ARITHMETICMATHSSIMPLIFICATION

QUESTIONS ON SIMPLIFICATION (PART-I)

Click Here to Buy Online Practice Sets exclusivly on Testbook.com

QUESTIONS ON SIMPLIFICATION (PART-I)

QUERY 1

If 2x + 2/x = 1, then find the value of x³ + 1/x³

A) -8/11
B) -11/8
C) 3/2
D) 2/3

JAYANTCHARAN CHARAN
Whenever such type of question is asked …..first make it simple by converting the equation like: x + 1/x = 1/2
Hence 2x + 2/x = 1
=> x + 1/x = 1/2

Cubing it on both sides: (x + 1/x)³ = (1/2)³
=> x³ + (1/x)³ + 3*x*(1/x)(x+1/x) = (1/2)³
=> x³ + (1/x)³ = 1/8 – 3/2
Hence the expression to be find = -11/8 (option ‘A’)

QUERY 2

If a² = 2, then find the value of a+1

A) a – 1
B) 2/(a -1)
C) (a + 1)/(3 – 2a)
D) (a – 1)/(3 – 2a)

JAYANTCHARAN CHARAN
Just put a+1 equal to all the options one by one till the result is a² = 2
a) a+1 = a ­- 1 impossible

b) a+1 = 2/(a – 1)
=> a² ­- 1 = 2
=> a² = 3 (no solution)

c) a + 1 = (a + 1)/(3 ­- 2a)
=> a = 3/2 (no solution)

We see first 3 solutions are not there; so last one (option ‘D’) 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 (y² – x²)/(y² + 2xy + x²)

A) 730/77
B) 73/77
C) 73/770
D) 74/77

Mukesh Kumar Bansal
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 ‘B’) Ans

QUERY 4

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

A) 5
B) 10
C) 15
D) 12

PRAKASH G
x² + y² – 2x + 6y + 10 = 0
=> (9x² – 2x + 1) + (y² + 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 x² + y² = 10 (option ‘B’)

QUERY 5

If x4 + 1/x4 = 119; then x3 – 1/x3 = ?

A) 11
B) 63
C) 36
D) 30

RONNIE BANSAL
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 = ?
x3 – 1/x3 = (x – 1/x)3 + 3*x*1/x(x – 1/x)
=> 3³ + 3*3
=> 27 + 9
36 (option ‘C’)

CHANDAN DUTTA on the above solution
How much time do you allot to one to solve and answer this question ?

If I were you I would allot maximum ONE Minute. Many steps one has to pass through mental calculation to get the answer looking to the constrain of allotted time.

RONNIE BANSAL
Of course such questions are to be solved within seconds. But here the steps are only to make the things clear; this doesn mean one has to go every step. Sometimes we have to solve a sum with the help of given answer options; sometimes other techniques have to be kept in mind. So many things are there actually. Also the net has it’s own limitations.

QUERY 6

If X : Y=5 : 6, then (3x² – 2y²) : (y² – X²) is?

A) 11:3
B) 3:11
C) 6:7
D) 1:2

RONNIE BANSAL
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*5² – 2*6²] : (6² – 5²)
=> (75 – 72) : (36 – 25)
=> 3 : 11 (option ‘B)

QUERY 7

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

A) 6
B) 9
C) 12
D) 66/7

JAYANTCHARAN CHARAN
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 ‘C’)

QUERY 8

If x + y + z = 1, xy + yz + zx = -1, xyz = -1, then what is the value of x³ + y³ + z³ = ?

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

MADHAV JHA
(x+y+z)² = x² + y² + z² + 2(xy+yz+zx)
=> x² + y² + z² = 3
we know,
x³ + y² + z² – 3xyz = (x+y+z) (x² + y² + z² – xy -yz -zx)
Now just put all the given & derived values, we get:
x² + y² + z² = 1 (option ‘D’)

TRICK
Ronnie Bansal
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 x² + y² + z² 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. (option ‘D’)

QUERY 9

If x1/3 + y1/3 = z1/3; then (x + y – z)³ + 27xyz is equal to?

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

SANDES….. HIGHLY INFLAMMABLE
The given equation = x1/3 + y1/3 = z1/3

Cubing both sides
(x1/3 + y1/3)3 = z

=> x + y + 3(x2/3)(y1/3) + 3(x1/3)(y2/3) = z

=> x + y + 3[(x2/3)(y1/3) + (x1/3)( y2/3)] = z

=> x + y + 3(x1/3)(y1/3)(x1/3 + y1/3) = z

putting x1/3 + y1/3 = z1/3

=> x + y+ 3(x1/3)(y1/3)(z1/3) = z

=> x + y – z = -3(x1/3)(y1/3)(z1/3)

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

=> (x + y – z)3 + 27xyz = 0 (option ‘A’)

QUERY 10

x² + y² + 1/x² + 1/y² = 4 then the value of x² + y² is?

A) 4
B) 3
C) 1
D) 2

Hemant Singh
The given equation is = (x² + 1/x²) + (y² + 1/y²) = 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 x² = 1 and y² = 1

Hence x² + y² = 2 (option ‘D’)

Previous post

QUESTIONS ON REMAINDERS & DIVISIBILITY (PART-V)

Next post

QUESTIONS ON SIMPLIFICATION (PART-II)

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:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)