Problem Set 2 : Square Root and Cube Root - Quantitative Aptitude Video Lecture | CSAT Preparation - UPSC

let us solve some more questions on
square root in this questions at number
two this is the first question what
should come in place of these question
marks in the following equation and this
question was asked in Bank Bo exam of
1999 well let us replace these question
marks with X so we have X upon root 128
is equal to root 162 upon X so moving
this X to the left hand side of equal
two and moving this route 128 to the
right hand side we get x squared is
equal to root 162 times root 128 now let
us divide both 162 and 128 by 2 so that
we can express them in multiples of 2 so
162 when / - you get 81 so 162 can be
written as 2 times 81 this is equal to
under root of 2 times 81 and 128 if it
is divided by 2 you get 64 so 64 times 2
is 128 so you have root of 2 times 64 so
this is equal to root 2 times root 81
times root 2 times root 64 root 2 times
root 2 gives you 2 times root 81 gives
you 9 9 times 9 is 81 times root 64 is 8
8 reads are 64 this is equal to 2 times
9 times 8 so 9 times 8 is 72 so you have
2 times 72 which is equal to 144
therefore you have X square is equal to
144 and therefore taking under root on
both the sides of equal to you have x
equals to root 144 which is equal to 12
so the answer is 12 let's move on to
question number 2 which was
in the be SRB exam of 1998 so let us
rewrite the whole equation so you have
28 root x plus 1 4 to 6 is equal to 3/4
of - eight seven - so you have 3 pi/4 x
- 8 7 - now let us reduce the right hand
side of equals 2 so 4 times 1 is 4 4
times 7 is 28 4 times 1 is 4 4 times 8
is 32 3 times 7 1 8 is equal to 2 1 5 4
on the left hand side you have 28 root x
let us move this one 4 to 6 to the right
hand side of equals 2 so it becomes
minus 1 4 - 6
therefore root X is equal to 2 1 5 4 - 1
4 2 6 you will get 728 and this 28 comes
down so this is divided by 28
so root X is equal to 7 28 divided by 28
let us reduce the right hand side again
so you can divide this let us divide
this by 4 so 4 sevens are 28 4 times 1
is 4 4 it's a 32 4 to side so 182 by 7
and when you divide 182 by 7 you get 26
so therefore you have root X is equal to
26
so taking square on both the sides of
equal to you get x equals to 26 Square +
26 square let us calculate it 26 times
26 you get 6 6 a 36 3 carry 6 to the 12
plus 3 is 15 6 2 s are 12 1 carry 2
those are 4 plus 1 is 5 and this is 6 5
plus 2 is 7 5 plus 1 is 6 so you get 676
as the value of x there
or the correct answer is option B
two-thirds of a square of a number
is 96 what is the number well you have
2/3 of the square of a number if the
number is X you have X square as 96 so
you have to calculate the value of x let
us move X to the left hand side and the
other terms to the right hand side of
equals 2 so you have X square X equal to
96 times 3 divided by 2 so 2 times 1 is
2 2 times 4 is 8 and 2 times 8 is 16 so
X square is equal to 48 times 3 which is
equal to 144 therefore X is equal to
under root of 144 which is equal to 144
is the square of 12 therefore under root
144 is equal to 12 so the answer is
option D let's move on to question
number 4 now it was asked in the SSC
examination of 2002 so let us simplify
this what you have to do is to make the
denominator equal to the numerator so
that you can cancel out most of the
terms so therefore this is equal to you
have this big square root you can write
the zero point zero 3 square plus zero
point 2 1 square plus 0.06 5 square upon
so you want to make 0.003 square as 0.03
square so you can write 0.03 upon 10
square so 0.03 divided by 10 gives you
zero point zero zero three so + again
you can do the same thing with zero
point zero two one so you can write zero
point two 1/10
whole square plus again 0.06 5/10 whole
square now what you can do is again
write the same thing now 0.03 by 10
whole square can be written as zero
point zero 3 square upon 10 square plus
the rest of the things in the same way
the 10 square is common to all the terms
in the denominator so you can write the
denominator like this you can have only
one 10 square so there it is
the numerator and the denominator have
common terms and you can cancel them so
this gets cancelled out and this gets
cancelled out what remains is under root
of 1 by 1 by 10 square this 10 square
can go up so it is equal to under root
of 10 square and under root of 10 square
is nothing but 10 so the answer is 10
and I hope there is the option right
let us solve the last question now well
you have to simplify this thing so let's
simplify it let's put this question mark
as X so X is equal to let us write it
root 3 by 1 minus 1 by root 3 the whole
square so it is equal to root 3 times
root 3 is 3 minus 1 times 1 is 1 upon
root 3 the whole square which is equal
to 2 by root 3 square which is equal to
2 square upon root 3 Square and it is
equal to 4 2 square is 4 by root 3
square is 3 so the answer is 4 by 3