Let the number be x.
442 < x < 452
? 1936 < x < 2025 ...(i)
From equation (i), the required number will be any number between 1936 and 2025. Since one part of the number is the square of 6 means one factor is 36.
? L, C, M of 36 and 5 = 180
? Number will be multiple of 180 i,e.,
180 x 11 = 1980 the only value which satisfies the equation (i).
Let first number = P
And second number = Q
According to the question
P - Q = 10 ....(i)
And 3P - 4Q = 18...(ii)
on multiply eq (i) by 3 and then subtract eq (ii) from it
3P - 3Q = 54
3P - 4Q = 18
Q = 36
On putting the value of Q in the eq (i)
P = 18 + Q = 18 +36
? P = 54
Required sum = P + Q = 54 +36 =90
Number of trees that can be planted on one side of road = (1760 / 20) + 1
= 88 + 1
=89
? Trees on the both side = 2 x 89=178
Let the third number = N
Then the first number = 3N
And second number = 3N/2
According to the question [N + 3N + 3N/2] / 3=154
? (2N +6N +3N) / 6 = 154
? N = (154 x 6) / 11 = 84
? Required difference = 3N - N = 2N
= 2 x 84 = 168
Let the number be P and Q
According to the question
(P + Q) = 2.5 (P - Q)
? P + Q = 2.5P - 25Q
? 3.5Q = 1.5P
? P / Q = 7/3 .....(i)
Now PQ = 84
? Q2 = (84 x 3) / 7
? Q2 = 36
? Q = 6
&ther4; P = (7/3) x 6 = 14
Sum of the number = P + Q = 14 + 6 = 20
let the total number of sweet = N
According to the question
(N/250) - (N/300) = 1
? (6N - 5N) / 1500=1
? N = 1500
By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
? Req. smallest number =42735.
let the two-digit number be 10p + q
Now according to the question
p + q = 10
and (p +10q) + 36 = (q + 10p)
? -9q + 9p = 36
? -q + p = 4 ....(ii)
on adding eq (i) and (ii) we get
2p = 14 ? p =7
? p = 7 and q = 4
? Required number is the 73
So, the number is a multiple of prime number.
The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.
Let the two digit number be 10x + y.
Now, according to the question,
x + y = 10............(i)
if digits reversed , number get decreased by 36,
and ( x +10y) = ( 10x + y ) - 36
? - 9y + 9x = 36
? x - y = 4 ..................... (ii)
On adding Eqs. (i) and (ii), we get
2x = 14
? x = 7
? x = 7 and y = 3
? Required number is 73.
So,the number is a multiple of a prime number.
Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
? 2x + 4 = y+1
? 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
? 5x +5 = 3y - 6
? 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
? 2 x 2 -y = -3
? -y = -3 - 4
? y = 7
? Original fraction = x/y =2/7
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