Question

The value of
(6.43)² +33.3074+(5.18)²
____________________
6.43x41.3449-10.36x 2.59x 5.18

lies between ??

1) 0.1 and 0.5
2) 0.5 and 1
3) 1 and 1.2
4) 1.2 and 1.5

{ Asked in TCS NQT Exam}

​

Answer 1

Answer:

0.5 and 1

Step-by-step explanation:

Let’s assume A = 6.43 and B = 5.18

Numerator

A. 6.43*6.43 = A2

B. 5.18*5.18 = B2

C. 33.3074 = A*B

Numerator = A2

+ AB + B2

As per BODMAS,

Denominator:

A. 6.43 * 41.3449 = 6.43*6.43*6.43 = A3

B. 10.36*2.59*5.18 = 5.18*2*2.59*5.18 = B3

Denominator = A3

- B3

, this can also be written as (A2

+ AB + B2

)(A-B)

Numerator/Denominator = (A2

+ AB + B2

) / (A2

+

AB + B2

)(A-B) = 1/(A-B) = 1/(6.43 - 5.18) = 1/1.25 =

0.8

0.8 is between 0.5 and 1.

Answer 2

SOLUTION :

TO CHOOSE THE CORRECT OPTION

The value of

$\displaystyle \sf{ \frac{ {(6.43)}^{2} + 33.074 + {(5.18)}^{2} }{6.43 \times 41.3449 - 10.36 \times 2.59 \times 5.18} \: }$

lies between

$\sf{1. \: \: \: \: 0.1 \: \: \: and \: \: \: \: 0.5 \: }$

$\sf{2. \: \: \: \: 0.5 \: \: \: and \: \: \: \: 1 \: }$

$\sf{3. \: \: \: \: 1 \: \: \: and \: \: \: \: 1.2 \: }$

$\sf{4. \: \: \: \: 1.2 \: \: \: and \: \: \: \: 1.5 \: }$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

$\sf { {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2}) \: }$

CALCULATION

$\displaystyle \sf{ \frac{ {(6.43)}^{2} + 33.074 + {(5.18)}^{2} }{6.43 + 41.3449 - 10.36 + 2.59 \times 5.18} \: }$

$= \displaystyle \sf{ \frac{ {(6.43)}^{2} +( 6.43 \times 5.18) + {(5.18)}^{2} }{6.43 \times 6.43 \times 6.43 - (5.18 \times 2 \times 2.59 \times 5.18)} \: }$

$= \displaystyle \sf{ \frac{ {(6.43)}^{2} +( 6.43 \times 5.18) + {(5.18)}^{2} }{ {(6.43)}^{3} - (5.18 \times 5.18 \times 5.18)} \: }$

$= \displaystyle \sf{ \frac{ {(6.43)}^{2} +( 6.43 \times 5.18) + {(5.18)}^{2} }{ {(6.43)}^{3} - {(5.18)}^{3} } \: }$

$= \displaystyle \sf{ \frac{ {(6.43)}^{2} +( 6.43 \times 5.18) + {(5.18)}^{2} }{ {(6.43} - {5.18)} \bigg[ \: {(6.43)}^{2} +( 6.43 \times 5.18) + {(5.18)}^{2} \bigg] } \: }$$= \displaystyle \sf{ \frac{1 }{ {(6.43} - {5.18)} } \: }$

$= \displaystyle \sf{ \frac{1 }{ 1.25 } \: }$

$= \displaystyle \sf{0.8 }$

Hence the value of

$\displaystyle \sf{ \frac{ {(6.43)}^{2} + 33.074 + {(5.18)}^{2} }{6.43 \times 41.3449 - 10.36 \times 2.59 \times 5.18} \: }$

$\sf{lies \: \: between \: \: \: \: 0.5 \: \: and \: \: 1 \: }$

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The angles of the triangle are

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then the value of x is

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