Discussion Room : Quant


Find the number of number between 100 to 400 which are divisible. by either 2,3,5,7

@deepalis727
According to me
numbers divisible only by 2 i.e 100 102 104 ... 400
numbers divisible only by 3 Now we are left with only odd numbers i.e 101 103 105 ... 399 we'll check for only multiples of 3 here, because even multiples of 3 have been taken cared by 2 the numbers will be 105 111 ... 399
numbers divisible only by 5 The numbers left will be 101 103 107 109 113 115 ... 395 We'll check for only multiples of 5 115 125 ... 395
numbers divisible only by 7 Now the numbers left we'll be 101 103 ... 397 And once again we'll check for only multiples of 7
The sum of multiples individually checked in 1, 2, 3, and 4 will give the answer


@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

@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)

@zabeer
Two trains start simultaneously from A and B towards B and A respectively at 8 a.m. They cross each other at 12 noon. Train starting from B, thereafter, takes 6 hrs to reach A. On a particular day the train starting from A, reduced its speed and arrived at B, 200 min late. At what time did the trains cross each other on that day?
A)1:00 pm
B)2:00 pm
C)1:30 pm
D) None of these


@zabeer
Find the volume of the solid obtained when the region bounded by y = √x, y = −x and the line x = 9 is revolved around the xaxis.Sir, according to me that will be a cone having radius=6 unit and height=9 unit. So the required volume will be 108π. But tell me the right approach....


A steel container is sold at Rs 120 cash or Rs 25 as cash down payment and Rs.25 a month for 4 months .The rate of interest per annum charged under the installment plan is .
(a) 26.09%
(b) 24.09%
(c) 20 %
(d) 23.25%Sir i have tried this way. Worth of Rs 95 after 4 months = 25 * [100 * 4+(R * 4 * 3)/100]
Please guide me

A,B and C start running along the circumference of a circle(circumference=1200 m) at the same time from the same point.A and B run in the clockwise direction with constant speeds of 8 m/s and 6 m/s respectively.C runs in the anticlockwise direction with a constant speed of 5 m/s.What is the time difference between the first two instances when C is equidistant from A and B along the circumference?
a)120 seconds b) 50 secs c)60 secs d) 100 secs
What's the procedure to solve this question?

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%

Working together B and C take 50% more number of days than A, B and C together take and A and B working together, take 8/3 more number of days than A, B and C take together. If A, B and C all have worked together till the completion of the work and B has received Rs. 120 out of total earnings of Rs, 450, then in how many days did A, B and C together complete the whole work?