CBSE Class 5 Mathematics Revision Worksheet Set 13

Fill in the blanks :

Question 1. A number is divisible by 10, if its one's place is ____________________________
Answer: 0
In simple words: Any number ending in 0 can be divided completely by 10.

Exam Tip: Always look at the last digit (ones place) when checking for divisibility by 10.

Question 2. A number is divisible by__________________, if its one's place is 0 or 5.
Answer: 5
In simple words: Numbers ending in 0 or 5 are always multiples of 5.

Exam Tip: Check the ones digit; it must be either 0 or 5.

Question 3. A number is divisible by ______________________, if its one's place is 0,2,4, 6 or 8.
Answer: 2
In simple words: Even numbers are always divisible by 2.

Question 4. ______________________is neither prime not composite.
Answer: 1
In simple words: The number 1 has only one factor (itself), so it is neither prime nor composite.

Exam Tip: Memorize that 1 is unique and does not fit into prime or composite categories.

Question 5. _____________________________is the smallest prime no.
Answer: 2
In simple words: 2 is the smallest number greater than 1 that has only two factors: 1 and itself.

Exam Tip: Remember that 2 is also the only even prime number.

Question 6. ___________________________is the only even prime no.
Answer: 2
In simple words: All other even numbers are divisible by 2, making them composite. Only 2 is prime.

Exam Tip: A frequent trick question in exams; always answer 2.

Question 7. A number is divisible by 9, if the sum of the digits is divisible by ____________.
Answer: 9
In simple words: Add up all digits in the number; if the total is in the 9 times table, the number is divisible by 9.

Exam Tip: This rule is identical in structure to the divisibility rule for 3.

Question 8. Prime numbers have only _______________________factors.
Answer: two
In simple words: A prime number can only be divided by 1 and by itself.

Exam Tip: The two distinct factors are always 1 and the number itself.

Question 9. A number is divisible by 3 if the _______________________ is divisible by 3.
Answer: sum of the digits
In simple words: Sum all the digits of the number; if that sum is a multiple of 3, the whole number is divisible by 3.

Exam Tip: You can repeat the digit-sum process if the number is large until you get a single digit.

Question 10. Co primes are numbers which have only _______________as their common factors.
Answer: 1
In simple words: Two numbers are co-prime if their HCF is 1.

Exam Tip: Consecutive numbers (like 8 and 9) are always co-prime.

Question 11. LCM of 9 and 10 is ___________________________
Answer: 90
In simple words: Since 9 and 10 are co-prime, their LCM is simply their product \(9 \times 10 = 90\).

Exam Tip: For co-prime numbers, LCM = product of the numbers.

Question 12. HCF of 17 and 19 is ___________________________
Answer: 1
In simple words: Both 17 and 19 are prime numbers, so their only common factor is 1.

Exam Tip: The HCF of any two distinct prime numbers is always 1.

Question 13. LCM of 6 and 18 is ______________________
Answer: 18
In simple words: Since 18 is a multiple of 6, the larger number is automatically the LCM.

Exam Tip: If one number is a multiple of the other, the larger number is the LCM.

Question 14. HCF of 9 and 45 is ___________________________
Answer: 9
In simple words: Since 45 is a multiple of 9, the smaller number is the HCF.

Exam Tip: If one number is a multiple of the other, the smaller number is the HCF.

Question 15. A number is divisible by _____________________, if it is divisible by both 2 and 3.
Answer: 6
In simple words: The prime factors of 6 are 2 and 3, so a number divisible by both must be divisible by 6.

Exam Tip: Always check both conditions: even number AND sum of digits divisible by 3.

Answer the following:

Question 1. Without actual division , check whether the following are divisible by 2

(a) 7422 (b)39582 (c)14659 (d)52314
Answer:
(a) 7422 ends in 2, which is even. Therefore, it is divisible by 2.
(b) 39582 ends in 2, which is even. Therefore, it is divisible by 2.
(c) 14659 ends in 9, which is odd. Therefore, it is not divisible by 2.
(d) 52314 ends in 4, which is even. Therefore, it is divisible by 2.
In simple words: Numbers ending in 0, 2, 4, 6, or 8 are divisible by 2.

Exam Tip: Only look at the units digit.

Question 2. Check whether the following are divisible by 3
(a) 741352 (b)2034198 (c)317925 (d)3412920
Answer:
(a) Sum of digits = \(7 + 4 + 1 + 3 + 5 + 2 = 22\). Since 22 is not divisible by 3, 741352 is not divisible by 3.
(b) Sum of digits = \(2 + 0 + 3 + 4 + 1 + 9 + 8 = 27\). Since 27 is divisible by 3, 2034198 is divisible by 3.
(c) Sum of digits = \(3 + 1 + 7 + 9 + 2 + 5 = 27\). Since 27 is divisible by 3, 317925 is divisible by 3.
(d) Sum of digits = \(3 + 4 + 1 + 2 + 9 + 2 + 0 = 21\). Since 21 is divisible by 3, 3412920 is divisible by 3.
In simple words: Add all digits; if the sum is a multiple of 3, the number is divisible by 3.

Exam Tip: Sum the digits carefully and check against the 3 multiplication table.

Question 3. Check whether the following are divisible by 4
(a) 4137156 (b)135764 (c)34560 (d)167435
Answer:
(a) Number formed by last two digits = 56. Since \(56 \div 4 = 14\), 4137156 is divisible by 4.
(b) Number formed by last two digits = 64. Since \(64 \div 4 = 16\), 135764 is divisible by 4.
(c) Number formed by last two digits = 60. Since \(60 \div 4 = 15\), 34560 is divisible by 4.
(d) Number formed by last two digits = 35. Since 35 is not divisible by 4, 167435 is not divisible by 4.
In simple words: Check if the number formed by the last two digits can be divided by 4.

Exam Tip: Focus strictly on the tens and units digits.

Question 4. Check whether the following are divisible by 5
(a) 187620 (b)258732 (c)361245 (d)705204
Answer:
(a) 187620 ends in 0, so it is divisible by 5.
(b) 258732 ends in 2, so it is not divisible by 5.
(c) 361245 ends in 5, so it is divisible by 5.
(d) 705204 ends in 4, so it is not divisible by 5.
In simple words: A number is divisible by 5 only if its last digit is 0 or 5.

Exam Tip: Quickest divisibility rule to verify on exams.

Question 5. Check whether the following are divisible by 6
(a) 4234156 (b)1027863 (c)924658 (d)1850421
Answer:
(a) Ends in 6 (even, divisible by 2). Sum of digits = \(4+2+3+4+1+5+6 = 25\) (not divisible by 3). Therefore, not divisible by 6.
(b) Ends in 3 (odd, not divisible by 2). Therefore, not divisible by 6.
(c) Ends in 8 (even, divisible by 2). Sum of digits = \(9+2+4+6+5+8 = 34\) (not divisible by 3). Therefore, not divisible by 6.
(d) Ends in 1 (odd, not divisible by 2). Therefore, not divisible by 6.
In simple words: A number must be divisible by both 2 and 3 to be divisible by 6.

Exam Tip: Always check if it is even first; if not, immediately rule out 6.

Question 6. Check whether the following are divisible by 8
(a) 1967000 (b)2587382 (c)3614944 (d)191640
Answer:
(a) Last 3 digits = 000. Divisible by 8, so 1967000 is divisible by 8.
(b) Last 3 digits = 382. \(382 \div 8 = 47\) remainder 6. Not divisible by 8.
(c) Last 3 digits = 944. \(944 \div 8 = 118\). Divisible by 8, so 3614944 is divisible by 8.
(d) Last 3 digits = 640. \(640 \div 8 = 80\). Divisible by 8, so 191640 is divisible by 8.
In simple words: Check if the number formed by the last three digits is divisible by 8.

Question 7. Check whether the following are divisible by 9
(a) 739602 (b)2034198 (c)674132 (d)7413552
Answer:
(a) Sum of digits = \(7+3+9+6+0+2 = 27\). Divisible by 9, so 739602 is divisible by 9.
(b) Sum of digits = \(2+0+3+4+1+9+8 = 27\). Divisible by 9, so 2034198 is divisible by 9.
(c) Sum of digits = \(6+7+4+1+3+2 = 23\). Not divisible by 9, so 674132 is not divisible by 9.
(d) Sum of digits = \(7+4+1+3+5+5+2 = 27\). Divisible by 9, so 7413552 is divisible by 9.
In simple words: Sum the digits; if the sum is a multiple of 9, the number is divisible by 9.
Exam Tip: Works exactly like the test for 3, but the sum must be a multiple of 9.

Question 8. Check whether the following are divisible by 10
(a) 395725 (b)6042190 (c)814231 (d)42950
Answer:
(a) Ends in 5. Not divisible by 10.
(b) Ends in 0. Divisible by 10.
(c) Ends in 1. Not divisible by 10.
(d) Ends in 0. Divisible by 10.
In simple words: The number must end in 0.

Exam Tip: Easiest rule: look only at the units place for a 0.

Question 9. Check whether the following are divisible by 11
(a) 666666 (b)1251206 (c)7917624 (d)6510625
Answer:
(a) Sum of digits at odd places = \(6+6+6 = 18\). Sum of digits at even places = \(6+6+6 = 18\). Difference = \(18 - 18 = 0\). Divisible by 11.
(b) Sum of odd places = \(6+2+5+1 = 14\). Sum of even places = \(0+1+2 = 3\). Difference = \(14 - 3 = 11\). Divisible by 11.
(c) Sum of odd places = \(4+6+1+7 = 18\). Sum of even places = \(2+7+9 = 18\). Difference = \(18 - 18 = 0\). Divisible by 11.
(d) Sum of odd places = \(5+6+1+6 = 18\). Sum of even places = \(2+0+5 = 7\). Difference = \(18 - 7 = 11\). Divisible by 11.
In simple words: Subtract the sum of digits in even positions from the sum of digits in odd positions; if the result is 0 or a multiple of 11, it is divisible.

Exam Tip: Label positions from right to left (1st, 2nd, 3rd, etc.) before summing.

Question 10. Find LCM
(a) 36,20 (b)36, 48, 64
(c)24, 36 (d)39, 65, 78
(e) 60,72,96 (f)48,60
(g)12,15,45 (h)42,36
(i) 14,21 (j)25,65
(k)48,60 (l)35,40
(m) 50,60 (n)36, 49
(o)26,13,91 (p)60,84,96
Answer:
(a) Prime factors of 36 = \(2^2 \times 3^2\). Prime factors of 20 = \(2^2 \times 5\). LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\).
(b) 36 = \(2^2 \times 3^2\), 48 = \(2^4 \times 3\), 64 = \(2^6\). LCM = \(2^6 \times 3^2 = 64 \times 9 = 576\).
(c) 24 = \(2^3 \times 3\), 36 = \(2^2 \times 3^2\). LCM = \(2^3 \times 3^2 = 8 \times 9 = 72\).
(d) 39 = \(3 \times 13\), 65 = \(5 \times 13\), 78 = \(2 \times 3 \times 13\). LCM = \(2 \times 3 \times 5 \times 13 = 390\).
(e) 60 = \(2^2 \times 3 \times 5\), 72 = \(2^3 \times 3^2\), 96 = \(2^5 \times 3\). LCM = \(2^5 \times 3^2 \times 5 = 32 \times 9 \times 5 = 1440\).
(f) 48 = \(2^4 \times 3\), 60 = \(2^2 \times 3 \times 5\). LCM = \(2^4 \times 3 \times 5 = 16 \times 15 = 240\).
(g) 12 = \(2^2 \times 3\), 15 = \(3 \times 5\), 45 = \(3^2 \times 5\). LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\).
(h) 42 = \(2 \times 3 \times 7\), 36 = \(2^2 \times 3^2\). LCM = \(2^2 \times 3^2 \times 7 = 4 \times 9 \times 7 = 252\).
(i) 14 = \(2 \times 7\), 21 = \(3 \times 7\). LCM = \(2 \times 3 \times 7 = 42\).
(j) 25 = \(5^2\), 65 = \(5 \times 13\). LCM = \(5^2 \times 13 = 25 \times 13 = 325\).
(k) 48 = \(2^4 \times 3\), 60 = \(2^2 \times 3 \times 5\). LCM = \(2^4 \times 3 \times 5 = 240\).
(l) 35 = \(5 \times 7\), 40 = \(2^3 \times 5\). LCM = \(2^3 \times 5 \times 7 = 8 \times 35 = 280\).
(m) 50 = \(2 \times 5^2\), 60 = \(2^2 \times 3 \times 5\). LCM = \(2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300\).
(n) 36 = \(2^2 \times 3^2\), 49 = \(7^2\). LCM = \(36 \times 49 = 1764\).
(o) 26 = \(2 \times 13\), 13 = \(13\), 91 = \(7 \times 13\). LCM = \(2 \times 7 \times 13 = 182\).
(p) 60 = \(2^2 \times 3 \times 5\), 84 = \(2^2 \times 3 \times 7\), 96 = \(2^5 \times 3\). LCM = \(2^5 \times 3 \times 5 \times 7 = 32 \times 3 \times 35 = 3360\).
In simple words: Find the prime factorization of each number and take the highest power of every prime factor present.

Exam Tip: Common division method using prime numbers is often faster and less prone to errors than listing multiples.

Question 11. Find the HCF
(a) 16,24 (b)28,36 (c)40,24 (d)52,68 (e) 36,18 (f)42,72,18 (g)16,32,40 (h)42,72,18 (i) 6,10,28 (j)66,44 (k)38,60 (l)45,18,36 (m) 26,43
Answer:
(a) Factors of 16: 1, 2, 4, 8, 16. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. HCF = 8.
(b) Factors of 28: 1, 2, 4, 7, 14, 28. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. HCF = 4.
(c) Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. HCF = 8.
(d) 52 = \(2^2 \times 13\), 68 = \(2^2 \times 17\). HCF = \(2^2 = 4\).
(e) Since 36 is a multiple of 18, HCF = 18.
(f) 42 = \(2 \times 3 \times 7\), 72 = \(2^3 \times 3^2\), 18 = \(2 \times 3^2\). HCF = \(2 \times 3 = 6\).
(g) 16 = \(2^4\), 32 = \(2^5\), 40 = \(2^3 \times 5\). HCF = \(2^3 = 8\).
(h) (Duplicate of f) HCF = 6.
(i) 6 = \(2 \times 3\), 10 = \(2 \times 5\), 28 = \(2^2 \times 7\). HCF = 2.
(j) 66 = \(2 \times 3 \times 11\), 44 = \(2^2 \times 11\). HCF = \(2 \times 11 = 22\).
(k) 38 = \(2 \times 19\), 60 = \(2^2 \times 3 \times 5\). HCF = 2.
(l) 45 = \(3^2 \times 5\), 18 = \(2 \times 3^2\), 36 = \(2^2 \times 3^2\). HCF = \(3^2 = 9\).
(m) 26 = \(2 \times 13\), 43 is prime. HCF = 1.
In simple words: Find the greatest common factor that divides all the given numbers evenly.

Exam Tip: Prime factorization method is foolproof for large numbers.

Please refer to attached file for CBSE Class 5 Maths Revision Worksheets (1)