Samacheer Class 7 Maths - Number System Intext Questions

5 \(\frac { 1 }{ 5 } \) = \(\frac { 26 }{ 5 } \) = \(\frac{26 \times 2}{5 \times 2}\) = \(\frac { 52 }{ 10 } \) = 5.2

Comparing the digits of 0.567 and 0.576 from left to right, we have the tenths place same comparing the hundredths place 7 > 6.
⇒ 0.576 > 0.567

(i) 0.83 + 0.04
0.83 = \(\frac { 83 }{ 100 } \) and 0.04 = \(\frac { 4 }{ 100 } \)Shading the regions
0.83 and 0.04
The sum is the total shaded region.
S = 0.83 + 0.04 = 0.87
(ii) 0.35 – 0.09
0.35 = \(\frac { 35 }{ 100 } \) and 0.09 = \(\frac { 9 }{ 100 } \)Shading the regions 0.35 by shading 35 boxes out of 100. Striking off 9 boxes out of 35 shaded boxes to subtract 0.09 from 0.35.
The left over shaded boxes represent the required value.
∴ 0.35 – 0.09 = 0.26Try These (Text book Page No. 7)

(i) 1.2 + 3.5Here 1.2 is represented in blue colour and 3.5 is represented in Green colour. Sum of 1.2 and 3.5 is 4.7.
(ii) 3.5 – 2.3Representing 3.5 using 3 squares and 5 rectangular strips. Crossing out 2 squares from 3 squares and 3 rectangular strips from 5 to get the difference. So 3.5 – 2.3 = 1.2.Try These (Text book Page No. 9)

2.1 × 3.2 = 6.72 and 21 × 32 = 672.
In both the cases the digits ambers are the same. But the place value differs.Try These (Text book Page No. 13)

(i) 17.237 ÷ 10
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 17237 }{ 1000 } \)
= 1.7237
(ii) 17.237 ÷ 100
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 17237 }{ 100000 } \)
= 0.17237
(iii) 17.237 ÷ 1000
= \(\frac { 17237 }{ 1000 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 17237 }{ 1000000 } \)
= 0.017237
Try These (Text book Page No. 21)

(i) \(\frac { 9.25 }{ 0.25 } \)(ii) \(\frac { 8.6 }{ 4.3 } \)(iii) \(\frac { 44.1 }{ 0.21 } \)(iv) \(\frac { 9.6 }{ 1.2 } \)Think (Text book Page No. 22)

Price of 30 tablets = ₹ 22.63 = ₹ \(\frac { 2263 }{ 100 } \)
∴ Price of 1 tablet= \(\frac { 2263 }{ 100 } \) × \(\frac { 1 }{ 30 } \)
= \(\frac { 2263 }{ 30 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 2263 }{ 3 } \) × \(\frac { 1 }{ 1000 } \)
= 754.33 × \(\frac { 1 }{ 1000 } \)
= \(\frac { 754.33 }{ 1000 } \)
= 0.75433
Price of each tablet is ₹ 0.7543
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(i) 992; tenths place
Underlining the digit to be rounded 5. 9 92. Since the digit next to the underlined digit is 9 greater than 5, we add 1 to the underlined digit.
Hence the rounded number is 6.0.
(ii) 21.805; hundredth place
Underlining the digit to be rounded 21.805 since the digit next to the underlined digit is 5, we add 1 to the underlined digit.
Hence the rounded number is 21.81.
(iii) 35.0014; thousandth place
Underlining the digit to be rounded 35.00 1 4. Since the digit next to the underlined digit is 4 less than 5 the underlined digit remains the same.
Hence the rounded number is 35.001.

(i) 123.37
Rounding 123.37 upto one places of decimal means round to the nearest tenths place. Underling the digit in the tenths place of 123.37 gives 123. 3 7. Since the digit next to the tenth place value is 7 which is greater than 5, we add 1 to the underlined digit to get 123.4. Hence the rounded value of 123.37 upto one places of decimal is 123.4.
(ii) 19.99
Rounding 19.99 upto one places of decimal means round to the nearest tenth place. Underling the digit in the tenths place of 19.99 gives 19. 9 9. Since the digit next to the tenth place value is 9 which is greater than 5, we add 1 to the underlined digit to get 20.
Hence the rounded value of 19.99 upto one places of decimal is 20.0.
(iii) 910.546
Rounding 910.546 upto one places of decimal means round to the nearest tenths place underlining the digit in the tenths place of 910. 5 46 gives 910.546. Since the digit next to the tenth place value is 4, which is less than 5 the underlined digit remains the same. Hence the rounded value of 910.546 upto one places of decimal is 910.5.

(i) 87.755
Rounding 87.755 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.7 5 5 gives 87.755. Since the digit next to the hundredth place value is 5, we add 1 to the underlined digit.
Hence the rounded value of 87.755 upto two places of decimal is 87.76.
(ii) 301.513
Rounding 301.51 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 301.5 1 3 gives 301.5 1 3. Since the digit next to the underlined digit 3 is less than 5, the underlined digit remains the same.
∴ The rounded value of 301.513 upto 2 places of decimal is 301.51.
(iii) 79.997
Rounding 79.997 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 79.9 9 7 gives 79.997. Since the digit next to the underlined digit 7 is greater than 5, we add 1 to the underlined number.
Hence the rounded value of 79.997 upto 2 places of decimal is 80.00.

(a) 24.4003
Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.40 0 3. In 24.40 0 3 the digit next to the thousandths value is 3 which is less than 5.
∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400.
(b) 1251.2345
Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.23 4 5, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235.
(c) 61.00203
Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00 2 03 gives 61.00 2 03. In 61.00 2 03, the digit next to the thousandths place value is 0, which is less than 5.
Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002.
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Here 0.51 = \(\frac { 51 }{ 100 } \) and 0.25 = \(\frac { 25 }{ 100 } \)First we shade the region 0.51 and then 0.25.
The sum is the total shaded area. 0.51 + 0.25 = 0.76

(i) 25.8 + 18.53.
Using place value grid.Therefore 25.8 + 18.53 = 44.33
(ii) 17.4 + 23.435
Lets use the place value grid.Therefore 17.4 + 23.435 = 40.835

Here 0.46 = \(\frac { 46 }{ 100 } \) and 0.13 = \(\frac { 13 }{ 100 } \)Shading the region 0.46 and then crossing out 0.13 from the shaded area. The left out shaded region without cross marks is the difference. So 0.46 – 0.13 = 0.33

Weight of Potato = 3.350 kg
Weight of Tomato = 2.250 kg
Total weight of Potato and Tomato = (3.350 + 2.250 kg)
= 5.600 kg
Weight of potato, tomato and onions = 10.250
Weight of potato and tomato = 5.600
∴ Weight of onions = (10.250 – 5.600) Kg = 4.650 Kg
Weight of onions = 4.650 Kg

Perimeter of an equilateral triangle = (Side + Side + Side) Sq. units.Given side = 3.8
∴ Perimeter = 3.8 + 3.8 + 3.8
Perimeter of the triangle = 11.4 cm
Objective Type Questions

(i) 0.5 × 3
5 × 3 = 15
∴ 0.5 × 3 = 1.5
(ii) 3.75 × 6375 × 6 = 2250
3.75 × 6 = 22.50(iii) 50.2 × 4502 × 4 = 2008
50.2 × 4 = 200.8
(iv) 0.03 × 9
3 × 9 = 27
0.03 × 9 = 0.27
(v) 453.03 × 745303 × 7 = 317121
453.03 × 7 = 3171.21
(vi) 4 × 0.7
4 × 7 = 28
4 × 0.7 = 2.8

Base of the parallelogram b = 6.8 cm
Height of the parallelogram h = 3.5 cm
Area of the parallelogram A = b × h sq.units = 6.8 × 3.5 cm 2
Area of the parallelogram = 23.80 cm 2

Length of the rectangle l = 23.7 cm
Breadth of the rectangle b= 15.2 cm
Area of the rectangle A = l × b sq.units
= 23.7 × 15.2 cm 2
Area of the rectangle = 360.24 cm 2

(i) 3.6 × 0.336 × 3 = 108
3.6 × 0.3 = 1.08
(ii) 52.3 × 0.1
523 × 1 = 523
52.3 × 0.1 = 5.23
(iii) 537.4 × 0.25374 × 2 = 10748
537.4 × 0.2 = 107.48(iv) 0.6 × 0.06
6 × 6 = 36
0.6 × 0.06 = 0.036
(v) 62.2 × 0.23622 × 23 = 14306
62.2 × 0.23 = 14.306
(vi) 1.02 × 0.05
102 × 5 = 510
1.02 × 0.05 = 0.0510
(vii) 10.05 × 1.051005 × 105 = 105525
10.05 × 1.05 = 10.5525
(viii) 101.01 × 0.01
10101 × 1 = 10101
101.01 × 0.01 = 1.0101
(ix) 100.01 × 1.1
1001 × 11 = 110011
100.01 × 1.1 = 110.011
Objective Type Questions

(iii) 53.0 cm
Hint:
53 × 10 = 530
5.3 × 10 = 53.0
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(i) 0.6 ÷ 3
= \(\frac { 6 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 6 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= 2 × \(\frac { 1 }{ 10 } \)
= \(\frac { 2 }{ 10 } \)
= 0.2
(ii) 0.90 ÷ 5
= \(\frac { 90 }{ 100 } \) × \(\frac { 1 }{ 5 } \)
= \(\frac { 90 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= 18 × \(\frac { 1 }{ 100 } \) = \(\frac { 18 }{ 100 } \)
= 0.18
(iii) 4.08 ÷ 4
= \(\frac { 408 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 408 }{ 4 } \) x \(\frac { 1 }{ 100 } \)
= 102 × \(\frac { 1 }{ 100 } \)
= \(\frac { 102 }{ 100 } \)
= 1.02
(iv) 21.56 ÷ 7
= \(\frac { 2156 }{ 100 } \) × \(\frac { 1 }{ 7 } \)
= \(\frac { 2156 }{ 7 } \) × \(\frac { 1 }{ 100 } \)
= 308 × \(\frac { 1 }{ 100 } \)
= \(\frac { 308 }{ 100 } \)
= 3.08
(v) 0.564 ÷ 6
= \(\frac { 564 }{ 1000 } \) × \(\frac { 1 }{ 6 } \)
= \(\frac { 564 }{ 6 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 94 }{ 1000 } \)
= 0.094
(vi) 41.36 ÷ 4
= \(\frac { 4136 }{ 100 } \) × \(\frac { 1 }{ 4 } \)
= \(\frac { 4136 }{ 4 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 1034 }{ 100 } \)
= 10.34
(vii) 298.2 ÷ 3
= \(\frac { 2982 }{ 10 } \) × \(\frac { 1 }{ 3 } \)
= \(\frac { 2982 }{ 3 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 994 }{ 10 } \)
= 99.4

(i) 5.7 ÷ 10
= \(\frac { 57 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 57 }{ 100 } \)
= 0.57
(ii) 93.7 ÷ 10
= \(\frac { 937 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 937 }{ 100 } \)
= 9.37
(iii) 0.9 ÷ 10
= \(\frac { 9 }{ 10 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 9 }{ 100 } \)
= 0.09
(iv) 301.301 ÷ 10
= \(\frac { 301301 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 301301 }{ 10000 } \)
= 30.1301
(v) 0.83 ÷ 10
= \(\frac { 83 }{ 100 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 83 }{ 1000 } \)
= 0.083
(vi) 0.062 ÷ 10
= \(\frac { 62 }{ 1000 } \) × \(\frac { 1 }{ 10 } \)
= \(\frac { 62 }{ 1000 } \)
= 0.0062

(i) 0.7 ÷ 100
= \(\frac { 7 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 7 }{ 1000 } \)
= 0.007
(ii) 3.8 ÷ 100
= \(\frac { 38 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 38 }{ 1000 } \)
= 0.038
(iii) 49.3 ÷ 100
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493
(iv) 463.85 ÷ 100
= \(\frac { 46385 }{ 100 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 46385 }{ 1000 } \)
= 4.6385
(v) 0.3 ÷ 100
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 3 }{ 1000 } \)
= 4.6385
(vi) 27.4 ÷ 100
= \(\frac { 274 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 274 }{ 1000 } \)
= 0.274

(i) 18.9 ÷ 1000
= \(\frac { 189 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 189 }{ 10000 } \)
= 0.0189
(ii) 0.87 ÷ 1000
= \(\frac { 87 }{ 100 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 87 }{ 100000 } \)
= 0.00087
(iii) 49.3 ÷ 1000
= \(\frac { 493 }{ 10 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 493 }{ 1000 } \)
= 0.493
(iv) 0.3 ÷ 1000
= \(\frac { 3 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3 }{ 10000 } \)
= 0.0003
(v) 382.4 ÷ 1000
= \(\frac { 3824 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 3824 }{ 10000 } \)
= 0.3824
(vi) 93.8 ÷ 1000
= \(\frac { 938 }{ 10 } \) × \(\frac { 1 }{ 1000 } \)
= \(\frac { 938 }{ 10000 } \)
= 0.0938

Weight of the sweet = 9.55 Kg
Weight of sweet for 5 children = \(\frac { 955 }{ 100 } \) Kg
Weight of sweet for 1 child = \(\frac{\left(\frac{955}{100}\right)}{5}\) = \(\frac { 955 }{ 100 } \) × \(\frac { 1 }{ 5 } \) = \(\frac { 955 }{ 5 } \) × \(\frac { 1 }{ 100 } \)
= \(\frac { 191 }{ 100 } \) = 1.91
Each child will get 1.91 kg sweet.

For 1.2 litre of petrol the distance covered = 76.8 Km = \(\frac { 768 }{ 10 } \) KmFor 1 litre of petrol distance covered = 64 Km

Length of ribbon 1 = 13.92 m
Length of ribbon 2 = 11.50 m
Length of ribbon 3 = 10.64 mTotal Length of the ribbons = 13.92 m + 11.5 m + 10.64 m = 36.06 m
Totla length of the ribbons = 36.06 m

Total weight of ghee bought = 10 kg 35 g
Weight of ghee used = 8 kg 59 g
Weight of ghee left = 10.35 kg – 8.59 kg = 1.76 kg∴ Weight of ghee left= 1 kg 76 g = 1.76 kg

Capacity of 1 milk can= 2.53 l
∴ Capacity of 8 milk cans= 2.53 l × 8 = 20.24 lTo fill 8 cans 20.24 l of milk is required.

Total weight of orange = 22.5 kg
Weight of orange required for 1 family = 2.5 kg
∴ Number of families sharing orange = 22.5 kg ÷ 2.5 kg
= \(\frac { 22.5 }{ 2.5 } \) = \(\frac { 22.5 }{ 2.5 } \) × \(\frac { 10 }{ 10 } \) = \(\frac { 225 }{ 25 } \) = 9
∴ 9 families can share the oranges.

For 10 cakes sugar required = 3.924 kg
For 1 cake sugar required = 3.924 ÷ 10 = \(\frac { 3.924 }{ 10 } \) = 0.3924 kg
For 1 cake sugar required = 0.3924 kg.

In 1 round distance covered = 23.761 m
∴ In 100 rounds distance = 23.761 × 100
= 2376.1 m
∴ In 100 round he covers 2376.1 m.

Distance travelled by Anbu:
From place A to place B = 8.3 km
Distance from place B to place C = 15.6 km
∴ Total distance travelled by Anbu = 8.3 + 15.6
= 23.9 km
Distance travlled by Mala:
Distance travelled place A to D = 7.5 km
Distance from place D to place C = 16.9 km
Total distance travelled by mala = (7.5 + 16.9) km = 24.4 km
24.4 > 23.9
∴ Mala travelled more distance. She travelled (24.4 – 23.9) km more i.e she travelled 0.5 km more

Payment for the taxi for an hour = ₹ 97.75
Total hours the taxi was used = 35 hrs
∴ Total payment for the taxi for the week= 97.75 × 35
= 3421.25
Total payment for a week = ₹ 3421.25

In 6 hours the distance travelled = 2781.20 km
In 1 hour the distance travelled = \(\frac { 2781.20 }{ 6 } \) kmAverage speed of the aroplane = 463.53 km/hr.