Quiz Discussion

Solve 14 × 627 ÷ \(\sqrt {\left( {1089} \right)} \)   = (?)3 + 141

• 1] 5√5
• 2] (125)2
• 3] 25
• 4] 5
• 5] None of these

The value of \({ \text{3}}\frac{1}{2} - \left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right]\)       = ?

• 1]

  1/2

• 2]

  \(2\frac{1}{2}\)

• 3]

  \(3\frac{1}{2}\)

• 4]

  \(9\frac{1}{2}\)

\({ \text{If }}x = \frac{1}{{2 + \frac{1}{2}}}{ \text{ then }}\frac{1}{x} = ?\)

• 1]

  2/5

• 2]

  5/2

• 3]

  3/5

• 4]

  1/2

If \(\left( {x - \frac{1}{x}} \right){ \text{ = }}\sqrt {21} { \text{,}} \)    then the value of \(\left( {{x^2} + \frac{1}{{{x^2}}}} \right) \left( {x + \frac{1}{x}} \right)\) is = ?

• 1] 42
• 2] 63
• 3] 115
• 4] 120
• 5] 125

Supply the two missing figures in order indicated by x and y in the given equation, the fractions being in their lowest terms.

• 1] 3, 1
• 2] 3, 3
• 3] 4, 1
• 4] 5, 3

\(4\frac{4}{5} \div 6\frac{2}{5} = ?\)

• 1]

  3/4

• 2]

  5/7

• 3]

  5/8

• 4]

  7/11

• 5]

  None of these

The value of \(\frac{5}{{1\frac{7}{8}{ \text{of 1}}\frac{1}{3}}} \times \frac{{2\frac{1}{{10}}}}{{3\frac{1}{2}}}{ \text{ of 1}}\frac{1}{4} = ?\)

• 1]

  \(1\frac{1}{2}\)

• 2]

  0.05

• 3]

  1

• 4]

  2

If \(\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1\)     and \(\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0\)     where a, b, c, p, q, r are non-zero real numbers, then \(\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}}\)    is equal to = ?

• 1] 0
• 2] 1
• 3] 3
• 4] 9

What is \(\frac{{\frac{7}{8} \times \frac{7}{8} + \frac{5}{6} \times \frac{5}{6} + \frac{7}{8} \times \frac{5}{3}}}{{\frac{7}{8} \times \frac{7}{8} - \frac{5}{6} \times \frac{5}{6}}}\)     equal to ?

• 1]

  41/24

• 2]

  1/24

• 3]

  41

• 4]

  None of these

\(\frac{{ \root 3 \of 8 }}{{\sqrt {16} }} \div \sqrt {\frac{{100}}{{49}}} \times \root 3 \of {125} \)     is equal to = ?

• 1]

  7

• 2]

  \(1\frac{3}{4}\)

• 3]

  \(\frac{7}{{1000}}\)

• 4]

  \(\frac{4}{7}\)