Simplification - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option A

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

Answer: Option D

According to question,
(0.98)³ + (0.02)³ + 3 × 0.98 × 0.02 - 1
= (0.98)³ + (0.02)³ + 3 × 0.98 × 0.02(0.98 + 0.02) - 1
= (0.98 + 0.02)³ - 1   [∴ (a + b)³ = a³ + b³ + 3ab(a + b)]
= (1)³ - 1
= 0

Alternate
a = 0.98, b = 0.02, c = -1
a + b + (-c) = 0
So,
a³ + b³ + (-c)³ - 3abc = 0

Answer: Option C

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

Answer: Option C

$$\eqalign{ & {\text{According to question,}} \cr & \Rightarrow \sqrt {\frac{8}{3}} \cr & \Rightarrow \sqrt {\frac{{8 \times 3}}{{3 \times 3}}} \cr & \Rightarrow \sqrt {\frac{{24}}{9}} \cr & \Rightarrow \frac{{4.898}}{3} \cr & \Rightarrow 1.633 \cr} $$

Answer: Option B

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

Answer: Option D

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

Answer: Option C

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

Answer: Option D

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

Answer: Option C

According to question
$$\root 3 \of {{{\left( {333} \right)}^3} + {{\left( {333} \right)}^3} + {{\left( {334} \right)}^3} - 3 \times 333 \times 333 \times 334} $$
As we know that
a³ + b³ + c³ - 3abc
= $$\frac{1}{2}$$ (a + b + c)[(a - b)² + (b - c)² + (c - a)²]
= $$\frac{1}{2}$$ (333 + 333 + 334)[(333 - 333)² + (333 - 334)² + (334 - 333)²]
= $$\frac{1}{2}$$ × 1000[0 + 1 + 1]
= $$\frac{1}{2}$$ × 1000 × 2
= $$\root 3 \of {1000} $$
= 10

Answer: Option B

$$\eqalign{ & {\text{4}}x = p\left( {x + 3} \right) + q\left( {x - 1} \right) \cr & \Rightarrow 4x = px + 3p + qx - q \cr & \Rightarrow 4x = \left( {p + q} \right)x + \left( {3p - q} \right) \cr & \Rightarrow p + q = 4\,and\,3p - q = 0 \cr & \Rightarrow p + q = 4\,and\,p = \frac{q}{3} \cr & \Rightarrow \frac{q}{3} + q = 4 \cr & \Rightarrow \frac{{4q}}{3} = 4 \cr & \Rightarrow q = 3 \cr & p + q = 4\,\,\,and \cr & \Rightarrow p = 1 \cr & \Rightarrow q = 3 \cr} $$