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  1. The sum of the numerator and denominator of a fraction is $12$. If $1$ is added to both the numerator and the denominator the fraction becomes $3 \over 4$. Find the fraction.

  2. $4$ men and $6$ boys can finish a piece of work in $5$ days while $3$ men and $4$ boys can finish it in $7$ days. Find the time taken by $1$ man alone or that by $1$ boy alone.

  3. If one diagonal of a trapezium divides the other diagonal in the ratio $1 : 3$. Prove that one of the parallel sides is three times the other.

  4. The diagonals of a trapezium $ABCD$ in which ADBC intersect at $O$ and $BC2AD$. Find the ratio of the areas of $\triangle AOD$ to $\triangle BOC$.

  5. If $\sin(a - b) = \frac 1 2$ and $\cos(a+b) = \frac 1 2$, $0 < a + b \leq 90$ $A > B$, find $a$ and $b$.

Also, this isn't homework. Its from the last year question paper. These ones I failed to solve.

Proof: SUMMATIVE ASSESSMENT –I (2011) (pdf).

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up vote 2 down vote accepted

For the first problem, for the fraction $p/q$ you've got $p+q=12$ and $(p+1)/(q+1)=3/4$. If you multiply the second equation by (q+1), you'll get a system of two linear equations which is not hard to solve.

For the second question, it is implicitly assumed that the total work done per day is a linear function of the number of men and the number of boys (which is a highly unrealistic assumption, of course). So if you denote the "number of pieces of work per day" one man can do as $m$, and the corresponding number for boys as $b$, then the first equation reads $4m+6b=1/5$ (because if they finish the piece in five days, under the linearity assumption they finish 1/5 of the piece each day). The second condition can be translated the same way, leaving you again with a system of linear equations. Note however that the number of days taken by one man is $1/m$, and the number of days taken by one boy is $1/b$.

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1) Denote the fraction by $\frac ab$. You know that $a+b=12$ and that $\frac{a+1}{b+1}=\frac 34$. So $4a+4=3b+3$, i.e. $4a-3b=-1$. Continue from here...

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