Discussion Room : Quant
Tanu21 last edited by
4 dice are rolled, what is the probability that sum of numbers on 4 dice is 18?
For a regular hexagon,
Diagonal = 2L and W = √3 L
So in our diagram,
BC = xz = 12
Bx = Cz = 6√3
Shaded area = Area of Rectangle (BxzC) - Area of Right Triangle Czy
= 12 * 6√3 - 1/2 * 6 * 6√3
= 72√3 - 18√3
What is the OA ?
a + b + c + d = 18
Number of ways = (n - 1) C (k - 1) = 18C3 = 816 (If a, b, c and d can be any natural number)
But in our case a, b, c and d can only be natural numbers less than 6.
So cases like 12 + 2 + 2 + 2 = 18 should be removed.
Say a = A + 6 (To find cases where a is > 6)
A + 6 + b + c + d = 18
A + b + c + d = 12
Number of ways = 11C3 = 165
similarly for b, c and d.
So we have 816 - 4 * 165 = 156 favourable cases.
What's the OA ? Not sure about the calculations.. logic should be as explained.
This concept and various question types are explained in detailed in the below link - Please have a look, you won't have any issues with similar questions :)
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