Number System - Aptitude MCQ Questions and Solutions with Explanations

Let the numbers are a, b
⇒ a + b = 75.....(i)
⇒ a - b = 25.....(ii)
⇒ After solving (i) and (ii)
⇒ a = 50,     b = 25
⇒ ab = 50 × 25 = 1250
So, their product is 1250

$$1\frac{1}{2}$$ = $$\frac{3}{2}$$
$$1\frac{1}{3}$$ = $$\frac{4}{3}$$
Square = $$\frac{9}{4}$$   $$\frac{16}{9}$$
Reciprocal ratio
$$\frac{4}{9}$$ : $$\frac{9}{16}$$
64 : 81

$$\root 3 \of 7 ,\root 4 \of {13} ,\sqrt 5 $$
Taking L.C.M. of 3, 4 and 2.

$${7^{\frac{1}{3} \times 12}}$$	$${13^{\frac{1}{4} \times 12}}$$	$${5^{\frac{1}{2} \times 12}}$$
74	133	56
↓	↓	↓
2401	2197	15625

Let us assume any such number which when divided by 5 leaves remainder as 3.
Let it be 8
So, now
$$\eqalign{ & = \frac{{{{\left( 8 \right)}^2} + {{\left( 8 \right)}^3}}}{5} \cr & = \frac{{64 + 512}}{5} \cr & = \frac{{576}}{5} = 1\,\,\left[ {{\text{Remainder}}} \right] \cr} $$

$$\eqalign{ & = \frac{{\left( {0.0539 - 0.002} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{\left( {0.0519} \right) \times 0.4 + 0.56 \times 0.07}}{{0.04 \times 0.25}} \cr & = \frac{{0.02076 + 0.0392}}{{0.04 \times 0.25}} \cr & = \frac{{0.05996}}{{0.01}} \cr & = 5.996 \cr} $$

A + B = 3(B + C)
A + B = 3B + 3C
A = 2B + 3C.....(i)
A + B + C = A + 30
B + C = 30.....(ii)
B = 5C (given)
5C + C = 30
C = Rs. 5
B = 30 - 5 = Rs. 25
A + 25 = 3(25 + 5)
A = 90 - 25
A = Rs. 65

Let the natural number (n) be 1
∴ n (n + 1) (n + 2)
= 1 (1 + 1) (1 + 2)
= 6

Let the number = 10x + y
Interchange number = 10y + x
Difference
= 10x + y - 10y - x
= 9 (x - y)
The difference is always divisible by 9

Unit's digit in 3 x 38 x 537 x 1256
= Unit's digit in 3 x 8 x 7 x 6
= 4 x 2
= 8
Note : Always multiply only unit digit of first number, to second and product's unit digits number with 3^rd number. Again product of last's unit digit to fourth and so on.

Square root
$$ \Rightarrow \sqrt {\frac{{\left( {0.064 - 0.008} \right)\left( {0.16 - 0.04} \right)}}{{\left( {0.016 + 0.08 + 0.04} \right){{\left( {0.4 + 0.2} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left[ {{{\left( {0.4} \right)}^3} - {{\left( {0.2} \right)}^3}} \right]\left[ {{{\left( {0.4} \right)}^2} - {{\left( {0.2} \right)}^2}} \right]}}{{\left[ {{{\left( {0.4} \right)}^2} + {{\left( {0.2} \right)}^2} + \left( {0.2 \times 0.4} \right)} \right]{{\left[ {\left( {0.4} \right) + 0.2} \right]}^3}}}} $$
$$ \Rightarrow {\text{Let }}0.4{\text{ }} = {\text{ }}a,{\text{ }}0.2{\text{ }} = {\text{ }}b$$
$$ \Rightarrow \sqrt {\frac{{\left( {{a^3} - {b^3}} \right)\left( {{a^2} - {b^2}} \right)}}{{\left( {{a^2} + {b^2} + ab} \right){{\left( {a + b} \right)}^3}}}} $$
$$ \Rightarrow \sqrt {\frac{{\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\left( {a + b} \right)\left( {a - b} \right)}}{{{{\left( {a + b} \right)}^3}\left( {{a^2} + {b^2} + ab} \right)}}} $$
$$\eqalign{ & \Rightarrow \sqrt {\frac{{{{\left( {a - b} \right)}^2}}}{{{{\left( {a + b} \right)}^2}}}} \cr & \Rightarrow \frac{{a - b}}{{a + b}} \cr & \Rightarrow \frac{{0.4 - 0.2}}{{0.4 + 0.2}} \cr & \Rightarrow \frac{{0.2}}{{0.6}} \cr & \Rightarrow \frac{1}{3} \cr} $$