≡
Online Test      LOGIN      SIGN UP
Forgot your password?

  • Class 6 Maths Study Material

An Educational platform for Preparation and Practice Class 6. Kidsfront provide unique pattern of learning Maths with free online comprehensive study material in the form of QUESTION & ANSWER for each Chapter of Maths for Class 6. This study material help Class 6, Maths students in learning every aspect of Playing with Numbers . Students can understand Playing with Numbers concept easily and consolidate their learning by doing Online Practice Tests on Maths, Playing with Numbers chapter repeatedly till they excel in Class 6, Playing with Numbers . Free ONLINE PRACTICE TESTS on Class 6, Playing with Numbers comprise of Hundreds of Questions on Playing with Numbers , prepared by the highly professionals team. Every repeat test of Playing with Numbers will have new set of questions and help students to prepare themselves for exams by doing unlimited Online Test exercise on Playing with Numbers . Attempt ONLINE TEST on Class 6,Maths, Playing with Numbers in Academics section after completing this Playing with Numbers Question Answer Exercise.


Unique pattern

  • Topic wise: Playing with Numbers preparation in the form of QUESTION & ANSWER.
  • Evaluate preparation by doing ONLINE TEST of Class 6, Maths, Playing with Numbers .
  • Review performance in PRACTICE TEST and do further learning on weak areas.
  • Attempt repeat ONLINE TESTS of Maths Playing with Numbers till you excel.
  • Evaluate your progress by doing ONLINE MOCK TEST of Class 6, Maths, All TOPICS.


Playing with Numbers
The least number which when divided by 6, 15 and 18 leaves a remainder 5 in each case is

a) 100
b) 95
c) 90
d) 80


✔Answer
✔Solution
Correct Answer Is : 95
Solution Is : LCM of 6, 15 and 18 = 90 Now, the required least number that leaves remainder 5 in each case = 90 + 5 = 95
The HCF and LCM of the numbers 30 and 45 are in the ratio

a) 2:3
b) 6:1
c) 7:2
d) 7:5


✔Answer
✔Solution
Correct Answer Is : 6:1
Solution Is : Factors of 30 = 2 x 3 x 5 Factors of 45 = 3 x 3 x 5 LCM of 30 and 45 = 2 x 3 x 3 x 5 = 90 and HCF of 30 and 45 = 3 x 5 = 15 Required ratio of LCM and HCF = 90 : 15 = 6 : 1
The HCF of two numbers is 16 and their LCM is 160. If one of the number is 80, then the other number is

a) 20
b) 28
c) 32
d) 38


✔Answer
✔Solution
Correct Answer Is : 32
Solution Is : Let the other number be n. We know that, Product of the numbers = HCF x LCM Therefore, 80 x n = 16 x 160 or, n = 32
The number, divisible by 9, is

a) 327
b) 3527
c) 8253
d) 8564


✔Answer
✔Solution
Correct Answer Is : 8253
Solution Is : If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9. The sum of the digits of 8253 = 8 + 2 + 5 + 3 = 18, divisible by 9.
The number, divisible by 5, is

a) 3527
b) 3614
c) 4399
d) 37595


✔Answer
✔Solution
Correct Answer Is : 37595
Solution Is : A number which has either 0 or 5 in its ones place is divisible by 5. Since, the number 37595 has 5 is in ones place thus it is divisible by 5.
Sum of two prime numbers (except 2) is always

a) Divisible by 2.
b) Divisible by 4.
c) Odd
d) Prime


✔Answer
✔Solution
Correct Answer Is : Divisible by 2.
Solution Is : We know that all the prime numbers except 2 like 3, 5, 7, 11, and so on are odd numbers. So, the sum of any two prime numbers other than 2 is an even number. For example, 3 + 5 = 8 and 7 + 11 = 18. An even number is always divisible by 2.
All the factors of 63 are

a) 1, 3, 7, 9, 21 and 63.
b) 3, 7, 9, 21 and 63.
c) 1, 3, 7, 9 and 21.
d) 3, 7, 9 and 21.


✔Answer
✔Solution
Correct Answer Is : 1, 3, 7, 9, 21 and 63.
Solution Is : A factor of a number is an exact divisor of that number. Factors of 63 are 1, 3, 7, 9, 21 and 63.
The numbers other than 1, with only factors namely 1 and the number itself, are known as

a) Composite number
b) Even number
c) prime number
d) Odd number


✔Answer
Correct Answer Is : prime number
Solution Is :
The sum of all prime numbers from 1 to 20 is

a) 75
b) 76
c) 77
d) 78


✔Answer
✔Solution
Correct Answer Is : 77
Solution Is : Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19 Therefore, the sum of all prime numbers from 1 to 20 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77
1023 and 1078 both are divisible by 11. Therefore 1023 + 1078 will be divisible by

a) 3
b) 7
c) 13
d) 11


✔Answer
Correct Answer Is : 11
Solution Is :
PREVIOUS

Other Topics

Kids Learning

Academics

Competitive Exams

Govt. Jobs Exams

Copyright © 2024 Kidsfront.com | Terms Of Use

script type="text/javascript">