Decimal Fractions

decimal fractions

Step by Step solutions of Concise Mathematics ICSE Class-7 Maths chapter 4- Decimal Fractions by Selina is provided.

Table Of Contents
  1. Q1: Convert the following into fractions in their lowest terms:
  2. Q2: Convert into decimal fractions:
  3. Q3: Write the number of decimal places in:
  4. Q4: Write the following decimals as word statements:
  5. Q5: Convert the given fractions into like fractions:
  6. Q1: Add:
  7. Q2: Subtract the first number from the second:
  8. Q3: Simplify:
  9. Q4: Find the difference between 6.85 and 0.685.
  10. Q5: Take out the sum of 19.38 and 56.025 from 200.111.
  11. Q6: Add 13.95 and 1.003, and from the result, subtract the sum of 2.794 and 6.2.
  12. Q7: What should be added to 39.587 to give 80.375?
  13. Q8: What should be subtracted from 100 to give 19.29?
  14. Q9: What is excess of 584.29 over 213.95?
  15. Q10: Evaluate:
  16. Q11: What is the excess of 75 over 48.29?
  17. Q12: If A = 237.98 and B = 83.47. Find:
  18. Q13: The cost of one kg of sugar increases from ₹28.47 to ₹32.65. Find the increase in cost.
  19. Q1: Multiply:
  20. Q2: Multiply each number by 10, 100 and 1000:
  21. Q3: Evaluate:
  22. Q4: Divide:
  23. Q5: Divide each of given numbers by 10, 100, 1000 and 10000:
  24. Q6: Evaluate:
  25. Q7: Evaluate:
  26. Q8: Evaluate:
  27. Q9: Find the cost of 36.75 kg wheat at rate of ₹12.80 per kg.
  28. Q10: The cost of a pen is ₹56.15. Find the cost of 16 such pens.
  29. Q11: Evaluate
  30. Q12: Fifteen identical articles weigh 31.50 kg. Find the weigh of each article.
  31. Q13: The product of two numbers is 211.2. If one of these two numbers is 16.5, find the other number.
  32. Q14: One dozen identical articles cost ₹45.96. Find the cost of each article.
  33. Q15: Find whether the given division forms a terminating or a non-terminating decimal:
  34. Q1: The weight of an object is 306 kg. Find the total weight of 48 similar objects.
  35. Q2: Find the cost of 17.5m cloth at the rate of ₹112.50 per metre.
  36. Q3: One kilogram of oil costs ₹73.40. Find the cost of ₹9.75 kilograms of the oil.
  37. Q4: Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.
  38. Q5: 18.5 m of oil costs ₹ 666. Find the cost of 3.8 m cloth.
  39. Q6: Find the value of:
  40. Q7: Evaluate:
  41. Q1: Which is greater: 5.038 or 5.3?
  42. Q2: Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Saria bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
  43. Q3: Two kg of milk contains 0.315 kg of cream. The cream in 20 kg milk is:
  44. Q4: The distance walked by a boy is 86.4 km in 4.8 hours. The distance covered by him in one hour is:
  45. Q5: The number seven and 7 thousandth is:
  46. Q6: (56.56div1.4) is equal to:
  47. Q7: (left(2+frac{1}{2}right)divfrac{3}{5}) is equal to:
  48. Q8: Total cost of two pens at ₹5.30 each and four notebooks at ₹20.50 each is:
  49. Q9: (2.5+3.8div0.02) is equal to:
  50. Q10: By what decimal number should 0.0001 be divided to get 0.01?
  51. Q11: (3frac{1}{5}timesleft(frac{1}{2}+frac{3}{8}right)divfrac{21}{40}) is equal to:
  52. Q12: 5.80, 0.95, 1.87 and 1.92 in descending order are:
  53. Q13: (3-frac{1}{4} of left(15.8-3right)) is equal to:
  54. Q14: Statement 1: (0.05=0.050=0.005=0.00500) Statement 2: Any number of zeros put at the end (i.e. on the right side) of a decimal number does not change its value. Which of the following options is correct?
  55. Q15: Assertion (A): Representation of 6.25 as a vulgar fraction is (6frac{1}{4}). Reason (R): A fraction is said to be a vulgar fraction if the denominator is a whole number but not of the form ({10}^n, nin N).
  56. Q16: Assertion (A): If the product of two decimal numbers is 17.55 and one of them is 6.5, then other one is 2.7. Reason (R): In division of decimal numbers, the dividend is always exactly divisible and no remainder is left after certain steps. Also quotient is always reduced to a terminating decimal.
  57. Q17: Assertion (A): (3.10divleft(0.1times0.1right)=3.1). Reason (R): In division of a decimal number by ({10}^n, nin N), shift the decimal point to the right by as many digits equivalent to n in the power of 10 in the divisor.
  58. Q18: Assertion (A): 9, 9.56, 9.2, 9.005 are all unlike decimals, hence addition operations can't be performed. Reason (R): A whole number can also be expressed as a decimal number by putting a decimal after its unit's digit and after it as many zeroes required to perform addition operations with other like or unlike decimal numbers.

Exercise: 4-D

Q1: The weight of an object is 306 kg. Find the total weight of 48 similar objects.

Step 1: Weight of one object = 306 kg
Step 2: Number of objects = 48
Step 3: Total weight = Weight of one object × Number of objects \[ = 306 \times 48 \] Step 4: Multiply 306 and 48:

    306
  ×  48
  ──────
   2448     (306 × 8)
  12240     (306 × 40)
  ──────
  14688

Answer: Total weight of 48 objects = 14688 kg.


Q2: Find the cost of 17.5m cloth at the rate of ₹112.50 per metre.

Step 1: Length of cloth = 17.5 m
Step 2: Rate per metre = ₹112.50
Step 3: Total cost = Length × Rate per metre \[ = 17.5 \times 112.50 \]Step 4: Multiply 17.5 and 112.50:

     1125    
    × 175    
  ─────────
     5625     (1125 × 5)
    78750     (1125 × 70)
   112500     (1125 × 100)
  ─────────
   196875

Step 5: Place the decimal back (1 + 2 = 3 decimal places): \[ 17.5 \times 112.50 = 196.875 \]Answer: Total cost of 17.5 m cloth = ₹1968.75.


Q3: One kilogram of oil costs ₹73.40. Find the cost of ₹9.75 kilograms of the oil.

Step 1: Cost of 1 kg oil = ₹73.40
Step 2: Quantity of oil = 9.75 kg
Step 3: Total cost = Cost per kg × Quantity \[ = 73.40 \times 9.75 \]Step 4: Multiply 73.40 and 9.75:

    7340
  ×  975
  ─────────
    36700    (7340 × 5)
   513800    (7340 × 70)
  6606000   (7340 × 900)
  ─────────
  7156500

Step 5: Place the decimal back (2 + 2 = 4 decimal places): \[ 73.40 \times 9.75 = 715.6500 = 715.65 \]Answer: Total cost of 9.75 kg oil = ₹715.65.


Q4: Total weight of 8 identical objects is 51.2 kg. Find the weight of each object.

Step 1: Total weight of 8 objects = 51.2 kg
Step 2: Number of objects = 8
Step 3: Weight of each object = Total weight ÷ Number of objects \[ = \frac{51.2}{8} \]Step 4: Divide 51.2 by 8:

    6.4
   ─────
8 | 51.2
   -48
   ──────
     32
    -32
   ──────
     0

Answer: Weight of each object = 6.4 kg.


Q5: 18.5 m of oil costs ₹ 666. Find the cost of 3.8 m cloth.

Step 1: Cost of 18.5 m oil = ₹666
Step 2: Cost per metre = Total cost ÷ Length \[ = \frac{666}{18.5} = \frac{6660}{185} \]Step 3: Divide 6660 by 18.5:

       36
     ────────
185 | 6660
     -555    (185 × 3)
     ─────
      1110
     -1110    (185 × 6)
     ─────
        0

Cost per metre = ₹36Step 4: Find cost of 3.8 m oil = Cost per metre × 3.8 \[ = 36 \times 3.8 \]Step 5: Multiply 36 and 3.8:

     36
  × 3.8
  ─────
   28.8     (36 × 0.8)
  108.0     (36 × 3)
  ─────
  136.8

Answer: Cost of 3.8 m oil = ₹136.8.


Q6: Find the value of:

i. 0.5 of ₹7.60 + 1.62 of ₹30

Step 1: Calculate \(0.5 \times 7.60\):

  0.5 × 7.60 = 3.80

Step 2: Calculate \(1.62 \times 30\):

    1.62
  ×  30
  ──────
    0.00     (1.62 × 0)
   48.60      (1.62 × 30)
  ──────
   48.60

Step 3: Add the two results: \[ 3.80 + 48.60 = 52.40 \]Answer: 52.40

ii. 2.3 of 7.3 kg + 0.9 of 0.48 kg

Step 1: Calculate \(2.3 \times 7.3\):

    2.3
  × 7.3
  ─────
   0.69    (2.3 × 0.3)
  14.00    (2.3 × 7)
  ─────
  16.79

Step 2: Calculate \(0.9 \times 0.48\):

   0.48
   ×0.9
  ─────
  0.432    (0.48 × 0.9)

Step 3: Add the two results: \[ 16.79 + 0.432 = 17.222 \]Answer: 17.222 kg

iii. 6.25 of 8.4 – 4.7 of 3.24

Step 1: Calculate \(6.25 \times 8.4\):

   6.25
  × 8.4
   ─────
   2.50    (6.25 × 0.4)
 -50.00    (6.25 × 8)
  ─────
  52.50

Step 2: Calculate \(4.7 \times 3.24\):

   3.24
  × 4.7
  ─────
  2.268    (3.24 × 0.7)
 12.960    (3.24 × 4)
  ─────
 15.228

Step 3: Subtract the two results: \[ 52.50 – 15.228 = 37.272 \]Answer: 37.272

iv. 0.98 of 235 – 0.09 of 3.2

Step 1: Calculate \(0.98 \times 235\):

     0.98
   × 235
  ───────
     4.9    (0.98 × 5)
    29.4    (0.98 × 30)
   196.0    (0.98 × 200)
  ───────
   230.30

Step 2: Calculate \(0.09 \times 3.2\):

   3.2
  ×0.09
  ─────
   0.288    (0.09 × 32)

Step 3: Subtract the two results: \[ 230.30 – 0.288 = 230.012 \]Answer: 230.012


Q7: Evaluate:

i. \(5.6 – 1.5 \text{ of } 3.4\)

Step 1: Calculate \(1.5 \times 3.4\):

  1.5
× 3.4
─────
  0.6    (1.5 × 0.4)
  4.5    (1.5 × 3)
─────
 5.10

Step 2: Subtract from 5.6: \[ 5.6 – 5.10 = 0.50 \]Answer: 0.50

ii. \(4.8 \div 0.04 \text{ of } 5\)

Step 1: Calculate \(0.04 \times 5\):

  0.04
×   5
─────
 0.20

Step 2: Divide \(4.8 \div 0.20\):

4.8 ÷ 0.20 = 24

Answer: 24

iii. \(0.72 \text{ of } 80 \div 0.2\)

Step 1: Calculate \(0.72 \times 80\):

  0.72
×  80
─────
  0.0    (0.72 × 0)
 57.60    (0.72 × 80)
─────
 57.6

Step 2: Divide \(57.6 \div 0.2\):

57.6 ÷ 0.2 = 288

Answer: 288

iv. \(0.72 \div 80 \text{ of } 0.2\)

Step 1: Calculate \(80 \times 0.2\):

   80
× 0.2
 ────
   16

Step 2: Divide \(0.72 \div 16\):

0.72 ÷ 16 = 0.045

Answer: 0.045

v. \(6.45 \div (3.9 – 1.75)\)

Step 1: Subtract inside the bracket: \[ 3.9 – 1.75 = 2.15 \]Step 2: Divide \(6.45 \div 2.15\):

6.45 ÷ 2.15 = 3

Answer: 3

vi. \(0.12 \text{ of } (0.104 – 0.02) + 0.36 \times 0.5\)

Step 1: Calculate inside bracket: \[ 0.104 – 0.02 = 0.084 \]Step 2: Calculate \(0.12 \times 0.084\):

  0.12
 ×0.084
────────
0.00048   (0.12 × 0.004)
0.00960   (0.12 × 0.080)
────────
0.01008

Step 3: Multiply 0.36 by 0.5: \[ 0.36 \times 0.5 = 0.18 \]Step 4: Add 0.01008 and 0.18: \[ 0.01008 + 0.18 = 0.19008 \]Answer: 0.019008


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