Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Choose 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 of those given below 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
(2) Read the following statements and choose the correct alternative from those given below them.

(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) Statement (ii) and (iii) are true
(B) Only statement (ii) is true
(C) Statements (ii) and (iv) are true
(D) Statements (i), (iii) and (iv) are true.
(3) If 193 = 6859, find $\sqrt[3]{0.006859}$.
(A) 1.9
(B) 19
(C) 0.019
(D) 0.19

(1) We have, in ☐ PQRS, m∠ P = m∠ R = 108m∠ Q = m∠ S =72.

Since, both the pair of opposite angles of the quadrilateral are equal.

So, PQRS is a parallelogram.

$⇒$ Side PQ and side SR are parallel. (Opposite sides of parallelogram are equal.)

Hence, the correct answer is option (B).

Note: Since, side PQ and side QR; side SR and side SP; and side PS and PQ are pair of adjacent sides. So, side PQ and side QR; side SR and side SP; and side PS and PQ cannot be the pair of parallel sides. Thus, the only pair left side PQ and side SR are parallel.

(2) Since, the diagonals of rectangle and parallegram bisect each other but are not perpendicular and the one of the diagonal of kite bisect the other.

Also, the diagonals of a rhombus are perpendicular bisectors of each other.

Hence, the correct answer is option (B).

(3) We have, 193 = 6859

$\sqrt[3]{0.006859}\phantom{\rule{0ex}{0ex}}=\sqrt[3]{\frac{6859}{1000000}}\phantom{\rule{0ex}{0ex}}=\sqrt[3]{{\left(\frac{19}{100}\right)}^{3}}\phantom{\rule{0ex}{0ex}}=\frac{19}{100}\phantom{\rule{0ex}{0ex}}=0.19$

Hence, the correct answer is option (D).

#### Question 2:

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

(1)

(2)

#### Question 3:

m α n, n = 15 when m = 25. Hence
(1) Find m when n = 87
(2) Find n when m = 155

Since, m α n

$⇒m=kn$, where k is proportional constant.

We have, = 15 when = 25.

$⇒25=k×15\phantom{\rule{0ex}{0ex}}⇒k=\frac{25}{15}\phantom{\rule{0ex}{0ex}}⇒k=\frac{5}{3}$
So, $m=\frac{5}{3}n$      ...(i)

(1) We have, n = 87

By using equation (i), we get

$m=\frac{5}{3}×87=5×29=145$

(2) We have, = 155

By using equation (i), we get

$155=\frac{5}{3}n\phantom{\rule{0ex}{0ex}}⇒n=\frac{3}{5}×155=3×31=93$

#### Question 4:

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

Since, varies inversely with x.

We have, = 30 and = 12,

(1) We have, x = 15

By using equation (i), we get

$y=\frac{360}{15}=24$

(2) We have, = 18

By using equation (i), we get

$18=\frac{360}{x}\phantom{\rule{0ex}{0ex}}⇒x=\frac{360}{18}=20$

#### Question 5:

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

Steps of construction:

(1) Make a line ‘l’ and take point O on it.

(2) With O as centre and a certain radius, draw an arc intersecting the line ‘l’ at two points P and Q.

(3) With P and Q as centres and a radius greater than OP, draw two arcs, which cut each other at R.

(4) Join OR and produce it to a certain point. Then OR is perpendicular to ‘l’.

(5) With O as centre and radius of 3.5 cm, draw an arc intersecting the  ray OR at point M.

(6) With M as centre and a certain radius, draw an arc intersecting ray OM at two points T and S.

(7) With S and T as centres and radius greater than MS, draw two arcs, which cut each other at N.

(8) Join MN, then MN is parallel to line 'l'.

Thus, line 'm' is parallel to line 'l'.

#### Question 6:

Fill in the blanks in the following statement.
The number $\left(256{\right)}^{\frac{5}{7}}$ is .............th root of ..........th power of ...............

The number $\left(256{\right)}^{\frac{5}{7}}$ is   7 th root of    5 th power of   256 .

#### Question 7:

Expand.
(1) (5x − 7) (5x − 9)
(2) (2x − 3y)3
(3) (a + $\frac{1}{2}$)3

(1) (5x − 7) (5x − 9)

$=5x\left(5x-9\right)-7\left(5x-9\right)\phantom{\rule{0ex}{0ex}}=25{x}^{2}-45x-35x+63\phantom{\rule{0ex}{0ex}}=25{x}^{2}-80x+63$

(2) (2x − 3y)3

(3) (a + $\frac{1}{2}$)3

#### Question 16:

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

Given: DC = 5.5 cm, ∠D = 45, and AD = 4 cm

In parallelogram ABCD,

BC = AD = 4 cm and DC = AB = 5.5 cm (Opposite sides of parallegram are always equal)

And ∠D + ∠C = 180   (Adjacent angles of a paralllegram is always supplementary)

or, 45 + ∠C = 180

or, ∠C = 135

To construct: A parallelogram ABCD.

Steps of construction:

(i) Draw a line segment DC = 5.5 cm.

(ii) At point D, make an angle ∠XDC = 45.

(iii) With D as centre and radius of 4 cm, draw an arc to intersect the ray DX at point A.

(iv) At point C, make an angle ∠YCD = 135.

(v) With C as centre and radius of 4 cm, draw an arc to intersect the ray CY at point B.

(vi) Join AB.

Hence, ABCD is the required parallelogram.

#### Question 17:

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

In the figure, line ∥ line and line ∥ line q. Find the measures ∠ a, ∠ b, ∠ and ∠ d.

Since, line l || line m   (Given)

So, ∠ a = 78$°$   (Corresponding angles)

Also,

Since, line p || line q   (Given)

So, ∠ b = ∠ a    (Alternate interior angles)

or, ∠ b = 78$°$

And,

∠ d = ∠ b    (Vertically opposite angles)

or, ∠ d = 78$°$

Again,

Since, line p || line q   (Given)

So, ∠ d + ∠ c = 180$°$   (Consecutive interior angles)

or, 78$°$ + ∠ c = 180$°$

or, ∠ c = 102$°$

#### Question 8:

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

Given: $∆$ABC is an obtuse angled tringle.

To construct: All medians of $∆$ABC and their point of concurrence (Centroid).

Steps of construction:

(i) Draw any obtuse angled triangle ABC, $\angle$C is an obtuse angle.

(ii) With A as centre and radius more than half of AC, draw arcs on both sides of AC.

(iii) With C as centre and radius more than half of AC, draw arcs on both sides of AC intersecting the arcs of step (ii) at points P and Q.

(iv) Join PQ intersecting AC at point Z.

(v) With A as centre and radius more than half of AB, draw arcs on both sides of AB.

(vi) With B as centre and radius more than half of AB, draw arcs on both sides of AB intersecting the arcs of step (ii) at points R and S.

(vii) Join RS intersecting AB at point X.

(viii) With B as centre and radius more than half of BC, draw arcs on both sides of BC.

(xi) With C as centre and radius more than half of BC, draw arcs on both sides of AC intersecting the arcs of step (ii) at points D and E.

(x) Join DE intersecting BC at point Y.

(xi) Join AY, BZ and CX intersecting at point G.

Thus, AY, BZ and CX are the required medians of $∆$ABC and point is their point of concurrence (Centroid.

#### Question 9:

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

Given: In $∆$ABC, BC = 5.5 cm, AB = 4 cm and ∠ABC = 90$°$.

To construct: Orthocentre of $∆$ABC.

Steps of construction:

(i) Draw a line segment BC = 5.5 cm.

(ii) At point B, make an $\angle$XBC = 90$°$.

(iii) With B as centre and radius = 4 cm, draw an arc intersecting the ray BX at point A.

(iv) Join AC.

(v) Now, with B as centre and any convenient radius, draw arcs intersecting AC at points D and E.

(vi) With D and E as centres and radius more than half of DE, draw arcs intersecting each other point F.

(vii) Join BF intersecting AC at point G.

Thus, AB, BC and BG are altitudes of $∆$ABC and point of their concurrence is the required orthocentre.

#### Question 10:

Identify the variation and solve. It takes 5 hours to travel from one town to the other  if 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?

If the speed of the bus is reduced, then the time taken to treavel the same distance will increase.

So, this is an inverse variation.

Let the speed of the bus be x km/h and the time taken to travel the given distance be y hours.

We have,

 Speed (x km/h) 48 40 Time (y hours) 5 y

Since, this is an inverse variation.

$⇒{x}_{1}{y}_{1}={x}_{2}{y}_{2}\phantom{\rule{0ex}{0ex}}⇒48×5=40×y\phantom{\rule{0ex}{0ex}}⇒y=\frac{48×5}{40}\phantom{\rule{0ex}{0ex}}⇒y=6$

So, it will take 6 hours for the same travel.

#### Question 11:

Seg 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).

Given: In ∆ABC, AD and BE are medians and point G is the centroid such that AG = 5 cm, and GE = 2 cm.

GD = ? and BE = ?

In ∆ABC,

Since, G is the centroid and AD is median.

So,

Similarly,

Since, G is the centroid and BE is median.

So,

#### Question 12:

Convert the following rational numbers into decimal form.
(1)$\frac{8}{13}$

(2)$\frac{11}{7}$

(3)$\frac{5}{16}$

(4)$\frac{7}{9}$

(1) $\frac{8}{13}$

$130.615384680-7820-1370-6550-39110-10460-5280-782\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

So, $\frac{8}{13}=0.\overline{)615384}$

(2) $\frac{11}{7}$

$71.571428511-740-3550-4910-730-2820-1460-564035\overline{)5}\phantom{\rule{0ex}{0ex}}$

So, $\frac{11}{7}=1.\overline{)571428}$

(3) $\frac{5}{16}$

$160.312550-4820-1640-3280-800\phantom{\rule{0ex}{0ex}}$

So, $\frac{5}{16}=0.3125$

(4) $\frac{7}{9}$

$90.7770-6370-637\phantom{\rule{0ex}{0ex}}$

So, $\frac{7}{9}=0.\overline{)7}$

#### Question 13:

Factorise.
(1) 2y2−11y+5
(2) x2−2x−80
(3) 3x2−4x+1

(1) 2y2 − 11+ 5

$=2{y}^{2}-10y-y+5\phantom{\rule{0ex}{0ex}}=2y\left(y-5\right)-1\left(y-5\right)\phantom{\rule{0ex}{0ex}}=\left(y-5\right)\left(2y-1\right)$

(2) x− 2− 80

$={x}^{2}-10x+8x-80\phantom{\rule{0ex}{0ex}}=x\left(x-10\right)+8\left(x-10\right)\phantom{\rule{0ex}{0ex}}=\left(x-10\right)\left(x+8\right)$

(3) 3x− 4+ 1

$=3{x}^{2}-3x-x+1\phantom{\rule{0ex}{0ex}}=3x\left(x-1\right)-1\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(3x-1\right)$

#### Question 14:

The marked price of T.V. Set is ₹50000. The shop keeper sold it at 15% discount. Find the price of it for the customer.

We have,

MP = ₹50000 and Discount = 15%

So, the price of T.V. for the customer will be ₹42500.

#### Question 15:

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

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

We have,

Amount of flat ₹ 88,00,000

Agent's commission = 2 %

Rajabhau pay amount commission to agent = 2 % of ₹ 88,00,000

$=\frac{2×88,00,000}{100}=₹1,76,000$

Vasant Rao pay amount commission to agent = 2 % of ₹ 88,00,000

$=\frac{2×88,00,000}{100}=₹1,76,000$

So, the total commission of agent from both parties = ₹ 1,76,000 + ₹ 1,76,000 = ₹ 3,52,000

Hence, the agent got ₹ 3,52,000 in total.

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