Numbers - 02

Explanation:-

91 is divisible by 7. So, it is not a prime number.

Explanation:-

Given that a^b + a^c + a^d = 0
Hence, out of these three terms some are positive and some are positive and some are negative. Since all of a, b, c and d are negative integers, only that term which hass an even power can be positive. Hence, the absolute value of at least one of a, b, c or d is even. Hence, a X b X c X d is also even

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{\left( {3k - 9} \right)}}{{4k}} = \left( {\frac{2}{3}} \right) \times \left( {\frac{{3k}}{{4k}}} \right) = \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}\to {A} \frac{1}{2} \cr & \therefore 6k - 18 = 4k \cr & \therefore k = 9 \cr & \Rightarrow {\text{Numerator}} = 3k = 27 \cr}$$

Explanation:-

Let the factors of the number be 2, a, b, c, d, e, ......
Then, as the number can be expressed as a cube as well as a square of an integer, it has to have 26 = 64 as its factor.

Explanation:-

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

Explanation:-

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

Explanation:-

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

Explanation:-

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.

Explanation:-

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

Explanation:-

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