LCPS Grade 4 Mathematics Curriculum Guide 2016-2017 Quarter 2: 45 days Days Unit Standard Content Strand Topic All year Unit 1-Classroom Routines NUMBER TALKS, Problem Solving, Elapsed Time (hour, ½ hour, ¼ hour) 27 4.4b Estimation 4.4a Unit 4-Whole Number Operations & Applications (multiplication and division) Computation and Estimation. 529 #14-17 in class rotations Week 6 10/8- Quiz 2.3, 3.2 Practice: Unit 1 Practice Test 10/9- Quizizz Practice- Ask your students to join game with this code 820292 at 10/10- Unit Test Practice-Jigsaw 10/11- Angle Relationships Foldable Practice- Angle Relationships Wkst (Color Angles).
1.1
1.
Soln:
Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Teachers who design math lessons for self-contained special education classes often need to keep several balls in the air at once. They may need to consider a broad range of age and cognitive ability in their group, understand specific exceptionalities and learning styles, and implement a variety of unique learning objectives.
A man can enter the stadium in 4 ways. Again the man can leave the stadium in 9 ways.
So, total no.of ways with which a man enters and then leaves the stadium = 4 * 9 = 36ways.
2.
Soln:
There are 6 choices for a student to enter the hostel. There are 5 choices for a student to leave the hostel as different door is to be used.
So, total no.of ways = 6 * 5 = 30.
3.
Soln:
There are 7 choices for 1st son, 6 choices for 2nd son and 5 choices for 3rd son.
Now, by the basic principle of counting, the total number of ways of choice = 7 * 6 * 5 = 210.
4.
Soln:
A man can go from city A to city B in 5 ways. As he has to return by a different road, so he can return from city B to city A in 4 ways.
So, total no.of ways by which a man can go from city A to city B and returns by a different road = 5 * 4 = 20 ways.
5.
Soln:
A person can go from city A to city B in 5 ways. Again, he can go from city B to city C in 4 ways. So, a person can go from city A to city C in 5 * 4 = 20ways. The person has to return from C to A without driving on the same road twice, So, he can return from city C to city B in 3 ways and from city B to city A in 4 ways.
So, he can return from city C to city A in 3 * 4 = 12 ways.
So, Total no.of ways by which a person can go from city A to city C and return from city C to city A = 20 * 12 = 240 ways.
6.
Soln:
Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.
There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred's place respectively.
So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120
Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.
the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.
The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.
So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.
Old Unit4 Agendasmrs. Colville's Math Class 1
7.
Old Unit4 Agendasmrs. Colville's Math Classroom
Soln:
The numbers formed must be of three digits and less than 500, so the digit in the hundred's place should be 1,2,3 or 4. So, there are 4 choices for the digit in the hundred's place. There are 5 choice for the digit in the ten's place. There are 4 choices for the digit in the unit's place.
Old Unit4 Agendasmrs. Colville's Math Classes
So, no of ways by which 3 digits numbers les than 500 can be formed = 4.5.4 = 80.
8.
Soln:
The numbers formed should be even. So, the digit in the unit's place must be 2 or 4. So, the digit in unit's place must be 2 or 4. So, for the digit in unit's place, there are 2 choices. So, after fixing the digit in the unit's place, remaining 4 figures can be arranged in P(4,4) ways.
Ie. $frac{{left( 4 right)!}}{{left( {4 - 4} right)!}}$ = $frac{{4!}}{{0!}}$ = $frac{{4{rm{*}}3{rm{*}}2{rm{*}}1}}{1}$ = 24 ways.
Old Unit4 Agendasmrs. Colville's Math Class 9
So, total no.of ways by which 5 even numbers can be formed = 2 * 24 = 48.
9.
Soln:
The numbers formed must be of 4 digits. The digit in the thousand's place must always be 4. For this, there is only one choice. After that, n = 6 – 1 = 5, r = 4 – 1 = 3. Then remaining 5 figures can be placed in remaining 3 places in:
Or, P(5,3) ways = $frac{{5!}}{{left( {5 - 3} right)!}}$ = $frac{{5!}}{{2!}}$ = $frac{{5{rm{*}}4{rm{*}}3{rm{*}}2{rm{*}}1}}{{2{rm{*}}1}}$ = 60 ways.
So, Total no.of ways by which 4 digits numbers between 4,000 and 5,000 can be formed = 1 * 60 = 60.
So, total no.of ways with which a man enters and then leaves the stadium = 4 * 9 = 36ways.
2.
Soln:
There are 6 choices for a student to enter the hostel. There are 5 choices for a student to leave the hostel as different door is to be used.
So, total no.of ways = 6 * 5 = 30.
3.
Soln:
There are 7 choices for 1st son, 6 choices for 2nd son and 5 choices for 3rd son.
Now, by the basic principle of counting, the total number of ways of choice = 7 * 6 * 5 = 210.
4.
Soln:
A man can go from city A to city B in 5 ways. As he has to return by a different road, so he can return from city B to city A in 4 ways.
So, total no.of ways by which a man can go from city A to city B and returns by a different road = 5 * 4 = 20 ways.
5.
Soln:
A person can go from city A to city B in 5 ways. Again, he can go from city B to city C in 4 ways. So, a person can go from city A to city C in 5 * 4 = 20ways. The person has to return from C to A without driving on the same road twice, So, he can return from city C to city B in 3 ways and from city B to city A in 4 ways.
So, he can return from city C to city A in 3 * 4 = 12 ways.
So, Total no.of ways by which a person can go from city A to city C and return from city C to city A = 20 * 12 = 240 ways.
6.
Soln:
Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.
There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred's place respectively.
So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120
Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.
the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.
The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.
So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.
Old Unit4 Agendasmrs. Colville's Math Class 1
7.
Old Unit4 Agendasmrs. Colville's Math Classroom
Soln:
The numbers formed must be of three digits and less than 500, so the digit in the hundred's place should be 1,2,3 or 4. So, there are 4 choices for the digit in the hundred's place. There are 5 choice for the digit in the ten's place. There are 4 choices for the digit in the unit's place.
Old Unit4 Agendasmrs. Colville's Math Classes
So, no of ways by which 3 digits numbers les than 500 can be formed = 4.5.4 = 80.
8.
Soln:
The numbers formed should be even. So, the digit in the unit's place must be 2 or 4. So, the digit in unit's place must be 2 or 4. So, for the digit in unit's place, there are 2 choices. So, after fixing the digit in the unit's place, remaining 4 figures can be arranged in P(4,4) ways.
Ie. $frac{{left( 4 right)!}}{{left( {4 - 4} right)!}}$ = $frac{{4!}}{{0!}}$ = $frac{{4{rm{*}}3{rm{*}}2{rm{*}}1}}{1}$ = 24 ways.
Old Unit4 Agendasmrs. Colville's Math Class 9
So, total no.of ways by which 5 even numbers can be formed = 2 * 24 = 48.
9.
Soln:
The numbers formed must be of 4 digits. The digit in the thousand's place must always be 4. For this, there is only one choice. After that, n = 6 – 1 = 5, r = 4 – 1 = 3. Then remaining 5 figures can be placed in remaining 3 places in:
Or, P(5,3) ways = $frac{{5!}}{{left( {5 - 3} right)!}}$ = $frac{{5!}}{{2!}}$ = $frac{{5{rm{*}}4{rm{*}}3{rm{*}}2{rm{*}}1}}{{2{rm{*}}1}}$ = 60 ways.
So, Total no.of ways by which 4 digits numbers between 4,000 and 5,000 can be formed = 1 * 60 = 60.
10.
Soln:
For the three digits numbers, there are 5 ways to fill in the 1st place, there are 4 ways to fill in the 2nd place and there are 3 ways to fill in the 3rd place. By the basic principle of counting, number of three digits numbers = 5 * 4 * 3 = 60.
Again, for three digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten's place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 5 = 1 * 4 * 3 = 12.
