@floamiya said:Find The Mines : There are how many mines in the figure and at which place.Digits are indicating the number of mines in adjacent squaresM B A - Maths By Amiyawww.facebook.com/MathsByAmiya
mines at places
P J K L M H E C
MDH funda
600*28 = 600*4 + 800*x
=> 600*24=800*x
=> x=18
@floamiya said:Find The Mines : There are how many mines in the figure and at which place.Digits are indicating the number of mines in adjacent squaresM B A - Maths By Amiyawww.facebook.com/MathsByAmiya
A,B,O,R,Q,N,I,F,G,D
@Maxray2 said:At A also, no?
Not at A. atleast one point is required there. three 1's are there. check again.
All possible two factor products are formed from the numbers 1,2,3,4, €Ś200. The number of factors out of the total obtained which are a multiple of 5 is?a) 5040 b) 7180 c) 8150 d) None
no oA
no oA
@shadowwarrior said:All possible two factor products are formed from the numbers 1,2,3,4, €Ś200. The number of factors out of the total obtained which are a multiple of 5 is?a) 5040 b) 7180 c) 8150 d) Noneno oA
C(200, 2) - C(160, 2) = 7180
Total pairs = C(200, 2)
Now, those pairs whose products is not divisible by 5 will have only those numbers which are not multiples of 5, and there are 160 (200 - 200/5) such numbers
So, no of such pairs = C(160, 2)
Hence C(200, 2) - C(160, 2) pairs will have the products as multiple of 5
@floamiya said:Find The Mines : There are how many mines in the figure and at which place.Digits are indicating the number of mines in adjacent squaresM B A - Maths By Amiyawww.facebook.com/MathsByAmiya
Total 10 mines in : A R O Q N I G F B D
@chillfactor said:C(200, 2) - C(160, 2) = 7180Total pairs = C(200, 2)Now, those pairs whose products is not divisible by 5 will have only those numbers which are not multiples of 5, and there are 160 (200 - 200/5) such numbersSo, no of such pairs = C(160, 2)Hence C(200, 2) - C(160, 2) pairs will have the products as multiple of 5
yeah........didn't read the question correctly....
i thought the question was:
All possible two factor products are formed from the numbers 1,2,3,4, €Ś200. The number of products out of the total obtained which are a multiple of 5 is?a) 5040 b) 7180 c) 8150 d) None
in this case, repeated ones(like 10 = 1*10 and 5*2) ones will have to be filtered out....
i was thinking of a way to do that....
i thought the question was:
All possible two factor products are formed from the numbers 1,2,3,4, €Ś200. The number of products out of the total obtained which are a multiple of 5 is?a) 5040 b) 7180 c) 8150 d) None
in this case, repeated ones(like 10 = 1*10 and 5*2) ones will have to be filtered out....
i was thinking of a way to do that....@12sahil12 said:A garrison of 600 men had provisions for 28 days.After 4 days, a reinforcement of 200 men arrived.The food will now last for how many days???PLease give a detailed solution.Any help to solve such ratio problems ?????????
4 DAYS + 18 DAYS => 22 days
@12sahil12 said:A garrison of 600 men had provisions for 28 days.After 4 days, a reinforcement of 200 men arrived.The food will now last for how many days???PLease give a detailed solution.Any help to solve such ratio problems ?????????
What's the OA ??
@12sahil12 said:A garrison of 600 men had provisions for 28 days.After 4 days, a reinforcement of 200 men arrived.The food will now last for how many days???PLease give a detailed solution.Any help to solve such ratio problems ?????????
18 days
600*28-600*4/800=18
@12sahil12 said:A garrison of 600 men had provisions for 28 days.After 4 days, a reinforcement of 200 men arrived.The food will now last for how many days???PLease give a detailed solution.Any help to solve such ratio problems ?????????
Suppose one man consumes 1 unit per day.
Hence, in 28 days 600 men would consume 600*28= 16800 units
So units consumed in first four days= 600*4 = 2400 units
Remainig provision= 16800-2400 = 14400 units.
This should suffice 800 men for (14400/800)= 18 days :)
@adwaitjw said:Suppose one man consumes 1 unit per day.Hence, in 28 days 600 men would consume 600*28= 16800 unitsSo units consumed in first four days= 600*4 = 2400 unitsRemainig provision= 16800-2400 = 14400 units.This should suffice 800 men for (14400/800)= 18 days
@adwaitjw ane tips for solving such problems..I get confused and den get a wrong solution which matches 1 of the options as well..:P
and do temme best Books for DI/LR/verbal ...????????????????????????
and do temme best Books for DI/LR/verbal ...????????????????????????
@shadowwarrior said:All possible two factor products are formed from the numbers 1,2,3,4, €Ś200. The number of factors out of the total obtained which are a multiple of 5 is?a) 5040 b) 7180 c) 8150 d) Noneno oA
sol:- there are 40 multiples of 5(1 to 200)
so...40*160 + 40*39/2
= 6400 + 780 = >7180
There are 30 number (1,2,3 €Ś30). In how many ways three of them can be selected such that sum of the selected numbers is multiple of three.
Could solve this Q during CAT time. no more :(
Approach plz !
Could solve this Q during CAT time. no more :(
Approach plz !