Practice MCQ Questions and Answer on Number System

Solution: Given, x + y + z = 0 Cubing both side, (x + y + z)3 = 0 x3 + y3 + z3 - 3xyz = 0 [using formula] x3 + y3 + z3 = 3xyz

Solution: Let numbers are = 3x, 4x (3x)3 + (4x)3 = 5824 27x3 + 64x3 = 5824 91x3 = 5824 x3 = $$\frac{5824}{91}$$ x3 = 64 x = 4 So, numbers are 3x = 3 × 4 = 12 4x = 4 × 4 = 16 Sum = 12 + 16 = 28

Solution: Clearly, each of the 38 terms $$\left[ {\frac{1}{4}} \right], \left[ {\frac{1}{4} + \frac{1}{{50}}} \right], \left[ {\frac{1}{4} + \frac{2}{{50}}} \right], $$      $$ ..... $$ $$ , \left[ {\frac{1}{4} + \frac{{37}}{{50}}} \right]$$   has a value lying between 0 and 1, While each one of the 12 terms $$\left( {\frac{1}{4} + \frac{{38}}{{50}}} \right),$$   $$\left( {\frac{1}{4} + \frac{{39}}{{50}}} \right),$$   $$.....,$$ $$\left( {\frac{1}{4} + \frac{{49}}{{50}}} \right)$$   has a value lying between 1 and 2. Hence, the given expression = (0 × 38) + (1 × 12) = 12

Solution: Let number are a, b = 3x, 5x So, 3x + 5x = 8x = 80 x = 10 So, number is : 3 × 10 = 30 5 × 10 = 50

Solution: $$\eqalign{ & {\text{Remainder of}}\frac{{{{67}^{99}}}}{7} \cr & {\text{or, }}R = \frac{{{{\left( {63 + 4} \right)}^{99}}}}{7} \cr} $$ 63 is divisible by 7 for any power, so required remainder will depend on the power of 4 Required remainder $$\eqalign{ & \frac{{{4^{99}}}}{7} = = R = = \frac{{{4^{\left( {96 + 3} \right)}}}}{7} \cr & \frac{{{4^3}}}{7} \Rightarrow \frac{{64}}{7} \Rightarrow \frac{{\left( {63 + 1} \right)}}{7} = = R \Rightarrow 1 \cr & \cr & {\text{Note}}: \cr & \frac{4}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4} \right)}}{7} = \frac{{16}}{7}{\text{remainder}} = 2 \cr & \frac{{\left( {4 \times 4 \times 4} \right)}}{7} = \frac{{64}}{7} = 1 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4} \right)}}{7} = \frac{{256}}{7}{\text{remainder}} = 4 \cr & \frac{{\left( {4 \times 4 \times 4 \times 4 \times 4} \right)}}{7} = 2 \cr} $$ If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e. $$\frac{{{{\text{A}}^{\text{x}}}}}{{\text{B}}}$$

Solution: $$\eqalign{ & \Leftrightarrow {n^{65}} - n \cr & = n\left( {{n^{64}} - 1} \right) \cr & = n\left( {{n^{32}} - 1} \right)\left( {{n^{32}} + 1} \right) \cr & = n\left( {{n^{16}} - 1} \right)\left( {{n^{16}} + 1} \right)\left( {{n^{32}} + 1} \right) \cr} $$   $$ = n\left( {{n^8} - 1} \right)\left( {{n^8} + 1} \right)\left( {{n^{16}} + 1} \right)$$     $$\left( {{n^{32}} + 1} \right)$$   $$ = n\left( {{n^4} - 1} \right)\left( {{n^4} + 1} \right)\left( {{n^8} + 1} \right)$$     $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$   $$ = n\left( {{n^2} - 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$     $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$   $$ = \left( {n - 1} \right)n\left( {n + 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$       $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{64}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$ Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd. Thus, the given number is definitely a multiple of 24

Solution: Quotient = 16, Divisor = (25 × 16) = 400 5 × Remainder = Divisor ⇒ Remainder = $$\frac{400}{5}$$ = 80 Divided = (400 × 16) + 80 = 6480

Solution: p 1 ⇒ $$\frac{1}{p}$$ > 1 ⇒ $$\frac{2}{p}$$ > 2 ⇒ $$\frac{2}{p}$$ - p > 2 - p > 0 [∵ p 1] Hence, $$\left( {\frac{2}{p} - p} \right)$$ is a positive number

Solution: In this type of question, We need to find out the LCM of the given numbers. LCM of 12, 15, 18 and 20; $$\eqalign{ & 12 = 2 \times 2 \times 3; \cr & 15 = 3 \times 5; \cr & 18 = 2 \times 3 \times 3; \cr & 20 = 2 \times 2 \times 5; \cr} $$ Hence, LCM = $$2 \times 2 \times 3 \times 5 \times 3$$ Since, the soldiers are in the form of a solid square. Hence, LCM must be a perfect square. To make the LCM a perfect square, We have to multiply it by 5, hence, The required number of soldiers = $$2 \times 2 \times 3 \times 3 \times 5 \times 5$$ = 900

Solution: The greatest 4 digit number = 9999 ∴ Number to be added = 307 - 175 = 132