Std 8 Mathematics · Miscellaneous Exercise 2

Miscellaneous Exercise 2

Solutions to all 13 questions of Miscellaneous Exercise 2, covering Chapters 10 to 17, with graphs and figures. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

1Choose the correct alternative answer.

(1) Find the circumference of a circle whose area is 1386 cm². (A) 132 cm² (B) 132 cm (C) 42 cm (D) 21 cm²

πr² = 1386 ∴ 227 × r² = 1386 ∴ r² = 1386 × 722 = 441 ∴ r = 21 cm
Circumference = 2πr = 2 × 227 × 21 = 132 cm (a length, so the unit is cm, not cm²)

Answer: (B) 132 cm

(2) The side of a cube is 4 m. If it is doubled, how many times will the volume of the new cube be, compared with the original cube? (A) Two times (B) Three times (C) Four times (D) Eight times

Original volume = 4³ = 64 cubic m. New side = 8 m, new volume = 8³ = 512 cubic m.
512 ÷ 64 = 8. (Doubling the side multiplies the volume by 2 × 2 × 2 = 8.)

Answer: (D) Eight times

2Pranalee ran a 100 m distance 20 times. The time taken, in seconds, for each attempt was: 18, 17, 17, 16, 15, 16, 15, 14, 16, 15, 15, 17, 15, 16, 15, 17, 16, 15, 14, 15. Find the mean of the times taken for running.

Time (seconds) x_i	Tally marks	Frequency f_i	f_i × x_i
14	\|\|	2	28
15	~~\|\|\|\|~~\|\|\|	8	120
16	~~\|\|\|\|~~	5	80
17	\|\|\|\|	4	68
18	\|	1	18
		N = 20	Σf_ix_i = 314

Mean = Σf_ix_iN = 31420 = 15.7

Answer: 15.7 seconds

3△DEF and △LMN are congruent in the correspondence EDF ↔ LMN. Write the pairs of congruent sides and congruent angles.

The correspondence EDF ↔ LMN means E ↔ L, D ↔ M, F ↔ N.

Congruent sides: side ED ≅ side LM, side DF ≅ side MN, side EF ≅ side LN
Congruent angles: ∠E ≅ ∠L, ∠D ≅ ∠M, ∠F ≅ ∠N

Answer: ED ≅ LM, DF ≅ MN, EF ≅ LN; ∠E ≅ ∠L, ∠D ≅ ∠M, ∠F ≅ ∠N

4The cost of a machine is ₹ 2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years.

P = 250000, R = 4 (depreciation), N = 3
A = 250000 × (96100)³ = 250000 × (2425)³ = 250000 × 1382415625 = 16 × 13824 = 221184

Answer: ₹ 2,21,184

5In □ABCD, side AB ∥ side DC, seg AE ⊥ seg DC. If l(AB) = 9 cm, l(AE) = 10 cm and A(□ABCD) = 115 cm², find l(DC).

Area of a trapezium = 12 × (sum of parallel sides) × height
115 = 12 × (9 + l(DC)) × 10 = 5 × (9 + l(DC))
∴ 9 + l(DC) = 23 ∴ l(DC) = 14

Answer: l(DC) = 14 cm

6The diameter and height of a cylindrical tank are 1.75 m and 3.2 m respectively. How much is the capacity of the tank in litres? (π = 22/7)

r = 1.75 ÷ 2 = 0.875 m, h = 3.2 m
V = 227 × 0.875 × 0.875 × 3.2 = 22 × 0.125 × 0.875 × 3.2 = 7.7 cubic m
1 cubic m = 1000 litres ∴ 7.7 × 1000 = 7700 litres

Answer: 7700 litres

7The length of a chord of a circle is 16.8 cm and the radius is 9.1 cm. Find its distance from the centre.

The perpendicular from the centre bisects the chord, so half-chord = 16.8 ÷ 2 = 8.4 cm.
Distance² = 9.1² − 8.4² = (9.1 + 8.4)(9.1 − 8.4) = 17.5 × 0.7 = 12.25
Distance = √12.25 = 3.5 cm

Answer: 3.5 cm

8The table shows the number of female and male workers under the employment guarantee scheme in villages A, B, C and D (females: 150, 240, 90, 140; males: 225, 160, 210, 110). (1) Show the information by a subdivided bar diagram. (2) Show it by a percentage bar diagram.

(1) Subdivided bar diagram

Village	A	B	C	D
Females	150	240	90	140
Males	225	160	210	110
Total	375	400	300	250

Each bar shows the total workers; the lower part is females.

(2) Percentage bar diagram

Village	A	B	C	D
Females %	150375 × 100 = 40	240400 × 100 = 60	90300 × 100 = 30	140250 × 100 = 56
Males %	60	40	70	44

Every bar is 100 units high.

9Solve the following equations.

(1) 17(x + 4) + 8(x + 6) = 11(x + 5) + 15(x + 3)

17x + 68 + 8x + 48 = 11x + 55 + 15x + 45
25x + 116 = 26x + 100
116 − 100 = 26x − 25x
x = 16

Answer: x = 16

(2) 3y2 + y + 44 = 5 − y − 24

Multiplying each term by 4,
6y + y + 4 = 20 − (y − 2)
7y + 4 = 22 − y
8y = 18
y = 188 = 94

Answer: y = 94

(3) 5(1 − 2x) = 9(1 − x)

5 − 10x = 9 − 9x
5 − 9 = 10x − 9x
x = −4

Answer: x = −4

10Complete the activity: (1) Draw rhombus ABCD and diagonal AC. (2) Show the congruent parts by identical marks. (3) State by which test and in which correspondence △ADC and △ABC are congruent. (4) Give reasons to show ∠DCA ≅ ∠BCA and ∠DAC ≅ ∠BAC. (5) State which property of a rhombus is revealed.

Rhombus ABCD with diagonal AC. All four sides are congruent (single marks).

(1), (2) Draw rhombus ABCD and diagonal AC. Mark AB ≅ BC ≅ CD ≅ DA with the same mark.
(3) In △ADC and △ABC: side AD ≅ side AB, side DC ≅ side BC (sides of a rhombus), side AC ≅ side AC (common). ∴ △ADC ≅ △ABC by the SSS test in the correspondence ADC ↔ ABC.
(4) ∠DCA ≅ ∠BCA and ∠DAC ≅ ∠BAC because they are corresponding angles of congruent triangles.
(5) Diagonal AC divides ∠A and ∠C into two equal parts. The diagonals of a rhombus bisect its opposite angles.

Answer: SSS test, ADC ↔ ABC; the diagonal of a rhombus bisects the opposite angles.

11A farm is a quadrilateral PQRS with l(PQ) = 170 m, l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m and l(PR) = 260 m. Find the area of the field in hectares (1 hectare = 10,000 sq m).

Diagonal PR divides the farm into △PQR and △PSR (drawn to scale).

△PQR: sides 170, 250, 260. s = 170 + 250 + 2602 = 340
A = √340 × 170 × 90 × 80 = √41,61,60,000 = 20400 sq m
△PSR: sides 240, 100, 260. Since 100² + 240² = 10000 + 57600 = 67600 = 260², it is right angled at S.
A = 12 × 100 × 240 = 12000 sq m
Area of the farm = 20400 + 12000 = 32400 sq m = 32400 ÷ 10000 hectare = 3.24 hectare

Answer: 3.24 hectare

12In a library, 50% of the books are Marathi. English books are 1/3 of the Marathi books. Mathematics books are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library?

Let the total number of books be x.
Marathi = 50100x = x2
English = 13 × x2 = x6
Mathematics = 25% of English = 14 × x6 = x24
x = x2 + x6 + x24 + 560
Multiplying by 24: 24x = 12x + 4x + x + 13440 ∴ 7x = 13440 ∴ x = 1920

Answer: 1920 books

Check: Marathi 960, English 320, Mathematics 80, others 560; 960 + 320 + 80 + 560 = 1920. ✓

13Divide the polynomial (6x³ + 11x² − 10x − 7) by the binomial (2x + 1). Write the quotient and the remainder.

	3x²	+ 4x	− 7
2x + 1	6x³	+ 11x²	− 10x	− 7
	6x³	+ 3x²
	−	−
	0	+ 8x²	− 10x	− 7
		8x²	+ 4x
		−	−
		0	− 14x	− 7
			−14x	− 7
			+	+
			0

Quotient = 3x² + 4x − 7 Remainder = 0

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