In questions 1 to 38, out of the four options, only one is correct. Write the correct answer. Q1: The product of the place values of two 2's in 428721 is (a) 4 (b) 40000 (c) 400000 (d) 40000000
Solution:
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
Q2: 3 × 10000 + 7 × 1000 + 9 × 100 + 0 ×10 + 4 is the same as (a) 3794 (b) 37940 (c) 37904 (d) 379409
Solution:
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
Q3: If 1 is added to the greatest 7- digit number, it will be equal to (a) 10 thousand (b) 1 lakh (c) 10 lakh (d) 1 crore
Solution:
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
Q5: When rounded off to nearest thousands, the number 85642 is (a) 85600 (b) 85700 (c) 85000 (d) 86000
Solution:
Ans: (d) To round 85642 to the nearest thousand, look at the hundreds digit (6). Since 6 ≥ 5, increase the thousands part by 1: 85 → 86, and set lower places to 0. Rounded value = 86000.
Q6: The largest 4-digit number, using any one digit twice, from digits 5, 9, 2 and 6 is (a) 9652 (b) 9562 (c) 9659 (d) 9965
Solution:
Ans: (d) To get the largest 4-digit number, place the largest digit(s) at the left. Use 9 twice: 9 9 6 5 → 9965.
Q7: In Indian System of Numeration, the number 58695376 is written as (a) 58,69, 53, 76 (b) 58,695,376 (c) 5,86,95,376 (d) 586,95,376
Solution:
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
Q8: One million is equal to (a) 1 lakh (b) 10 lakh (c) 1 crore (d) 10 crore
Solution:
Ans: (b) One million is equal to ten lakh. 1,000,000 = 10,00,000
Q9: The greatest number which on rounding off to nearest thousands gives 5000, is (a) 5001 (b) 5559 (c) 5999 (d) 5499
Solution:
Ans: (d) The greatest number which on rounding off to nearest thousands gives 5000, is 5499.
Q10: Keeping the place of 6 in the number 6350947 same, the smallest number obtained by rearranging other digits is (a) 6975430 (b) 6043579 (c) 6034579 (d) 6034759
Solution:
Ans: (c) Keep 6 fixed in its place (first digit). Arrange remaining digits 3,5,0,9,4,7 to get the smallest number: put smallest non-zero in second place if leading zero not allowed in that position; here zero can occupy third position. The smallest arrangement is 6 0 3 4 5 7 9 → 6034579.
Q11: Which of the following numbers in Roman numerals is incorrect? (a) LXXX (b) LXX (c) LX (d) LLX
Solution:
Ans: (d) As we know that, the symbol L can never be repeated. Therefore, LLX is incorrect.
Q12: The largest 5-digit number having three different digits is (a) 98978 (b) 99897 (c) 99987 (d) 98799
Solution:
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.
Q13: The smallest 4-digit number having three different digits is (a) 1102 (b) 1012 (c) 1020 (d) 1002
Solution:
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.
Q14: Number of whole numbers between 38 and 68 is (a) 31 (b) 30 (c) 29 (d) 28
Solution:
Ans: (c) Whole numbers strictly between 38 and 68 are 39,40,...,67. Count = 67 - 39 + 1 = 29.
Q15: The product of successor and predecessor of 999 is (a) 999000 (b) 998000 (c) 989000 (d) 1998
Solution:
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
Q16: The product of a non-zero whole number and its successor is always (a) an even number (b) an odd number (c) a prime number (d) divisible by 3
Solution:
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
Q17: A whole number is added to 25 and the same number is subtracted from 25. The sum of the resulting numbers is (a) 0 (b) 25 (c) 50 (d) 75
Solution:
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
Q18: Which of the following is not true? (a) (7 + 8) + 9 = 7 + (8 + 9) (b) (7 × 8) × 9 = 7 × (8 × 9) (c) 7 + 8 × 9 = (7 + 8) × (7 + 9) (d) 7 × (8 + 9) = (7 × 8) + (7 × 9)
Solution:
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
Q19: 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? (a) 9 (b) 10 (c) 11 (d) 12
Solution:
Ans: (b)
Ten dots can form a straight line, a triangular arrangement (e.g., 4th triangular number is 10) and a rectangular (for example 2×5) arrangement.
Q20: Which of the following statements is not true? (a) Both addition and multiplication are associative for whole numbers. (b) Zero is the identity for multiplication of whole numbers. (c) Addition and multiplication both are commutative for whole numbers. (d) Multiplication is distributive over addition for whole numbers.
Solution:
Ans: (b) Zero is the additive identity, not multiplicative. The multiplicative identity is 1. So (b) is not true.
Q21: Which of the following statements is not true? (a) 0 + 0 = 0 (b) 0 - 0 = 0 (c) 0 × 0 = 0 (d) 0 ÷ 0 = 0
Solution:
Ans: (d) Zero divided by zero is not defined.
Q22: The predecessor of 1 lakh is (a) 99000 (b) 99999 (c) 999999 (d) 100001
Solution:
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
Q23: The successor of 1 million is (a) 2 millions (b) 1000001 (c) 100001 (d) 10001
Solution:
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
Q24: Number of even numbers between 58 and 80 is (a) 10 (b) 11 (c) 12 (d) 13
Solution:
Ans: (a) Even numbers between 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.
Q25: Sum of the number of primes between 16 to 80 and 90 to 100 is (a) 20 (b) 18 (c) 17 (d) 16
Solution:
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
Q26: Which of the following statements is not true? (a) The HCF of two distinct prime numbers is 1 (b) The HCF of two co prime numbers is 1 (c) The HCF of two consecutive even numbers is 2 (d) The HCF of an even and an odd number is even.
Solution:
Ans: (d) HCF of an even and an odd number can be 1 or another odd number dividing both; it is not necessarily even. For example, HCF(6,9)=3 (odd). So (d) is false.
Q27: The number of distinct prime factors of the largest 4-digit number is (a) 2 (b) 3 (c) 5 (d) 11
Solution:
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
Q28: The number of distinct prime factors of the smallest 5-digit number is (a) 2 (b) 4 (c) 6 (d) 8
Solution:
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
Q29: If the number 7254*98 is divisible by 22, the digit at * is (a) 1 (b) 2 (c) 6 (d) 0
Solution:
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
Q30: The largest number which always divides the sum of any pair of consecutive odd numbers is (a) 2 (b) 4 (c) 6 (d) 8
Solution:
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.
Q31: A number is divisible by 5 and 6. It may not be divisible by (a) 10 (b) 15 (c) 30 (d) 60
Solution:
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.
Q32: The sum of the prime factors of 1729 is (a) 13 (b) 19 (c) 32 (d) 39
Solution:
Ans: (d) The prime factors of 1729 = 7 × 13 × 19 Therefore, the sum of prime numbers = 7 + 13 + 19 = 39
Q33: The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is (a) 6 (b) 4 (c) 16 (d) 8
Solution:
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.
Q34: The number of common prime factors of 75, 60, 105 is (a) 2 (b) 3 (c) 4 (d) 5
Solution:
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.
Q35: Which of the following pairs is not coprime? (a) 8, 10 (b) 11, 12 (c) 1, 3 (d) 31, 33
Solution:
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.
Q36: Which of the following numbers is divisible by 11? (a) 1011011 (b) 1111111 (c) 22222222 (d) 3333333
Solution:
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.
Q37: LCM of 10, 15 and 20 is (a) 30 (b) 60 (c) 90 (d) 180
Solution:
Ans: (b) Factors of 10, 15 and 30 is, Then, LCM of 10, 15 and 30 is 2 × 2 × 3 × 5 × 1 = 60
Q38: LCM of two numbers is 180. Then which of the following is not the HCF of the numbers? (a) 45 (b) 60 (c) 75 (d) 90
Solution:
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.
In questions 39 to 98 state whether the given statements are true (T) or false (F). Q39: In Roman numeration, a symbol is not repeated more than three times.
Solution:
True. As per the rule of the Roman numerals, a symbol is not repeated more than three times.
Q40: In Roman numeration, if a symbol is repeated, its value is multiplied as many times as it occurs.
Solution:
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.
Q41: 5555 = 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1
Solution:
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
Q45: XXIX = 31
Solution:
False. Where, X = 10 IX = 9 So, XXIX = 10 + 10 + 9 = 29
Q46: LXXIV = 74
Solution:
True. Where, L = 50 X = 10 IV = 4 So, LXXIV = 50 + 10 + 10 + 4 = 74
Q47: The number LIV is greater than LVI.
Solution:
False. Where, L = 50 IV = 4 VI = 6 So, LIV = 50 + 4 = 54 LVI = 50 + 6 Therefore, 54 < 56 Hence, LIV < LVI
Q48: The numbers 4578, 4587, 5478, 5487 are in descending order.
Solution:
False. In the question, the arrangement of the numbers in ascending order. Descending order of the given number = 5487, 5478, 4587, 4578.
Q49: The number 85764 rounded off to nearest hundreds is written as 85700.
Solution:
False. The number 85764 rounded off to nearest hundreds is written as 85800.
Q50: Estimated sum of 7826 and 12469 rounded off to hundreds is 20,000.
Solution:
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.
Q51: The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875403.
Solution:
False. The largest six digit telephone number that can be formed by using digits 5, 3, 4, 7, 0, 8 only once is 875430.
Q52: The number 81652318 will be read as eighty one crore six lakh fifty two thousand three hundred eighteen.
Solution:
False. The given number 8,16,52,318 will be read as eight crore sixteen lakh fifty two thousand three hundred and eighteen.
Q53: The largest 4-digit number formed by the digits 6, 7, 0, 9 using each digit only once is 9760.
Solution:
True. Arrange digits in descending order: 9,7,6,0 → 9760.
Q54: Among kilo, milli and centi, the smallest is centi.
Solution:
False. Among kilo, milli and centi, the smallest is milli.
Q55: Successor of a one-digit number is always a one-digit number.
Solution:
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.
Q56: Successor of a 3-digit number is always a 3-digit number.
Solution:
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.
Q57: Predecessor of a two-digit number is always a two-digit number.
Solution:
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.
Q58: Every whole number has its successor.
Solution:
True.
For any whole number n, n + 1 exists as its successor.
Q59: Every whole number has its predecessor.
Solution:
False. Consider the whole number 0, Then, its predecessor = 0 - 1 = -1 - 1 is an integer.
Q60: Between any two natural numbers, there is one natural number.
Solution:
False. Consider the two natural numbers 4 and 8. Then, natural numbers between 4 and 8 are 5, 6, 7.
Q61: The smallest 4-digit number is the successor of the largest 3-digit number.
Solution:
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
Q62: Of the given two natural numbers, the one having more digits is greater.
Solution:
True. As per the rule, Of the given two natural numbers, the one having more digits is greater.
Q63: Natural numbers are closed under addition.
Solution:
True. We know that, sum of two natural numbers is always natural number. Therefore, natural numbers are closed under addition.
Q64: Natural numbers are not closed under multiplication.
Solution:
False. We know that, multiplication of two natural numbers is always natural number. Therefore, natural numbers are closed under multiplication.
Q65: Natural numbers are closed under subtraction.
Solution:
False. Difference of two natural numbers are not always a natural number. Therefore, natural numbers are not closed under subtraction.
Q66: Addition is commutative for natural numbers.
Solution:
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
Q67: is the identity for addition of whole numbers.
Solution:
False. Zero (0) is the identity for addition of whole numbers. Consider any whole number i.e. 8. Then, 8 + 0 = 8
Q68: is the identity for multiplication of whole numbers.
Solution:
True. Consider any whole number i.e. 6. 6 × 1 = 6
Q69: There is a whole number which when added to a whole number, gives the number itself.
Solution:
True. Zero (0) is a whole number which when added to a whole number, gives the number itself.
Q70: There is a natural number which when added to a natural number, gives the number itself.
Solution:
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.
Q71: 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:
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.
Q72: Any non-zero whole number divided by itself gives the quotient 1.
Solution:
True. Consider any non-zero whole number i.e. 5 5 is divided by itself = 5/5 = 1
Q73: The product of two whole numbers need not be a whole number.
Solution:
False. The product of two whole number is always a whole number. Because, we know that, whole numbers are closed under multiplication.
Q74: A whole number divided by another whole number greater than 1 never gives the quotient equal to the former.
Solution:
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.
Q75: Every multiple of a number is greater than or equal to the number.
Solution:
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
Q76: The number of multiples of a given number is finite.
Solution:
False. The number of multiples of a given number is infinite. Because, we know that numbers are infinite.
Q77: Every number is a multiple of itself.
Solution:
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.
Q78: Sum of two consecutive odd numbers is always divisible by 4.
Q79: If a number divides three numbers exactly, it must divide their sum exactly.
Solution:
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
Q80: If a number exactly divides the sum of three numbers, it must exactly divide the numbers separately.
Solution:
False. A number may divide the sum but not each addend separately (e.g., 3 divides 1+2=3 but divides neither 1 nor 2).
Q81: If a number is divisible both by 2 and 3, then it is divisible by 12.
Solution:
False. Let us consider the number 6, it is actually divisible by 2 and 3, But 6 is not divisible by 12.
Q82: 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:
False. Divisibility by 6 requires divisibility by both 2 and 3; knowing only last two digits gives divisibility by 2 and 4 related tests but not guarantees divisibility by 3 of the whole number.
Q83: 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:
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.
Q84: If the sum of the digits of a number is divisible by 3, then the number itself is divisible by 9.
Solution:
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.
Q85: All numbers which are divisible by 4 may not be divisible by 8.
Solution:
True. Consider the number 20, it is divisible by 4 but not divisible by 8.
Q86: The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.
Solution:
False. The Highest Common Factor of two or more numbers is lower than their Lowest Common Multiple.
Q87: LCM of two or more numbers is divisible by their HCF.
Solution:
True. As per the rule, LCM of two or more numbers is divisible by their HCF.
Q88: LCM of two numbers is 28 and their HCF is 8.
Solution:
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.
Q89: LCM of two or more numbers may be one of the numbers.
Solution:
True. Consider the two numbers 2 and 4. Then, LCM of 2 and 4 is 4.
Q90: HCF of two or more numbers may be one of the numbers.
Solution:
True. If one number divides the other(s), HCF can be the smaller number (e.g., HCF of 6 and 12 is 6).
Q91: Every whole number is the successor of another whole number.
Solution:
False. We know that, 0 is the whole number. 0 is no the successor of another whole number.
Q92: Sum of two whole numbers is always less than their product.
Solution:
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.
Q93: If the sum of two distinct whole numbers is odd, then their difference also must be odd.
Solution:
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.
Q94: Any two consecutive numbers are coprime.
Solution:
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.
Q95: If the HCF of two numbers is one of the numbers, then their LCM is the other number.
Solution:
True. If HCF(a,b) = a, then a divides b. LCM(a,b) = b in that case.
Q96: The HCF of two numbers is smaller than the smaller of the numbers.
Solution:
False. The HCF of two numbers is either greater than or equal to the smaller of the numbers.
Q97: The LCM of two numbers is greater than the larger of the numbers.
Solution:
False. The LCM of two numbers may be equal to or greater than the larger of the numbers.
Q98: The LCM of two coprime numbers is equal to the product of the numbers.
Solution:
True. For coprime numbers a and b, LCM(a,b) = a × b because they share no common factors.
In questions 99 to 100, fill in the blanks to make the statements true. Q99: (a) 10 million = _____ crore.
Solution:
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.
(b) 10 lakh = _____ million.
Solution:
10 lakh = 1 million.
Q100: (a) 1 metre = _____ millimetres.
Solution:
1 metre = 1000 millimetres We know that, 1 metre = 100 centimetre 1 centimetre = 10 millimeter Then, 100 cm = 10 × 100 = 1000 millimetres
(b) 1 centimetre = _____ millimetres.
Solution:
1 centimetre = 10 millimetres.
(c) 1 kilometre = _____ millimetres.
Solution:
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
Q103: Height of a person is 1m 65cm. His height in millimetres is_______.
Solution:
1650 : 1 m 65 cm = (1000 + 650) mm = 1650 mm
Q104: Length of river 'Narmada' is about 1290km. Its length in metres is_______.
Solution:
1290000 : 1290 km = (1290 × 1000) m = 1290000 m
Q105: The distance between Sringar and Leh is 422km. The same distance in metres is_____.
Solution:
422000 : 422 km= (422 × 1000) m = 422000 m
Q106: Writing of numbers from the greatest to the smallest is called an arrangement in _____ order.
Solution:
Descending
Q107: 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:
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.
Q108: By adding 1 to the greatest_____ digit number, we get ten lakh.
Solution:
6 : Greatest 6-digit number = 999999 By adding 1 to 999999, we get 1000000.
Q109: 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:
5, 23, 78, 401
Q110: In Roman Numeration, the symbol X can be subtracted from_____, M and C only.
Solution:
L
Q111: The number 66 in Roman numerals is_____.
Solution:
LXVI : 66 = LXVI
Q112: The population of Pune was 2,538,473 in 2001. Rounded off to nearest thousands, the population was __________.
Solution:
2,538,000
Q113: The smallest whole number is_____.
Solution:
0 : 0 is the smallest whole number.
Q114: Successor of 106159 is _____.
Solution:
106160 : Successor of 106159 is 106159 + 1, i.e., 106160
Q115: Predecessor of 100000 is_____.
Solution:
99999 : Predecessor of 100000 is 100000-1, i.e., 99999
Q116: 400 is the predecessor of _____.
Solution:
401 : 400 is the predecessor of 400 + 1, i.e., 401
Q117: _____ is the successor of the largest 3 digit number.
Solution:
1000 : Largest 3 digit number = 999 Successor of 999 is 999 + 1, i.e., 1000
Q118: If 0 is subtracted from a whole number, then the result is the _____ itself .
Solution:
Number
Q119: The smallest 6 digit natural number ending in 5 is _____.
Solution:
100005
Q120: Whole numbers are closed under _____ and under_____.
Solution:
Addition, multiplication
Q121: Natural numbers are closed under _____ and under_____.
Solution:
Addition, multiplication
Q122: Division of a whole number by _____ is not defined.
Solution:
0
Q123: Multiplication is distributive over _____ for whole numbers.
Solution:
Addition
Q124: 2395 × _____ = 6195 × 2395
Solution:
6195 : Since, multiplication is commutative for whole numbers.
Q125: 1001 × 2002 = 1001 × (1001+_____ )
Solution:
1001
Q126: 10001 × 0 = _____
Solution:
0
Q127: 2916 × _____ = 0
Solution:
0
Q128: 9128 × _____ = 9128
Solution:
1 : Since, 1 is the multiplicative identity for whole numbers.
Q129: 125 + (68+17) = (125 + _____ ) + 17
Solution:
68 : Since, addition is associative for whole numbers.
Q130: 8925 ×1 = _____
Solution:
8925
Q131: 19 × 12 + 19 = 19 × (12 + _____)
Solution:
1 : Since, multiplication is distributive over addition for whole numbers.
Q132: 24 × 35 = 24 × 18 + 24 × _____
Solution:
17
Q133: 32 × (27 × 19) = (32 × _____ ) × 19
Solution:
27 : Since, multiplication is associative for whole numbers.
Q138: The number of factors of a prime number is_____.
Solution:
2
Q139: A number for which the sum of all its factors is equal to twice the number is called a _____ number.
Solution:
Perfect
Q140: The numbers having more than two factors are called _____ numbers.
Solution:
Composite
Q141: 2 is the only _____ number which is even.
Solution:
Prime
Q142: Two numbers having only 1 as a common factor are called _____ numbers.
Solution:
Co-prime
Q143: Number of primes between 1 to 100 is _____.
Solution:
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.
Q144: If a number has _____ in ones place, then it is divisible by 10.
Solution:
0
Q145: A number is divisible by 5, if it has _____ or _____ in its ones place.
Solution:
0, 5
Q146: A number is divisible by _____ if it has any of the digits 0, 2, 4, 6, or 8 in its ones place.
Solution:
2
Q147: If the sum of the digits in a number is a _____ of 3, then the number is divisible by 3.
Solution:
Multiple
Q148: 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:
11
Q149: The LCM of two or more given numbers is the lowest of their common_____.
Solution:
Multiple
Q150: The HCF of two or more given numbers is the highest of their common _____.
Solution:
Factors
Q151:Given below are two columns - Column I and Column II. Match each item of Column I with the corresponding item of Column II.
Q155: As per the census of 2001, the population of four states are given below. Arrange the states in ascending and descending order of their population. (a) Maharashtra 96878627 (b) Andhra Pradesh 76210007 (c) Bihar 82998509 (d) Uttar Pradesh 166197921
Solution:
Ascending order ➝ (b), (c), (a), (d) Descending order ➝ (d), (a), (c), (b)
Q156: The diameter of Jupiter is 142800000 metres. Insert commas suitably and write the diameter according to International System of Numeration.
Solution:
The diameter of Jupiter is 142,800,000 metres.
Q157: India's population has been steadily increasing from 439 millions in 1961 to 1028 millions in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in Indian System of Numeration, using commas suitably.
Solution:
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
Q158: Radius of the Earth is 6400km and that of Mars is 4300000m. Whose radius is bigger and by how much?
Solution:
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.
Q159: In 2001, the poplulations of Tripura and Meghalaya were 3,199,203 and 2,318,822, respectively. Write the populations of these two states in words.
Solution:
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.
Q160: In a city, polio drops were given to 2,12,583 children on Sunday in March 2008 and to 2,16,813 children in the next month. Find the difference of the number of children getting polio drops in the two months.
Solution:
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
Q161: A person had Rs 1000000 with him. He purchased a colour T.V. for Rs 16580, a motor cycle for Rs 45890 and a flat for Rs 870000. How much money was left with him?
Solution:
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
Q162: Out of 180000 tablets of Vitamin A, 18734 are distributed among the students in a district. Find the number of the remaining vitamin tablets.
Solution:
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
Q163: Chinmay had Rs 610000. He gave Rs 87500 to Jyoti, Rs 126380 to Javed and Rs 350000 to John. How much money was left with him?
Solution:
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
Q164: Find the difference between the largest number of seven digits and the smallest number of eight digits.
Solution:
The smallest number of eight digits = 10000000 The largest number of seven digits = 9999999 The required difference = 10000000 - 9999999 = 1
Q165: A mobile number consists of ten digits. The first four digits of the number are 9, 9, 8 and 7. The last three digits are 3, 5 and 5. The remaining digits are distinct and make the mobile number, the greatest possible number. What are these digits?
Solution:
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.
Q166: A mobile number consists of ten digits. First four digits are 9,9,7 and 9. Make the smallest mobile number by using only one digit twice from 8, 3, 5, 6, 0.
Solution:
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.
Q167: In a five digit number, digit at ten's place is 4, digit at unit's place is one fourth of ten's place digit, digit at hunderd's place is 0, digit at thousand's place is 5 times of the digit at unit's place and ten thousand's place digit is double the digit at ten's place. Write the number.
Solution:
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.
Q168: Find the sum of the greatest and the least six digit numbers formed by the digits 2, 0, 4, 7, 6, 5 using each digit only once.
Solution:
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
Q169: A factory has a container filled with 35874 litres of cold drink. In how many bottles of 200 ml capacity each can it be filled?
Solution:
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.
Q170: The population of a town is 450772. In a survey, it was reported that one out of every 14 persons is illiterate. In all how many illiterate persons are there in the town?
Solution:
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
Q171: Find the LCM of 80, 96, 125, 160.
Solution:
we have, The LCM of 80, 96, 125 and 160 is = 2 × 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 = 12000
Q172: Make the greatest and the smallest 5-digit numbers using different digits in which 5 appears at ten's place.
Solution:
By using the digit 5 at ten's place, the greatest 5-digit number is 98756, and the smallest 5-digit number is 10253
Q173: How many grams should be added to 2kg 300g to make it 5kg 68g?
Solution:
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
Q174: A box contains 50 packets of biscuits each weighing 120g. How many such boxes can be loaded in a van which cannot carry beyond 900kg?
Solution:
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.
Q175: How many lakhs make five billions?
Solution:
50000 lakhs make 5 billions.
Q176: How many millions make 3 crores?
Solution:
30 millions make 3 crores.
Q177: Estimate each of the following by rounding off each number to nearest hundreds: (a) 874 + 478 (b) 793 + 397 (c) 11244 + 3507 (d) 17677 + 13589
Solution:
(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
Q178: Estimate each of the follwoing by rounding off each number to nearest tens: (a) 11963 - 9369 (b) 76877 - 7783 (c) 10732 - 4354 (d) 78203 - 16407
Solution:
(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
Q179: Estimate each of the following products by rounding off each number to nearest tens: (a) 87 × 32 (b) 311×113 (c) 3239 × 28 (d) 1385 × 789
Solution:
(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
Q180: The population of a town was 78787 in the year 1991 and 95833 in the year 2001. Estimate the increase in population by rounding off each population to nearest hundreds.
Solution:
78787 rounded off to the nearest hundreds = 78800 95833 rounded off to the nearest hundreds = 95800 The estimated increase in population = 95800 - 78800 = 17000
Q181: Estimate the product 758 × 6784 using the general rule.
Solution:
758 can be rounded off to 800 and 6784 can be rounded off to 7000 ∴ Estimated product = 800 × 7000 = 5600000
Q182: A garment factory produced 216315 shirts, 182736 trousers and 58704 jackets in a year. What is the total production of all the three items in that year?
Solution:
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
Q183: Find the LCM of 160, 170 and 90.
Solution:
We have, The LCM of 160,170 and 190 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 24480
Q184: A vessel has 13 litres 200 mL of fruit juice. In how many glasses each of capacity 60 mL can it be filled?
Solution:
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.
Q185: Determine the sum of the four numbers as given below: (a) successor of 32 (b) predecessor of 49 (c) predecessor of the predecessor of 56 (d) successor of the successor of 67
Solution:
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
Q186: A loading tempo can carry 482 boxes of biscuits weighing 15kg each, whereas a van can carry 518 boxes each of the same weight. Find the total weight that can be carried by both the vehicles.
Solution:
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
Q187: In the marriage of her daughter, Leela spent Rs 216766 on food and decoration,Rs 122322 on jewellery, Rs 88234 on furniture and Rs 26780 on kitchen items. Find the total amount spent by her on the above items.
Solution:
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
Q188: A box contains 5 strips having 12 capsules of 500mg medicine in each capsule. Find the total weight in grams of medicine in 32 such boxes.
Solution:
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
Q189: Determine the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.
Solution:
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.
Q190: A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of a third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Solution:
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.
Q191: Find a 4-digit odd number using each of the digits 1, 2, 4 and 5 only once such that when the first and the last digits are interchanged, it is divisible by 4.
Solution:
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.
Q192: Using each of the digits 1, 2, 3 and 4 only once, determine the smallest 4-digit number divisible by 4.
Solution:
By using the digits 1, 2, 3 and 4 only once, the smallest 4-digit number which is divisible by 4 is 1324.
Q193: Fatima wants to mail three parcels to three village schools. She finds that the postal charges are Rs 20, Rs 28 and Rs 36, respectively. If she wants to buy stamps only of one denomination, what is the greatest denomination of stamps she must buy to mail the three parcels?
Solution:
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.
Q194: Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeepeer wants to buy an equal number of biscuits, of each brand, what is the minimum number of packets of each brand, he should buy?
Solution:
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
Q195: The floor of a room is 8m 96cm long and 6m 72cm broad. Find the minimum number of square tiles of the same size needed to cover the entire floor.
Solution:
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
Q196: In a school library, there are 780 books of English and 364 books of Science. Ms. Yakang, the librarian of the school wants to store these books in shelves such that each shelf should have the same number of books of each subject. What should be the minimum number of books in each shelf?
Solution:
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
Q197: In a colony of 100 blocks of flats numbering 1 to 100, a school van stops at every sixth block while a school bus stops at every tenth block. On which stops will both of them stop if they start from the entrance of the colony?
Solution:
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 30th block, then aL 60th block and lastly at 90th block.
Q198: Test the divisiblity of following numbers by 11 (a) 5335 (b) 9020814
Solution:
(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.
Q199: Using divisiblity tests, determine which of the following numbers are divisible by 4? (a) 4096 (b) 21084 (c) 31795012
Solution:
(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.
Q200: Using divisiblity test. determine which of the following numbers are divisible by 9? (a) 672 (b) 5652
Solution:
(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.
1. What are natural numbers and whole numbers, and how are they different?
Ans. Natural numbers start from 1 and include all positive integers (1, 2, 3, ...), while whole numbers include 0 plus all natural numbers (0, 1, 2, 3, ...). The key distinction is that zero belongs to whole numbers but not natural numbers. Understanding this difference is crucial for CTET and TET exams, especially when classifying number systems and solving problems involving the number line.
2. How do I identify rational and irrational numbers in NCERT problems?
Ans. Rational numbers can be expressed as p/q where p and q are integers and q ≠ 0 (like 3/4 or 0.5), while irrational numbers cannot be written this way (like π or √2). To classify numbers, check if a decimal terminates or repeats (rational) or continues non-repeating (irrational). This distinction is frequently tested in CTET Paper 2 Mathematics pedagogy questions.
3. Why do we need to understand integers and their properties for teaching mathematics?
Ans. Integers encompass negative numbers, zero, and positive whole numbers, forming the foundation for understanding algebraic concepts and real-world applications like temperature and debt. For TET candidates, grasping integer properties enables effective pedagogy, helping students understand number operations on the number line, order of integers, and their practical significance in classroom teaching.
4. What's the easiest way to remember the number system hierarchy for my exam?
Ans. The hierarchy flows from smallest to largest: Natural Numbers ⊂ Whole Numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real Numbers. Each set contains the previous one with additional elements. Using visual number line diagrams, flashcards, and mind maps helps candidates retain this classification structure for quick recall during CTET and state TET examinations.
5. How do I explain even and odd numbers to students using the number system concept?
Ans. Even numbers are divisible by 2 (0, 2, 4, 6, ...) while odd numbers leave remainder 1 when divided by 2 (1, 3, 5, ...). Both categories exist within integers and natural numbers. For pedagogy, demonstrating this with concrete examples, grouping objects, and using the divisibility rule makes the concept accessible and helps students develop number sense effectively.
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