Discussion Room : Quant
Supriya Shinde last edited by zabeer
x/|x| < x . Which of the following must be true about x ?
a. x > 1
b. x > -1
c. |x| < 1
d. |x| = 1
e. |x|^2 > 1
x/|x| is 1 when x > 0
x/|x| is -1 when x < 0
x/|x| < x is not possible when x > 0
now if 0 > x > -1 then x/|x| < x
example, take x = -0.3, -0.3/|-0.3| = -1 < -0.3
So every x which satisfies x/|x| < x should have x > -1
Is OA B ?
@vinaycat2017 please share the steps so that we can continue from the point you got stuck. will benefit you better this way :)
Let side of regular octagon be a.
AB = a + 2x and BC = a + 2y
a + 2x = 22
a + 2y = 26
22 - 2x = 26 - 2y
2x = 2y - 4
x = y - 2
a = 26 - 2y
x^2 + y^2 = a^2
(y - 2)^2 + y^2 = (26 - 2y)^2
solve for y and substitute to get a.
Sir , Let centre of Triangle PQR be 'O'. Considering triangle ORF OR=Sqrt(144-36)=6sqrt 3.
Now cant proceed further...................
Pulkit Sancheti last edited by
Shyama and Ramesh walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Ramesh’s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Ramesh (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
Speed of escalator = x steps/sec
Speed of Shyama = (3 + x) steps/sec
Speed of Ramesh = (2 + x) steps/sec
Time taken by Shyama = 25/3
Time taken by Ramesh = 20/2 = 10
Distance covered is same
So, (3 + x) * 25/3 = (2 + x) * 10
x = 3 steps/sec
total number of steps (distance covered) = 50
So it would take 50 steps if the escalator was off. (OA matching with this?)
@Pulkit-Sancheti instead of tackling this one question, we will finish the topic itself :)
Please go through below concept notes covering the fundas in escalator problems
Collection of 30 questions from escalator topic - https://www.mbatious.com/topic/933/cat-question-bank-escalators-time-speed-distance
If you find any concept tough to understand, please let us know.. will discuss!
Supriya Shinde last edited by
For a certain integer x, the units digit of (x+2)^2 is 9. Which of the following could be the units digit of |x+1|?
if (x + 2)^2 ends with 9, then x should be 1, 5, -5 or -9.
4 is a possible value for |x + 1| when x = -5
vinaycat2017 last edited by zabeer
The investments made by X and Y are in the ratio 3 :2.
If 5% of total profit is donated and X gets & 8,550 as his share of profit then what is the amount of total profit.
(d) 12, 020
Total profit = 100P
X's share = 3/5 * 95P = 8550
P = 8550 * 5 / (3 * 95) = 150
Total Profit = 100P = 15000.
deepalis727 last edited by
Find the number of number between 100 to 400 which are divisible. by either 2,3,5,7
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
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
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)
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?
D) None of these