Miscellaneous Exercise 1 8th Std Maths Answers
Solution & Step-by-Step Answer:
(B) side PQ and side SR
Hint:
In ₹PQRS,
m∠P + m∠S = 108°+ 72
= 180°
Since, interior angles are supplementary.
∴ side PQ || side SR
ii. Read the following statements and choose the correct alternative from those given below them. [Chapter 8]
a. Diagonals of a rectangle are perpendicular bisectors of each other.
b. Diagonals of a rhombus are perpendicular bisectors of each other.
c. Diagonals of a parallelogram are perpendicular bisectors of each other.
d. Diagonals of a kite bisect each other.
(A) Statements (b) and (c) are true
(B) Only statement (b) is true
(C) Statements (b) and (d) are true
(D) Statements (a), (c) and (d) are true.
Solution:
(B) Only statement (b) is true
iii. If 19³ = 6859, find . [Chapter 3]
(A) 1.9
(B) 19
(C) 0.019
(D) 0.19
Solution:
(D) 0.19
Hint:
Solution & Step-by-Step Answer:
i. 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 = (2 × 3 × 3) × (2 × 3 × 3) × (2 × 3 × 3) = (2 × 3 × 3)³ = (18)³
ii. 4096 = (4 × 4) × (4 × 4) × (4 × 4)
= (4 × 4)
= 16³
Solution & Step-by-Step Answer:
Given that, m ∝ n ∴ m = kn …(i) where, k is the constant of variation. When m = 25, n = 15 ∴ Substituting, m = 25 and n = 15 in (i), we get m = kn ∴ 25 = k × 15 ∴ k = ∴ k = Substituting k = in (i), we get m = kn ∴ m = …(ii)
i. When n = 87, m = ?
Substituting n = 87 in (ii), we get
m =
m = × 87
m = 5 × 29
m = 145
ii. When m = 155, n = ?
∴ Substituting m = 155 in (ii), we get
m =
∴ 155 =
∴
∴ n = 31 × 3
∴ n = 93
Solution & Step-by-Step Answer:
Given that, ∴ where, k is the constant of variation. ∴ y × x = k …(i) When x = 12, y = 30 ∴ Substituting, x = 12 and y = 30 in (i), we get y × x = k ∴ 30 × 12 = k ∴ k = 360 Substituting, k = 360 in (i), we get y × x = k ∴ y × x = 360 ….(ii)
i. When x = 15,y = ?
∴ Substituting x = 15 in (ii), we get
y × x = 360
∴ y × 15 = 360
∴ y =
∴ y = 24
ii. When y = 18, x = ?
∴ Substituting y = 18 in (ii), we get
y × x = 360
∴18 × x = 360
∴ x =
∴ x = 20
Solution & Step-by-Step Answer:
Steps of construction:
Line n is parallel to line l at a distance of 3.5 cm from it.
Solution & Step-by-Step Answer:
The number is 7th root of 5th power of 256.
Solution & Step-by-Step Answer:
i. (5x – 7) (5x – 9) = (5x)² + (-7 -9) 5x + (-7) × (-9). …[∵ (x + a) (x + b) = x² + (a + b)x + ab] = 25x² + (-16) × 5x + 63 = 25x² – 80x + 63
ii. Here, a = 2x and b = 3y
(2x – 3y)³
= (2x)³ – 3 (2x)² (3y) + 3 (2x) (3y)² – (3y)³
…[∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8x³ – 3 (4x²) (3y) + 3 (2x) (9y²) – 27y³
= 8x³ – 36x²y + 54xy² – 27p³
iii. Here, A= a and B =
…[(A + B)³ = A³ + 3A²B + 3AB² + B³]
Solution & Step-by-Step Answer:
The point of concurrence of the medians PS, RU and QV is G.
Solution & Step-by-Step Answer:
Here, point B is the orthocentre of ∆ABC.
Solution & Step-by-Step Answer:
Let, v represent the speed of the bus and t represent the time required to travel from one town to the other. The speed of the bus varies inversely with the time required to travel from one town to the other. ∴ ∴ where, k is the constant of variation. ∴ v × t = k …(i) It takes 5 hours to travel from one town to the other if speed of the bus is 48 km/hr. i.e., when v = 48, t = 5 ∴ Substituting v = 48 and t = 5 in (i), we get v × t = k ∴ 48 × 5 = k ∴ k = 240 Substituting k = 240 in (i), we get v × t = k ∴ v × t = 240 …(ii) Since, the speed of the bus is reduced by 8 km/hr, ∴ Speed of the bus in second case (v) = 48 – 8 = 40 km/hr ∴ When v = 40, t = ? ∴ Substituting v = 40 in (ii), we get v × t = 240 ∴ 40 × t = 240 ∴ ∴ t = 6 ∴ The problem is of inverse variation and the bus would take 6 hours to travel the distance if its speed is reduced by 8 km/hr.
Solution & Step-by-Step Answer:
The centroid of a triangle divides each median in the ratio 2:1. i. Point G is the centroid and seg AD is the median.
ii. Point G is the centroid and seg BE is the median.
∴ l(BG) × 1 = 2 × 2
∴ l(BG) = 4 cm
Now, l(BE) = l(BG) + l(GE)
∴ l(BE) = 4 + 2
∴ l(BE) = 6 cm
Solution & Step-by-Step Answer:
i.
ii.
iii.
iv.
Solution & Step-by-Step Answer:
i. 2y² – 11y + 5 = 2y² – 10y – y + 5 = 2y(y – 5) – 1(y – 5) = (y – 5)(2y – 1)
ii. x² – 2x – 80
= x² – 10x + 8x – 80
= x (x – 10) + 8 (x – 10)
= (x – 10)(x + 8)
iii. 3x² – 4x + 1
= 3x² – 3x – x + 1
= 3x(x – 1) – 1(x – 1)
= (x – 1) (3x – 1)
Solution & Step-by-Step Answer:
Here, marked price = Rs 50,000, discount = 15% Let the discount percent be x ∴x = 15% i. Discount = 500 × 15 = Rs 7,500
ii. Selling price = Marked price – Discount
= 50,000 – 7,500
= Rs 42,500
∴The price of the T.V. set for the customer is Rs 42,500.
Solution & Step-by-Step Answer:
Here, selling price of the flat = Rs 88,00,000 Rate of commission = 2% Commission = 2% of selling price = × 88,00,000 = 2 × 88,000 = Rs 1,76,000 ∴ Total commission = Commission from Rajabhau + Commission from Vasantrao = Rs 1,76,000 + Rs 1,76,000 = Rs 3,52,000 ∴ The agent got a commission of Rs 3,52,000.
Solution & Step-by-Step Answer:
Opposite sides of a parallelogram are congruent. ∴ l(AD) = l(BC) = 4 cm and l(DC) = l(AB) = 5.5 cm
Solution & Step-by-Step Answer:
i. line l|| line m and line p is a transversal. ∴m∠a = 78° …(i) [Corresponding angles]
ii. line p || line q and line m is a transversal.
∴m∠d = m∠a …[Corresponding angles]
∴m∠d = 78° …(ii)[From (i)]
iii. m∠b = m∠d …[Vertically opposite angles]
∴m∠b = 78° …[From (ii)]
iv. line l|| line m and line q is a transversal.
∴m∠c + m∠d = 180° …[Interior angles]
∴m∠c + 78° = 180° … [From (ii)]
∴m∠c =180° – 78°
∴m∠c = 102°
∴m∠a = 78°, m∠b = 78°, m∠c = 102°, m∠d = 78°
Maharashtra Board Class 8 Maths Solutions