Solution:
4 | x             y = (5 × 1 + 4) = 9
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5 | y - 1         x = (4 × y + 1) = (4 × 9 + 1) = 37
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   | 1 - 4

Now, 37 when divided successively by 5 and 4, we get

5 | 37
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4 | 7 - 2
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   | 1 - 3
Respective remainders are 2 and 3

Solution:
$$2.3\overline {49} = \frac{{2349 - 23}}{{990}} = \frac{{2326}}{{990}}$$

Solution:
$$\eqalign{ {\text{ }}\frac{2}{5},\frac{5}{6},\frac{{11}}{{15}}{\text{ }}\frac{7}{8} \cr \frac{7}{8}\,{\text{is largest among }}\frac{2}{5},\frac{5}{6},\frac{7}{8} \cr {\text{now compare}} \cr } $$

Solution:
Prime numbers between 0 and 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
Their number is 15

Solution:
Let P = x + $${r_{1}}$$ and Q = y + $${r_{2}}$$, where each of x and y are divisible by the common divisor.
Then, P + Q = (x + $${r_{1}}$$) + (y + $${r_{2}}$$)
                    = (x + y) + ($${r_{1}}$$ + $${r_{2}}$$)
(P + Q) leaves remainder $${r_{3}}$$ when divided by the common divisor.
⇒ [(x + y) + ($${r_{1}}$$ + $${r_{2}}$$) - $${r_{3}}$$] is divisible by the common divisor.
Since (x + y) is divisible by the common divisor, so divisor = $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$

Solution:
In this type of questions go through option, we take option B.
$$\frac{{34}}{{13}}$$ remainder is 8.
$$\frac{{34}}{7}$$ remainder is 6.
It satisfy the condition so answer is option B.

Solution:
LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes Hence, all the six bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes = $$1 + \frac{{60}}{2}$$
= 31