NCERT Solutions Class 6 Maths Chapter 1

Q.1 Fill in the blanks:

(a) 1 lakh = _______ ten thousand.

(b) 1 million = _______ hundred thousand.

(c) 1 crore = _______ ten lakh.

(d) 1 crore = _______ million.

(e) 1 million = _______ lakh.

Ans-

(a) 1 lakh = __10___ ten thousand.

(b) 1 million = __10__ hundred thousand.

(c) 1 crore = __10__ ten lakh.

(d) 1 crore = _10__ million.

(e) 1 million = _10__ lakh.

Q.2 Place commas correctly and write the numerals:

(a) Seventy three lakh seventy five thousand three hundred seven.

(b) Nine crore five lakh forty one.

(c) Seven crore fifty two lakh twenty one thousand three hundred two.

(d) Fifty eight million four hundred twenty three thousand two hundred two.

(e) Twenty three lakh thirty thousand ten.

Ans-

(a) 73,75,307

(b) 9,05,00,041

(c) 7,52,21,302

(d) 58,423,202

(e) 23,30,010

Q.3 Insert commas suitably and write the names according to Indian System of Numeration:

(a) 87595762 (b) 8546283

(c) 99900046 (d) 98432701

Ans-

(a) 8,75,95,762

Eight crore seventy five lakh ninety five thousand seven hundred sixty two

(b) 85,46,283

Eighty five lakh forty six thousand two hundred eighty three

(c) 9,99,00,046

Nine crore ninety nine lakh forty six

(d) 9,84,32,701

Q.4 Nine crore eighty four lakh thirty two thousand seven hundred one

Insert commas suitably and write the names according to International System of Numeration:

(a) 78921092 (b) 7452283

(c) 99985102 (c) 48049831

Ans-

(a) 78,921,092

Seventy eight million nine hundred twenty one thousand ninety two

(b) 7,452,283

Seven million four hundred fifty two thousand two hundred eighty three

(c) 99,985,102

Ninety nine million nine hundred eighty five thousand one hundred two

(d) 48, 049,831

Forty eight million forty nine thousand eight hundred thirty one

Q.5 A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Ans-

\begin{array}{l}{\text{Tickets sold on 1}}^{\text{st}}\text{day}=1094\\ {\text{Tickets sold on 2}}^{\text{nd}}\text{day}=1812\\ {\text{Tickets sold on 3}}^{\text{rd}}\text{day}=\text{2}0\text{5}0\\ {\text{Tickets sold on 4}}^{\text{th}}\text{day}=\text{2751}\\ \text{Total tickets sold}=1094+1812+2050+2751\\ \text{Now,}\\ 1094\\ 1812\\ 2050\\ \underset{¯}{+2751}\\ 7707\\ \\ \therefore \text{Total tickets sold in four days at book exhibition}=7,707\end{array}

Q.6 Shekhar is a famous cricket player. He has so far scored 6,980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Ans-

Runs scored = 6,980

Runs desired = 10,000

Therefore,

Runs needed = 10,000 – 6,980 = 3,020

Q.7 In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

Ans-

$\begin{array}{l}\text{Votes secured by successful candidate}=\text{5},\text{77},\text{5}00\\ \text{Votes secured by nearest rival}=\text{3},\text{48},\text{7}00\\ \text{Margin}=\text{5},\text{77},\text{5}00-\text{3},\text{48},\text{7}00\\ \text{Now,}\\ \text{5},\text{77},\text{5}00\\ \underset{¯}{-\text{3},\text{48},\text{7}00}\\ 2,28,800\\ \text{Hence successful candidate won the election by}\\ \text{2,28,800 votes}\text{.}\end{array}$

Q.8 Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.

Ans-

Let us form the greatest and the least number using the digits 6, 2, 7, 4, 3 only once.

So, we have

Greatest number = 76432

Least number = 23467

Difference = 76432 – 23467 = 52965

Therefore, the required difference is 52,965.

Q.9 A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Ans-

Screws manufactured in one day = 2,825

Days in January = 31

Screws produced in 31 days = 2825 × 31 = 87575

Therefore, screws produced in the month of January = 87,575

Q.10 A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?

Ans-

(7236 × 65) – (7236 × 56) = 7236 × (65 – 56)

= 7236 × 9

= 65,124

Therefore, student’s answer was greater than the correct answer by 65,124.

Q.11 To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?

Ans-

Cloth required for one shirt = 2 m 15 cm

= 200 cm + 15 cm

= 215 cm

Total cloth = 40 m = 40 × 100 = 4000 cm

Number of shirts that can be stitched out of 4000 cm =

$$\text{4}000÷\text{215}=18\frac{130}{215}$$

Therefore, 18 shirts can be stitched and 130 cm cloth will remain.

Q.12 Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?

Ans-

\begin{array}{l}\text{We know that}\\ \text{1 kg}=\text{1}000\text{g}\\ \text{Therefore,}\\ \text{4 kg 5}00\text{g}=(4\times 1000)\text{g}+500\text{g}\\ =4000+500=4500\text{g}\\ \text{and}\\ \text{8}00\text{kg}=\text{8}00\times \text{1}000\text{g}=\text{8,}00,000\text{g}\\ \text{Number of boxes that can be loaded in the van}\\ =\text{8}00000÷\text{45}00\\ =177\frac{3500}{4500}\\ \text{Hence},\text{177 boxes at maximum can be loaded in the van}.\end{array}

Q.13 The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.

Ans-

Distance between school and house = 1 km 875 m

Now, 1 km = 1000 m

So, 1 km 875 m = (1000 + 875) m = 1875 m

Distance covered in 1 day = 1875 × 2 = 3750 m

Distance covered in 6 days = 3750 × 6 = 22,500

Therefore, distance covered in 6 days = 22,500 m

= 22 km 500 m

Q.14 A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?

Ans-

We know that

1 litre = 1000 ml

Therefore,

Capacity of vessel = 4 litres and 500 ml

= (4000 + 500) ml

= 4500 ml

Capacity of a glass = 25 ml

Number of glasses that can be filled = 4500/25 = 180

Therefore, 180 glasses can be filled.

Estimate each of the following using general rule:

(a) 730 + 998 (b) 796 – 314 (c) 12,904 +2,888 (d) 28,292 – 21,496

Q.15 Make ten more such examples of addition, subtraction and estimation of their outcomes.

Ans-

(a) 730 + 998

We round off the given numbers to hundreds.

Number 730 rounds off to 700 and 998 rounds off to 1000.

Therefore,

Estimated sum = 700 + 1000 = 1700

(b) 796 – 314

We round off the given numbers to hundreds.

Number 796 rounds off to 800 and 314 rounds off to 300.

Therefore,

Estimated difference = 800 – 300 = 500

(c) 12904 + 2888

We round off the given numbers to thousands.

Number 12904 rounds off to 13000 and 2888 rounds off to 3000.

Therefore,

Estimated sum = 13000 + 3000 = 16000

(d) 28,292 − 21,496

We round off the given numbers to thousands.

Number 28,2962 rounds off to 28000 and 21,496 rounds off to 21000.

Therefore,

Estimated difference = 28000 − 21000 = 7000

Q.16 Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :

(a) 439 + 334 + 4,317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365

Make four more such examples.

Ans- 

(a) 439 + 334 + 4317

We round off the given numbers to hundreds.

Number 439 rounds off to 400, 334 rounds off to 300 and 4317 rounds off to 4300.

Therefore,

Estimated sum = 400 + 300 + 4300 = 5000

Now, we round off the given numbers to tens.

Number 439 rounds off to 440, 334 rounds off to 330 and 4317 rounds off to 4320.

Therefore,

Estimated sum = 440 + 330 + 4320= 5090

(b) 1,08,734 − 47,599

We round off the given numbers to hundreds.

Number 1,08,734 rounds off to 1,08,700 and 47,599 rounds off to 47,600.

Therefore,

Estimated difference = 1,08,700 – 47,600 = 61,100

Now, we round off the given numbers to tens.

Number 1,08,734 rounds off to 1,08,730 and 47,599 rounds off to 47,600.

Therefore,

Estimated difference = 1,08,730 – 47,600 = 61,130

(c) 8325 − 491

We round off the given numbers to hundreds.

Number 8325 rounds off to 8300 and number 491 rounds off to 500.

Therefore,

Estimated difference = 8300 – 500 = 7,800

Now, we round off the given numbers to tens.

Number 8325 rounds off to 8330 and number 491 rounds off to 490.

Therefore,

Estimated difference = 8330 – 490 = 7,840

(d) 4,89,348 − 48,365

We round off the given numbers to hundreds.

Number 4,89,348 rounds off to 4,89,300 and number 48,365 rounds off to 48,400.

Therefore,

Estimated difference = 4,89,300 – 48,400 = 4,40,900

Now, we round off the given numbers to tens.

Number 4,89,348 rounds off to 4,89,350 and number 48,365 rounds off to 48,370.

Therefore,

Estimated difference = 4,89,350 – 48,370 = 4,40,980

Q.17 Estimate the following products using general rule:

(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29

Make four more such examples.

Ans- 

(a) 578 × 161

We round off the given numbers to hundreds.

Number 578 rounds off to 600 and number 161 rounds off to 200.

Therefore,

Estimated product = 600 × 200 = 1,20,000

(b) 5281 × 3491

We round off 5281 to nearest thousands and so 5281 rounds off to 5000.

We round off 3491 to nearest hundreds and so 3491 rounds off to 3500.

Therefore,

Estimated product = 5000 × 3500 = 1,75,00,000

(c) 1291 × 592

We round off the given numbers to hundreds.

Number 1291 rounds off to 1300 and number 592 rounds off to 600.

Therefore,

Estimated product = 1300 × 600 = 7,80,000

(d) 9250 × 29

We round off 9250 to 9000 and 29 to 30.

Therefore,

Estimated product = 9,000 × 30 = 2,70,000

Q.18 Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Ans-

Worth of Books sold in 1^st week = ₹2,85,891

Worth of books sold in 2^nd week = ₹4,00,768

Total sale = Sale in 1^st week + Sale in 2^nd week

= ₹2,85,891 + ₹4,00,768 = ₹6,86,659

Thus, the sale for the two weeks together was ₹6,86,659.

Also, 4,00,768 – 2,85,891 = 1,14,877

Thus, the sale in 2^nd week was greater than the sale in 1^st week by ₹1,14,877.

Q.19 A merchant had ₹78,592 with her. She placed an order for purchasing 40 radio sets at ₹1200 each. How much money will remain with her after the purchase?

Ans-

Cost of one radio set = ₹1200

So, cost of 40 radio sets = ₹1200 × 40 = ₹48000

Merchant had ₹78,592 with her before purchase.

So, after purchase, she will have ₹(78,592 – 48,000),i.e. ₹30,592.

Therefore, ₹30,592 will remain with her after the purchase.