+y+9=55 solving gives y= 6+x hence y=36m x=30m t= 20sec d= 10m speed of A=d/t=0.5m/s B=d/(t-5)=0.67m/s C=d/(t-10)=1m/s
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/(d/(t-10))