Let the required number be 'p'.
From the given data,
p + 12 = 160 x 1/p
=> p + 12 = 160/p
=> p(p + 12) = 160
=> P^2 + 12p - 160 = 0
=> p^2 + 20p - 8p - 160 = 0
=> P(p + 20) - 8(p + 20) = 0
=> (p + 20)(p - 8) = 0
=> p = -20 or p = 8
As, given the number is a natural number, so it can't be negative.
Hence, the required number p = 8.
Let the required number be 'p'
We know, sum of angles in a triangle are 180 deg
According to given data,
p - p/7 = 180
7p - p = 180 x 7
=> 6p = 1260
=> p = 1260/6 = 210
A number is divisible by 9 only if the sum of the digits of the number is divisible by 9.
Here 86236 = 8 + 6 + 2 + 3 + 6 = 25
We must add 2 to 25 to become 27 which is divisible by 9.
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F of 551 and 1073 = 29;
First number = 551/29 = 19
Third number = 1073/29 = 37.
Required sum = 19 + 29 + 37 = 85.
Cp1 = p2 - p1 ,
Cp2 = p3 - p2
..
..
..
Cp23 = p24 - p23
Sum of series = (p2-p1) + (p3-p2) + .....(p23-p22) + (p24-p23)
All terms get cancelled, except p1 = 2 and p24 = 89
So Sum = -p1 + p24
Sum of series = -2 + 89 = 87
Total number of exams written by 100 students = 48 + 45 + 38 = 131
Now let us say x members are writing only 1 exam, y members are writing only 2 exams, z members are writing only 3 exams.
Therefore, x + 2y + 3z = 131 also x + y + z = 100.
Given that z = 5. So x + 2y = 116 and x + y = 95.
Solving we get y = 21.
So 21 members are writing exactly 2 exams.
Let the number be x.
Then, 2/3x-50 = 1/4x +50
=> 5/12x = 90
x = (90 x 12) / 5 = 216
Let the ten's digit be x and unit's digit be y.
Then, (10x + y) - (10y + x) = 63
=> 9 (x - y) = 63
=> x - y = 7.
Thus, none of the numbers can be deermined..
As per divisibility rule a number is divisible by 6 means it should be divisible by 2 and 3
case 1 : the divisibility rule for 2 is the number should be end with even number
in this case R =2 & R=4
case 2 : The divisibility rule for 3 is the sum of numbers should be divisible by 3
so if we take option (2) Q=1 & R=4 the sum is 39
if we take option (3) Q=1 & R=2 the sum is 37
So Option (2) is correct 39 is divided by 3
Let the number be 'x'. Then, from given data
x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264
16200 =
A perfect cube has a property of having the indices of all its prime factors divisible by 3.
Required number = = 9x5 = 45.
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