Practicing with Maths Mela Class 4 Solutions Chapter 13 The Transport Museum Question Answer NCERT Solutions improves a student’s confidence in the subject.
Class 4 Maths Chapter 13 The Transport Museum Question Answer Solutions
The Transport Museum Class 4 Maths Solutions
Class 4 Maths Chapter 13 Solutions
Mystery Matrix (NCERT Pg 184)
Question 1.
Fill the shaded boxes with 1-digit numbers (multiplicands and multipliers) such that you get the products given in the white boxes.
Fill the remaining white boxes with appropriate products.
Answer:
Question 2.
The product of the numbers in each row is given in the magenta boxes. The product of the numbers in each column is given in the gray boxes. Identify appropriate numbers to fill the blank boxes.
Answer:
Times-10 (NCERT Pg 184-185)
Question 3.
Match each problem with the appropriate pictorial representation and write the answer.
Answer:
(i) → (c), (ii) → (a), (iii) → (b)
Question 4.
Answer:
Constructing Tables (NCERT Pg 185-187)
Question 5.
(i) How many pebbles are there in this arrangement?
(ii) Recall the times-tables that we created in Grade 3. Now, construct a times-15 table. You may use the arrangement given below and split the columns into 10 and 5 for ease of counting, as shown on the previous page.
How can we find 1 × 15, 2 ×15, …………..with this?
(iii) What patterns do you see in this table? [see part (ii)]
(iv) Compare the times-15 table with the times- 5 table. What similarities and differences do you notice?
(v) What times-table is this? How did we get this?
(vi) Construct other times-tables for numbers from 11 to 20, as you did for 15.
(vii) As you compared the times- 5 table with the times- 15 table, compare the times -1 table with the times- 11 table, the times- 2 table with the times- 12 table and so on. Share your observations.
(i) Number of pebbles =5 ×15 = 75 × pebbles
This is a 5 × 15 arrangment. There is any easy way to find this product by splitting the arrangement.
So, 5 × 15 can be split as 5 × 10 and 5 × 5. Then, add them.
(ii)
(iii) The products increase by 15 each time. All numbers end with either 0 or 5 (since $15 \times$ any number ends in 5 or 0 ). And all the numbers can be divided by 3.
(iv) Similarities and Differences
- Both tables increase linearly.
- Every number in times-15 table is 3 times the corresponding value in times- 5.
- Both tables have multiple of 5.
Differences
- Times-5 table is increasing by 5 and gives smaller values.
- Times-15 tables has bigger jumps (increasing by 15 instead of 5).
(v) This table is times-10. We got this table by subtract the corresponding product values of tables times-15 and times-5.
vi) Do yourself.
(vii) Do yourself.
Making tables by Splitting into Equal Groups (NCERT Pg 187-188)
Question 6.
Here is an arrangement of wheels. To count the total number of wheels, Tara splits them into two equal groups.
3 × 14 =3 × 7 and 3× 7
=21+21= Double of 21
=42
(i) We have seen how to calculate 3 × 14 and 6 × 14 by splitting and doubling. Can we construct the times-14 table by splitting and doubling? Try!
(ii) What other times-tables can be constructed by splitting into equal groups and doubling? Give examples.
Answer:
(i) Do yourself.
(ii) Do yourself.
Multiples of 10 (NCERT Pg 188)
Question 7.
Find the answers of the following.
(i) 15 × 10= ……… Tens = ………
(ii) 16 × 10= ……… Tens = ………
(iii) 19 × 10= ……… Tens = ………
(iv) 20 ×10= ……… Tens = ………
10 × 10=
2 times (i.e. double of) 10 × 10=………
(i) 15 Tens =150
(ii) 16 Tens =160
(iii) 19 Tens =190
(iv) 20 Tens = 200
10 × 10=100
2 times (i.e. double of) 10 × 10=200
Question 8.
Discuss in grade what happens when we take several groups of 10.
Answer:
When we take several groups of 10, we multiply 10 by how many groups we have. Then, a zero is added at the end of the number. e.g.
- 4 × 10=40
- 9× 10=90
(NCERT Pg 189-190)
Question 9.
Think and answer the following problems.
(i) 30 x 10= ………..
(ii) 40 x 10= ………..
(iii) 70 x 10= ………..
(iv) 50 x 10= ………..
(v) 60 x 10= ………..
(vi) 80 x 10= ………..
Answer:
(i) 30 Tens =300
(ii) 40 Tens =400
(iii) 70 Tens =700
(iv) 50 Tens =500
(v) 60 Tens =600
(vi) 80 Tens =800
Question 10.
Answer the following questions.
(i) 21 x 10= ………..
(ii) 42 x 10= ………..
(iii) 65 x 10= ………..
(iv) 38 x 10= ………..
(v) 53 x 10= ………..
(vi) 87 x 10= ………..
Answer:
(i) 21 Tens =210
(ii) 42 Tens =420
(iii) 65 Tens =650
(iv) 38 Tens =380
(v) 53 Tens =530
(vi) 87 Tens =870
Question 11.
Solve the following problems. Share your thoughts.
(i) 24 x 40= ………..
(ii) 50 x 60= ………..
(iii) 13 x 30= ………..
(iv) 43 x 60= ………..
(v) 70 x 80= ………..
Answer:
(i) 24 x 40=960
I can also solve it as first, break 24 into ( 20+4 ).
Then, 20 x 40=800 and 4 x 40=160.
Then, add them i.e. 800+160=960
So, 24 x 40=960
Or 24 x 40=24 x 4 tens =960
(ii) 50 x 60=3000
50 x 60 means 5 tens x 6 tens
So, 5 x 6=30, then 30 x 100=3000
So, 50 x 60=3000
(iii) 13 x 30=390
First, break 13 into ( 10+3 ). Then, 10 x 30=300 and 3 x 30=90.
Then, add them i.e. 300+90=390
So, 13 x 30=390
Or 13 x 30=13 x 3 tens =390
(iv) 43 x 60=2580
First, break 43 into 40+3
Then, 40 x 60=2400 and 3 x 60=180
Then, add them i.e. 2400+180=2580
So, 43 x 60=2580
Or 43 x 60=43 x 6 tens =258 tens =2580
(v) 70 x 80=5600
70 x 80 means 7 tens x 8 tens.
So, 7 x 8=56, then 56 x 100=5600
(NCERT Pg 191)
Question 12.
Raahi wonders how many coaches will be needed for the 324 children from her school. Remember, each coach can seat only 14 children. We have to find 324 ÷ 14
(i) What do we do with the remaining 2 children? Discuss the grade.
(ii) Total number of coaches x=x x\qquadx
Answer:
(i) Even if there are only 2 children left, they still need a coach.
Since, we need 1 more coach just for them.
So, total coaches needed x=24x
(ii) Total number of coaches x=10+10+3x x=23x coaches
Let Us Solve (NCERT Pg 192)
Question 13.
Also, identify remainder (if any) in the division problems.
(i) 25 x 34x
(ii) 16 x 43x
(iii) 68 x 12x
(iv) 39 x 13x
(v) 125 ÷ 15x
(vi) 94 ÷ 11x
(vii) 440 ÷ 22x
(viii) 508 ÷ 18x
Answer:
(i) 25 x 34
=25 x (30+4)
=(25 x 30)+(25 x 4)
=750+100=850
(ii) 16 x 43
=16 x(40+3)
=(16 x 40)+(16 x 3)
=640+48=688
(iii) 68 x 12
=68 x(10+2)
=(68 x 10)+(68 x 2)
=680+136=816
(iv) 39 x 13
=39 x(10+3)
=(39 x 10)+(39 x 3)
=390+117=507
(v) 125÷15
It can be obtained as,
15 × 8 =120
Remainder =125-120=5
(vi) 94 ÷ 11
It can be obtained as,
11 × 8 =88
Remainder =94-88=6
(vii) 440 ÷ 22
22 × 20=440
Remainder =0
(viii) 508 ÷18
18 × 28 =504
Remainder =508-504=4
Multiples of 100 (NCERT Pg 192)
(i) 2 x 100=2 Hundreds =200
(ii) 3 x 100= ……….. Hundreds = ………..
(iii) 5 x 100= ……….. Hundreds = ………..
(iv) 8 x 100= ……….. Hundreds = ………..
(v) What happens when we put 10 Hundreds together?
(vi) (a) 12 x 100= ………..
(b) 15 x 100= ………..
(c) 27 x 100= ………..
(d) 70 x 100= ………..
Answer:
(ii) 3 Hundreds =300
(iii) 5 Hundreds =500
(iv) 8 Hundreds =800
(v) When we put 10 hundreds together we make 1000.
(vi) (a) 10 Hundreds +2 Hundreds
=1000+200=1200
(b) 10 Hundreds +5 Hundreds
=1000+500=1500
(c) 20 Hundreds +7 Hundreds
=2000+700=2700
(d) 70 Hundreds =7000
(NCERT Pg 193)
Question 15.
Answer the following questions. Share your thoughts.
Answer:
(i) (a) 3000
(b) 4000
(c) 5000
(d) 2400
(e) 5300
(f) 1900
(ii) (a) 80 x 50=8 x 10 x 5 x 10
=40 x 100=4000
(b) 40 x 50=4 x 10 x 5 x 10
=20 x 100=2000
(NCERT Pg 194-195)
Question 16.
Share what you notice about the answers to these problems.
(i) 11 x 100= ………..
(ii) 22 x 100= ………..
(iii) 11 x 200= ………..
(iv) 22 x 200= ………..
Answer:
(i) 11 hundreds =1100
(ii) 22 hundreds =2200
(iii) 11 x 2 hundreds =22 hundreds
=2200
(iv) 22 x 2 hundreds =44 hundreds
=4400
Any number x 100
Just add two zeroes to the number.
e.g. 11 x 100=1100
Any number × 200
Double the number or multiply the number by 2 then add two zeroes
e.g. 11 x 200 =(11 x 2) x 100
& =2200
Question 17.
Answer the following questions.
Answer:
To multiply a number by a multiple of 100 , first multiply by the factor of multiple then add two zeroes.
Question 18.
Find the answer in Set A. Examine the relationships between the problems and the answers in Set A carefully. Then use this understanding to find the answers in Set B.
Answer:
Let Us Solve (NCERT Pg 197)
Question 19.
Identify remainder (if any) in the division problems.
(i) 237 x 28
(ii) 140 x 16
(iii) 389 x 57
(iv) 807 ÷ 24
(v) 692 ÷ 33
(vi) 996 ÷ 45
Answer:
(i) 237 x28 Break into parts
237 x28 =(237 x20)+(237 x8)
=4740+1896=6636
(ii) 140 x 16
=(140 x 10)+(140 x 6)
=1400+840=2240
(iii) 389 x 57
=(389 x 50)+(389 x 7)
=19450+2723=22173
(iv) 807 ÷ 24
24 × 33 =792
Remainder =807-792=15
(v) 692÷33
33 × 20=660
Remainder =692-660=32
(vi) 996 ÷ 45
45 × 22 =990
Remainder =996-990=6
Dividing by 10 and 100 (NCERT Pg 197-198)
Question 20.
A farmer packs his rice in sacks of 10 kg each.
(i) If he has 60 kg of rice, how many sacks does he need?
(ii) If he has 600 kg of rice, how many sacks does he need?
(iii)
60 ÷ 10= ………..
600 ÷10= ………..
600 ÷ 100= ………..
(iv) If a sack of rice weighs 100 kg then how many sacks does he need for 600 kg of rice?
Answer:
Given that farmer packs his rice in sacks of 10 kg each.
(i) Number of sacks for 60 kg of rice
=60 ÷ 10=6 sacks
(ii) Number of sacks for 600 kg of rice
=600 ÷ 10=60 sacks
(iii) 60 ÷10=6
600 ÷ 10=60
600 ÷ 100=6
(iv) 6
Question 21.
Find the answers of the following questions. Share your thoughts in grade.
Answer:
Question 22.
Think and answer. Write the division statement in each case.
(i) Manku the monkey sees 870 bananas in the market. Each bunch has 10 bananas. How many bunches are there in the market?
(ii) Rukhma Bi wants to distribute ₹ 1000 equally among her 10 grandchildren on the occasion of Eid. How much money will each of them get?
Answer:
(i) Division statement
870 ÷ 10=87
There are 87 bunches of bananas in the market.
(ii) Division statement
1000 ÷ 10=100
Each grandchild will get ₹ 100
Let Us Solve (NCERT Pg 198-200)
Question 23.
The oldest long-distance train of the Indian Railways in the Punjab Mail which ran between Mumbai and Peshawar. Its first journey was on 12 October 1912. Do you know how many coaches it had on its first journey? It had 6 coaches 3 carrying 96 passengers and 3 for goods.
(i) How many people travelled in each coach on the first journey?
(ii) This train has been running for 106 years now. It runs between Mumbai, Maharashtra and Ferozepur, Punjab. It has 24 coaches. Each coach can carry 72 passengers. How many people can travel on this train?
Answer:
(i) We are told that 96 passengers were carried in 3 coaches.
Division statement for the people who travelled in each coach
96+3=32
32 people travelled in each coach on the first journey.
(ii) The train has 24 coaches and each coach carries 72 passengers.
Multiplication statement for number of people who can travel on this train
24 x72
Let us break in into parts
24 x 72 =(20×72)+(4 x72)
=1440+288
=1728
1728 people can travel on this train.
Question 24.
Amala and her 35 classmates, along with 6 teachers, are going on a school trip in Goa. They are using the double-decker ‘hop on hop off’ sightseeing bus to explore the city.
(i) 2 people can sit on every seat of the bus. There are 15 seats in the lower deck and 10 seats in the upper deck. How many seats will they need to occupy? Are there enough seats for everyone?
(ii) Find the total cost of the tickets for all children.
Ticket price
Adult – ₹ 899/-
Children – ₹ 359/-
(iii) What is the cost of the tickets for all teachers?
Answer:
(i) Total number of students
=Amala +35 Classmates =36
Total number of people
=36+6=36 Children and 6 teachers =42
Now, each seat holds 2 people.
Number of seats available in bus
=15 Seats (lower back) +10 Seats (upper deck) =25 Seats
So, total capacity of bus
=25 Seats x 2 People per seat
=50 People
They will need 42 seats to fit everyone.
Yes, there are enough seats for everyone.
(ii) Ticket price for children = ₹ 359
So, total cost of the tickets for all children
=36 x ₹ 359
=36 x(360-1)
=(36 x 360)-36
=12960-36=₹ 12924
(iii) Number of teachers =6
Ticket price for one adult =₹ 899
So, total cost of the tickets for all teachers
=6 x ₹ 899
=6 x(900-1)
=(6 x 900)-6
=5400-6
=₹ 5394
Question 25.
Kedar works in a brick kiln.
(i) The kiln makes 125 bricks in a day. How many bricks can be made in a month?
(ii) Each brick is sold in the market for ₹ 9 . How much money can they earn in a month?
Answer:
(i) Assuming a month has 30 days and the kiln makes 125 bricks per day. So, the number of bricks that is made by kiln in a month =125 x 30=3750 bricks 3750 bricks can be made in a month.
(ii) From part (i), we know 3750 bricks are made in a month.
Price per brick = ₹ 9
Total earnings in a month
=3750 x ₹ 9
=3750 x(10-1)
=(3750 x 10)-3750
=37500-3750
=₹ 33750
₹ 33750 can be earned in a month.
Question 26.
Chilika lake in Odisha is the largest saltwater lake in India. It is famous for the Irrawaddy dolphins. Boats can be hired to go see the dolphins. The trip from Puri includes a bus ride followed by a boat ride. Eight people will be going on the trip.
- A bus ticket from Puri to Satapada costs ₹ 60
- A two-hour boat ride for 8 people costs ₹ 1200
How much money do we need to spend on each person?
Answer:
Given,
- Number of people =8
- Bus ticket cost per person =₹ 60
- Boat ride cost for 8 people =₹ 1200
Step 1 Total bus fare for 8 people
₹ 60 x 8=₹ 480
Step 2 Total cost of the trip
Bus fare + Boat ride =480+1200=₹ 1680
Step 3 Cost per person
₹ 1680+8=₹ 210
Each person needs to spend ₹ 210 for the entire trip.
Question 27.
Find the multiplication and division sentences below. Shade the sentences. How many can you find? Some are done for you.
Answer:
We found 6 sentences.
Question 28.
Solve
(i) 35 x 76
(ii) 267 x 38
(iii) 498 x 9
(iv) 89 x 42
(v) 55 x 23
(vi) 345 x 17
(vii) 66 x 22
(viii) 704 x 11
(ix) 319 x 26
(x) 459 ÷ 3
(xi) 774 ÷ 18
(xii) 864 ÷ 26
(xiii) 304 ÷ 12
(xiv) 670 ÷ 9
(xv) 584 ÷ 25
(xvi) 900 ÷ 15
(xvii) 658 ÷ 32
(xviii) 974 ÷ 9
Answer:
(i) 35 x 76 =35 x 76
=(30+5) x 76
=30 x 76+5 x 76
=2280+380=2660
(ii) 267 x 38 =267 x 38
=267 x(30+8)
=(267 x 30)+(267 x 8)
=8010+2136=10146
(iii) 498 x 9 =498 x(10-1)
=(498 x 10)-498
=4980-498=4482
(iv) 89 x 42 =89 x(40+2)
=(89 x 40)+(89 x 2)
=3560+178=3738
(v) 55 x 23 =55 x(20+3)
=(56 x 20)+(55 x 3)
=1120+165=1285
(vi)345×17=345×17
=(300+45)x17
=300×17+45×17
=5100+45x(20-3)
=5100+(45×20)-(45×3)
=(5100+900)-135
=6000-135=5865
(vii)66×22=66×22
=(60+6)x22
=60×22+6×22
=1320+132=1452
(viii)704×11=704x(10+1)
=(704×10)+704
=7040+704=7744
(ix)319×26=319×26
=(300+19)x26
=300×26+19×26
=7800+(20-1)x26
=7800+(20×26)-26
=(7800+520)-26
=8320-26=8294
(x)459÷3
459÷3=153
So, quotient =153
(xi) 774 ÷ 18
So, quotient =774 ÷ 18=43
(xii) 864 ÷ 26
So, quotient =33 remainder =6
(xiii) 304 ÷ 12
So, quotient =25 remainder =4
(xiv) 670 ÷ 9
So, quotient =74 remainder =4
(xv) 584 ÷25
So, quotient =23 remainder =9
(xvi) 900 ÷15
So, quotient =900 ÷ 15=60
(xvii) 658 ÷ 32
So, quotient =20, remainder =18
(xviii) 974 ÷ 9
So, quotient =108, remainder =2
Chinnu’s Coins (NCERT Pg 201-202)
Question 29.
Five friends plan to visit an amusement park nearby. Each of them uses different notes and coins to buy the ticket. The cost of the ticket is ₹ 750
- Bujji has brought all notes of ₹ 200
- And Munna has brought all notes of ₹ 50
- Whereas Balu has brought all notes of ₹ 20
- And guess what, Chinnu has all coins of ₹ 5
- And Sansu has all coins of ₹ 2
(i) Find out how many notes/coins each child has to bring to buy the ticket.
(ii) Which of these children will not receive any change from the cashier?
(iii) How long would the cashier take to count Chinnu’s coins?
Answer:
(i) 
(ii) Only those who paid exactly ₹ 750 from above part (i).
Munna, Chinnu and Sansu will not receive any change from the cashier.
(iii) Chinnu uses ₹ 5 coins.
Cost of ticket =₹ 750
Number of coins =₹ 750 ÷ ₹ 5=150 coins
If we assume the cashier takes 2 seconds per coin to count.
So, time taken by cashier
=150 coins × 2 seconds
=300 seconds =5 minutes
The cashier will take 5 minutes (or 300 seconds) to count Chinnu’s coins. This answer may be vary because it depends on counting efficiency of cashier.
Question 30.
Observe the following multiplications. The answers have been provided.
(i) In each case, do you see any pattern in the two numbers and their product?
(Hint Look at the coloured digits!)
(ii) For what other multiplication problems will this pattern hold? Find 5 such examples.
Answer:
(i) Do yourself.
(ii) Do yourself.
Question 31.
Assume each vehicle is travelling with full capacity. How many people can travel in each of these vehicles? Match them up.
Answer:
(i) (b) 75 cycles
Each cycle carries 1 person,
75 x 1=75
(ii) (d) 52 Autos
Each auto rickshaw carries 3 people,
52 x 3=156
(iii) (f) 103 Cars
Each car carries 4 people, 103 x 4=412
(iv) (a) 20 Minibus
Each minibus carries 20 people, 20 x 20=400
(v) (c) 30 Aeroplanes
Each airplane carries approximately 152 people, 30 x 152=30 x(150+2)
=(30 x 150)+(30 x 2)
=4500+60=4560
(vi) (e) 15 Train Sleeper Coaches and each coach carries approximately 58 people, so 15 x 58=870, but closest option is 864