Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?
Anil in one day can do th of the work
Sunil in one day can do th of the work
Anil starts the job and Sunil joins him after three days.
So, Anil would have done th of the work by the time Sunil joins
After Sunil joins, they both would be doing th of work everyday
Now, it is known that Bimal joins them after some days and finishes 10 % of the work.
Now, Anil alone had done th of the work and Bimal completes th of the work
So, in total they would have done = th of the work (is the work done by Bimal)
Remaining work = , which would be done by Anil and Sunil in 10 days.
Combining everything,
Total number of days = 3 + 10 = 13 days
In an examination, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali’s score exceeded Rama’s score by
It is given that the scores of Anjali, Mohan and Rama after review were in the ratio 11: 10: 3
So, let their values be 11x, 10x and 3x respectively.
It is known that their score increased by 6 after review.
So, scores before review = 11x-6, 10x-6 and 3x-6 respectively
Now, from the data given
(11x - 6 + 10x - 6) x = 3x - 6
21x - 12 = 36x - 72
60 = 15x
x = 4
So, marks after revision are 44, 40 and 12 respectively.
Therefore, Anjali’s score exceeded Rama’s by 44 - 12 = 32 marks
In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was [TITA]
From the data,
A scored 72
A’s score was 10% less than B
So, Score of B = 80
We know that B was 25% more than C
So, C x = 80
C = 64
Now, we know that D scored 20% less than D
So, C = x D
64 = x D
D = 80 marks
The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is
It is given that the diagram is a square pyramid and all 4 sides of the triangle are of equal sides.
Now, we need to plug in a right triangle and apply Pythagoras theorem to find the vertical height of the triangle.
Draw a vertical from the apex of the pyramid and draw a line to the base along the equilateral triangle. Connect the points in the base to form a right triangle.
The hypotenuse would be the slant height of the pyramid, which is the altitude of the equilateral triangle.
Length of the base = 10 cms
Hypotenuse = Altitude of the equilateral triangle = x 20 = 10cms
Therefore, (Vertical height)2 = (10)2 - 102 = 200
Vertical height = = 10cms
Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at
By the time A and B meet for the first time, A covers 60% of the distance, while B covers 40% of the distance.
So, the speeds of A and b are in the ratio 60:40 or 3:2
Hence, the time they take to cover a particular distance will be in the ratio 2:3
We know that A covers 60% of the distance at 10:00 AM and covers 100% of the distance at 10:12 AM.
That means A takes 12 minutes to cover 40% of the track. So to cover the entire track he must have taken 12+12+6 = 30 minutes.
(because 40% + 40% + 20% = 100%)
Since the time taken by A and B to complete the track are in the ratio 2:3, the time taken by B to complete the track will be 45 minutes.
At 10:00 AM, B has covered 40% of the track. If we can find out what time does B take to complete the remaining 60% of the track, we can find the finish time of B.
Time required to complete 60% of the track = 60% of 45 = 27 minutes.
Hence, B complete one single round at 10:27 AM.
105 = (n+m) (n-m)
105 can be written as 3 x 5 x 7, which can be written in 4 ways
(1 x 105), (3 x 35), (5 x 21), (7 x 15) can be the possible values of m and n respectively
4 pairs
The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh’s salary during this period is closest to
The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6: 5: 7 in 2010
In 2015, their salaries are in the ratio 3: 4: 3 respectively.
It is also given that; Ramesh’s salary increases by 25% during 2010 - 2015
So, Ramesh’s salary in 2015 = x 6 = 7.5
Salary in 2015 = 3: 4: 3 (Given)
Salary in 2015 = 7.5: x: y (From data)
3 in order to jump to 7.5, must be multiplied by 2.5
So, multiplying 2.5 to all the other values,
Salary in 2015 = 7.5: 10: 7.5
Percentage increase in Rajesh’s salary during 2010 - 2015 = Percentage increase from 7 to 7.5 = x 100, which is close to 7%
A man makes complete use of 405 cc of iron, 783 cc of aluminium, and 351 cc of copper to make a number of solid right circular cylinders of each type of metal. These cylinders have the same volume and each of these has radius 3 cm. If the total number of cylinders is to be kept at a minimum, then the total surface area of all these cylinders, in sq cm, is
Firstly, the three materials are not reduced, mixed and then moulded.
Instead a cylinder is formed with one material and one material only.
The base area of the cylinders to be casted is the same (because the base radius is the same)…
Let’s us imagine that each sphere is casted into a single cylinder,
In this case we will have 3 cylinders, each made up of a different material and having different heights.
So, the heights of these cylinders will be
But we need to have cylinders with equal volume.
Since the base area is the same, this translates to the condition that the cylinders should have the same height. And they should be as high as possible, because the question also wants us to have as less number of cylinders as possible…
So, we need the height of each cylinder to be HCF of
this implies there will be a total of of 15+29+13 = 57 cylinders.
The total surface area of one such Cylinder is
The TSA of all the 57 cylinders is 57*18(π+1)18(π+1) = 1026(π+1)1026(π+1).