Mathematics Solutions Solutions for Class 8 Maths Chapter 19 - Miscellaneous Exercise 2

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Page No 119:

Question 1:

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

Answer:

(1) Area = 1386 cm2
$π r^{2} = 1386 \Rightarrow \frac{22}{7} r^{2} = 1386 \Rightarrow r^{2} = 441 \Rightarrow r = 21 cm$
Circumference = $2 π r = 2 \times \frac{22}{7} \times 21 = 132 cm$
Hence, the correct answer is option B.

(2) Side of a cube = 4 m
Volume = $a^{3} = 4^{3} = 64 m^{3}$
New side = $2 \times original side = 2 \times 4 = 8$ m
Volume of new cube = ${(8)}^{3} = 512 m^{3}$
New volume = 512 m³= $8 \times 64 = 8 \times old volume$
Thus, the new volume is 8 times the old volume.
Hence, the correct answer is option D.

Page No 119:

Question 2:

Pranalee was practising for a 100 m running race. She ran 100 m distance 20 times. The time required, in seconds, for each attempt was as follows.
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.

Answer:

Mean = $\frac{Sum of observations}{Number of observations}$
$Mean = \frac{18 + 17 + 17 + 16 + 15 + 16 + 15 + 14 + 16 + 15 + 15 + 17 + 15 + 16 + 15 + 17 + 16 + 15 + 14 + 15}{20}$
$\Rightarrow Mean = \frac{314}{20} = 15.7$ sec

Page No 119:

Question 3:

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

Answer:

âˆ†DEF ≅ âˆ†LMN
$\Rightarrow ∠ DEF = ∠ LMN, ∠ EFD = ∠ MNL, ∠ FDE = ∠ NLM And DE = LM, EF = MN, DF = LN$

Page No 119:

Question 4:

The cost of a machine is â‚¹2,50,000. It depreciates at the rate of 4% per annum. Find the cost of the machine after three years.

Answer:

Cost of a machine = â‚¹2,50,000
Rate of depreciation = 4%
Time period = 3 years
$A = P {(1 - \frac{r}{100})}^{n} \Rightarrow A = 250000 {(1 - \frac{4}{100})}^{3} \Rightarrow A = 250000 {(\frac{96}{100})}^{3} \Rightarrow A = 2, 21, 184$
Amount after three years = Rs 2,21,184

Page No 119:

Question 5:

In â˜ ABCD side AB âˆ¥ side DC, seg AE ⊥ seg DC. If l (AB) = 9 cm, l (AE) = 10 cm, A(â˜ ABCD) = 115 cm², find l (DC).

Answer:

AB = 9 cm
AE = 10 cm
side AB âˆ¥ side DC
So, ABCD is a trapezium
Area of trapezium = $\frac{1}{2} \times AE (AB + DC) = \frac{1}{2} \times 10 (9 + DC) = 115$
$\Rightarrow 9 + DC = \frac{115 \times 2}{10} = 23 \Rightarrow DC = 14 cm$

Page No 119:

Question 6:

The diameter and height of a cylindrical tank is 1.75 m and 3.2 m respectively. How much is the capacity of tank in litre? (π = $\frac{22}{7}$ )

Answer:

Diameter = 1.75 m
Radius = 0.875 m
Height = 3.2 m
Capacity = $π r^{2} h = \frac{22}{7} \times {(0.875)}^{2} \times 3.2 = 7.7$ m³
$1 m^{3} = 1000 L 7.7 m^{3} = 7.7 \times 1000 = 7700 L$

Page No 119:

Question 7:

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

Answer:

Length of the chord = 16.8 cm
Radius = 9.1 cm
Let the distance from the centre of the chord be d.
Perpendicular drawn from the centre of the circle to the chord bisects the chord.
So, AC = CB = $\frac{16.8}{2} = 8.4$ cm
In $△$ AOC,
Applying the Pythagoras theorem we have
${OC}^{2} + {AC}^{2} = {AO}^{2} \Rightarrow d^{2} + 8 . 4^{2} = 9 . 1^{2} \Rightarrow d^{2} = 82.81 - 70.56 \Rightarrow d^{2} = 12.25 \Rightarrow d = 3.5 cm$
Thus, the distance of chord from the centre is 3.5 cm.

Page No 119:

Question 8:

The following tables shows the number of male and female workers, under employment gurantee scheme, in villages A, B, C and D.

Village	A	B	C	D
No. of females	150	240	90	140
No. of males	225	160	210	110

(1) Show the information by a sub-divided bar-diagram.
(2) Show the information by a percentage bar diagram.

Answer:

(1) Sub-divided bar diagram

(2) Percentage bar diagram

Page No 120:

Question 9:

Solve the following equations.
(1) 17(x+4) + 8(x+6) = 11(x+5) +15(x+3)

(2) $\frac{3 y}{2}$ + $\frac{y + 4}{4}$ = 5− $\frac{y - 2}{4}$

(3) 5(1−2x) = 9(1−x)

Answer:

(1) 17(x+4) + 8(x+6) = 11(x+5) +15(x+3)
$\Rightarrow 17 x + 68 + 8 x + 48 = 11 x + 55 + 15 x + 45 \Rightarrow 25 x + 116 = 26 x + 100 \Rightarrow 26 x - 25 x = 116 - 100 \Rightarrow x = 16$

(2) $\frac{3 y}{2}$ + $\frac{y + 4}{4}$ = 5− $\frac{y - 2}{4}$
$3 y \times 2 + y + 4 = 20 - y - 2 \Rightarrow 6 y + y + 4 = 18 - y \Rightarrow 8 y = 14 \Rightarrow y = \frac{14}{8} = \frac{7}{4}$

(3) 5(1−2x) = 9(1−x)
$\Rightarrow 5 - 10 x = 9 - 9 x \Rightarrow - 10 x + 9 x = 9 - 5 \Rightarrow - x = 4 \Rightarrow x = - 4$

Page No 120:

Question 10:

Complete the activity according to the given steps.
(1) Draw rhombus ABCD. Draw diagonal AC.
(2) Show the congruent parts in the figure by identical marks.
(3) State by which test and in which correspondence âˆ† ADC and âˆ† ABC are congruent.
(4) Give reason to show ∠DCA â‰Œ ∠BCA, and ∠DAC â‰Œ ∠BAC
(5) State which property of a rhombus is revealed from the above steps.

Answer:

(1)

(2)

(3) In âˆ† ADC and âˆ† ABC
AC = AC (Common)
AD = AB (ABCD is a rhombus so all sides are equal)
DC = BC (ABCD is a rhombus so all sides are equal)
So, by SSS congruency âˆ† ADC is congruent to âˆ† ABC
(4) Since âˆ† ADC is congruent to âˆ† ABC so,
∠DCA â‰Œ ∠BCA and ∠DAC â‰Œ ∠BAC by CPCT (corresponding parts of congruent triangles)
(5) Since ∠DCA â‰Œ ∠BCA and ∠DAC â‰Œ ∠BAC so, AC acts as the angle bisector.
Thus, we can say that diagonals of a rhombus act as angle bisectors.

Page No 120:

Question 11:

The shape of a farm is a quadrilateral. Measurements taken of the farm, by naming its corners as P, Q, R, S in order are as follows. l(PQ) = 170 m, l(QR) = 250 m, l(RS) = 100 m, l(PS) = 240 m, l(PR) = 260 m.
Find the area of the field in hectare (1 hectare = 10,000 sq.m)

Answer:

In $∆$ PSR,
$s = \frac{100 + 260 + 240}{2} = \frac{600}{2} = 300 Ar (△ PSR) = \sqrt{300 (300 - 100) (300 - 260) (300 - 240)} = \sqrt{300 \times 200 \times 40 \times 600} = \sqrt{144000000} = 12000 sq m$
In $∆$ PQR,
$s = \frac{260 + 250 + 170}{2} = 340 Ar (△ PQR) = \sqrt{340 (340 - 260) (340 - 250) (340 - 170)} = \sqrt{340 \times 80 \times 90 \times 170} = \sqrt{416160000} = 20400 sq m$
Area of PQRS = Ar( $∆$ PSR) + Ar( $∆$ PQR)
= 12000 + 20400
= 32400 m²
1 hectare = 10,000 sq.m
$32400 m^{2} = 3.24 hec$

Page No 120:

Question 12:

In a library, 50% of total number of books is of Marathi. The books of English are $\frac{1}{3}$ rd of Marathi books. The books on mathematics are 25% of the English books. The remaining 560 books are of other subjects. What is the total number of books in the library?

Answer:

Let the total number of books be x.
50% books are in Marathi = $\frac{50}{100} x = \frac{x}{2}$
Books of English = $\frac{1}{3}$ rd of Marathi books = $\frac{1}{3} \times \frac{x}{2} = \frac{x}{6}$
Books on mathematics are 25% of the English books = $\frac{25}{100} \times \frac{x}{6} = \frac{x}{24}$
Remaining books = 560
So,
$\frac{x}{2} + \frac{x}{6} + \frac{x}{24} + 560 = x \Rightarrow \frac{17 x}{24} + 560 = x \Rightarrow x - \frac{17 x}{24} = 560 \Rightarrow \frac{7 x}{24} = 560 \Rightarrow x = \frac{560 \times 24}{7} = 1920$
Total books in the library are 1920.

Page No 120:

Question 13:

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

Answer:

Given polynomial = (6x³+11x²−10x−7)
Binomial = 2x+1

Thus, Quotient = $3 x^{2} + 4 x - 7$
Remainder = 0

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