Selina solutions

Selina solutions

Grade 7

Fractions (Including Problems) | Exercise 3(E)

Question 11

Q11)The heights of two vertical poles , above the earth’s surface are 14\frac{1}{4}m and 22\frac{1}{3}m respectively. How much higher is the second pole as compared with the height of the first pole ?

Solution :

Height of one pole above earth’s surface =14\frac{1}{4}m

The height of second pole above earth’s surface = 22\frac{1}{3}m

Now according to the question,

22\frac{1}{3}-14\frac{1}{4}

=\frac{67}{3}-\frac{57}{4}

=\frac{67\times4}{3\times4}-\frac{57\times3}{4\times3}

=\frac{268-171}{12}

=\frac{97}{12}

=8\frac{1}{12}m

Hence, the second pole is greater than the first pole by 8\frac{1}{12}m

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subject-cta

Question 11

Q11)The heights of two vertical poles , above the earth’s surface are 14\frac{1}{4}m and 22\frac{1}{3}m respectively. How much higher is the second pole as compared with the height of the first pole ?

Solution :

Height of one pole above earth’s surface =14\frac{1}{4}m

The height of second pole above earth’s surface = 22\frac{1}{3}m

Now according to the question,

22\frac{1}{3}-14\frac{1}{4}

=\frac{67}{3}-\frac{57}{4}

=\frac{67\times4}{3\times4}-\frac{57\times3}{4\times3}

=\frac{268-171}{12}

=\frac{97}{12}

=8\frac{1}{12}m

Hence, the second pole is greater than the first pole by 8\frac{1}{12}m

Still have questions? Our expert teachers can help you out

Book a free class now

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subject-cta
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