Balbharti Maharashtra State Board Class 7 Maths Solutions covers the 7th Std Maths Miscellaneous Problems Set 2 Answers Solutions.

## Maharashtra Board Miscellaneous Problems Set 2 Class 7 Maths Solutions

Question 1.

Angela deposited Rs 15000 in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to Rs 5400. For how many years had she deposited the amount?

Solution:

Here, P = Rs 15000, R = 9 p.c.p.a., I = Rs 5400

∴ T = 4

∴ Angela had deposited the amount for 4 years.

Question 2.

Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?

Solution:

Let us suppose that 8 men require x days to tar the road.

Number of days required by 10 men to tar the road = 4

The number of men and the number of days required to tar the road are in inverse proportion.

∴ 8 × x = 10 x 4

∴ \(x=\frac{10 \times 4}{8}\)

∴ x = 5

∴ 8 men will require 5 days to tar the road.

Question 3.

Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?

Solution:

Total amount invested = Rs 40,000 + Rs 60,000

= Rs 1,00,000

Profit earned = 30%

∴ Total profit = 30% of 1,00,000

= \(\frac { 30 }{ 100 }\) × 100000

= Rs 30000

Proportion of investment = 40000 : 60000

= 2:3 …. (Dividing by 20000)

Let Nasruddin’s profit be Rs 2x and Mahesh’s profit be Rs 3x.

∴ 2x + 3x = 30000

∴ 5x = 30000

∴ x = \(\frac { 30000 }{ 5 }\).

∴ x = 6000

∴ Nasruddin’s profit = 2x = 2 × 6000 = Rs 12000

Mahesh’s profit = 3x = 3 × 6000 = Rs 18000

∴ The profits of Nasruddin and Mahesh are Rs 12000 and Rs 18000 respectively.

Question 4.

The diameter of a circle is 5.6 cm. Find its circumference.

Solution:

Diameter of the circle (d) = 5.6 cm

Circumference = πd

= \(\frac{22}{7} \times 5.6\)

= \(\frac{22}{7} \times \frac{56}{10}\)

= 17.6 cm

∴ The circumference of the circle is 17.6 cm.

Question 5.

Expand:

i. (2a – 3b)²

ii. (10 + y)²

iii. \(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}\)

iv. \(\left(y-\frac{3}{y}\right)^{2}\)

Solution:

i. Here, A = 2a and B = 3b

∴ (2a – 3b)² = (2a)² – 2 × 2a × 3b + (3b)²

…. [(A – B)² = A² – 2AB + B²]

= 4a² – 12ab + 9b²

ii. Here, a = 10 and b = y

(10 + y)² = 102 + 2 × 10xy + y²

…. [(a + b)² = a² + 2ab + b²]

= 100 + 20y + y²

iii. Here, a = \(\frac { p }{ 3 }\) and b = \(\frac { q }{ 4 }\)

\(\left(\frac{p}{3}+\frac{q}{4}\right)^{2}=\left(\frac{p}{3}\right)^{2}+2 \times \frac{p}{3} \times \frac{q}{4}+\left(\frac{q}{4}\right)^{2}\)

…. [(a + b)² = a² + 2ab + b²]

\(\frac{p^{2}}{9}+\frac{p q}{6}+\frac{q^{2}}{16}\)

iv. Here, a = y and b = \(\frac { 3 }{ y }\)

\(\left(y-\frac{3}{y}\right)^{2}=y^{2}-2 \times y \times \frac{3}{y}+\left(\frac{3}{y}\right)^{2}\)

…. [(a – b)² = a² – 2ab + b²

= \(y^{2}-6+\frac{9}{y^{2}}\)

Question 6.

Use a formula to multiply:

i. (x – 5)(x + 5)

ii. (2a – 13)(2a + 13)

iii. (4z – 5y)(4z + 5y)

iv. (2t – 5)(2t + 5)

Solution:

i. Here, a = x and b = 5

(x – 5)(x + 5) = (x)² – (5)²

…. [(a + b)(a – b) = a² – b²]

= x² – 25

ii. Here, A = 2a and B = 13

(2a – 13)(2a + 13) = (2a)² – (13)²

…. [(A + B)(A – B) = A² – B²]

= 4a² – 169

iii. Here, a = 4z and b = 5y

(4z – 5y)(4z + 5y) = (4z)² – (5y)²

…. [(a + b)(a – b) = a² – b²]

= 16z² – 25y²

iv. Here, a = 2t and b = 5

(2t – 5)(2t + 5) = (2t)² – (5)²

…. [(a + b)(a – b) = a² – b²]

= 4t² – 25

Question 7.

The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?

Solution:

Diameter of the wheel (d) = 1.05 m

∴ Distance covered in 1 rotation of wheel = Circumference of the wheel

= πd

= \(\frac{22}{7} \times 1.05\)

= 3.3 m

∴ Distance covered in 1000 rotations = 1000 x 3.3 m

= 3300 m

= \(\frac { 3300 }{ 1000 }\) km …[1m = \(\frac { 1 }{ 1000 }\)km]

= 3.3 km

∴ The distance covered by the cart in 1000 rotations of the wheel is 3.3 km.

Question 8.

The area of a rectangular garden of length 40 m, is 1000 sq m. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of Rs 250 per metre.

Solution:

Length of the rectangular garden = 40 m

Area of the rectangular garden = 1000 sq. m.

∴ length × breadth = 1000

∴ 40 × breadth = 1000

∴ breadth = \(\frac { 1000 }{ 40 }\)

= 25 m

Now, perimeter of the rectangular garden = 2 × (length + breadth)

= 2 (40 + 25)

= 2 × 65

= 130 m

Length of one round of fence = circumference of garden – width of the entrance

= 130 – 4

= 126 m

∴ Total length of fencing = length of one round of wire × number of rounds = 126 × 3

= 378 m

∴ Total cost of fencing = Total length of fencing × cost per metre of fencing

= 378 × 250

= 94500

∴ The cost of fencing the garden is Rs 94500.

Question 9.

From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.

Solution:

In ∆ABC, ∠B = 90°, and l(BC) = 21, and l(AB) = 20

∴ According to Pythagoras’ theorem,

∴ l(AC)² = l(BC)² + l(AB)²

∴ l(AC)² = 21² + 20²

∴ l(AC)² = 441 + 400

∴ l(AC)² = 841

∴ l(AC)² = 29²

∴ l(AC) = 29

Perimeter of ∆ABC = l(AB) + l(BC) + l(AC)

= 20 + 21 + 29

= 70

∴ The length of hypotenuse AC is 29 units, and the perimeter of ∆ABC is 70 units.

Question 10.

If the edge of a cube is 8 cm long, find its total surface area.

Solution: ,

Total surface area of the cube = 6 × (side)²

= 6 × (8)²

= 6 × 64

= 384 sq. cm

The total surface area of the cube is 384 sq.cm.

Question 11.

Factorize: 365y^{4}z^{3} – 146y^{2}z^{4}

Solution:

= 365y^{4}z^{3} – 146y^{2}z^{4
}= 73 (5y^{4}z^{3} – 2y^{2}z^{4})

= 73y^{2} (5y^{2}z^{3} – 2z^{4})

= 73y^{2}z^{3}(5y^{2} – 2z)