Let the five consecutive number are N - 2, N - 1, N, N +1, N+2
Sum of numbers = 190
? N-2 + N -1 + N + N + 2 = 190
? 5N = 190
? N = 38
Sum of largest and smallest numbers = (N + 2) + (N - 2)
= 2N = 2 x 38
= 76
Let the unit digits of the number be N and 10s digit be M
? Number = 10M + N
According to the question
MN = 10 .......(i)
And 10N + M = (10M + N) - 36
? 10N + M - 10M - N = -36
? 9N - 9M = - 36
? N - M = -4 .......(2)
On Adding eq. (i) and (ii) we get
N + M = 10
N - M = - 4
2N = 6
? N = 3 and M = 7
? Required product of two digits = 3 x 7 = 21
Let d = divisor and q = quotient
First remove the last remainder:
d x q + 413 = 380606
d x q = 380193
So d cannot be even (choice d) or a multiple of 5 (choice c).
If we try choice a or b,
a quick division (by calculator) shows that
380193 = 451 x 843, the other two choices.
(In the absence of choices, we could find the prime factors of 380193,
which are 3 x 11 x 41 x 281 and get various candidate divisors from that.)
So one of them is the divisor and the other the quotient.
So we can just check the first remainder:
843 x 4 = 3372
3806 - 3372 = 434 ... a match!
And just to be sure:
451 x 8 = 3608
3806 - 3608 = 198
So the divisor is 843.
119 is divisible by 7, 187 is divisible by 11,247 is divisible by 13 and 551 is divisible by 19. So none of the given numbers is prime.
For even values of n, the number (10n - 1 ) consists of even numbers of nines and hence it will be divisible by 11.
On dividing 1056 by 23, we get 21 as remainder
? Required number to be added = (23 - 21 ) = 2.
Given Exp. = (a2+ b2+ ab) / (a3-b3)
= 1 / (a - b)
= 1 / (16 - 5)
= 1/11.
Let the number be N
According to the question 5N = 2N2 - 3
? 2N2 - 5N -3 = 0
? (N - 3) (2N + 1) = 0
N = 3 & -1/2
Thus the required number is 3
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
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 total number of sweet = N
According to the question
(N/250) - (N/300) = 1
? (6N - 5N) / 1500=1
? N = 1500
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