Reshebnik — A.ershova V.v.goloborodko K-4 Umnozhenie Drobei 6 Klass Resheniia

1×15×3=115the fraction with numerator 1 cross 1 and denominator 5 cross 3 end-fraction equals 1 over 15 end-fraction Convert to improper fractions first:

Following the rule, perform multiplication before addition/subtraction. 1×15×3=115the fraction with numerator 1 cross 1 and

2110×1514=21×1510×1421 over 10 end-fraction cross 15 over 14 end-fraction equals the fraction with numerator 21 cross 15 and denominator 10 cross 14 end-fraction Simplify (21 and 14 by 7; 15 and 10 by 5): Goloborodko for the 6th grade, here are the

3×57×11=1577the fraction with numerator 3 cross 5 and denominator 7 cross 11 end-fraction equals 15 over 77 end-fraction Simplify by dividing 6 and 18 by 6, and 5 and 25 by 5: Goloborodko for the 6th grade

3×32×2=94=214the fraction with numerator 3 cross 3 and denominator 2 cross 2 end-fraction equals nine-fourths equals 2 and one-fourth

For from the manual by A.P. Ershova and V.V. Goloborodko for the 6th grade, here are the step-by-step solutions for the standard level problems (Variant A1). Variant A1 Solutions 1. Find the product