Home/ Aachman/Answers
  1. Asked: In:

    Estimate the following products using general rule: (a) 578 x 161 (b) 5281 x 3491 (c) 1291 x 592 (d) 9250 x 29

    Aachman

    (a) 578 x 161 578 round off to 600 161 round off to 200 The estimated product = 600 x 200 = 1,20,000 (b) 5281 x 3491 5281 round of to 5,000 3491 round off to 3,500 The estimated product = 5,000 x 3,500 = 1,75,00,000 (c) 1291 x 592 1291 round off to 1300 592 round off to 600 The estimated product = 1Read more

    (a) 578 x 161
    578 round off to 600
    161 round off to 200
    The estimated product = 600 x 200 = 1,20,000
    (b) 5281 x 3491
    5281 round of to 5,000
    3491 round off to 3,500
    The estimated product = 5,000 x 3,500 = 1,75,00,000
    (c) 1291 x 592
    1291 round off to 1300
    592 round off to 600
    The estimated product = 1300 x 600 = 7,80,000
    (d) 9250 x 29
    9250 round off to 10,000
    229 round off to 30
    The estimated product = 10,000 x 30 = 3,00,000

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    See less
    • 6
  2. Asked: In:

    Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens): (a) 439 + 334 + 4317 (b) 1,08,737 – 47,599 (c) 8325 – 491 (d) 4,89,348 – 48,365

    Aachman

    (a) 439 round off to 400 334 round off to 300 4317 round off to 4300 Estimated sum = 5000 (b) 108734 round off to 108700 47599 round off to 47600 Estimated difference = 61100 (c) 8325 round off to 8300 491 round off to 500 Estimated difference = 7800 (d) 489348 round off to 489300 48365 round off toRead more

    (a) 439 round off to 400
    334 round off to 300
    4317 round off to 4300
    Estimated sum = 5000

    (b) 108734 round off to 108700
    47599 round off to 47600
    Estimated difference = 61100

    (c) 8325 round off to 8300
    491 round off to 500
    Estimated difference = 7800

    (d) 489348 round off to 489300
    48365 round off to 48400
    Estimated difference = 440900

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    See less
    • 6
  3. Asked: In:

    Estimate each of the following using general rule: (a) 730 + 998 (b) 796 – 314 (c) 12,904 + 2,888 (d) 28,292 – 21,496

    Aachman

    (a) 730 round off to 700 998 round off to 1000 Estimated sum = 1700 (b) 796 round off to 800 314 round off to 300 Estimated sum = 500 (c) 12904 round off to 13000 2888 round off to 3000 Estimated sum = 16000 (d) 28292 round off to 28000 21496 round off to 21000 Estimated difference= 7000 https://wwwRead more

    (a) 730 round off to 700
    998 round off to 1000
    Estimated sum = 1700

    (b) 796 round off to 800
    314 round off to 300
    Estimated sum = 500

    (c) 12904 round off to 13000
    2888 round off to 3000
    Estimated sum = 16000

    (d) 28292 round off to 28000
    21496 round off to 21000
    Estimated difference= 7000

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    See less
    • 6
  4. Asked: In:

    A vessel has 4 litres and 500 ml of curd. In how many glasses each of 25 ml capacity, can it be filled?

    Aachman

    Capacity of curd in a vessel = 4 litres 500 ml = 4 x 1000 ml + 500 ml = 4500 ml Capacity of one glass = 25 ml Number of glasses can be filled = 4500 ÷ 25 Therefore, 180 glasses can be filled by curd. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    Capacity of curd in a vessel = 4 litres 500 ml = 4 x 1000 ml + 500 ml = 4500 ml
    Capacity of one glass = 25 ml
    Number of glasses can be filled = 4500 ÷ 25
    Therefore, 180 glasses can be filled by curd.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    See less
    • 6
  6. Asked: In:

    The distance between the school and the house of a student’s house is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.

    Aachman

    Distance between school and home = 1.875 km Distance between home and school = + 1.875 km Total distance covered in one day = 3.750 km Distance covered in six days = 3.750 x 6 = 22.500 km Therefore, 22 km 500 m distance covered in six days. https://www.tiwariacademy.com/ncert-solutions/class-6/mathsRead more

    Distance between school and home = 1.875 km
    Distance between home and school = + 1.875 km
    Total distance covered in one day = 3.750 km
    Distance covered in six days = 3.750 x 6 = 22.500 km
    Therefore, 22 km 500 m distance covered in six days.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-1/

    See less
    • 5