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
Sum of the five consecutive odd number = N + (N + 2) + (N + 4) + (N + 6) + (N + 8) =175
? 5N + 20 =175
? N = (175 - 20) / 5 = 31
Sum of the second largest and square of smallest one = (31 + 6) + (31)2
= 37 + 961=998
Let the number be N
According to the question,
[N x (3/2)] - [N / (3/2)] =10
? 3N/2 - 2N/3 = 10
? (9N - 4N)/6 = 10
? 5N / 6 = 10
? N = (10 x 6)/5 =12
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
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).
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.
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