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Ncert exemplar Solutions Class 6 Mathematics Number System
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of NCERT Class 6 Mathematics Book, Unit 1, Number System. Here students can easily find step by step solutions of all the problems for Number System, 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. Here in this post students will get Unit 1 solutions.
Number System Unit 1 Solution :
Multiple Choice Questions :
(1) The product of the place values of two 2’s in 428721 is
Solution :
Place values of first 2 is = 2000
and Place values of 2nd 2 is = 20
The product = 20000 × 20
= 400000
The correct answer is – (C) 400000
(2) 3 × 10000 + 7 × 1000 + 9 × 100 + 0 ×10 + 4 is the same as
Solution :
= 3 × 10000 + 7 × 1000 + 9 × 100 + 0 × 10 + 4
= 30000 + 7000 + 900 + 0 + 4
= 37904
The correct answer is – (C) 37904
(3) If 1 is added to the greatest 7- digit number, it will be equal to
Solution :
We know, The greatest 7 digit number is 9999999.
= (9999999 + 1)
= 10000000
The appropriate is option – (D) 1 crore.
(4) The expanded form of the number 9578 is
Solution :
The expanded from of the number 9578 is
= (9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1)
The correct answer is – (B) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1
(5) When rounded off to nearest thousands, the number 85642 is
Solution :
The number 85642 is rounded off to nearest thousands.
85642 is greater than 85,500 hence it is rounded off 86,000
Thus, the answer is – (D) 86000
(6) The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is
Solution :
We have to make 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is
We use 9 as twice digit.
The largest 4-digit number is 9,965
So the answer is – (D) 9965
(7) In Indian System of Numeration, the number 58695376 is written as
Solution :
In Indian system of Number then number 58695376 is written as 5, 86, 95, 376
= 5,86,95,376
The answer is alternative – (C) 5,86,95,376
(8) One million is equal to
Solution :
We know that, One million is equal to 10 lakh = 10 lakh.
So, the correct option is – (B) 10 lakh.
(9) The greatest number which on rounding off to nearest thousands gives 5000, is
Solution :
The greatest number which on rounding off to nearest thousands gives 5000, is 5499.
The correct option is (C) – 5499.
(10) Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is
Solution :
The smallest number obtained by rearranging other digits is 6034579.
The correct option is – (C) 6034579
(11) Which of the following numbers in Roman numerals is incorrect?
Solution :
The incorrect Roman numeral is LLX.
Correct option (D) – LLX, is incorrect.
(12) The largest 5-digit number having three different digits is
Solution :
The largest 5 digit number having three different digit is 99987.
Thus, the correct option is (C) 99987.
(13) The smallest 4-digit number having three different digits is
Solution :
We have to use smallest number twice.
The smallest 4-digit number having three different digits is 1,102
The answer will be – (A) 1102
(14) Number of whole numbers between 38 and 68 is
Solution :
Number of whole numbers between 38 and 68 is 30.
Hence, correct option is – (B) 30.
(16) The product of successor and predecessor of 999 is
Solution :
The product of successor and predecessor of 999 is 998000.
= 1000 × 998
= 998000
Thus, the correct option is – (B) 998000
(16) The product of a non-zero whole number and its successor is always
Solution :
The product of a non-zero whole number and its successor is always an even number.
Thus, correct option is – (C) a prime number.
(17) A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is
Solution :
The sum of the resulting numbers is 50.
So, the correct option is (C) 50
(18) Which of the following is not true?
Solution :
The incorrect one is 7 + 8 × 9 = (7 + 8) × (7 + 9)
Correct option is – (C) 7 + 8 × 9 = (7 + 8) × (7 + 9)
(19) By using dot (.) patterns, which of the following numbers can be arranged in all the three ways namely a line, a triangle and a rectangle?
Solution :
The appropriate answer is option,
(B) 10
(20) Which of the following statements is not true?
Solution :
The incorrect statement is – Zero is the identity for multiplication of whole numbers.
(B) is the correct option.
(21) Which of the following statements is not true?
Solution :
The incorrect statement is 0 ÷ 0 = 0
(D) is the correct option.
(22) The predecessor of 1 lakh is
Solution :
The predecessor of 1 lakh is
= 100000 – 1
= 99999
The correct option is – (B) 99999
(23) The successor of 1 million is
Solution :
The successor of 1 million is
= 1000000 + 1
= 1000001
The appropriate option is (B) 1000001
(24) Number of even numbers between 58 and 80 is
Solution :
Number of even numbers between 58 and 80 is 10.
So, the correct option is – (A) 10
(25) Sum of the number of primes between 16 to 80 and 90 to 100 is
Solution :
Primes number between 16 to 80 is – 16
Primes number between 90 to 100 is 2
∴ Sum = 16 + 2 = 18
= 18
The correct option is – (B) 18
(26) Which of the following statements is not true?
Solution :
Here, the incorrect statement is – The HCF of an even and an odd number is even.
So, (D) is the correct option.
(27) The number of distinct prime factors of the largest 4 digit number is
Solution :
The number of distinct prime factors of the largest 4 digit number is 3.
So, the correct option is – (B) 3
(28) The number of distinct prime factors of the smallest 5-digit number is
Solution :
The number of distinct prime factors of the smallest 5-digit number is 2.
Thus, the correct option is – (A) 2
(29) If the number 7254*98 is divisible by 22, the digit at * is
Solution :
If the number 7254*98 is divisible by 22, the digit at * is 6.
So, the correct option is – (C) 6
(30) The largest number which always divides the sum of any pair of consecutive odd numbers is
Solution :
The largest number which always divides the sum of any pair of consecutive odd numbers is 2.
The correct option is – (A) 2
(31) A number is divisible by 5 and 6. It may not be divisible by
Solution :
A number is divisible by 5 and 6. It may not be divisible by 60.
Thus, the appropriative option is – (D) 60
(32) The sum of the prime factors of 1729 is
Solution :
Prime factors of 1729 7, 13, 19
Sum of prime factor 7 + 13 + 19
= 39
The correct option is – (D) 39.
(33) The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is
Solution :
The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is 8.
So, the correct option is – (D) 8
(34) The number of common prime factors of 75, 60, 105 is
Solution :
The number of common prime factors of 75, 60, 105 is 2.
Hence, the correct option is – (A) 2.
(35) Which of the following pairs is not coprime?
Solution :
8, 10 is not coprime.
So, the correct option is (A) 8, 10
(36) Which of the following numbers is divisible by 11?
Solution :
22222222 is divisible by 11.
So, the correct option is – (C) 22222222.
(37) LCM of 10, 15 and 20 is
Solution :
LCM of 10, 15 and 20 is 60.
So, the correct option is – (B) 60
(38) LCM of two numbers is 180. Then which of the following is not the HCF of the numbers?
Solution :
The appropriate answer is option – (C)
State True or False :
(39) In Roman numeration, a symbol is not repeated more than three times
Solution :
This statement is True
(40) In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs
Solution :
This statement is False.
(41) 5555 = 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1
Solution :
= 5555 = 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1
= 5000 + 500 + 50 + 5
= 5555
The given statement is True.
(42) 39746 = 3 × 10000 + 9 × 1000 + 7 × 100 + 4 × 10 + 6
Solution :
= 3 × 10000 + 9 × 1000 + 7 × 100 + 4 × 10 + 6
= 30000 + 9000 + 700 + 40 + 6
= 39746
Hence, the above statement is True.
(43) 82546 = 8 × 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6
Solution :
= 8 × 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6
= 8000 + 2000 + 500 + 40 + 6
= 10546 ≠ 82546
So. the given statement is False.
(44) 532235 = 5 × 100000 + 3 × 10000 + 2 × 1000 + 2 × 100 + 3 × 10 + 5
Solution :
= 5 × 100000 + 3 × 10000 + 2 × 1000 + 2 × 100 + 3 × 10 + 5
= 500000 + 30000 + 2000 + 200 + 30 + 5
= 532235
The above statement is True.
(45) XXIX = 31
Solution :
The given statement is False.
Because, = XXIX = 29
31 means = XXIX
(46) LXXIV = 74
Solution :
This statement is True.
Because, we know,
L = 50, X = 10, X = 10, IV = 4
(47) The number LIV is greater than LVI
Solution :
LIV = 54, LIV = 56
Therefore LIV < LIV
This statement is False.
(48) The numbers 4578, 4587, 5478, 5487 are in descending order
Solution :
The given statement is False.
Reason :
Descending means from greatest to lowest 4587 > 4578.
(49) The number 85764 rounded off to nearest hundreds is written as 85700
Solution :
The number 85764 rounded off to nearest hundred is written as 85800
Because – Difference between 85800 – 85764 = 36 is lower than difference between 85764 – 85700 = 64
Hence, the above statement is True.
(50) Estimated sum of 7826 and 12469 rounded off to hundreds is 20,000
Solution :
= Sum CP = (7826 + 12469)
= 20295
Rounded off to the nearest hundred is 20300.
The given statement is False.
(51) The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403
Solution :
The given statement is False.
Reason :
Largest telephone number would be 875430 not 875403.
52) The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen
Solution :
The above statement is False.
Reason :
The number 81652318 will be read as eight crore sixteen lakh fifty two thousand three hundred eighteen.
(53) The largest 4-digit number formed by the digits 6, 7, 0, 9 using each digit only once is 9760
Solution :
The given statement is True.
Reason :
The largest 4-digit number formed by this number will be 9760.
(54) Among kilo, milli and centi, the smallest is centi.
Solution :
1 kilo = 100000 centi
1 centi = 10 mili
Hence, the smallest is milli.
This statement is False.
(55) Successor of a one digit number is always a one digit number
Solution :
Successor of a one digit number is not always a one digit number.(may be 2 digit number)
So, the statement is False
(56) Successor of a 3-digit number is always a 3-digit number.
Solution :
Given statement is False.
Correct statement is
Successor of a 3-digit number is not always a 3-digit number. (may be 4 digit number)
(57) Predecessor of a two digit number is always a two digit number.
Solution :
Given statement is False.
Correct statement is –
Predecessor of a two digit number is not always a two digit number.(may be 1 digit number)
(58) Every whole number has its successor
Solution :
The statement is True.
(59) Every whole number has its predecessor.
Solution :
Given statement is False
Correct statement is – Not every whole number has its predecessor.
(60) Between any two natural numbers, there is one natural number
Solution :
The above statement is False.
Let, Because 2 and 3 is any two natural number.
Between 2 and 3 there are no natural number.
(61) The smallest 4 digit number is the successor of the largest 3 digit number.
Solution :
The largest 3-digit number is = 999
Successor of 999 = 999 + 1 = 1000 which the smallest 4-digit number
So, the statement is True.
(62) Of the given two natural numbers, the one having more digits is greater
Solution :
Let, 5 is a natural number and 41 is a natural number
= 41 is always greater.
So, the statement is True.
(63) Natural numbers are closed under addition.
Solution :
Let, 3 and 4 are two natural number.
3 + (3 + 4) = (3 + 3) + 4
= 3 + 7 = 6 + 4
⇒ 10 = 10
10 also belongs to natural number
∴ Therefore they are closed under addition.
Hence, the statement is True.
(64) Natural numbers are not closed under multiplication
Solution :
4, 5 are two natural number
= 4 × 5 = 20 is also a natural number
Therefore natural number is closed under multiplication.
Thus, the statement is False.
(65) Natural numbers are not closed under multiplication.
Solution :
4, 5 are two natural number
= 4 × 5 = 20 is also a natural number
Therefore natural number is closed under multiplication.
So, the statement is False.
(66) Addition is commutative for natural numbers
Solution :
For any natural number a and b
= a + b = b + a
= 4 + 5 = 5 + 4 = 9
Therefore, addition is commutative for natural number
So, the statement is True.
(67) 1 is the identity for addition of whole numbers
Solution :
Identity element is 1 where N + 1 = N
one whole number = 5
= 5 + 1 = 6
= 5 + 0 = 5
Therefore, in addition 0 is the identity element 1 is not the identity element in addition.
Thus, the statement is False.
(68) 1 is the identity for multiplication of whole numbers
Solution :
Identity element I
In multiplication, N.1 = N
One whole number 6
= 6.1 = 6
Therefore, 1 is the identity element for multiplication for whole numbers.
∴ The statement is True.
(69) There is a whole number which when added to a whole number, gives the number itself
Solution :
0 is a whole number
When it added to another whole number gives the number itself.
= 7 + 0 = 0, 8 + 0 = 0, 10 + 0 = 0
Thus, the statement is True.
(70) There is a natural number which when added to a natural number, gives the number itself
Solution :
0 is not the natural number.
So, when a natural number added to a natural number cannot be a natural number itself.
So, the statement is False.
(71) If a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.
Solution :
Let, 5 is a whole number and 6 is another whole number which is greater than 1
Division of 5 by 6 = 5/6 which is not zero.
So, the statement is True.
(72) Any non-zero whole number divided by itself gives the quotient 1
Solution :
Let, 5 is a whole number and 6 is another whole number which is greater than 1
Division of 5 by 6 = 5/6 which is not zero.
So, the statement is True.
(73) The product of two whole numbers need not be a whole number
Solution :
Product of two whole number is always a whole number,
Let, 5, 0 is two whole number.
So, the product of it is = 5 × 0 = 0
∴ The given statement is False.
(74) A whole number divided by another whole number greater than 1 never gives the quotient equal to the former
Solution :
A whole number 6
another whole number = 2 which is greater than 1
∴ Dividing we get 6/2 = 3 which is not equal to the former
So, the statement is True.
(75) Every multiple of a number is greater than or equal to the number
Solution :
Multiple of number 6 are = 6, 12, 18, 24, 30.
6 is equal and 12, 18, 24 are greater multiple.
Thus, the statement is True.
(76) The number of multiples of a given number is finite
Solution :
Because 3 is a number
Multiples of 3 is = 3,6,9,12, 15 ….. infinite
Hence, the given statemen is False.
(77) Every number is a multiple of itself
Solution :
Multiple of 2 is = 2 …
multiple of 3 is = 3…
multiple of 5 is = 5…
∴ Every number is multiple itself.
So, the statement is True.
(78) Sum of two consecutive odd numbers is always divisible by 4
Solution :
Two conceptive odd number is (1,3), (3, 5) (5, 7), (7, 9) …..
∴ Sum of 1, 3 = 1 + 3 = 4 is divisible by 4
Sum of 3, 5 = 3 + 5 = 8 is divisible by 4
Sum of 5, 7 = 5 + 7 = 12 is divisible by 4
Sum of 7, 9 = 7 + 9 = 16 is divisible by 4
∴ Sum of two consecutive odd number is always
Thus, the statement is True.
(79) If a number divides three numbers exactly, it must divide their sum exactly.
Solution :
3 divides 3, 6 and 9
it also divides (3 + 6 + 9) = 18
Thus, the statement is True.
(80) If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately
Solution :
Three number are 3, 5, 2
The sum of these three number is (3 + 5 + 2) = 10
= 10 is divisible by 2 but, 5 and is not divisible by 2
Hence, the statement is False.
(81) If a number is divisible both by 2 and 3, then it is divisible by 12
Solution :
Number 6 is divisible by 2 and 3
but 6 is not divisible by 12
Thus, the statement is False.
(82) A number with three or more digits is divisible by 6, if the number formed by its last two digits (i.e., ones and tens) is divisible by 6
Solution :
216 is a three digit number and it is divisible by 6.
But 16 is not divisible by 6.
Thus, the statement is False.
(83) A number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8
Solution :
A four digit number is 1224 which is divisible by 8
It the number formed with fast three digit the number will be 224
It is also divisible by 8
Thus, the statement is True.
(84) If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 9
Solution :
One number is 33
the sum of the digit of this number is (3 + 3) = 6
It is divisible by 3
But the number 33 is not divisible by 9.
Thus, the statement is False.
(85) All numbers which are divisible by 4 may not be divisible by 8
Solution :
36 is divisible by 4 but it is not divisible by 8
So, the statement is True.
(86) The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple
Solution :
The Highest common factor 3, 4, 5, is 1 and the lowest common multiple is = 60
1 is not greater than 60
∴ Therefore the H.C.F. of two more number is not greater than L.C.M. of these number.
Hence, the statement is False.
(87) LCM of two or more numbers is divisible by their HCF
Solution :
L.C.M. of 6 and 3 is = 18
H.C.F of 6 and 3 is = 12
18 is divisible by 3
So, the given statement is True.
(88) LCM of two numbers is 28 and their HCF is 8
Solution :
28 is not divisible by 8.
L.C.M of two more or number is divisible by their H.C.F So it would not be possible.
Thus, the statement is False.
(89) LCM of two or more numbers may be one of the numbers
Solution :
L.C.M of 2, and 4 is = 4
Therefore L.C.M of two are more number may be one of the number.
Hence, the statement is True.
(90) HCF of two or more numbers may be one of the numbers
Solution :
H.C.F of 2 and 4 = 2
Therefore H.C.F of two or more numbers may be one of the number.
So, the statement is True.
(91) Every whole number is the successor of another whole number
Solution :
O is a whole number and it is not successor of another whole number.
So, the statement is True.
(92) Sum of two whole numbers is always less than their product
Solution :
Sum of two whole number 5 and 0 is (5 + 0) = 5
and product of these two number = 5 × 0 = 0
So, the sum of two whole number may be greater than product.
Thus, the statement is False.
(93) If the sum of two distinct whole numbers is odd, then their difference also must be odd
Solution :
Let, Two distinct number 5, 8
Sum of these number = 5 + 8 = 13 which is odd number.
The distance between these two number 8 – 5 = 3 is also odd number.
Hence, the statement is True.
(94) Any two consecutive numbers are coprime
Solution :
Any two consecutive number is 3, 4
4 is the multiple of 3
So, 3 and 4 are co-prime.
∴ The given statement is True.
(95) If the HCF of two numbers is one of the numbers, then their LCM is the other number
Solution :
H.C.F of 23 and 4 is -= 2
and the L.C.M is = 4
∴ Therefore it is true
So, the statement is True.
(96) The HCF of two numbers is smaller than the smaller of the numbers.
Solution :
The H.C.F of 4 and 8 is 4
The H.C.F may be equal to one number always it will not be smaller than the smaller one.
Thus, the statement is False.
(97) The LCM of two numbers is greater than the larger of the numbers
Solution :
The L.C.M of two number 4 and 8 is = 8
Therefore, the L.C.M of two number may be equal to the larger one.
Always it will not be greater than the larger one.
Thus, the statement is False.
(98) The LCM of two coprime numbers is equal to the product of the numbers.
Solution :
The coprime number 5, 7
L.C.M of 5 and 7 = 35 = (5 × 7) their product.
∴ The L.C.M of two coprime number is equal to the product of the numbers,
Hence, the statement is True.
Fill in the blanks :
(99) – (a) 10 million = _____ crore.
(b) 10 lakh = _____ million.
Solution :
(a) 10 million = 1 crore
(b) 10 lakh = 1 Million
(100) (a) 1 metre = _____ millimetres.
(b) 1 centimetre = _____ millimetres.
(c) 1 kilometre = _____ millimetres.
Solution :
(a) 1 metre = 1000 millimetres.
(b) 1 centimetre = 10 millimetres.
(c) 1 kilometre = 1000000 millimetres.
(101) – (a) 1 gram = _____ milligrams
(b) 1 litre = _____ millilitres.
(c) 1 kilogram = _____ miligrams
Solution :
(a) 1 gram = 1000 milligrams.
(b) 1 litre = 1000 millilitres.
(c) 1 kilogram = 1000000 miligrams
(102) 100 thousands = _____ lakh
Solution :
100 thousands = 1 lakh.
(103) Height of a person is 1m 65cm. His height in millimetres is_______.
Solution :
Height of a person is 1m 65cm.
We know, 1 meter = 100 cm
1centimetre = 10 mm
1 meter = 100 x 10 = 1000 mm
65 cm = 6 x 10 = 650 mm
Height of a person is 1m 65 cm. His height in millimeters is 1650 mm.
(104) Length of river ‘Narmada’ is about 1290km. Its length in metres is_______.
Solution :
As we know,
1 km = 1000 metre
1290 × 1000 = 1290000
Length of river ‘Narmada’ is about 1290 km. Its length in metres is 1290000.
(105) The distance between Sringar and Leh is 422km. The same distance in metres is_____.
Solution :
The distance between Sringar and Leh is 422 km. The same distance in metres is 422000.
As we know, 1 km = 1000 metre
422 × 1000 = 422000 metre.
(106) Writing of numbers from the greatest to the smallest is called an
arrangement in _____ order.
Solution :
Writing of numbers from the greatest to the smallest is called an arrangement in descending order.
(107) By reversing the order of digits of the greatest number made by five different non-zero digits, the new number is the _____ number of five digits.
Solution :
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 five digits.
(108) By adding 1 to the greatest_____ digit number, we get ten lakh.
Solution :
By adding 1 to the greatest Six digit number, we get ten lakh.
(109) The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian System of Numeration as _____.
Solution :
The number five crore twenty three lakh seventy eight thousand four hundred one can be written, using commas, in the Indian System of Numeration as 5, 23, 78, 401.
(110) In Roman Numeration, the symbol X can be subtracted from_____, M and C only
Solution :
In Roman Numeration, the symbol X can be subtracted from L, M and C only.
(111) The number 66 in Roman numerals is_____.
Solution :
The number 66 in Roman numerals is LXVI
(112) The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was ______.
Solution :
The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was 2538, 000.
(113) The smallest whole number is_____.
Solution :
The smallest whole number is 0.
(114) Successor of 106159 is _____.
Solution :
Successor of 106159 is 106160.
(115) Predecessor of 100000 is_____.
Solution :
Predecessor of 100000 is previous number of
100000 = 99,999
(116) 400 is the predecessor of _____.
Solution :
400 is the predecessor of 401
(117) _____ is the successor of the largest 3 digit number.
Solution :
= 1000 is the successor of the largest 3 digit number
(118) If 0 is subtracted from a whole number, then the result is the _____ itself .
Solution :
If 0 is subtracted from a whole number, then the result is the number itself.
(119) The smallest 6 digit natural number ending in 5 is _____.
Solution :
The smallest 6 digit natural number ending in 5 is 100005.
(120) Whole numbers are closed under _____ and under_____.
Solution :
Whole numbers are closed under addition and under multiplication.
(121) Natural numbers are closed under _____ and under_____.
Solution :
Natural numbers are closed under addition and under multiplication.
(122) Division of a whole number by _____ is not defined.
Solution :
Division of a whole number by 0 is not defined.
(123) Multiplication is distributive over _____ for whole numbers
Solution :
Multiplication is distributive over addition for whole numbers.
(124) 2395 × _____ = 6195 × 2395
Solution :
= 2395 × 6195 = 6195 × 2395
(125) 1001 × 2002 = 1001 × (1001+_____ )
Solution :
= 1001 × 2002 = 1001 × (1001 + 1001)
(126) 10001 × 0 = _____
Solution :
= 10001 × 0 = 0
(127) 2916 × _____ = 0
Solution :
= 2916 × 0 = 0
(128) 9128 × _____ = 9128
Solution :
= 9128 × 1 = 9128
(129) 125 + (68+17) = (125 + _____ ) + 17
Solution :
= 125 + (68 + 17) = (125 + 68) + 17
(130) 8925 ×1 = _____
Solution :
= 8925 × 1 = 8925
= Multiplication of any number with 1 is number itself.
(131) 19 × 12 + 19 = 19 × (12 + _____)
Solution :
= 19 × 12 + 19 = 19 × (12 + 19)
(132) 24 × 35 = 24 × 18 + 24 × ____
Solution :
= 24 × 35 = 24 × 18 + 24 × 17
(133) 32 × (27 × 19) = (32 × _____ ) × 19
Solution :
= 32 × (27 × 19) = (32 × 27) × 19
(134) 786 × 3 + 786 × 7 = _____
Solution :
= 786 × 3 + 786 × 7 = 7860
(135) 24 × 25 = 24 × ____
Solution :
= 24 × 25 = 24 × 25
(136) A number is a _____ of each of its factor.
Solution :
= A number is a multiple of each of its factor.
(137) _____ is a factor of every number
Solution :
= 1 is a factor of every number.
(138) The number of factors of a prime number is _____.
Solution :
= The number of factors of a prime number is Two.
(139) A number for which the sum of all its factors is equal to twice the number is called a _____ number.
Solution :
A number for which the sum of all its factors is equal to twice the number is called a Perfect number.
(140) The numbers having more than two factors are called _____ numbers.
Solution :
The numbers having more than two factors are called Composite Number.
(141) 2 is the only _____ number which is even
Solution :
2 is the only Prime number which is even.
(142) Two numbers having only 1 as a common factor are called _____ numbers.
Solution :
Two numbers having only 1 as a common factor are called Co-prime numbers.
(143) Number of primes between 1 to 100 is _____.
Solution :
Number of primes between 1 to 100 is 25.
(144) If a number has _____ in ones place, then it is divisible by 10.
Solution :
If a number has 0 in ones place, then it is divisible by 10.
(145) A number is divisible by 5, if it has _____ or _____ in its ones place.
Solution :
A number is divisible by 5, if it has 5 or 0 in its ones place.
(146) A number is divisible by _____ if it has any of the digits 0, 2, 4, 6, or 8 in its ones place.
Solution :
A number is divisible by 2 if it has any of the digits 0, 2, 4, 6, or 8 in its ones place.
(147) If the sum of the digits in a number is a _____ of 3, then the number is divisible by 3.
Solution :
If the sum of the digits in a number is a multiple of 3, then the number is divisible by 3.
(148) If 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 is either 0 or divisible by _____, then the number is divisible by 11.
Solution :
If 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 is either 0 or divisible by 11, then the number is divisible by 11.
(149) The LCM of two or more given numbers is the lowest of their common _____.
Solution :
The LCM of two or more given numbers is the lowest of their common Multiple.
(150) The HCF of two or more given numbers is the highest of their common _____.
Solution :
The HCF of two or more given numbers is the highest of their common Factor.
Question no – (152)
Solution :
Given numbers, 8435, 4835, 13584, 5348, 25843
In descending order : 25843, 13584, 8435, 5348, 4835
Question no – (153)
Solution :
The greatest number = 67205602
The smallest number = 30040700
Question no – (154)
Solution :
Given numbers in expanded form,
(a) 74836 : Seventy four thousand eight eight hundred thirty six.
(b) 574021 : Five lakh seventy four thousand twenty one.
(c) 8907010 : Eighty nine lakh seven thousand ten.
Question no – (155)
Solution :
Population in ascending order :
(b) Andhra Pradesh (76210007)
(c) Bihar (82998509)
(a) Maharashtra (96878627)
(d) Uttar Pradesh (166197921)
Population in Descending order :
(d) Uttar Pradesh (166197921)
(a) Maharashtra (96878627)
(c) Bihar (82998509)
(b) Andhra Pradesh (76210007)
Question no – (156)
Solution :
The diameter of Jupiter is 14, 28, 00000 metres.
Question no – (157)
Solution :
In 1961 the population was 439 millions = 439, 000, 000
In 2001 the population 1028, millions = 1028, 000,000
In Indian system of multiplication increase of population,
1028,000,000
43,90,00,000
————————
58,90,00,000
Question no – (158)
Solution :
Radius of the Earth is 6400km
Radius of the of Mars is 4300000 m.
We know, 1 km = 1000 m
4300000 m = 4300000 m / 1000 = 4300 km
Radius of the Earth is bigger.
By 6400 km – 4300 km = 2100 km
Radius of the Earth by 2100 km.
Question no – (159)
Solution :
The populations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively.
The populations of Tripura = 3,199,203 = Three million, one hundred ninety-nine thousand, two hundred three
The populations of Meghalaya = 2,318,822 = Two million, three hundred eighteen thousand, eight hundred twenty two.
Question no – (160)
Solution :
The difference of the children getting polio drops is,
2,16,813
– 2,12,583
————————
4230
So, the difference of the number is 4,230
Question no – (161)
Solution :
= 1000000 – (16580 + 45890 + 870000)
= 1000000 – 932470
= 67530
Thus, Rs 67,530 was left with him.
Question no – (162)
Solution :
As per the question,
Total tablets of Vitamin A = 180000
Distributed tablets = 18734
∴ The number of remaining Vitamin tablet,
= (180000 – 18734)
= 1,6,12,66
Therefore, the remaining vitamin tablets will be 1,6,12,66
Question no – (163)
Solution :
According to the question,
Chinmay had = Rs 610000.
He gave = Rs 87500 to Jyoti,
Rs 126380 to Javed and Rs 350000 to John
∴ Chinmay had left with him,
= 610000 – (87500 + 126380 + 350000)
= 610000 – 563880
= 46120
Hence, Rs. 46120 was left with him.
Question no – (164)
Solution :
Largest number of seven digits = 99, 99,999
Smallest number of eight digits = 1, 00, 00,000
Difference between the largest number of seven digits and the smallest number of eight digits
1, 00, 00,000 – 99, 99,999
1, 00, 00,000
– 99, 99,999
——————————
1
Difference between the largest number of seven digits and the smallest number of eight digits is 1.
Question no – (165)
Solution :
The greatest possible number = 9987987355
Question no – (166)
Solution :
The smallest number = 99700335
Question no – (167)
Solution :
Five digit number
Digit at ten’s place is 4
Digit at unit’s place is one fourth of ten’s place digit = 4 x (¼) = 1
Digit at hundred’s place is 0
Digit at thousand’s place is 5 times of the digit at unit’s place = 5 x 1 = 5
Ten thousand’s place digit is double the digit at ten’s place = 2 x 4 = 8
The five digit number is 85,041
Question no – (168)
Solution :
The greatest number = 765420
The least number = 204567
∴ The Sum,
= (765420 + 204567)
= 969987
Therefore, the sum is 969987
Question no – (169)
Solution :
In container had cold drink = 35874 litre
In each bottle had cold drink = 200 ml
= 200/1000 litre
= 2/10 litre
∴ Number of bottles,
= 35874/2/10
= 35874 × 10/2
= 179370
Therefore, 179370 can be filled.
Question no – (170)
Solution :
As per the question
Population of a town = 450772
One out of every 14 persons is illiterate
∴ Illiterate persons in the town,
= 450772× 1/14
= 32693
Hence, there are 32693 illiterate persons in the town.
Question no – (171)
Solution :
Factors of 80 = 2 x 2 x 2 x 2 x 5
Factors of 96 = 2 x 2 x 2 x 2 x 2 x 3
Factors of 125 = 5 x 5 x 5
Factors of 160 = 2 x 2 x 2 x 2 x 2 x 5
LCM of 80, 96, 125, 160.
= (multiplication of common and uncommon factors) 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 3
= 12000
LCM of 80, 96, 125, and 160 is 12000
Question no – (172)
Solution :
The greatest digit = 98765
The smallest digit = 10235
Question no – (173)
Solution :
We have to subtract 2 kg 300 g from 5 kg 68 g
5 kg 068 g
– 2 kg 300 g
——————–
2 kg 768 g
2 kg 768 g added to 2 kg 300 g to make it 5 kg 68 g.
Question no – (174)
Solution :
The van can carry 50 boxes,
= (50 × 120)
= 6000 gm
= 6 kg
∴ 900/6 = 150 box
Therefore, 150 such boxes can be loaded in the van.
Question no – (175)
Solution :
We know,
1 billion = 100 crore = 10,000 Lakh
5 billion = 500 crore = 50,000 Lakh
Question no – (176)
Solution :
We know,
1 million = 10 Lakh.
3 crores = 300 lakh
= 30 x 10 lakh
= 30 millions
Question no – (177)
Solution :
(a) 874 + 478 = 1352
= nearest hundred 1400
(b) 793 + 397 = 1190
= rounding off nearest hundred 1200
(c) 11244 + 3507 = 14751
= rounding off nearest hundre3d 14800
(d) 17677 + 13589 = 31266
= rounding off nearest hundred 31300
Question no – (178)
Solution :
(a) 11963 – 9369
11963 by rounding off to nearest tens is 11960.
9369 by rounding off to nearest tens is 9370.
11960
– 9370
—————
2590
(b) 76877 – 7783
76877 by rounding off to nearest tens is 76,880
7783 by rounding off to nearest tens is 7780.
76880
– 7780
—————
69100
(c) 10732 – 4354
10732 by rounding off to nearest tens is 10,730
4354 by rounding off to nearest tens is 4350.
10730
– 4350
—————
6380
(d) 78203 – 16407
78203 by rounding off to nearest tens is 78200
16407 by rounding off to nearest tens is 16410.
78200
-16410
—————
61790
Question no – (179)
Solution :
(a) 87 × 32 = 2784
= By rounding off nearest tens 2780
(b) 311×113 = 35143
= By rounding off nearest tens 35140
(c) 3239 × 28 = 90692
= By rounding off nearest tens 90690
(d) 1385 × 789 = 10, 92, 765
= By rounding off nearest tens 10,92765
Question no – (180)
Solution :
Population in 1991 is = 78787
By rounding off to nearest hundred = 78800
Population in 2001 is = 95833
By rounding off to nearest hundred = 95800
∴ Increase of population,
= (95800 – 78800)
= 17000
Thus, the increase in population is 17000
Question no – (181)
Solution :
Product of (758 × 6784) is
= 51, 42, 272
Question no – (182)
Solution :
A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year.
The total production of all the three items in that year = 216315 + 182736 + 58704
216315
+ 182736
+ 58704
———————
4, 57,755
The total production of all the three items in that year is 4, 57,755.
Question no – (183)
Solution :
Factors of 160 = 16 x 10
Factors of 170 = 17 x 10
Factors of 90 = 9 x 10
LCM of 160, 170 and 90 = multiplication of common and uncommon factors.
= 10 x 9 x 16 x 17 = 24480
LCM of 160, 170 and 90 is 24480
Question no – (184)
Solution :
Fruit Juice vessel = 13 litre 200 ml
= (130000 + 200)
= 130200 ml
∴ 130200/60 = 2170
Therefore, 2170 glasses can be filled.
Question no – (185)
Solution :
Successor of 32 is = 33
Predecessor of 49 is = 48
Predecessor of the predecessor of 56 is = 54
Successor of the successor of 67 is = 69
∴ The sum,
= (33 + 48 + 54 + 69)
= 204
Question no – (186)
Solution :
A loading tempo can carry 482 boxes of biscuits weighing 15 kg each.
Total weight of 482 boxes = 482 x 15 = 7230 kg.
Van can carry 518 boxes each of the same weight.
Total weight of 518 boxes = 518 x 15 = 7770 kg.
The total weight that can be carried by both the vehicles = 7230 kg. + 7770 kg.
7230 kg.
+ 7770 kg.
——————-
15000 kg
The total weight that can be carried by both the vehicles is 15,000 kg.
Question no – (187)
Solution :
As per the question,
Leela spent Rs 216766 on food and decoration,
On jewellery = Rs 122322
On furniture = Rs 88234
On kitchen items = Rs 26780
∴ Total amount,
= (216766 + 122322 + 88234 + 26780)
= 4,54,102
Therefore, total Rs 4, 54, 102 spent by Leela by her on the above item
Question no – (188)
Solution :
Given in the question,
A box contains 5 strips having 12 capsules of 500 mg medicine
Total weight in grams of medicine in 32 such boxes = ?
∴ Total weight of medicine,
= (32 × 5 × 12 × 500) mg
= 9, 60, 000 mg
= 960 mg
So, total weight of 32 such boxes is 960 mg
Question no – (189)
Solution :
We have to first find LCM of 3,4,5 then adding 2 on LCM.
No common factors between 3,4,5
LCM = 3 x 4 x 5 = 60
Remainder left is always 2 = 60 + 2 = 62
The number is 62.
Question no – (190)
Solution :
In the given question,
A merchant has 120 litres of oil of one kind,
180 litres of another kind
240 litres of a third kind
∴ H.C.F of 120, 180, 240 is 60
Hence, 60 liters should be the greatest capacity of such a tin.
Question no – (191)
Solution :
Divisibility of 4 : The last 2 numbers are divisible by 4 or last two digit became 0.
4-digit odd number using each of the digits 1, 2, 4 and 5 = 4521
When we interchange 1 and last number it became 1524
Last 2 digit is 24 which is divisible by 4.
Hence, number is 4521
Question no – (192)
Solution :
Divisibility of 4: The last 2 numbers are divisible by 4 or last two digit became 0.
Smallest 4-digit number using digits 1, 2, 3 and 4 only once = 1324
Last 2 digit is 24 which is divisible by 4.
Hence, number is 1324.
Question no – (193)
Solution :
Postal charges are Rs 20, Rs 28 and Rs 36, respectively.
Greatest denomination of stamps she must buy to mail the three parcels =
We have to find LCM of Rs 20, Rs 28 and Rs 36, respectively.
Factors of 20 = 4 x 5
Factors of 28 = 4 x 7
Factors of 36 = 4 x 9
Common factor is 4.
LCM of Rs 20, Rs 28 and Rs 36 is 4.
Greatest denomination of stamps she must buy to mail the three parcels is Rs. 4.
Question no – (194)
Solution :
Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively.
A shopkeeper wants to buy an equal number of biscuits, of each brand.
We have to find LCM of 12, 15 and 21
Factors of 12 = 3 x 4
Factors of 15 = 3 x 5
Factors of 21 = 3 x 7
LCM of 12, 15 and 21 = 3 x 4 x 5 x 7 = 420
LCM of 12, 15 and 21 is 420.
Brands A contains 12 biscuits in packet = 420 / 12 = 35 packets
Brands B contains 15 biscuits in packet = 420 / 15 = 28 packets
Brands c contains 21 biscuits in packet = 420 / 21 = 20 packets
Question no – (195)
Solution :
In the given question,
Floor of a room is 8 m 96 cm long and 6 m 72 cm broad
∴ H.C.F of 896 and 672 = 224
are of square this = 224 × 224
are of the floor = 896 × 672 square. cm
∴ Number of this,
= 896 × 672/224 × 224
= 12
So, 12 tiles is needed to cover the entire floor.
Question no – (196)
Solution :
780 books of English and 364 books of Science.
Each shelf should have the same number of books of each subject.
We have to find HCM of 780 and 364.
Factor of 780 = 4 x 5 x 3 x 13
Factor of 364 = 13 x 4 x 7
Common factors is 13 x 4 = 52
HCF of 780 and 364 is 52
Minimum number of books in each shelf is 52.
Question no – (197)
Solution :
In a colony of 100 blocks of flats numbering 1 to 100.
School van stops at every sixth block.
School bus stops at every tenth block.
We have to find LCM of 6 and 10 block.
Factor of 6 = 2 x 3
Factor of 10 = 2 x 5
LCM of 6 and 10 block = 2 x 3 x 5 = 30 block
30 + 30 = 60 block
60 + 30 = 90 block.
Both of them stop at 30, 60 , 90 block.
Question no – (198)
Solution :
(a) 5335
Divisibility test of 11 = adding alternate number and their difference is 0 or multiple of 11.
5335
= 5 + 3 = 8
= 3 + 5 = 8
8 – 8 = 0
5335 is divisible by 11.
(b) 9020814
Divisibility test of 11 = adding alternate number and their difference is 0 or multiple of 11.
9020814
4 + 8 + 2 + 9 = 23
1 + 0 + 0 = 1
23 – 1 = 22 which is multiple of 11.
9020814 is divisible by 11.
Question no – (199)
Solution :
(a) 4096
Divisibility of 4: The last 2 numbers are divisible by 4 or last two digit became 0.
4096 = 96 which is divisible by 4
4096 is divisible by 4.
(b) 21084
Divisibility of 4: The last 2 numbers are divisible by 4 or last two digit became 0.
21084 = 84 which is divisible by 4
21084 is divisible by 4.
(c) 31795012
Divisibility of 4: The last 2 numbers are divisible by 4 or last two digit became 0.
31795012 = 12 which is divisible by 4
31795012 is divisible by 4.
Question no – (200)
Solution :
(a) 672
Divisibility of 9 = Adding all number and addition is divisible by 9.
672 = 6 + 7 + 2 = 15 which is not divisible by 9.
672 is not divisible by 9.
(b) 5652
Divisibility of 9 = Adding all number and addition is divisible by 9.
5652 = 5 + 6 + 5 + 2 = 18 which is divisible by 9.
5652 is divisible by 9.
Next Chapter Solution :
