Selina solutions

Selina solutions

Grade 7

Fractions (Including Problems) | Exercise 3(D)

Question 10

Q10) Simplify :

5\frac{3}{4}-\frac{3}{7}\times15\frac{3}{4}+2\frac{2}{35}\div1\frac{11}{25}

Solution :

5\frac{3}{4}-\frac{3}{7}\times15\frac{3}{4}+2\frac{2}{35}\div1\frac{11}{25}

=\frac{23}{4}-\frac{3}{7}\times\frac{63}{4}+\frac{72}{35}\div\frac{36}{25}

=\frac{23}{4}-\frac{3}{7}\times\frac{63}{4}+\frac{72}{35}\times\frac{25}{36}

=\frac{23}{4}-\frac{27}{4}+\frac{10}{7}

=\frac{23\times7}{4\times7}-\frac{27\times7}{4\times7}+\frac{10\times4}{7\times4} (L.C.M. of 4 and 7 is 28)

=\frac{161-189+40}{28}

=\frac{201-189}{28}

=\frac{12}{28}

=\frac{3}{7}

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta

Question 10

Q10) Simplify :

5\frac{3}{4}-\frac{3}{7}\times15\frac{3}{4}+2\frac{2}{35}\div1\frac{11}{25}

Solution :

5\frac{3}{4}-\frac{3}{7}\times15\frac{3}{4}+2\frac{2}{35}\div1\frac{11}{25}

=\frac{23}{4}-\frac{3}{7}\times\frac{63}{4}+\frac{72}{35}\div\frac{36}{25}

=\frac{23}{4}-\frac{3}{7}\times\frac{63}{4}+\frac{72}{35}\times\frac{25}{36}

=\frac{23}{4}-\frac{27}{4}+\frac{10}{7}

=\frac{23\times7}{4\times7}-\frac{27\times7}{4\times7}+\frac{10\times4}{7\times4} (L.C.M. of 4 and 7 is 28)

=\frac{161-189+40}{28}

=\frac{201-189}{28}

=\frac{12}{28}

=\frac{3}{7}

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

Learn from an expert tutor.

Book a free class now
subject-cta
linkedin instgram facebook youtube
2020 © Quality Tutorials Pvt Ltd All rights reserved
linkedin instgram facebook youtube