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Solving fourth-grade math on pages 34 and 35 of VBT Volume 2, Focus Practice, Exercise 113, providing step-by-step solutions for 4 exercises in Part 1 and 2 exercises in Part 2, along with answers closely aligned with the learning content. Students can refer to understand how to solve and practice skills related to fraction problems and divisibility.
Solving fourth-grade math on pages 34 and 35 of VBT Volume 2, Focus Practice, Exercise 113
1. Solving Exercise 1 - Part 1 - Math Exercise Book for Grade 4 Volume 2 Page 34
Problem Statement:
Among the numbers 6215; 6261; 617 ; 6281, the numbers divisible by 3 are:
A. 6215
B. 6261
C. 6217
D. 6281
Solution Method
Calculate the sum of the digits of the given numbers.
A number whose sum of digits is divisible by 3 is itself divisible by 3.
Answer
Since 6 + 2 + 1 + 5 = 14 and 14 is not divisible by 3, 6215 is not divisible by 3.
Since 6 + 2 + 6 + 1 = 15 and 15 is divisible by 3, 6261 is divisible by 3.
Since 6 + 2 + 1 + 7 = 16 and 16 is not divisible by 3, 6217 is not divisible by 3.
Since 6 + 2 + 8 + 1 = 17 and 17 is not divisible by 3, 6281 is not divisible by 3.
Choose B.
2. Solving Exercise 2 - Part 1 - Math Exercise Book for Grade 4 Volume 2 Page 34
Problem Statement:
Hoa has 8 marbles consisting of 4 green marbles, 3 red marbles, and 1 yellow marble. The fraction representing the number of green marbles in the total number of marbles of Hoa is:
A. 4/3
B. 3/8
Choose option D.
3. Solve Exercise 3 - Part 1 - Math workbook for grade 4 exercise book 2 page 34
Problem statement:
The fraction 7/8 is equal to which fraction:
A. 21/32
B. 35/32
C. 21/24
D. 35/48
Solution Method
Apply the fundamental property of fractions: If you multiply both the numerator and denominator of a fraction by the same nonzero natural number, you get a new fraction equivalent to the given fraction.
Answer
4. Solve Exercise 4 - Part 1 - Math workbook for grade 4 exercise book 2 page 34
Problem statement:
Among the following fractions, the fraction less than 1 is:
A. 8/7
B. 7/7
C. 8/8
D. 7/8
Solution Method
A fraction with a numerator smaller than the denominator is less than 1.
Answer
A fraction less than 1 is a fraction with a numerator smaller than the denominator.
So the fraction less than 1 is 7/8.
Choose option D.
1. Solve Exercise 1 - Part 2 - Math workbook for grade 4 exercise book 2 page 34
Problem statement:
Set up and then calculate:
78653 + 80694
527684 - 81946
526 x 205
76140 ÷ 324
Solution Method
Set up and calculate according to the rules learned about addition, subtraction, multiplication, and division of natural numbers.
Answer
2. Solve Exercise 2 - Part 2 - Math workbook for grade 4 exercise book 2 page 35
Problem statement:
Two squares ABCD and BMNC, each with sides of 3cm, are arranged to form rectangle AMND. Given that quadrilateral BMCD is a parallelogram. Calculate the area of parallelogram BMCD using different methods.
Solution Method
- Apply the following formulas:
+ Area of parallelogram = base length x corresponding height.
+ Area of square = side x side.
* Method 1: Area of parallelogram BMCD = DC x BC : 2.
* Method 2: Area of parallelogram BMCD = Area of triangle BCD + Area of triangle BCM.
* Method 3: Area of parallelogram BMCD = Area of half square ABCD + Area of half square BMNC.
Answer
Method 1:
Since quadrilaterals ABCD and BMNC are squares, and quadrilateral BNMC is a parallelogram, it follows that the height h is also the side length BC and the base length is also the side length DC.
The area of parallelogram BMCD is:
S = a x h = DC x BC = 3 x 3 = 9 (cm2)
Answer: 9cm2
Method 2:
The area of parallelogram BMCD equals the area of triangle BCD plus the area of triangle BCM.
Triangle BCD has height BC = 3cm and base length DC = 3cm
The area of triangle BCD is:
SBCD = base length x height : 2
= DC x BC : 2 = 3 x 3 : 2 = 4.5 cm2
Triangle BCM has height CB = 3cm and base length BM = 3cm
The area of triangle BCM is:
SBCM = base length x height : 2
= CB x BM : 2 = 3 x 3 : 2 = 4.5 cm2
The area of parallelogram BMCD is:
S = SBCD + SBCM = 4.5 + 4.5 = 9 cm2
Method 3:
The area of parallelogram BMCD equals the area of half square ABCD plus the area of half square BMNC.
The area of half square ABCD with side length 3cm is:
3 x 3 : 2 = 4.5 cm2
The area of half square BMNC with side length 3cm is:
3 x 3 : 2 = 4.5 cm2
The area of parallelogram BMCD is:
S = 4.5 + 4.5 = 9 cm2
You are currently viewing the instructions for solving grade 4 math problems on pages 34 and 35 of workbook 2. You can review the instructions for solving grade 4 math problems on page 33 of workbook 2 or preview the instructions for solving grade 4 math problems on page 35 of workbook 2 to better understand the lesson.
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