Std 8 Mathematics · Miscellaneous Exercise 1

Miscellaneous Exercise 1

Solutions to all 17 questions of Miscellaneous Exercise 1, covering Chapters 1 to 9, with figures. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

1Choose the correct alternative answer for each of the following questions.

(1) In □PQRS, m∠P = m∠R = 108°, m∠Q = m∠S = 72°. State which pair of sides is parallel.
(A) side PQ and side QR (B) side PQ and side SR (C) side SR and side SP (D) side PS and side PQ

m∠Q + m∠R = 72° + 108° = 180°. ∠Q and ∠R are interior angles made by transversal QR on sides PQ and SR, so side PQ ∥ side SR. (Options A, C and D are pairs of adjacent sides, which meet and cannot be parallel.)

Answer: (B) side PQ and side SR

(2) Which statements are true?
(i) Diagonals of a rectangle are perpendicular bisectors of each other. (ii) Diagonals of a rhombus are perpendicular bisectors of each other. (iii) Diagonals of a parallelogram are perpendicular bisectors of each other. (iv) Diagonals of a kite bisect each other.
(A) (ii) and (iii) (B) only (ii) (C) (ii) and (iv) (D) (i), (iii) and (iv)

(i) False — a rectangle’s diagonals bisect each other but are not perpendicular. (ii) True. (iii) False — a parallelogram’s diagonals only bisect each other. (iv) False — in a kite only one diagonal bisects the other.

Answer: (B) Only statement (ii) is true

(3) If 19³ = 6859, find ∛0.006859.
(A) 1.9 (B) 19 (C) 0.019 (D) 0.19

0.006859 = 68591000000 ∴ ∛0.006859 = ∛6859∛10⁶ = 19100 = 0.19

Answer: (D) 0.19

2Find the cube roots of the following numbers. (1) 5832 (2) 4096

(1) 58325832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= (2 × 3 × 3)³ = 18³
∛5832 = 18

(2) 40964096 = 2¹² (2 taken 12 times)
= (2⁴)³ = 16³
∛4096 = 16

3m ∝ n, and n = 15 when m = 25. Hence (1) find m when n = 87, (2) find n when m = 155.

m ∝ n ∴ m = kn
When n = 15, m = 25: 25 = 15k ∴ k = 2515 = 53
Equation of variation: m = 53n

(1) When n = 87: m = 53 × 87 = 5 × 29 = 145
(2) When m = 155: 155 = 53n ∴ n = 155 × 35 = 93

Answer: (1) m = 145 (2) n = 93

4y varies inversely with x. If y = 30 when x = 12, find (1) y when x = 15, (2) x when y = 18.

y ∝ 1x ∴ x × y = k
When x = 12, y = 30: k = 12 × 30 = 360

(1) When x = 15: 15 × y = 360 ∴ y = 24
(2) When y = 18: x × 18 = 360 ∴ x = 20

Answer: (1) y = 24 (2) x = 20

5Draw a line l. Draw a line parallel to line l at a distance of 3.5 cm.

Step 1: Draw line l and take two points A and B on it.

Step 2: Draw perpendiculars to line l at A and at B.

Step 3: On the perpendiculars, on the same side of line l, mark P and Q such that AP = BQ = 3.5 cm.

Step 4: Draw line PQ. Line PQ ∥ line l, at a distance of 3.5 cm from it.

Line PQ is parallel to line l at a distance of 3.5 cm.

6Fill in the blanks: The number 256⁵⁷ is ...... th root of ...... th power of ...... .

256⁵⁷ = (256⁵)¹⁷: the denominator 7 gives the root and the numerator 5 gives the power.

Answer: The number 256^5/7 is 7th root of 5th power of 256.

7Expand.

(1) (5x − 7)(5x − 9)

(5x − 7)(5x − 9)
Common term 5x; use (x + a)(x + b) = x² + (a + b)x + ab.
= (5x)² + (−7 − 9)(5x) + (−7)(−9)
= 25x² + (−16)(5x) + 63

Answer: 25x² − 80x + 63

(2) (2x − 3y)³

(2x − 3y)³
= (2x)³ − 3(2x)²(3y) + 3(2x)(3y)² − (3y)³
= 8x³ − 3 × 4x² × 3y + 3 × 2x × 9y² − 27y³

Answer: 8x³ − 36x²y + 54xy² − 27y³

(3) (a + 12)³

(a + 12)³
= a³ + 3(a)²(12) + 3(a)(12)² + (12)³
= a³ + 3a²2 + 3a × 14 + 18

Answer: a³ + 3a²2 + 3a4 + 18

8Draw an obtuse angled triangle. Draw all of its medians and show their point of concurrence.

Step 1: Draw an obtuse angled △ABC (here ∠A is obtuse).

Step 2: Find the midpoint P of side BC and draw seg AP.

Step 3: Find the midpoint Q of side CA and draw seg BQ.

Step 4: Find the midpoint R of side AB and draw seg CR.

Step 5: The three medians meet at one point G, the centroid.

Equal tick marks show the halves of each side. The centroid G lies inside the triangle.

━ median

9Draw △ABC such that l(BC) = 5.5 cm, m∠ABC = 90°, l(AB) = 4 cm. Show the orthocentre of the triangle.

Step 1: Draw seg BC of length 5.5 cm.

Step 2: At B, draw a ray perpendicular to BC. On it, mark A such that BA = 4 cm. Draw seg AC.

Step 3: The altitude from A to BC is seg AB itself, because AB ⊥ BC.

Step 4: The altitude from C to AB is seg CB itself, because CB ⊥ AB.

Step 5: Draw the third altitude BD ⊥ AC. All three altitudes pass through B.

All three altitudes meet at B, so the orthocentre is B.

Answer: The orthocentre O is the vertex B of the right angle.

10Identify the variation and solve. It takes 5 hours to travel from one town to the other if the speed of the bus is 48 km/hr. If the speed of the bus is reduced by 8 km/hr, how much time will it take for the same travel?

For the same distance, when the speed decreases the time increases. So this is inverse variation: s ∝ 1t, that is, s × t = k.
k = 48 × 5 = 240 (the distance in km)
New speed = 48 − 8 = 40 km/hr ∴ 40 × t = 240 ∴ t = 6

Answer: Inverse variation; the bus will take 6 hours.

11Seg AD and seg BE are medians of △ABC and point G is the centroid. If l(AG) = 5 cm, find l(GD). If l(GE) = 2 cm, find l(BE).

The centroid divides each median in the ratio 2 : 1 from the vertex.
AG : GD = 2 : 1 ∴ l(GD) = 5 ÷ 2 = 2.5 cm
BG : GE = 2 : 1 ∴ l(BG) = 2 × 2 = 4 cm ∴ l(BE) = 4 + 2 = 6 cm

Answer: l(GD) = 2.5 cm, l(BE) = 6 cm

12Convert the following rational numbers into decimal form.

(1) 813Remainders 8, 2, 7, 5, 11, 6, then 8 again, so 615384 repeats.
813 = 0.615384

(2) 11711 ÷ 7 = 1, remainder 4. Remainders 4, 5, 1, 3, 2, 6, then 4 again, so 571428 repeats.
117 = 1.571428

(3) 5165 ÷ 16 = 0.3125, remainder 0. Terminating decimal.
516 = 0.3125

(4) 7970 ÷ 9 = 7, remainder 7, again and again. 7 repeats.
79 = 0.7

13Factorise.

(1) 2y² − 11y + 5

2y² − 11y + 5
Product = 2 × 5 = 10; (−10) × (−1) = 10 and (−10) + (−1) = −11
= 2y² − 10y − y + 5
= 2y(y − 5) − 1(y − 5)

Answer: (y − 5)(2y − 1)

(2) x² − 2x − 80

x² − 2x − 80
Product = −80; (−10) × (8) = −80 and (−10) + (8) = −2
= x² − 10x + 8x − 80
= x(x − 10) + 8(x − 10)

Answer: (x − 10)(x + 8)

(3) 3x² − 4x + 1

3x² − 4x + 1
Product = 3 × 1 = 3; (−3) × (−1) = 3 and (−3) + (−1) = −4
= 3x² − 3x − x + 1
= 3x(x − 1) − 1(x − 1)

Answer: (x − 1)(3x − 1)

14The marked price of a T.V. set is ₹ 50000. The shopkeeper sold it at 15% discount. Find the price of it for the customer.

Discount = 50000 × 15100 = 500 × 15 = ₹ 7500
Price for the customer = 50000 − 7500 = 42500

Answer: ₹ 42,500

15Rajabhau sold his flat to Vasantrao for ₹ 88,00,000 through an agent. The agent charged 2% commission from both of them. Find how much commission the agent got.

Commission from one person = 88,00,000 × 2100 = ₹ 1,76,000
The agent charged both the seller and the buyer.
Total commission = 1,76,000 + 1,76,000 = ₹ 3,52,000

Answer: ₹ 3,52,000

16Draw a parallelogram ABCD such that l(DC) = 5.5 cm, m∠D = 45°, l(AD) = 4 cm.

Step 1: Draw seg DC of length 5.5 cm.

Step 2: At D, draw a ray making an angle of 45° with seg DC. On it, mark A such that DA = 4 cm.

Step 3: With centre A and radius 5.5 cm, draw an arc. With centre C and radius 4 cm, draw an arc cutting it. Name the point B. (Opposite sides of a parallelogram are equal.)

Step 4: Draw seg AB and seg CB. □ABCD is the required parallelogram.

Figure drawn to scale. ∠A = ∠C = 135° and ∠B = 45°.

17In the figure, line l ∥ line m and line p ∥ line q. Find the measures of ∠a, ∠b, ∠c and ∠d.

Line l ∥ line m and line p ∥ line q.

Step 1: Line l ∥ line m, transversal p: ∠a and the 78° angle are corresponding angles. ∴ m∠a = 78°

Step 2: Line p ∥ line q, transversal m: ∠d and ∠a are corresponding angles. ∴ m∠d = 78°

Step 3: ∠b and ∠d are vertically opposite angles. ∴ m∠b = 78°

Step 4: Line l ∥ line m, transversal q: ∠c and ∠d are interior angles on the same side. ∴ m∠c = 180° − 78° = 102°

Answer: m∠a = 78°, m∠b = 78°, m∠c = 102°, m∠d = 78°

← 9. Discount and Commission 10. Division of Polynomials →