# RS Aggarwal Class 6 Solutions Chapter 3 - Whole Numbers Ex 3A (3.1)

## RS Aggarwal Class 6Chapter 3 - Whole Numbers Ex 3A (3.1) Solutions Free PDF

RS Aggarwal Chapter 3 solutions of Class 6 is of great use if you want to excel in your school examination. These solutions are highly beneficial for all students of CBSE Board for the preparation of their final exam. If students practice these solutions well then it will give you a fruitful outcome as it is the first step towards higher classes and complex theories. Download RS Aggarwal Class 6 Solutions Chapter 3 – Whole Number Ex 3A (3.1) and make the most out of it.

It is considered as one of the best reference material for Class 6 students from exam point of view. Kickstart your preparation by practicing the solutions on a daily basis as a part of revision before the exam. It is highly advisable for students to solve all the questions mentioned in the exercises with full focus to score good marks in their exam. The solutions provided below are created in accordance with the latest CBSE guidelines.

## Download PDF of RS Aggarwal Class 6 Solutions Chapter 3 – Whole Numbers System Ex 3A (3.1)

Q1: Perform the following subtractions. Check your result by the corresponding additions.

(i) 6237 – 694

(ii) 21205 – 10899

(iii) 100000 – 78987

(iv) 1010101 – 656565

Sol:

(i) Subtraction: 6237 – 694 = 5543

Addition : 5543 + 694 = 6237

(ii) Subtraction: 21205 – 10899 = 10306

Addition : 10306 + 10899 = 21205

(iii) Subtraction: 100000 – 78987 = 21013

Addition : 21013 + 78987 = 100000

(iv) Subtraction: 1010101 – 656565 = 353536

Addition : 353536 + 656565 = 1010101

Q2: Replace each * by the correct digit in each of the following:

Sol:

917 – *5* = 5*8

$\Rightarrow$ 917 – 359 = 558

(ii) 6172 – **69 = 29**

$\Rightarrow$ 6172 – 3569 = 2903

(iii) 5001003 – **6989 = 484****

$\Rightarrow$ 5001003 – 155987 = 4845016

(iv) 1000000 – ****1 = *7042*

$\Rightarrow$ 1000000 – 29571 = 970429

Q3 : Find the difference:

(i) 463 – 9

(ii) 5632 – 99

(iii) 8640 – 999

(iv) 13006 – 9999

Sol:

(i) 463 – 9

= 463 – 10 + 1

= 464 – 10

= 454

(ii) 5632 – 99

= 5632 – 100 + 1

= 5633 – 100

= 5533

(iii) 8640 – 999

= 8640 – 1000 + 1

= 8641 – 1000

= 7641

(iv) 13006 – 9999

= 13006 – 10000 + 1

= 13007 – 10000

= 3007

Q4: Find the difference between the smallest number of 7 digits and the largest number of 4 digits.

Sol:

Smallest seven digit number = 1000000

Largest four digit number = 9999

Therefore, Their difference = 1000000 – 9999

= 1000000 – 10000 + 1

= 1000001 – 10000

= 990001

Q5: Ravi opened his account in a bank by depositing Rs. 136000. Next day he withdrew Rs. 73129 from it. How much money was left in his account.

Sol:

Money deposited by Ravi = Rs. 1,36,000

Money withdrawn by Ravi = Rs. 73,129

Money left in his account = Money deposited – Money withdrawn

= Rs (136000 – 73129)

= Rs. 62871

Therefore, Rs. 62,871 is left in Raviâ€™s account.

Q6: Mrs. Saxena withdrew Rs. 100000 from her bank account. She purchased a TV set for Rs. 38750, a refrigerator for Rs. 23890 and jewelry worth Rs. 35560. How much money was left with her?

Sol:

Money withdrawn by Mrs. Saxena = Rs. 1,00,000

Cost of the TV set = Rs. 38,750

Cost of the refrigerator = Rs. 23,890

Cost of the jewellery = Rs. 35,560

Total money spent = Rs. (38750 + 23890 + 35560) = Rs. 98200

Now, money left = Money withdrawn – money spent

= Rs. (100000 – 98200)

= Rs. 1800

Therefore, Rs. 1,800 is left with Mrs. Saxena.

Q7: The population of a town was 110500. In one year it increased by 3608 due to new births. However, 8973 persons died or left the town during the year. What was the population at the end of the year?

Sol:

Population of the town = 110500

Increased population = 110500 + 3608 = 114108

Number of persons who died or left the town = 8973

Population at the end of the year = 114108 – 8973 = 105135

Therefore, The population at the end of the year will be 105135.

Q8: Find the whole number n when:

(i) n + 4 = 9

(ii) n + 35 = 101

(iii) n – 18 = 39

(iv) n – 20568 = 21403

Sol:

(i) n + 4 = 9

$\Rightarrow$ n = 9 – 4 = 5

(ii) n + 35 = 101

$\Rightarrow$ 101 – 35 = 66

(iii) n – 18 = 39

$\Rightarrow$ 18 + 39 = 57

(iv) n – 20568 = 21403

$\Rightarrow$ n = 21403 + 20568 = 41971

Q1:Fill in the blanks to make each of the following a true statement:

(i) 246 $\times$ 1 = _____

(ii) 1369 $\times$ 0 = _____

(iii) 593 $\times$ 188 = 188 $\times$ _____

(iv) 286 $\times$ 753 = _____ $\times$ 286

(v) 38 $\times$ (91 $\times$ 37) = _____ $\times$ (38 $\times$ 37)

(vi) 13 $\times$ 100 $\times$ _____ = 1300000

(vii) 59 $\times$ 66 + 59 $\times$ 34 = 59 $\times$ (_____ + _____)

(viii) 68 $\times$ 95 = 68 $\times$ 100 – 68 _____

Sol:

(i) 246 $\times$ 1 = 246

(ii) 1369 $\times$ 0 = 0

(iii) 593 $\times$ 188 = 188 $\times$ 593

(iv) 286 $\times$ 753 = 753 $\times$ 286

(v) 38 $\times$ (91 $\times$ 37) = 91 $\times$ (38 $\times$ 37)

(vi) 13 $\times$ 100 $\times$ 1000 = 1300000

(vii) 59 $\times$ 66 + 59 $\times$ 34 = 59 $\times$ (66 + 34)

(viii) 68 $\times$ 95 = 68 $\times$ 100 – 68 $\times$ 5

Q2 : State the property used in each of the following statements:

(i) 19 $\times$ 17 = 17 $\times$ 19

(ii) 16 $\times$ 32 is a whole number

(iii) (29 $\times$ 36) $\times$ 18 = 29 $\times$ (36 $\times$ 18)

(iv) 1480 $\times$ 1 = 1480

(v) 1732 $\times$ 0 = 0

(vi) 72 $\times$ 98 + 72 $\times$ 2 = 72 $\times$ (98 + 2)

(vii) 63 $\times$ 126 – 63 $\times$ 26 = 63 $\times$ (126 – 26)

Sol:

(i) Commutative law in multiplication.

(ii) Closure property.

(iii) Associativity of multiplication

(iv) Multiplicative identity

(v) Property of zero

(vi) Distributive law of multiplication over addition

(vii) Distributive law of multiplication over subtraction

Q3: Find the value of each of the following using various properties:

(i) 647 $\times$ 13 + 647 $\times$ 7

(ii) 8759 $\times$ 94 + 8759 $\times$ 6

(iii) 7459 $\times$ 999 + 7459

(iv) 9870 $\times$ 561 – 9870 $\times$ 461

(v) 569 $\times$ 17 + 569 $\times$ 13 + 569 $\times$ 70

(vi) 16825 $\times$ 16825 – 16825 $\times$ 6825

Sol:

(i) 647 $\times$ 13 + 647 $\times$ 7

= 647 $\times$ (13 + 7)

= 647 $\times$ 20

= 12940 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

(ii) 8759 $\times$ 94 + 8759 $\times$ 6

= 8759 $\times$ (94 + 6)

= 8759 $\times$ 100

= 875900 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

(iii) 7459 $\times$ 999 + 7459

= 7459 $\times$ (999 + 1)

= 7459 $\times$ 1000

= 7459000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

(iv) 9870 $\times$ 561 – 9870 $\times$ 461

= 9870 $\times$ (561 – 461)

= 9870 $\times$ 100

= 987000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

(v) 559 $\times$ 17 + 569 $\times$13 + 569 $\times$ 70

= 569 $\times$ (17 + 13 + 70)

= 569 $\times$100

= 569000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

(vi) 16825 $\times$ 16825 – 16825 – 6825

= 16825 $\times$ (16825 – 6825)

= 16825 $\times$ 10000

= 168250000 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ( By using distributive property)

Q4:Determine each of the following products by suitable rearrangements :

(i) 2 $\times$ 1658 $\times$ 50

(ii) 4 $\times$ 927 $\times$ 25

(iii) 625 $\times$ 20 $\times$ 8 $\times$ 50

(iv) 574 $\times$ 625 $\times$ 16

(v) 250 $\times$ 60 $\times$ 50 $\times$ 8

(vi) 8 $\times$ 125 $\times$ 40 $\times$ 25

Sol:

(i) 2 $\times$ 16858 $\times$ 50

= (2 $\times$ 50) $\times$ 1658

= 100 $\times$ 1658

= 165800

(ii) 4 $\times$ 927 $\times$ 25

= (4 $\times$ 25) $\times$ 927

= 100 $\times$ 927

= 92700

(iii) 625 $\times$ 20 $\times$ 8 $\times$ 50

= (20 $\times$ 50) $\times$ 8 $\times$ 625

= 1000 $\times$ 8 $\times$ 625

= 8000 $\times$ 625

= 5000000

(iv) 574 $\times$ 625 $\times$ 16

= 574 $\times$ (625 $\times$ 16)

= 574 $\times$ 10000

= 5740000

(v) 250 $\times$ 60 $\times$ 50 $\times$ 8

= (250 $\times$ 8) $\times$ Â (60 $\times$ 50)

= 2000 $\times$ 3000

= 6000000

(vi) 8 $\times$ 125 $\times$ 40 $\times$ 25

= (8 $\times$ 125) $\times$ Â (40 $\times$ 25)

= 1000 $\times$ 1000

= 1000000

Q5: Find each of the following products, using distributive laws:

(i) 740 $\times$ 105

(ii) 245 $\times$ 1008

(iii) 947 $\times$ 96

(iv) 996 $\times$ 367

(v) 472 $\times$ 1097

(vi) 580 $\times$ 64

(vii) 439 $\times$ 997

(viii) 1553 $\times$ 198

Sol:

(i) 740 $\times$ Â 105

= 740 $\times$ (100 + 5)

= 740 $\times$ 100 + 740 $\times$ 5 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 74000 + 3700

= 77700

(ii) 245 $\times$ 1008

= 245 $\times$ (1000 + 8)

= 245 $\times$ 1000 + 245 $\times$ 8 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 245000 + 1960

=246960

(iii) 947 $\times$ 96

= 947 $\times$ (100 – 4)

= 947 $\times$ 100 – 947 $\times$ 4 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 94700 – 3788

= 90912

(iv) 996 $\times$ 367

= 367 $\times$ (1000 – 4)

= 367 $\times$ 1000 – 367 $\times$ 4 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 367000 $\times$ 1468

= 365532

(v) 472 $\times$ 1097

= 472 $\times$ (1000 + 97)

= 472 $\times$ 1000 + 472 $\times$ 97 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 472000 + 45784

= 517784

(vi) 580 $\times$ 64

= 580 $\times$ (60 + 4)

= 580 $\times$ 60 + 580 $\times$ 4 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 34800 + 2320

= 37120

(vii) 439 $\times$ 997

= 439 $\times$ (1000 – 3)

= 439 $\times$ 1000 – 439 $\times$ 3 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 439000 – 1317

= 437683

(viii) 1553 $\times$ 198

= 1553 $\times$ (200 – 2)

= 1553 $\times$ 200 – 1553 $\times$ 2 Â Â Â Â Â Â Â Â Â (Using distributive law of multiplication over addition)

= 310600 – 3106

= 307494

Q6: Find each of the following products, using distributive laws:

(i) 3576 $\times$ 9

(ii) 847 $\times$ 99

(iii) 2437 $\times$ 999

Sol:

Distributive property of multiplication over addition states that a(b + c) = ab + ac

Distributive property of multiplication over subtraction states that a(b – c) = ab – ac

(i) 3476 $\times$ 9

= 3576 $\times$ (10 – 1)

= 3576 $\times$ 10 – 3576 $\times$ 1

= 35760 – 3576

= 32184

(ii) 847 $\times$ 99

= 847 $\times$ (100 – 1)

= 847 $\times$ 100 – 847 $\times$ 1

= 84700 – 847

= 83853

(iii) 2437 $\times$ 999

= 2437 $\times$ (1000 – 1)

= 2437 $\times$ 1000 – 2437 $\times$ 1

= 2437000 – 2437

= 2434563

Q7: Find the products:

(i)

458 $\times$ 67 = 30686

(ii)

3709 $\times$ 89 = 330101

(iii)

4617 $\times$ 234 = 1080378

(iv)

15208 $\times$ 542 = 8242736

Q8: Find the product of the largest 3 digit number and the largest 5 digit number.

Sol:

Largest three digit number = 999

Largest five digit number = 99999

Therefore, Product of two numbers = 999 $\times$ 99999

= 999 $\times$ (100000 – 1)

= 99900000 – 999

= 99899001

Q9: A car moves at a uniform speed of 75 km/h. How much distance will it cover in 98 hours?

Sol:

Uniform speed of a car = 75 km/h

Distance = Speed $\times$ time

= 75 $\times$ 98

= 75 $\times$ (100 – 2)

= 75 $\times$ 100 – 75 $\times$ 2

= 7500 – 150

= 7350 km

Therefore, The distance covered in 98 h is 7350 km.

Q10: A dealer purchased 139 VCRs. If the cost of each set is Rs. 24350, find the cost of all the sets together.

Sol:

Cost of 1 VCR set = Rs. 24350

Cost of 139 VCR sets = 139 $\times$ 24350

= 24350 $\times$ (140 – 1)

= 24350 $\times$ 140 – 24350 $\times$ 1

= 3409000 – 24350

= Rs. 3384650

Therefore, The cost of all the VCR sets is Rs. 33,84,650

Q11: A housing society constructed 187 houses. If the cost of construction for each house is Rs. 450000, what is the total cost for all the houses?

Sol:

Cost of construction of 1 house = Rs. 450000

Cost of construction of 197 such houses = 197 $\times$ 450000

= 450000 $\times$ (200 – 3)

= 450000 $\times$ 200 – 450000 $\times$ 3

= 90000000 – 1350000

= 88650000

= Therefore, the total cost of construction of 197 houses is Rs. 8,86,50,000

Q12: 50 chairs and 30 blackboards were purchased for a school. If each chair costs Rs. 1065 and each blackboard cost Rs. 1645, find the total amount of the bill.

Sol:

Cost of a chair = Rs. 1065

Cost of a blackboard = Rs. 1645

Cost of 50 chairs = 50 $\times$ 1065 = Rs. 53250

Cost of 30 blackboards = 30 $\times$ 1645 = Rs. 49350

Therefore, Total amount of the bill = cost of 50 chairs + cost of 30 blackboards

= Rs (53250 + 49350)

= Rs. 1,02,600

Q13: There are six sections of Class VI in a school and there are 45 students in each section. If the monthly charges from each student be Rs. 1650, find the total monthly collection from Class VI.

Sol:

Number of students in 1 section = 45

Number of students in 6 sections = 45 $\times$ 6 = 270

Monthly charges from 1 student = Rs. 1650

Therefore, Total monthly collection from class VI = Rs. 1650 $\times$ 270 = Rs. 4,45,500

Q14: The product of two whole number is zero. What do you conclude?

Sol: If the product of two whole number is zero, then one of them is definitely zero.

Example: 0 $\times$ 2 = 0 and 0 $\times$ 15 = 0

If the product of whole number is zero, then both of them may be zero.

I.e. 0 $\times$ 0 = 0

Now, 2 $\times$ 5 = 10. Here, the product will be non-zero because the numbers to be multiplied are not equal to zero.

Q15: Fill in the blanks:

(i) Sum of two odd numbers is an ______ number.

(ii) Product of two odd numbers is an _____ number.

(iii) $a \neq 0$ and a $\times$ a = a $\Rightarrow a = ?$

Sol:

Sum of two odd numbers is an even number. Example: 3 + 5 = 8, which is an even number.

(ii) Product of two odd numbers is an odd number. Example: 5 $\times$ 7 = 35, which is an odd number.

(iii) $a \neq 0$ and a $\times$ a = a

Given: a $\times$ a = a

$\Rightarrow a = \frac{a}{a} = 1, \;a \neq 0$

Q1: Divide and check your answer by the corresponding multiplication in each of the following:

(i) 1936 $\div$ 16

(ii) 19881 $\div$ 47

(iii) 257796 $\div$ 341

(iv) 612846 $\div$ 582

(v)34419 $\div$ 149

(vi) 39039 $\div$ 1001

Sol:

(i) 1936 $\div$ 36

Dividend = 1936, Divisor = 36, Quotient = 53, Remainder = 28

Check: Divisor $\times$ Quotient + Remainder = 36 $\times$ 53 + 28

= 1936

=Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(ii) $19881 \div 47$

Dividend = 19881, Divisor = 47, Quotient = 423, Remainder = 0

Check: Divisor $\times$ Quotient + Remainder = 47 $\times$ 423 + 0

= 19881

= Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(iii) 257796 $\div$ 341

Dividend = 257796, Divisor = 341, Quotient = 756, Remainder = 0

Check: Divisor $\times$ Quotient + Remainder = 341 $\times$ 756 + 0

= 257796

= Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(iv) 612846 $\div$ 582

Dividend = 612846, Divisor = 582, Quotient = 1053, Remainder = 0

Check: Divisor $\times$ Quotient + Remainder = 582 $\times$ 1053 + 0

= 612846

= Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(v)34419 $\div$ 149

Dividend = 34419, Divisor = 149, Quotient = 231, Remainder = 0

Check: Divisor $\times$ Quotient + Remainder = 149 $\times$ 231 + 0

= 34419

= Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(vi) 39039 $\div$ 1001

Dividend = 39039, Divisor = 1001, Quotient = 39, Remainder = 0

Check: Divisor $\times$ Quotient + Remainder = 1001 $\times$ 39 + 0

= 39039

= Divident

Hence, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

Q2: Divide and find out the quotient and remainder. Check your answer.

(i) 6971 $\div$ 47

(ii) 4178 $\div$ 35

(iii) 36195 $\div$ 153

(iv) 93575 $\div$ 400

(v) 23025 $\div$ 1000

(vi) 16135 $\div$ 875

Sol:

(i) 6971 $\div$ 47

Quotient = 148 and remainder = 15

Check : Divisor $\times$ Quotient + Remainder = 47 $\times$ 148 + 15

= 6971

= Divident

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(ii) 4178 $\div$ 35

Dividend = 119 and Remainder = 13

Check: Divisor $\times$ Quotient + remainder = 35 $\times$ 119 + 13

= 4178

= Dividend

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(iii) 36195 $\div$ 153

Quotient = 236 and remainder = 87

Check : Divisor $\times$ Quotient + Remainder = 153 $\times$ 236 + 87

= 36195

= Divident

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(iv) 93575 $\div$ 400

Quotient = 233 and remainder = 375

Check : Divisor $\times$ Quotient + Remainder = 400 $\times$ 233 + 375

= 93575

= Divident

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(v) 23025 $\div$ 1000

Quotient = 23 and remainder = 25

Check : Divisor $\times$ Quotient + Remainder = 1000 $\times$ 23 + 25

= 23025

= Divident

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

(vi) 16135 $\div$ 875

Quotient = 18 and remainder = 385

Check : Divisor $\times$ Quotient + Remainder = 875 $\times$ 18 + 385

= 16135

= Divident

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified.

Q3: Find the value of

(i) 65007 $\div$ 1

(ii) 0 $\div$ 879

(iii) 981 + 5720 $\div$ 10

(iv) 1507 – (625 $\div$ 25)

(v) 32277 $\div$ (628 – 39)

(vi) (1573 $\div$ 1573) – (1573 $\div$ 1573)

Sol:

(i) 65007 $\div$ 1 = 65007

(ii) 0 $\div$ 879 = 0

(iii) 981 + 5720 $\div$ 10

= 981 + (5720 $\div$ 10) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Following DMAS property)

= 981 + 572

= 1553

(iv) 1507 – (625 $\div$ 25) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Following BODMAS property)

= 1507 – 25

= 1482

(v) 32277 $\div$ (628 – 39) Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Following BODMAS property)

= 32277 $\div$ (609 )

= 53

(vi) (1573 $\div$ 1573) – (1573 $\div$ 1573) Â Â Â Â Â Â Â Â Â Â (Following BODMAS property)

= 1 – 1

= 0

Q4: Find a whole number n such that n $\div$ n = n

Sol:

Given : n $\div$ n = n

$\Rightarrow \frac{n}{n} = n$ $\Rightarrow n = n^{2}$

I.e. the whole number n is equal to $n^{2}$.

Therefore, The given whole number must be 1.

Q5: The product of two numbers is 504347. If one of the numbers is 317, find the other.

Sol:

Let x and y be the two numbers.

Product of the two numbers = x $\times$ y = 504347

If x = 317, we have

317 $\times$ y = 504347

$\Rightarrow$ y = 504347 $\div$ 317

y = 1591

Therefore, The other number is 1591.

Q6: On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.

Sol:

Dividend = 59761, quotient = 189, remainder = 37 and divisor = ?

Dividend = Divisor $\times$ quotient + remainder

$\Rightarrow$ 59761 = Divisor $\times$ 189 + 37

$\Rightarrow$ 59761 – 37 = Divisor $\times$ 189

$\Rightarrow$ 59724 = divisor $\times$ 189

Hence, divisor = 316

Q7: On dividing 55390 by 299, the remainder is 75. Find the quotient using the division algorithm.

Sol:

Here, Dividend = 55390, Divisor = 299 and remainder = 75.

We have to find the quotient.

Now, Dividend = Divisor $\times$ quotient + remainder

$\Rightarrow$ 55390 = 299 $\times$ quotient + 75

$\Rightarrow$ 55390 – 75 = 299 $\times$ quotient

$\Rightarrow$ 55315 = 299 $\times$ quotient

$\Rightarrow$ quotient Â = 55315 $\div$ 299

Hence, quotient = 185.

Q8: What least number must be subtracted from 13601 to get a number exactly divisible by 87?

Sol:

First, we will divide 13601 by 87.

Remainder = 29

So, 29 must be subtracted from 13601 to get a number exactly divisible by 87.

i.e., 16301 – 29 = 13572

Now, we have:

Therefore, 29 must be subtracted from 13601 to make it divisible by 87.

Q9: What least number must be added to 1056 to get a number exactly divisible by 23?

Sol:

First we will divide 1056 by 23.

Required number = 23 – 21 = 2

So, 2 must be added to 1056 to make if exactly divisible by 23.

i.e., 1056 + 2 = 1058

Now, we have:

Therefore, 1058 is exactly divisible by 23.

Q10: Find the largest 4 digit number divisible by 16.

Sol:

We have to find the largest four digit number divisible by 16.

The largest four digit number = 9999

Therefore, dividend = 9999

Divisor = 16

Here, we get remainder = 15

Therefore, 15 must be subtracted from 9999 to get the largest four digit number that is divisible by 16.

i.e. 9999 – 15 = 9984

Thus, 9984 is the largest four digit number that is divisible by 16.

Q11: Divide the largest 5 digit number by 653. Check your answer by the division algorithm.

Sol:

Largest five digit number = 99999

Dividend = 99999, Divisor = 653, Quotient = 153 and Remainder = 90

Check : Divisor $\times$ Quotient + Remainder

= 653 $\times$ 153 + 90

= 99909 + 90

= 99999

= Dividend

Therefore, Dividend = Divisor $\times$ Quotient + Remainder

Verified

Q12: Find the least 6 digit number exactly divisible by 83.

Sol:

Least six digit number = 100000

Here, dividend = 100000 and divisor = 83

In order to find a number exactly divisible by 83, we have to subtract the remainder from the dividend.

i.e. 100000 – 68 = 99932

So, 99932 is the largest six digit number exactly divisible by 83.

Q13: 1 dozen bananas cost Rs. 29. How many dozens can be purchased for Rs. 1392?

Sol:

Cost of 1 dozen bananas = Rs. 29

Number of dozens purchased for Rs. 1392 = 1392 $\div$ 29

Hence, 48 dozens of bananas can be purchased with Rs. 1392.

Q14: 19625 trees have been equally planted in 157 rows. Find the number of trees in each row.

Sol:

Number of trees planted in 157 rows = 19625

Trees planted in 1 row = 19625 $\div$ 157

Therefore, 125 trees are planted in each row.

Q15: The population of a town is 517530. If one out of every 15 is reported to be literate, find how many literate persons are there in the town.

Sol:

Population of the town = 517530

$\left ( \frac{1}{15} \right )$ of the population is reported to be literate i.e. $\left ( \frac{1}{15} \right ) \times 517530 = 517530 \div 15$

Therefore, There are 34502 illiterate persons in the given town.

### Key Features of RS Aggarwal Class 6 Solutions Chapter 3 – Whole Numbers Ex 3A (3.1)

• It acts as a step to step guide to solve all your problems and help you to clear your examination with flying colours.
• These solutions are prepared by our subject experts to give you an in-depth understanding of all the concepts.
• Practicing the RS Aggarwal Maths solutions will help you gain a strong foothold in mathematics.
• These solutions are extremely helpful while solving the tough questions that might be asked in the exam.

#### Practise This Question

Consider the following set
20y, 20, 15q, -25s, 25a, 12z, y
Which of the following represent the group of like term?