ARITHMETICMATHSSIMPLIFICATION

QUESTIONS ON SIMPLIFICATION (PART-VII)

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QUESTIONS ON SIMPLIFICATION (PART-VII)

QUERY 61

16612_620662531343195_3890421441722864539_n

SHIV KISHOR
Let 1.331 = a
Then the given expression
= [(1/a + 1/a² + ….1/a7)/(1/a2 +1/a3 + …..1/a8)]2/3

= [(1/a +1/a2 + ….1/a7)/{(1/a)(1/a + 1/a2 + …..1/a7)}]2/3

= [1/(1/a)]2/3

= a2/3

= (1.331)2/3

= 1.21 (option ‘4’)


QUERY 62

If √x = √3 – √5, then the value of x² – 16x + 6

A) 0
B) -2
C) 2
D) 4

MAHA GUPTA
√x = √3 – √5

=> x = 8 – 2√15 —- squaring both sides

=> x – 8 = -2√15

=> x² + 64 – 16x = 60 —-again squaring both sides

=> x² – 16x = -4

Now the find expression
x² – 16x + 6
= -4 + 6               —-putting x² – 16x = -4
= 2 (option ‘C’)


QUERY 63

1907330_286165148226571_8870986184854288870_n

SHIV KISHOR

In such a question convert the expression on the very right under the root sign in whole square of an expression, and so on
10312495_578937902221242_2350599461009047153_n

(Option ‘b’)

QUERY 64

If x = b + c – 2a, y = c + a – 2b, z = a + b – 2c, then value of x2 + y2 – z2 + 2xy

A) 1
B) 3
C) 0
D) 7

MAHA GUPTA
Adding all the equations
x + y + z = b + c – 2a + c + a – 2b + a + b – 2c
=> x + y + z = 0

Now the given expression can be re-written as
(x + y)² – z²
= (x + y + z)(x + y – z)
= 0 (option ‘C’)               —-putting x + y + z = 0


QUERY 65

x + y + z = 0; then find x²/yz + y²/zx + z²/xy = ?

A) 3
B) 0
C) 5
D) 9

MAHA GUPTA
As x + y + z = 0, let x = 2, y = -1 and z = -1

Putting these values in the find expression
(2²)/(-1)(-1) + (-1)²/-1*2 + (-1)²/2(-1)
= 4 -1/2 – 1/2 = 3 (option ‘A’)


QUERY 66

x + y = 9; if x and y are real numbers find the maximum value of xy

A) 20
B) 81/4
C) 81
D) 21

MAHA GUPTA
Though this question can be solved by AM, GM method also, here we do need not that. If sum of two numbers is there, the maximum value of their multiplication can only be when if both of them are equal. Here the sum is 9; so both of them should be 9/2.

Therefore maximum value of xy = (9/2)*(9/2) = 81/4 (option ‘B’)

ADDITIONAL
If the numbers are integers, they must be 4 and 5 for xy to be the maximum as the numbers needs to have as near as possible. So xy = 4*5 = 20


QUERY 67

If 9√x = √12 + √147, find the value of x.

A) 2
B) 3
C) 4
D) 5

RAHUL VERMA
√12 + √147 = 2√3 + 7√3 = 9√3

Comparing it with LHS of the equation
x = 3 (option ‘B’)


QUERY 68

If √x = √3 – √5, then the value of x² – 16x + 6 = ?

A) 2
B) 10
C) 12
D) -2

AJAY SHEORAN
10369606_909125405779848_3957414357626593437_n

(Option ‘A’)

QUERY 69

10364167_522694007836156_341056448888632554_n

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

SUBHASH CHANDRA
10429259_1440478152869540_2224753659521054732_n
10367590_238684179660733_1515224163016100598_n

(option ‘A’)

QUERY 70

If 9x4 – 12x3 + 10x2 + px + 1 is a perfect square, find the value of p

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

MAHA GUPTA
By looks we can see that the given expression can be converted into the form: [3x² – 2x + 1]²

See how
[3x² – 2x +1]²
= (3x²)² + 4x² + 1² + 2(3x²)(-2x) + 2(-2x)1 + 2*1*(3x²)             —- Applying (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
= 9x² -12x² + 10x² + (- 4x) + 1

Now comparing this with the given expression
p = -4 (option ‘C’)

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Maha Gupta

Maha Gupta

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