abhinayrao91 Saysprofit and loss ...lod 2...question 50 plz?
plz write ques here.....bcoz i dnt hv tht book rite nw.....thnks in adv
plz write ques here.....bcoz i dnt hv tht book rite nw.....thnks in adv
gudda1122 Saysyaar just calc prop i hv done this this twice...thts why i m damm sure ki ans 8 hi h
In fact you are correct. I did a silly mistake and got it wrong. Yes the answer is 8. Case closed! 😉
There are 5 numbers.The H.C.F of each pair is 2 and the L.C.M. of the 5 numbers is 420.What is the product of the 5 numbers?
a)840 b)1680 c)6720 d)789 e)can't be determined
There are 5 numbers.The H.C.F of each pair is 2 and the L.C.M. of the 5 numbers is 420.What is the product of the 5 numbers?
a)840 b)1680 c)6720 d)789 e)can't be determined
i think ans of this ques is c)6720///i m nt sure...nor i can see the ans sheet 2 confirm...so just let me knw if this is wrong........thnks
Arun Sharma Probability LOD 1, Q. 37
The probability that a student will pass in Mathematics is 3/5 and the probability that he will pass in English is 1/3. If the probability that he will pass in both Mathematics and English is 1/8, what is the probability that he will pass in at least one subject?
Solution:
P(M U E) = P(M) + P(E) - P(M intersection E)
= 3/5 + 1/3 - 1/8
= 97/120 (which is the correct answer)
Doubt:
Why is this approach incorrect?
P(atleast pass in one subject) = +
= +
= +
= 79/120
Is this because they are independent events? I don't know...
There are 5 numbers.The H.C.F of each pair is 2 and the L.C.M. of the 5 numbers is 420.What is the product of the 5 numbers?
a)840 b)1680 c)6720 d)789 e)can't be determined
The H.C.F of pair of each number being 2 means that the H.C.F of all the numbers is 2 only.
Going by the formula hcf*lcm=product of the numbers
2*420=840
Ans= 840
The H.C.F of pair of each number being 2 means that the H.C.F of all the numbers is 2 only.
Going by the formula hcf*lcm=product of the numbers
2*420=840
Ans= 840
dude.....ur formula is correct bt there is sligth mistake in tht...
formula sayslcm*(hcf)^n-1=product
where n is the no of digit used...what u hv applied is correct for 2 digts but nt for 5 digts
dude.....ur formula is correct bt there is sligth mistake in tht...
formula sayslcm*(hcf)^n-1=product
where n is the no of digit used...what u hv applied is correct for 2 digts but nt for 5 digts
Oops sorry didnt notice it..i wonder where my mind was wandering while posting it..:splat:
vish1206 SaysOops sorry didnt notice it..i wonder where my mind was wandering while posting it..:splat:
waise thnks yaar bcoz i solved this ques wid reasoning......i didn't remeber formula...bt due 2 u i hv revised tht......thnks
gudda1122 Sayswaise thnks yaar bcoz i solved this ques wid reasoning......i didn't remeber formula...bt due 2 u i hv revised tht......thnks
Can u please share the process
vish1206 SaysCan u please share the process
kkk...reasoning as follows....
1)it is being said tht hcf of every pair is 2...hence every num will contain 2....
2)we have given tht lcm is 420..when v factorize it we will se tht it can b written as 5*7*3*2^2
3)this means there is 5,7,3....once in the num...we r nt sure tht in which num wud contain tht....bt we r sure tht they will b only once...bt we hv 2^2 in lcm...this means higest power of 2 in all the nums is 2....bt it is given tht we hv hcf as 2...this means only one num will hv 2^2.
4)also hcf is 2...therefore 3,5,7 will b there only once...else hcf wuld hv been changed....then we just mul all the nums.....
i hope i hv cleared it
bcoz i m vry bad in this...
Ten books are arranged in a row on a bookshelf. A student has to select three out of these ten books
in such a way that no two books selected by him must have been lying adjacently. In how many
ways can he make the selection?
....please someone explain it in detail.
Ten books are arranged in a row on a bookshelf. A student has to select three out of these ten books
in such a way that no two books selected by him must have been lying adjacently. In how many
ways can he make the selection?
....please someone explain it in detail.
Say, arrangement is like this:
a XX b XX c XX d
Where, XX represents a book to be selected and a, b, c and d represent the number of books at those positions.
So, a+b+c+d = 7
Now, a and d can be 0 but b and c must be at least 1.
So, a+b'+c'+d = 5 ... b' = b+1 and c' = c+1
Now, a+b'+c'+d = 5 can be done in 8C3 ways.
So, totally we have 8C3 (i.e. 56) ways to do this.
ur ans is correct.pls show me d process.
rahul005 Saysur ans is correct.pls show me d process.
i have done tht on previous page........
thnks this will give me alot of confidence
Arun Sharma Probability LOD 1, Q. 37
The probability that a student will pass in Mathematics is 3/5 and the probability that he will pass in English is 1/3. If the probability that he will pass in both Mathematics and English is 1/8, what is the probability that he will pass in at least one subject?
Solution:
P(M U E) = P(M) + P(E) - P(M intersection E)
= 3/5 + 1/3 - 1/8
= 97/120 (which is the correct answer)
Doubt:
Why is this approach incorrect?
P(atleast pass in one subject) = +
= +
= +
= 79/120
Is this because they are independent events? I don't know...
Anybody have any idea?
A+B+C=12...
G1-G2=9..9 would appear if the units tens digit are reverse..and consecutive..
1,5,6
Ans:7
Couldn't understand.. Can you please explain with some more deatailed explanation !!
Thanks
Hi puys..
My doubt is from Geometry(Pg-360)
A circle with radius 2 is placed against a right angle.Another circle is placed in the gap between the circle and the right angle.what is the radius of the smaller circle?
Although the solution is printed in the book...but found the construction part confusing.
I haven't read the solution from the book..
But, the construction would be that the tip of the right angle would have been connected to the centres..
Let the small radius be r
So, r*sq_root(2) + r + 2 = 2*sq_root(2)
Solving for r, we get r equal to 6-4*sq_root(2)
A no. when divided by11 gives remainder 3,when divided by 16 gived remainder 4 but when divided by 11 gives no remainder.What is d least such no?
pls show d process
A no. when divided by11 gives remainder 3,when divided by 16 gived remainder 4 but when divided by 11 gives no remainder.What is d least such no?
pls show d process
dude i cant give u whole solu here........bt i m writeing some hints here...i hope they will work 4 u......
the smallest num will be
k(lcm of all the numbers(16,11))+smallest number
i think u hv done some typeing mistake.........