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Preparation for Algebra Class 5 Problem Set 56 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 56 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 16 Preparation for Algebra

Question 1.
Use a letter for ‘any number’ and write the following properties in short.

(1) The sum of any number and zero is the number itself.
Answer:
a + 0 = a

(2) The product of any two numbers and the product obtained after changing the order of those numbers is the same.
Answer:
a x b = b x a

(3) The product of any number and zero is zero.
Answer:
a x 0 = 0

Question 2.
Write the following properties in words :

(1) m – 0 = m
Answer:
Subtracting zero from any number, gives the number itself.

(2) n ÷ 1 = n
Answer:
Dividing any number by 1, gives the number itself.

Preparation for Algebra Problem Set 56 Additional Important Questions and Answers

Use a letter for any number and write the following properties in short.

Question 1.
The product of any number and 1 is the number itself.
Answer:
a x 1 = a

Question 2.
The division of any two different numbers and the divisions obtained after changing the order of those numbers is not the same.
Answer:
a ÷ b ≠ b + a

Write the following properties in words:

Question 1.
p x 0 = 0
Answer:
The product of any number and zero is zero.

(4) a + b = b + a
Answer:
The sum of any two numbers and the sum obtained after changing the order of these numbers is the same.

Using brackets write three pairs of numbers whose

(1) Sum is 9
Answer:
5 + 4 = 9,
7 + 2 = 9,
8 + 1 = 9

(2) difference is 9
Answer:
12 – 3 = 9,
11 – 2 = 9,
10 – 1 = 9

(3) multiplication is 16 and
Answer:
4 x 4 = 16,
8 x 2 = 16,
16 x 1 = 16

(4) division is 16.
Answer:
32 ÷ 2 = 16,
48 ÷ 3 = 16,
64 ÷ 4 = 16,

Fill in the blanks.

(1) 4 + 2 = 7 – ……….
(2) 4 + 2 = 3 x ……….
(3) 4 + 2 = 12 ÷ ……….
Answer:
(1) 1
(2) 2
(3) 2

Match the columns:

(A)

Answer:
(1 – c),
(2 – a),
(3-d),
(4-b)

(B)

Answer:
(1-d),
(2 – c),
(3 – a),
(4 – b)

Say whether right or wrong.

(1) (6 + 5) = (5 + 6)
(2) (8 + 5) > 10
(3) (8 + 5) < 10
(4) 108 > 108
(5) 108 = 108
(6) 108 < 108
(7) (6 x 3) = (20 – 2)
(8) 40 + 8 > 5
(9) (3 x 7) = (7 x 3)
(10) (5 + 0) = (5 x 1)
(11) (6 + 5) = 10
(12) (30 + 5) < (30 – 25)
Answer:
Right : (1), (2), (5), (7), (9), (10)
Wrong : (3), (4), (6), (8), (11), (12)

Fill in the blanks with the right symbol from <, > or =

(1) (24 ÷ 5) ……… (9 – 5)
(2) (4 + 2) ……… (5 x 1)
(3) (7 x 3) ……… (20 + 2)
(4) (8 x 2) (5 x 3)
(5) (5 x 6) ……… (25 + 5)
(6) (6 x 7) (9 x 5)
Answer:
(1) =
(2) >
(3) <
(4) >
(5) =
(6) <

Fill in the blanks in the expressions with the proper numbers.

(1) (4 x 4) = (………. x 2)
(2) (2 x 7) > (4 x ……….)
(3) (30 + 5) < ( x 3)
(4) (5 + 0)> (4 x ……….)
(5) (36 +3) = ( + )
(6) (9 – ……….) < (4 + 1)
(7) (8 + 9) < (3 x ……….)
(8) (0 + 3) > (4 x ……….)
(9) (28 ÷ 2) = (7 x ……….)
Answer:
(1) 8
(2) 3
(3) 9
(4) 1
(5) 7 + 5
(6) 5,
(7) 6
(8) 0
(9) 2

Use a letter for any number and write the following properties in short:

(1) Dividing zero by any non zero number is zero.
Answer:
0 + a = 0

(2) The difference of any two different numbers and the difference obtained after changing the order of those numbers is not same.
Answer:
a – b ≠ b – a

(3) Dividing non zero number by itself gives us 1.
Answer:
a ÷ a = 1

Write the followîng properties in words:

(1) a x 1 = a
Answer:
The product of any number and 1 is the number itself.

(2) a – a = 0
Answer:
Difference of the same two numbers is zero.

Maharashtra Board Class 5 Maths Solutions

• Three Dimensional Objects and Nets Problem Set 51  Class 5 Maths Solutions
• Pictographs Problem Set 52 Class 5 Maths Solutions
• Patterns Problem Set 53 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 54 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 55 Class 5 Maths Solutions
• Preparation for Algebra Problem Set 56 Class 5 Maths Solutions

Multiplication and Division Class 5 Problem Set 16 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 4 Multiplication and Division Problem Set 16 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 4 Multiplication and Division

Question 1.
From a total of 10,000 rupees, Anna donated 7,000 rupees to a school. The remaining amount was to be divided equally among six students as the ‘all-round student’ prize. What was the amount of each prize?
Solution:
1 0 0 0 0 Total rupees
–
7 0 0 0 rupees donated
_______
3 0 0 0 remained
_______
This amount was divided among 6 students 500

Answer:
Amount of the prize is ₹ 500.

Question 2.
An amount of 260 rupees each was collected from 50 students for a picnic. If 11,450 rupees were spent for the picnic, what is the amount left over?
Solution:
₹ 2 6 0 Collected from 1 student
x
5 0 No. of students
0 0 0
+
1 3 0 0 0
1 3 0 0 0 Rupees, collected amount
–
1 1 4 5 0 Rupees spent
1 5 5 0 Rupees left over

Answer:
1,550 Rupees leftover

Question 3.
A shopkeeper bought a sack of 50kg of sugar for 1750 rupees. As the price of sugar fell, he had to sell it at the rate of 32 rupees per kilo. How much less money did he get than he had spent?
Solution:
₹ 3 2 Sale price of 1 kg
x
5 0 kg sold
0 0
+
1 6 0 0
1 6 0 0 Amount received
₹ 1 7 5 0 Purchased price
–
₹ 1 6 0 0 Obtained price
₹ 1 5 0 Lesshegot

Answer:
₹ 150 less he got than he had spent

Question 4.
A shopkeeper bought 7 pressure cookers at the rate of 1870 rupees per cooker. He sold them all for a total of 14,230 rupees. Did he get less or more money than he had spent?
Solution:
₹ 1 8 7 0 Purchase price of 1 cooker
x
7 No. of cookers
₹ 1 3 0 9 0 Purchase price
₹ 1 4 2 3 0 Sell price
–
₹ 1 3 0 9 0 Purchase price
₹ 0 1 1 4 0 he got more

Answer:
₹ 1,140 he got more

Question 5.
Fourteen families in a Society together bought 8 sacks of wheat, each weighing 98 kilos. If they shared all the wheat equally, what was the share of each family?
Solution:
9 8 Kilo weight of 1 sack
x
8 No. of sacks
7 8 4 Kilo

Answer:
Share of each family = 56 kilo

Question 6.
The capacity of an overhead water tank is 3000 litres. There are 16 families living in this building. If each family uses 225 litres every day, will the tank filled to capacity be enough for all the families? If not, what will the daily shortfall be?
Solution:
₹ 2 2 5 Litres uses 1 family
x 1 6 No. of families
1 3 5 0
+
2 2 5 0
3 6 0 0 Litres required
–
3 0 0 0 Litres capacity
6 0 0 Litres daily shortfall

Answer:
600 litres is daily shortfall

Multiplication and Division Problem Set 16 Additional Important Questions and Answers

Solve the following word problems:

(1) A farmer brought 250 trays of tomatoes seedlings. Each tray had 48 seedlings. He planted all the seedlings in his field, putting 25 in a row. How many rows of tomatoes did he plant?
Solution:
₹ 2 5 0 Tray of tomatoes seedlings
x
4 8 Seedlings in 1 tray
2 0 0 0
+
1 0 0 0 0
1 2 0 0 0 Total no. of seedlings


Answer:
The number of rows is 480.

Q.l. Solve the following :
Multiply :
(1) 438 x 76
(2) 594 x 208
(3) 3467 x 926
(4) 3581 x 87
(5) 425 x 87
(6) 579 x 49
Answer:
(1) 33,288
(2) 1,23,552
(3) 32,10,442
(4) 3,11,547
(5) 36,975
(6) 28,371

Solve the following and write the quotient and remainder:

(1) 1345 ÷ 37
(2) 9682 ÷ 83
(3) 6371 ÷ 42
(4) 72534 ÷ 23
(5) 1284 ÷ 32
(6) 63240 ÷ 37
Answer:
(1) Quotient 36, Remainder 13
(2) Quotient 116, Remainder 54
(3) Quotient 151, Remainder 29
(4) Quotient 3153, Remainder 15
(5) Quotient 40, Remainder 4
(6) Quotient 1709, Remainder 7

Fill in the blanks :

(1) 88 x 17; 17 is called …………………………. and 88 is called
(2) Product of the greatest three-digit number and smaller two-digit number is ………………………… .
(3) Multiplicand and multiplier are interchanged the product remains the ………………………… .
(4) While multiplying by tens digit, we have to put in the units place ………………………… .
Answer:
(1) multiplier, multiplicand
(2) 99,900
(3) same
(4) zero

Maharashtra Board Class 5 Maths Solutions

• Multiplication and Division Problem Set 14 Class 5 Maths Solutions
• Multiplication and Division Problem Set 15 Class 5 Maths Solutions
• Multiplication and Division Problem Set 16 Class 5 Maths Solutions

Symmetry Class 6 Maths Chapter 7 Practice Set 21 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 7 Symmetry Class 6 Practice Set 21 Answers Solutions.

Std 6 Maths Practice Set 21 Solutions Answers

Question 1.
Along each figure shown below, a line l has been drawn. Complete the symmetrical figures by drawing a figure on the other side such that the line l becomes the line of symmetry.

Solution:

Maharashtra Board Class 6 Maths Chapter 7 Symmetry Practice Set 21 Intext Questions and Activities

Question 1.
In the figures below, the line l divides the figure in two parts. Do these parts fall on each other? Verify? (Textbook pg. no. 42)

Solution:
Yes, the two parts of both the figures fall on each other on folding along the line l.

6th Std Maths Digest Pdf Download

• Bar Graphs Practice Set 18 Class 6 Maths Solution
• Bar Graphs Practice Set 19 Class 6 Maths Solution
• Symmetry Practice Set 20 Class 6 Maths Solution
• Symmetry Practice Set 21 Class 6 Maths Solution
• Divisibility Practice Set 22 Class 6 Maths Solution
• HCF-LCM Practice Set 23 Class 6 Maths Solution
• HCF-LCM Practice Set 24 Class 6 Maths Solution
• HCF-LCM Practice Set 25 Class 6 Maths Solution

Decimal Fractions Class 5 Problem Set 37 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 9 Decimal Fractions Problem Set 37 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 9 Decimal Fractions

Write the following mixed fractions in decimal form and read them aloud.

\(\text { (1) } 9 \frac{1}{10}\)
Answer:
9.1, Nine point one.

\(\text { (2) } 9 \frac{1}{100}\)
Answer:
9.01, Nine point zero one.

\(\text { (3) } 4 \frac{53}{100}\)
Answer:
4.53, Four point five three.

\(\text { (4) } \frac{78}{100}\)
Answer:
0.78, Zero point seven eight.

\(\text { (5) } \frac{5}{100}\)
Answer:
0.05, Zero point zero five.

\(\text { (6) } \frac{5}{10}\)
Answer:
0.5, Zero point five.

\(\text { (7) } \frac{2}{10}\)
Answer:
0.2, Zero point two.

\(\text { (8) } \frac{20}{100}\)
Answer:
0.20, Zero point two zero.

Place value of the digits in decimal fractions

We can determine the place value of the digits in decimal fractions in the same way that we determine the place values of digits in whole numbers.

Example (1)
In 73.82, the place value of 7 is 7 × 10 = 70, and of 3, it is 3 × 1 = 3.
Similarly, the place value of 8 is 8 × \(\frac{1}{10}=\frac{8}{10}\) = 0.8 and the place value of 2 is 2 × \(\frac{1}{100}=\frac{2}{100}\) = 0.02

Example (2)
Place values of the digits in 210.86.

Decimal Fractions Problem Set 37 Additional Important Questions and Answers

\(\text { (1) } 3 \frac{3}{100}\)
Answer:
3.03, Three point zero three.

\(\text { (2) } 3 \frac{33}{100}\)
Answer:
3.33, Three point three three.

\(\text { (3) } 30 \frac{41}{100}\)
Answer:
30.41, Thirty point four one.

\(\text { (4) } 11 \frac{11}{100}\)
Answer:
11.11, Eleven point one one.

Q.3. Write the following numbers using the decimal point:
(1) Sixty-eight point seven six .
Answer:
68.76

(2) Nine point five zero one
Answer:
9.501

(3) Eighty-four point zero three.
Answer:
84.03

(4) Eighty-four point zero zero seven.
Answer:
84.007

(5) Two hundred ftinety-eight point zero seven.
Answer:
298.07

Maharashtra Board Class 5 Maths Solutions

• Decimal Fractions Problem Set 36 Class 5 Maths Solutions
• Decimal Fractions Problem Set 37 Class 5 Maths Solutions
• Decimal Fractions Problem Set 38 Class 5 Maths Solutions
• Decimal Fractions Problem Set 39 Class 5 Maths Solutions
• Decimal Fractions Problem Set 40 Class 5 Maths Solutions
• Decimal Fractions Problem Set 41 Class 5 Maths Solutions
• Decimal Fractions Problem Set 42 Class 5 Maths Solutions

Percentage Class 6 Maths Chapter 12 Practice Set 30 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 12 Percentage Class 6 Practice Set 30 Answers Solutions.

Std 6 Maths Practice Set 30 Solutions Answers

Question 1.
Shabana scored 736 marks out of 800 in her exams. What was the percentage she scored?
Solution:
Total marks of the examination = 800
Marks scored by Shabana = 736
Suppose Shabana scored A% marks.

∴ A = 92%
∴ Shabana scored 92% marks.

Question 2.
There are 500 students in the school in Dahihanda village. If 350 of them can swim, what percent of them can swim and what percent cannot?
Solution:
Total number of students in the school = 500
Number of students who can swim = 350
Suppose A% students can swim.

∴ A = 70%
Percentage of students who cannot swim = 100% – Percentage of students who can swim .
= 100% – 70% = 30%
∴ 70% of the students can swim and 30% cannot swim.

Question 3.
If Prakash sowed jowar on 75% of the 19500 sq. m. of his land, on how many sq. m. did he actually plant jowar?
Solution:
Total area of the land = 19500 sq. m.
Percentage of area in which Prakash sowed jowar = 75%
Suppose Prakash planted jowar in A sq. m.

∴ A = 14,625 sq. m.
∴ Prakash planted jowar in 14,625 sq.m.

Question 4.
Soham received 40 messages on his birthday. Of these, 90% were birthday greetings. How many other messages did he get besides the greetings?
Solution:
Total messages received by Soham on his birthday = 40
Percentage of messages received for birthday greetings = 90%
Suppose Soham got A number of birthday greetings.

∴ A = 36
∴ Number of messages received other than birthday greetings
= total messages received – total number of birthday greetings
= 40 – 36 = 4
∴ The number of messages received other than birthday greetings is 4.

Question 5.
Of the 5675 people in a village 5448 are literate. What is the percentage of literacy in the village?
Solution:
Number of people in the village = 5675
Number of people who are literate = 5448
Suppose the percentage of literacy in the village is A%.

∴ A = 96%
∴ The percentage of literacy in the village is 96%.

Question 6.
In the elections, 1080 of the 1200 women in Jambhulgaon cast their vote, while 1360 of the 1700 in Wadgaon cast theirs. In which village did a greater proportion of women cast their votes?
Solution:
Total number of women in Jambhulgaon = 1200
Number of women in Jambhulgaon who voted = 1080
Suppose A% women cast their vote in Jambhulgaon village.


∴ A = 90%
In Jambhulgaon, the percentage of women who voted in the elections was 90%.
Total number of women in Wadgaon = 1700 Number of women in Wadgaon who voted = 1360
Suppose B% women cast their vote in Wadgaon.

∴ B = 80%
∴ In Wadgaon, the percentage of women who voted in the elections was 80%.
∴ A greater proportion of women cast their votes in Jambhulgaon.

Maharashtra Board Class 6 Maths Chapter 12 Percentage Practice Set 30 Intext Questions and Activities

Question 1.
There are 9 squares in the figure alongside. The letters ABCDEFGHI are written in squares. Give each of the letters a unique number from 1 to 9 so that every letter has a different number.
Besides, A + B + C = C + D + E = E + F + G = G + H + I should also be true. (Textbook pg. no. 64)

Solution:

(This is one of the possible solutions of the above riddle. There are more solutions possible.)

6th Std Maths Digest Pdf Download

• Equations Practice Set 26 Class 6 Maths Solution
• Equations Practice Set 27 Class 6 Maths Solution
• Ratio-Proportion Practice Set 28 Class 6 Maths Solution
• Ratio-Proportion Practice Set 29 Class 6 Maths Solution
• Percentage Practice Set 30 Class 6 Maths Solution
• Profit-Loss Practice Set 31 Class 6 Maths Solution
• Profit-Loss Practice Set 32 Class 6 Maths Solution
• Profit-Loss Practice Set 33 Class 6 Maths Solution
• Profit-Loss Practice Set 34 Class 6 Maths Solution
• Banks and Simple Interest Practice Set 35 Class 6 Maths Solution

Integers Class 6 Maths Chapter 3 Practice Set 8 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 3 Integers Class 6 Practice Set 8 Answers Solutions.

Std 6 Maths Practice Set 8 Solutions Answers

Question 1.
Subtract the numbers in the top row from the numbers in the first column and write the proper number in each empty box:

Solution:

Maharashtra Board Class 6 Maths Chapter 3 Integers Practice Set 8 Intext Questions and Activities

Question 1.
A Game of Integers. (Textbook pg. no. 20)
The board for playing this game is given in the back cover of the textbook. Place your counters before the number 1. Throw the dice. Look at the number you get. It is a positive number. Count that many boxes and move your counter forward. If a problem is given in that box, solve it. If the answer is a positive number, move your counter that many boxes further. It it is negative, move back by that same number of boxes.

Suppose we have reached the 18th box. Then the answer to the problem in it is -4 + 2 = -2. Now move your counter back by 2 boxes to 16. The one who reaches 100 first, is the winner.

Solution:
(Students should attempt this activity on their own)

6th Std Maths Digest Pdf Download

• Basic Concepts in Geometry Practice Set 1 Class 6 Maths Solution
• Angles Practice Set 2 Class 6 Maths Solution
• Angles Practice Set 3 Class 6 Maths Solution
• Integers Practice Set 4 Class 6 Maths Solution
• Integers Practice Set 5 Class 6 Maths Solution
• Integers Practice Set 6 Class 6 Maths Solution
• Integers Practice Set 7 Class 6 Maths Solution
• Integers Practice Set 8 Class 6 Maths Solution

Symmetry Class 6 Maths Chapter 7 Practice Set 20 Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 7 Symmetry Class 6 Practice Set 20 Answers Solutions.

Std 6 Maths Practice Set 20 Solutions Answers

Question 1.
Draw the axes of symmetry of each of the figures below. Which of them has more than one axis of symmetry?

Solution:

Figures (i), (ii) and (iv) have more than one axis of symmetry.

Question 2.
Write the capital letters of the English alphabet in your notebook. Try to draw their axes of symmetry. Which ones have an axis of symmetry? Which ones have more than one axis of symmetry?
Solution:
Alphabets having axis of symmetry:

Alphabets having more than one axis of symmetry:

Question 3.
Use color, a thread and a folded paper to draw symmetrical shapes.
Solution:
Take any color, a thread and a folded square paper.
Step 1:
Take a folded square paper which is folded along one of its axis of symmetry.

Step 2:
Open the paper. Draw a square in one comer. Place the thread in the square drawn and apply colour on it as shown in the figure.

Step 3:
Remove the thread. You will see a white patch where the thread was.

Step 4:
Fold the paper and press it along the axis of symmetry. When you unfold the paper, you will see an imprint on the other side of the fold which is identical to the color patch you had made earlier.

Question 4.
Observe various commonly seen objects such as tree leaves, birds in flight, pictures of historical buildings, etc. Find symmetrical shapes among them and make a collection of them.
Solution:
Some of the symmetrical objects seen in daily life are shown below:

Maharashtra Board Class 6 Maths Chapter 7 Symmetry Practice Set 20 Intext Questions and Activities

Question 1.
Do you recognize this picture?
Why do you think the letters written on the front of the vehicle are written the way they are? Copy them on a paper. Hold the paper in front of a mirror and read it.
Do you see letters written like this anywhere else?
(Textbook pg. no. 40)

Solution:

1. The name written in reverse alphabets on the vehicle reads
   as ‘AMBULANCE’ when viewed in the mirror.
   In the case of an emergency, it helps a driver to quickly notice an ambulance by looking into his rear view mirror and read the reverse alphabets which appear perfectly normal in a mirror
2. Other than ambulance, we see letters written in reverse on school bus.

6th Std Maths Digest Pdf Download

• Bar Graphs Practice Set 18 Class 6 Maths Solution
• Bar Graphs Practice Set 19 Class 6 Maths Solution
• Symmetry Practice Set 20 Class 6 Maths Solution
• Symmetry Practice Set 21 Class 6 Maths Solution
• Divisibility Practice Set 22 Class 6 Maths Solution
• HCF-LCM Practice Set 23 Class 6 Maths Solution
• HCF-LCM Practice Set 24 Class 6 Maths Solution
• HCF-LCM Practice Set 25 Class 6 Maths Solution

Fractions Class 5 Problem Set 18 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 18 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 5 Fractions

Convert the given fractions into like fractions.

Solution :
8 is the multiple of 4 So, make 8, the common denominator \(\frac{3}{4}=\frac{3 \times 2}{4 \times 2}=\frac{6}{8}\).Thus 6/8 and 5/8are the required like fractions.

Solution :
The number 35 is a multiple of both 5 and 7 So, making 35 as the common denominater \(\frac{3}{5}=\frac{3 \times 7}{5 \times 7}=\frac{21}{35}, \frac{3}{7}=\frac{3 \times 5}{7 \times 5}=\frac{15}{35}\) Therefore, 21/35 and 15/35 are required like fractions.

Solution :
Here 10 is the multiples of 5. So make 10 as the common denominator \(\frac{4}{5}=\frac{4 \times 2}{5 \times 2}=\frac{8}{10}\). Thus 8/10 and 3/10 are required like fractions.

Solution :
Least common multiple of 9 and 6 is 18. So, make, 18 as the common denominator. \(\frac{2}{9}=\frac{2 \times 2}{9 \times 2}=\frac{4}{18}, \frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}\). Thus, 4/18 and 3/18 are the required like fractions.

Solution :
Least common multiple of 4 and 3 is 12 So, make 12 as common denominator \(\frac{1}{4}=\frac{1 \times 3}{4 \times 3}=\frac{3}{12}, \frac{2}{3}=\frac{2 \times 4}{3 \times 4}=\frac{8}{12}\). so, \(\frac{3}{12}, \frac{8}{12}\) are required like fractions.

Solution :
Least common multiple of 6 and 5 is 30 So, make 30 as common denominator \(\frac{5}{6}=\frac{5 \times 5}{6 \times 5}=\frac{25}{30}, \frac{4}{5}=\frac{4 \times 6}{5 \times 6}=\frac{24}{30}\) So, \(\frac{25}{30}, \frac{24}{30}\) are required like fractions.

Solution :
Least common multiple of 8 and 6 is 24 So, make 24 as common denominator \(\frac{3}{8}=\frac{3 \times 3}{8 \times 3}=\frac{9}{24}, \frac{1}{6}=\frac{1 \times 4}{6 \times 4}=\frac{4}{24}\) So, \(\frac{9}{24}, \frac{4}{24}\) are required like fractions.

Solution :
Least common multiple of 6 and 9 is 18 So, make 18 as common denominator \(\frac{1}{6}=\frac{1 \times 3}{6 \times 3}=\frac{3}{18}, \frac{4}{9}=\frac{4 \times 2}{9 \times 2}=\frac{8}{18}\) So, 3/18 and 8/18 are the required like fractions.

Comparing like fractions
Example (1) A strip was divided into 5 equal parts. It means that each part is 1/5 . The coloured part is \(\frac{3}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\).

The white part is \(\frac{2}{5}=\frac{1}{5}+\frac{1}{5}\). The coloured part is bigger than the white part. This tells us that 3/5 is greater than 2/5. This is written as 3/5 > 2/5.

Example (2) This strip is divided into 8 equal parts. 3 of the parts have one colour and 4 have another colour. Here, 3/8 < 8/4.

In like fractions, the fraction with the greater numerator is the greater fraction.

Comparing fractions with equal numerators
You have learnt that the value of fractions with numerator 1 decreases as the denominator increases.

Even if the numerator is not 1, the same rule applies so long as all the fractions have a common numerator. For example, look at the figures below. All the strips in the figure are alike.
2 of the 3 equal parts of the strip
2 of the 4 equal parts of the strip
2 of the 5 equal parts of the strip
The figure shows that 2/3 > 2/4 > 5/2.

Of two fractions with equal numerators, the fraction with the greater denominator is the smaller fraction.

Comparing unlike fractions
Teacher : Suppose we have to compare the unlike fractions 3/5 and 4/7. Let us take an example to see how this is done. These two boys are standing on two blocks. How do we decide who is taller?

Sonu : But the height of the blocks is not the same. If both blocks are of the same height, it is easy to tell who is taller.

Nandu : Now that they are on blocks of equal height, we see that the boy on the right is taller.

Teacher : The height of the boys can be compared when they stand at the same height. Similarly, if fractions have the same denominators, their numerators decide which fraction is bigger.

Nandu : Got it! Let’s obtain the same denominators for both fractions.

Sonu : 5 × 7 can be divided by both 5 and 7. So, 35 can be the common denominator.

To compare unlike fractions, we convert them into their equivalent fractions so that their denominators are the same.

Fractions Problem Set 18 Additional Important Questions and Answers

Question 1.
\(\frac{5}{9}, \frac{17}{36}\)
Solution :
36 is the multiple of 9 So, make 36 the common denominator \(\frac{5}{9}=\frac{5 \times 4}{9 \times 4}=\frac{20}{36}\), Thus 20/36 and 17/36 are the required like fractions.

Question 2.
\(\frac{5}{6}, \frac{7}{9}\)
Solution:
Least common multiple of 6 and 9 is 18 So, make 18 as the common denominator \(\frac{5}{6}=\frac{5 \times 3}{6 \times 3}=\frac{15}{18}, \quad \frac{7}{9}=\frac{7 \times 2}{9 \times 2}=\frac{14}{18}\) So, 15/18 and 14/18 are the required like fractions.

Question 3.
\(\frac{7}{11}, \frac{3}{5}\)
Solution:
Least common multiple of 11 and 5 is 55 So, make 55 as the common denominator. \(\frac{7}{11}=\frac{7 \times 5}{11 \times 5}=\frac{35}{55}, \frac{3}{5}=\frac{3 \times 11}{5 \times 11}=\frac{33}{55}\). Thus 35/55 and 33/55 are required like fractions.

Maharashtra Board Class 5 Maths Solutions

• Fractions Problem Set 17 Class 5 Maths Solutions
• Fractions Problem Set 18 Class 5 Maths Solutions
• Fractions Problem Set 19 Class 5 Maths Solutions
• Fractions Problem Set 20 Class 5 Maths Solutions
• Fractions Problem Set 21 Class 5 Maths Solutions
• Fractions Problem Set 22 Class 5 Maths Solutions
• Fractions Problem Set 23 Class 5 Maths Solutions

Addition and Subtraction Class 5 Problem Set 10 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 3 Addition and Subtraction Problem Set 10 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 3 Addition and Subtraction

Question 1.
Subtract the following:

(1) 64293
–
28547
________
________
Solution:
6 4 2 9 3
–
2 8 5 4 7
Answer:
3 5 7 4 6

(2) 37058
–
23469
________
________
Solution:
3 7 0 5 8
–
2 3 4 6 9
Answer:
1 3 5 8 9

(3) 71540
–
58628
________
________
Solution:
7 1 5 4 0
–
5 8 6 2 8
Answer:
1 2 9 1 2

(4) 50432
–
48647
________
________
Solution:
5 0 4 3 2
–
4 8 6 4 7
Answer:
0 1 7 8 5

Subtraction of six and seven-digit numbers

You have learnt to carry out subtractions of five-digit numbers. Using the same method, we can do subtractions of numbers with more than five digits. Study the following examples.

Subtract :
Example (1) 65,07,843 – 9,25,586

Example (2) 34,61,058 – 27,04,579

As shown in the above example, learn to subtract by keeping the borrowed numbers in your mind without writing them down.

Subtraction by another method Before subtracting one number from another, if we add 10 or 100 to both of them, the difference remains the same. Let us use this fact.

Example : Subtract : 724 – 376
As we cannot subtract 6 from 4, we shall add a ten to both the numbers. For the upper number, we untie one tens. We add those ten units to 4 units to get 14 units.

We write the tens added to the lower number below it, in the tens place.

We subtract 6 units from 14.

Now, we cannot subtract ( 7 + 1) i.e. 8 tens from 2 tens. So, we add one hundred to both the numbers. For the upper number, we untie the hundred and add the ten tens to 2 tens. To add the hundred to the lower number, we write it below, in the hundreds place. 12 tens minus 8 tens is 4 tens. And 7 hundreds minus (3 + 1) i.e. 4 hundreds is 3 hundreds. Hence, the difference is 348.

Example (1)

Example (2)

Roman Numerals Problem Set 5 Additional Important Questions and Answers

Question 1.
Subtract the following:

(1) 81,345 – 35,667
Solution:
8 1 3 4 5
–
3 5 6 6 7
Answer:
4 5 6 7 8

(2) 64,723 – 52,378
Solution:
6 4 7 2 3
–
5 2 3 7 8
Answer:
1 2 3 4 5

Maharashtra Board Class 5 Maths Solutions

• Addition and Subtraction Problem Set 7 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 8 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 9 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 10 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 11 Class 5 Maths Solutions 
• Addition and Subtraction Problem Set 12 Class 5 Maths Solutions
• Addition and Subtraction Problem Set 13 Class 5 Maths Solutions

Angles Class 5 Problem Set 24 Question Answer Maharashtra Board

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 24 Textbook Exercise Important Questions and Answers.

Std 5 Maths Chapter 6 Angles

Complete the following table.

Answer:

The protractor

A protractor is used to measure an angle and also to draw an angle according to a given measure.

The picture opposite shows a protractor.

A protractor is semi-circular in shape.

The semi-circular edge of a protractor is divided into 180 equal parts. Each part is ‘one degree’. ‘One degree’ is written as ‘1°’.

The divisions on a protractor, i.e., the degrees can be marked in two ways. The divisions 0, 10, 20, 30,…180 are marked anticlockwise or from right to left; the divisions 0, 10, 20, 30,…180 are also marked clockwise, or serially from left to right.

The centre of the circle of which the protractor is a half part, is called the centre of the protractor. A diameter of that circle is the baseline or line of reference of the protractor.

Measuring angles

Observe how to measure ∠ABC given alongside, using a protractor.

1. First, put the centre of the protractor on the vertex B of the angle. Place the baseline of the protractor exactly on arm BC. The arms of the angle do not reach the divisions on the protractor.

2. At such times, set the protractor aside and extend the arms of the angle. Extending the arms of the angle does not change the measure of the angle.

3. The angle is measured starting from the zero on that side of the vertex on which the arm of the angle lies. Here, the arm BC is on the right of the vertex B. Therefore, count the divisions starting from the 0 on the right. See which mark falls on arm BA. Read the number on that mark. This number is the measure of the angle.

The measure of∠ABC is 40°.

We can measure the same ∠ABC by positioning the protractor differently.

1. First put the centre point of the protractor on vertex B of the angle. Align the baseline of the protractor with arm BA.
2. Find the 0 mark on the side of BA. Count the marks starting from the 0 on the side of point A. See which mark falls on arm BC. Read the number at that point.

Observe that here, too, the measure of ∠ABC is 40°.

See how the angles given below have been measured with the help of a protractor.

Chapter 6 Angles Problem Set 24 Additional Important Questions and Answers

Question 1.
Observing the adjacent figure, answer the following questions. Write the name of (1) all angles (2) Vertex of the angles (3) aims of the angles.
Solution:
(1) ZXOY, ZYOZ, ¿XOZ
(2) Vertex of the angles is ‘O’
(3) arms of Z)OY are OX and UY
arms of ZYOZ are UY and OZ
arms of ZXOZ are OX and OZ

Maharashtra Board Class 5 Maths Solutions

• Angles Problem Set 24 Class 5 Maths Solutions
• Angles Problem Set 25 Class 5 Maths Solutions
• Angles Problem Set 26 Class 5 Maths Solutions
• Angles Problem Set 27 Class 5 Maths Solutions