Tuesday, 28 August 2018

CBSE Class 6 - Chapter 1 - Number System - Multiple Choice Questions Answers (#cbsenotes)(#eduvictors)

Number System 

CBSE Class 6 - Chapter 1 - Number System - Multiple Choice Questions Answers (#cbsenotes)(#eduvictors)

Multiple Choice Questions (MCQs)


Question 1: The product of the place values of two 2’s in 428721 is

(a) 4
(b) 40000
(c) 400000
(d) 40000000


Question 2:  3 × 10000 + 7 x 1000 + 9 x 100 + 0 x 10 + 4 is the same as

(a) 3794
(b) 37940
(c) 37904
(d) 379409



Question 3: 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




Question 4: The expanded form of the number 9578 is

(a) 9 × 10000 + 5 × 1000 + 7 × 10 + 8 × 1
(b) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1
(c) 9 × 1000 + 57 × 10 + 8 × 1
(d) 9 × 100 + 5 × 100+ 7 × 10 + 8 × 1


Question 5: 5 When rounded off to nearest thousands, the number 85642 is

(a) 85600
(b) 85700
(c) 85000
(d) 86000


Question 6: 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


Question 7: 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


Question 8: One million is equal to

(a) 1 lakh
(b) 10 lakh
(c) 1 crore
(d) 10 crore


Question 9: The greatest number which on rounding off to nearest thousands gives 5000, is

(a) 5001
(b) 5559
(c) 5999
(d) 5499


Question 10: 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



Answers:
1: (c) 400000
First 2 is at ten's place i.e. 2 × 10 = 20
the place value of second 2 = 2 × 10000 = 20000

Product = 20 × 20000 = 400000

2:  (c) 37904
= 3 × 10000 + 7 × 1000 + 9 × 100 + 0 × 10 + 4
= 30000 + 7000 + 900 + 0 + 4
= 37904

3: (d) 1 crore
Greatest 7-digit number = 9999999
= 9999999 + 1 = 10000000  = 1 crore

4: (b) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1

5: (d) 86000

6: (d) 9965
Given digits are 5,9, 2 and 6.
Descending order of the given digits is 9, 6, 5, 2.
Since, we can use any one digit twice, so the required largest 4-digit number is 9965.

7: (c) 5, 86, 95, 376 

8: (b) 10 lakh 
∵ 1 million = 1000000 = 10 lakh


9: (d) 5499
On rounding off to nearest thousands, we have
5001→ 5000 
5559 → 6000 
5999 → 6000 
5499 → 5000

Out of 5001 and 5499, the greatest number is 5499, which gives 5000 on rounding off to
nearest thousands.

10: (c) 6034579 
The digits in the given number 6350947 are 6, 3, 5, 0, 9, 4 and 7. Keeping the digit 6 at ten
lakh’s place, the rest of the digits fill other places like lakh, ten thousands, thousand, hundreds, tens and ones place by decreasing order of remaining number, i.e. 0,
3, 4, 5, 7, 9.
Hence, the required smallest number is 6034579.


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