Simplification - 05

Explanation:-

Let the three consecutive numbers are x, x + 1, x + 2
According to question
x² + (x + 1)² + (x + 2)² = 110
⇒ x² + x² + 1 + 2x + x² + 4 + 4x = 110
⇒ 3x² + 6x + 5 = 110
⇒ 3x² + 6x - 105 = 0
⇒ x² + 2x - 35 = 0
⇒ x² + 7x - 5x - 35 = 0
⇒ x(x + 7) -5(x + 7) = 0
⇒ (x + 7)(x - 5) = 0
⇒ x = 5 & -7
[∴ (-) value can not considered]
∴ Smallest number is = 5

Explanation:-

$$\eqalign{ & {\text{3}}x + y = 19\,.....(i) \cr & x - y = 9\,.....(ii) \cr & {\text{Adding (i) and (ii),}} \cr & {\text{we get }} \cr & {\text{4}}x = 28 \cr & x = 7 \cr & {\text{Putting }}x\,{\text{ = 7}}\,{\text{in (i),}} \cr & {\text{we get }} \cr & 3x + y = 19 \cr & 21 + y = 19 \cr & y = 19 - 21 = - 2 \cr & {\text{Answer is 7, }} - {\text{2}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{4 gal}} \cr & {\text{ = (4}} \times {\text{4)qt}} \cr & {\text{ = 16 qt}} \cr & {\text{ = }}\left( {16 \times 12} \right){\text{pt}} \cr & {\text{ = 32 pt}} \cr & \therefore {\text{Number of botteles}} \cr & {\text{ = }}\frac{{32}}{{\frac{1}{2}}} \cr & {\text{ = 64}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Number of bananas required daily}} \cr & {\text{ = }}\frac{{798}}{7} \cr & = 144 \cr & \therefore \,\,\,{\text{Required number}} \cr & {\text{ = }}\left[ {114 \times \left( {\mathop {31}\limits^{{\text{Jan}}} + \mathop {29}\limits^{{\text{Feb}}} + \mathop {31}\limits^{{\text{Mar}}} } \right)} \right] \cr & = \left( {114 \times 91} \right) \cr & = 10374 \cr}$$

Explanation:-

71 × 29 + 27 × 15 + 8 × 4
= 2059 + 405 +32
= 2496

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & \,{\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02 - 1 \cr & = {\left( {0.98} \right)^3} + {\left( {0.02} \right)^3} + 3 \times 0.98 \times 0.02\left( {0.98 + 0.02} \right) - 1 \cr & = {\left( {0.98 + 0.02} \right)^3} - 1\,\,\,\left[ {\because {{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] \cr & = {\left( 1 \right)^3} - 1 \cr & = 0 \cr & {\text{Alternate}} \cr & a = 0.98 \cr & b = 0.02 \cr & c = - 1 \cr & a + b + \left( { - c} \right) = 0 \cr & So, \cr & {a^3} + {b^3} + {( - c)^3} - 3abc \cr & = 0 \cr}$$

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & \,\,\,\,\,\,\frac{2}{3} \times \frac{3}{{\frac{5}{6} \div \frac{2}{3}{\text{of 1}}\frac{1}{4}}} \cr & \Rightarrow \frac{2}{3} \times \frac{3}{{\frac{5}{6} \div \left( {\frac{2}{3} \times \frac{5}{4}} \right)}} \cr & \Rightarrow \frac{2}{3} \times \frac{3}{{\frac{5}{6} \div \frac{5}{6}}} \cr & \Rightarrow \frac{2}{3} \times \frac{3}{1} \cr & \Rightarrow 2 \cr}$$

Explanation:-

$$\eqalign{ & {\text{Length of reel}} \cr & {\text{ = (32}} \times {\text{120 + 80)}}\,{\text{cm}} \cr & {\text{ = 3920}}\,{\text{cm}} \cr & {\text{ = 39}}{\text{.20}}\,{\text{m}} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let the number be = }}x \cr & {\text{According to question}} \cr & {x^2} = {\left( {75.15} \right)^2} - {\left( {60.12} \right)^2} \cr & \Rightarrow {x^2} = {\left( {75.15 + 60.12} \right)^2}{\left( {75.15 - 60.12} \right)^2} \cr & \Rightarrow {x^2} = 135.27 \times 15.03 \cr & \Rightarrow {x^2} = 2033.1081 \cr & \Rightarrow x = 45.09 \cr}$$

Explanation:-

$$\eqalign{ & {\text{According to question}} \cr & \sqrt {192} - \frac{1}{2}\sqrt {48} - \sqrt {75} \cr & \Rightarrow 8\sqrt 3 - \frac{4}{2}\sqrt 3 - 5\sqrt 3 \cr & \Rightarrow 8\sqrt 3 - 2\sqrt 3 - 5\sqrt 3 \cr & \Rightarrow \sqrt 3 \cr & \Rightarrow 1.732 \cr}$$