f) Now clear the numbers set in the extreme left.
g) After clearing, set the numerator of the second fraction in the extreme left of the
abacus. That is, your numerator of the second fraction 4 should appear in the first
column from the extreme left and the denominator 5 in the third column from the
extreme left.
h) By placing your left hand on the 5 and right hand on the denominator 35, ask :
how many 5’s in 35? By saying 7, clear the 5 and set 7 in the same column. Now
multiply 7 and 4. By saying 7×4 = 28, add the number 28 with the value 30 in the
numerator section. Add 2 with 3 and 8 with 0. You get the value 58.
i) At this stage, check the numerator and the denominator sections. If the value of
the numerator section is less than the value of the denominator section, leave it.
If the value of the numerator section is greater than the value of the denominator
section, you have to do some additional calculations.
j) In the result, you should get a fraction in such a way that the numerator is always
less than the denominator. In this problem, your numerator value is 58 and the
denominator value is 35. Now ask : how many 35’s in 58? By saying 1, add the
number 1 (which is nothing but 3535 ) with the whole number set already in the
whole number section, thus making the whole number as 13. At the same time,
the due 35 must be subtracted from the numerator. Now you have 23 in the
numerator and 35 in the denominator which is in the right form.
k) At this stage, check all the three sections. You have 13 as whole number, 23 as
numerator and 35 as denominator. The answer is 13 3523.