Samacheer Kalvi Class 7 Maths Term 3 Chapter 1 Number System Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 3 chapter 1 Number System. Here students can easily find all the solutions for Number System Exercise 1.1, 1.2, 1.3, 1.4 and 1.5. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Number System Exercise 1.1 Solutions :
(1) Round each of the following decimals to the nearest whole number.
(i) 8.71
(ii) 26.01
(iii) 69.48
(iv) 103.72
(v) 49.84
(vi) 101.35
(vii) 39.814
(viii) 1.23
Solution :
(i) 8.71
Given, number 8.71
Underline the digit to be rounded – 8.71
We observe that the digit after the underlined digit place value is 7 which is more than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 9.00
Hence, the rounded value of 8.71 to the nearest whole number is 9.
(ii) 26.01
Underline the digit to be rounded – 26.01
Since the digit next to the underlined digit is 0 which is less than 5, the underlined digit 6
remains the same. We get 26.00
Hence, the rounded value of 26.01 to the nearest whole number is 26.
(iii) 69.48
Given number 69.48
Underline the digit to be rounded – 69.48
Since the digit next to the underlined digit is 4 which is less than 5, the underlined digit 9 remains the same. We get 69.00
Hence, the rounded value of 69.48 to the nearest whole number is 69.
(iv) 103.72
Given, number 103.72
Underline the digit to be rounded – 103.72
We observe that the digit after the underlined digit place value is 7 which is more than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 104
Hence, the rounded value of 103.72 to the nearest whole number is 104
(v) 49.84
Given, number 49.84
Underline the digit to be rounded – 49.84
We observe that the digit after the underlined digit place value is 8 which is more than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 50
Hence, the rounded value of 49.84 to the nearest whole number is 50.
(vi) 101.35
Given number 101.35
Underline the digit to be rounded – 101.35
Since the digit next to the underlined digit is 3 which is less than 5, the underlined digit 1 remains the same. We get 101
Hence, the rounded value of 101.35 to the nearest whole number is 101
(vii) 39.814
Given, number 39.814
Underline the digit to be rounded – 39.814
We observe that the digit after the underlined digit place value is 8 which is more than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 40
Hence, the rounded value of 39.814 to the nearest whole number is 40
(viii) 1.23
Given number 1.23
Underline the digit to be rounded – 1.23
Since the digit next to the underlined digit is 2 which is less than 5, the underlined digit 1 remains the same. We get 1
Hence, the rounded value of 1.23 to the nearest whole number is 1
(2) Round each decimal number to the given place value.
(i) 5.992 to tenth place
(ii) 21.805 to hundredth place
(iii) 35.0014 to thousandth place
Solution :
(i) Given number 5.992
Underline the digit to be rounded 5.992
Since the digit right to the tenth place is 9, we add 1 to the tenth place (underlined digit) of 5.992 and we get 6.00.
(ii) Given number 21.805
Underline the digit to be rounded 21.805
Since the digit right to the hundredth place is 5, we add 1 to the hundredth place (underlined digit) of 21.805 and we get 21.810.
(iii) Given number 35.0014
Underline the digit to be rounded 35.0014
Since the digit right to the thousandth place is 5, we remain same thousandth place (underlined digit) of 35.0014 and we get 35.001.
(3) Round the following decimal numbers upto 1 places of decimal
(i) 123.37
(ii) 19.99
(iii) 910.546
Solution :
(i) Given number 123.37
Since one place of decimal means round to the nearest tenth place.
Underline the digit to be rounded 123.37
Hence, the digit right to the tenth place is 7, we add 1 to the tenth place (underlined digit) of 123.37 and we get 123.40.
(ii) Given number 19.99
Since one place of decimal means round to the nearest tenth place.
Underline the digit to be rounded 19.99
Hence, the digit right to the tenth place is 9, we add 1 to the tenth place (underlined digit) of 19.99 and we get 20.00.
(iii) Given number 910.546
Since one place of decimal means round to the nearest tenth place.
Underline the digit to be rounded 910.546
Hence, the digit right to the tenth place is 4, we remain same the tenth place (underlined digit) of 910.546 and we get 910.500.
(4) Round the following decimal numbers upto 2 places of decimal
(i) 87.755
(ii) 301.513
(iii) 79.997
Solution :
(i) Given number 87.755
Since 2 place of decimal means round to the nearest hundredth place.
Underline the digit to be rounded 87.755
Thus, the digit right to the hundredth place is 5, we add 1 to the hundredth place (underlined digit) of 87.755 and we get 87.76
(ii) Given number 301.513
Since 2 place of decimal means round to the nearest hundredth place.
Underline the digit to be rounded 301.513
Hence, the digit right to the hundredth place is 3, we remain same the hundredth place (underlined digit) of 301.513 and we get 301.510
(iii) Given number 79.997
Since 2 place of decimal means round to the nearest hundredth place.
Underline the digit to be rounded 79.997
Therefore, the digit right to the hundredth place is 7, we add 1 to the hundredth place (underlined digit) of 79.997 and we get 80.00
(5) Round the following decimal numbers upto 3 place of decimal
(i) 24.4003
(ii) 1251.2345
(iii) 61.00203
Solution :
(i) Given number 24.4003
Since 3 place of decimal means round to the nearest thousandth place.
Underline the digit to be rounded 24.4003
Since the digit right to the thousandth place is 3, we remain same the thousandth place (underlined digit) of 24.4003 and we get 24.4000
(ii) 1251.2345
Since 3 place of decimal means round to the nearest thousandth place.
Underline the digit to be rounded 1251.2345
Since the digit right to the thousandth place is 5, we add 1 to the thousandth place (underlined digit) of 1251.2345 and we get 1251.2345
(iii) 61.00203
Since 3 place of decimal means round to the nearest thousandth place.
Underline the digit to be rounded 61.00203
Since the digit right to the thousandth place is 0, we remain same the thousandth place (underlined digit) of 61.00203 and we get 61.00200
Number System Exercise 1.2 Solutions :
(2) Add the following by using place value grid.
(i) 25.8 + 18.53
(ii) 17.4 + 23.435
Solution :
(i) 25.8 + 18.53
Decimal No. 25.8
Tens = 2
Ones = 5
Tenth = 8
Hundredth = 0
Decimal No. 18.53
Tens = 1
Ones = 8
Tenth = 5
Hundredth = 3
Decimal No. 44.33
Tens = 4
Ones = 4
Tenth = 3
Hundredth = 3
Therefore, 25.8 + 18.53 = 44.33
(ii) 17.4 + 23.435
Decimal No. 17.4
Tens = 1
Ones = 7
Tenth = 4
Hundredth = 0
Thousandth = 0
Decimal No. = 23.435
Tens = 2
Ones = 3
Tenth = 4
Hundredth = 3
Thousandth = 5
Decimal No = 40.835
Tens = 4
Ones = 0
Tenth = 8
Hundredth = 3
Thousandth = 5
Therefore, 17.4 + 23.435 = 40.835
(4) Subtract the following by using place value grid.
(i) 6.567 from 9.231
(ii) 3.235 from 7
Solution :
(i) 6.567 from 9.231
= 9.231 – 6.567 = ?
Let us use the place value grid.
Decimal No = 9.231
Ones = 9
Tenth = 2
Hundredth = 3
Thousandth = 1
Decimal No = 6.567
Ones = 6
Tenth = 5
Hundredth = 6
Thousandth = 7
Decimal No = 2.664
Ones = 2
Tenth = 6
Hundredth = 6
Thousandth = 4
Therefore, 9.231 – 6.567 = 2.664
(ii) 3.235 from 7
7 – 3.235 = ?
Let us use the place value grid.
Decimal No = 7
Ones = 7
Tenth = 0
Hundredth = 0
Thousandth = 0
Decimal No = 3.235
Ones = 3
Tenth = 2
Hundredth = 3
Thousandth = 5
Decimal No = 3.765
Ones = 3
Tenth = 7
Hundredth = 6
Thousandth = 5
Therefore, 7 – 3.235 = 3.765
(5) Simplify : 23.5 − 27.89 + 35.4 − 17
Solution :
23.5 − 27.89 + 35.4 − 17
= -4. 39 + 18.4
= 14.01 …(Simplified)
(6) Sulaiman bought 3.350 kg of Potato, 2.250 kg of Tomato and some Onions. If the weight of the total items are 10.250 kg, then find the weight of Onions?
Solution :
Given, Sulaiman bought potato = 3.350 kg
Sulaiman bought tomato = 2.250 kg
Let Sulaiman bought onion = x kg
Since, total weight of items = 10.250 kg
Thus, 3.350 + 2.250 + x = 10.250
5.600 + x = 10.250
x = 10.250 – 5.600
x = 4.650 kg
Hence, the weight of onion is 4.650 kg.
(7) What should be subtracted from 7.1 to get 0.713?
Solution :
Let x be a number that subtract from 7.1
Thus, 7.1 – x = 0 . 713
x = 7.1 – 0.713
x = 6.387
(8) How much is 35.6 km less than 53.7 km?
Solution :
53.7 – 35.6 = 18.1
Hence, 35.6 km less than 53.7 km by 18.1 km.
(9) Akilan purchased a geometry box for ₹ 25.75, a pencil for ₹ 3.75 and a pen for ₹ 17.90. He gave ₹ 50 to the shopkeeper. What amount did he get back?
Solution :
Cost of geometry box = ₹25.75
Cost of pencil = ₹3.75
Cost of pen = ₹ 17.90
Thus, total cost
= 25.74 + 3.75 + 17.90
= ₹47.4
Since, He gave ₹ 50 to the shopkeeper.
Return back,
= 50 – 47.4
= ₹2.6
Hence, Akilan ₹2.6 get back from shopkeeper.
(10) Find the perimeter of an equilateral triangle with a side measuring 3.8 cm.
Solution :
Given, side of equilateral triangle is 3.8 cm
Thus, the perimeter of an equilateral triangle,
= 3.8 + 3.8 + 3.8
= 11.4 cm
Objective Type Question Solutions :
(11) 1.0 + 0.83 = ?
(i) 0.17
(ii) 0.71
(iii) 1.83
(iv) 1.38
Solution :
Correct Option → (iii)
= 1.0 + 0.83
= 1.83
(12) 7.0 – 2.83 = ?
(i) 3.47
(ii) 4.17
(iii) 7.34
(iv) 4.73
Solution :
Correct Option → (ii)
= 7.0 – 2.83
= 4.17
(13) Subtract 1.35 from 3.51
(i) 6.21
(ii) 4.86
(iii) 8.64
(iv) 2.16
Solution :
Correct Option → (iv)
= 3.51 – 1.35
= 2.16
(14) Sum of two decimals is 4.78. If one decimal is 3.21, then the other one is
(i) 1.57
(ii) 1.75
(iii) 1.59
(iv) 1.58
Solution :
Correct Option → (i)
Then the other one will be 1.57
(15) The difference of two decimals is 86.58 and one of the decimal is 42.31 Find the other one
(i) 128.89
(ii) 128.69
(iii) 128.36
(iv) 128.39
Solution :
Correct Option → (i)
The other one will be 128.89.
Number System Exercise 1.3 Solutions :
(1) Find the product of the following
(i) 0.5 × 3
(ii) 3.75 × 6
(iii) 50.2 × 4
(iv) 0.03 × 9
(v) 453.03 × 7
(vi) 4 × 0.7
Solution :
(i) 0.5 × 3
= 1.5
(ii) 3.75 × 6
= 22.50
(iii) 50.2 × 4
= 200.8
(iv) 0.03 × 9
= 0.27
(v) 453.03 × 7
= 3171.21
(vi) 4 × 0.7
= 2.8A
(2) Find the area of the parallelogram whose base is 6.8 cm and height is 3.5 cm.
Solution :
Given, base of parallelogram = 6.8 cm
Height of parallelogram = 3.5 cm
Thus, area of the parallelogram
= base × height
= 6.8 × 3.5
= 23.80 sq.cm
Hence, the area of the parallelogram is 23.80 sq.cm.
(3) Find the area of the rectangle whose length is 23.7 cm and breadth is 15.2 cm.
Solution :
Given, length of rectangle = 23.7 cm
breadth of rectangle = 15.2 cm.
Thus, area of rectangle = length × breadth
= 23.7 × 15.2
= 360.24 sq.cm
Thus, the area of the rectangle is 360.24 sq.cm.
(4) Multiply the following
(i) 2.57 × 10
(ii) 0.51 × 10
(iii) 125.367 × 100
(iv) 34.51 × 100
(v) 62.735 × 100
(vi) 0.7 × 10
(vii) 0.03 × 100
(viii) 0.4 × 1000
Solution :
(i) 2.57 × 10
= 25.7
(ii) 0.51 × 10
= 5.1
(iii) 125.367 × 100
= 12536.7
(iv) 34.51 × 100
= 3451
(v) 62.735 × 100
= 6273.5
(vi) 0.7 × 10
= 7
(vii) 0.03 × 100
= 3
(viii) 0.4 × 1000
= 400
(5) A wheel of a baby cycle covers 49.7 cm in one rotation. Find the distance covered in 10 rotations.
Solution :
Given, a wheel of a baby cycle covers 49.7 cm in one rotation.
Thus, for 1 rotation = 49.7 cm
Therefore, for 10 rotation
= 49.7 × 10
= 497 cm
Hence, a wheel of a baby cycle covers 497 cm in 20 rotations.
(6) A picture chart costs ₹ 1.50. Radha wants to buy 20 charts to make an album. How much does she have to pay?
Solution :
Given, a picture chart costs ₹ 1.50.
Since, Radha wants to buy 20 charts to make an album.
Thus, coat of 1 chart = ₹1.50
So, cost of 20 chart
= 20 × 1.50
= ₹30
Hence, Radha have to pay ₹30.
(7) Find the product of the following.
(i) 3.6 × 0.3
(ii) 52.3 × 0.1
(iii) 537.4 × 0.2
(iv) 537.4 × 0.2
(v) 62.2 × 0.23
(vi) 1.02 × 0.05
(vii) 10.05 × 1.05
(viii) 101.01 × 0.01
(ix) 100.01 × 1.1
Solution :
(i) 3.6 × 0.3
= 1.08
(ii) 52.3 × 0.1
= 5.23
(iii) 537.4 × 0.2
= 1007.48
(iv) 0.6 × 0.06
= 0.036
(v) 62.2 × 0.23
= 14.306
(vi) 1.02 × 0.05
= 0.051
(vii) 10.05 × 1.05
= 10.5525
(viii) 101.01 × 0.01
= 1.0101
(ix) 100.01 × 1.1
= 110.011
Objective Type Question Solutions :
(8) 1.07 × 0.1 = ____
(i) 1.070
(ii) 0.107
(iii) 10.70
(iv) 11.07
Solution :
= 1.07 × 0.1
= 0.107
So, the correct answer of this question is option (ii)
(9) 2.08 × 10 = ______
(i) 20.8
(ii) 208.0
(iii) 0.208
(iv) 280.0
Solution :
= 2.08 × 10
= 20.8
Hence, option (i) is the correct answer.
(10) A frog jumps 5.3 cm in one jump. The distance travelled by the frog in 10 jumps is ___
(i) 0.53 cm
(ii) 530 cm
(iii) 53.0 cm
(iv) 53.5 cm
Solution :
The distance travelled by the frog in 10 jumps will be 53.0 cm.
Number System Exercise 1.4 Solutions :
(1) Simplify the following.
(i) 0.6 ÷ 3
(ii) 0.90 ÷ 5
(iii) 4.08 ÷ 4
(iv) 21.56 ÷ 7
(v) 0.564 ÷ 6
(vi) 41.36 ÷ 4
(vii) 298.2 ÷ 3
Solution :
(i) 0.6 ÷ 3
= 0.2
(ii) 0.90 ÷ 5
= 0.18
(iii) 4.08 ÷ 4
= 1.02
(iv) 21.56 ÷ 7
= 3.08
(v) 0.564 ÷ 6
= 0.094
(vi) 41.36 ÷ 4
= 10.34
(vii) 298.2 ÷ 3
= 99.4
(2) Simplify the following.
(i) 5.7 ÷ 10
(ii) 93.7 ÷ 10
(iii) 0.9 ÷ 10
(iv) 301.301 ÷ 10
(v) 0.83 ÷ 10
(vi) 0.062 ÷ 10
Solution :
(i) 5.7 ÷ 10
= 5.7/10
= 57/100
= 0.57 ….(Simplified)
(ii) 93.7 ÷ 10
= 93.7/10
= 937/100
= 9.37 ….(Simplified)
(iii) 0.9 ÷ 10
= 0.9/10
= 9/100
= 0.09 ….(Simplified)
(iv) 301.301 ÷ 10
= 301301/10000
= 30.1301 ….(Simplified)
(v) 0.83 ÷ 10
= 0.83/10
= 83/1000
= 0.083 ….(Simplified)
(vi) 0.062 ÷ 10
= 0.062/10
= 62/10000
= 0.0062 ….(Simplified)
(3) Simplify the following.
(i) 0.7 ÷ 100
(ii) 3.8 ÷ 100
(iii) 49.3 ÷ 100
(iv) 463.85 ÷ 100
(v) 0.3 ÷ 100
(vi) 27.4 ÷ 100
Solution :
(i) 0.7 ÷ 100
= 0.7/100
= 7 /1000
= 0.007 …(Simplified)
(ii) 3.8 ÷ 100
= 38/1000
= 0.038 …(Simplified)
(iii) 49.3 ÷ 100
= 493/1000
= 0.493 …(Simplified)
(iv) 463.85 ÷ 100
= 46385/10000
= 4.6385 …(Simplified)
(v) 0.3 ÷ 100
= 3/1000
= 0.003 …(Simplified)
(vi) 27.4 ÷ 100
= 274/1000
= 0.274 …(Simplified)
(4) Simplify the following.
(i) 18.9 ÷ 1000
(ii) 0.87 ÷ 1000
(iii) 49.3 ÷ 1000
(iv) 0.3 ÷ 1000
(v) 382.4 ÷ 1000
(vi) 93.8 ÷ 1000
Solution :
(i) 18.9 ÷ 1000
= 18.9/1000
= 189/10000
= 0.0189 …(Simplified)
(ii) 0.87 ÷ 1000
= 87/100000
= 0.00087 …(Simplified)
(iii) 49.3 ÷ 1000
= 493/1000
= 0.0493 …(Simplified)
(iv) 0.3 ÷ 1000
= 3/10000
= 0.0003 …(Simplified)
(v) 382.4 ÷ 1000
= 3824/1000
= 0.3824 …(Simplified)
(vi) 93.8 ÷ 1000
= 938/10000
= 0.0938 …(Simplified)
(5) Simplify the following.
(i) 19.2 ÷ 2.4
(ii) 4.95 ÷ 0.5
(iii) 19.11 ÷ 1.3
(iv) 0.399 ÷ 2.1
(v) 5.4 ÷ 0.6
(vi) 2.197 ÷ 1.3
Solution :
(i) 19.2 ÷ 2.4
= 192/10 ÷ 24/10
= 192/10 * 10/24
= 192/24
= 8 …(Simplified)
(ii) 4.95 ÷ 0.5
= 495/100 ÷ 5/10
= 495/100 * 10/5
= 99/10
= 9.9 …(Simplified)
(iii) 19.11 ÷ 1.3
= 1911/100 ÷ 13/10
= 1911/100 * 10/13
= 147/10
= 14.7 …(Simplified)
(iv) 0.399 ÷ 2.1
= 399/1000 ÷ 21/10
= 399/1000 * 10/21
= 19/100
= 0.19 …(Simplified)
(v) 5.4 ÷ 0.6
= 54/10 ÷ 6/10
= 54/10 * 10/6
= 9 …(Simplified)
(vi) 2.197 ÷ 1.3
= 2197/1000 ÷ 13/10
= 2197/1000 * 10/13
= 169/100
= 1.69 …(Simplified)
(6) Divide 9.55 kg of sweet among 5 children. How much will each child get?
Solution :
Given, total sweet = 9.55 kg
Children = 5
Thus, 9.55 ÷ 5
= 955/100 ÷ 5
= 955/100 * 1/5
= 191/100
= 1.91
Hence, each child get 1.91 kg sweet.
(7) A vehicle covers a distance of 76.8 km for 1.2 litre of petrol. How much distance will it cover for one litre of petrol?
Solution :
Given, A vehicle covers a distance of 76.8 km for 1.2 litre of petrol.
Thus, for 1.2 litre = 76.8 km
So, for 1 litre = x km
Therefore, x = 76.8 ÷ 1.2
= 768/10 ÷ 12/10
= 768/10 × 10/12
= 64 km
Therefore, it will cover 64 km in one litre petrol.
(8) Cost of levelling a land at the rate of 15.50 sq.ft is 10075. find the area of the land.
Solution :
Given, cost of levelling a land at the rate of ₹ 15.50 sq. ft is ₹ 10,075.
Thus, cost of 1 sq.ft = ₹15.50
Let cost of x sq.ft = ₹ 10,075
Therefore, x = 10,075 ÷ 15.50
= 10,075 ÷ 155/10
= 10,075 × 10/155
= 65 × 10
= 650
Therefore, the area of land will be 650 sq. ft
(9) The cost of 28 books are ₹ 1506.4. Find the cost of one book.
Solution :
Given, the cost of 28 books are ₹ 1506.4
Thus, cost of 28 books = ₹1506.4
So, cost of 1 book = ₹X
Therefore, X = 1506.4 ÷ 28
= 15064/10 × 1/28
= 538/10
= 53.8
Therefore, the cost of 1 book will be ₹53.8
(10) The product of two numbers is 40.376. One number is 14.42. Find the other number.
Solution :
Let x be other number
According to question,
14.42 × x = 40.376
x = 40.376 ÷ 14.42
x = 40376/1000 ÷ 1442/100
x = 40376/1000 × 100/1442
x = 28/10
x = 2.8
Therefore, the other number is 2.8
Objective Type Question Solutions :
(11) 5.6 ÷ 0.5 = ?
(i) 11.4
(ii) 10.4
(iii) 0.14
(iv) 11.2
Solution :
Correct Option → (iv)
= 5.6 ÷ 0.5
= 11.2
(12) 2.01 ÷ 0.03 = ?
(i) 6.7
(ii) 67.0
(iii) 0.67
(iv) 0.067
Solution :
Correct Option → (ii)
= 2.01 ÷ 0.03
= 67.0
(13) 0.05 ÷ 0.5 = ?
(i) 0.01
(ii) 0.1
(iii) 0.10
(iv) 1.0
Solution :
Correct Option → (ii)
= 0.05 ÷ 0.5
= 0.1
Number System Exercise 1.5 Solutions :
(1) Malini bought three ribbons of length 13.92 m, 11.5 m and 10.64 m. Find the total length of the ribbons?
Solution :
Given, Malini bought three ribbons of length 13.92 m, 11.5 m and 10.64 m.
Thus, total length
= 13.92 + 11.5 + 10.64
= 36.06 m
Therefore, the total length of the ribbons will be 36.06 m.
(2) Chitra has bought 10 kg 35 g of ghee for preparing sweets. She used 8 kg 59 g of ghee. How much ghee will be left?
Solution :
Given, Chitra has bought 10 kg 35 g of ghee for preparing sweets.
Total ghee = 10 kg 35 g
Used ghee = 8 kg 59 g
We know, 1 kg = 1000 g
Thus, 10 kg + 35 g – (8 kg + 59 g)
= 10000 g + 35 g – (8000 g+59 g)
= 10035 g – 8059 g
= 1976 g
= 1000 g +976 g
= 1 kg 976 g
Therefore, 1 kg 976 g ghee will left.
(3) If the capacity of a milk can is 2.53 l, then how much milk is required to fill 8 such cans?
Solution :
Given, the capacity of a milk can is 2.53 l,
Thus, for 1 can = 2.53 l
So, for 8 can = x l
Therefore,
x = 8 * 2.53
= 20.24 l
Therefore, we required 20.24 l milk to fill 8 such cans.
(4) A basket of orange weighs 22.5 kg. If each family requires 2.5 kg of orange, how many families can share?
Solution :
Given, a basket of orange weighs 22.5 kg.
Total orange = 22.5 kg
Supposed, each family requires 2.5 kg of orange
Therefore, 22.5 ÷ 2.5
= 225 ÷ 25
= 9
Therefore, there are 9 family share orange.
(5) A baker uses 3.924 kg of sugar to bake 10 cakes of equal size. How much sugar is used in each cake?
Solution :
Given, a baker uses 3.924 kg of sugar to bake 10 cakes of equal size.
Thus, sugar can use for 10 cake = 3.924 kg
So sugar can use for 1 cake = x kg
Therefore,
x = 3.924 ÷ 10
= 0.3924 kg
Therefore, 0.3924 kg sugar will used for 1 cake.
(6) Evaluate :
(i) 26.13 × 4.6
(ii) 3.628 + 31.73 – 2.1
Solution :
(i) 26.13 * 4.6
= 120.198
(ii) 3.628 + 31.73 – 2.1
= 35.358 – 2.1
= 33.258
(7) Murugan bought some bags of vegetables. Each bag weighs 20.55 kg. If the total weight of all the bags is 308.25 kg, how many bags did he buy?
Solution :
Given, Murugan bought some bags of vegetables. Each bag weighs 20.55 kg.
Since, total weight of all bags is 308.25 kg
Thus, 308.25 ÷ 20.55
= 30825/100 ÷ 2055/100
= 30825/100 × 100/2055
= 15
Therefore, Murugan bought 15 bags.
(8) A man walks around a circular park of distance 23.761 m. How much distance will he cover in 100 rounds?
Solution :
Given, a man walks around a circular park of distance 23.761 m.
Since, for 1 round = 23.761 m
Now, for 100 round = x m
Therefore,
x = 23.761 × 100
= x = 2376.1 m
Therefore, a man in a 100 rounds 2376.1 m distance covered.
(9) How much 0.0543 is greater than 0.002?
Solution :
= 0.0543 – 0.002
= 0.0523
Therefore, 0.0543 is greater than 0.002 by 0.0523
(10) A printer can print 15 pages per minute. How many pages can it print in 4.6 minutes?
Solution :
Given, a printer can print 15 pages per minute.
Thus, pages print in 1 m = 15
So, pages print in 4.6 m = x
Therefore,
x = 15 × 4.6
= x = 69
Hence, pages can be printed in 4.6 m is 69.
Challenge Problem Solutions :
(11) The distance travelled by Prabhu from home to Yoga centre is 102 m and from Yoga centre to school is 165 m. What is the total distance travelled by him in kilometres (in decimal form)?
Solution :
Distance of from home to Yoga centre = 102 m
Distance of Yoga centre to school = 165 m
Thus, total distance
= 102 + 165
= 267 m
We know, 1 m = 1/1000 km
Now, 267 m = 267/1000 km
= 0.267 km
Hence, Prabhu traveled 0.267 km total distance.
(12) Anbu and Mala travelled from A to C in two different routes. Anbu travelled from place A to place B and from there to place C. A is 8.3 km from B and B is 15.6 km from C. Mala travelled from place A to place D and from there to place C. D is 7.5 km from A and C is 16.9 km from D. Who travelled more and by how much distance?
Solution :
Since, Anbu travelled from place A to place B and from there to place C.
Distance from A to B = 8.3 km
Distance from B to C = 15.6 km
Thus, Anbu total distance
= AB + BC
= 8.3 + 15.6
= 23.9 km
Also, Mala travelled from place A to place D and from there to place C.
Distance from A to D = 7.5 km
Distance from D to C = 16.9 km
Thus, Mala total distance
= AD + DC
= 7.5 + 16.9
= 24.4 km
Therefore,
= 24.4 – 23.9
= 0.5 km
Therefore, Mala travelled more distance by 0.5 km.
(13) Ramesh paid ₹ 97.75 per hour for a taxi and he used 35 hours in a week. How much he has to pay totally as taxi fare for a week?
Solution :
Given, Ramesh paid ₹ 97.75 per hour for a taxi
Thus, paid for 1 hour = ₹97.75
Paid for 35 hours = ₹X
Therefore,
X = 97.75 × 35
= ₹3421.25
Hence, Ramesh pay totally ₹3421.25 as taxi fare for a week.
(14) An Aeroplane travelled 2781.20 kms in 6 hours. Find the average speed of the aeroplane in km/hr.
Solution :
Given, an Aero plane travelled 2781.20 kms in 6 hours.
Thus, for 6 hours = 2781.20 km
So, for 1 hour = x km
Therefore,
x = 2781.20/6
x = 463.53 km
Thus, the average speed of the aero plane in km/hr is 463.53 km.
(15) Kumar’s car gives 12.6 km mileage per litre. If his fuel tank holds 25.8 litres then how far can he travel?
Solution :
Given, for 1 liter = 12.6 km
So, for 25.8 litre = x km
Therefore,
x = 12.6 * 25.8
= 325.08 km
Therefore, Kumar’s car will travel 325.08 km in 25.8 litres.
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