Robert Bosch Divisibility Quiz 4

Let the number be x.
Then, \frac{2}{3}of \frac{1}{4} of x = 45
2x = 3*4*45
x = 270
So, required number = (frac{1}{10}*270)= 27

1323 = (7 x 7) x (3 x 3 x 3)
Number of factors = (2+1) x (3+1) = 3 x 4 = 12

Let the ten's digit be x and the unit's digit be y.
Then, (10x + y) - (10y + x) = 81
=> 9(x - y) = 81
=> x - y = 9.

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 63
9x = 63
x = 7.
Therefore, Required difference = (2x + x) - (2x - x) = 2x = 2*7 =14

Let the ten’s and unit digits be x and 4/x respectively.
Then, (10\times x+ \frac{4}{x})+27 = 10\times \frac{4}{x} + x
10x^{2}+4+27x = 40 + x^{2}
9x^{2}+27x-36 = 0
x^{2}+3x-4 = 0
x^{2}+4x-x-4 = 0
x(x+4)-1(x+4) = 0
(x-1)(x+4) = 0
x=1 or x=-4
Hence , the number is 1.

Let the ten's digit be x and the unit's digit be y.
Then, x + y = 30 and x - y = 6 or y - x = 6.
Solving x + y = 30 and x - y = 6, we get: x = 18, y = 12.
Solving x + y = 30 and y - x = 6, we get: x = 12, y = 18.
So, the number is either 1812 or 1218.

Let the numbers be a, b and c.
Then, a2 + b2 + c2 = 250 and (ab + bc + ca) = 257.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 250 + 2(257)
(a + b + c) = \sqrt{764} = 27.64.

Let the numbers be x and y.
According to the question:
x + y = 37
x - y = 13
By adding both the equations:
2x = 50
x = 25
Substituting the value of x = 25, any of the given equations:
x + y = 37 , y = 37 - 25 = 12
Now, products will be = xy = 25 x 12 =300

Let the number be x.
Then, x+8=-15/x
x^{2}+8x+15=0
(x+3)(x-5)
x=-3 or x=-5

Let the numbers be x and y.
Then, xy = 8550 and x/y = 25
(xy)/(x/y) = 8550/25
y2 = 342
y = 18.50
x = 15y = (15 x 18.50) = 277.5
Sum of the numbers = x + y = 277.50 + 18.50 = 296