Arithmetic Chapter 1 Revision
Question 1. Convert the following to a fraction:
1. 7.028
Solution:
Let x = 7.02828…
1000x= 7028.28…
Multiply ‘x’ by 10 to align the repeating parts 10x = 70.2828…..
Subtract the second equation from the first to eliminate 1000x – 10x = 7028.2828……..70.2828…
990x = 6958
Finding the greatest common divisor (GCD) of 6958 and 990
The GCD of 6958 and 990 is ‘2’ \(\frac{6958 \div 2}{990 \div 2}=\frac{3479}{495}\)
So the recurring decimal 7.02828 as a fraction is \(\frac{3479}{495}\)
∴ \(7.0 \dot{2}=17 \frac{14}{495}\)
2. 3.432
Solution:
Let x = 3.43232··…
Multiply x by 1000
1000x = 3432.32…..
Multiply x by 10 to align the repeating parts.
10x = 34.32.32……
Subtract the second equation from the First to eliminate the repeating part
4000x – 10x = 3432.323232……..34.3232…
990x = 3398
x = \(\frac{3398}{990}\)
Finding the Greatest common divisor (GCD) of 3398 and 990.
The GCD of 3398 and 990 are 2
∴ \(\frac{3398 \div 2}{990 \div 2}=\frac{1649}{495}\)
So the recurring decimal \(3 \cdot 4 \overline{32}\) as a fraction is \(\frac{1699}{495}\)
∴ \(3.4 \dot{3} \dot{2}=3 \frac{214}{495}\)
Class 7 Arithmetic Problems With Solutions
Question 2. Convert the following percentage into decimal fractions:
- 0.03
- 1.26
Solution:
1. 0.03
Decimal Fraction = \(\frac{0.03}{100}\) = 0.0003
0.03% = 0.0003
2. 1·26
Decimal Fraction = \(\frac{1.26}{100}\)
= 0.0126
1.26% = 0.0126
Question 3. Express the Following in percentage:
- ₹5 out of ₹25
- 0.3
Solution:
percentage = \(\left(\frac{\text { Part }}{\text { Whole }}\right) \times 100\)
Here part is 5 and the whole is 25
percentage = (\(\frac{5}{25}\))x100
= (0.2) x 100
percentage = 20%
∴ ₹5 out of ₹25 is 20%.
2. 0.3
To Convert the decimal 0.3 to a percentage, we multiply by 100.
percentage = 0.3× 100 = 30%
Question 4. Find the values of the following.
- 18% of 3600
- 12 \(\frac{1}{2}\) % of ₹12.08
Solution:
1. 18% of 3600
⇒ 3600 x \(\frac{18}{100}\)
⇒ 36 x 18
⇒ 648
∴ 18% of 3600 = 648
2. 12\(\frac{1}{2}\) % of 12.08.
⇒ 12.08 x \(\frac{25}{2 \times 100}\)
⇒ 12.08 x \(\frac{1}{8}\)
⇒ 12.08 × 0.125
⇒ 1.51
∴ 12 \(\frac{1}{2}\)% of ₹12.08 is ₹1.51
Class 7 Maths Chapter 1 Solved Exercises
Question 5. If the product of two numbers is 150 and their quotient is \(\frac{3}{2}\) then find the numbers.
Solution:
Let’s denote the two numbers as x and y.
Given
- The product of two numbers is 150:xy = 150
- The quotient of the two numbers is \(\frac{3}{2}\):\(\frac{x}{y}\)= \(\frac{3}{2}\)
From the second condition. \(\frac{x}{y}\) = \(\frac{3}{2}\)
x = \(\frac{3}{2}\) y
Now Substitute the ‘x’ value in 1-condition.
xy= 150
(\(\frac{3}{2}\)y)y = 150
⇒ \(\frac{3}{2}\)y² = 150
y² = 150 x \(\frac{2}{3}\)
y² = 100
y = √100
y = 10
Now that we have found y = 10, we can find x.
x = \(\frac{3}{2}\) x 10
x = 15
So the two numbers are x = 15 and y = 10.
Question 6. If the sum of two numbers is so and their HCF is 16 then Find the numbers.
Solution:
Let’s denote the two numbers as x and y.
Given
- The sum of the two numbers is 80: x+y=80.
- Their highest Common Factor (HCF) is 16.
Let’s express x and y as
x = 16a
y = 16b
where a and b are integers.
Now we substitute these expressions into the equation for their sum:
16a + 16b = 80
16(a+b)=80
(a+b) = \(\frac{80}{16}\)
a+b = 5
since a+b=5, and a and b are integers the
Possible values for a and b are a = 1, and b=4 (or) a=2, and b=3.
So the numbers are 16×1 = 16, and 16×4 = 64,08 16×2=32, and 16×3=48.
∴ The two numbers are 16 and 64 (or) 32 and 48.
Arithmetic Formulas For Class 7 WBBSE
Question 7. Find the square roots of the following numbers:
1. 11025
Solution:
To Find the square root use prime factorization.
First, let’s see if it’s a perfect square by trying
Some divisors.
11025 ÷ 25 = 441
So, we have;
11025 = 25X441
441 = 21×21
25 = 5×5
Thus
11025 = 5² x 21²
Take the square root: \(\sqrt{11025}=\sqrt{(5 \times 21)^2}\)
= \(5 \times 21\)
∴ \(\sqrt{11025}\) = 105
Question 8. Find the square root of the following numbers.
1. 15376
Solution:
First, let’s see if it’s a perfect square by trying Some divisors.
15376 ÷ 16 = 961.
15376 = 16×961
961 = 31×31
16 = 4×4.
Thus 15376 = 4² x 31²
So we can take the square root: \(\sqrt{15376}=\sqrt{(4 \times 31)^2}\)
= \(4 \times 31\)
∴ \(\sqrt{15376}\) = 124
Question 9. Parthababu pays 157. of his salary for house rent. If he pays 4500 per month for rent, then find his monthly salary.
Solution:
Let’s denote Partha babu’s monthly salary as ‘s’.
Given,
House rent = 15% of salary.
House rent = 4500.
15% of S = 4500
⇒ \(\frac{15}{100}\) X S = 4500
S = \(\frac{4500}{0.15}\)
S = 30000
∴ so Parthababu’s monthly salary is 30000.
WBBSE Maths Study Material Class 7
Question 10. A General wishing to arrange his soldiers 632 in number into a solid square found that there were 7 Soldiers over. How many were there in the front?
Solution:
Let’s denote the number of soldiers on each side of the Square as 2.
The total number of soldiers is 632 and there are 7 extra soldiers.
x² +7=632 2
x = 632-7
x² = 625
x = √625
x = 25
So there are 25 soldiers on each side of the square.
∴ 25 Soldiers on each FRONT SIDE
Question 11. 15 men working 6 hours a day can do a piece of work in 20 days. How many men working 8 hours a day Can do it in 25 days?
Solution:
Given:
15 men work 6 hours a day to complete the work in 20 days.
Total man-hours = Number of men x Number of hours per day X Number of days.
Total man-hours = 15×6×20
Now, The total man-hours For the second scenario
The number of days is 25
The number of hours per day is ‘8’
we need to find out how many men are required.
Let’s denote the number of men required as ‘x’
15×6×20 = x x 8 x 25
1800 = 200x
x = \(\frac{1800}{200}\)
x = 9
So, 9 men are required to complete the work in 25 days working 8 hours a day.
Question 12. Simplify \(\frac{2.8 \text { of } 2.2 \ddot{2}}{1.3 \dot{3}}+\left\{\frac{4.4-2.83}{1.3+2.629}\right\}\)of 8.2
Solution:
⇒ \(\frac{2.8 \times 2.27}{1.36}+\left\{\frac{4.4-2.83}{1.3+2.629}\right\} \times 8.2\)
⇒ \(\frac{6.356}{1.36}+\left\{\frac{1.57}{3.929}\right\} \times 8.2\)
⇒ \(4.67352+\{0.39959\} \times 8.2\)
⇒ 4.67352+3.276638
⇒ \(7.950158 \simeq 8\)
∴ \(\frac{2.80 f 2.27}{1.36}+\left\{\frac{4.4-2.83}{1.3+2.629}\right\}\) of 8.2 is = 8
Class 7 Maths Arithmetic Solutions WBBSE
Question 13. Write the ratio of the three angles of an isosceles Yright-angled triangle.
Solution:
Given the total sum of angles in any triangle is 180°
Let’s denote the two equal angles as ‘X’
The equation for the sum of angles in the triangle is
90+X+X = 180°
2x+90° = 180°
2x = 180-90°
2x = 90°
x = 90°/2
X = 45°
∴ The three angles in the triangle are 90°, 45° and 45°
The ratios of these angles are 90:45:45
Dividing each term by 45 is 2:1:1
∴ The ratio of the three angles in a right-angled
Isosceles triangle is 1:2:1
Question 14. The ratio of textbooks and story books in a School library is 3:5. If a number of textbooks is 864, then find the number of story books.
Solution:
Given:
The ratio of textbooks to storybooks= 3:5
If the number of textbooks is 864
we can set up a proportion to find the number of Storybooks:
Let x’ be the number of storybooks \(\frac{\text { Number of textbooks }}{\text { Number of story books }}=\frac{3}{5}\)
⇒ \(\frac{864}{x}=\frac{3}{5}\)
864 x 5 = 3x
4320 = 3x
x = \(\frac{4320}{3}\)
x = 1440
∴ So the number of story books is x = 1440.
Question 15. verify which of the following numbers are in proportion:
1. 4,6,7,8
Solution:
Given numbers are 4, 6, 7, 8
Let’s check the ratios:
1. Ratio of 4 to 6:\(\frac{4}{6}\) = \(\frac{2}{3}\)
2. Ratio of 6 to 7:\(\frac{6}{7}\)
3. Ratio of 7 t0 8: \(\frac{7}{8}\)
“If the ratios of consecutive pairs are equal, then the numbers are in proportion”
∴ \(\frac{6}{7}\) = 0.8571
∴ \(\frac{7}{8}\) = 0.875
The ratio of 6 to 7 is approximately 0.8571 and the ratio of 7 to 8 is 0.875. These two ratios are not equal.
The ratios of consecutive pairs are not equal, the numbers 4, 6, 7, and 8 are not in proportion.
Class 7 Maths Arithmetic Solutions WBBSE
Question 16. Verify which of the following numbers are in Proportion.
1. 8,12,6,9
Solution:
The given numbers are 8, 12, 6 and 9.
Let’s check the ratios:
1. Ratio of 8 to 12:\(\frac{8}{12}\) = \(\frac{2}{3}\)
2. Ratio of 12 to 6:\(\frac{12}{6}\) = 2
3. Ratio of 6 to 9:\(\frac{6}{9}\) = \(\frac{2}{3}\)
“If the ratios of consecutive pairs are equal, then the numbers are in proportion?”
- \(\frac{8}{12}\) = \(\frac{2}{3}\)
- \(\frac{12}{6}\) = 2
- \(\frac{6}{9}\) = \(\frac{2}{3}\)
The ratios of Consecutive pairs are equal \(\frac{2}{3}\) the numbers 8, 12, 6, and 9 are in proportion.
Question 17. Verify which of the following numbers core in proportion.
1. \(\frac{1}{2}, \frac{1}{6}, \frac{1}{3}\) and \(\frac{1}{9}\)
Solution:
The given numbers are \(\frac{1}{2}, \frac{1}{6}, \frac{1}{3}\) and \(\frac{1}{9}\)
Let’s check the ratios:
1. Ratio of \(\frac{1}{2}\) to \(\frac{1}{6}\):
⇒ \(\frac{\frac{1}{2}}{\frac{1}{6}}=\frac{1}{2} \times \frac{6}{1}=3 \text {. }\)
2. Ratio of \(\frac{1}{6}\) to \(\frac{1}{3}\):
⇒ \(\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{6} \times \frac{3}{1}=\frac{1}{2}\)
3. Ratio of \(\frac{1}{3}\) to \(\frac{1}{9}\):
⇒ \(\frac{\frac{1}{3}}{\frac{1}{9}}=\frac{1}{3} \times \frac{9}{1}=3\)
“If the ratios of consecutive pairs are equal, then the numbers are in proportion”
∴ All the ratios are equal to 3 the numbers \(\frac{1}{2}, \frac{1}{6}, \frac{1}{3}\) and \(\frac{1}{9}\) are in proportion.
2. Simplify: \(\frac{3}{5} \div 4 \frac{1}{3} \times 7 \frac{1}{5}+999 \frac{324}{325} \times 324\)
Solution:
⇒ \(\frac{3}{5} \div 4 \frac{1}{3} \times 7 \frac{1}{5}+999 \frac{324}{325} \times 324\)
⇒ \(\frac{3}{5} \div \frac{13}{3} \times \frac{36}{5}+\frac{324999}{325} \times 324\)
⇒ 0·6÷4·3 × 7·2+ 999.9969 x 324
⇒ 0·139534 × 7·2 + 323999.0030
⇒ 1.0046 + 323999.0030
⇒ 324000.0076.
∴ \(\frac{3}{5} \div 4 \frac{1}{3} \times 7 \frac{1}{5}+999 \frac{324}{325} \times 324=324000\)
Arithmetic Formulas For Class 7 WBBSE
Question 18. Find the greatest number that divides 175,220, and 325 to keep equal remainder in all cases.
Solution:
To find the greatest number that divides, 175, 220, and 325 while keeping equal remainders in all cases
we need to find the greatest common divisor (GCD) of the differences between these numbers.
Let ‘d’ be the Common difference then we have:
175-d = 220-d = 325-d.
Subtracting each pair of numbers we get,
220-175 = 45
325-220 = 105
325 175 = 150
Now, we need to find the greatest common divisor of these differences: 45, 105, and 150.
GCD(45,105,150) = 15
∴ The greatest number that divides 175,220, and 325 to keep equal remainder in all cases is 15.