# HBSE 6th Class Maths Solutions Chapter 1 Knowing Our Numbers InText Questions

Haryana State Board HBSE 6th Class Maths Solutions Chapter 1 Knowing Our Numbers InText Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 1 Knowing Our Numbers InText Questions

Try These (Page 2)

Question 1.
Can you instantly find the greatest number in each row and also which is the smallest number ?
1. 387, 4972, 18, 59785, 750
2. 1473,89423, 100, 5000, 310
3. 1834, 75284, 111, 2333, 450
4. 2853,7691,9999, 12002, 1245
1. G-59785, S-18.
2. G-89423, S- 100.
3. G-75284, S–111.
4. G-12002, S-124.

Try These (Page 3)

Question 1.
Find the greatest and the smallest number :
(a) 4536, 4892, 4370, 4452.
(b) 15623, 15073, 15189, 15800.
(c) 25286, 25245, 25270, 25210.
(d) 6895, 23787, 24569, 24659.
1. G-4892, S-4370.
2. G-15800, S-15073.
3. G-25286, S-25210.
4. G-24659, S-6895.

Try These (Page 3-4)

Question 1.
Use the given digits without repetition and make the greatest and smallest four-digit numbers.
(a) 2, 8, 7, 4 (b) 9, 7, 4, 1
(c) 4, 7, 5, 0 (d) 1, 7, 6, 2,
(e) 5, 4,0,3.
(a) G-8742, S-2478.
(b) G-9741, S-1479.
(c) G-7540, S-4057.
(d) G-7621, S-1267.
(e) G-5430, S-3045.

Question 2.
Now make the greatest and the smallest four-digit numbers by using any one digit twice.
(a) 3,8,7
(b) 9, 0, 5
(c) 0,4,9
(d) 8, 5, 1
(a) G-8873, S-3378.
(b) G-9950, S-5009.
(c) G-9940, S-4009.
(d) G-8851, S-1158.

Question 3.
Make the greatest and the smallest four-digit numbers using any four different digits, with condition as given
(a) Digit 7 is always at one’s place.
Greatest 9867
Smallest 1027
(b) Digit 4 is always at ten’s place.
Greatest 9847
Smallest 1042
(c) Digit 9 is always at hundred’s place.
Greatest 8976
Smallest 1902
(d) Digit 1 is always at thousand’s place.
Greatest 1987
Smallest 1023

Question 4.
Take two digits, say 2 and 3. From them make four-digit npmbers, using both the digits equal number of times.
(a) Which is the largest number ?
(b) Which is the smallest number ?
(c) How many different numbers can you make in all ?
(a) Largest number : 3322
(b) Smallest number : 2233
(c) Different numbers : 3322, 3232, 3223, 2323, 2332, 2233 i.e. 6 in all.

Try These (Page 6)

Question 1.
Arrange the following numbers in ascending order :
(a) 847, 9754, 8320, 571
(b) 9801, 25751, 36501, 38802
(a) 571, 847, 8320, 9754
(b) 9801, 25751, 36501, 38802.

Question 2.
Arrange the following numbers in descending order;
(a) 5000, 7500, 85400, 7861
(b) 1971, 45321, 88715, 92547
(a) 85400, 7861, 7500, 5000
(b) 92547, 88715, 45321, 1971.

Try These (Page 8)

Question 1.
Read and expand the following numbers :
(i) 50000 (ii) 41000 (iii) 47300 (iv) 57630
(v) 29485 (vi) 29085 (vii) 20085 (iv) 20005.

 Number Number Name Expansion (i) 50,000 fifty thousand 5 x 10,000 (ii) 41,000 fourty one thousand 4 x> 10,000 + 1 x 1000 (iii) 47,300 fourty seven thousand three hundred 4 x 10,000 + 7 x 1000 + 3 x 100 (iv) 57,630 fifty seven thousand six hundred thirty 5 x 10,000 + 7 x 1000 + 6 x 100 + 3 x 10 (v) 29,485 twenty nine thousand four hundred eighty five 2 x 10,000 + 9 x 1000 + 4x 100 + 8 x 10 + 5 (vi) 29,085 twenty nine thousand eighty five 2 x 10,000 + 9 x 1000 + 8 x 10 + 5          . (vii) 20,085 twenty thousand eighty-five 2x 10,000 + 8x 10 + 5 (viii) 20,005 twenty thousand five 2 x 10,000 + 5

Try These (Page 9)

Question 1.
Read and expand the following numbers :
(i) 4,57,928 (ii) 4,07,928 (iii) 4,00,829 (iv) 4,00,029

 Number Number Name Expansion (i) 4,57,928 four lakh, fifty-seven thousand nine hundred twenty-eight 4 x 1,00,000 + 5 x 10,000 + 7 x 1000 + 9 x 100 + 2 x 10 + 8 (ii) 4,07,928 fourty lakh, seven thousand nine hundred twenty-eight 4 x 1,00,000 + 7 x 1000 + 9 x 100 + 2 x 10 + 8 (iii) 4,00,829 four lakh eight hundred twenty nine 4 x 1,00,000 + 8 x 100 + 2 x 10 + 9 (iv) 4,00,029 four lakh, twenty-nine 4 x 1,00,000 + 2 x 10 + 9

Try These (Page 10)

Question 1.
What is 10 – 1 = ?
9.

Question 2.
What is 100 -.1 = ?
99.

Question 3.
What is 10,000 – 1 – ?
9999.

Question 4.
What is 1,00,000 – 1 = ?
99999.

Question 5.
What is 1,00,00,000 – 1 = ?
9999999.

Try These (Page 14)

Question 1.
You have the following digits 4, 5, 6, 0, 7 and 8. Using them make 5 numbers each with 6 digits.
(a) Put commas for ease of reading.
(b) Arrange them in ascending and descending order.
(a) 8,76,540; 7,86,540; 6,87,540; 5,78,640; 4,58,760
(b) Ascending order : 4,58,760; 5,78,640; 6,87,540; 7,86,540; 8,76,540
Descending order : 8,76,540; 7,86,540; 6,87,540; 5,78,640; 4,58,760

Question 2.
Take the digits 4, 5, 6, 7, 8 and 9. Make any 3 numbers each with 8 digits. Put commas for ease of reading.
4,56,78,469; 5,64,78,965; 6,45,87,695.

Question 3.
From the digits 3, 0 and 4 make 5 numbers each with 6 digits. Use commas.
3,04,034; 4,03,403; 3,40,430; 3,03,404; 3,34,400.

Try These (Page 16)

Question 1.
How many centimetres make a kilometre ?
1 kilometre (km)
= 1000 x 100 centimetres (cm)
= 1,00,000 cm.

Question 2.
How many milligrams make one kilogram ?
1 kilogram
= 1000 x 1000 milligrams (mg)
= 10,00,000 milligrams (mg)

Question 3.
A box of medicine tablets contains 2,00,000 tablets each weighing 20 mg. What is the total weight of all the tablets in the box in grams ? in kilo¬grams ?
Total weight of all the tablets
= 20 x 2,00,000 mg
= 40,00,000 mg
= 4,000 g (∵ 1000 mg = 1 g)
= 4kg (∵ 1000 g= lkg)

Kilo shows 1000 times greater,
Milli shows 1000 times smaller,
Centi shows 100 times smaller.

Try These (Page 17)

Question 1.
A bus started its journey and reached different places with a speed of 60 km/hr. The journey is shown below :
(i) Find the total distance covered by the bus from A to D.
(ii) Find the total distance covered by the bus from D to G.

(iii) Find the total distance covered by the bus if it starts from A and returns back to A.
(iv) Can you find the difference of distances from C to D and D to E.
(v) Find out the time taken by bus to reach.
(a) AtoB
(b) C to D
(c) E to G
(d) Total journey.
Solution:
(i) Total distance covered by the bus from A to D
= 4170 km + 3410 km + 2160 km
= 9,740 km.

(ii) Total distance covered by the bus from D to G
= 8140 km + 4830 km + 2550 km
= 15,520 km.

(iii) Total distance covered by the bus
= 9740 km + 15520 km + 1290 km = 26,550 km

(iv) Difference of distances from C to D and D to E
= 8140 km – 2160 km = 5980 km.

(v) (a) A to B = 4170 km, 4170 + 60 = 69% Hrs
(b) C to D = 2160 km,
2160 H- 60 = 36 Hrs
(c) E to G = 4830 km + 2550 km, 7380 + 60 = 123 Hrs
(d) Total journey
= 4170 + 3410 + 2160 + 8140 + 4830 + 2550 + 1290
= 26550 km = 442 1/2 Hrs.

Question 2.

 Things Price Raman.s shop The sales during the last year (i) Apples Rs. 40/- kg 2457 kg (ii) Oranges Rs. 30/- kg 3004 kg (iii) Combs Rs. 3 for one 22760 (iv) Tooth brushes Rs. 10 for one 25367 (u) Pencils Re 1 for one 38530 (vi) Note-books Rs. 6 for one 40002 (vii) Soap cakes Rs. 8 for one 20005

(a) Can you find the total weight of apples and oranges Raman sold last year ?
(b) Can you find the total money Raman got by selling apples ?
(c) Can you find the total money Raman got by selling apples and oranges together ?
(d) Make a table showing how much money Raman received from selling each item. Arrange the entries of amount of money received in decreasing order. Find the item which brought him the highest amount. How much is this amount ?
Solution:
(a) Wt. of apples = 2457 kg
Wt. of oranges = 3004 kg
.’. Total weight = 2457 kg + 3004 kg
= 5461 kg.
(b) Total money got by selling apples = Rs. 40 x 2457
= Rs. 98280.

(c) Total money got by selling apples and oranges together
= Rs. 40 x 2457 + Rs. 30 x 3004 = Rs. 98280 + Rs. 90120 = Rs. 188400.

(d)

 Things Price Sale Amount (i) Apples Rs. 40/- kg 2457 kg Rs. 40 x 2457 = Rs. 98,280 (ii) Oranges Rs. 30/– kg 3004 kg Rs. 30 x 3004 = Rs. 90,120 (iii) Combs Rs. 3 for one 22760 Rs. 3 x 22760 = Rs. 68,280 (iv) Tooth-brushes Rs. 10 for one 25367 Rs. 10 x 25367 = Rs. 2,53,670 (v) Pencils Re 1 for one 38530 Rs. 1 x 38530 = Rs. Rs. 38,530 (vi) Note-books Rs. 6 for one 40002 Rs. 6 x 40002 = Rs. 2,40,012 (vii) Soap cakes Rs. 8 for one 20005 Rs. 8 x 20005 = Rs. 1,60,040

Arranging the entries of amount of money in decreasing order :
Rs. 2,53,670; Rs. 2,40,012; Rs. 1,60,040; Rs. 98,280; Rs. 90,120; Rs. 68,280; Rs. 38,530.
Tooth brushes brought him the highest amount which is Rs. 2,53,670.

Try These (Page 23)

Question 1.
Round these numbers to the nearest tens.
28,32,52,41, 39,48,64,59,99,215,1453, 2936
(i) 28 is rounded off to 30, correct to the nearest tens.
(ii) 32 is rounded off to 30, correct to the nearest tens.
(iii) 52 is rounded off to 50, correct to the nearest tens.
(iv) 41 is rounded off to 40, correct to the nearest tens.
(v) 39 is rounded off to 40, correct to the nearest tens.
(vi) 48 is rounded off to 50, correct to the nearest tens.
(vii) 64 is rounded off to 60, correct to the nearest tens.
(viii) 59 is rounded off to 60, correct to the nearest tens.
(ix) 99 is rounded off to 100, correct to the nearest tens.
(x) 215 is rounded off to 220, correct to the nearest tens.
(xi) 1453 is rounded off to 1450, correct to the nearest tens.
(xii) 2936 is rounded off to 2940, correct to the nearest tens.

(b) Estimating to the nearest hundreds: Check if the following rounding off to 100
is correct or not. Correct those which are wrong.
(i) 841→ 800 correct
(ii) 9537 → 9500 correct
(iii) 49730 → 49700 correct
(iv) 2546 → 2500 correct
(v) 286 → 300 correct
(vi) 5750 → 5800 correct
(vii) 168 → 200 correct
(viii) 149 → 100 correct
(ix) 9870 → 9800 wrong
→ 9900 correct.

(c) Rounding off to nearest thousands :
Check the following rounding offs. Correct those which are wrong.
(0 2,573 → 3,000 correct
(ii) ‘ 53,552 → 53,000 wrong
→ 54,000 correct
(iii) 6,404 → 6,000 correct
(iv) 65,437 → 65,000 correct
(v) 7,805 → 7,000 → wrong
→ 8,000 correct
(vi) 3,499 → 4,000 wrong
→ 3,000 correct.

Try These (Page 24)

Question 1.
Round off the given numbers to the nearest tens, hundreds and thousands.

 Given Number Approximated to Nearest Rounded form (i)            75847 Tens 75850 (ii)           75847 Hundreds 75800 (iii)          75847 Thousands 76000 (iv)          75847 Ten thousands 80000

Note : There are no rigid rules when you want to estimate the outcotnes on numbers. The procedure depends on the degree of accuracy required, how quickly the estimate is needed and most importantly, how sensible the guessed answer would be.

Try These (Page 28)

Question 1.
Estimate the following products :
(i) 87 x 313
(ii) 9×795
(iii) 898 x 785
(iv) 958 x 387.
Solution:
(i) We try approximating 87 to 90 arid 313 to 300. We get the product 90 x 300 = 27,000 is a quick as well as good estimate of the product.
(ii) We try approximating 9 to 10 and 795 to 800. We get the product 10 x 800 = 8,000 is a quick as well as good estimate of the product.
(iii) We try approximating 898 to 900 and 785 to 800. We get the product 900 x 800 =
7,20,0 is a quick as well as good estimate of the product.
(iv) We try approximating 958 to 1,000 and 387 to 400. We get the product 1,000 x 400 = 4,00,000 is a quick as well as good estimate of the product.

Try These (Page 29)

Question 1.
Write the expressions for each of the following using brackets :
(a) Four multiplied by the sum of nine and two.
(b) Divide difference of eighteen and six by four. ,
(c) Forty five divided by three times of the sum of three and two.
(a) 4 x (9 + 2),
(b) (18 – 6) + 4,
(c) 45 ÷ 3(3 + 2).

Question 2.
Write 3 different situations for (5 + 8) x 6.
(a) 1st situation: Sohani and Reeta work for 6 days; Sohani works 5 hours a day and Reeta 8 hours a day. How many hours do both of them work in a day ?
(b) Ilnd situation : Suman bought 5 note-books from the market and the cost was Rs. 6 per note-book. Her sister Seema also- bought 8 note-books of the same type. Find the total money they paid.
(c) Illrd situation : Ram covered a
distance of 5 km in one hour and Shyam covered a distance of 8 km in one hour. How much distance did they cover in 6 hours ?

Question 3.
Write 5 such situations for the following where brackets would be necessary.
(a) 7 (8 – 3) is 7 times (8 – 3) and so the answer is 7 x 5 = 35.
(b) (7 + 2) (10 – 3) is (9) (7) or 9 x 7 = 63.
(c) 5 (10 – 3) is 5 times (10 – 3) and so the answer is 5 x 7 = 35.
(d) (7 – 2) (10 + 3) is (5) (13) or 5 x 13 = 65.
(e) 18 (5 – 2) is 18 times (5 – 2) and so the answer is 18 x 3 = 54.

Try These (Page 29-30)

1. 7 x (100 + 9)
7 x 100 + 7.x 9 700 + 63 = 763

2.  102 x 103 = (100 + 2) x (100 + 3)
= 100 x 100 + 2 x 100 + 100 x 3 + 2 x 3
= 10,000 + 200 + 300 + 6
= 10,000 + 500 + 6
= 10,506.

3. 17 x 109 = (10 + 7) x (100 + 9)
= 10 x 100 + 7 x 100 + 10 x 9 + 7 x 9
= 1000 + 700 + 90 + 63
= 17790 + 63
= 1853.

Try These (Page 32)

Question 1.
Write in Roman Numerals :
1. 73
2. 92.
Solution:
1. 73 = 70 +3 = (50+ 20)+ 3 = LXX + III = LXXIII
2. 92 = 90 + 2 = (100 – 10) + 2 = XC + II = XCII.