Three or more categories of heirs cannot share their portions of the estate.
Generally, the maximum number of categories that cannot share their portions is three. But we have decided to say “three or more” because the rule of solving for three categories is applicable to four, five and so on (if such higher order problems exist). Remember how to determine the LCM of 3 or more numbers? Select any two, find their LCM. Call it ‘X’. Then find the LCM of ‘X’ and the third number. Name this ‘Y’, and proceed like that until all the numbers are exhausted. That is how to solve Level 3 problems. There is no clear-cut rule on which categories to start with. One has the liberty to select any two. But for simplification, the following may be very handy.
Rule W: Examine the categories that cannot share their portions. Select any two that are SIMILAR and resolve them using the appropriate rule. Let the solution be ‘X’. Next, consider ‘X’ along with the next category; resolve them using the appropriate rule. Continue like that until all the categories have been resolved. Thereafter, multiply the end result by the base number to obtain the new base number.
From the above, it will be deduced that Level 3 has no any new rule. Depending on the problem at hand, the appropriate rules from (K) to (V) are applied.
Example 32
Heirs | 10 Daughters | 2 Wives | Mother | 6 Full sisters |
Shares | 2/3 | 1/8 | 1/6 | Residue |
Base number | 24 | |||
Portions | 16 | 3 | 4 | 1 |
Question 1: What categories are not able to share their portions?
Answer: “Daughters,” “wives” and “full sisters” categories.
Question 2: What is the relationship between the number of heads and number of portions of these categories?
Answer: “Daughters” category: 10 and 16 converge.
“Wives” category: 2 and 3 are parallel.
“Full sisters” category: 6 and 1 are parallel.
Question 3: Since two categories are parallel, consider them first. What is the relationship between their numbers of heads?
Answer: 2 (number of heads of wives) and 6 (number of heads of full sisters); one is a multiple of the other.
Question 4: What do we do?
Answer: Select the higher one.
Question 5: Which is…
Answer: 6
Question 6: What rule is that?
Answer: Rule L, “If the number of heads in one category is a multiple of the number of heads in the other category, use the higher number to multiply the base number.” But we do NOT multiply by the base number, until all the categories have been resolved.
Question 7: What is the adjust of “daughters” category that converges?
Answer: Common divisor of 10 and 16 is 2. Therefore, adjust of “daughters” category is 10 ÷ 2 = 5.
Question 8: What is the relationship between 6 and 5; solutions of questions 5 and 7 respectively?
Answer: They are parallel.
Question 9: What do we do?
Answer: Apply Rule P, “In a situation whereby there is a parallel relationship between the ADJUST of the category whose number of heads and number of portions have a common divisor and the NUMBER OF HEADS of the category that has no common divisor, multiply the adjust with the number of head.” That is 5 × 6 = 30.
Question 10: Anymore category to resolve?
Answer: No.
Question 11: What next?
Answer: Apply Rule W, “…Thereafter, multiply the end result by the base number to obtain the new base number.”
Question 12: So, what is the new base number?
Answer: 5 × 6 = 30 × 24 = 720
Question 13: Determine the new portion of each category of heir.
Answer: 10 daughters: 720 × 2/3 = 480 portions
2 wives: 720 × 1/8 = 90 portions
Mother: 720 × 1/6 = 120 portions
6 full sisters: 720 – (480 + 90 + 120) = 30 portions
The complete table is
Heirs | 10 Daughters | 2 Wives | Mother | 6 Full sisters |
Shares | 2/3 | 1/8 | 1/6 | Residue |
Base number | 24 | |||
Portions | 16 | 3 | 4 | 1 |
New base number | 720 | |||
New portions | Each = 48 | Each = 45 | 120 | Each = 5 |
Example 33
Heirs | 2 Wives | Mother | 6 Uterine sisters | 2 Consanguine brothers |
Shares | ¼ | 1/6 | 1/3 | Residue |
Base number | 12 | |||
Portions | 3 | 2 | 4 | 3 |
Using the procedure above but with less explanation and incorporating the original technical Arabic terms (for familiarisation), this problem can be solved as follows:
“Wives” category: Number of heads, 2, and number of portions, 3, are tabayin (parallel).
Mother has no problem. Note that if a category consists of only ONE heir, he/she simply takes whatever is allocated to the category even if it’s 1 portion. That is why all along, mother do not use to have problem for the fact that one cannot have two mothers!
“Uterine sisters” category: Number of heads, 6, and number of portions, 4, are tawafuq (converge).
“Consanguine brothers” category: Number of heads, 2, and number of portions, 3, are tabayin (parallel).
Considering the two that are tabayin, number of heads of wives, 2 and the number of heads of consanguine brothers, 2 are tamathul (same). So, one is chosen (Rule K).
As for the “uterine sisters” category, the wafq (adjust) is 3. Now, 2 (selected number of heads) and 3 (wafq) are tabayin, so we multiply them (Rule Q). This gives 2 × 3 = 6. Finally,
New base number = 2 × 3 = 6 × 12 = 72.
New portion of 2 wives: 72 × ¼ = 18
New portion of mother: 72 × 1/6 = 12
New portion of 6 uterine sisters: 72 × 1/3 = 24
New portion of 2 consanguine brothers: 72 – (18 + 12 + 24) = 18
The table will now look like this.
Heirs | 2 Wives | Mother | 6 Uterine sisters | 2 Consanguine brothers |
Shares | ¼ | 1/6 | 1/3 | Residue |
Base number | 12 | |||
Portions | 3 | 2 | 4 | 3 |
New base number | 72 | |||
New portions | Each = 9 | 12 | Each = 4 | Each = 9 |
Example 34
Heirs | 2 Granddaughters; 4 grandsons |
2 Grandmothers | 12 Daughters | 4 Wives |
Shares | Residue | 1/6 | 2/3 | 1/8 |
Base number | 24 | |||
Portions | 1 | 4 | 16 | 3 |
New base number | 1440 | |||
New portions | Each granddaughter = 6; each grandson = 12 |
Each = 120 | Each = 80 | Each = 45 |
2 grandmothers can share their 4 portions.
“Grandchildren” category: 10 and 1 are tabayin.
“Daughters” category: 12 and 16 are tawafuq.
“Wives” category: 4 and 3 are tabayin.
Number of heads of “grandchildren” and “wives” categories, 10 and 4 respectively are tawafuq. Apply Rule N.
Wafq of 10 and 4 = 10 ÷ 2 = 5 × 4 = 20; or 4 ÷ 2 = 2 × 10 = 20.
The wafq of “daughters” category is 3. How?
Actually, common divisors of 12 and 16 are 2 and 4.
Using 2, wafq of 12 daughters = 12 ÷ 2 = 6
With 4, wafq of 12 daughters = 12 ÷ 4 = 3
Recall that only the Highest Common Divisor (HCD) is considered. That is why the wafq of division by 4 is chosen.
Now, what is the relationship between the two adjusts (wafqan) 20 and 3? They are tabayin. So, we multiply them (Rule U).
Finally, new base number = 20 × 3 = 60 × 24 = 1440
New portion of 2 grandmothers: 1440 × 1/6 = 240
New portion of 12 daughters: 1440 × 2/3 = 960
New portion of 4 wives: 1440 × 1/8 = 180
New portion of 2 granddaughters and 4 grandsons: 1440 – (240 + 960 + 180) = 60
Had it being 2 was chosen to be the common divisor of 12 and 16, the wafq of 12 daughters should have been 6 (as above). But then the new base number would be = 20 × 6 = 120 × 24 = 2880 which is double of 1440. Not that 2880 is wrong, however the principle of base number is that the minimum value is used.
Rule W says, “Select any two SIMILAR categories and resolve them using the appropriate rule.” What happens if intentionally or otherwise, DISSIMILAR categories are selected first and resolved, will the new base number still be the same? Yes! Let’s prove it.
“Grandchildren” category: 10 and 1 are tabayin.
“Daughters” category: 12 and 16 are tawafuq.
“Wives” category: 4 and 3 are tabayin.
Instead of considering the two categories that are tabayin as before, let’s resolve the “grandchildren” and “daughters” categories first. Since the number of heads and number of portions of grandchildren is tabayin, the emphasis shifts to the number of heads, 10. The wafq of 12 daughters is 3 (as explained earlier). Now, what is the relationship between 10 and 3? Tabayin. So, we multiply them (Rule Q).
10 × 3 = 30
The number of heads and number of portions of 4 wives is also tabayin. Again, the number of heads, 4, is considered. What is the relationship between 30 and 4? Tawafuq. Common divisor of 30 and 4 is 2. Therefore,
New base number = 30 ÷ 2 = 15 × 4 = 60 × 24 = 1440
Alternatively, 4 ÷ 2 = 2 × 30 = 60 × 24 = 1440
As a result, selecting and resolving similar or dissimilar categories of heirs that cannot share their portions do not make any difference. But choosing and resolving similar categories first simplify the problem.
Exercise 2
A deceased leaves behind two wives, five daughters and three full brothers. How will the estate be shared among them?
Solution
Quick links
- Introduction
- Male heirs
- Female heirs
- Non heirs
- Impediments to inheritance
- Exclusion
- Exclusion – Part 2
- Exclusion – Part 3
- Partial exclusion
- Note on difference of opinion
- Inheritance of children
- Inheritance of spouses
- Inheritance of parents
- Inheritance of grandparents
- Inheritance of siblings
- Residuaries (‘Asabah)
- Partial exclusion
- Inheritance arithmetic (“inherithmetic”)
- Procedure of solving inheritance problems
- Levels of inheritance problems (Level one)
- Level one – continued
- Lowest Common Multiple (LCM)
- Highest Common Factor (HCF)
- Prime numbers
- Increment of base number (‘Awl)
- Level two – Part 1
- Level two – Part 2
- Level two – Part 3
- Level two – Part 4
- YOU ARE HERE: Level three
- Inheritance of grandfather along with siblings
- Inheritance of grandfather along with siblings in the presence of other heirs
- Special cases
- Summary of rules
- Further reading
- Solutions to exercises
Your Questions, Our Answers
We have received a number of emails from those who visited this website or downloaded and read INHERITANCE IN ISLAM. Almost all of them were questions on either aspects of inheritance not covered in the book or clarifications needed regarding specific cases. Hence, we thought it wise to reproduce the emails so that others may benefit as well. As always, we welcome suggestions, criticisms and of course, more questions!