WBBSE Class 7 Math Solution Arithmetic Chapter 4 Approximation Of Values Exercise 4 Solved Problems
Approximate value: In our daily transactions it is not always possible to find the exact value of a quantity or to ascertain accurate values, lengths, weights, etc. of a thing.
⇒ So in calculations involving values, length, weights etc, we have to take the approximate values just a little above or a little below the accurate values. This is called the approximate value of a quantity or the nearest value or measurement of a thing.
⇒ In order to find the approximate value up to certain places of the digit, if that digit is any number between 5 to 9, then 1 is added to the digit at that certain place or else the same digit is kept.
⇒ The sign for approximate value is “≈” (almost equal)
Read and Learn More WBBSE Solutions for Class 7 Maths
Question 1. Determine the approximate value of 3746 up to tens, hundreds, and thousands of places.
Solution:
Given
3746
| Thousands | Hundreds | Tens | Units |
| 3 | 7 | 4 | 6 |
⇒ The number at the tens place is 4 and on its immediate right side is 6 which is between 5 to 9.
⇒ So up to tens place, the approximate value is (374 + 1) ten = 3750.
⇒ The digit on the hundreds place is 7 and on its immediate right side is 4. So the approximate value upto hundreds place is 3700.
⇒ The digit on the thousands place is 3 and on its right side we have 7, so the approximate value upto thousands place is 4000.
We can write:
3746 ≈ 3750 (up to tens place).
3746 ≈ 3700 (upto hundreds place)
3746 ≈ 4000 (upto thousands of places)
Wbbse Class 7 Maths Solutions
Question 2. Find the approximate value of 3-853672 up to 1, 2, 3, 4, and 5 decimal places and whole numbers.
Solution:
Given 3-853672
⇒ 3.853672 ≈ 3.8537
[6th place after decimal is 2]
⇒ 3.853672 ≈ 3.853672
[5th place after decimal is 7 hence I is added to 4th decimal place value 6, 6 + 1 = 7]
⇒ 3.853672 ≈ 3.854
[4th place after decimal is 6 hence 1 is added to 3rd decimal place value 3, 3+1 = 4]
⇒ 3.853672 ≈ 3.85
[3rd place after decimal is 3]
⇒ 3.853672 ≈ 3.9
[2nd place after decimal is 5 hence 1 is added to 1 decimal place value 4, 8+ 1 = 9]
⇒ 3.853672 ≈ 4
[1 place after decimal is 8 hence 1 is added to whole number 3, 3+1 = 4]
Wbbse Class 7 Maths Solutions
Question 3. Find the approximate value of \(\frac{22}{7}\) upto 2, 3, 4 and 5 decimal place.
Solution:
Given \(\frac{22}{7}\)
\(\frac{22}{7}\) ≈ 3.142857……
\(\frac{22}{7}\) ≈ 3.14 [two decimal places]
\(\frac{22}{7}\) ≈ 3.143 [3 decimal places]
\(\frac{22}{7}\) ≈ 3.1429 [4 decimal places]
\(\frac{22}{7}\) ≈3-14286 [5 decimal places]
Wbbse Class 7 Maths Solutions
Question 4. Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km. Find one distance traveled by light in 1 second in kilometers approximately.
Solution:
Given
Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km.
1 mile = 1.6093 km
186000 miles = (186000 x 1.6093) km
= 299,329.8 km
≈ 299,330 km [1 place after the decimal is 8. Hence 1 is added to the whole number at the unit place i.e., 9 + 1 = 10]
Question 5. Divide 22 among 8 boys and 7 girls equally. Find how much each would get (approximated up to 2 places of the decimal). Also, find the total money received by 8 boys and that received by 7 girls. Then find the total money received by 8 boys and 7 girls and how much this total amount is more or less than ₹ 22.
Solution:
⇒ Total number of boys and girls is (8 + 7) or 15.
⇒ ₹ 22 is divided among 15 boys and girls.
⇒ Each gets = ₹ \(\frac{22}{15}\)
= \(\frac{2200}{15}\) paise
≈ 146.67 paise
⇒ The total money received by 8 boys is (146-67 x 8) paise
= 1173.36 paise
≈ ₹ 11.73.
⇒ The total money received by 7 girls is (146-67 x 7) paise
= 1026.69 paise
≈ ₹ 10.27
⇒ Total money received by 8 boys and 7 girls is (11.73 + 10.27) = ₹ 22.00 = ₹ 22
⇒ This amount is equal to 22.
Wbbse Class 7 Maths Solutions
Question 6. If \(\frac{2}{7}\) radian = 180°, find the value of 1 radian in degree upto 2 places of decimals.
Solution:
Given
\(\frac{2}{7}\) radian = 180°
1 radian = \(\left(\frac{7}{22} \times 180\right)^{\circ}\)
= \(\left(\frac{1260}{22}\right)^{\circ}\)
= 57.27°
The value of 1 radian in degree upto 2 places of decimals = 57.27°
Question 7. Simplify the following up to 4 decimal places. \(0 \cdot 004872+13 \cdot \dot{725}+0 \cdot \dot{3} 6\)
Solution: \(0 \cdot 004872+13 \cdot \dot{725}+0 \cdot \dot{3} 6\)
=0.004872+13.725725………..+0.363636……
= 14.094233……
= 14.0942
WBBSE Class 7 Math Solution Approximation Of Values
Approximation Of Values Exercise 4.1
Question 1. Let’s write the approximate values of the following fractions correct up to two, three and four places of decimal.
1. \(\frac{13}{17}\)
Solution:
Given \(\frac{13}{17}\)
⇒ \(\frac{13}{17}\) = 0.76470588.
≈ 0. 76 (Two decimal places)
≈ 0.765 (Three decimal places)
≈ 0.7647 (Four decimal places)
2. \(\frac{19}{29}\)
Solutions:
Given \(\frac{19}{29}\)
⇒ \(\frac{19}{29}\) = 0.6551724
≈ 0.66 (Two decimal places)
≈ 0.655 (Three decimal places)
≈ 0. 6552 (Four decimal places)
Approximation Of Values Exercise 4.2
Question 1. Let’s write the approximate values of the following numbers up to 1 0’s, 1000s, and 1 0,000’s places.
Question 2. Let us divide Rs. 3 among 7 children and calculate how much Paise each wou,d get (approximately upto 2 places of decimal). Let’s then find total money received by 7 children and then try to find how more or less it will be than Rs. 3.
Solution:
Total amount- Rs. 3 = 300p
Total children = 7
Each will receive = \(\frac{300}{7}\) p
= 42.857≈ 42.86 p
∴ Total money received by 7 children = 42.86 x 7 p = 300.02 p
∴ 300.02 p – 300p = 02p more.
Question 3. 1 divided Rs. 22 among 8 boys and 7 girls equally. Let’s find how much each would get. (approximated up to 2 places of decimal). Also let’s find the total money received by 8 boys and that received by 7 girls. Then let’s find total money received by 8 boys and 7 girls and how much this total amount is more or less than Rs. 22.
Solution :
Total amount = Rs. 22 = 2200p.
Total number of boys & girls = 8 + 7=15
Each will get = \(\frac{2200}{15}\)
p = 1 46.666p
= 146.67 p (approx)
∴ 8 boys will receive = 146.67 x 8 = 1 173.36 p = Rs. 11.73
7 girls will received = 146.67 x 7 = 1026.69 p = Rs. 10.27p
Total amount = Rs. (11.73 + 10.27) = Rs. 22.
It is equal.
Question 4. Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km, let’s find the distance travelled by light in 1 second in Kilometres approximately (up to 3 decimal places of approximation).
Solution :
Given
Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km,
Distance travelled by light in 1 second
= 1 86000 mile = 1 86000 x 1 .6093 km.
= 299329.8 km = 299330 km.
Question 5. Let’s write the approximate value up to 2 decimal places of 0.997.
Solution :
0.997 = 1.(up to 2 decimal) (approx).
Question 6. Let us fill in the gaps
Solution :
WBBSE Class 7 Math Solution Question 7. Let us write the approximate values of the following fractions, up to 2, 3 & 4 decimal places
1. \(\frac{22}{7}\)
Solution :
Given \(\frac{22}{7}\)
⇒ \(\frac{22}{7}\) = 3.14 (upto 2 decimal places)
= 3.143 (up to 3 decimal places)
= 3.1429 (up to 4 decimal places)
\(\frac{22}{7}\) = 3.1429 (up to 4 decimal places)
2. \(\frac{3}{14}\)
Solution :
Given \(\frac{3}{14}\)
⇒ \(\frac{3}{14}\) = 0.21 (up to 2 decimal places)
= 0.214 (up to 3 decimal places)
= 0.2143 (4 decimal places)
\(\frac{3}{14}\) = 0.2143 (4 decimal places)
3. \(\frac{1}{5}\)
Solution :
Given \(\frac{1}{5}\)
⇒ \(\frac{1}{5}\) = 0.20 (upto 2 decimal places)
= 0.200 (up to 3 decimal places)
= 0.2000(upto 4 decimal places)
\(\frac{1}{5}\) = 0.2000(upto 4 decimal places)
4. \(\frac{47}{57}\)
Solution :
Given
\(\frac{47}{57}\)⇒ \(\frac{47}{57}\) = 0.82 (up to 2 decimal places)
= 0.825 (up to 3 decimal places)
= 0.8246 (up to 4 decimal places)
\(\frac{47}{57}\) = 0.8246 (up to 4 decimal places)
Question 8. Let’s write the approximate values of the following numbers up to lacs, thousands and hundreds place.
Solution:
Question 9. Practical application of approximation.
1. 11 hours 9 min 40 sec – approximately the value up to min.
Solution: 11 hours 10 mins.
2. If the price of the shoe is written as Rs. 99.99, let us find what will be
its approximate value.
Solution: Rs. 100.
3. The length of a line segment is 1.59 cm, Let us find what is the
approximate value of the length.
Solution: 1. 6cm.
4. The weight Of poppy seed (post) is 102gm. Let’s find
approximately the weight for which money is charged.
Solution: 100