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JSS3: MATHEMATICS - 1ST TERM

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  1. Binary Number System I | Week 1
    5Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1Topic
    |
    1 Quiz
  5. Factorisation I | Week 5
    4Topics
    |
    1 Quiz
  6. Factorisation II | Week 6
    3Topics
    |
    1 Quiz
  7. Factorisation III | Week 7
    3Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1Topic
    |
    1 Quiz
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The quotient of two numbers is the result obtained by dividing one number by another.

a. Find two-thirds of the sum of 23 and 19.

Solution

\( \frac{2}{3} \: \times \: \scriptsize \left (23 \: + \: 19 \right) \)

= \( \frac{2}{3} \: \times \: \frac{42}{1} \)

= \( \scriptsize 2 \: \times \: 14 = 28 \)

b. Divide 88 by the sum of 2 and the product of 3 and 3.

Solution

product of 3 and 3

3 x 3 = 9

Add 2 to it = \( \scriptsize 2 + (3 \: \times \: 3) \\ \scriptsize = 2 + 9 = 11\)

Divide 88 by the sum which is 11 = \( \frac{88}{11} \\ \scriptsize = 8 \)

c. Add the square of 9 to the product of the square roots of 25 and 4, and divide the result by the square root of 100.

Solution

\( \frac{9^2 \: + \: \left(\sqrt{25}\: \times \: \sqrt{4}\right)}{\sqrt{100}} \\ = \frac{81 \: + \: \left (5 \: \times \: 2 \right) }{10} \\ = \frac{81 \: + \: 10}{10} \\ = \frac{91}{10} \scriptsize = 9.1\)

d. Find four-fifth of the difference between the product of 0.25 and 128 and the difference between 32.8 and 25.8

\( \frac{4}{5} \scriptsize \: \times \: \left(0.25 \: \times \: 128 \right) \: – \: \left(32.8 \: – \: 25.8 \right) \)

\( \frac{4}{5} \scriptsize \: \times \: 32 \:\: – \: 7 \)

\( \frac{4}{5} \scriptsize \: \times \: 25 \)

\( \scriptsize = 4 \: \times \: 5 = 20 \)

e. Divide the square root of 81 by the difference between the product of  

-1 and -3 and the square root of 4.

Solution

\( \frac {\sqrt{81}}{(-1 \; \times \: -3) \; – \: \sqrt{4}} \\ = \frac{9}{+3 \: – \: 2}\\ = \frac{9}{1} \\ \scriptsize = 9\)
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