thanku jha, for replying to it...and forgive me if i ask some stupid question..but dont u think 65% and 35% add up to 100% of customers..and also if 65% used store coupon and 35% used manufacturer's coupon ..there would be no space left for any person using neither of these two coupons?
Please clarify
Yar both the circles are not disjoint. They have 15% of overlap as u will c once u draw it. It is like a+x= 65, and b+x = 35. Then a+b+x+n=100. where n is neither outside the circles. Tell me if u need ne thing more.
thanku jha, for replying to it...and forgive me if i ask some stupid question..but dont u think 65% and 35% add up to 100% of customers..and also if 65% used store coupon and 35% used manufacturer's coupon ..there would be no space left for any person using neither of these two coupons?
Please clarify
That's the exact point being tested here.Do not assume that every customer used a coupon for sure.read the statements to confirm whether there were any who didnt use any coupon???Then u can see whether the 2 sets are overlapping too.
1. Customers can use a manufacturer's coupon and a store coupon to obtain a discount when buying soap powder in a certain store. In one week, 65 percent of customers used the store coupon when purchasing the soap powder, and 35 percent used the manufacturer's coupon. What percent of customers used both the manufacturer's and the store coupon when purchasing the soap powder?
a) 15 percent of customers used neither coupon when purchasing the soap powder. b) 50 pecent of customers used the store coupon but not the manufacturer's coupon when purchasing the soap powder.
I am stuck with this question...need ur inputs guys..
jha16june Says
Hey answer is C. It is a simple application of V'diagram. I m sure u can solve it now.
hey jha...y shud answer be C..i am sure u r confused with answer option..each statement is individually sufficient to answer the question..Ans D..
st 1 tells u ...15% of customers use both..
st 2 tells that out of 65, 50 use only store coupon or the remaining 15% used manufacturers coupon as well..sufficient...
Total customers - 100% store coupon - 65% manufacture coupon - 35% both - x% none - n%
ie 100 = 65+35 - x + n
(1) => n = 15 . This gives x = 15
(2) => 65 - x = 50 => x = 15.
Hope things are clear.
My pick is D.
from the question we have n(s)=65,n(m)=35. we want n(s n m)? n(s n m) = n(s) + n(m) -n(s U m) is the formula a) n(s U m)' = 15 ===> n(s U m) =85 therefore we can calculate n(s n m)
b) n(s - m) = 50% ==> n(s n m) = n(s) - n(s - m) = 65 - 50 =15 hence sufficient. Hope that helps..
equation m-n = 11x implies that only 11x is an integer; x can have a value 10/11 which makes 11x an integer.
I have picked up this question from kaplan800..but not convinced with its answer and explanation.... i think, question is asking whether x is an integer or not. but in the book, they have solved to justify whether m-n = 11x or not.
per my understanding of the question answer has to be (D), but as per book it is (C).
what do u think abt the ques...what is the real question: is x an integer? or is m-n=11x?
yes I too got ans as C.
1. m =aa where a is a digit. eg. m=11 and n could be any 2 digit value such as 10.=> m-n
11x where x is an integer.so this is insufficient
2. m+n=22 or 33 or 44... here m = 12 and n =11 also holds true.but it does not satisfy m-n=11x condition.
but when you consider both 1&2 together, you'll see that only valid value would be m=11 n=11 m=22 n=11 or 22.. => x has to be an integer in m-n=11x
equation m-n = 11x implies that only 11x is an integer; x can have a value 10/11 which makes 11x an integer.
I have picked up this question from kaplan800..but not convinced with its answer and explanation.... i think, question is asking whether x is an integer or not. but in the book, they have solved to justify whether m-n = 11x or not.
per my understanding of the question answer has to be (D), but as per book it is (C).
what do u think abt the ques...what is the real question: is x an integer? or is m-n=11x?
Hi montaques, the first statement says The tens difit and the units digit of m are the same. So let's say m is 88.n can be anything, say 60. Then m-n = 11x implies x = 28/11 (not an integer). If n = 66, x can be an integer. So statement 1 is not sufficient.
The 2 statements together mean that m has both the digits same and m+n is a multiple of 11.Hence n has to be a multiple of 11 too. So if both m and n are multiples of 11 and m-n = 11x , then x has to be an integer.
So C is the answer! Hope this clears your confusion..
2. A certain jar contains only b black marbles, w white marbles and r red marbles. If one marble is to be chosen at random from the jar, is the probability that the marble chosen will be red greater than the probability the marble chosen will be white?
if x and y are psoitive integers, is xy a multiple of 8?
(1) the greatest common divisor of x and y is 10
(2) the lease common multiple of x and y is 100
my ans is c from 1 alone..... gcd is 10. so numbers are 10a,10b. their product is 100ab.since 100 is not divisible by 8, 100ab may or may not be divisible by 8.so 1 alone is not sufficient. from 2 alone lcm is 100..so product of numbers is a multiple of 100,which may or may not be divisible by8.
after combinding use the formula PRODUCT OF 2 NUMBERS=LCM X GCD xy=1000 ,which is divisible by 8.
if x and y are psoitive integers, is xy a multiple of 8?
(1) the greatest common divisor of x and y is 10
(2) the lease common multiple of x and y is 100
my ans is c from 1 alone..... gcd is 10. so numbers are 10a,10b. their product is 100ab.since 100 is not divisible by 8, 100ab may or may not be divisible by 8.so 1 alone is not sufficient. from 2 alone lcm is 100..so product of numbers is a multiple of 100,which may or may not be divisible by8.
after combinding use the formula PRODUCT OF 2 NUMBERS=LCM X GCD xy=1000 ,which is divisible by 8.
what is the greated possible area of a triangular region with one vertex at the center of circle of radius 1 and the other two vertices on the circle?
one vertex is at the centre and the other 2 are on the circle means 2 sides of the triangle are = radius of circle. when 2 sides are given AREA OF TRIANGLE=1/2 .ab sin x. where x is angle between the sides. max value of sin x is 1 GREATEST POSSIBLE AREA OF TRINGLE = 1/2 .1. 1. 1= 1/2
St 1 : HCF of x and y is 10... i.e x= 10m and y= 10n, where m and n are integers.. hence xy = 100mn...
so, xy may or may not be a multiple of 8.. eg x =10 ; y= 30 not sufficient..
st 2: LCM of x and y is 100 i.e 100=kx and 100=ly, where k and l are integers.. hence xy=10000/kl...
so, xy may or may not be a multiple of 8. eg x=2 and y=50.. (This gives lcm as 50)
st 1 & st 2 combined : now, HCF * LCM = product of nos hence 10*100=xy or xy = 1000 hence xy is a multiple of 8.. sufficient..
Ans C
Hey Bhawin, your answer is correct but argument is grossly flwed.
for st1: x= 10m, y = 10n, where m & n are integers is not sufficient to express the concept. m and n will have to be relatively coprime. so x=10*2 and y=10*3 have gcd as 10 but xy is divisible by 8; but x=10*3 and y=10*5, have gcd as 10 but xy is not divisible by 8; So st 1 is inconclusive.
for st2: lcm(x,y) is 100 =2^2*5^2 => x=2^a * 3^b, and y = 2^c * 3^d, where max(a,c)=2 and max(b,d)=2. For xy to be divisible by 8 we need a+c>=3. For (a,c)=(2,0) xy is not divisible by 8 but for (a,c) = (2,1) it holds true. So st 2 alsone is not sufficient.
st1 and st2 taken together, xy = 100*10=1000, which gives a guarantee.
if x and y are psoitive integers, is xy a multiple of 8?
(1) the greatest common divisor of x and y is 10
(2) the lease common multiple of x and y is 100
(1) the greatest common divisor of x and y is 10
Case 1: consider 20 and 30 , Not Divisible by 8 Case 2: Consider 40 and 30 , Divisible by 8
Condition 1 not sufficient when taken alone
(2) the lease common multiple of x and y is 100
Case 1 : Consider 25 and 4 , LCM 100 , divisible by 8 only case here can be that of co priomes coz if we have a number which is the factor of other number eg 50 and 2 , then lcm wd be 50
Cant think of ne such other pair right now so iguess We can find the answer with only option 2
what is the greated possible area of a triangular region with one vertex at the center of circle of radius 1 and the other two vertices on the circle?
Hmmmmm... say base is 2b and height is h then So we need to maximise bh subject to b^2 + h^2 = 1, age old solution, nature likes symmetry. take b = h = 1/root(2). So THE area = 1/2.