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Formation of Algebraic Expressions

What is the o...

Question

What is the output produced by the following program?

$void main( ) {int x =1, y =2, z =3; y = z++; x = x > y < z && x < y < z; printf ("%d", x + y + z);}$

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Solution

The correct option is D 8
option (a)

$x = ((x > y) < z) & & ((x < y) < z); = ((1 > 3) < 4) & & ((1 < 3) < 4) = (0 < 4) & & (1 < 4) = 1 & & 1 = 1 x + y + z = 1 + 3 + 4 = 8$

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Similar questions

If x, y and z are variables, verify the cyclic symmetry of the following expressions.

(1) x(y + z) + y(z + x) + z(x + y)

(2) xy(x − y) + yz(y − z) + zx(z − x)

(3) x²y(x + y) + y²z(y + z) + z²x(z + x)

(4) x³(x + y) + y³(y + z) + z³(z + x)

(5) xy²(x − y) + yz²(y − z) + zx²(z − x)

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\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

If $x^{2}$ + $y^{2}$ = $z^{2}$ . Then which of the following is/are true?

Let x, y, z be positive reals. Which of the following implies x = y = z?

(I) x $^{3}$ + y $^{3}$ + z $^{3}$ = 3xyz

(II) x $^{3}$ + y $^{2}$ z + yz $^{2}$ = 3xyz

(III) x $^{3}$ + y $^{2}$ z + z $^{2}$ x = 3xyz

(IV) (x + y + z) $^{3}$ = 27xyz

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