120 rupees is the original price
25 rupees down payment means 95 rupees is what we are "Borrowing" from the shop.
Amount paid in instalments = 25 * 4 = 100
So 100 - 95 = 5 is the interest
Now this interest gained (Rs 5) is through the interest gained in the available amount month by month.
5 = 95R/(12 * 100) + 70R/(12 * 100) + 45R/(12 * 100) + 20R/(12 * 100)
5 = 230R/(12 * 100)
R = 6000/230 = 26.09%
@Ishita 120 is correct. Can you share your detailed approach so that others will benefit :)
@Ishita 22.5 is the right answer.
A does 3x and B does x amount of work in a given amount of time.
as work is inversely proportional to the time taken, A takes t days and B takes 3t days to complete a given amount of work.
we know 3t - t = 60 days
t = 30 days
So A takes 30 days and B takes 30 + 60 = 90 days
Together, 1/30 + 1/90 = 4/90 of work will be completed and it will take 90/4 = 22.5 days to finish the work together.
If two objects A and B start simultaneously from opposite points and, after meeting, reach their destinations in ‘a’ and ‘b’ hours respectively (i.e. A takes ‘a hrs’ to travel from the meeting point to his destination and B takes ‘b hrs’ to travel from the meeting point to his destination), then the ratio of their speeds is given by:
Sa/Sb = √(b/a)
i.e. Ratio of speeds is given by the square root of the inverse ratio of time taken.
Two trains A and B starting from two points and travelling in opposite directions, reach their destinations 9 hours and 4 hours respectively after meeting each other. If the train A travels at 80kmph, find the rate at which the train B runs.
Sa/Sb = √(4/9) = 2/3
This gives us that the ratio of the speed of A : speed of B as 2 : 3.
Since speed of A is 80 kmph, speed of B must be 80 * (3/2) = 120 kmph
A and B start from Opladen and Cologne respectively at the same time and travel towards each other at constant speeds along the same route. After meeting at a point between Opladen and Cologne, A and B proceed to their destinations of Cologne and Opladen respectively. A reaches Cologne 40 minutes after the two meet and B reaches Opladen 90 minutes after their meeting. How long did A take to cover the distance between Opladen and Cologne?
Sa/Sb = sqrt(90/40) = 3/2
This gives us that the ratio of the speed of A : speed of B as 3:2. We know that time taken is inversely proportional to speed. If ratio of speed of A and B is 3:2, the time taken to travel the same distance will be in the ratio 2:3. Therefore, since B takes 90 mins to travel from the meeting point to Opladen, A must have taken 60 (= 90*2/3) mins to travel from Opladen to the meeting point
So time taken by A to travel from Opladen to Cologne must be 60 + 40 mins = 1 hr 40 mins
[Credits: Veritasprep]
@vinaycat2017
Let the total distance be d
truck started it journey with 100 kmph and continued so for distance d/3
At d/3 it unloaded 150 cartons, so speed is increased by 20 + 20 + 400/100 = 44% (using successive percentage change formula)
So d/3 distance covered at 100 kmph
next d/3 distance covered at 144 kmph
and last d/3 distance covered at 172 kmph
average speed = 3abc/(ab + bc + ca) = 132 (approx)
d = 132 * 9.82 = 1296 km (approx)
@deepalis727
there are 299 numbers between 100 and 400 (excluding 100 and 400)
Out of 299 numbers [299/2] = 149 numbers are divisible by 2
Out of remaining 150 numbers [150/3] = 50 numbers are divisible by 3
Out of remaining 100 numbers [100/5] = 20 numbers are divisible by 5
Out of remaining 80 numbers [80/7] = 11 numbers are divisible by 7
so 149 + 50 + 20 + 11 = 230 numbers are divisible by 2, 3, 5 or 7
If it is including 100 and 400, then we have 232 numbers which are divisible by 2, 3, 5 or 7
@deepalis727 Cannot be determined due to insufficient data
@manaswini Rohan's speed is 9 times than sohan.. so when Sohan completes 3 rounds, Rohan completes 3 * 9 = 27 rounds. 27 times ?
Funda : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a – b|
A and B will meet at |1 - 2| = 1 point.
A and C will meet at |1 - 3| = 2 points
A and D will meet at |1 - 4| = 3 points
A and E will meet at |1 - 5| = 4 points
A and F will meet at |1 - 6| = 5 points
A and G will meet at |1 - 7| = 6 points
So A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags.
similarly B and C will meet at |2 - 3| = 1 point
B and D will meet at |2 - 4| = 2 points
B and E will meet at |2 - 5| = 3 points
B and F will meet at |2 - 6| = 4 points
B and G will meet at |2 - 7| = 5 points
So B will put 1 + 2 + 3 + 4 + 5 = 15 flags
Similarly find for C, D, E and F.
We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags
Let the distance PQ be d and speed of river be R and speed of boat be B
d/(B + R) = 4
d/(B - R) = 6
Solve for B.
4(B + R) = 6(B - R)
4B + 4R = 6B - 6R
10R = 2B
R = B/5
d/(B + B/5) = d/(6B/5) = 5d/6B = 4
d = 24B/5
If we put d = 24 units, then B = 5 and R = 1
So first hour boat will cover 6 units, second hour boat will cover 5 * 1.4 + 1 = 8 units and third hour 10 units.
So 24 units will be covered in 3 hours
