A set of mixing bowls consists of 6 bowls of varying diameters. Will the 4 bowls with the smallest diameters fit side by side on a kitchen shelf x centimeters in length?
(1) The average (arithmetic mean) diameter of all 6 bowls is 15 centimeters. (2) The length of the shelf is 50 centimeters.
I think C should be the answer, when we know that sum of all the diameters is 15x6 = 90 and the length of shelf i.e. x = 50, the 4 smallest bowls can be arranged in the shelf.
A set of mixing bowls consists of 6 bowls of varying diameters. Will the 4 bowls with the smallest diameters fit side by side on a kitchen shelf x centimeters in length?
(1) The average (arithmetic mean) diameter of all 6 bowls is 15 centimeters. (2) The length of the shelf is 50 centimeters.
My take 'E'.
a1+a2+a3+a4+a5+a6 = 90.
length of shelf = 50.
a1+a2+a3+a4 40. the minimum value a5+a6 take is 30.1 (considering an approximation) . the maximum value a5+a6 can take is around 70.
Hence we cannot say for sure..that the smallest 4 bowls will fit..
a1+a2+a3+a4 40. the minimum value a5+a6 take is 30.1 (considering an approximation) . the maximum value a5+a6 can take is around 70.
Hence we cannot say for sure..that the smallest 4 bowls will fit..
Yes the OA is 'e'; but how did u derive at this "the minimum value a5+a6 take is 30.1 (considering an approximation) . the maximum value a5+a6 can take is around 70." plz elaborate....
Yes the OA is 'e'; but how did u derive at this "the minimum value a5+a6 take is 30.1 (considering an approximation) . the maximum value a5+a6 can take is around 70." plz elaborate....
lets take all the values are integer....... let four small values are 14,15,16,17 (these are four highest possible values) they will make total of 62 leaving 38 as sum of two remaing values which are individually greater than each above four so they can be 18 n 20. so in this case four small values are > 50. now if we take 10,11,12,13 as four smaller values they are =46 so we can deduce the answer hence E.
Yes the OA is 'e'; but how did u derive at this "the minimum value a5+a6 take is 30.1 (considering an approximation) . the maximum value a5+a6 can take is around 70." plz elaborate....
I followed this analogy:
Since it is given that they are all different... and tht the average is 15, the maximum values of the least 4 diameters could be 15,14.99,14.98,14.97 their sum equals...59.94 and hence a5+a6 = 30.06...... this is the least value for a5+a6..
Considering a1 =1, a2=2, a3=3, a4=5... the minimum values.. and hence a5+a6 - can take a maximum of around 70...
hence i concluded tht a solution cant be reached..
lets take all the values are integer....... let four small values are 14,15,16,17 (these are four highest possible values) they will make total of 62 leaving 38 as sum of two remaing values which are individually greater than each above four so they can be 18 n 20. so in this case four small values are > 50. now if we take 10,11,12,13 as four smaller values they are =46 so we can deduce the answer hence E.
If x,y and z are integers and xy+z is an odd integer , is x an even integer? 1. xy+xz is even 2. y+xz is odd
My Take 'A'. Given : xy+z is odd => xy is even and z is odd or xy is odd and z is even stmt1: xy+xz is even => xy and xz are even or --- (1) xy and xz are odd --- (2)
if z is odd .. xy must be even - and hence from (1) xz must be even => x is even. if z is even .. xy must be odd - and hence from (1) xz must be odd -> but when 'z' is even xz can never be odd.. so with hypothesis (2) cannot exist..
stmt 2: y+xz is odd => y is odd and xz is even or --- (3) y is even and xz is odd --- (4) Cannot say based on this statement. alone....
My Take 'A'. Given : xy+z is odd => xy is even and z is odd or xy is odd and z is even stmt1: xy+xz is even => xy and xz are even or --- (1) xy and xz are odd --- (2)
if z is odd .. xy must be even - and hence from (1) xz must be even => x is even. if z is even .. xy must be odd - and hence from (1) xz must be odd -> but when 'z' is even xz can never be odd.. so with hypothesis (2) cannot exist..
stmt 2: y+xz is odd => y is odd and xz is even or --- (3) y is even and xz is odd --- (4) Cannot say based on this statement. alone....
hope this helps
Stem 1: xy+xz is even => x(y+z) is even this is true when either x is even/odd and y+z is odd/even or both x and (y+z) is even
if x is even and y+z is odd, xy+z is always even The same can be proved for other cases and one can say that xy+z is even
Statement 2: y+xz is odd For diff conditions of x, y and z, xy+z can be even or odd. hence B is insufficient.
1) If 500 is a multiple of 100 that is closest to x and 400 is a multiple of 100 that is closest to y,which multiple of 100 is closest to x+y? a) xb) y 2) If x and y are positive integers such that x= 8y +12 ,what is the GCD of x and y? a) x=12u , where u is an integer b) y=12z , where z is an integer
1.
stmt 1: xstmt 2: y combining both statements:
stmt 1..the maximum value x can take is 450, to coincide with hypothesis. stmt2.. the maximum value y can take is 350, to coincide with hypothesis.
so 450 or 800 so depending on this value, the vlaue varies.. for e,g if x+y = 810, multiple is 800 and if x+y is 890.. multiple will be 900... Hence even both are insufficient.... hence 'E'.
1) If 500 is a multiple of 100 that is closest to x and 400 is a multiple of 100 that is closest to y,which multiple of 100 is closest to x+y? a) xb) y 2) If x and y are positive integers such that x= 8y +12 ,what is the GCD of x and y? a) x=12u , where u is an integer b) y=12z , where z is an integer
2. My take is 'B'.
stmt 1: x = 12u => x is multiple of 12. 2 y = 3 (u-1) => y is a multiple of 3..
depending on the divisors of 'y'..(6,12...) - the gcd varies...=> insufficient.
stmt 2: y = 12z => y is a multiple of 12. x = 12 (z+1) => x is a multiple of 12. => x is 12 units away from y.. and since 12 is common factor of both x and y=> 12 is the gcf for x and y...