The DAV Class 6 Maths Book Solutions Pdf and **DAV Class 6 Maths Chapter 1 Brain Teasers **Solutions of Natural Numbers and Whole Numbers offer comprehensive answers to textbook questions.

## DAV Class 6 Maths Ch 1 Brain Teasers Solutions

Question 1. A.

Tick (✓) the correct answer.

(a) Which of the following is meaningless?

(i) XLVI

(ii) ICVII

(iii) XML

(iv) XLIX

Solution:

(ii) ICVII is meaningless

(b) The greatest 2-digit number exactly divisible by 17 is –

(i) 68

(ii) 91

(iii) 85

(iv) 97

Solution:

(iii) 85

So, 85 is the greatest 2-digit number, exactly divisible by 17.

(c) The smallest 5-digit number formed by using the digits 3, 0, 1 (Repetition of digits allowed) is-

(i) 10003

(ii) 10013

(iii) 13000

(iv) 00013

Solution:

(i) 10003

(d) The estimated value of 36 + 71 – 55 is –

(i) 40

(ii) 50

(iii) 70

(iv) 150

Solution:

(ii) 36 + 71 – 55

We round off 36 to the nearest ten which is 40. Similarly, 71 and 55 can be rounded off to the nearest ten as 70 and 60 respectively.

∴ 36 + 71 – 55 can be written as

40 + 70 – 60 = 110 – 60 = 50

(e) Which of the following is not a natural number?

(i) 3 + 5 – 2

(ii) 4 × 0

(iii) 8 ÷ 8

(iv) 6 – 3 + 1

Solution:

(ii) 4 × 0 = 0, which is not a natural number.

B. Answer the following questions.

(a) How many millions make 3 crores?

Solution:

1 crore = 10000000

∴ 3 crores = 30000000

= 30 × 1000000

= 30 × 1 million

∴ 30 millions make 3 crores.

(b) Which whole number does not have a predecessor?

Solution:

Zero (0).

(c) What is the estimated value of 786 × 1385?

Solution:

786 × 1385

786 can be rounded off to the nearest hundred as 800 and

1385 can be rounded off to the nearest hundred as 1400

∴ 786 × 1385 can be written as 800 × 1400 = 11,20,000

(d) What is the value of 125 × 4 × 25 × 8?

Solution:

125 × 4 × 25 × 8

We can arrange these numbers as follows:

(125 × 8) × (4 × 25) = 1000 × 100

= 1,00,000

(e) What is the difference between the place value and face value of 8 in 38, 46, 197?

Solution:

Place value of 8 in 38, 46, 197 is = 8,00,000

Face value of 8 in 38, 46, 197 is = 8

∴ Difference between place value and face value = 8,00,000 – 8 = 7,99,992.

Question 2.

Write the greatest 6-digit number using three different digits.

Solution:

The largest. 6-digit number using different digits using 9, 8 and 7 is = 999987

hence the required number = 999987.

Question 3.

Find the smallest and the greatest of each 7-digit and 8-digit number using the digits 5, 0, 4, 1. (digits can be repeated).

Solution:

Smallest 7-digit number = 10,00,045

Greatest 7-digit number = 55,55,4 10

Smallest 8-digit number = 1,00,00,045

Greatest 8-digit number = 5,55,55,410

Question 4.

Find the difference between the largest and the smallest 7-digit numbers formed by using the digits in the number 6427310 (digits should not repeat)

Solution:

Largest 7-digit number = 7643210

Smallest 7-digit number = 1023467

∴ Difference between largest and smallest number = 7643210 – 1023467

= 6619743

Question 5.

Using distributive property, simplify:

223 × 25 × 6 – 223 × 10 × 15.

Solution:

223 × 25 × 6 – 223 × 10 × 15 = 223 (25 × 6 – 10 × 15)

= 223 (150 – 150)

= 223 × 0 = 0

Question 6.

Complete the series:

1, 1, 2, 3, 5, 8, 13, 21, 34 , ___, ___

Solution:

Here, every number is the sum of its two predecessors

∴ the Ist required number = 34 + 21 = 55

and IInd required number = 34 + 55 = 89

Hence, the two required numbers of the series are 55 and 89.

Question 7.

Fill in the blanks in the following magic square:

Solution:

Question 8.

Form the greatest 6-digit number using the digits of prime numbers between 80 and 100.

Solution:

Prime numbers between 80 and 100 are 83, 89, 97.

∴ The digits present in the above prime numbers are 3, 7, 8, 9

∴ The greatest 6 digit number using 3, 7, 8, 9 = 998873

Question 9.

Find the number which is:

(a) The successor of the successor of 304998

(b) The predecessor of the predecessor of the smallest 6-digit number

Solution:

(a) The successor of the successor of

304998 = 304998 + 2 = 305000

(b) The predecessor of the predecessor of the smallest 6 digit 100000

= 100000 – 2 = 99998

Question 10.

Fill in the blanks using Roman Numerals.

(a) CXIX – ___ = XXV

Solution:

XCIV

(b) ___ + XLVI = LXX

Solution:

XXIV

Question 11.

Arrange the following in ascending order.

LVII, XC, XV, LXIV, LXXI, XXIX

Solution:

Ascending order is

XV < XXIX < LVII < LXIV < LXXI < XC

Question 11.

Estimate the following:

(a) 234 + 649 186

Solution:

234 + 649 – 186

Rounding off to the nearest hundred

= 200 + 600 – 200

= 800 – 200

= 600

(b) 9483 – 6321 – 2178

Solution:

9483 – 6321 – 2178

Rounding off to the nearest thousand

9000 – 6000 – 2000

= 9000 – 8000

= 1000

(c) 3284 × 639

Solution:

3284 × 639

Rounding off 3284 to nearest thousand and 639 to nearest hundred

3000 × 600 = 18,00,000

(d) 12345 × 6789

Solution:

12345 × 6789

Rounding off 12345 to the nearest ten thousand and 6789 to the nearest thousand

10000 × 7000 = 7,00,00,00

### DAV Class 6 Maths Chapter 1 HOTS

Question 1.

How many times does the digit 7 occurs if we write all the numbers from 1 to 200?

Solution:

Numbers which have 7 at their ones Place only are 7, 17, 27, 37, 47, 57, 67, 87, 97, 107, 117, 127, 137, 147, 157, 167, 187, 197,

i.e., 7 occurs 18 tunes

Numbers which have 7 at their tens place only are 70, 71, 72, 73, 74, 75, 76, 78, 79, 170, 171, 172, 173, 174, 175, 176, 178, 179, Le., 7 occurs 18 times.

Numbers which have 7 at both their ones and tens place 77, 177, i.e., 7 occurs 4 times.

∴ Total number of times 7 occurs = 18 + 18 + 4 = 40

Question 2.

Write all the 2-digit numbers which when added to 27 get reversed.

Solution:

By hit and trial method, we have 14, 25, 36, 47, 58 and 69 numbers, which when added to 27 get reversed

14 + 27 = 41

25 + 27 = 52

36 + 27 = 63

47 + 27 = 74

58 + 27 = 85

69 + 27 = 96

### DAV Class 6 Maths Chapter 1 Enrichment Questions

Question 1.

Get 100 using four 9’s and some of the symbols like +, -, x, ÷.

Solution:

Using two 9’s to get greatest 2-digit number = 99

Now, we are left with two 9’s.

If we use these two 9’s as 9 ÷ 9 we get 1.

By adding two results,

we get 99 + (9 + 9) = 99 + 1 = 100

Question 2.

A 2-digit number is three times the sum of its digits. Find the number.

Solution:

By hit and trial method, we get

10 × [2] + [7] = 20 + 7 = 27

Sum of digits = 2 + 7 = 9

Three times the sum of digits = 3 × 9 = 27.

Additional Questions:

Question 1.

Convert the following Roman Numerals into Hindu Arabic Numerals.

(i) XLIX

(ii) XCVII

(iii) LXXIX

(iv) XCI

Solution:

(i) XLIX = 40 + 9 = 49

(ii) XCVII = 90 + 7 = 97

(iii) LXXIX = 50 + 10 + 10 + 9 = 79

(iv) XCI = 90 + 1 = 91

Question 2.

Add the following in 2 different ways

(i) 469, 35, 31, 5

Solution:

(i) 469, 35, 31, 5

1st way (469 + 31) + (35 + 5)

= 500 + 40

= 540

IInd way (469 + 5) + (35 + 31)

= 474 + 66 = 540

(ii) 625 + 35 + 65 + 75

Solution:

625 + 35 + 65 + 75

Ist way (625 + 75) + (35 + 65)

= 700 + 100 = 800

IInd way (625 + 35) + (75 + 65)

= 660 + 140 = 800.

Question 3.

Re-arrange and multiply the following

(a) 250 × 8 × 625 × 4

Solution:

250 × 8 × 625 × 4

= (250 × 4) × (8 × 625)

= 1000 × 5000

= 5000000

(b) 200 × 61 × 60 × 10

Solution:

200 × 61 × 60 × 10

= (200 × 10) × (61 × 10)

= 2000 × 610

= 1220000

Question 4.

Find the product using distributive property

(i) 97 × 628

(ii) 352 × 102

(iii) 282 × 999

(iv) 109 × 262

(u) 449 × 995

(vi) 380 × 1005

Solution:

(i) 97 × 628 = (100 – 3) × 628

= 100 × 628 – 3 × 628

= 62800 – 1884

= 60916

(ii) 352 × 102 = 352 × (100 + 2)

= 35200 + 352 × 2

= 35200 + 704

= 35904

(iii) 282 × 999 = 282 × (1000 – 1)

= 282 × 1000 – 282 × 1

= 282000 – 282

= 281718

(iv) 109 × 262 = (100 + 9) × 262

= 100 × 262 + 9 × 262

= 26200 + 2358

= 28558

(v) 449 × 995 = 449 × (1000 – 5)

= 449 × 1000 – 449 × 5

= 449000 – 2245

= 446755

(vi) 380 × 1005 = 380 × (1000 + 5)

= 380 × 1000 + 380 × 5

= 380000 + 1900

= 381900

Question 5.

Solve the following

(i) 650 × 18 + 650 × 48 + 650 × 30 + 650 × 4

Solution:

650 × 18 + 650 × 48 + 650 × 30 + 650 × 4

= 650 × (18 + 48 + 30 + 4)

= 650 × 100

= 65000

(ii) 659 × 45 + 659 × 50 + 659 × 5

Solution:

659 × 45 + 659 × 50 + 659 × 5

= 659 × (45 + 50 + 5)

= 659 × 100

= 65900

Question 6.

Find the smallest 7-digit number using four different digits.

Solution:

The smallest 7-digit number using 0, 1, 2, 3 is = 1000023

Question 7.

How many 5-digit numbers in all do we have?

Solution:

Total number of 5-digit numbers is = 9 × 10 × 10 × 10 × 10 = 90000

Hence required number = 90000

Question 8.

How many times does the digit 7 occur in the unit place when we write the natural numbers from 1 to 100?

Solution:

Numbers are 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 = 10 times

Hence the required answer = 10 times.