Quiz Discussion

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

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

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

Solve \({ \text{1}}\frac{4}{5} + 20 - 280 \div 25 = ?\)

• 1]

  \(8\frac{1}{5}\)

• 2]

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

• 3]

  \(11\frac{1}{2}\)

• 4]

  \(10\frac{3}{5}\)

• 5]

  \(12\frac{1}{5}\)

The value of \({ \text{5}}\frac{1}{3} \div 1\frac{2}{9} \times \frac{1}{4}  \left( {10 + \frac{3}{{1 - \frac{1}{5}}}} \right)\)   = ?

• 1]

  15

• 2]

  67/25

• 3]

  128/11

• 4]

  128/99

64^12 / 4^15 = 64^x

• 1]

  6

• 2]

  5

• 3]

  4

• 4]

  7

24.962 ÷ (34.11 ÷ 20.05) + 67.96 - 89.11 = ?

• 1] 884
• 2] 346
• 3] 252
• 4] 424
• 5] 366

If \(\sqrt {{ \text{4096}}}\) = 64, then the value of \(\sqrt {{ \text{40}}{ \text{.96}}}\)   + \(\sqrt {{ \text{0}}{ \text{.4096}}}\)   + \(\sqrt {{ \text{0}}{ \text{.004096}}}\)   + \(\sqrt {{ \text{0}}{ \text{.00004096}}}\)     up to two place of decimals is = ?

• 1] 7.09
• 2] 7.10
• 3] 7.11
• 4] 7.12

1 - [5 - {2 + (- 5 + 6 - 2) 2}] is equal to:

• 1] -4
• 2] 2
• 3] 0
• 4] 2

The difference of \({ \text{1}}\frac{3}{{16}}\) and its reciprocal is equal to = ?

• 1]

  \(1\frac{1}{8}\)

• 2]

  4/3

• 3]

  15/16

• 4]

  None of these

5² + 13²  – 11² = (?)³ – 52

• 1]

  6

• 2]

  3

• 3]

  4

• 4]

  5