RS Aggarwal Class 6 Maths Test Paper Chapter 2 - Factors and Multiples

 

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

Rs Aggarwal Class 6 Math Solution Test Paper Chapter 2 Factors and Multiples is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. 

Class 6 RS Aggarwal Maths Chapter 2 - Factors and Multiples Solution Test Paper

TEST PAPER-2

A. 1. Test the divisibility of 5869473 by 11. 

Solution:  

A number is divisible by 11 if the the difference of the sums of the digits at the odd places and that at the even places (starting from ones place) is either 0 or a multiple of 11.

Sum of the digits at even places = 7 + 9 + 8 = 24

Sum of the digits in odd places = 3 + 4 + 6 + 5 = 18

Difference = 24 - 18 = 6

Since 6 is not divisible by 11, 5869473 is not divisible by 11.

   2. Test the divisibility of 67529124 by 8.

Solution:  

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

Since the digits at the hundred’s, ten’s and unit places are 124, which is not divisible by 8, 67529124 is not divisible by 8. 

   3. On dividing 5035 by 31, the remainder is 13. Find the quotient. 

Solution: 

Remainder is 13

∴ Number exactly divisible by 31 =  5035 − 13

                                            = 5022

Quotient = `frac\{Dividend}{Divisor}`

        =`frac\{5022}{31}` = 162

   4. The HCF of two numbers is 15 and their product is 1650. Find their LCM.

Solution:  

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

Product of the two numbers = 1650

H.C.F. = 15

Required L.C.M. = `frac\{1650}{15}`

                          =110 

                 

   5. Find the least 5-digit number which is exactly divisible by 20, 25, 30.

Solution: 

Least 5 digit number = 10000

First, we find LCM of 20, 25and 30.

LCM of 20, 25and 30 = 5 x 2 x 2 x 3 x 5 = 300

But we want the least five digit number which is divisible by 20, 25, 30.

So, we will multiply the L.C.M. by a number that makes it the least five digit number divisible by 20, 25, 30.

300×31 = 9300

300×32 = 9600

300×33 = 9900

300×34 = 10200

 So, the least five digit number divisible by 20, 25, 30 is 10200.

   6. Find the largest number which divides 630 and 940 leaving remainders 6 and 4                        respectively. 

Solution: 

First, we subtract 6 and 4 from 630 and 940 respectively

so, 630 - 6 = 642

      940 - 4 = 936

Now, we find HCF of 642 and 936

624 = 2 × 2 × 2 × 2 × 3 × 13

936 = 2 × 2 × 2 × 3 × 3 × 13

H.C.F. of 624 and 936 = 8 × 3 × 13

                                     = 312

So, 312 is the greatest number that divides 630 and 940, leaving 6 and 4 as the respective remainders.

   7. Find the least number which when divided by 16, 36 and 40 leaves 5 as remainder in             each case.

Solution: 

On subtracting 5 from each number:

16 − 5 = 11

36 − 5 = 31

40 − 5 = 35

The required number will be the least common multiple of 11, 31 and 35.

L.C.M. of 11, 31 and 35 = 11×31×35

                                       = 11935

This is because they do not have any factor in common.

So, 11935 is the required number.

   8. Write all prime numbers between 50 and 100.

Solution: 

The prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 .

   9. Write seven consecutive composite numbers less than 100 having no prime number             between them. 

Solution: 

Seven consecutive composite numbers less than 100 having no prime number between them are 90, 91, 92, 93, 94, 95 and 96.

  10. Can two numbers have 12 as their HCF and 512 as their LCM ? Justify your answer.

Solution: 

No, they cannot have 512 as their L.C.M.

We know that the H.C.F. is one of the factors of the L.C.M. Here, 3, which is a factor of 12, is not a factor of 512. 

B. Mark (✔) against the correct answer in each of the following:

  11. Which of the following are co-primes? 

      (a) 91 and 72     (b) 34 and 51     (c) 21 and 36      (d) 15 and 20

Solution: 

The correct option is (a).

The H.C.F. of 72 and 91 is 1.

So, they are co-primes.

Option (b) is not correct because 34 and 51 have 17 as their H.C.F.

Option (c) is not correct because 21 and 56 have 3 as their H.C.F.

Option (d) is not correct because 15 and 20 have 5 as their H.C.F.

12. The LCM of two co-prime numbers is their

      (a) sum        (b) difference      (c) product      (d) quotient

Solution: 

The correct option is (c).

The L.C.M of two co-prime numbers is their product.

13. The number which is neither prime nor composite is 

      (a) 0       (b) 1       (c) 2       (d) 3

Solution: 

The correct option is (b).

1 is neither prime nor composite.

Option (a) is incorrect because composite numbers are defined for positive numbers, but 0 is neither a positive number nor a negative number.

Option (c) ​is not correct because 2 is a prime number.

Option (d) is not correct because 3 is a prime number.​

14. What least number should be replaced for * so that the number 67301*2 is exactly divisible by 9?

      (a) 5       (b) 6       (c) 7       (d) 8

Solution: 

The correct option is (d).

6 + 7 + 3 + 0 + 1 + * + 2 = 19 + *

8 is the least number that should be added to 19 such that number will be divisible by 9.

Sum of the digits:

6 + 7 + 3 + 0 + 1 + 8 + 2 = 27

27 is divisible by 9.

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

      (a) 67821     (b) 78134      (c) 87432       (d) none of these

Solution: 

The correct option is (c).

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

Since the ones digit of 87432 is 2, it is divisible by 2.

Now, Sum of digits = 8 + 7 + 4 + 3 + 2 = 24

and 24 is divisible by 3.

Hence, 87432 is divisible by 6 because it is divisible by both 2 and 3.

Option (a) is incorrect because 67821 is not divisible by 2.

Option (b)  is incorrect because 78134 is not divisible by 3.

7 + 8 + 1 + 3 + 4 = 23

23 is not divisible by 3.

16. Which of the following is a prime number?

      (a) 143         (b) 131          (c) 147           (d) 161

Solution: 

The correct option is (b).

To find a prime number between 100 and 200, we have to check whether the given number is divisible by any prime number less than 15. If yes, it is not prime, otherwise it is.

By examining, we find that 131 is a prime number.

17. `frac\{289}{391}` when reduced to lowest term is

(a)  `frac\{13}{17}`      (b)  `frac\{17}{19}`      (c)  `frac\{17}{23}`      (d)  `frac\{17}{21}`

Solution: 

The correct option is (c).

18. Every counting number has an infinite number of

     (a) factors      (b) multiples       (c) prime factor       (d) none of these

Solution: 

The correct option is (c).

Every counting number has an infinite number of multiples. 

If n is a counting number, its multiples are 1n, 2n, 3n.... 

C. 19. Fill in the blanks.

(1) 1 is neither ……..nor …….

Solution: - prime   nor   composite 

(ii) The smallest prime number is ……..

Solution:  - 2

(iii) The smallest composite number is .......

Solution:  - 4

(iv) The HCF of two consecutive odd numbers is ……..

Solution: - 1

(v) Two perfect numbers are ...... and.......

Solution:  - 6  and 28

D. 20.  Write 'T' for True and 'F' for false statement.

    (i) Every prime number is odd.

    (ii) Every even number is composite.

    (iii) The sum of two odd numbers is always odd.

    (iv) The sum of two even numbers is always even.

    (v) The HCF of two given numbers is always a factor of their LCM.

Solution:

(i) F

2 is an even prime number.

(ii) F

2 is an even number, but it is not composite.

(iii) F

The sum of two odd numbers is always even. For example, 11 and 13 are odd numbers, but their sum, i.e. 23, is an even number.

(iv) T

The sum of two even numbers is always even. For example, 6 and 14 are even numbers, and their sum, i.e. 20, is an even number.

(v) T

For example, 4 and 6 are two numbers whose H.C.F is 2 and L.C.M. is 12, but 2 is a factor of 12.