Number System - Aptitude MCQ Questions and Solutions with Explanations

5^{(x + 3)} = 25^{(3x - 4)}
Or, 5^{(x + 3)} = 5^{2 × (3x - 4)}
Now, comparing the powers,
(x + 3) = 2 × (3x - 4)
Or, x + 3 = 6x - 8
Or, 6x - x = 8 + 3
Or, 5x = 11
Hence, x = $$\frac{{11}}{5}$$

HCF of 551 and 1073 is 29. So second number is 29
First number = $$\frac{{551}}{{29}}$$ = 19
Third number = $$\frac{{1073}}{{29}}$$  = 37
So,
Required Sum
= 19 + 29 + 37
= 85

$$\eqalign{ & {\text{Let}}\,{\text{two}}\,{\text{numbers}}\,{\text{are}}\,x\,{\text{and}}\,y. \cr & {\text{Now,}}\,{\text{Sum}}\,{\text{of}}\,{\text{the}}\,{\text{numbers}}, \cr & x + y = 12 \cr & {\text{Product}}\,{\text{of}}\,{\text{the}}\,{\text{number}}, \cr & xy = 35 \cr & {\text{Then}}\,{\text{sum}}\,{\text{of}}\,{\text{the}}\,{\text{reciprocals}} \cr & = {\frac{1}{x}} + {\frac{1}{y}} \, \cr & = \frac{{ {x + y} }}{{xy}} \cr & = \frac{{12}}{{35}} \cr} $$

Given,
HCF = 44
LCM = 264
First Number = When divided by 2, the quotient is 44
So,
First Number = 2 × 44 = 88
Second number = N. (let).

Now, using formula,
HCF × LCM of two numbers = First number × Second number
44 × 264 = 88 × N
N = 132
Second Number = 132

NOTE :
Dividend = Divisor × Quotient + Remainder

1369 = 37 × 37
It means there are 37 rows and 37 columns.
And each row will have 37 students, Therefore, the last row also have 37 students.

$$\eqalign{ & {\text{Let}}\,{\text{total}}\,{\text{property}}\,{\text{of}}\,{\text{Ram}} = x \cr & {\text{His}}\,{\text{widow}}\,{\text{gets}} \cr & \frac{1}{3}\,of\,x = \frac{x}{3} \cr & {\text{His}}\,{\text{daughter}}\,{\text{gets}}\, \cr & = \frac{3}{5}\,of\,\left[ {x - {\frac{x}{3}} } \right] \cr & = \frac{3}{5} \times {\frac{{2x}}{3}} \cr & = \frac{{2x}}{5} \cr & \cr & {\text{His}}\,{\text{son}}\,{\text{gets}}\,{\text{his}}\,{\text{rest}}\,{\text{property}} \cr & {\text{His}}\,{\text{son's}}\,{\text{property}} \cr & = x - \left[ { {\frac{x}{3}} + {\frac{{2x}}{5}} } \right] \cr & 6400 = x - {\frac{{11x}}{{15}}} \cr & 6400 = \frac{{ {15x - 11x} }}{{15}} \cr & 4x = 6400 \times 15 \cr & x = 1600 \times 15 \cr & x = 24000 \cr & {\text{Total}}\,{\text{Property}} = 24000 \cr} $$

Ratio of the numbers = x : y
HCF of the numbers = z
So, z is the common factor of the numbers
Then, First number = xz
Second Number = yz
Now,
First Number × Second Number = HCF and LCM of the numbers
xzyz = z × LCM
LCM = xyz

Let the Original number is (X + 10Y)
According to question,
Sum of the digit of the number = 12
X + Y = 12 ----------- (1) When digit interchanged, number become 36 more than original number, so,
10X + Y = 10Y + X + 36
X - Y = 4 ------------------- (2)
On solving equation (1) and (2), we get
X = 8
Y = 4
So, Original number,
= X + 10Y = 8 + 10 × 4 = 48

Let Price of each capsule were Rs. X, originally.
So, Number of capsule purchased originally,
= $$ \frac{{176}}{{\text{X}}}$$
If price of capsule is Rs. 3 less then 6 more capsule were purchased. In this case, the number of capsules,
= $$ \frac{{176}}{{{\text{X}} - 3}}$$

According to question,
$$\eqalign{ & \frac{{176}}{{{\text{X}} - 3}} - \frac{{176}}{{\text{X}}} = 6 \cr & \frac{{176{\text{X}} - 176{\text{X}} + 528}}{{{\text{X}} \times \left( {{\text{X}} - 3} \right)}} = 6 \cr & 6{{\text{X}}^2} - 18{\text{X}} - 528 = 0 \cr & {{\text{X}}^2} - 3{\text{X}} - 88 = 0 \cr & {{\text{X}}^2} - 11{\text{X}} + 8{\text{X}} - 88 = 0 \cr & \left( {{\text{X}} - 11} \right)\left( {{\text{X}} + 8} \right) = 0 \cr} $$
Either,
X = 11
Or, X = -8 (Price cannot be negative)
Thus, X = Rs. 11 Hence,
Number of the capsule purchased = $$\frac{{176}}{{11}}$$ = 16

Check the option. 32 and 128 are nither of a square and nor cube root. 64 is the square of 8 and a cube of 4