# DAV Class 6 Maths Chapter 1 Brain Teasers Solutions

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.

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.