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11th April 2016, 04:21 PM
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Simplification Question for SSC CGL

Hi I would like to have some questions on Simplification topic for the Staff Selection Commission’s Combined Group Level Examination?
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  #2  
11th April 2016, 04:21 PM
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Join Date: May 2012
Re: Simplification Question for SSC CGL

Some questions on Simplification topic for the Staff Selection Commission’s Combined Group Level Examination are as fillows:

QUERY 1

If 2x + 2/x = 1, then find the value of x^3 + 1/x^3

Answer:
whenever such type of question is asked …..first make it simple by converting the equation like: x+1/x = 1/2

Cubing it on both sides: (x + 1/x)^3 = (1/2)^3
= x^3 + (1/x)^3 + 3*x*1/x(x+1/x) = (1/2)^3
= x^3 + 1/x^3 = 1/8 – 3*1/2
Hence the expression to be find = -11/8

QUERY 2

If a^2 = 2, then find the value of a+1

a – 1
2/(a -1)
(a + 1)/(3 – 2a)
(a – 1)/(3 – 2a)

Answer:
Just put a+1 equal to all the options one by one till the result is a^2 = 2

1) a+1 = a – 1 impossible

2) a+1 = 2/(a – 1)
a^2 – 1 = 2; a^2 = 3 (no solution)

3) a + 1 = (a + 1)/3 – 2a
a = 3/2 (no solution)

We see first 3 solutions are not there; so last one (option ‘4’) is the desired option…

To me it seems the fastest method in such a question

QUERY 3

If 1.5x = 0.04y, then the value of (y^2 – x^2) / (y^2 + 2xy + x^2)

730/77
73/77
73/770
74/77

Answer:
1.5 X = 0.04 Y
therefore x/y = 2/75

The given expression = (y+x)(y-x)/(y+x)^2
= (y – x)/(y + x)
= (1 – x/y)/(1 + x/y)
put the value of x/y=2/75

= (1 – 2/75)/(1 + 2/75)

= 73/75*75/77

= 73/77 (option ‘2’) Ans

QUERY 4

If x^2 + y^2 – 2x + 6y + 10 = 0, then the value of x^2 + y^2 is

Answer:
x^2 + y^2 – 2x + 6y + 10 = 0
>> (9x^2 – 2x + 1) + (y^2 + 6y + 9) = 0
>> (x – 1)^2 + (y + 3)^2 = 0
Here we see that both the term oh LHS are square of a number, hence both are positive. We also know that if sum of the two positive numbers is zero both are zero.
Therefore here both (x – 1)^2 = 0 and (y + 3)^2 = 0
>> x=1 , y=-3
Hence x^2 + y^2 = 10

QUERY 5

If x^4 + 1/x^4 = 119; then x^3 – 1/x^3 = ?

Answer:
x4 + 1/x4 = 119
(x2 + 1/x2)2 – 2 = 119
(x2 + 1/x2)2 = 121
x2 + 1/x2 = 11
(x – 1/x)2 + 2 = 11
(x – 1/x)2 = 9
x – 1/x = 3

Now to be valued:
x3 – 1/x3 = ?
(x – 1/x)3 + 3.x.1/x(x – 1/x)
33 + 3.3
27 + 9
36


QUERY 6

If X : Y=5 : 6, then (3x^2 – 2y^2) : (y^2 – X^2) is?

11:3
3:11
6:7

Answer:
From the ratio of X and Y if X = 5 then Y = 6
Now putting these value of X and Y in in the other expression of ratios [3 (5)^2 – 2 (6)^2] : (6^2 – 5^2)
==> (75 – 72) : (36 – 25)
==> 3 : 11 (option ‘3)

QUERY 7

If a : b : c = 2 : 3 : 4 and 2a – 3b + 4c = 33, then the value of c is

6
9
12
66/7

Answer:
Put a, b, c as 2k, 3k, 4k in the given equation
Then 11k = 33
==> k = 3
==> But ‘c’ = 4k
==> ‘c’ = 4 x 3 = 12 (option ‘3’)

QUERY 8

If x + y + z = 1, xy + yz + zx = -1, xyz = -1, then what is the value of x^3 + y^3 + z^3 = ?

0
-1
-2
1

Answer:
(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx)
=> x^2 + y^2 + z^2 = 3
we know,
x^3 + y^3 + z^3 – 3xyz = (x+y+z) (x^2 + y^2 + z^2 – xy -yz -zx)
Now just put all the given & derived values, we get:
x^3 + y^3 + z^3 = 1 (option ‘4’)

TRICK
By looking at the answer option we can easily derive that values of x, y, & z must be in integers. Now xyz = -1 (given) ; means none of them can be ‘0’; and either ‘all three are -1’ or ‘only one of them is -1 and two are 1 each’. If we see x^3 + y^3 + z^3 it’s not possible that x, y, z all are -1.. So two of them is 1 each and the third is -1. Therefore answer 1 is possible only.

QUERY 9

If x^(1/3) + y^(1/3) = z^(1/3); then (x + y – z)^3 + 27xyz is equal to?

Answer:
The given equation = [x^(1/3) + y^(1/3)]^3 = z

==> x + y + 3x^2/3 y^1/3 + 3x^1/3 y^2/3 = z

==> x + y + 3x^(1/3) y^(1/3) [x^(1/3) + y^(1/3)] = z

==> x+y+ 3x^(1/3) y^(1/3) z^(1/3) = z [putting x^(1/3) + y^(1/3) = z^(1/3)]

==> x + y – z = -3 x^1/3 y ^1/3 z ^1/3

==> (x + y – z )^3 = -27 xyz (cubing both sides)

==> (x + y – z)^3 + 27xyz = 0.

QUERY 10

x^2 + y^2 + 1/x^2 + 1/y^2 = 4 then the value of x^2 + y^2 is?

Answer:
The given equation is = (x^2 + 1/x^2) + (y^2 + 1/y^2) = 4

==> [(x – 1/x)^2 + 2] + [(y – 1/y)^2 + 2] = 4

==> (x – 1/x)^2 + (y – 1/y)^2 + 4 = 4

==> (x – 1/x)^2 + (y – 1/y)^2 = 0

We see that both the left hand terms are positive and their sum is equal to zero
So (x – 1/x) = 0 and (y – 1/y) = 0

Solving both x^2 = 1 and y^2 = 1

Hence x^2 + y^2 = 2


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