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Class 6 Whole Numbers NCERT Solutions

Chapter 2: Whole Numbers

NCERT Solutions

Question 1.Write the next three natural numbers after 10999.

Sol:

10,999 + 1 = 11,000
11,000 + 1 = 11,001
11,001 + 1 = 11,002
Hence, the next three natural numbers after 10,999 are 11,000; 11,001; 11,002.


Question 2.Write the three whole numbers occurring just before 10001.

Sol:

10,001 – 1 = 10,000
10,000 – 1 = 9,999
9,999 – 1 = 9,998
Hence, the three whole numbers occurring just before 10,001 are 10,000; 9,999; 9,998

Question 3.Which is the smallest whole number?

Sol:

0 is the smallest whole number.


Question 4.How many whole numbers are there between 32 and 53?

Sol:

Since, 53 – 32 – 1 = 20
∴ There are 20 whole numbers between 32 and 53.


Question 5.Write the successor of :
(A) 2440701
(B) 100199
(C) 1099999
(D) 2345670

Sol:

 (A) Successor of 2440701 is
2440701 + 1 = 2440702
(B) Successor of 100199 is
100199 + 1 = 100200
(C) Successor of 1099999 is
1099999 + 1 = 1100000
(D) Successor of 2345670 is
2345670 + 1 = 2345671


Question 6.Write the predecessor of :
(A) 94
(B) 10000
(C) 208090
(D) 7654321

Sol:

 (A) The predecessor of 94 is, 94 – 1 = 93
(B) The predecessor of 10000 is, 10000 – 1 = 9999
(C) The predecessor of 208090 is,
208090 – 1 = 208089
(D) The predecessor of 7654321 is,
7654321 – 1 = 7654320


Question 7.In each of the following pairs of numbers, a state which the whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(A) 530, 503
(B) 370, 307
(C) 98765, 56789
(D) 9830415, 10023001

Sol:

 (A) 503 appears on the left side of 530 on the number line. So, 530 > 503
(B) 307 appears on the left side of 370 on the number line. So, 370 > 307
(C) 56789 appears on the left side of 98765 on the number line. So, 98765 > 56789
(D) 9830415 appears on the left side of 10023001 on the number line. So, 9830415 < 10023001


Question 8. Which of the following statements are true (T) and which are false (F)?
(A) Zero is the smallest natural number.
(B) 400 is the predecessor of 399.
(C) Zero is the smallest whole number.
(D) 600 is the successor of 599.
(E) All-natural numbers are whole numbers.
(F) All whole numbers are natural numbers.
(G) The predecessor of a two-digit number is never a single-digit number.
(H) 1 is the smallest whole number.
(I) The natural number 1 has no predecessor.
(J) The whole number 1 has no predecessor.
(K) The whole number 13 lies between 11 and 12.
(L) The whole number 0 has no predecessor.
(M) The successor of a two-digit number is always a two-digit number.

Sol:

 (A) False: Since zero is not a natural number.
(B) False: Since 400 is the successor of 399.
(C) True
(D) True
(E) True
(F) False: All whole numbers are not natural numbers as 0 is a whole number but not a natural number.
(G) False: Since, the predecessor of a two-digit number i.e., 10 is a single-digit number i.e., 9.
(H) False: 0 is the smallest whole number.
(I) True
(J) False: Since 0 is the predecessor of 1 in whole numbers.
(K) False: Since, 13 lies between 12 and 14.
(L) True
(M) False: Since, the successor of two-digit number i.e., 99 is a three-digit number i.e., 100.

Question 9.Find the sum by suitable rearrangement:
(A) 837 + 208 + 363
(B) 1962 + 453 + 1538 + 647

Sol:

 (A) 837 + 208 + 363 = (837 + 363) + 208
= 1200 + 208 = 1408
(B) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100 = 4600


Question 10.Find the product by suitable rearrangement:
(A) 2 × 1768 × 50
(B) 4 × 166 × 25
(C) 8 × 291 × 125
(D) 625 × 279 × 16
(E) 285 × 5 × 60
(F) 125 × 40 × 8 × 25

Sol:

 (A) 2 × 1768 × 50 = (2 × 50) × 1768
= 100 × 1768 = 176800
(B) 4 × 166 × 25 = (4 × 25) × 166
= 100 × 166 = 16600
(C) 8 × 291 × 125 = (8 × 125) × 291
= 1000 × 291 = 291000
(D) 625 × 279 × 16 = (625 × 16) × 279
= 10000 × 279 = 2790000
(E) 285 × 5 × 60 = 285 × (5 × 60)
= 285 × 300 = 85500
(F) 125 × 40 × 8 × 25 = (125 × 8) × (40 × 25)
= 1000 × 1000 = 1000000

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