RS Aggarwal Class 6 Maths Chapter 2 - Factors and Multiples EXERCISE 2F

 

RS Aggarwal 2021-2022 for Class 6 Maths Chapter 2 - Factors and Multiples 

EXERCISE 2F

OBJECTIVE QUESTIONS

Mark (✓) against the correct answer in each of the following: 

1. Which of the following numbers is divisible by 3?  

    (a) 24357806 (b) 35769812 (c) 83479560 (d) 3336436

Solution: (c) is correct

By the divisibility rule of 3 :-

A number is divisible by 3 if the sum of its digits is divisible by 3.

(a) Consider the number 24357806.

Sum of its digits = 2 + 4 + 3 + 5+ 7 + 8 + 0 + 6 = 35, which is not divisible by 3.

So, 2357806 is not divisible by 3.

(b) Consider the number 35769812.

Sum of its digits = 3 + 5 + 7 + 6 +9 + 8 + 1 + 2 = 41, which is not divisible by 3.

So, 35769812 is not divisible by 3.

(c) Consider the number 83479560.

Sum of its digits = 8 + 3 + 4+ 7 + 9 + 5 + 6 + 0 = 42, which is divisible by 3.

So, 2357806 is divisible by 3. (d) Consider the number 3336433.

Sum of its digits = 3 + 3 +3 + 6 +4 + 3 +3 = 25, which is not divisible by 3.

So, 3336433 is not divisible by 3.

2. Which of the following numbers is divisible by 9? 

    (a) 8576901 (b) 96345210 (c) 67594310 (d) none of these

Solution: (a) is correct

By the divisibility rule of 9 :-

A number is divisible by 9 if the sum of its digits is divisible by 9.

(a) Consider the number 8576901. 

Sum of its digits = 8 + 5 +7 + 6 + 9+ 0 + 1 = 36, which is divisible by 9.

So, 8576901 is divisible by 9.

(b) Consider the number 96345210.

Sum of its digits = 9 + 6 + 3+ 4 + 5+ 2 + 1 + 0 = 30, which is not divisible by 9.

So, 96345210 is not divisible by 9.
 
(c) Consider the number 67594310. 

Sum of its digits = 6 + 7 + 5 + 9 + 4 + 3 + 1 + 0 = 35, which is not divisible by 9.

So, 67594310 is not divisible by 9.

3. Which of the following numbers is divisible by 4?

    (a) 78653234 (b) 98765042 (c) 24689602 (d) 87941032

Solution: (d) is correct

By the divisibility rule of 4 :-

 A number is divisible by 4 if the number formed by its digits in the tens and ones places is divisible by 4.

(a) 78653234

Consider the number 78653234.

Here, the number formed by the tens and the ones digit is 34, which is not divisible by 4.

Therefore, 78653234 is not divisible by 4. (b) 98765042

Consider the number 98765042.

Here, the number formed by the tens and the ones digit is 42, which is not divisible by 4.

Therefore, 98765042 is not divisible by 4. (c) 24689602

Consider the number 24689602.

Here, the number formed by the tens and the ones digit is 02, which is not divisible by 4.

Therefore, 24689602 is not divisible by 4 (d) 87941032

Consider the number 87941032.

Here, the number formed by the tens and ones digit is 32, which is divisible by 4.

Therefore, 87941032 is divisible by 4.

4. Which of the following numbers is divisible by 8?

   (a) 96354142 (b) 37450176 (c) 57064214 (d) none of these

Solution: (d) is correct

By the divisibility rule of 8 :-

A number is divisible by 8 if the number formed by its digits in hundreds, tens and ones places is divisible by 8.

(a) 96354142

Consider the number 96354142.

Here, the number formed by the digits in hundreds, tens and ones places is 142, which is clearly divisible by 8.

Therefore, 96354142 is not divisible by 8.

 (b) 37450176

Consider the number 37450176.

The number formed by the digits in hundreds, tens and ones places is 176, which is clearly divisible by 8.

Therefore, 37450176 is divisible by 8.

(c) 57064214

Consider the number 57064214.

Here, the number formed by the digits in hundreds, tens and ones places is 214, which is clearly not divisible by 8.

Therefore, 57064214 is not divisible by 8.

5. Which of the following numbers is divisible by 6?

    (a) 8790432 (b) 98671402 (c) 85492014 (d) none of these

Solution: (a) and (c) is correct

By the divisibility rule of 6 :-

A number is divisible by 6, if it is divisible by both 2 and 3.

(a) 8790432

Consider the number 8790432.

The number at the ones place is 2.

Therefore, 8790432 is divisible by 2.

Now, the sum of its digits (8+7+9+0+2+3+2) is 33. Since 33 is divisible by 3, we can say that 8790432 is also divisible by 3.

Since 8790432 is divisible by both 2 and 3, it is also divisible by 6.

(b) 98671402

Consider the number 98671402.

The number at the ones place is 2.

Therefore, 98671402 is divisible by 2.

Now, the sum of its digits (9+8+6+7+1+4+0+2) is 37. Since 37 is not divisible by 3, we can say that 98671402 is also not divisible by 3.

Since 98671402 is not divisible by both 2 and 3, it is not divisible by 6.

(c) 85492014

Consider the number 85492014.

The number at the ones place is 4.

Therefore, 85492014 is divisible by 2.

Now, the sum of its digits (8+5+4+9+2+0+1+4) is 33. Since 33 is divisible by 3, we can say that 85492014 is also divisible by 3.

Since 85492014 is divisible by both 2 and 3, it is also divisible by 6.

6. Which of the following numbers is divisible by 11?

    (a) 3333333 (b) 1111111 (c) 2222222 (d) none of these

Solution: (c)  is correct

By the divisibility rule of 11 :-

A number is divisible by 11, if the difference of the sum of its digits in odd places and the sum of the digits in even places (starting from ones place) is either 0 or a multiple of 11.

(a) 3333333

Consider the number 3333333.

Sum of  odd places digits (3 + 3 + 3 + 3) = 12

Sum of  even places digits (3 + 3 + 3) = 9

There difference is = 12 − 9 = 3, which is not equal to '0' or a multiple of 11.

So, 3333333 is not divisible by 11.

(b) 1111111

Consider the number 1111111.

Sum of  odd places digits (1 + 1 + 1 + 1) = 4

Sum of  even places digits (1 + 1 + 1) = 3

There difference is  = 4 − 3 = 1, which is not equal to '0' or a multiple of 11.

So, 1111111 is also not divisible by 11.

(c) 22222222

Consider the number 22222222.

Sum of  odd places digits  (2 + 2 + 2 + 2)= 8

Sum of  even places digits  (2 + 2 + 2 + 2) = 8

There difference is = 8 − 8 = 0, which is  equal to '0' .

So, 22222222 is also divisible by 11.

7. Which of the following is a prime number?

    (a) 81 (b) 87 (c) 91 (d) 97

Solution: (d) is correct

(a) 81 is not a prime number because 81 has more than two factor, like 3, 9, 27 etc.

(b) 87 is not a prime number because 87 has more than two factor, like 1, 3, 29, 87 etc.

(c) 91 is not a prime number because 91 has more than two factor, like 1, 7, 13, 91 etc..

(d) 97 is a prime number because 91 has only two factor, 1 and 97.

8. Which of the following is a prime number?

    (a) 117 (b) 171 (c) 179 (d) none of these

Solution: (c) is correct

(a) 117 is not a prime number because 117 has more than two factor, like 3, 13, 39 etc.

(b) 171 is not a prime number because 171 has more than two factor, like 3, 9, 19, 57 etc..

(c) 179 is prime number because 179 has only two factor, 1 and 179

9. Which of the following is a prime number?

    (a) 323 (b) 361 (c) 263 (d) none of these

Solution: (c) is correct

(a) 323 is not a prime number because 323 has more than two factor, like 1, 17, 19 etc.

(b) 361 is not a prime number because 361 has more than two factor, like 1, 19, 361 etc..

(c) 263 is prime number because 263 has only two factor, 1 and 263

10. Which of the following are co-primes

    (a) 8, 12 (b) 9, 10 (c) 6, 8 (d) 15, 18

Solution: (b) is correct

(a) 8, 12 are not co-primes as they have a common factor 4.

(b) 9, 10 are co-primes as they do not have a common factor.

(c) 6, 8 are not co-primes as they have a common factor 2.

(d)15,18 are not co-primes as they have a common factor 3.

11. Which of the following is a composite number

    (a)23 (b) 29 (c) 32 (d) none of these

Solution: (c) is correct

(a) 23 is not a composite number because 23 has only two factor, like 1 and 23.

(b) 29 is not a composite number because 29 has only two factor, like 1 and 29.

(c) 32 is a composite number because 32 has more than two factor, like 1, 2, 4, 8, 16 etc.

12. The HCF of 144 and 198 is

    (a) 9 (b) 12 (c) 6 (d) 18

Solution: (d) is correct

144 = 2 × 2 × 2 × 2 × 3 × 3

198 = 2 × 3 × 3 × 11 

Here, HCF is 2 × 3 × 3 = 18

13. The HCF of 144, 180 and 192 is 

    (a) 12 (b) 16 (c) 18 (d) 8

Solution: (a) is correct

144 = 2 × 2 × 2 × 2 × 3 × 3

180 = 2 × 2 × 3 × 3 × 5

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Here, HCF is 2 × 2 × 3 = 12

14. Which of the following are co-primes? 

    (a) 39, 91 (b) 161, 192 (c) 385, 462 (d) none of these

Solution: (b) is correct

(a) 39 and 91 are not co-primes as 39 and 91 have a common factor, i.e. 13.

(b) 161 and 192 are co-primes as 161 and 192 have no common factor other than 1.

 (c) 385 and 462 are not co-primes as 385 and 462 have common factors 7 and 11.

15. `frac\{289}{391}`  when reduced to the lowest terms is

    (a) `frac\{11}{23}`         (b) `frac\{13}{31}`         (c) `frac\{17}{31}`         (d) `frac\{17}{23}` 

Solution: (d) is correct

Because HCF of 289 and 391 is 17

and `frac\{289÷17}{391÷17}` = `frac\{17}{23}`

16. The greatest number which divides 134 and 167 leaving 2 as remainder in each case is

    (a) 14 (b) 17 (c) 19 (d) 33

Solution: (d)  is correct

First, we will subtract 2 from each of the numbers.

167 − 2 = 165

134 − 2 = 132

Now, any of the common factors of 165 and 132 will be the required divisor.

On factorising:

165 =  3 × 5 × 11

132 = 2 × 2 × 3 × 11

Their common factors are 11 and 3.

So, 3 × 11 = 33 is the required divisor.

17. The LCM of 24, 36, 40 is

    

    (a) 4 (b) 90 (c) 360 (d) 720

Solution: (c) is correct

224,36,40212,18,20  26,   9, 10   33,  9,   5     31,  3,   5     51,  1,   5        1,  1,   1    $\begin{array}{l}2\overline{)24,36,40}\\ 2\overline{)12,18,20}\\ 2\overline{)\text{6, 9, 10 }}\\ 3\overline{)\text{3, 9, 5 }}\\ 3\overline{)\text{1, 3, 5 }}\\ 5\overline{)\text{1, 1, 5 }}\\ \text{ 1, 1, 1 }\end{array}$

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5
            = 360           

18. The LCM of 12, 15, 20, 27 is

 

   (a) 270 (b) 360 (c) 480 (d) 540

Solution: (c) is correct

212,15,20,2726,15,10,27  33,15, 5, 27   31, 5, 5,  9   31, 5, 5,  3   51, 5, 5,  1        1, 1, 1, 1    $\begin{array}{l}2\overline{)12,15,20,27}\\ 2\overline{)6,15,10,27}\\ 3\overline{)\text{3,15, 5, 27 }}\\ 3\overline{)\text{1, 5, 5, 9 }}\\ 3\overline{)\text{1, 5, 5, 3 }}\\ 5\overline{)\text{1, 5, 5, 1 }}\\ \text{ 1, 1, 1, 1 }\end{array}$

L.C.M. = 2 × 2 × 3 × 3 × 3 × 5 = 540

19. The smallest number which when diminished by 3 is divisible by 14,  28, 36 and 45 is

    (a)12             (b)57 (b) 1260 (c) 1263 (d) none of these

Solution: (c) is correct

L.C.M. of the three numbers = 2 × 2 × 3 × 3 × 5 × 7
                                              = 1260
∴ Required number = 1260 + 3 = 1263

20. The HCF of two co-primes is 

    (a) the smaller number (b) the larger number

   (c) 1        (d) none of these

Solution: (c) is correct

H.C.F. of two co-primes is 1.

This is because two co-prime numbers do not have any common factor.

For example, 15 and 16 are co-primes.

Their H.C.F. is 1.

21. If a and b are co-primes, then their LCM is 

    

    (a) 1 (b) `frac\{a}{b}` (c) ab (d) none of these

Solution: (c) is correct

If a and b are co-primes then their LCM will be ab.

For example, 10 and 9 are co-primes.

L.C.M. of 10 and 9 is 10 × 9. 

22. The product of two numbers is 2160 and their HCF is 12. The LCM of these numbers is

    (a) 12 (b) 25920 (c) 180 (d) none of these

Solution: (c) is correct

Here, H.C.F. = 12

Product of two numbers = 2160

 We know:

L.C.M. × H.C.F. = Product of the two numbers

 LCM = `frac\{2160}{H.C.F}`

          = `frac\(2160}{12}`

         = 180

23. The H.C.F of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, the other number is

    (a) 290 (b) 435 (c) 5     (d) none of these

Solution: (b) is correct

One of the numbers = 725.

H.C.F. = 145

L.C.M. = 2175

We know:

L.C.M. × H.C.F. = Product of the two numbers

∴ Product of the two numbers = 145 × 2175

                                               = 315375

∴ Other number = `frac\{315375}{725}`

                           =  435

24. The least number divisible by each of the numbers 15, 20, 24,32 and 36 is 

    (a) 1660 (b) 2880 (c) 1440         (d) none of these

Solution: (c) is correct

215,20,24,32,36215,10,12,16,18  215, 5,  6 , 8, 9   215, 5,  3,  4, 9   215, 5,  3,  2, 9  315, 5,  3,  1, 9 3 5,  5, 1,  1, 3   5 5,  5, 1,  1, 1        1,  1,  1,  1,  1 $\begin{array}{l}2\overline{)15,20,24,32,36}\\ 2\overline{)15,10,12,16,18}\\ 2\overline{)\text{15, 5, 6 , 8, 9 }}\\ 2\overline{)\text{15, 5, 3, 4, 9 }}\\ 2\overline{)\text{15, 5, 3, 2, 9 }}\\ 3\overline{)\text{15, 5, }}\end{array}$

$\begin{array}{l}\overline{)\text{ 3, 1, 9 }}\\ 3\overline{)\text{ 5, 5, 1, 1, 3 }}\\ \text{5}\overline{)\text{ 5, 5, 1, 1, 1 }}\\ \text{ 1, 1, 1, 1, 1 }\end{array}$

L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5

            = 1440 

25. Three bells toll together at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?

    (a)1 hour (b) 90 (c) 360 (d) 3 hours

Solution: (d) is correct


The L.C.M. of 9, 12 and 15 will give us the minutes after which the bells will next toll together.

29,12,1529,6,15  39, 3, 15   33, 1,  5     51,  1,   5        1,  1,   1    $\begin{array}{l}2\overline{)9,12,15}\\ 2\overline{)9,6,15}\\ 3\overline{)\text{9, 3, 15 }}\\ 3\overline{)\text{3, 1, 5 }}\\ 5\overline{)\text{1, 1, 5 }}\\ \text{ 1, 1, 1 }\end{array}$

L.C.M. = 2 × 2 × 3 × 3 × 5
            = 180

So,the bells will toll together after 180 min.

On converting into hours:

180/60 = 3 hours