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Std 7 Mathematics · Part Two

Miscellaneous Problems: Set 2

Step-by-step solutions to Miscellaneous Problems Set 2, covering Chapters 9 to 15: simple interest, proportion, circle, area, Pythagoras’ theorem, expansion formulae and statistics. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

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Questions 1 to 11

1Angela deposited 15000 rupees in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to 5400 rupees. For how many years had she deposited the amount?

Principal (P) = ₹15000, Rate (R) = 9 p.c.p.a., Simple interest (I) = ₹5400, Time (N) = ?
I = P × R × N100
5400 = 15000 × 9 × N100 = 1350 × N
N = 54001350 = 4

Angela had deposited the amount for 4 years.

2Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?

Fewer men will take more days, so the number of men and the number of days are in inverse proportion.

Number of men × number of days is constant.
10 × 4 = 8 × x
x = 408 = 5

8 men would take 5 days.

3Nasruddin and Mahesh invested ₹40,000 and ₹60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?

Total investment = 40,000 + 60,000 = ₹1,00,000
Profit = 30% of 1,00,000 = 30100 × 1,00,000 = ₹30,000
Ratio of investments = 40,000 : 60,000 = 2 : 3
Nasruddin’s share = 25 × 30,000 = ₹12,000
Mahesh’s share = 35 × 30,000 = ₹18,000

Nasruddin’s profit = ₹12,000, Mahesh’s profit = ₹18,000.

Check: each partner gets 30% of what he invested — 30% of 40,000 = 12,000 and 30% of 60,000 = 18,000.

4The diameter of a circle is 5.6 cm. Find its circumference.

Diameter (d) = 5.6 cm
Circumference = πd = 227 × 5.6
= 22 × 0.8 = 17.6

The circumference of the circle is 17.6 cm.

5Expand. (i) (2a − 3b)2 (ii) (10 + y)2 (iii) (p3 + q4)2 (iv) (y − 3y)2

(a + b)2 = a2 + 2ab + b2    (a − b)2 = a2 − 2ab + b2

(i) (2a − 3b)2

= (2a)2 − 2 × 2a × 3b + (3b)2
= 4a2 − 12ab + 9b2

(ii) (10 + y)2

= 102 + 2 × 10 × y + y2
= 100 + 20y + y2

(iii) (p3 + q4)2

= (p3)2 + 2 × p3 × q4 + (q4)2
= p29 + pq6 + q216

(iv) (y − 3y)2

= y2 − 2 × y × 3y + (3y)2
= y2 − 6 + 9y2

6Use a formula to multiply. (i) (x − 5)(x + 5) (ii) (2a − 13)(2a + 13) (iii) (4z − 5y)(4z + 5y) (iv) (2t − 5)(2t + 5)

(a + b)(a − b) = a2 − b2

(i) (x − 5)(x + 5)

= x2 − 52 = x2 − 25

(ii) (2a − 13)(2a + 13)

= (2a)2 − 132 = 4a2 − 169

(iii) (4z − 5y)(4z + 5y)

= (4z)2 − (5y)2 = 16z2 − 25y2

(iv) (2t − 5)(2t + 5)

= (2t)2 − 52 = 4t2 − 25

7The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?

In one rotation, the wheel covers a distance equal to its circumference.
Circumference = πd = 227 × 1.05 = 22 × 0.15 = 3.3 m
Distance in 1000 rotations = 3.3 × 1000 = 3300 m

The cart will cover 3300 m, that is, 3.3 km.

8The area of a rectangular garden of length 40 m, is 1000 sqm. 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 250 rupees per metre.

Area of rectangle = length × breadth
1000 = 40 × breadth
Breadth = 100040 = 25 m

Perimeter = 2 × (length + breadth) = 2 × (40 + 25) = 2 × 65 = 130 m

Length of fencing for one round, leaving a 4 m entrance = 130 − 4 = 126 m
Length for 3 rounds = 126 × 3 = 378 m
Cost of fencing = 378 × 250 = ₹94,500

The cost of fencing the garden is ₹94,500.

9From the given figure, find the length of hypotenuse AC and the perimeter of ΔABC. (In the figure, ∠B is a right angle, AB = 20 and BC = 21.)
BCA212029

Right-angled ΔABC, m∠B = 90°

In ΔABC, ∠B = 90°, so AC is the hypotenuse. By Pythagoras’ theorem:

AC2 = AB2 + BC2
= 202 + 212 = 400 + 441 = 841
AC = √841 = 29

Perimeter of ΔABC = AB + BC + AC = 20 + 21 + 29 = 70

The figure gives no unit, so the answers are in the same unit as the sides: hypotenuse AC = 29 units and perimeter = 70 units.

10If the edge of a cube is 8 cm, find its total surface area.

Total surface area of a cube = 6 × (edge)2
= 6 × 82 = 6 × 64 = 384

The total surface area of the cube is 384 cm2.

11Factorise. 365y4z3 − 146y2z4

365y4z3 = 73 × 5 × y2 × y2 × z3
146y2z4 = 73 × 2 × y2 × z3 × z
Common factor = 73y2z3
∴ 365y4z3 − 146y2z4 = 73y2z3(5y2 − 2z)

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Multiple Choice Questions

1If the average of the numbers 33, 34, 35, x, 37, 38, 39 is 36, what is the value of x? (i) 40 (ii) 32 (iii) 42 (iv) 36

Average = Sum of the numbersNumber of numbers
There are 7 numbers, so their sum = 36 × 7 = 252.
33 + 34 + 35 + x + 37 + 38 + 39 = 252
216 + x = 252, so x = 36

  • Answer: (iv) 36
2The difference of the squares, (612 − 512) is equal to ________. (i) 1120 (ii) 1230 (iii) 1240 (iv) 1250

a2 − b2 = (a + b)(a − b)
612 − 512 = (61 + 51)(61 − 51) = 112 × 10 = 1120

  • Answer: (i) 1120
3If 2600 rupees are divided between Sameer and Smita in the proportion 8 : 5, the share of each is ________ and ________ respectively. (i) ₹1500, ₹1100 (ii) ₹1300, ₹900 (iii) ₹800, ₹500 (iv) ₹1600, ₹1000

Sum of the ratio terms = 8 + 5 = 13
Sameer’s share = 813 × 2600 = 8 × 200 = ₹1600
Smita’s share = 513 × 2600 = 5 × 200 = ₹1000

  • Answer: (iv) ₹1600, ₹1000
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