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Sia ? 6 years, 5 months ago
Let one part = <m:omath><m:r>x</m:r></m:omath>
Then, the other part = 184 – <m:omath><m:r>x</m:r></m:omath>
One-third of the first part {tex} = \frac{1}{3}x = \frac{x}{3}{/tex}
One-seventh of the other part {tex} = \frac{1}{7}(184 - x){/tex}
According to the problem, {tex}\frac{x}{3} = \frac{1}{7}(184 - x) + 8{/tex}
{tex}\frac{x}{3} - \frac{1}{7}(184 - x) = 8{/tex}
{tex}\frac{{7x - 552 + 3x}}{{21}} = 8{/tex}
{tex}10x - 552 = 8 \times 21{/tex}
10x = 168 + 552 = 720
{tex}x = \frac{{720}}{{10}} = 72{/tex}
Hence, the first part is 72 and the second part is (184 – 72) = 112
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Sia ? 6 years, 5 months ago
Let the smaller number be x.
Then, the larger number = x + 15
{tex}\because {/tex} Sum of two numbers is 95
∴ x + (x + 15) = 95
∴ 2x + 15 = 95
∴ 2x = 95 – 15 . . . . [Transposing 15 to R.H.S.]
∴ 2x = 80
∴ {tex}x = \frac{{80}}{2}{/tex}. . . . [Dividing both sides by 2]
∴ x = 40
∴ x + 15 = 40 + 15 = 55
Hence, the desired numbers are 40 and 55.
Verification
55 = 40 + 15
40 + 55 = 95
Posted by Rakshit Bisht 7 years, 1 month ago
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Sonali Aggarwal 7 years, 1 month ago
A trapezium is a quadrilateral with exactly one pair of parallel sides.
Remember, a trapezium has only one pair of opposite sides to be parallel which clearly says it is not a parallelogram.
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