Back to Course
JSS3: MATHEMATICS - 1ST TERM
0% Complete
0/0 Steps
-
Binary Number System I | Week 15 Topics|1 Quiz
-
Binary Number System II | Week 26 Topics|1 Quiz
-
Word Problems I | Week 34 Topics|1 Quiz
-
Word Problems with Fractions II | Week 41 Topic|1 Quiz
-
Factorization I | Week 54 Topics|1 Quiz
-
Factorization II | Week 63 Topics|1 Quiz
-
Factorization III | Week 73 Topics|1 Quiz
-
Substitution & Change of Subject of Formulae | Week 82 Topics|1 Quiz
-
Simple Equations Involving Fractions | Week 93 Topics|1 Quiz
-
Word Problems | Week 101 Topic|1 Quiz
Lesson 3,
Topic 3
In Progress
Solving Combined Products With Sums & Differences
Lesson Progress
0% Complete
Topic Content:
- Solving Combined Products With Sums & Differences
Worked Example 3.3.1:
Subtract the product of 9 and 40 from 450
Solution
You solve them in stages/steps:
Step 1: Product of 9 and 40 ⇒ 9 × 40 = 360
Step 2: Subtract the product from 450 ⇒ 450 – 360 = 90
Worked Example 3.3.2:
Add the negative difference between 27.9 and 35.7 to the product of 27.9 and 35.7
Solution
Step 1: Negative difference ⇒ 27.9 – 35.7 = -7.8
Step 2: Product ⇒ 27.9 × 35.7 = 996.03
Full Equation: (27.9 × 35.7) + (27.9 – 35.7)
= 996.03 + (- 7.8)
⇒ (+ × – = -)
= 996.03 – 7.8
Ans = 988.23
Worked Example 3.3.3:
Find the difference between the product of \(\scriptsize 1 \frac{5}{8} \; and \; \frac{4}{5}\) and the difference between \(\scriptsize 1 \frac{5}{8} \; and \; \frac{4}{5}\)
Solution
\(\left [ \scriptsize 1 \frac{5}{8} \: \times \; \frac{4}{5} \right] \: – \: \left[ \scriptsize 1 \frac{5}{8} \: – \: \frac{4}{5} \right] \\ \left [ \frac{13}{8} \: \times \: \frac{4}{5} \right] \: – \: \left[\frac{13}{8} \: – \: \frac{4}{5} \right] \) \( \frac{13}{10} \: – \: \frac {65 \: – \: 32}{40} \\ = \frac{13}{10} \: – \: \frac{33}{40} \\ = \frac{ 52 \: – \: 33}{40}\\ = \frac{19}{40} \)