Three times the first of three consecutive odd integers is $3$ more than twice the third. The third integer is

$A$ fraction becomes $4$ when $1$ is added to both the numerator and denominator; and it becomes $7$ when $1$ is subtracted from both the numerator and denominator. The numerator of the given fraction is

$\frac{901}{29} \times \frac{91}{301} \div \frac{51}{599} = ?$

In a division sum,the divisor is $12$ times the quotient and $5$ times the remainder. If the remainder is $48$,then the dividend is:

What least number must be subtracted from $1294$ so that the remainder when divided by $9, 11, 13$ will leave in each case the same remainder $6$?

If the sum of one-half and one-fifth of a number exceeds one-third of that number by $7 \frac{1}{3},$ the number is