Quiz Discussion

If x = a + m, y = b + m, z = c + m, then the value of \(\frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}\)       is = ?

• 1]

  1

• 2]

  \(\frac{{x + y + z}}{{a + b + c}}\)

• 3]

  \(\frac{{a + b + c}}{{x + y + z}}\)

• 4]

  Not possible to find

(9.0 / 2 * 27 / 9 ) / (18/7.5 * 5.0 / 4) = ?

• 1]

  5.5

• 2]

  4.0

• 3]

  4.5

• 4]

  3.0

Solve this 9 3/7 - 6 4/7 - ? = 14 4/7

• 1]

  3/7

• 2]

  4/7

• 3]

  

• 4]

  

The simplified value of \(\frac{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)\left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}}{{\left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) + \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right)}} = ?\)

• 1]

  100

• 2]

  200/101

• 3]

  200

• 4]

  202/100

Simplify : \({ \text{10}}\frac{1}{8}{ \text{ of }}\frac{{12}}{{15}} \div \frac{{35}}{{36}}{ \text{ of }}\frac{{20}}{{49}}\)

• 1]

  \(17\frac{5}{{12}}\)

• 2]

  \(17\frac{8}{{17}}\)

• 3]

  \(20\frac{3}{{25}}\)

• 4]

  \(20\frac{{103}}{{250}}\)

If a - b = 3 and a2 + b2 = 29, find the value of ab = ?

• 1] 10
• 2] 12
• 3] 15
• 4] 18

Solve : 128.43 + 30.21 + ? = 173

• 1] 35.66
• 2] 29.66
• 3] 43.66
• 4] 24.66
• 5] 14.36

\(\left\{ {\left( {64 - 38} \right) \times 4} \right\} \div 13 = ?\)

• 1] 4
• 2] 1
• 3] 8
• 4] 2
• 5] 5

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

( √5 – 2)^2 = x – √80

• 1]

  9

• 2]

  7

• 3]

  6

• 4]

  10