NCERT Exemplar Solutions: Number System | Mathematics (Maths) Class 6 PDF Download

Ans: (c)
The product of the place values of two 2’s in 428721 is
There are two 2’s in the given number.
So, the first 2 is in the tenth place,
Then, the product is = 2 × 10
= 20
The other 2 is in place value of ten thousand.
Then, = 2 × 10000
= 20000
Therefore, the product of place values = 20 × 20000
= 400000

Ans: (c)
3 × 10000 = 30000
The place value is 30000
7 × 1000 = 7000
The place value is 7000
9 × 100 = 900
The place value is 900
0 × 10 = 0
4 = 4
Therefore, the sum of all place value is = 30000 + 7000 + 900 + 0 + 4
= 37904

Ans: (d)
We know that, the greatest number is 99,99,999
Then, 1 is added to 99,99,999 = 99,99,999 + 1
= 1,00,00,000
= 1 crore

Ans: (b)
Consider the given number 9578,
The place value of 8 is ones = 8 × 1
The place value of 7 is tens = 7 × 10
The place value of 5 is thousand = 5 × 100
The place value of 9 is ten thousand = 9 × 1000

Ans: (d)
When rounded off to nearest thousands, the number 85642 is = 86000

Ans: (d)
Using 9 as twice from 5, 9, 2 and 6, then then number is 9965.

Ans: (c)
In Indian System of Numeration, the number 58695376 is written as 5 crore, eighty six lakh, ninety five thousand, three hundred and seventy six = 5,86,95,376

India Place Value Chart

Ans: (b)
One million is equal to ten lakh.
1,000,000 = 10,00,000

Ans: (d)
The greatest number which on rounding off to nearest thousands gives 5000, is 5499.

Ans: (c)
Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is 6034579.

Ans: (d)
As we know that, the symbol L can never be repeated.
Therefore, LLX is incorrect.

Ans: (c)
In the given, options there are three numbers used 9, 8 and 7
To get the largest of 5- digit we have to arrange the numbers in descending order.
Then, from the given options 99987 is the largest of 5 – digit number.

Ans: (d)
In the given, options there are three numbers used 0, 1 and 2
To get the largest of 4- digit we have to arrange the numbers in ascending order.
Then, from the given options 1002 is the smallest of 4 – digit number.

Ans: (c)
Number of whole numbers between 38 and 68 is 29.

Ans: (b)
The number which comes immediately before a particular number is called its predecessor.
The successor of a whole number is the number obtained by adding 1 to it.
So, Successor of 999 = 999 + 1 = 1000
Predecessor = 999 – 1 = 998
Then, product of successor and predecessor of 999 is = 1000 × 998
= 998000

Ans: (a)
The product of a non-zero whole number and its successor is always an even number.
For example: – 4 × 5 = 20, 7 × 8 = 56

Ans: (c)
Let us assume the number be x.
From the question it is given that, number is added to 25 = x + 25
The same number is subtracted to from 25 = 25 – x
Then, the sum of the resulting numbers is = (x + 25) + (25 – x)
= x + 25 + 25 – x
= 50 + x – x
= 50 + 0
= 50

Ans: (c)
Consider the left hand side = 7 + 8 × 9
= 7 + (8 × 9)
= 7 + 72
= 79
Now, consider the right hand side = (7 + 8) × (7 + 9)
= 15 × 16
= 240
By comparing LHS and RHS
LHS ≠ RHS
79 ≠ 240

Ans: (b)

Ans: (b)

Ans: (d)
Zero divided by zero is not defined.

Ans: (b)
The number which comes immediately before a particular number is called its predecessor.
The predecessor of 1 lakh is = 1,00,000 – 1
= 99,999

Ans: (b)
The successor of a whole number is the number obtained by adding 1 to it.
We know that, 1 million = 10,00,000
Then, successor = 10,00,000 + 1
= 10,00,001

Ans: (a)
Even numbers between 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

Ans: (c)
Prime numbers between 16 to 80 = 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79.
Then, number of primes between 16 to 80 = 16
Prime numbers between 90 to 100 = 97
Then, number of primes between 90 to 100 = 1
Therefore, Sum of the number of primes between 16 to 80 and 90 to 100 is,
= 16 + 1
= 17

Ans: (d)
The HCF of an even and an odd number is odd number.

Ans: (b)
The largest 4 – digit number = 9999
Prime factors of 9999 = 3 × 3 × 11 × 101
So, 9999 = 3² × 11 × 101
Therefore, distinct prime factors are = 3, 11 and 101
Number of distinct prime factors of the largest 4-digit number is = 3

Ans: (a)
The smallest 5 – digit number = 10000
Prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5
So, 10000 = 2⁴ × 5⁴
Therefore, distinct prime factors are = 2 and 5
Number of distinct prime factors of the smallest 5-digit number is = 2

Ans: (c)
Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, then the original number.
7 – 2 + 5 – 4 + * – 9 + 8 = (5 + *)
For the given number 7254 * 98 to be divisible by 11,
(5 + *) must also be divisible by 11
So, 5 + * = 11
Therefore * = 11 – 5
= 6

Ans: (b)
1 + 3 = 4 = 4/4 = 1
3 + 5 = 8 = 8/4 = 2
The largest number which always divides the sum of any pair of consecutive odd numbers is 4.

Ans: (d)
The LCM of 6 and 5 is 30.
So, 30 is divisible by 10, 15 and 30 in the given options.
But, 30 is not divisible by 60.

Ans: (d)
The prime factors of 1729 = 7 × 13 × 19
Therefore, the sum of prime numbers = 7 + 13 + 19
= 39

Ans: (b)
Let us assume an odd natural number be 5.
Then, predecessor of 5 = 5 -1 = 4
Successor of 5 = 5 + 1 = 6
Then, the product of predecessor and successor = 4 × 6
= 24
24 is divided by 4 = 24/4 = 6
Therefore, the greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is 4.

Ans: (a)
Prime factors of,
75 = 3 × 5 × 5
60 = 2 × 2 × 3 × 5
105 = 3 × 5 × 7
So, common prime factors in the given three numbers are 3 and 5.
Therefore, the number of common prime factors of 75, 60, 105 is 2.

Ans: (a)
First of all, both the numbers are even.
Then, common factor of both numbers is 2 other than 1.
Therefore, 8 and 10 are not coprime.

Ans: (c)
To check the divisibility of a number by 11, the rule is, find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is either 0 or divisible by 11, then the number is divisible by 11.
So, 2 – 2 + 2 – 2 + 2 – 2 + 2 – 2 = 0
Therefore, 22222222 is divisible by 11.

Ans: (b)
Factors of 10, 15 and 30 is,

Then, LCM of 10, 15 and 30 is 2 × 2 × 3 × 5 × 1 = 60

Ans: (c)
Factors of 180.

Then, factors of 180 = 2 × 2 × 3 × 3 × 5
180 is not divided by 180.
Therefore, 75 is not the HCF of the number 180.

True.
As per the rule of the Roman numerals, a symbol is not repeated more than three times.

False.
If a symbol is repeated, its value is added as many times as it occurs: i.e. II is equal 2, XX is 20 and XXX is 30.

True.
Left Hand Side = 5555
Right Hand Side = 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1
= 5000 + 500 + 50 + 5
= 5555
Left Hand Side = Right Hand Side

True.
Left Hand Side = 39746
Right Hand Side = 3 × 10000 + 9 × 1000 + 7 × 100 + 4 × 10 + 6
= 30000 + 9000 + 700 + 40 + 6
= 39746
Left Hand Side = Right Hand Side

False.
Left Hand Side = 82546
Right Hand Side = 8 × 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6
= 8000 + 2000 + 500 + 40 + 6
= 10,546
Left Hand Side ≠ Right Hand Side

True.
Left Hand Side = 532235
Right Hand Side = 5 × 100000 + 3 × 10000 + 2 × 1000 + 2 × 100 + 3 × 10 + 5
= 5,00,000 + 30,000 + 2000 + 200 + 30 + 5
= 5,32,235
Left Hand Side = Right Hand Side

False.
Where, X = 10
IX = 9
So, XXIX = 10 + 10 + 9
= 29

True.
Where, L = 50
X = 10
IV = 4
So, LXXIV = 50 + 10 + 10 + 4
= 74

False.
Where, L = 50
IV = 4
VI = 6
So, LIV = 50 + 4 = 54
LVI = 50 + 6
Therefore, 54 < 56
Hence, LIV < LVI

False.
In the question, the arrangement of the numbers in ascending order.
Descending order of the given number = 5487, 5478, 4587, 4578.

False.
The number 85764 rounded off to nearest hundreds is written as 85800.

True.
The number 7826 rounded off to nearest hundreds is written as 7800.
The number 12469 rounded off to nearest hundreds is written as 12500
So, sum of numbers after rounded off to hundreds = 7800 + 12500 = 20,300
Therefore, 20,300 is nearest to 20,000.

False.
The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.

False.
The given number 8,16,52,318 will be read as eight crore sixteen lakh fifty two thousand three hundred and eighteen.

True.

False.
Among kilo, milli and centi, the smallest is milli.

False.
The successor of a whole number is the number obtained by adding 1 to it.
Example: – consider the number 9 it is a one digit, then its successor = 9 + 1 = 10.

False.
The successor of a whole number is the number obtained by adding 1 to it.
Example: – consider 3-digit number 999 it is a one digit, then its successor = 999 + 1 = 1000.

False.
The number which comes immediately before a particular number is called its predecessor.
Example: – consider 2-digit number 10, then its predecessor = 10 – 1 = 9.

True.

False.
Consider the whole number 0,
Then, its predecessor = 0 – 1 = -1
– 1 is an integer.

False.
Consider the two natural numbers 4 and 8.
Then, natural numbers between 4 and 8 are 5, 6, 7.

True.
The successor of a whole number is the number obtained by adding 1 to it.
The largest 3-digit number = 999
Then, its successor = 999 + 1
= 1000

True.
As per the rule, Of the given two natural numbers, the one having more digits is greater.

True.
We know that, sum of two natural numbers is always natural number.
Therefore, natural numbers are closed under addition.

False.
We know that, multiplication of two natural numbers is always natural number.
Therefore, natural numbers are closed under multiplication.

False.
Difference of two natural numbers are not always a natural number.
Therefore, natural numbers are not closed under subtraction.

True.
Let us assume ‘a’ and ‘b’ are the two natural numbers.
Then commutative for natural numbers is a + b = b + a.
Consider the two natural numbers 2 and 4.
Where, a = 2, b = 4
a + b = b + a
2 + 4 = 4 + 2
6 = 6

False.
Zero (0) is the identity for addition of whole numbers.
Consider any whole number i.e. 8.
Then, 8 + 0 = 8

True.
Consider any whole number i.e. 6.
6 × 1 = 6

True.
Zero (0) is a whole number which when added to a whole number, gives the number itself.

False.
We know that, ‘0’ is not a natural number.
Therefore, there is no any natural number which when added to a natural number, gives the number itself.

True.
As per the standard rule, if a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.

True.
Consider any non-zero whole number i.e. 5
5 is divided by itself = 5/5 = 1

False.
The product of two whole number is always a whole number.
Because, we know that, whole numbers are closed under multiplication.

True.
As per the standard rule, a whole number divided by another whole number greater than 1 never gives the quotient equal to the former.

True.
As per the standard rule, every multiple of a number is greater than or equal to the number.
2 × 1 = 2
2 × 3 = 6

False.
The number of multiples of a given number is infinite.
Because, we know that numbers are infinite.

True.
We know that, 1 is the identity for multiplication of whole numbers
Therefore, any number is multiplied by 1 we get the number itself.
Hence, every number is a multiple of itself.

True.
For example, 1 + 3 = 4 = 4/4 = 1
11 + 13 = 24 = 24/4 = 6

True.
As per the standard rule, if a number divides three numbers exactly, it must divide their sum exactly.
Let us consider one number i.e. 2, it divides 4, 6 and 8.
Then, sum of three numbers = 4 + 6 + 8
= 18 is exactly divisible by 2

False.

False.
Let us consider the number 6, it is actually divisible by 2 and 3, But 6 is not divisible by 12.

False.

True.
As per the rule of divisibility test, a number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

False.
As per the rule of divisibility test, the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.

True.
Consider the number 20, it is divisible by 4 but not divisible by 8.

False.
The Highest Common Factor of two or more numbers is lower than their Lowest Common Multiple.

True.
As per the rule, LCM of two or more numbers is divisible by their HCF.

False.
From the question it is given that, LCM of two numbers is 28 and their HCF is 8.
But, 28 is not exactly divide by 8.

True.
Consider the two numbers 2 and 4.
Then, LCM of 2 and 4 is 4.

True.

False.
We know that, 0 is the whole number.
0 is no the successor of another whole number.

False.
For example:-
2 + 3 = 5
2 × 3 = 6
From the above example, we can say that sum of two whole numbers is not always less than their product.

True.
Consider the two odd numbers 2 and 5.
Then, sum = 2 + 5 = 7 it is an odd number.
Now, difference = 2 – 5 = 3 it also an odd number.

True.
Co-prime number is a set of numbers or integers which have only 1 as their common factor i.e. their highest common factor (HCF) will be 1. Co-prime numbers are also known as relatively prime or mutually prime numbers. It is important that there should be two numbers in order to form co-primes.

True.

False.
The HCF of two numbers is either greater than or equal to the smaller of the numbers.

False.
The LCM of two numbers may be equal to or greater than the larger of the numbers.

True.

10 million = 1 crore
We know that, 1 million = 10 lakh
Then, 10 million = 10 × 10 = 100 lakh = 1,00,00,000
Therefore, 10 million = 1 crore.

10 lakh = 1 million.

1 metre = 1000millimetres
We know that, 1 metre = 100 centimetre
1 centimetre = 10 millimeter
Then, 100 cm = 10 × 100 = 1000 millimetres

1 centimetre = 10 millimetres.

1 kilometre = 10,00,000 millimetres.
We know that, 1 km = 1000 meters.
1 metre = 100 centimetre
1000 metre = 1000 × 100
= 1,00,000 centimetre
1 cm = 10 millimetres
Then, 1,00,000 centimetre = 10 × 1,00,000 = 10,00,000 millimetres

(a) 1000,
(b) 1000,
(c) 10,00,000

1

1650 : 1 m 65 cm = (1000 + 650) mm = 1650 mm

1290000 : 1290 km = (1290 × 1000) m = 1290000 m

422000 : 422 km= (422 × 1000) m = 422000 m

Descending

Smallest
By reversing the order of digits of the greatest number made by five different non-zero digits, the new number is the smallest number of these digits.

6 : Greatest 6-digit number = 999999
By adding 1 to 999999, we get 1000000.

5, 23, 78, 401

L

LXVI : 66 = LXVI

2,538,000

0 : 0 is the smallest whole number.

106160 : Successor of 106159 is 106159 + 1, i.e., 106160

99999 : Predecessor of 100000 is 100000-1, i.e., 99999

401 : 400 is the predecessor of 400 + 1, i.e., 401

1000 : Largest 3 digit number = 999
Successor of 999 is 999 + 1, i.e., 1000

Number

100005

Addition, multiplication

Addition, multiplication

0

Addition

6195 : Since, multiplication is commutative for whole numbers.

1001

0

0

1 : Since, 1 is the multiplicative identity for whole numbers.

68 : Since, addition is associative for whole numbers.

8925

1 : Since, multiplication is distributive over addition for whole numbers.

17

27 : Since, multiplication is associative for whole numbers.

7860 : 786 × 3 + 786 × 7 = 786 × (3 + 7)
= 786 × 10 = 7860

100

Multiple

1

2

Perfect

Composite

Prime

Co-prime

25 : Prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
So, there are 25 primes between 1 to 100.

0

0, 5

2

Multiple

11

Multiple

Factors

(i) ➝ (d) ; (ii) ➝ (f) ; (iii) ➝ (b) ; (iv) ➝ (e) ; (v) ➝ (c)

Descending order of given numbers is,
25843, 13584, 8435, 5348, 4835

The greatest number is 67205602 and the smallest number is 30040700.

(a) 74836 = 7 × 10000 + 4 × 1000 + 8 × 100 + 3 × 10 + 6 × 1
(b) 574021 = 5 × 100000 + 7 × 10000 + 4 x 1000 + 0 × 100 + 2 × 10 + 1 × 1
(c) 8907010 = 8 × 1000000 + 9 × 100000 + 0 × 10000 + 7 × 1000 + 0 × 100 + 1 × 10 + 0 × 1

Ascending order ➝ (b), (c), (a), (d)
Descending order ➝ (d), (a), (c), (b)

The diameter of Jupiter is 142,800,000 metres.

Total increase in population from 1961 to 2001 = 1028 millions – 439 millions = 589 millions
According to Indian System of Numeration, the increase in population = 58,90,00,000

Radius of the Earth = 6400 km = 6400 × 1000 m = 6400000 m
And radius of the Mars = 4300000 m
∴ Radius of the Earth is greater than the radius of the Mars by (6400000 – 4300000) m = 2100000 m.

The population of Tripura = 3,199,203 i.e., Three million one hundred ninety-nine thousand two hundred three.
And the population of Meghalaya = 2,318,822, i.e., Two million three hundred eighteen thousand eight hundred twenty two.

Number of children getting polio drops in March 2008 = 2,12,583
Number of children getting polio drops in April 2008 = 2,16,813
∴ The required difference of the number of children getting polio drops in the two months = 2,16,813 – 2,12,583 = 4,230

Total amount a person had = ₹ 1000000
The amount he spent on a colour T.V. = ₹ 16580
The amount he spent on a motorcycle = ₹ 45890
The amount he spent on a flat = ₹ 870000
∴ Total amount he spent = ₹ (16580 + 45890 + 870000) = ₹ 932470
Thus, the amount left with him = ₹ 1000000 – ₹ 932470 = ₹ 67530

Total tablets of Vitamin A = 180000
Number of tablets distributed among the students in a district = 18734
∴ The number of remaining vitamin tablets = 180000 – 18734 = 161266

Chinmay had total amount = ₹ 610000
The amount he gave to Jyoti = ₹ 87500
The amount he gave to Javed = ₹ 126380
The amount he gave to John = ₹ 350009
Total amount given by Chinmay = ₹ (87500 + 126380 + 350000) = ₹ 563880
Thus, the amount left with him
= ₹ 610000 – ₹ 563880 = ₹ 46120

The smallest number of eight digits = 10000000
The largest number of seven digits = 9999999
The required difference = 10000000 – 9999999 = 1

A mobile number consists of 10 digits.
If the first four digits of the number are 9, 9, 8 and 7 and the last three digits of the number are 3, 5 and 5.
Thus, for the greatest possible number, the remaining distinct digits are 6, 4 and 2.

A mobile number consists of 10 digits.
If the first four digits are 9, 9, 7 and 9.
Thus, the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0 is 9979003568.

A number consists of 5 digits.
Now, the digit at ten’s place = 4,
the digit at unit’s place
= (1/4) x 4 = 1,
the digit at hundred’s place = 0,
the digit at thousand’s place = 5 × 1 = 5
the digit at ten thousand’s place = 2 × 4 = 8
Therefore, the number is 85041.

By using the digits 2, 0, 4, 7, 6, 5
The greatest number formed = 765420,
and the least number formed = 204567
∴ The required sum = 765420 + 204567 = 969987

Quantity of cold drink in a container = 35874 litres = 35874 × 1000 ml = 35874000 ml
The capacity of one bottle = 200 ml
∴ The required number of bottles = 35874000 ÷ 200 = 179370
Therefore, 179370 bottles can be filled by cold drink.

Total population of a town = 450772
Since, one out of every 14 persons is illiterate.
∴ The number of illiterate persons in the town = 450772 ÷ 14 = 32198
Therefore, 32198 persons are illiterate in the town

we have,
The LCM of 80, 96, 125 and 160 is
= 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 = 12000

By using the digit 5 at ten’s place, the greatest 5-digit number is 98756, and the smallest 5-digit number is 10253

5 kg 68 g = (5 × 1000 + 68) g = 5068 g and 2 kg 300 g = (2 × 1000 + 300) g = 2300 g
∴ The required number of grams should be added = 5068 g – 2300 g = 2768 g or 2 kg 768 g

The total weight of a box containing 50 packets of biscuits each weighing 120 g
= 50 × 120 g = 6000 g
The capacity of a van = 900 kg = 900 × 1000 g = 900000 g
∴ The required number of boxes = 900000 + 6000 = 150
Therefore, 150 boxes can be loaded in the van.

50000 lakhs make 5 billions.

30 millions make 3 crores.

(a) 874 rounded off to the nearest hundreds = 900
478 rounded off to the nearest hundreds = 500
Estimated sum = 900 + 500 = 1400
(b) 793 rounded off to the nearest hundreds = 800
397 rounded off to the nearest hundreds = 400
Estimated sum = 800 + 400 = 1200
(c) 11244 rounded off to the nearest hundreds =11200
3507 rounded off to the nearest hundreds = 3500
Estimated sum = 11200 + 3500 = 14700
(d) 17677 rounded off to the nearest hundreds =17700
13589 rounded off to the nearest hundreds =13600
Estimated sum = 17700 + 13600 = 31300

(a) 11963 rounded off to the nearest tens = 11960
9369 rounded off to the nearest tens = 9370
Estimated difference = 11960 – 9370 = 2590
(b) 76877 rounded off to the nearest tens = 76880
7783 rounded off to the nearest tens = 7780
Estimated difference = 76880 – 7780 = 69100
(c) 10732 rounded off to the nearest tens =10730
4354 rounded off to the nearest tens = 4350
Estimated difference = 10730 – 4350 = 6380
(d) 78203 rounded off to the nearest tens = 78200
16407 rounded off to the nearest tens =16410
Estimated difference = 78200 – 16410 = 61790

(a) 87 rounded off to the nearest tens = 90
32 rounded off to the nearest tens = 30
Estimated product = 90 × 30 = 2700
(b) 311 rounded off to the nearest tens = 310
113 rounded off to the nearest tens = 110
Estimated product = 310 x 110 = 34100
(c) 3239 rounded off to the nearest tens = 3240
28 rounded off to the nearest tens = 30
Estimated product = 3240 × 30 = 97200
(d) 1385 rounded off to the nearest tens = 1390
789 rounded off to the nearest tens = 790
Estimated product = 1390 × 790 = 1098100

78787 rounded off to the nearest hundreds = 78800
95833 rounded off to the nearest hundreds = 95800
The estimated increase in population = 95800 – 78800 = 17000

758 can be rounded off to 800 and 6784 can be rounded off to 7000
∴ Estimated product = 800 × 7000 = 5600000

Number of shirts produced by the factory = 216315
Number of trousers produced by the factory =182736
Number of jackets produced by the factory = 58704
Total production of the factory = 216315 + 182736 + 58704 = 457755

We have,
The LCM of 160,170 and 190 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 24480

Quantity of fruit juice in a vessel = 13 L 200 mL = (13 × 1000 + 200) mL = 13200 mL
Capacity of one glass = 60 mL
∴ The required number of glasses = 13200 ÷ 60 = 220
Therefore, 220 glasses can be filled by fruit juice.

Since, successor of 32 is 33, predecessor of 49 is 48, predecessor of the predecessor of 56 is 54 and successor of the successor of 67 is 69.
∴ The required sum = 33 + 48 + 54 + 69 = 204

Total weight can be carried by a tempo = (482 × 15) kg = 7230 kg and the total weight can be carried by a van = (518 × 15) kg = 7770 kg Thus, the total weight that can be carried by both the vehicles = (7230 + 7770) kg = 15000 kg

Amount spent by Leela on food and decoration = ₹ 216766
Amount spent by her on jewellery = ₹ 122322
Amount spent by her on furniture = ₹ 88234
Amount spent by her on kitchen items = ₹ 26780
∴ Total amount spent by her = ₹ (216766 + 122322 + 88234 + 26780)
= ₹ 454102

Quantity of medicine in one capsule = 500 mg
∴ Quantity of medicine in 12 capsules or 1 strip = (500 × 12) mg = 6000 mg = 6 g
Quantity of medicine in 5 strips or 1 box = (6 × 5) g = 30 g
Quantity of medicine in 32 boxes = (30 × 32)g = 960 g

We have,
Since, the LCM of 3,4 and 5 is 2 × 2× 3 × 5 = 60.
∴ The required number is 6is the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

A merchant has 3 kinds of oil with different quantities like 120 litres, 180 litres and 240 litres.
Since, he wants to sell the oil by filling the three kinds of oil in tins of equal capacity, so the greatest capacity of such a tin is the HCF of 120,180 and 240.

Now 120 = 2 × 2 × 2 × 3 × 5
180 = 2 × 2 × 3 × 3 × 5
240 = 2  × 2  × 2 × 2 × 3 × 4
∴ The required greatest capacity of a tin = (2 × 2 × 3 × 5) litres = 60 litres

By using the digits 1, 2, 4 and 5 only once, we get a 4 digit odd number 4521.
When we interchanged its first and last digits we get a new number, i.e., 1524 which is divisible by 4.
Thus, the required number is 4521.

By using the digits 1, 2, 3 and 4 only once, the smallest 4-digit number which is divisible by 4 is 1324.

The postal charges to mail three parcels are  ₹ 20, ₹ 28 and ₹ 36 respectively.
Also, Fatima wants to buy stamps only of one denomination.
So, to find the greatest denomination of stamps, we find the HCF of 20, 28 and 36.
Now 20 = 2 × 2 ×5
28 = 2 × 2 × 7
36 = 2 × 2 × 3 × 3
∴ The HCF of 20, 28 and 36 is 2 × 2 = 4
So, ₹ 4 is the greatest denomination of stamps, she must buy to mail the three parcels.

A shopkeeper has three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively.
Also, a shopkeeper wants to buy equal number of biscuits of each brand for that we need to find the LCM of 12, 15 and 21.
∴ The LCM of 12,15 and 21 = 2 × 2 × 3 × 5 × 7 = 420 Thus, the required number of packets of 420.
brand A = 420/12 = 35,
brand B = 420/15 = 28 and
brand C = 420/21 = 20

Length of floor of a room = 8 m 96 cm = 896 cm
Breadth of floor of the room = 6 m 72 cm = 672 cm
To find the minimum number of square tiles of same size needed to cover the entire floor, we find the LCM of 896 cm and 672 cm.

∴ LCM of 672 and 896 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 = 2688
∴  Number of square tiles =
= 4 x 3 = 12

Number of books of English = 780
Number of books of Science = 364

780 = 2 × 2 × 3 × 5 × 13
364 = 2 × 2 × 7 × 13
∴  The HCF of 780 and 364 = 2 × 2 × 13 = 52
Thus, the minimum number of books in each shelf = 52

There are 100 blocks in a colony numbering 1 to 100.
A school van stops at every sixth block and a school bus stops at every tenth block.
We have to find the common stops at which they both stop if they start from a same entrance.
∴  We need to find the LCM of 6 and 10
The LCM of 6 and 10 is 2 × 3 × 5 = 30. Firstly both will stop at 30^th block, then aL 60^th block and lastly at 90^th block.

(a) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number 5335 is (5 + 3) – (3 + 5) = 8 – 8 = 0, which is divisible by 11.
∴ 5335 is divisible by 11.
(b) We have the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of the number 9020814 is
(4 + 8 + 2 + 9) – (1 + 0 + 0) = 23 – 1 = 22, which is divisible by 11.
∴ 9020814 is divisible by 11.

(a) We have, 4096
Since, the last two digits 96 is divisible by 4.
∴ 4096 must be divisible by 4.
(b) We have, 21084
Since, the last two digits 84 is divisible by 4.
∴ 21084 must be divisible by 4.
(c) We have, 31795012
Since, the last two digits 12 is divisible by 4.
∴ 31795012 must be divisible by 4.

(a) We have, 672
Since, the sum of all the digits of 672 is 15, which is not divisible by 9.
∴ 672 is not divisible by 9.
(b) We have, 5652
Since, the sum of all the digits of 5652 is 18, which is divisible by 9.
∴ 5652 must be divisible by 9.