Quiz Discussion

Assume that $$\sqrt {13}$$ = 3.605(approximately) and $$\sqrt {130}$$  = 11.40(approximately) Find the value of: $$\sqrt {1.3}$$ + $$\sqrt {1300}$$  + $$\sqrt {0.013}$$

Course Name: Quantitative Aptitude

• 1] 36.164
• 2] 36.304
• 3] 37.304
• 4] 37.164
Solution
No Solution Present Yet

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# Quiz
1
Discuss

(36.14)^2 – (21.28)^2 =?

• 1]

850

• 2]

860

• 3]

865

• 4]

853

Solution
2
Discuss

98643 – 21748 = 51212 + ?

• 1]

24383

• 2]

24713

• 3]

25683

• 4]

26243

Solution
3
Discuss

The value of $$\frac{{{x^2} - {{\left( {y - z} \right)}^2}}}{{{{\left( {x + z} \right)}^2} - {y^2}}}{ \text{ + }}\frac{{{y^2} - {{\left( {x - z} \right)}^2}}}{{{{\left( {x + y} \right)}^2} - {z^2}}} +\frac{{{z^2} - {{\left( {x - y} \right)}^2}}}{{{{\left( {y + z} \right)}^2} - {x^2}}}$$   is = ?

• 1] -1
• 2] 0
• 3] 1
• 4] None of these
Solution
4
Discuss

5907 – 1296 / 144 = x * 8.0

• 1]

700.05

• 2]

705.25

• 3]

751.54

• 4]

737.25

Solution
5
Discuss

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
Solution
6
Discuss

Simplify : $$\sqrt {3 + \frac{{33}}{{64}}} \div \sqrt {9 + \frac{1}{7}} \times 2\sqrt {3\frac{1}{9}}$$   = ?

• 1]

45/256

• 2]

$$1\frac{{17}}{{28}}$$

• 3]

$$4\frac{3}{8}$$

• 4]

$$2\frac{3}{{16}}$$

Solution
7
Discuss

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

Solution
8
Discuss

If $$x = \sqrt 3 { \text{ + }}\sqrt 2 { \text{,}}$$ then the value of $${x^3} - \frac{1}{{{x^3}}}$$  is?

• 1]

$$10\sqrt 2$$

• 2]

$$14\sqrt 2$$

• 3]

$$22\sqrt 2$$

• 4]

$$8\sqrt 2$$

Solution
9
Discuss

25 * 3.250 + 50.40 / 24.0 = ?

• 1]

75

• 2]

80

• 3]

77

• 4]

83

Solution
10
Discuss

The expression $$\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}$$  is equal to = ?

• 1]

$$\frac{8}{{{x^8} + 1}}$$

• 2]

$$\frac{8}{{{x^8} - 1}}$$

• 3]

$$\frac{8}{{{x^7} - 1}}$$

• 4]

$$\frac{8}{{{x^7} + 1}}$$

# Quiz