Class 6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.5

1. Add the following:

Question (i)

Solution:

2. Subtract the following:

Question (i)

Solution:

3. Simplify the following:

Question (i)

Solution:

4. An iron pipe of length \(6 \frac{2}{3}\) metres long was cut into two pieces. One piece is \(4 \frac{3}{7}\) metre long. What is the length of other pieces?
Solution:
Total length of an iron pipe
= \(6 \frac{2}{3}\) m = \(\frac{20}{3}\) metre
Length of one piece
= \(4 \frac{3}{7}\) metre = \(\frac{31}{7}\) metre
∴ Length of other piece

5. Ashok bought \(\frac{7}{10}\)kg of mangoes and Taran \(\frac{11}{15}\)kg of apples. How much fruit did he buy in all?
Solution:

6. Avi did \(\frac{3}{5}\) of his homework on Saturday and \(\frac{1}{10}\) of the same homework on Sunday. How much of the homework did he do over the weekend?
Solution:
Homework he did weekend
\(\frac{3}{5}+\frac{1}{10}=\frac{3 \times 2}{5 \times 2}+\frac{1}{10}=\frac{6}{10}+\frac{1}{10}\)
∴ Home work done in weekend
\(\frac{6+1}{10}=\frac{7}{10}\)

7. Charan spent \(\frac {1}{4}\) of his pocket money on a movie and \(\frac {3}{8}\) on a new pen and \(\frac {1}{8}\) on a pencil. What fraction of his pocket money did he spend?
Solution:
Total money he spent
= \(\frac{1}{4}+\frac{3}{8}+\frac{1}{8}=\frac{1}{4} \times \frac{2}{2}+\frac{3}{8}+\frac{1}{8}\)
= \(\frac{2}{8}+\frac{3}{8}+\frac{1}{8}=\frac{2+3+1}{8}\)
= \(\frac{6}{8}=\frac{3}{4}\)
So, pocket money spent = \(\frac {3}{4}\)

8. Simar lives at a distance of 4 km from the school. Prabhjot lives at a distance of \(\frac {2}{3}\) km less than Simar’s distance from the school. How far does Prabhjot live from the school?
Solution:
Distance of Simar from school = 4 km Distance of Prabhjot from school

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.3

1. Write the fraction for the shaded part and check whether these fractions are equivalent or not?

Question (i)

Solution:

2. Find four equivalent fractions of the followings:

Question (i)
(i) \(\frac {1}{4}\)
(ii) \(\frac {3}{5}\)
(iii) \(\frac {7}{9}\)
(iv) \(\frac {5}{11}\)
(v) \(\frac {2}{3}\)
Solution:

3. Write the lowest equivalent fraction (simplest form) of :

Question (i)
(i) \(\frac {10}{25}\)
(ii) \(\frac {27}{54}\)
(iii) \(\frac {48}{72}\)
(iv) \(\frac {150}{60}\)
(v) \(\frac {162}{90}\)
Solution:

4. Are the following fractions equivalent or not?

Question (i)
\(\frac{5}{12}, \frac{25}{60}\)
Solution:
We have

By cross product,
5 × 60 = 300 and 12 × 25 = 300
Since two cross products are same
So, the given fractions are equivalent.

Question (ii)
\(\frac{6}{7}, \frac{36}{42}\)
Solution:
We have

By cross product,
6 × 42 = 252 and 7 × 36 = 252
Since two cross products are same
So, the given fractions are equivalent.

Question (iii)
\(\frac{7}{9}, \frac{56}{72}\)
Solution:
We have

By cross product,
7 × 72 = 504 and 9 × 56 = 504
Since two cross products are same
So, the given fractions are equivalent.

5. Replace [ ] 1 in each of the following by the correct number.

Question (i)
\(\frac{2}{7}\) = 12 / [ ]
Solution:
Observe the numerators we have 12 ÷ 2 = 6
So, we multiply both numerator and denominator of \(\frac {2}{7}\) by 6
We get \(\frac{2}{7}=\frac{2 \times 6}{7 \times 6}=\frac{12}{42}\)
Hence, the correct number in [ ] 1 is 42

Question (ii)
\(\frac{5}{8}\) = 35 / [ ]
Solution:
Observe the numerators we have 35 ÷ 5 = 7
So, we multiply both numerator and denominator of \(\frac {5}{8}\) by 7
We get \(\frac{5}{8}=\frac{5 \times 7}{8 \times 7}=\frac{35}{56}\)
Hence, the correct number in [ ] is 56.

Question (iii)
\(\frac{24}{36}\) = 6 / [ ]
Solution:
Observe the numerators we have 24 ÷ 6 = 4
So, we divide both numerator and denominator of \(\frac {24}{36}\) by 4
We get \(\frac{24}{36}=\frac{24 \div 4}{36 \div 4}=\frac{6}{9}\)
Hence, the correct number in [ ] is 9

Question (iv)
\(\frac{30}{48}\) = 8 / [ ]
Solution:
Observe the denominators we have 48 ÷ 8 = 6
So, we divide both numerator and denominator of \(\frac {30}{48}\) by 6
We get \(\frac{30}{48}=\frac{30 \div 6}{48 \div 6}=\frac{5}{8}\)
Hence, the correct number in ⊇ is 5

Question (v)
\(\frac{7}{4}\) = 42 / [ ]
Solution:
Observe the numerators we have 42 ÷ 7 = 6
So, we multiply both numerator and denominator of \(\frac {7}{4}\) by 6
We get \(\frac{7}{4}=\frac{7 \times 6}{7 \times 6}=\frac{42}{24}\)
Hence, the correct number in [ ] is 24

6. Find the equivalent fraction of \(\frac {3}{5}\), having

Question (i)
numerator 18
Solution:
(i) Equivalent fraction of \(\frac {3}{5}\), having numerator 18 is
\(\frac{3}{5}\) = 18 / [ ]
Observe the numerators, we have 18 ÷ 3=6
So, we multiply both numerator and denominator of \(\frac {3}{5}\) by 6
∴ \(\frac{3}{5}=\frac{3 \times 6}{5 \times 6}=\frac{18}{30}\)
Thus, required equivalent fraction of \(\frac{3}{5}=\frac{18}{30}\)

Question (ii)
denominator 20
Solution:
Equivalent fraction of \(\frac {3}{5}\), having denominator 20 is \(\frac{3}{5}\) = [ ] / 20
Observe the denominators, we have 20 ÷ 5 = 4
So, we multiply both numerator and denominator of \(\frac {3}{5}\) by 4
Thus, required equivalent fraction of \(\frac{3}{5}=\frac{12}{20}\)
∴ \(\frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
Thus, required equivalent fraction of
\(\frac{3}{5}=\frac{12}{20}\)

Question (iii)
numerator 24.
Solution:
Equivalent fraction of \(\frac {3}{5}\) , having numerator 24 is
\(\frac{3}{5}\) = 24 / [ ]
Observe the numerators, we have 24 ÷ 3 = 8
So, we multiply both numerator and denominator of \(\frac {3}{5}\) by 8
∴ \(\frac{3}{5}=\frac{3 \times 8}{5 \times 8}=\frac{24}{40}\)
Thus, required equivalent fraction of
\(\frac{3}{5}=\frac{24}{40}\)

7. Find the equivalent fraction of \(\frac {24}{40}\), having

Question (i)
(i) numerator 6
(ii) numerator 48
(iii) denominator 20
Solution:
(i) Equivalent fraction of \(\frac {24}{40}\), numerator 6 is
\(\frac{24}{40}\) = 6 / [ ]
Observe the numerators, we have 24 ÷ 6 = 4
So, we divide both numerator and denominator of \(\frac {24}{40}\) by 4
∴ \(\frac{24}{40}=\frac{24 \div 4}{40 \div 4}=\frac{6}{10}\)
Thus, required equivalent fraction of
\(\frac{24}{40}=\frac{6}{10}\)

Question (ii)
numerator 48
Solution:
Equivalent fraction of \(\frac {24}{40}\), having numerator 48 is
\(\frac{24}{40}\) = 48 / [ ]
Observe the numerators, we have 48 ÷ 24 = 2
So, we multiply both numerator and denominator of \(\frac {24}{40}\) by 2
∴ \(\frac{24}{40}=\frac{24 \times 2}{40 \times 2}=\frac{48}{80}\)
Thus, required equivalent fraction of
\(\frac{24}{40}=\frac{48}{80}\)

Question (iii)
denominator 20
Solution:
Equivalent fraction of \(\frac {24}{40}\), having denominator 20 is
\(\frac{24}{40}\) = [ ] / 20
Observe the denominators, we have 40 ÷ 20 = 2
So, we divide both numerator and denominator of \(\frac {24}{40}\) by 2
\(\frac{24}{40}=\frac{24 \div 2}{40 \div 2}=\frac{12}{20}\)
∴ Thus, required equivalent fraction of
\(\frac{24}{40}=\frac{12}{20}\)

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.4

1. Find the different set of like fractions:

Solution:

2. Write any three like fractions of:

Question (i)
(i) \(\frac {2}{5}\)
(ii) \(\frac {1}{4}\)
(iii) \(\frac {11}{6}\)
Solution:

3. Encircle unit fractions:

Solution:
\(\frac{1}{8}, \frac{1}{9}, \frac{1}{7}\)

4. Fill in the boxes with >, < or =

Question (i)

Solution:

5. Compare using >, < or =

Question (i)

Solution:

6. Compare using >, < or =

Question (i)

Solution:

7. Arrange the following fractions in ascending order:

Question (i)

Solution:
We know that in fractions having the same denominator the greater the numerator, the greater the value of the fractional numbers. Therefore the given fractions in:
(i) Ascending order is : \(\frac{3}{10}, \frac{5}{10}, \frac{7}{10}\)
(ii) Ascending order is : \(\frac{1}{7}, \frac{4}{7}, \frac{6}{7}\)
(iii) Ascending order is : \(\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\)
We know that in fractions having the same numerator, the fractions with smaller denominator is greater:
(iv) Ascending order is : \(\frac{5}{9}, \frac{5}{7}, \frac{5}{3}\)
(v) Ascending order is : \(\frac{3}{13}, \frac{3}{11}, \frac{3}{7}\)
(vi) First find L.C.M. of denominators 4, 6, 12. Now, we convert the given fractions into fractions with denominator 12, we have:

(vii) First find L.C.M. of denominators 7, 35, 14, 28. Now, we convert the given fraction into fractions with denominators 140, we have

(viii) First find HCF of 3, 9, 12, 15. Now we convert the given fractions into fractions with denominator 180, we have:

8. Arrange the following fractions in descending order:

Question (i)
\(\frac{5}{9}, \frac{7}{9}, \frac{1}{9}\)
Solution:
If two or more fractions having the same denominator then fraction with greater numerator is greater fraction:
(i) Descending order is : \(\frac{7}{9}, \frac{5}{9}, \frac{1}{9}\)

Question (ii)
\(\frac{3}{11}, \frac{5}{11}, \frac{2}{11}, \frac{7}{11}\)
Solution:
Descending order is : \(\frac{7}{11}, \frac{5}{11}, \frac{3}{11}, \frac{2}{11}\)
If two or more fractions having the same numerator then the fraction with a small denominator is greater.

Question (iii)
\(\frac{2}{7}, \frac{2}{13}, \frac{2}{9}\)
Solution:
Descending order is : \(\frac{2}{7}, \frac{2}{9}, \frac{2}{13}\)

Question (iv)
\(\frac{1}{5}, \frac{1}{3}, \frac{1}{8}, \frac{1}{2}\)
Solution:
Descending order is : \(\frac{1}{2}, \frac{1}{3}, \frac{1}{5}, \frac{1}{8}\)

Question (v)
\(\frac{1}{6}, \frac{5}{12}, \frac{5}{18}, \frac{2}{3}\)
Solution:
First find L.C.M. of denominators 6, 12, 18, 3.
Now, we convert the given fractions into a fraction with denominator 36, we have:

Question (vi)
\(\frac{3}{4}, \frac{9}{20}, \frac{11}{15}, \frac{17}{30}\)
Solution:
First find L.C.M. of denominator 4, 20,15, 30. Now, we convert the given fractions into a fraction with denominator 60, we have:

9. Kasvi covered \(\frac {1}{3}\) of her journey by car, \(\frac {1}{5}\) by rickshaw and \(\frac {2}{15}\) on foot. Find by which means, she covered the major part of her journey.
Solution:
Journey covered by car = \(\frac {1}{3}\)
Journey covered by Rickshaw = \(\frac {1}{5}\)
Journey covered on foot = \(\frac {2}{15}\)

We observe that \(\frac {5}{15}\) i.e. \(\frac {1}{3}\) is the greatest.
Hence the major part of journey was covered by car.

10. Father distributed his property among his three sons. The eldest one got \(\frac {3}{10}\), the middle got \(\frac {1}{6}\) and the youngest got \(\frac {1}{5}\) part of the property. State how the property was distributed in ascending order.
Solution:
Property eldest son got = \(\frac {3}{10}\) part
Property middle son got = \(\frac {1}{6}\) part
and Property youngest son got = \(\frac {1}{5}\) part

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2

1. Find the common factors of the followings:

Question (i)
16 and 24
Solution:
The factors of 16
= 1, 2, 4, 8, 16
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
Common factors of 16 and 24
= 1, 2, 4, 8

Question (ii)
25 and 40
Solution:
The factors of 25
= 1, 5, 25
The factors of 40
= 1, 2, 4, 5, 8, 10, 20, 40
Common factors of 25 and 40
= 1, 5

Question (iii)
24 and 36
Solution:
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 36
= 1, 2, 3, 4, 6, 12, 18, 36
Common factors of 24 and 36
= 1, 2, 3, 4, 6, 12

Question (iv)
14, 35 and 42
Solution:
The factors of 14
= 1, 2, 7, 14
The factors of 35
= 1, 5, 7, 35
The factors of 42
= 1,2,3, 6, 7, 21, 42
Common factors of 14, 35 and 42
= 1, 7

Question (v)
15, 24 and 35.
Solution:
The factors of 15
= 1, 3, 5, 15
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 35
= 1, 5, 7, 35
Common factors of 15, 24 and 35.
= 1

2. Find first three common multiples of the followings:

Question (i)
3 and 5
Solution:
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45
The multiples of 5
= 5, 10, 15, 20, 25, 30, 35, 40,45
First three common multiples of 3 and 5
= 15, 30 and 45

Question (ii)
6 and 8
Solution:
The multiples of 6
= 6, 12, 18, 24, 30, 36, 42, 48 54, 60, 66, 72
The multiples of 8
= 8, 16, 24, 32, 40, 48, 56, 64, 72
First three common multiples of 6 and 8
= 24, 48 and 72

Question (iii)
2, 3 and 4.
Solution:
The multiples of 2
= 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
The multiples of 4
= 4, 8, 12, 16, 20, 24, 28, 32, 36
First three common multiples of 2, 3 and 4
= 12, 24 and 36

3. Which of the following numbers are divisible by 2 or 4?

Question (i)
52314
Solution:
52314 is divisible by 2 as it is even number.
52314 is not divisible by 4 because the last two digits i.e. 14 which is not divisible by 4

Question (ii)
678913
Solution:
678913 is not divisible by 2. As it is an odd number.
678913 is not divisible by 4 because the last two digits i.e. 13 is not divisible by 4.

Question (iii)
4056784
Solution:
4056784 is divisible by 2. As it is an even number.
4056784 is also divisible by 4 because the last two digits i.e. 84 which is divisible by 4.

Question (iv)
21536
Solution:
21536 is divisible by 2. As it is an even number.
21536 is divisible by 4. As number formed by their last two digits is divisible by 4.

Question (v)
412318.
Solution:
412318 is divisible by 2. As it is an even number.
412318. is not divisible by 4. As number formed by their last two digits is not divisible by 4.

4. Which of the following numbers are divisible by 3 or 9?

Question (i)
654312
Solution:
654312 is divisible by 3.
As sum of its digits = 6 + 5 + 4 + 3 + 1 + 2 = 21, which is divisible by 3.
654312 is not divisible by 9.
As sum of its digits = 21, which is not divisible by 9.

Question (ii)
516735
Solution:
516735 is divisible by 3.
As sum of its digits = 5 + 1 + 6 + 7 + 3 + 5 = 27, which is divisible by 3.
516735 is also divisible by 9.
As sum of its digits = 27, which is divisible by 3.

Question (iii)
423152
Solution:
423152 is divisible by 3.
As sum of its digits = 4 + 2 + 3 + 1 + 5 + 2=17, which is not divisible by 3.
423152 is also not divisible by 9.
As sum of its digits = 17, which is not divisible by 9.

Question (iv)
704355
Solution:
704355 is divisible by 3.
As sum of its digits = 7 + 0 + 4 + 3 + 5 + 5 = 24, which is divisible by 3.
704355 is not divisible by 9.
As sum of its digits = 24, which is not divisible by 9.

Question (v)
215478.
Solution:
215478 is divisible by 3.
As sum of its digits = 2 + 1 + 5 + 4 + 7 + 8 = 27, which is divisible by 3.
215478 is divisible by 9.
As sum of its digits = 27, which is divisible by 9.

5. Which of the following numbers are divisible by 5 or 10?

Question (i)
456803
Solution:
456803 is not divisible by 5
As its last digit is not 0 or 5.
456803 is not divisible by 10
As its last digit is not 0.

Question (ii)
654130
Solution:
654130 is divisible by both 5 and 10
As its last digit is 0.

Question (iii)
256785
Solution:
256785 is divisible by 5
As its last digit is 5.
256785 is not divisible by 10
As its last digit is not 0.

Question (iv)
412508
Solution:
412508 is not divisible by 5
As its last digit is not 0 or 5.
412508 is not divisible by 10
As its last digit is not 0.

Question (v)
872565.
Solution:
872565 is divisible by 5
As its last digit is 5.
872565 is not divisible by 10
As its last digit is not 0.

6. Which of the following numbers are divisible by 8?

Question (i)
457432
Solution:
457432 is divisible by 8, because its last three digits are 432, which is divisible by 8.

Question (ii)
5134214
Solution:
5134214 is not divisible by 8, because its last three digits are 214, which is not divisible by 8.

Question (iii)
7232000
Solution:
7232000 is divisible by 8, because its last three digits are 000, which is divisible by 8.

Question (iv)
5124328
Solution:
5124328 is divisible by 8, because its last three digits are 328, which is divisible by 8.

Question (v)
642516.
Solution:
642516 is not divisible by 8, because its last three digits are 516, which is not divisible by 8.

7. Which of the following numbers are divisible by 6?

Question (i)
425424
Solution:
425424 is divisible by 2 because, it has 4 in its units place.
Sum of digits = 4 + 2 + 5+4 + 2 + 4 = 21
Sum of digits of 425424 is divisible by 3.
∴ 425424 is divisible by 2 as well as 3
Hence, 425424 is divisible by 6.

Question (ii)
617415
Solution:
617415 is not divisible by 2 because, it has 5 in its units place.
∴ 617415 is not divisible by 6.

Question (iii)
3415026
Solution:
3415026 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 3 + 4 + 1 + 5 + 0 + 2 + 6 = 21
Sum of digits of 3415026 is divisible by 3
So, 3415026 is divisible by 3
∴ 3415026 is divisible by 2 as well as 3
Hence, 3415026 is divisible by 6.

Question (iv)
4065842
Solution:
4065842 is divisible by 2 because, it has 2 in its units place.
Sum of digits = 4 + 0 + 6 + 5 + 8 + 4 + 2 = 29
Sum of digits of 4065842 is not divisible by 3.
So, 4065842 is not divisible by 3.
∴ 4065842 is divisible by 2 but not by 3.
Hence, 4065842 is not divisible by 6.

Question (v)
725436.
Solution:
725436 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 7 + 2 + 5 + 4 + 3 + 6 = 27
Sum of digits of 725436 is divisible by 3.
So, 725436 is divisible by 3.
∴ 725436 is divisible by 2 as well as 3
Hence, 725436 is divisible by 6.

8. Which of the following numbers are divisible by 11?

Question (i)
4281970
Solution:
4281970 is divisible by 11.

Since sum of its digits in odd places = 4 + 8 + 9 + 0 = 21 and
sum of its digits in even places = 2 + 1 + 7 = 10
Their difference = 21 – 10=11, which is odd places digits divisible by 11.

Question (ii)
8049536
Solution:
8049536 is divisible by 11.

Since sum of its digits in odd places = 8 + 4 + 5 + 6 = 23
and sum of its digits in even places = 0 + 9 + 3 = 12
Difference = 23 – 12 = 11, which is divisible by 11.

Question (iii)
1234321
Solution:
1234321 is divisible by 11.

Since sum of its digits in odd places = 1 + 3 + 3 + 1 = 8
and sum of its digits in even places = 2 + 4 + 2 = 8
Difference = 8 – 8 = 0.

Question (iv)
6450828
Solution:
6450828 is not divisible by 11.

Since sum of its digits in odd places = 6 + 5 + 8 + 8 = 27
and sum of its digits in even places = 4 + 0 + 2 = 6
Difference = 27 – 6 = 21, which is not divisible by 11.

Question (v)
5648346.
Solution:
5648346 is divisible by 11.

Since sum of its digits in odd places = 5 + 4 + 3 + 6 = 18 and
sum of its digits in even places = 6 + 8 + 4 = 18.
Difference = 18 – 18 = 0.

9. State True or False:

Question (i)
If a number is divisible by 24, then it is also divisible by 3 and 8.
Solution:
True

Question (ii)
60 and 90 both are divisible by 10 then their sum is not divisible by 10.
Solution:
False

Question (iii)
If a number is divisible by 8 then it is also divisible by 16.
Solution:
False

Question (iv)
If a number is divisible by 15 then it is also divisible by 3.
Solution:
True

Question (v)
144 and 72 are divisible by 12 then their difference is also divisible by 12.
Solution:
True

10. If a number is divisible by 5 and 9 then by which other numbers will that number be always divisible?
Solution:
If a number is divisible by 5 and 9. Then the number is also divisible by their product i.e. 5 × 9 = 45.

11. Which of the following pairs are co-prime?

Question (i)
25, 35
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1.
Given numbers are 25 and 35 Factors of 25 = 1, 5, 25
Factors of 35 = 1, 5, 7, 35
Since 25 and 35 have 1 and 5 two common factors
∴ 25 and 33 are not co-prime.

Question (ii)
16,21
Solution:
Given numbers are 16 and 21
Factors of 16 = 1, 2, 4, 8, 16
Factors of 21 = 3, 7, 21
There is only 1 common factors 16 and 21 are co-prime
∴ 16 and 21 are co-prime.

Question (iii)
24, 41
Solution:
Given numbers are 24 and 41
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 41 = 1, 41
There is only one (1) common factors.
∴ 24 and 41 are co-prime.

Question (iv)
48,33
Solution:
Given numbers are 48 and 33
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 33 = 1, 3, 11
There are two common factors 1 and 3.
∴ 48 and 33 are not co-prime.

Question (v)
20, 57.
Solution:
Given numbers are 20 and 57
Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 57 = 1, 3, 19, 57
There is only only one (1) common factors.
∴ 20 and 57 are co-prime.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1

1. Write down all the factors of each of the following:

Question (i)
18
Solution:
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
So, 1, 2, 3, 6, 9 and 18 are factors of 18

Question (ii)
24
Solution:
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
So, 1, 2, 3, 4, 6, 8, 12 and 24 are factors of 24

Question (iii)
45
Solution:
45 = 1 × 45
45 = 3 × 15
45 = 5 × 9
So, 1, 3, 5, 9, 15 and 45 are factors of 45

Question (iv)
60
Solution:
60 = 1 × 60
60 = 2 × 30
60 = 3 × 20
60 = 4 × 15
60 = 5 × 12
60 = 6 × 10
So, the factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

Question (v)
65.
Solution:
65 = 1 × 65
65 = 5 × 13
So, 1, 5, 13 and 65 are the factors of 65

2. Write down the first six multiples of each of the following:

Question (i)
6
Solution:
First six multiples of 6 are:
6, 12, 18, 24, 30 and 36

Question (ii)
9
Solution:
First six multiples of 9 are:
9, 18, 27, 36, 45 and 54

Question (iii)
11
Solution:
First six multiples of 11 are:
11, 22, 33, 44, 55 and 66

Question (iv)
15
Solution:
First six multiples of 15 are:
15, 30, 45, 60, 75 and 90

Question (v)
24.
Solution:
First six multiples of 24 are:
24, 48, 72, 96, 120 and 144

3. List all the numbers less than 100 that are multiples of:

Question (i)
17
Solution:
Multiples of 17 less than 100 are:
17, 34, 51, 68 and 85

Question (ii)
12
Solution:
Multiples of 12 less than 100 are:
12, 24, 36,48, 60, 72, 84 and 96

Question (iii)
21.
Solution:
Multiples of 21 less than 100 are:
21, 42, 63 and 84

4. Which of the following are prime numbers?

Question (i)
39
Solution:
Given number = 39
We find that 39 is divisible by 3.
∴ It has more than two factors.
∴ So, 39 is not a prime number

Question (ii)
129
Solution:
Given number =129
It is divisible by 1 and itself So, it has exactly two factors.
∴ 129 is a prime number

Question (iii)
177
Solution:
Given number = 177
We find that 177 is divisble by 3
∴ It has more than two factors.
So, 177 is not a prime number

Question (iv)
203
Solution:
Given number = 203
It is divisible by 1 and itself
So, 203 is a prime number

Question (v)
237
Solution:
Given number = 237
We find that 237 is divisible by 3
∴ It has more than two factors.
So, 237 is not a prime number

Question (vi)
361.
Solution:
Given number = 361
We find that 361 is divisible by 19
∴ It has more than two factors.
So, 361 is not a prime number

5. Express each of the following as sum of two odd prime numbers:

Question (i)
16
Solution:
16 = 3 + 13
= 5 + 11

Question (ii)
28
Solution:
28 = 11+ 17

Question (iii)
40.
Solution:
40 = 3 + 37
= 11 + 29
= 17 + 23

6. Write all the prime numbers between the given numbers:

Question (i)
1 to 25
Solution:
Prime numbers between 1 to 25 are:
2, 3, 5, 7, 11, 13, 17, 19, 23

Question (ii)
85 to 105
Solution:
Prime numbers between 85 to 105 are:
89, 97, 101, 103

Question (iii)
120 to 140.
Solution:
Prime numbers between 120 to 140 are:
127, 129, 131, 137, 139

7. Is 36 a perfect number?
Solution:
Factors of 36 are:
2, 3, 4, 6, 9, 12, 18, 36
Sum of all the factors of 36
= 2 + 3 + 4 + 6 + 9 + 12+18 + 36
= 90
= 2 × 45
But sum of all factors of a number = 2 × Number
Thus, 36 is not a perfect number

8. Find the missing factors:

Question (i)
(i) 5 × …. = 30
(ii) …. × 6 = 48
(iii) 7 × …. = 63
(iv) …. × 8 = 104
(v) …. × 7 = 105.
Solution:
(i) 5 × 6 =30
(ii) 8 × 6 = 48
(iii) 7 × 9 = 63
(iv) 13 × 8 = 104
(v) 15 × 7 = 105.

9. List all 2-digit prime numbers, in which both the digits are prime numbers.
Solution:
All 2-digit numbers, in which both the digits are prime numbers are:
23, 37, 53, 73

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 2 Whole Numbers MCQ Questions

Multiple Choice Questions

Question 1.
The smallest whole number is:
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(a) 0

Question 2.
The smallest natural number is:
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 3.
The successor of 38899 is:
(a) 39000
(b) 38900
(c) 39900
(d) 38800.
Answer:
(b) 38900

Question 4.
The predecessor of 24100 is:
(a) 24999
(b) 24009
(c) 24199
(d) 24099.
Answer:
(d) 24099.

Question 5.
The statement 4 + 3 = 3 + 4 represents:
(a) Closure
(b) Associative
(c) Commutative property
(d) Identity.
Answer:
(c) Commutative property

Question 6.
Which of the following is the additive identity?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(a) 0

Question 7.
The multiplicative identity is ………………. .
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 8.
15 × 32 + 15 × 68 = …………….. .
(a) 1400
(b) 1600
(c) 1700
(d) 1500
Answer:
(d) 1500

Question 9.
The largest 4 digit number divisible by 13 is:
(a) 9997
(b) 9999
(c) 9995
(d) 9991.
Answer:
(a) 9997

Question 10.
The successor of 3 digit largest number is:
(a) 100
(b) 998
(c) 1001
(d) 1000
Answer:
(d) 1000

Question 11.
Which of the following is shown on the given number line?

(a) 2 + 5
(b) 5 + 2
(c) 7 – 2
(d) 7 – 5.
Answer:
(d) 7 – 5

Question 12.
The whole number which comes just before 10001 is:
(a) 10000
(b) 10002
(c) 9999
(d) 9998.
Answer:
(a) 10000

Question 13.
The smallest natural number is:
(a) 1
(b) 0
(c) 9
(d) 10
Answer:
(a) 1

Question 14.
Which is the smallest whole number?
(a) 1
(b) 0
(c) -1
(d) 9
Answer:
(b) 0

Question 15.
Which is the successor of 100199?
(a) 100198
(b) 100197
(c) 100200
(d) 100201.
Answer:
(c) 100200

Question 16.
Which is the predecessor of 10000?
(a) 10001
(b) 9999
(c) 10002
(d) 9998.
Answer:
(b) 9999

Question 17.
How many whole numbers are there between 32 and 53?
(a) 21
(b) 22
(c) 19
(d) 20.
Answer:
(d) 20

Fill in the blanks:

1. 25 …………… 205
2. 10001 …………. 9999
3. 15 × 0 = …………….
4. 0 ÷ 25 = ………….
5. 1 ÷ 1 = ……………

Answer:

1. <
2. >
3. 0
4. 0
5. 1

Write True or False:

Question 1.
Zero is smallest natural number. (True/False)
Answer:
False

Question 2.
All natural numbers are whole numbers. (True/False)
Answer:
True

Question 3.
All whole numbers are, natural numbers. (True/False)
Answer:
False

Question 4.
The naitural number 1 has no predecessor. (True/False)
Answer:
True

Question 5.
500 is the predecessor of 490. (True/False)
Answer:
False

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3

1. If the product of two whole numbers is zero. Can we say that one or both of them will be zero? Justify through examples.
Solution:
One of them can be Zero i.e. 0 × 5 = 0
Both of them can be Zero i.e. 0 × 0 = 0.

2. If the product of two whole numbers is 1. Can we say that one or both of them will be 1? Justify through examples.
Solution:
Both of them will be 1.
Example: 1 × 1 = 1.

3. Observe the pattern in the following and fill in the blanks:

Solution:

4. Observe the pattern and fill in the blanks:

Solution:

5. Represent numbers from 24 to 30 according to rectangular, square or triangular pattern.
Solution:
Numbers from 24 to 30 are 24, 25, 26, 27, 28, 29, 30.

6. Study the following pattern:

Question (i)

Hence find the sum of
(a) First 12 odd numbers
(b) First 50 odd numbers.
Solution:
(a) Sum of first 12 odd numbers = 12 × 12 = 144
(b) Sum of first 50 odd numbers = 50 × 50 = 2500

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

1. Find the sum by suitable arrangement of terms:

Question (a)
837 + 208 + 363
Solution:
837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408

Question (b)
1962 + 453 + 1538 + 647.
Solution:
1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600

2. Find the product by suitable arrangement of terms:

Question (a)
2 × 1497 × 50
Solution:
= (2 × 50) × 1497
= 100 × 1497
= 149700

Question (b)
4 × 263 × 25
Solution:
= (4 × 25) × 263
= 100 × 263
= 26300

Question (c)
8 × 163 × 125
Solution:
= (8 × 125) × 163
= 1000 × 163
= 163000

Question (d)
963 × 16 × 25
Solution:
= 963 × (16 × 25)
= 963 × 400
= 385200

Question (e)
5 × 171 × 60
Solution:
= (5 × 60) × 171
= 300 × 171
= 51300

Question (f)
125 × 40 × 8 × 25
Solution:
= (125 × 40) × (8 × 25)
= 5000 × 200
= 1000000

Question (g)
30921 × 25 × 40 × 2
Solution:
= 30921 × (25 × 40) × 2
= 30921 × 1000 × 2
= 61842000

Question (h)
4 × 2 × 1932 × 125
Solution:
4 × 2 × 1932 × 125
= 1932 × (4 × 2 × 125)
= 1932 × 1000
= 1932000

Question (i)
5462 × 25 × 4 × 2.
Solution:
= 5462 × 2 × 25 × 4
= 10924 × 100
= 1092400

3. Find the value of each of the following using distributive property:

Question (a)
(649 × 8) + (649 × 2)
Solution:
(649 × 8) + (649 × 2)
= 649 × (8 + 2)
= 649 × 10
= 6490

Question (b)
(6524 × 69) + (6524 × 31)
Solution:
(6524 × 69) + (6524 × 31)
= 6524 × (69 + 31)
= 6524 × 100
= 652400

Question (c)
(2986 × 35) + (2986 × 65)
Solution:
(2986 × 35) + (2986 × 65)
= 2986 × (35 + 65)
= 2986 × 100
= 298600

Question (d)
(6001 × 172) – (6001 × 72).
Solution:
(6001 × 172) – (6001 × 72)
= 6001 × (172 – 72)
= 6001 × 100
= 600100

4. Find the value of the following:

Question (a)
493 × 8 + 493 × 2
Solution:
(a) 493 × 8 + 493 × 2
= 493 × (8 + 2) = 493 × 10
= 4930

Question (b)
24579 × 93 + 7 × 24579
Solution:
24579 × 93 + 7 × 24579
= 24579 × (93 + 7)
= 24579 × 100
= 2457900

Question (c)
3845 × 5 × 782 + 769 × 25 × 218
Solution:
3845 × 5 × 782 + 769 × 25 × 218
= 769 × 5 × 5 × 782 + 769 × 5 × 5 × 218
= 769 × 5 × 5 × (782 + 218)
= 769 × 25 × 1000
= 19225 × 1000
= 19225000

Question (d)
3297 × 999 + 3297.
Solution:
3297 × 999 + 3297
= (3297) × (999 + 1)
= 3297 × 1000
= 3297000

5. Find the product, using suitable properties:

Question (a)
738 × 103
Solution:
= 738 × (100 + 3)
= (738 × 100) + (738 × 3)
[Using a × (b + c) = (a × b) + (a × c)]
= 73800 + 2214
= 76014

Question (b)
854 × 102
Solution:
= 854 × (100 + 2)
= (854 × 100) + (854 × 2)
[Using a × (b + c) = (a × b) + (a × c)]
= 85400 + 1708
= 87108

Question (c)
258 × 1008
Solution:
= 258 × (1000 + 8)
= (258 × 1000) + (258 x 8)
[Using a × (b + c) = (a × b) + (a × c)]
= 258000 + 2064
= 260064

Question (d)
736 × 93
Solution:
= 736 × (100 – 7)
= 736 × 100 = 736 × 7
[Using a × (b – c) = (a × b) – (a × c)]
= 73600 – 5152
= 68448

Question (e)
816 × 745
Solution:
= (800 + 16) × 745
= 800 × 745 + 16 × 745
[Using a × (b + c) = (a × b) + (a × c)]
= 596000 + 11920
= 607920

Question (f)
2032 × 613
Solution:
= 2032 × (600 + 13)
= 2032 × 600 + 2032 × 13
[Using a × (b + c) = (a × b) + (a × c)]
= 1219200 + 26416
= 1245616

6. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹ 78 per litre, how much he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 Litres
Petrol filled on Tuesday = 50 Litres
Total Petrol filled = 40 + 50 = 90 Litres
Cost per litre = ₹ 78
Total Cost = 90 × ₹ 78
= ₹ 7020

7. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 35 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 Litres
Milk supplied in the evening = 68 Litres
Total milk supplied = 32 + 68
= 100 Litres
Cost Per litre = ₹ 35
∴ Total cost of milk per day = 100 × ₹ 35
= ₹ 3500

8. We know that 0 × 0 = 0. Is there any other whole number which when multiplied by itself gives the product equal to the number itself? Find out the number.
Solution:
Yes, there is a whole number 1.
Here 1 × 1 = 1.

9. Fill in the blanks:

Question (i)
(a) 15 × 0 = ………….. .
(b) 15 + 0 = ………….. .
(c) 15 – 0 = ………….. .
(d) 15 ÷ 0 = ………….. .
(e) 0 × 15 = ………….. .
(f) 0 + 15 = ………….. .
(g) 0 ÷ 15 = ………….. .
(h) 15 × 1 = ………….. .
(i) 15 ÷ 1 = ………….. .
(j) 1 ÷ 1 = ………….. .
Solution:
(a) 0,
(b) 15,
(c) 15,
(d) Not defined,
(e) 0,
(f) 15,
(g) 0,
(h) 15,
(i) 15,
(j) 1.

Question 10.
The product of two Whole numbers is zero. What do you conclude. Explain with example.
Solution:
We conclude that one number must be zero such 25 × 0 = 0.

Question 11.
Match the following:

1. 537 × 106 = 537 ×100 + 537 × 6	(a)  Commutativity under multiplication
2. 4 × 47 × 25 = 4 × 25 × 47	(b) Commutativity under addition
3. 70 + 1923 + 30 = 70 + 30 + 1923	(c)  Distributivity of multiplication over addition.

Solution:

1. 537 × 106 = 537 × 100 + 537 × 6	(c)  Distributivity of multiplication over addition.
2. 4 × 47 × 25 = 4 × 25 × 47	(a)  Commutativity under multiplication
3. 70 + 1923 + 30 = 70 + 30 + 1923	(b) Commutativity under addition

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1

1. Answer the following questions:

Question (a)
Write the smallest whole number.
Solution:
The smallest Whole number = 0

Question (b)
Write the smallest natural number.
Solution:
The smallest natural number = 1

Question (c)
Write the successor of 0 in whole numbers.
Solution:
Successor of 0 = 0 + 1 = 1

Question (d)
Write the predecessor of 0 in whole numbers.
Solution:
Predecessor of 0 is whole number is not possible.

Question (e)
Write the Largest whole number.
Solution:
Largest whole number is not possible.

2. Which of the following statements are True (T) and which are False (F)?

Question (a)
Zero is the smallest natural number.
Solution:
False

Question (b)
Zero is the smallest whole number.
Solution:
True

Question (c)
Every whole number is a natural number.
Solution:
False

Question (d)
Every natural number is a whole number.
Solution:
True

Question (e)
1 is the smallest whole number.
Solution:
False

Question (f)
The natural number 1 has no predecessor in natural numbers.
Solution:
True

Question (g)
The whole number 1 has no predecessor in whole numbers.
Solution:
False

Question (h)
Successor of the largest two-digit number is smallest three-digit number.
Solution:
True

Question (i)
The successor of a two-digit number is always a two-digit number.
Solution:
False

Question (j)
300 is the predecessor of 299.
Solution:
False

Question (k)
500 is the successor of 499.
Solution:
True

Question (l)
The predecessor of a two-digit number is never a single-digit number.
Solution:
False

3. Write the successor of each of following:

Question (a)
100909
Solution:
Successor of 100909
= 100909 + 1
= 100910

Question (b)
4630999
Solution:
Successor of 4630999
= 4630999 + 1
= 4631000

Question (c)
830001
Solution:
Successor of 830001
= 830001 + 1
= 830002

Question (d)
99999.
Solution:
Successor of 99999
= 99999 + 1
= 100000

4. Write the predecessor of each of following:

Question (a)
1000
Solution:
Predecessor of 1000 = 1000 – 1
= 999

Question (b)
208090
Solution:
Predecessor of 208090 = 208090 – 1
= 208089

Question (c)
7654321
Solution:
Predecessor of 7654321 = 7654321 – 1
= 7654320

Question (d)
12576.
Solution:
Predecessor of 12576 = 12576 – 1
= 12575

5. Represent the following numbers on the number line: 2, 0, 3, 5, 7, 11, 15.
Solution:
Draw a line. Mark a point on it. Label it ‘O’. Mark a second point to the right of 0. Label it 1. The distance between these points labelled as 0 and 1 is called unit distance. On this line, mark a point to the right of 1 and at unit distance from 1 and label it 2. In this way go on labeling points at unit distance as 3, 4, 5, …………… on the line.

6. How many whole numbers are there between 22 and 43?
Solution:
Whole numbers between 22 and 43 are 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42
∴ There are 20 whole numbers between 22 and 43.
Or [(43 – 22) – 1 = 21 – 1 = 20].

7. Draw a number line to represent each of following on it.

Question (a)
3 + 2
Solution:
We draw a number line and move 3 steps from 0 to the right and mark this point as A.
Now, starting from A we move 2 steps towards right and move at B.

OA = 4, AB = 2, OB = 5
Hence, OB = 3 + 2 = 5.

Question (b)
4 + 5
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 4 steps to the right and mark this point as A.
Now, starting from A we move 5 steps towards right and arrive at B.

OA = 4, AB = 5, OB = 9
Hence, OB = 4 + 5 = 9.

Question (c)
6 + 2
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 6 steps to the right and mark this point as A.
Now, starting from A we move 2 steps towards right and arrive at B.

OA = 6, AB = 2, OB = 8
Hence, OB = 6 + 2 = 8.

Question (d)
8 – 3
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 8 steps to the right and arrive at A.
Now, starting from A we move 3 steps to the left of A and arrive at B.

OA = 8, AB = 3, OB = 5
Hence, OB = 8 – 3 = 5.

Question (e)
7 – 4
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 7 steps to the right and arrive at A.
Now, starting from A we move 4 steps to the left of A and arrive at B.

OA = 7, AB = 4, OB = 3
Hence, OB = 7 – 4 = 3.

Question (f)
7 – 2
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 7 steps to the right and arrive at A.
Now, starting from A, we move 2 steps to the left of A and arrive at B.

OA = 7, AB = 2, OB = 5
Hence, OB = 7 – 2 = 5.

Question (g)
3 × 3
Solution:
We draw a number line.
Starting from 0 we move 3 units to the right of 0 to arrive at A.
We make two more such same moves starting from A (total 3 moves of 3 units each) to reach finally at C which represents 9.

Hence, 3 × 3 = 9.

Question (h)
2 × 5
Solution:
We draw a number line.
We start from 0 move 5 units at a time to right.
We make 2 such moves. We shall reach at 10.

So, 2 × 5 = 10.

Question (i)
3 × 5
Solution:
We draw a number line.
We start from 0, move 5 units at a time to right.
We make 3 such moves. We shall reach at 15.

So, 3 × 5 = 15

Question (j)
9 ÷ 3
We draw a number line.
Starting from 0, we move 9 units to the right of 0 to arrive at A.
Now, from A take moves of 3 units to the left of A till we reach at ‘O’. We observe that there are 3 moves.

So, 9 ÷ 3 = 3.

Question (k)
12 ÷ 4
We draw a number line.
Starting from 0, we move 12 units to the right of 0 to arrive at A.
Now, from A take moves of 4 units to the left of A till we reach at ‘O’. We observe that there are 3 moves.

So, 12 ÷ 4 = 3.

Question (l)
10 ÷ 2
Solution:
We draw a number line.
Starting from 0, we move 10 units to the right of 0 to arrive at A.
Now, from A take moves of 2 units to the left c A till we reach at ‘O’. We observe that there are 5 moves.

So, 10 ÷ 2 = 5.

8. Fill in the blanks with the appropriate symbol < or > :

Question (i)
(a) 25 ……………. 205
(b) 170 …………… 107
(c) 415 …………… 514
(d) 10001 ………….. 9999
(e) 2300014 ………….. 2300041
(f) 99999 …………… 888888.
Solution:
(a) 25 < 205 (b) 170 > 107
(c) 415 < 514 (d) 10001 > 9999
(c) 2300014 < 2300041
(f) 99999 < 888888.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 1 Knowing Our Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 1 Knowing Our Numbers MCQ Questions

Multiple Choice Questions

Question 1.
The number of digits are:
(a) 9
(b) 10
(c) 8
(d) Infinite.
Answer:
(b) 10

Question 2.
The greatest 4 digit number using 1, 5, 2, 9 once is:
(a) 9215
(b) 9512
(c) 5912
(d) 9521.
Answer:
(b) 9512

Question 3.
The smallest 4 digit number using 2, 0, 3, 7 once is:
(a) 0237
(b) 2037
(c) 7320
(d) 7023.
Answer:
(b) 2037

Question 4.
Which of the following are in ascending order?
(a) 217, 271, 127, 721
(b) 217, 127, 721, 271
(c) 127, 217, 271, 721
(d) 721, 271, 217, 127.
Answer:
(c) 127, 217, 271, 721

Question 5.
The face value of digit 4 in 23468 is:
(a) 4
(b) 400
(c) 40
(d) 468.
Answer:
(a) 4

Question 6.
The place value of digit 2 in 4123 is:
(a) 23
(b) 2
(c) 20
(d) 200.
Answer:
(c) 20

Question 7.
The difference between place value and face value of 5 in 76542 is:
(a) 537
(b) 45
(c) 0
(d) 495
Answer:
(d) 495

Question 8.
5 × 10000 + 3 × 100 + 2 × 10 + 2 = …………..
(a) 5322
(b) 53022
(c) 50322
(d) 53202.
Answer:
(c) 50322

Question 9.
Four lakh two thousand three hundred fifty-one = …………..
(a) 42351
(b) 402351
(c) 420351
(d) 4002351.
Answer:
(b) 402351

Question 10.
How many four-digit numbers are there?
(a) 9999
(b) 9900
(c) 9000
(d) 9990.
Answer:
(c) 9000

Question 11.
Seventeen million twenty-four thousand fifty-four = …………….
(a) 172454
(b) 170024054
(c) 170240054
(d) 17024054.
Answer:
(d) 17024054.

Question 12.
1 Crore = …………….. million.
(a) 1
(b) 10
(c) 100
(d) 1000.
Answer:
(b) 10

Question 13.
Rounded off 7213 to nearest thousands.
(a) 7200
(b) 7000
(c) 7210
(d) 7213.
Answer:
(b) 7000

Question 14.
Rounded off 45553 to nearest hundreds.
(a) 45500
(b) 45550
(c) 45600
(d) 45650.
Answer:
(c) 45600

Question 15.
Solve : (9 – 4) × 6 = …………….. .
(a) 30
(b) 54
(c) 78
(d) 64.
Answer:
(a) 30

Question 16.
Which of the following number does not have symbol in Roman numerals?
(a) 0
(b) 1
(c) 10
(d) 1000.
Answer:
(a) 0

Question 17.
How many symbols are used in Roman Numerals?
(a) 5
(b) 8
(c) 9
(d) 7.
Answer:
(d) 7

Question 18.
Which of the following are meaningless?
(a) LXIX
(b) XC
(c) IL
(d) LI.
Answer:
(c) IL

Question 19.
CLXVI = ………..
(a) 164
(b) 144
(c) 176
(d) 166.
Answer:
(d) 166

Question 20.
XCIX + XLVI = …………….
(a) CVL
(b) CLV
(c) CXLV
(d) CXLIV.
Answer:
(c) CXLV

Question 21.
Using the digits 4, 5, 7 and 0 without repetition which of the following is the smallest four-digit number?
(a) 0457
(b) 4057
(c) 4507
(d) 4075.
Answer:
(b) 4057

Question 22.
Using the digits 2, 8, 7 and 4 without repetition which of the following is the greatest four-digit number?
(a) 2874
(b) 8742
(c) 8472
(d) 8274.
Answer:
(b) 8742

Question 23.
Which is the smallest four digits number made from the digits 3, 8, 7 by using one-digit twice?
(a) 3378
(b) 3783
(c) 3873
(d) 3837.
Answer:
(a) 3378

Question 24.
Make the greatest four-digit number from the digits 9, 0, 5 by using one-digit twice.
(a) 9005
(b) 9905
(c) 9950
(d) 9050.
Answer:
(c) 9950

Question 25.
Take two digits, 2 and 3, from diem make smallest four digit number, using both the digits equal number of time.
(a) 3232
(b) 2323
(c) 3223
(d) 2233.
Answer:
(d) 2233.

Question 26.
Take two digits, 2 and 3 from them make greatest four-digit number, using both the digits equal number of time.
(a) 3232
(b) 3322
(c) 3223
(d) 2323.
Answer:
(b) 3322

Question 27.
The greatest number from 4536, 4892, 4370, 4452 is:
(a) 4536
(b) 4892
(c) 4370
(d) 4452.
Answer:
(b) 4892

Question 28.
Out of 15623, 15073, 15189, 15800 the smallest number is:
(a) 15623
(b) 15073
(c) 15189
(d) 15800.
Answer:
(b) 15073

Question 29.
The ascending order of the numbers 847, 9754, 8320, 571 is:
(a) 847, 9754, 8320, 571
(b) 9754, 8320, 847, 571
(c) 571, 847, 8320, 9754
(d) 571, 8320, 847, 9754.
Answer:
(c) 571, 847, 8320, 9754

Fill in the blanks:

Question 1.
1 lakh = ten thousands.
Answer:
Ten

Question 2.
1 million = ……………… hundred thousand.
Answer:
Ten

Question 3.
1 crore = ……………….. million.
Answer:
Ten

Question 4.
1 crore = …………… ten lakh.
Answer:
Ten

Question 5.
1 million = ……………. lakh.
Answer:
Ten

Write True/False:

Question 1.
The number of digits are 10. (True/False)
Answer:
True

Question 2.
The greatest four-digit number is 1000. (True/False)
Answer:
False

Question 3.
The place value of digit 5 in 3564 is 50. (True/False)
Answer:
False

Question 4.
0 does not have symbol in Roman numbers. (True/False)
Answer:
True

Question 5.
IL is meaningless. (True/False)
Answer:
True