ARITHMETICMATHSSIMPLIFICATION

QUESTIONS ON SIMPLIFICATION (PART-II)

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QUESTIONS ON SIMPLIFICATION (PART-II)

QUERY 11

If x² + y² + z² + 2= 2(y – x), then what is the value of x³ + y³ + z³ = ?

A) 0
B) 1
C) 2
D) 3

MADHAV JHA
In such question one needs to be very vigil; by looking at the whole question we see that any formula involving x³ + y³ + z³ is not possible here. So we need to think elsewhere. If we see the given equation, it can be converted in 3 term easy quadratic equation like this:
(x² + 1+ 2x) + (y² + 1 – 2y) + z² = 0

=> (x+1)² + (y -1)² + z² = 0

Here we see that the left hand side is the sum of squares of three positive terms, and also equal to zero.
This is only possible when each of these terms is equal to zero.

so, x = -1, y = 1 & z = 0.
Now putting these values in x³+y³+z³ we get 0 (option ‘A’) which is the answer

QUERY 12

If x + y = 2z, then the value of x/(x – z) + z/(y – z)

A) 1/2
B) 2
C) 1
D) 3

RONNIE BANSAL
We can write x+y = 2z as x – z = z – y
Therefore the second term of the question z/(y – z) = -z/(z – y) = -z/(x – z)            [z – y = x – z (shown above)]

Hence x/(x – z) + z/(y – z) = x/(x – z) – z/(x – z)

=> (x – z)/(x – z)

=> 1. So option ‘C’ is correct

QUERY 13

If x² + y² – 2x + 6y + 10 = 0 then the value of x² + y² is ?

A) 15
B) 20
C) 10
D) 12

[email protected]
x² + y² – 2x + 6y + 10 = 0
=> (x² – 2x + 1) + (y² + 6y +9|
=> (x – 1)² + (y + 3)² = 0
=> x – 1 =0 and y + 3 =0
=> x = 1 and y = -3

Therefore x² + y² = 1 + 9 = 10 (option ‘C’)

QUERY 14

If x² = y+z,   y² = z+x,   z² = x+y; then the value of 1/(x+1) + 1/(y+1) + 1/(z+1) is ?

A) 0
B) 1
C) 15
D) 2

Hemant Singh
x² = y+z
=> x = (y+z)/x
=> x+1 = {(y+z)/x}+1
=> x+1 = (x+y+z)/x
=> 1/(x+1) = x/(x+y+z) ……..I

similarly
1/(y+1) = y/(x+y+z) and ……..II
1/(z+1) = z/(x+y+z) ……..III

on adding 1, 2 and 3

1/(x+1) + 1/(y+1) + 1/(z+1) = (x+y+z)/(x+y+z) = 1 (option ‘B’)

TRICK
This type of questions will surely appear in SSC. Though it is always good to know the basic method for solving these questions, but in the exam there isn’t that much time and therefore everything should be done in minimum time possible. So just put very small number as value of the variables such as 0, 1, 2 etc because they are framed in such a way that you will get the answer by putting these values surely almost always. If done like that you’ll see if 2 is taken as the value of each variable 1 is the answer of the question.

 

QUERY 15

If x + 1/x = √3; then the value of x18 + x12 + x6 + 1 is ?

A) 0
B) 1
C) 2
D) 3

Vaibhav Vats
x + 1/x = √3
squaring both sides and taking LCM we get:
x4 + 1 = x2………………..(1)
multiply both sides by x² (We have to multiply it by x2 so that the degree of the equation is same as the degree of the last term in x in the expression is, i.e. 6)
So the equation will be (x4*x2) + (1*x²) = (x²*x²)

=> x6 + x² = x4

=> x6 = x4 – x²

Now the given expression can be re-written as:
(x6)3 + (x6)2 + x6 + 1

= (x4 – x2)3 + (x4 – x2)2 + (x4 – x2) + 1 (by putting the value of x6)

from (1) we get x4 – x2 = -1

put it in the above and solve, answer = 0 (option ‘A’)

QUERY 16

If a/b = b/c = c/d; then (b³ + d³ + c³)/(a³ + b³ + c³) = ?

A) d/a
B) d/c
C) a/b
D) a/d

Vijay Bharath Reddy
Let a/b = b/c = c/d = k

=> c = dk, b = dk², a = dk³

Now the given expression
= (b³ + d³ + c³) /(a³ + b³ + c³)

= [(d3k6) + d3 + (d3k3)]/[(d3k9) + (d3k6) + (d3k3)]

= [d3(k6 + 1+ k3)]/[d3(k9 + k6 + k3)]

= (k6 + 1+ k3)/(k9 + k6 + k3)

= (k6 + 1+ k3)/[k3(k6 + k3 + 1)

= 1/k3

= (b/a)(c/b)(d/c)

= d/a (option ‘A’)

QUERY 17

If a/b = b/c = c/d; then √[(a – b + c) (b – c + d)] /[√(ab) – √(bc) + √(cd)]

A) -2
B) 0
C) 1
D) 2

Vijay Bharath Reddy
Let a/b = b/c = c/d = k
then,
c = dk
b = ck = dk²
a = bk = dk³

Now, by replacing a, b, c with these values in the given expression we will get the answer as 1 (option ‘C’)

QUERY 18

a4 + a2b2 + b4 = 8,
and a2 + ab + b2 = 4; then ab = ?

A) 0
B) 1
C) 2
D) 3

RONNIE BANSAL
If we see all the three terms of the LHS of the first equation are square of all the terms of LHS of equation second. So better to apply this formula in the second equation: (x + y + z)2 = x2 + y2 + z2 +2xy + 2yz + 2za)

Therefore a2 + ab + b2 = 4

Squaring both sides

(a2 + ab + b2)2 = 42

=> (a4 + a2b2 + b4) + 2a³b + 2ab3 + 2a²b²=16

by putting a4 + a2b2 + b4 = 8

8 +2ab(a2 + b² + ab) =16 

by putting a2 + ab + b2 = 4

2ab*4 = 8 (by putting a2 + ab + b2 = 4)

=> ab = 1 (option ‘B’)

QUERY 19

If a, b are rational numbers and (a – 1)√2 + 3 = b√2 + a; then the value of a+b is

A) -3
B) 5
C) -5
D) 3

JAYANTCHARAN CHARAN
Bringing root terms of the given equation on one side we get.
√2(a – 1 – b) = a – 3

Since a and b both are rational; a – 1 – b must be equal to zero. And if it’s so a – 3 also is zero.

Therefore
a – 1 – b = 0                                          (i)
a – 3 = 0                                               (ii)
Solving both we get
a + b = 5 (option ‘B’)

QUERY 20

10009 ÷ 1024 = ?

A) 900
B) 1000
C) 2000
D) 1500

MAHA GUPTA
10009 ÷ 1024
= (103)9 ÷ 1024
= 1027 ÷ 1024
= 10(27 – 24)
= 103
= 1000 (option ‘B’)

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

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