Arithmetic Chapter 4 Approximation Of Values
Question 1. Determine the approximate value of 5378 up to tens, hundreds, and thousands of places.
Solution:
5378
5-Thousand
3-Hundred
7-Tens
8-units
The number at the tens place is 7 and on its immediate right side is 8 which is between 5 to 9.
So up to tens place the approximate value is (537+1)ten
= 5380
The digit on the hundreds place is 3 and on its immediate right side is 7 So the approximate value upto hundreds place is 5300
The digit on the thousands place is 5 and on its. right side we have 3, so the approximate value. up to a thousand places is 6000
5378 ≈ 5380 (upto tens place).
5378 ≈ 5400 (upto Hundreds place)
5378 ≈ 5000 (upto Thousands place).
Read and Learn More Class 7 Maths Solutions
Class 7 Maths Arithmetic Chapter 4 Solutions
Question 2. Find the approximate value of 6.748564 up to whole number and 1, 2, 3, 4, and 5 decimal places
Solution:
6.748564 ≈ The number is up to 6th place is 4
6.748564 16.748571 [5th place after the decimal is 6 Hence 1 is added to 5th decimal place value 6, 6+1=7]
6.748564 ≈ 6.7486 [ 4th place after decimal is 6 Hence 1 is added to 3rd place value 5, 5+1=6
6.748564 ≈ 6.749[3rd place after decimal is 5 Hence 1 is added to 2nd place value 8, 8+1=9]
6.748564 ≈ 6.75 [2nd place after decimal is 4 Hence 1 is added to 1 decimal place value 44+1=5]
6.748564 ≈ 6.8
6.748564 ≈ 6.74856
≈ 6.7486
≈ 6.749
≈ 6.75
≈ 6.8
≈ 7
Question 3. Find the approximate value of \(5 \frac{7}{37}\) upto 2,3,4 and 5 decimal place
Solution:
⇒ \(5 \frac{7}{37} \Rightarrow \frac{192}{37}\)
⇒ \(5 \frac{7}{37}\) = 5.18 [TWO decimal place]
⇒ \(5 \frac{7}{37}\)= 5.189 [3 decimal place]
⇒ \(5 \frac{7}{37}\) = 5.1891 [4 decimal place]
⇒ \(5 \frac{7}{37}\) = 5.18918 [5 decimal place]
Class 7 Arithmetic Chapter 4 Questions
Question 4. Divide ₹37 among 5 boys and 3 girls equally. Find how much each would get. (approximated upto 2 places of decimal) Also, find the total money received by 5 boys and 3 girls and how much this total amount is more or less than ₹37
Solution:
The total number of boys and girls is (5+3) or 8
₹37 is divided among 8$ boys and girls
Each gets = ₹\(\frac{37}{8}\)
= 4.625 paise
The total money received by 5 boys is
= (4.625×5) paise
= 23.125 paise
The total money received by 3 girls is
= (4.625×3) Paise
= 13.875 Paise
Total money received by 5 boys and 3 girls is (23.125+13-87)
=₹37
This amount is equal to ₹37
WBBSE Class 7 Maths Solutions
Question 5. Simplify the following up to 3 decimal places. 21-3574 + 5.72 + 3.602
Solution:
21.3574+ 5.727272… +3.602602.
⇒ 30.687202.
Rounded the decimal upto
3 digits i.e. 30.687
⇒ \(\begin{array}{r}
21.3574 \\
+5.7272 \\
+3.602602 \\
\hline 30.687201 \\
\hline
\end{array}\)
Question 6. Find the difference between 47.286 and 28.6 upto 2 places of decimal
Solution:
Given:
First number = 47.286
Second number = 28.6
The difference = (First number) – (second number)
= 47.286 28.6
= 18.686
Rounded to two decimals places the difference is 18.69
Class 7 Arithmetic Textbook Solutions
Question 7. If 1 inch = 2.54 cm then find the value of 1cm into an inch upto 3 decimal places.
Solution:
Given:
1 inch = 2.54cm.
So to find how many inches are in 1 centimeter
we’ll divide by 2:54
⇒ \(\frac{1}{2.54} \approx 0.3937 .\)
The Rounded to three decimals places, 1 centimeter is approximately equal to 0.394 inches.
Question 8. Write the approximate value of the following numbers upto lacs thousands and hundreds of places.
Solution:
Question 9. Find the values of the following correct to 3 places of decimals.
- \(7 \frac{2}{7}\)
- \(3 \frac{8}{45}\)
- 8.0645
Solution:
1. \(7 \frac{2}{7}\)
⇒ \(\frac{51}{7}\)
⇒ 7.285714….
Rounded to three decimals places it’s approximately 7.286
2. \(3 \frac{8}{45}\)
⇒ \(3 \frac{8}{45}\)
⇒ \(\frac{143}{45}\)
= 3.17778 ≈ 3.178
3. 8.0645
Rounded to three decimal places It’s approximately 8.064
Class 7 Maths Arithmetic Problems
Question 10. Simplify and Find the approximate value upto 1 decimal place.
⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3}\)
Solution:
⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3}\)
⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \times \frac{2}{3}\)
⇒ \(\frac{23}{3} \times \frac{4}{5} \div \frac{3}{4} \times \frac{2}{3}\)
⇒ \(\frac{92}{15} \div \frac{6}{4}\)
⇒ \(\frac{\frac{92}{15}}{\frac{6}{4}}\)
⇒ \(\frac{92}{15} \times \frac{4}{6}=\frac{368}{90}=4.0\)
⇒ \(7 \frac{2}{3} \times \frac{4}{5} \div \frac{3}{4} \text { of } \frac{2}{3} \text { is } 4.0\)