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Brilliant’s Composite Mathematics Class 6 Solutions Chapter 3 Operations on Whole Numbers
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 3, Operations on Whole Numbers. Here students can easily find step by step solutions of all the problems for Operations on Whole Numbers, Exercise 3.1, 3.2, 3.3 and 3.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Operations on Whole Numbers Exercise 3.1 Solution
Question no – (1)
Solution :
(i) 25 + 58 = 58 + 25 …[Commutative Property]
(ii) 93 + 136 = 93 + 236 ….[Commutative Property]
(iii) 275 + 925 = 275 + 925 ….[Commutative Property]
(iv) 1016 + 1732 = 1016 + 1732 ….[Commutative Property]
(v) 56 + (137 + 963) = (56 + 137) + 963 ….[Associative law of addition]
(vi) 10197 + 0 = 0 + 10197 = 10197 ….[Additive property of zero]
(viii) (222944) + 956 222 + (956 + 944) ….[Associative property]
Question no – (2)
Solution :
No, the sum of two odd number is even number.
Question no – (3)
Solution :
No, the sum of two odd number is even number.
Question no – (4)
Solution :
Greatest number of 4 digit = 9999
Least number of 4 digit = 1000
Closure,
= 9999 + 1000
= 10999
Commutative,
= 100 + 9999
= 10999
Question no – (5)
Solution :
We know,
Greater number of 4 digit is = 9999
Least number of 5 digit is = 10000
∴ adding = 9999 + 10000
= 19999
= 9999 + 10000
= 10000 999
= 19999 [Commutative]
Question no – (6)
Solution :
(i) 178 + 923584;
= 923584 + 178
= 923762
(ii) 39784 + 5628;
= 5028 + 39784
= 45412
(iii) 591 + 112 + 15409;
= 591 + (112 + 15409)
= (591 + 112) + 15409
= 16,112
(iv) 2359 + 641 + 9000;
= 2359 + (641 + 9000)
= (2359 + 641) + 9000
= 12000
Question no – (7)
Solution :
(i) 653 + 747 + 856;
= (653 + 856) + 747
= 2256
(ii) 2983 + 1647 + 17 + 453;
= (2983 + 17) + (1647 + 435)
= 5100
(iii) 8265 + 1935 + 8773 + 3227;
= (8265 + 8773) + (1935 + 3227)
= 22200
(iv) 486 + 78+ 14 + 122 + 53.
= (486 + 14) + (122 + 78) + 53
= 753
Question no – (8)
Solution :
= 22 + 28 + 9 + 15 + 16 = 90
= 22 + 29 + 6 + 13 + 20 = 90
= 10 + 18 + 26 + 14 + 22 = 90
= 20 + 10 + 11 + 17 + 23 = 90
= 9 + 11 + 18 + 25 + 27 = 90
= 15 + 17 + 24 + 26 + 8 = 90
= 16 + 23 + 30 + 7 + 14 + 90
Now,
| 22 | 29 | 6 | 13 | 20 |
| 28 | 10 | 12 | 19 | 21 |
| 9 | 11 | 18 | 25 | 27 |
| 15 | 17 | 24 | 26 | 8 |
| 16 | 23 | 30 | 7 | 14 |
Question no – (9)
Solution :
(i) 5 * 3
= 10 + 9
= 19
(ii) 105 * 10
= 210 + 30
= 240
(iii) 219 * 25
= 2 × 219 + 3 × 25
= 438 + 75
= 513
(iv) 53 * 123
= 106 + 369
= 475
(v) 15 * 16
= 30 + 48
= 78
(vi) 23 * 33
= 46 + 99
= 145
Question no – (10)
Solution :
(i) Zero added to a number 23 is the number 32.
= Given statement is False
(ii) The sum of two odd numbers is always an odd number.
= Given statement is False
(iii) The sum of two even numbers may be an odd number.
= Given statement is False
(iv) (15 + 17) + 19 = (17 + 15) + 19.
= Given statement is True
(v) If a, b, c are any three whole numbers, then (a + b) + c = a + (b + c).
= Given statement is True
Question no – (11)
Solution :
0 + 0 = 0. Is there some other whole number q such that q + q = q
No, such whole number does not exists.
Operations on Whole Numbers Exercise 3.2 Solution
Question no – (1)
Solution :
(i) 59 – 45 = 14
addition,
45 + 14
= 59
(ii) 127 – 37 = 90
addition,
= 90 + 37
= 127
(iii) 453 – 356 = 97
addition,
97 + 365
= 453
(iv) 9652 – 1732 = 7920
addition,
7932 + 1732
= 9652
(v) 100000 – 99755 = 245
addition,
= 99755 + 245
= 100000
(vi) 3030301 – 767676 = 2262625
addition,
2262625 + 767676
= 3030301
Question no – (2)
Solution :
(i) (327 – *5) = 3 * 2
= (327 – 15)
= 312
(ii) (5376 – ** 57) = 25**
= (5376 = 2857)
= 2519
(iii) (6000107 – ** 8978) = 506 ****
= (6000107 = 938978)
= 5061129
(iv) (3941 – ** 78) = 18 **
= (3941 – 2078)
= 1863
(v) (14335 – ** 47) = 92**
= (14335 – 5147)
= 9188
(vi) (1000000 – **** 1) = * 7042 *
= (100000 = 29571)
= 970429
Question no – (3)
Solution :
As we know that
Greatest 4 digit number is = 9999
Smallest 5 digit number is = 10000
∴ The difference will be,
= 10000 – 9999
= 1
Question no – (4)
Solution :
(i) Given, n + 5 = 8;
= n = 8 – 5
= 3
∴ Here, ‘n’ represented = 3
(ii) Given, n + 3 = 19;
= n + 19 – 3
= 16
∴ Here, ‘n’ represented = 16
(iii) Given, n – 5 = 14.
= n = 14 + 5
= 19
∴ Here, ‘n’ represented = 19
Question no – (5)
Solution :
As we know that,
Largest number of 5 digit = 99999
Smallest number of 6 digit = 100000
∴ The difference will be,
= 10000 – 99999
= 1
Question no – (6)
Solution :
In the given question,
Prateek deposits = Rs 25, 000
Later he withdrew = Rs 5, 425
∴ Money left in his account,
= (25000 – 5425)
= 19575 Rs
Thus, Rs 19575 is left in his account.
Question no – (7)
Solution :
In the question we get,
The total number of men, women and children in a town is 99726.
If men is 36542 and women is 28543
∴ Number of children,
= 99726 – (36542 + 28543)
= 99726 – 65085
= 34641
Hence, the number of children will be 34641
Question no – (8)
Solution :
In the question we get,
Amit depositing on Monday = Rs. 45000
On Tuesday he withdrew = Rs. 35450
∴ Money left in his account,
= (45000 – 35450)
= 9550 Rs.
Thus, in the account of Amit Rs. 9550 was left.
Question no – (9)
Solution :
Given in the question,
The population of a town was = 65600.
In one year it was increased by = 24196.
However persons died or left = 7936
∴ Population at the and of the year,
= (65600 + 24196) – 7936
= 89796 – 7936
= 81860
Hence, the population at the end of the year will be 81860
Question no – (10)
Solution :
According to the question,
Priyanka went to the market with = Rs 12560.
She spent Rs 795 on shoes, Rs 632 on shirts and Rs 939 on jean.
∴ Money left with Priyanka,
= 12560 – (795 + 632 + 939)
= 12560 2366
= 10194 Rs.
Hence, 10194 Rs is left with Priyanka.
Question no – (11)
Solution :
From the question we get,
Number should be added to 4739 so that the sum may be 6000
∴ Number should added,
= 6000 – 4739
= 1261
So, 1261 number should be added.
Question no – (12)
Solution :
In the given question,
A cooler is marked for = Rs 2540
the dealer allows a discount of = Rs 219
∴ Net selling price of the colon,
= 2540 – 219
= 2321 Rs.
Thus, the net selling price of the cooler will be 2321 Rs
Operations on Whole Numbers Exercise 3.3 Solution
Question no – (1)
Solution :
(i) 129 × 0 = 0
(ii) 534 × 1 = 534
(iii) 643 × 199 = 199 × 643
(iv) 956 × 299 = (956 ) × 299
(v) 13 × (55 +23) = (3 × 55) + (13 × 23)
(vi) 76 × 31+ 76 × 69 = 76 × (31 + 69)
(vii) 136 × 94 = (136 × 100) 136 × (6)
Question no – (2)
Solution :
(i) 23 x 17 is a whole number
= Closure property
(ii) 35 × 46 = 46 × 35
= Commutative law
(iii) 15 × 1 = 15
= Property of unity
(iv) 138 × 0 = 0
= Property of zero
(v) (26 × 19) × 105.26 × (19 × 105)
= Associative property
(vi) 56 × (18 + 12) = (56 × 18) + (56 × 12)
= Distributive property of multiplication
(vii) 96 × (123 -11) = (96 × 123) – (96 × 11)
= Distributive property of multiplication over subtraction
(viii) 239 × 49 + 239 × 24 + 239 × 27 = 239 × (49 + 24 + 27)
= Distributive property
Question no – (4)
Solution :
(i) 5 × 1956 × 20
= (5 × 20) × 1956
= 100 × 1956
= 195600
(ii) 389 × 60 × 15
= 398 × (60 × 15)
= 398 × 900
= 358200
(iii) 45 × 80 × 25
= 45 × (25 × 80)
= 45 × 2000
= 90000
(vi) 325 × 60 × 50 × 8
= 325 × 8 × (60 × 50)
= 2600 × 300
= 780000
Question no – (6)
Solution :
As we know that,
Largest 4 digit number = 9999
Smallest 3 digit number = 999
∴ The Product will be,
= 999 × 9999 = (1000 – 1) × 9999
= 9999000 – 9999
= 9989001
Hence, the product will be 9989001
Question no – (7)
Solution :
First, the Sum,
= 53 + 47
= 100
Now, the difference,
= 53 ÷ 47
= 6
∴ The product will be,
= 100 × 6
= 600
Hence, the product will be 600
Question no – (8)
Solution :
Let, Two odd number are = 11, 13
∴ 11
13
———————————-
33
1110
———————————-
1433 (Odd number)
Therefore, It is True.
Question no – (9)
Solution :
Let, Two even numbers, = 10, 12
= 12 × 10
= 120 (Even)
Again, two ever numbers = 2, 4
= 2 × 4
= 8 (Ever)
So, Yes product of two even number always a even number.
Question no – (10)
Solution :
If the product of two whole numbers is zero, then, at least one whole number is Zero.
Question no – (11)
Solution :
We know, that 0 × 0 = 0.
Yes, such number is,
= 1 × 1 = 1
Question no – (12)
Solution :
According to the question,
A shopkeeper bought 55 chairs and 55 tables.
Table costs Rs 1250 and a chair costs Rs 750,
∴ The cost of all the items together,
= 55 × (1250 + 750)
= 56 × 2000
= 110000 Rs.
Hence, the cost of all the items together will be 110000 Rs.
Question no – (13)
Solution :
From the question we get,
A dealer purchased 52 colour and 48 black and white T.V sets.
Colour T.V set is = Rs 7290
Black and white set is = Rs 4125
∴ The cost of all the set together,
= 52 × 7290 + 48 × 4125
= 379080 + 198000
= 577080 Rs.
Therefore, the cost of all the set together will be Rs. 577080
Question no – (14)
Solution :
In the question we get,
Section of class in a school total students = 35
Month by charges from each student = Rs 450.
Now,
Hence, Monthly collection will be Rs. 15750
Question no – (15)
Solution :
According to the question,
A Car moves at a uniform speed of = 67 km per hour.
How many distance it will cover in 94 hours = ?
∴ Distance will be,
Hence, the car will cover 6298 km.
Operations on Whole Numbers Exercise 3.4 Solution
Question no – (1)
Solution :
We know that, a ÷ b ≠ b ÷ a
Example,
= 6 ÷ 3 = 6/3 = 2
= 3 ÷ 6 = 3/6 = 1/2
Question no – (2)
Solution :
(i) Given, 2371 ÷ 35
Now,
∴ Quotient = 67
∴ Remainder = 26
(ii) 7419 ÷ 25
Now,
∴ Quotient = 269
∴ Remainder = 19
(iii) 46519 ÷ 127
Now,
Quotient = 366
Remainder = 37
(iv) 57691 ÷ 357
Now,
∴ Quotient = 161
∴ Remainder = 214
(v) 96397 ÷ 400
Now,
∴ Quotient = 240
∴ Remainder = 397
(vi) 16135 ÷ 735
Now,
∴ Quotient = 21
∴ Remainder = 700
Question no – (3)
Solution :
(i) 276 ÷ 7
= 276 = 38 × 7 + 0
(ii) 9275 ÷ 23
= 9275 = 403 × 23 + 6
(iii) 3765 ÷ 135
= 3765 = 27 × 136 + 120
(iv) 32796 ÷ 875
= 32796 = 875 × 37 + 421
(v) 4697 ÷ 746
= 4697 = 6 × 746 + 221
(vi) 700965 ÷ 937
= 700965 = 748 × 937 + 89
Question no – (4)
Solution :
(i) 3716 ÷ 1
= 3716/1
= 3716
(ii) 0 ÷ 271
= 0/271
= 0
(iii) 989 + (5760 ÷ 20)
= 989 + 5760/20
= 989 + 288
= 1277
(iv) (1775 ÷ 71) – 21
= 1775/71 – 21
= 25 – 21
= 4
(v) 32807 ÷ (648 – 29)
= 32807 ÷ 619
= 53
(vi) 2573 ÷ 2573 – 4563 ÷ 4563
= 2573/2573 – 4563/4563
= 1 – 1
= 0
Question no – (5)
Solution :
Yes → n ÷ 1 = n
Question no – (6)
Solution :
According to the question,
∴ 10 should be subtract tor perfect divided.
Question no – (7)
Solution :
In the given question we get,
Quotient = 63
Remainder = 23,
Divided = 45
∴ Required Number,
= 63 × 45 + 23
= 2835 + 23
= 2858
Thus, the required number will be 2858
Question no – (8)
Solution :
According to the question,
∴ The least number is to be added,
= 65 – 15
= 50
Hence, the least number will be 50
Question no – (9)
Solution :
As we know that,
Greatest number of 6 digit = 999999
Greatest number of 3 digit = 999
Now,
∴ Quotient = 1001,
∴ Remainder = 0
Question no – (10)
Solution :
In the question we get,
The price of 35 TV sets is 313460
Now,
Therefore, one TV will costs Rs. 8956
Question no – (11)
Solution :
In the question,
146 apples can be put in a basket.
Baskets will be required to put = 5694 apples
Now,
Thus, 39 baskets will be required.
Question no – (13)
Solution :
In the question,
least number be subtracted from = 9969
Difference is exactly divisible by = 129
Now,
Therefore, we need subject by 36
Question no – (14)
Solution :
Given in the question,
The cost of 25 refrigerators is Rs 1,63,500.
The cost of each refrigerator = ?
Now,
∴ The cost of each refrigerator will be Rs. 6540
Question no – (15)
Solution :
In the question we get,
The product of two numbers is = 30135
One number is = 123
Other number = ?
Now,
Hence, the other number will be 245
Question no – (16)
Solution :
In the given question,
Least number must be added to = 1065
Get a number exactly divisible by = 33
Required number = ?
Now,
∴ least number must be added,
= 33 – 9
= 24
Question no – (17)
Solution :
From the question we get,
Least number must be subtracted from = 12706
To get a number exactly divisible by = 59
Required number = ?
Now,
Thus, 21 must be subtract tor exactly divisible.
Question no – (18)
Solution :
According to the question,
The largest 3 digit number exactly divisible by = 26
We know that,
Larger 3 digit number – 999
Now,
∴ 999 – 11 = 988 is exactly divisible by 26,
Question no – (19)
Solution :
In the question we get,
9525 trees have been equally planted in = 127 rows.
The number of trees in each row = ?
Now,
∴ Trees in each row will be 75
Next Chapter Solution :
