Practicing with Maths Mela Class 5 Solutions Chapter 4 We the Travellers Question Answer NCERT Solutions improves a student’s confidence in the subject.
Class 5 Maths Chapter 4 We the Travellers Question Answer Solutions
We the Travellers Class 5 Maths Solutions
Class 5 Maths Chapter 4 Solutions
Making Sums Equal (NCERT Pg 42)
Question 1.
In each of the following, there are two groups of numbers. Look carefully at the numbers in each group and their sums. Interchange pairs of numbers between the two groups to make their sums equal. Try to do this using the least number of moves. You could write each number on a small piece of paper.
Answer:
(i) If we interchange 2 and 5 in the sums given below,
Then,
If we interchange 1 and 5 in the original sums given below,
Similarly, we can consider interchanging of different digits and get different sums.
Now, since, 21-19=2.
∴ We need to decrease the second sum by 1 and increase the first sum by 1.
For this, we interchange 2 and 3.
Then,
(ii) Since, 47 – 39 = 8
∴ We need to decrease the first sum by 4 and increase the second sum by 4. For this, we interchange 5 and 9.
Then,
(iii) Since, 76 – 68 = 8
∴ We need to decrease the second sum by 4 and increase the first sum by 4.
For this, we first interchange 15 and 17. Then,
Again, we interchange 19 and 21. Then,
(iv) Since, 330 – 314 = 16
So, we need to decrease the second sum by 8 and increase the first sum by 8.
For this, we first interchange 78 and 84.
Then,
Again, we interchange 80 and 82.
Then,
(NCERT Pg 43)
Question 2.
Find the sum of 49 and 89.
Answer:
We have,
So, 49 + 89 = 138
Let Us Solve (NCERT Pg 43)
Question 3.
Add the following numbers. Wherever possible, find easier ways to add the pairs of numbers.
(i) 15 + 79
(ii) 46 + 99
(iii) 38 + 35
(iv) 5 + 89
(v) 76 + 28
(vi) 69 + 20
Answer:
(i) We have,
Easier way Add 1 to 79 to make it 80 and then add 15 to 80 to get 95. Finally, subtract 1 from 95 to get the final answer 94.
(ii) We have,
Easier way Add 1 to 99 to make it 100 and then add 46 to 100 to get 146. Finally, subtract 1 from 146 to get final answer 145.
(iii) We have,
(iv) We have,
Easier way Add 1 to 89 to make it 90 and then add 5 to 90 to get 95. Finally, subtract 1 from 95 to get final answer 94.
(v) We have,
(vi) We have,
Easier way Add 1 to 69 to make it 70 and then add 20 to 70 to get 90. Finally, subtract 1 from 90 to get final answer 89.
Relationship Between Addition and Subtraction (NCERT Pg 44)
Question 4.
Find the relationship between the numbers in the given statements and fill in the blanks appropriately.
(i) If 46 + 21 = 67, then
67 – 21 = ………..
67 – 46 = ………..
(ii) If 198 – 98 = 100, then
100 + ……. = 198.
198 – ……. = 98.
(iii) If 189 + 98 = 287, then
287 – 98 = ………..
287 – 189 = ……….
(iv) If 872 – 672 = 200,
200 + ……… = 872.
872 – ……… = 672.
Answer:
(i) If 46 + 21 = 67
then
67-21 =4 6
67-46 =2 1
Addition and subtraction are inverse operations. If we know the sum of two numbers, subtracting one of those numbers from the sum will give you the other number.
(ii) If 198 – 98 = 100 then 100 + 98 = 198
198 – 100 = 98
If we subtract anumber from another to get a differènce, adding the subtracted number back to the difference will result in the original number.
Also, if we subtract the difference from the original number, we get the subtracted number.
(iii) Do same as part (i).
Answer:
189, 98
(iv) Do same as part (ii).
Answer:
672, 200
Question 5.
In each of the following, write the subtraction and addition sentences that follow from the given sentence.
(i) If 78 + 164 = 242, then
(ii) If 462 + 839 = 1301, then
(iii) If 921 – 137 = 784, then
(iv) If 824 – 234 = 590, then
Answer:
(i) Given, if 78 + 164 = 242, then
242 – 78 = 164
242 – 164 = 78
(ii) Given, if 462 + 839 = 1301, then
1301 – 462 = 839
1301 – 839 = 462
(iii) Given, if 921 – 137 = 784, then
784 + 137 = 921
921 – 784 = 137
(iv) Given, if 824-234=590, then
590+234=824
824-590=234
Let Us Solve (NCERT Pg 45)
Question 6.
What is the difference between 82 and 37?
Answer:
We have,
Also, we need to check 37+45=82
∴ Yes, 37 + 45 = 82
Question 7.
Subtract
(i) 57 – 11 = ………..
(ii) 23 – 19 = ………..
(iii) 49 – 21 = ………..
(iv) 56 – 18 = ………..
(v) 93 – 35 = ………..
(vi) 84 – 23 = ………..
(vii) 70 – 43 = ………..
(viii) 65 – 47 = ………..
Answer:
Sums of Consecutive Numbers (NCERT Pg 45-46)
Question 8.
Consider the following sum of consecutive numbers:
Box 3
(i) In each of the boxes above, state whether the sums are even or odd. Explain why this is happening.
(ii) What is the difference between two successive sums in each box? Is it the same throughout?
(iii) What will be the difference between two successive sums for-
(a) 5 consecutive numbers
(b) 6 consecutive numbers
Answer:
(i) In box 1, the sums 3, 5, 7 and 9, all are odd because the sum of any two consecutive numbers is always odd (i.e. odd + even = odd).
In box 2 , the sums 6 and 12 are even but 9 and 15 are odd because the sum of any three consecutive numbers is odd and even, if the middle term is odd and even, respectively. Even and odd (i.e. odd + even + odd = even and even + odd + even = odd).
In box 3 , the sums 10,14,18,22, all are even because sum of any four consecutive number is always even (i.e. odd + even + odd + even = odd + odd = even).
(ii) In box 1, we have
5 – 3 = 2
7 – 5 = 2
9 – 7 = 2
In box 2, we have
9 – 6 = 3
12 – 9 = 3
15 – 12 = 3
In box 3, we have
14 – 10 = 4
18 – 14 = 4
22 – 18 = 4
Hence, the difference between two successive sums in each box is same throughout.
(ii) We observe from above part (ii) that, when sum of 2,3 and 4 consecutive numbers are considered, the difference between any two successive sums in each case is 2,3 and 4 , respectively (i.e. same throughout).
So,
(a) The difference between two successive sums for 5 consecutive numbers will be 5.
(b) The difference between two successive sums for 6 consecutive numbers will be 6.
Question 9.
Use your understanding to find the following sums without adding the numbers directly.
(i) 67 + 68 + 69 =
(ii) 24 + 25 + 26 + 27
(iii) 48 + 49 + 50 + 51 + 52
(vi) 237 + 238 + 239 + 240 + 241 + 242
Answer:
Let Us Solve (NCERT Pg 48)
Question 10.
Find the following sums. Try not to write TTh, Th, H, T, and O at the top. Just align the digits properly, at least for the smaller numbers.
(i) 238 + 367
(ii) 1,234 + 12,345
(iii) 12 + 123
(iv) 46,120 + 12,890
(v) 878 + 8,789
(vi) 1,749 + 17,490
Answer:
Question 11.
The great Indian road trip!
Nazrana and her friends planned a road trip across India, starting from Delhi. They first drove to Mumbai, then Goa, then Hyderabad and finally Puri. Look at the distances marked on the map and help them find the total distance travelled.
Answer:
Given, the distance travelled between Delhi and Mumbai = 1,600 km, the distance travelled between Mumbai and Goa = 590 km, the distance travelled between Goa and Hyderabad = 670 km and the distance travelled between Hyderabad and Puri = 1,055 km
So, the total distance travelled
= (1,600 + 590 + 670 + 1,055) km
We have,
So, the total distance travelled = 3,915 km
Question 12.
Find 2 numbers among 5,205, 6,220, 7,095,8,455, and 4,840 , whose sum is closest to the following.
(i) 10,000
(ii) 15,000
(iii) 13,000
(iv) 16,000
Answer:
Here, the possible sums are
Clearly,
(i) 10,045 is closest to 10,000.
So, 5,205 and 4,840 give the sum closest to 10,000.
(ii) The nearest neighbours of 15,000 are 14,675 and 15,550.
Since, the nearest thousands of 14,675 and 15,550 are 15,000 and 16,000, respectively.
So, 14,675 is closest to 15,000, which is the sum of 6,220 and 8,455.
(iii) The neighbours of 13,000 are 12,300 and 13,295.
Since, the nearest thousands of 12,300 and 13,295 are 12,000 and 13,000, respectively.
So, 13,295 is closest to 13,000, which is the sum of 8,455 and 4,840.
(iv) 15,550 is closest to 16,000.
So, 7,095 and 8,455 give the sum closest to 16,000.
Let Us Solve (NCERT Pg 50-51)
Question 13.
Subtract the following. Try not to write TTh, Th, H, T, and O at the top. Align the digits carefully.
(i) 4,578-2,222
(ii) 15,324-11,780
(iii) 5,423-423
(iv) 123-12
(v) 77,777-777
(vi) 826-752
Answer:
Question 14.
Mary’s train journey to Delhi.
Mary is on a train journey. She starts from Kolkata with ₹ 12,540.
She spends ₹ 3,275 on food and other expenses during her trip to Varanasi. In Varanasi, her uncle gives her a gift worth ₹ 4,900. She then travels to Delhi, spending ₹ 2,645 on the train ticket. She spends ₹ 1,275 on souvenirs in Delhi. How much money is Mary left with at the end of the Delhi trip?
Delhi ₹ 1,275
Varanasi ₹ 2,645
Kolkata ₹ 3,275
Answer:
Given, the total amgunt, which Mary had = ₹ 12,540
and the amount spent by her on food and other expenses = ₹ 3,275
So, remaining money = ₹ 12,540 – ₹ 3,275
= ₹ 9,265
Given, she received a gift from her uncle of ₹ 4,900.
So, remaining money =₹ 9,265 + ₹ 4,900
= ₹ 14,165
Given, the amount spent by her on the train tickets = ₹ 2,645
So, remaining money = ₹ 14,165 – ₹ 2,645
= ₹ 11,520
Also, given the amount spent by her on souvenirs = ₹ 1,275
So, the remaining money
= ₹ 11,520 – ₹ 1,275
= ₹ 10,245
Therefore, Mary is left with ₹ 10,245 at the end of the Delhi trip.
Question 15.
Members of a school council have raised ₹ 70,500. They plan to setup a Maths Lab with some games and models worth ₹ 39,785 , buy library books worth ₹ 9,545 and purchase sports equipment worth ₹ 19,548.
(i) Estimate whether the school council has raised enough money to make the purchases. Share your thoughts in the class.
(ii) Check your estimate with calculations.
Answer:
Given, the amount raised by school council = ₹ 70,500
The amount spent on Maths Lab with some games and models = ₹ 39,785
The amount spent on buying library books = ₹ 9,545
and the amount spent on sports equipment = ₹ 19,548
(i) In order to estimate this, we need to round off the expenses to the nearest 1,000s. Here, ₹ 39,785, ₹ 9,545 and ₹ 19,548 are rounded off to ₹ 40,000, ₹ 10,000 and ₹ 20,000, respectively.
So, the estimated amount = ₹ 40,000 + ₹ 10,000 + ₹ 20,000 = ₹ 70,000
Therefore, the school council has raised enough money, as ₹ 70,000 < ₹ 70,500.
(ii) Here, the total amount spent on expenses is ₹ 39,785 + ₹ 9,545 + ₹ 19,548
i.e. ₹ 68,878, which is less than ₹ 70,500. Therefore, the school council has raised enough money.
Question 16.
A truck can carry 8,250 kg of goods.
A factory loads 3,675 kg of cement and 2,850 kg of steel on it.
(i) What is the total weight loaded onto the truck?
(ii) How much more weight can the truck carry before reaching its maximum capacity?
Answer:
Given, total weight that a truck can carry = 8,250 kg
The weight of cement loaded onto truck = 3,675 kg
and the weight of steal loaded onto truck = 2,850 kg
(i) The total weight loaded onto truck
= (3,675+2,850) kg
= 6,525 kg
(ii) The remaining capacity of the truck
= 8,250 kg – 6,525 kg
= 1,725 kg
So, the truck can carry 1,725 kg more weight before reaching its maximum capacity.
Quick Sums and Differences (NCERT Pg 51-52)
Question 17.
Sukanta likes the numbers 10,100 , 1,000 , and 10,000. He wants to figure out what number he should add to a given number such that the sum is 100 or 1,000. Help him fill in the blanks with an appropriate number.
(i) 32 +……. = 100…..
Sukanta’s friend Piku shows him an interesting way to solve the problems.
(ii) 59 + ….. = 100
Try this method for the number 59.
(iii) Now, use this method to solve the following.
(a) 877 + ……… = 1,000
(b) 666 +……… = 1,000
(c) 4,103 + ……… = 10,000
(d) 5,555 + ……… = 10,000
(iv) Will this method work if the units digit is 0? What do you think? What other methods can you use to find the missing number to fill in the blanks? Share your thoughts in the class.
(a) 180+……… = 1,000
(b) 760+……… = 1,000
(c) 400+……… = 1,000
Answer:
(i) Yes, this method will always work.
We can subtract the initial numbers from 1,000 to get the missing number.
(a) We have, 1,000 – 180 = 820
So, 180 + 820 = 1,000
(b) We have, 1,000 – 760 = 240
So, 760 + 240 = 1,000
(c) We have, 1,000 – 400 = 600
So, 400+6 0 0=1,000
Question 18.
(i) Namita likes the number 9. She wants to subtract 9 or 99 from any number. Find a way to quickly subtract 9 or 99 from any number.
(a) 67 – 9 = ………..
(b) 83 – 9 = ………..
(c) 144-9 = ………..
(d) 187-99 = ………..
(e) 247-99 = ………..
(f) 763-99 = ………..
(ii) Now, use the above solutions to find answers to the following problems.
Do not calculate again.
Namita wonders if she can get 9 or 99 as the answer to any subtraction problem. Find a way to get the desired answer.
(a) 32 – ………. = 9
(b) 56 – ………. = 9
(c) 877 -………. = 99
(d) 666 – ………. = 99
Answer:
(i) (a) We have, 67 – 9 = 67 – 10 + 1
= 57 + 1 = 58
(b) We have, 83-9=83-10+1
= 73 + 1 = 74
(c) We have, 144-9=144-10+1
= 134 + 1 = 135
(d) We have, 187-99=187-100+1
= 87 + 1 = 88
(e) We have, 247 – 99 = 247 – 100 + 1
= 147 + 1 = 148
(f) We have, 763 – 99 = 763 – 100 + 1
= 663 + 1 = 664
(ii) Here, we need to subtract 9 or 99 from the initial number.
(a) We have, 32 – 9 = 23
So, 32 – 23 = 9
(b) We have, 56-9 = 47
So, 56 – 47 = 9
(c) We have, 877-99 = 778
So, 877 – 778 = 99
(d) We have, 666 – 99 = 567
So, 666 – 567 = 99
Let Us Think and Solve (NCERT Pg 53)
Question 19.
(i) List all palindrome numbers between 100 and 200.
(ii) List all palindrome numbers between 900 and 1,200.
(iii) List all palindrome numbers between 25,000 and 27,000.
Answer:
(i) All the palindrome numbers between 100 and 200 are 101, 111, 121, 131, 141,151,161,171,181 and 191.
(ii) All the palindrome numbers between 900 and 1,200 are 909,919,929,939,949, 959,969,979,989,999,1,001 and 1,111.
(iii) All the palindrome numbers between 25,000 and 27,000 are 25,052,25,152, 25,252, 25,352, 25,452, 25,552, 25,652, 25,752,25,852,25,952,26,062,26,162, 26,262, 26,362, 26,462, 26,562, 26,662, 26,762,26,862 and 26,962.
Question 20.
(i) In a 3 × 3 grid, arrange the numbers 1 to 9 such that each row and each column has numbers in an increasing (inc) order. Each number should be used only once.
(ii) This time, fill the grid such that each row and column has numbers in decreasing (dec) order.
(iii) Now, fill the grids below with numbers (1-9) based on the inc (increasing) and dec (decreasing) conditions, as indicated below.
Answer:
(i) We have,
(ii) We have,
(iii) (a) We have,
(b) We have,
(c) We have,
Even and Odd Numbers (NCERT Pg 54)
Question 21.
Circle the numbers that are even.
(i) 297
(ii) 498
(iii) 724
(iv) 100
(v) 199
(vi) 789
(vii) 49
(viii) 6,893
(ix) 846
(x) 111
(xi) 222
(xii) 1,023
Answer:
We know that the numbers which have 0,2,4,6 and 8 as their unit place digits, are called even numbers. So, here, 498, 724,100,846 and 222 are even numbers.
Question 22.
Observe the given arrangement.
(i) Add 2 to 18. What changes or does not change in the arrangement?
(ii) Add 2 to 23. What changes or does not change in the arrangement?
Answer:
(i) We have the below paired arrangement for 18+2 i.e. 20.
Since, 18 is an even number is perfectly paired by using 9 pairs of circles. and 20 is also an even number.
So, 20 is also perfectly paired by using 10 pairs of circles.
Therefore, only the number of pairs changes from 9 to 10.
(ii) We have the below paired arrangement for 23+2 i.e. 25.
Since, 23 is an odd number, which is paired by using 11 pairs of circles and 1 circle left unpaired.
In the same way, as 25 is an odd number, So, 25 is paired by using 12 pairs of circles and 1 circle left unpaired.
Therefore, only the number of pairs changes from 11 to 12.
Question 23.
What do you notice about the sums in each of the following cases?
Do you think it will be true for all pairs of such numbers? Explain your observations. You may use the paired arrangement to explain your thinking.
(i) 12 and 6 are a pair of even numbers. Choose 5 such pairs of even numbers. Add the numbers in each of the pairs.
(ii) 13 and 9 are a pair of odd numbers. Choose 5 such pairs of odd numbers. Add the numbers in each of the pairs.
(iii) 7 and 12 are a pair of odd and even numbers. Choose 5 such pairs of odd and even numbers. Add the numbers in each of the pairs.
Answer:
(i) We can choose the following 5 pairs of even numbers: 22,2 ; 4,8 ; 10,20 ; 2,14 ; 12,10 Now, 12 + 6 = 18 (even)
22 + 2 = 24 (even)
4 + 8 = 12 (even)
10 + 20 = 30 (even)
2 + 14 = 16 (even)
12 + 10 = 22 (even)
We observe that the sum of any two even numbers is always an even number.
(ii) We can choose the following 5 pairs of odd numbers: 13,19 ; 3,5 ; 7,11 ; 1,9 ; 15,17
Now, 13 + 9 = 22 (even)
13 + 19 =32 (even)
3 + 5 = 8 (even)
7 + 11 = 18 (even)
1 + 9 = 10 (even)
15 + 17 = 32 (even)
We observe that the sum of any two odd numbers is always an even number.
(iii) We can choose the following 5 pairs of an odd an even number:
3,4 ; 9,2 ; 1,6 ; 13,8 ; 17,12
Now, 7 + 12 = 19 (odd)
3 + 4 = 7 (odd)
9 + 2 = 11 (odd)
1 + 6 = 7 (odd)
13 + 8 = 21 (odd)
17 + 12 = 29 (odd)
We observe that the sum of an odd number and an even number is always an odd number.
Let Us Think (NCERT Pg 54-55)
Question 24.
Jincy opened her piggy bank. She found 8 coins of ₹ 1,9 coins of ₹ 2 and 5 coins of ₹ 5. She wants to buy stickers worth ₹ 38. What possible combination of coins can she use to pay the exact amount?
Answer:
Given, the number of coins of ₹ 1 = 8, the number of coins of ₹ 2 = 9 and the number of coins of ₹ 5 = 5 The possible combinations of coins, which Jincy can use to pay ₹ 38 are given below:
Question 25.
(i) Raghu is fond of his grandfather’s torch. He starts playing with it. He presses the switch once and the light turns ON. He presses it a second time and the light turns OFF. He presses the switch a third time and the light turns ON. He keeps doing this several times. Will the torch be ON or OFF after the 23rd press? How do you know?
(ii) For what number of presses will the torch be ON? For what number of presses of the switch will the torch be OFF?
Answer:
Given, on first press, the light turns ON, on second press, the light turns OFF, on third press, the tight turns ON, on fourth press, the light turns OFF and so on.
(i) Here, we observe that the light turns ON for all the odd number presses and OFF for all the even number presses. Since, 23 is an odd number. So, the torch will be ON after the 23rd press.
(ii) The torch will be ON for any odd number of presses, i.e. 1,3,5,7,9……… and the torch will be OFF for any even number of presses, i.e. 2,4,6,8,……….
Question 26.
Mountain climbing
Priyanka Mohite is the first Indian woman to climb five Himalayan peaks above 8,000 metres. In addition to that, she has also climbed mountain peaks in other parts of the world. Read the table below and answer the questions that follow.
(i) Which is the highest peak she climbed?
(ii) What is the difference in height between the highest and lowest peaks she has climbed, as per the table.
(iii) What is the difference between heights of Mount Elbrus and Mount Kanchenjunga?
(iv) If Priyanka was 20 years old when she summited Mount Everest in 2013, in which year was she born?
Answer:
On arranging the heights in increasing order, we get
5,642 < 5,895 < 8,091 < 8,485 < 8,516 < 8,586 < 8,848.
(i) Clearly, the maximum height is 8,848 m, which corresponds to Mount Everest.
(ii) Here, the maximum height is 8,848 and the height of the lowest peak is 5,642.
So, the difference in height between the highest and lowest peaks she has climbed is given by 8,848 – 5,642 = 3,206 m
(iii) Given, the heights of Mount Elbrus and Mount Kanchenjunga are 5,642 m and 8,586 m, respectively. So, the difference between their heights = 8,586 – 5,642 = 2,944 m
(iv) Given, Priyanka was 20 years old in 2013. So, she was born in 2013-20 i.e. 1993.
Math Metric Mela (NCERT Pg 56)
Question 27.
A grand Math Metric Mela was held at the district level to celebrate young math whizzes. Every participating student was to receive a certificate of participation. The organisers got certificates printed for each district before the Mela. The number of certificates printed and the number of students who attended the competition in each district are as follows.
(i) For each district, find out if the number of certificates were sufficient?
(ii) If insufficient, calculate how many certificates fell short.
(iii) If extra, calculate how many certificates were in excess.
Answer:
(i) Since, 18,225 > 18,104
∴ The number of certificates printed for Chittoor were sufficient.
Since, 19,043 < 19,265.
∴ The number of certificates printed for Jaunpur were not sufficient.
Since, 20,863 >19,974.
∴ The number of certificates printed for Raigad were sufficient.
(ii) The district Jaunpur needs to print 19,265-19,043.
i.e. there were 222 certificates fell short.
(iii) The district chittor has 18,225-18,104 i.e. there were 121 more certificates.
The district Raigad has 20,863-19,974 i.e. there were 889 more certificates.
Let Us Do (NCERT Pg 56)
Question 28.
Add
(i) 2,009+7,388
(ii) 26,444+71,111
(iii) 777+888
(iv) 1,234+1,234
(v) 56+56,789
(vi) 777+77,777
(vii) 5,922+9,221
(viii) 4,321+8,765
(ix) 50,050+55,000
Answer:
(i) We have, 2,009 + 7,388 = 9,397
(ii) We have, 26,444 + 71,111 = 97,555
(iii) We have, 777 + 888 = 1,665
(iv) We have, 1,234 + 1,234 = 2,468
(v) We have, 56 + 56,789 = 56,845
(vi) We have, 777 + 77,777 = 78,554
(vii) We have, 5,922 + 9,221 = 15,143
(viii) We have, 4,321 + 8,765 = 13,086
(ix) We have, 50,050 + 55,000 = 1,05,050
Question 29.
Subtract
(i) 458 – 226
(ii) 7,777 – 4,449
(iii) 65,447 – 47,299
(iv) 1,234 – 123
(v) 12,345 – 1,234
(vi) 56,789 – 56
(vii) 87,326 – 11,111
(viii) 878 – 52
(ix) 749 – 222
Answer:
(i) We have, 458 – 226 = 232
(ii) We have, 7,777 – 4,449 = 3,328
(iii) We have, 65,447 – 47,299 = 18,148
(iv) We have, 1,234 – 123 = 1,111
(v) We have, 12,345 – 1,234 = 11,111
(vi) We have, 56,789 – 56 = 56,733
(vii) We have, 87,326 – 11,111 = 76,215
(viii) We have, 878 – 52 = 826
(ix) We have, 749 – 222 = 527
Question 30.
Ambrish saved ₹ 92,375 over a year to buy cows and goats. He buys a cow for ₹ 26,000 and a goat for ₹ 17,000. He also buys a milking machine for ₹ 19,873. Does-he have enough money to buy these? How much more or less does he have than he needs?
Answer:
Given, the total money, which Ambrish has
= ₹ 92,375,
the cost of a cow = ₹ 26,000,
the cost of a goat = ₹ 17,000
and the cost of a milking machine
= ₹ 19,873
So, the total money-used in buying a cow, a goat and a milking
= ₹ 26,000 + ₹ 17,000 + ₹ 19,873
= ₹ 62,873
Clearly, yes, he has enough money to buy these.
As 92,375 – 62,873 = 29,502
So, he has ₹ 29,502 more than he needs.
Question 31.
A factory produces 54,000 nuts and bolts in a day. An order is placed for 85,300 nuts and bolts. How many more nuts and bolts does the factory need to produce to complete the order?
Answer:
Given, the number of nuts and bolts produced by a factory in a day = 54,000 and the number of nuts and bolts, which are ordered = 85,300
So, the number of nuts bolts needed to produce to complete the order
= 85,300 – 54,000
= 31,300.
Question 32.
Virat Kohli has scored 27,599 runs. He has 6,758 runs less than Sachin
Tendulkar. How many runs has Sachin Tendulkar scored?
Answer:
Given, the runs scored by Virat Kohli
= 27,599
Also, Virat Kohli has 6,758 runs less than Sachin Tendulkar.
So, the runs scored by Sachin Tendulkar
= 27,599 + 6,758
= 34,357