Latest Elitmus Aptitude Question SOLUTION: x=2^30 ,y=3^20,z=6^10 which of the following is closest to xyz a)(x-1)yz b)x(y-1)z c)xy(z-1) d)can't be determined

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(#M40030577) ELITMUS Numerical Ability question quant Keep an EYE

x=2^30 ,y=3^20,z=6^10
which of the following is closest to xyz
a)(x-1)yz b)x(y-1)z c)xy(z-1) d)can't be determined

Asked In Elitmus keshav (10 years ago)

Solved Mahesh Read Solution (18)

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Read Solution (18) :

Mahesh  (10 years ago)

The question is trying to test the reasoning of which number is the biggest number.

Lets consider the following example, to understand that.

Assume a multiplicatin like 19 * 3,
If I change, 3 to 2 the effect will be much bigger, because it is affecting the other bigger multiple
If I change 19 to 18, the effect is much smaller.

So, of the three values, if we can subtract 1 from the biggest value, the effect will be smaller,.

So, the entire question is about finding the biggest number

To find the highest,

3^20 can be written as (3^4) ^ 5 which is (81)^5, this is bigger than 2^30, which can be written as (2^6)^5 that is 64 ^ 5
(81)^5 is also bigger than 6^ 10 which can be written as (36)^5

So, 3^20 is the biggest number and the effect will be smaller when we subtract one from it.

As the effect is smaler that number is closer to XYZ.

So, x(y-1)z is the answer
option B is the answer

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saikumar  (10 years ago)

x=2^30=8^10
y=3^20=9^10
z=6^10=6^10

as we observe the value of y is maximum among the all the three values
in The product xyz subtracting '1' from x,y,z will cause a slight change in the product

for example : 4*5*6=120
now (4-1)*5*6=90
4*(5-1)*6=96
4*5*(6-1)=100

so from the example we can conclude that reducing one from the maximum component will result in less change (closely resembles the actual product)

so reducing 1 from y will be closer to xyz

Thank you

Is this solution Helpfull? Yes (20) | No   

RAKESH  (10 years ago)

sorry for last answer
in options a,b,c
yz,zx,xy is subtracted from xyz respectively,so result will be closest to xyz
when smallest of xy,yz,zx is subtracted from xyz
xy is greatest & zx is lowest
hence x(y-1)z is closest to xyz
option(b)

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siva  (10 years ago)

ANS is option:b
X=2^30=(2^3)^10=8^10
Y=3^20=(3^2)^10=9^10
Z=6^10 =6^10
so X=8,y=9,z=6
condition XYZ=?(close equal to) xyz=8*9*6=432
a)(x-1)yz=xyz-yz=432-54=378
b)x(y-1)z=xyz-z=432-6=426
c)xy(z-1)=xyz-xy=432-72=360
so 426 is closet to 432 so ans is :b

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dilbar  (10 years ago)

X has the highest power while z has the lowest power...so xyz-yz is greater than other 2 options...so its closer to xyz

Is this solution Helpfull? Yes (3) | No (1)  

RAKESH  (10 years ago)

i think ans is (d)can't be determined
as in options a,b,c
yz,zx,xy is subtracted from xyz respectively,so result cant be closest to xyz
as x,y,z are large,values of xy,yz,zx is not so small.

Is this solution Helpfull? Yes (2) | No (6)  

vinod  (10 years ago)

otion is b
let us take 2^3 3^2 6^1
then by option verification b is correct........

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VIPUL KUMAR  (10 years ago)

if we solve it then we get xyz=(2^40)*(3^30)
now let
xy=(2^30)*(3^20)=a
yz=(2^10)*(3^30)=b
xz=(2^40)*(3^10)=c
now if we compare a and c,then we get
a=c*{(3^10)/(2^10)}
here factor {(3^10)/(2^10)} is greater than 1 so we get a>c

similarly when we compare b and c then we get
b=c*{(3^20)/(2^30)}=c*{(9^10)/(8^10)}
here factor {(9^10)/(8^10)} is greater than 1 so we get b>c

hence c i.e. xz is smallest among all of them.

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ABHIJIT BARMAN  (10 years ago)

ANS b) is for sure

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kunan  (10 years ago)

shud be
(b)

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PANKAJ PANWAR  (10 years ago)

ans is b, which is xyz-xz..
reason is xz have lowest value than yz and xy.

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tejadeepthi  (10 years ago)

xy=2^30*3^20;
yz=3^20*(3^10*2^10)=3^30*2^10;
xz=2^30*(3^10*2^10)=3^10*2^40;
i think xz is smaller than remaining ones.so option b is the answer

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Ankita Mathur  (10 years ago)

a) (x-1)yz

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Shaibal Dutta  (10 years ago)

if we watch crfully thn we cn say dt y>x>z .... so if we minus 1 from z dn d chng of d multipliction is little bit... so ansr should be c.

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vinod  (10 years ago)

The value which having the highest when we decrease by 1 then the change before and after will be in the less percentage so that 2nd option is correct answer.....

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Ajoy Singh  (10 years ago)

we have to find smallest one among yz,xz,xy

now yz=3^20.2^10.3^10=2^10.3^30=2^10.(3^3)^10=2^10.27^10=54^10;
(as 6^10=3^10.2^10)
xz=2^30.2^10.3^10=2^40.3^10=(2^4)^10.3^10=16^10.3^10=48^10;
xy=2^30.3^20=(2^3)^10.(3^2)^10=8^10.9^10=72^10;

so xz is smallest
so b is the correct ans

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Himanshu Srivastava  (9 years ago)

xyz-min(xy,yz,zx)
is near about xyz
min(xy,yz,zx)=zx
xyz-zx will be near about xyz
so option b is correct

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vikram singh  (7 years ago)

why everyone apply long process
its simple
x=2^30= 8^10
y=3^20=9^10
z=6^10
so in x,y,z
y is big so if we substract 1 in y
does'nt matter lot
so option is B is corect

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