ML Aggarwal Archives - Noon Academy

ML Aggarwal

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 8 cm and RF = 9 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given dimensions, Consider the ∆PQR So, PE/EQ = 3.9/3 = 39/30 = 13/10 Then, PF/FR = 8/9 By comparing both the results, 13/10 ≠ 8/9 Therefore, PE/EQ ≠ PF/FR So, EF is not parallel...

Find the values of x if
$\mathbf{p}\text{ }+\text{ }\mathbf{7}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{q}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{and}\text{ }{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }+\text{ }\mathbf{q}\text{ }=\text{ }\mathbf{0}$
.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Given quadratic equation: \[\mathbf{p}\text{ }+\text{ }\mathbf{7}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{q}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\text{ }\mathbf{and}\text{...

Find the values of x if
$\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0}$
and
${{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}$
.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Given quadratic equation: \[{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{px}\text{ }\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\] And, \[\mathbf{p}\text{ }+\text{ }\mathbf{1}\text{ }=\text{...

In the given figure, DE || BC. (i) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x. (ii) If DB = x – 3, AB = 2x, EC = x – 2 and AC = 2x + 3, find the value of x.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) From the figure, it is given that, Consider the ∆ABC, AD/DB = AE/EC x/(x – 2) = (x + 2)/(x – 1) By cross multiplication we get, X(x – 1) = (x – 2) (x + 2) x2 – x = x2 – 4 -x = -4 x =...

Use the substitution y =
$\mathbf{3x}\text{ }+\text{ }\mathbf{1}$
to solve for x:
$\mathbf{5}{{\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right)}^{\mathbf{2}}}~+\text{ }\mathbf{6}\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right)\text{ }\text{ }\mathbf{8}\text{ }=\text{ }\mathbf{0}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Given equation, \[\mathbf{5}{{\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right)}^{\mathbf{2}}}~+\text{ }\mathbf{6}\left( \mathbf{3x}\text{ }+\text{ }\mathbf{1} \right)\text{ }\text{...

In the figure (iii) given below, if XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, find PY and XY.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, XY || QR, PX = 1 cm, QX = 3 cm, YR = 4.5 cm and QR = 9 cm, So, PX/QX = PY/YR 1/3 = PY/4.5 By cross multiplication we get, (4.5 × 1)/3 = PY PY = 45/30 PY = 1.5 Then, ∠X =...

(i)
$\surd \left( \mathbf{3x}\text{ }+\text{ }\mathbf{4} \right)\text{ }=\text{ }\mathbf{x}$
(ii)
$\surd \left[ \mathbf{x}\left( \mathbf{x}\text{ }\text{ }\mathbf{7} \right) \right]\text{ }=\text{ }\mathbf{3}\surd \mathbf{2}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\surd \left( \mathbf{3x}\text{ }+\text{ }\mathbf{4} \right)\text{ }=\text{ }\mathbf{x}\] On squaring on both sides, we get \[\begin{array}{*{35}{l}} 3x\text{ }+\text{ }4\text{ }=\text{...

$\mathbf{1}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{6} \right)\text{ }+\text{ }\mathbf{1}/\left( \mathbf{x}\text{ }\text{ }\mathbf{10} \right)\text{ }=\text{ }\mathbf{3}/\left( \mathbf{x}\text{ }\text{ }\mathbf{4} \right)$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Given equation, \[1/\left( x\text{ }+\text{ }6 \right)\text{ }+\text{ }1/\left( x\text{ }\text{ }10 \right)\text{ }=\text{ }3/\left( x\text{ }\text{ }4 \right)\] Taking L.C.M for the R.H.S of the...

(i)
$\mathbf{a}/\left( \mathbf{ax}\text{ }\text{ }\mathbf{1} \right)\text{ }+\text{ }\mathbf{b}/\left( \mathbf{bx}\text{ }\text{ }\mathbf{1} \right)\text{ }=\text{ }\mathbf{a}\text{ }+\text{ }\mathbf{b}$
,
$\mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }\ne \text{ }\mathbf{0},\text{ }\mathbf{ab}\text{ }\ne \text{ }\mathbf{0}$
(ii)
$\mathbf{1}/\left( \mathbf{2a}\text{ }+\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{2x} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{2a}\text{ }+\text{ }\mathbf{1}/\mathbf{b}\text{ }+\text{ }\mathbf{1}/\mathbf{2x}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{a}/\left( \mathbf{ax}\text{ }\text{ }\mathbf{1} \right)\text{ }+\text{ }\mathbf{b}/\left( \mathbf{bx}\text{ }\text{ }\mathbf{1} \right)\text{ }=\text{ }\mathbf{a}\text{ }+\text{...

(i)
$\left( \mathbf{x}\text{ }+\text{ }\mathbf{1} \right)/\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\text{ }+\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{2} \right)/\left( \mathbf{x}\text{ }+\text{ }\mathbf{2} \right)\text{ }=\text{ }\mathbf{3}$
(ii)
$\mathbf{1}/\left( \mathbf{x}\text{ }\text{ }\mathbf{3} \right)\text{ }\text{ }\mathbf{1}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{5} \right)\text{ }=\text{ }\mathbf{1}/\mathbf{6}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\begin{array}{*{35}{l}} \left( x\text{ }+\text{ }1 \right)/\left( x\text{ }\text{ }1 \right)\text{ }+\text{ }\left( x\text{ }\text{ }2 \right)/\left( x\text{ }+\text{ }2 \right)\text{...

(i)
$\mathbf{8}/\left( \mathbf{x}\text{ }+\text{ }\mathbf{3} \right)\text{ }\text{ }\mathbf{3}/\left( \mathbf{2}\text{ }\text{ }\mathbf{x} \right)\text{ }=\text{ }\mathbf{2}$
(ii)
$\mathbf{x}/\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)\text{ }+\text{ }\left( \mathbf{x}\text{ }\text{ }\mathbf{1} \right)/\mathbf{x}\text{ }=\text{ }\mathbf{2}{\scriptscriptstyle 1\!/\!{ }_2}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[8/\left( x\text{ }+\text{ }3 \right)\text{ }\text{ }3/\left( 2\text{ }\text{ }x \right)\text{ }=\text{ }2\] Taking L.C.M, we have \[\left[ 8\left( 2\text{ }\text{ }x \right)\text{ }\text{...

(i)
$\mathbf{3x}\text{ }\text{ }\mathbf{8}/\mathbf{x}\text{ }=\text{ }\mathbf{2}$
(ii)
$\left( \mathbf{x}\text{ }+\text{ }\mathbf{2} \right)/\left( \mathbf{x}\text{ }+\text{ }\mathbf{3} \right)\text{ }=\text{ }\left( \mathbf{2x}\text{ }\text{ }\mathbf{3} \right)/\left( \mathbf{3x}\text{ }\text{ }\mathbf{7} \right)$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{3x}\text{ }\text{ }\mathbf{8}/\mathbf{x}\text{ }=\text{ }\mathbf{2}\] Taking L.C.M, we have \[\begin{array}{*{35}{l}} (3{{x}^{2}}~~8)/x\text{ }=\text{ }2 \\ 3{{x}^{2}}~~8\text{...

(i)
$\mathbf{2}/{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5}/\mathbf{x}\text{ }+\text{ }\mathbf{2}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{x}\text{ }\ne \text{ }\mathbf{0}$
(ii)
${{\mathbf{x}}^{\mathbf{2}}}/\mathbf{15}\text{ }\text{ }\mathbf{x}/\mathbf{3}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}$
.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[2/{{x}^{2}}~\text{ }5/x\text{ }+\text{ }2\text{ }=\text{ }0\] Taking L.C.M for the given expression, \[\begin{array}{*{35}{l}} (2\text{ }\text{ }5x\text{ }+\text{...

(i)
${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\left( \mathbf{1}\text{ }+\text{ }\surd \mathbf{2} \right)\mathbf{x}\text{ }+\text{ }\surd \mathbf{2}\text{ }=\text{ }\mathbf{0}$
(ii)
$\mathbf{x}\text{ }+\text{ }\mathbf{1}/\mathbf{x}\text{ }=\text{ }\mathbf{2}\left( \mathbf{1}/\mathbf{20} \right)$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\left( \mathbf{1}\text{ }+\text{ }\surd \mathbf{2} \right)\mathbf{x}\text{ }+\text{ }\surd \mathbf{2}\text{ }=\text{ }\mathbf{0}\] Let us expand the given...

(i)
$\surd \mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{10x}\text{ }+\text{ }\mathbf{7}\surd \mathbf{3}\text{ }=\text{ }\mathbf{0}$
(ii)
$\mathbf{4}\surd \mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{2}\surd \mathbf{3}\text{ }=\text{ }\mathbf{0}$
.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\surd \mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~+\text{ }\mathbf{10x}\text{ }+\text{ }\mathbf{7}\surd \mathbf{3}\text{ }=\text{ }\mathbf{0}\] Let us factorize the given expression,...

$\mathbf{a}{}^\text{2}\mathbf{x}{}^\text{2}\text{ }+\text{ }\left( \mathbf{a}{}^\text{2}\text{ }+\text{ }\mathbf{b}{}^\text{2} \right)\mathbf{x}\text{ }+\text{ }\mathbf{b}{}^\text{2}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{a}\ne \mathbf{0}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Let us simplify the expression, \[\begin{array}{*{35}{l}} a{}^\text{2}x{}^\text{2}\text{ }+\text{ }\left( a{}^\text{2}\text{ }+\text{ }b{}^\text{2} \right)x\text{ }+\text{ }b{}^\text{2}\text{...

(i)
$\mathbf{a}{}^\text{2}\mathbf{x}{}^\text{2}\text{ }+\text{ }\mathbf{2ax}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{a}\text{ }\ne \text{ }\mathbf{0}$
(ii)
$\mathbf{x}{}^\text{2}\text{ }\text{ }\left( \mathbf{p}\text{ }+\text{ }\mathbf{q} \right)\mathbf{x}\text{ }+\text{ }\mathbf{pq}\text{ }=\text{ }\mathbf{0}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{a}{}^\text{2}\mathbf{x}{}^\text{2}\text{ }+\text{ }\mathbf{2ax}\text{ }+\text{ }\mathbf{1}\text{ }=\text{ }\mathbf{0},\text{ }\mathbf{a}\text{ }\ne \text{ }\mathbf{0}\] Let us...

In the figure (ii) given below, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm. Find CQ and BQ.

CBSE, Class 10, Maths, ML Aggarwal, Similarity

Solution:- From the figure, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm ∠BQP = ∠BCA … [because alternate angles are equal] Also, ∠B = ∠B … [common for both the triangles] Therefore, ∆ABC ~ ∆BPQ...

$\mathbf{2x}{}^\text{2}\text{ }\text{ }\mathbf{9x}\text{ }+\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}$
, when (i) x ∈ N (ii) x ∈ Q

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Let us factorize the expression, \[\begin{array}{*{35}{l}} 2x{}^\text{2}\text{ }\text{ }9x\text{ }+\text{ }10\text{ }=\text{ }0 \\ 2{{x}^{2}}~\text{ }4x\text{ }\text{ }5x\text{ }+\text{ }10\text{...

${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{4x}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}$
, when x ∈ N

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Let us factorize the expression, \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{ }4x\text{ }\text{ }12\text{ }=\text{ }0 \\ {{x}^{2}}~\text{ }6x\text{ }+\text{ }2x\text{ }\text{ }12\text{ }=\text{ }0 ...

(i)
$\mathbf{3}{{\left( \mathbf{x}\text{ }\text{ }\mathbf{2} \right)}^{\mathbf{2}}}~=\text{ }\mathbf{147}$
(ii)
$\mathbf{1}/\mathbf{7}\left( \mathbf{3x}\text{ }\text{ }\mathbf{5} \right){}^\text{2}\text{ }=\text{ }\mathbf{28}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{3}{{\left( \mathbf{x}\text{ }\text{ }\mathbf{2} \right)}^{\mathbf{2}}}~=\text{ }\mathbf{147}\] Firstly let us expand the given expression, \[\begin{align} & \begin{array}{*{35}{l}}...

In the figure (i) given below if DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm. Find
(i) AE : EC
(ii) DE.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, DE || BG, AD = 3 cm, BD = 4 cm and BC = 5 cm (i) AE: EC So, AD/BD = AE/EC AE/EC = AD/BD AE/EC = ¾ AE: EC = 3: 4 (ii) consider ∆ADE and ∆ABC ∠D = ∠B ∠E = ∠C Therefore,...

(i)
$\mathbf{6p}{}^\text{2}\text{ }+\text{ }\mathbf{11p}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}$
(ii)
$\mathbf{2}/\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~-\mathbf{1}/\mathbf{3x}\text{ }=\text{ }\mathbf{1}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i)\[\mathbf{6p}{}^\text{2}\text{ }+\text{ }\mathbf{11p}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}\] Let us factorize the given expression, \[\begin{array}{*{35}{l}} 6{{p}^{2}}~+\text{...

(i)
$\mathbf{3x}{}^\text{2}\text{ }=\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{4}$
(ii)
$\mathbf{x}\left( \mathbf{6x}\text{ }\text{ }\mathbf{1} \right)\text{ }=\text{ }\mathbf{35}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{3x}{}^\text{2}\text{ }=\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{4}\] Let us simplify the given expression, \[3{{x}^{2}}~\text{ }x\text{ }\text{ }4\text{ }=\text{ }0\] Now, let us...

A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Height of pole (PQ) = 6m Height of a woman (MN) = 1.5m So, shadow NR = 3m Therefore, pole and woman are standing in the same line PM ||MR ∆PRQ ~ ∆MNR...

(i)
$\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}$
(ii)
$\mathbf{21}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{8x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i)\[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }\text{ }\mathbf{12}\text{ }=\text{ }\mathbf{0}\] Let us simplify the given expression, \[\begin{array}{*{35}{l}}...

A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, Height of a tower PQ = 15m It’s shadow QR = 24 m Let us assume the height of a telephone pole MN = x It’s shadow NO = 16 m Given, at the same time,...

In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, consider ∆ABC, So, ∠A = 90o And AD ⊥ BC ∠BAC = 90o Then, ∠BAD + ∠DAC = 90o … [equation (i)] Now, consider ∆ADC ∠ADC = 90o So, ∠DCA + ∠DAC = 90o … [equation (ii)] From...

(i)
$\mathbf{x}{}^\text{2}\text{ }\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}$
(ii)
$\mathbf{x}\left( \mathbf{2x}\text{ }+\text{ }\mathbf{5} \right)\text{ }=\text{ }\mathbf{3}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[\mathbf{x}{}^\text{2}\text{ }\text{ }\mathbf{3x}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0}\] Let us simplify the given expression, \[\begin{array}{*{35}{l}} {{x}^{2}}~\text{...

In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, AF, BE and CD are parallel lines. Consider the ∆AEF and ∆CED ∠AEF and ∠CED [because vertically opposite angles are equal] ∠F = ∠C [alternate angles are equal] Therefore,...

(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, AB, EF and CD are parallel lines. (i) Consider the ∆EFG and ∆CGD ∠EGF = ∠CGD [Because vertically opposite angles are equal] ∠FEG = ∠GCD [alternate angles...

In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. (i) We have to prove that, EF = FC From the figure, D is the midpoint of...

The altitude BN and CM of ∆ABC meet at H. Prove that (i) CN × HM = BM × HN (ii) HC/HB = √[(CN × HN)/(BM × HM)] (iii) ∆MHN ~ ∆BHC

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the ∆ABC, Where, the altitude BN and CM of ∆ABC meet at H. and construction: join MN (i) We have to prove that, CN × HM = BM × HN In ∆BHM and ∆CHN ∠BHM = ∠CHN [because vertically...

In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. (i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN. Sol

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is give that, Consider the ∆RLQ and ∆PLN, ∠RLQ = ∠NLP [vertically opposite angles are equal] ∠RQL = ∠LNP [alternate angle are equal] Therefore, ∆RLQ ~ ∆PLN So, QR/PN...

In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, ABCD is a parallelogram, Then, E is a point on AD and produced and BE intersects CD at F. We have to prove that ∆ABE ~ ∆CFB Consider ∆ABE and ∆CFB ∠A = ∠C [opposite...

In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that BE/DE = AC/BC

CBSE, Class 10, Maths, ML Aggarwal, Similarity

Solution:- From the figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC We have to prove that, BE/DE = AC/BC Consider the ∆ABC and ∆DEB, ∠C = 90o ∠A + ∠ABC = 90o [from the figure equation (i)] Now in ∆DEB ∠DBE +...

In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the ∆ABC, So, we have to prove that, AB × AE = AC × AD Now, consider the ∆ADB and ∆AEC, ∠A = ∠A [common angle for both triangles] ∠ADB = ∠AEC [both angles are equal to 90o] ∆ADB...

In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/ODUsing the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, ABCD is a trapezium in which AB || DC, The diagonals AC and BD intersect at O. So we have to prove that, AO/OC = BO/OD Consider the ∆AOB and ∆COD, ∠AOB = ∠COD …...

Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Consider the two triangles, ∆MNO and ∆XYZ From the question it is given that, two triangles are similar triangles So, ∆MNO ~ ∆XYZ If two triangles are similar, the corresponding angles...

In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prove that BM x NP = CN x MP.

CBSE, Class 10, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. We have to prove that, BM x NP = CN x MP Consider the...

In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, ∠P = ∠P (common angle for both triangles) ∠PQR = ∠PRS [from the question] So, ∆PQR ~ ∆PRS Then, PQ/PR = PR/PS = QR/SR Consider PQ/PR = PR/PS PQ/8 = 8/4 PQ = (8 × 8)/4 PQ...

In the figure (2) given below, ∠ADE = ∠ACB.
(i) Prove that ∆s ABC and AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, (i) ∠A = ∠A (common angle for both triangles) ∠ACB = ∠ADE [given] Therefore, ∆ABC ~ ∆AED (ii) from (i) proved that, ∆ABC ~ ∆AED So, BC/DE = AB/AE = AC/AD AD = AB –...

(a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.

CBSE, Class 10, Maths, ML Aggarwal, Similarity

Solution:- From the question it is give that, AP = 2PB, CP = 2PD (i) We have to prove that, ∆ACP is similar to ∆BDP and AC || BD AP = 2PB AP/PB = 2/1 Then, CP = 2PD CP/PD = 2/1 ∠APC = ∠BPD [from...

(a) In the figure (i) given below, ∠P = ∠RTS. Prove that ∆RPQ ~ ∆RTS.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the given figure, ∠P = ∠RTS So we have to prove that ∆RPQ ~ ∆RTS In ∆RPQ and ∆RTS ∠R = ∠R (common angle for both triangle) ∠P = ∠RTS (from the question) ∆RPQ ~ ∆RTS (b) In the figure...

In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- (i) We have to prove that, ∆ACO ~ ∆BDO. So, from the figure Consider ∆ACO and ∆BDO Then, ∠AOC = ∠BOD [from vertically opposite angles] ∠A = ∠B Therefore, ∆ACO = ∆BDO Given, BD = 2.4 cm,...

In the figure given below, AB || DE, AC = 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, AB||DE AC = 3 cm CE = 7.5 cm BD = 14 cm From the figure, ∠ACB = ∠DCE [because vertically opposite angles] ∠BAC = ∠CED [alternate angles] Then, ∆ABC ~...

Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- Let us assume that, ∆ABC ~ ∆DEF ∆ABC is BC = 6cm ∆ABC ~ ∆DEF So, AB/DE = BC/EF = AC/DF Consider AB/DE = BC/EF AB/8 = 6/4 AB = (6 × 8)/4 AB = 48/4 AB = 12 Now, consider BC/EF = AC/DF 6/4 =...

If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question it is given that, ∆ABC ~ ∆PQR Perimeter of ∆ABC = 32 cm Perimeter of ∆PQR = 48 cm So, AB/PQ = AC/PR = BC/QR Then, perimeter of ∆ABC/perimeter of ∆PQR = AC/PR 32/48 =...

If
$2/3$
and
$-3$
are the roots of the equation
$\mathbf{px}{}^\text{2}\text{ }+\text{ }\mathbf{7x}\text{ }+\text{ }\mathbf{q}\text{ }=\text{ }\mathbf{0}$
px² + 7x + q = 0, find the values of p and q.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

Let us substitute the given value x = \[2/3\] in the expression, we get \[\begin{array}{*{35}{l}} px{}^\text{2}\text{ }+\text{ }7x\text{ }+\text{ }q\text{ }=\text{ }0 \\ p{{\left( 2/3...

(i) If
$\sqrt{2}$
is a root of the equation
$\mathbf{kx}{}^\text{2}\text{ }+\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}$
, find the value of k. (ii) If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) If \[\sqrt{2}\] is a root of the equation \[\mathbf{kx}{}^\text{2}\text{ }+\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0}\], find the value of k. Let us...

If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution; Now, we have to find out the perimeter of ΔABC Let ΔABC ~ ΔDEF So, AB/DE = AC/DF = BC/EF Consider, AB/DE = AC/DE 4/6 = AC/12 By cross multiplication we get, AC = (4 × 12)/6 AC = 48/6 AC =...

(i) If
$\mathbf{1}/\mathbf{2}~$
is a solution of the equation
$\mathbf{3x}{}^\text{2}\text{ }+\text{ }\mathbf{2kx}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}$
, find the value of k. (ii) If
$2/3$
is a solution of the equation
$\mathbf{7x}{}^\text{2}\text{ }+\text{ }\mathbf{kx}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}$
, find the value of k.

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) If \[\mathbf{1}/\mathbf{2}~\] is a solution of the equation \[\mathbf{3x}{}^\text{2}\text{ }+\text{ }\mathbf{2kx}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}\] Let us substitute the...

It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

From the question it is given that, ΔDEF ~ ΔLMN So, AB/ED = AC/EF = BC/DF Consider AB/ED = AC/EF 5/12 = 7/EF By cross multiplication, EF = (7 × 12)/5 EF = 16.8 cm Now, consider AB/ED = BC/DF 5/12 =...

If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the figure, two line segments are intersecting each other at P. In ΔBCP and ΔDPE 5/10 = 6/12 Dividing LHS and RHS by 2 we get, ½ = ½ Therefore, ΔBCD ~ ΔDEP

It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?

CBSE, Class 10, maths, Maths, ML Aggarwal, Similarity

Solution:- From the question is given that, ∆DEF ~ ∆RPQ ∠D = ∠R and ∠F = ∠Q not ∠P No, ∠F ≠ ∠P

In each of the following, determine whether the given numbers are solutions of the given equation or not: (i)
${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\surd \mathbf{3x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }\surd \mathbf{3},\text{ }-\mathbf{2}\surd \mathbf{3}$
(ii)
${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\surd \mathbf{2x}\text{ }\text{ }\mathbf{4}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }-\surd \mathbf{2},\text{ }\mathbf{2}\surd \mathbf{2}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{3}\surd \mathbf{3x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{x}\text{ }=\text{ }\surd \mathbf{3},\text{...

In each of the following, determine whether the given numbers are roots of the given equations or not; (i)
${{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{2},\text{ }-\mathbf{3}$
(ii)
$\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{13x}\text{ }\text{ }\mathbf{10}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{5},\text{ }-\mathbf{2}/\mathbf{3}$

Class 10, Exercise 1, ICSE, Maths, ML Aggarwal, Quadratic Equations in One Variable

(i) \[{{\mathbf{x}}^{\mathbf{2}}}~\text{ }\mathbf{5x}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{2},\text{ }-\mathbf{3}\] Let us substitute the given values in the...

Two different coins are tossed simultaneously. Find the probability of getting :
(iii) no tail
(iv) at most one tail.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Let E be an event of getting no tails. Favourable outcomes = HH Number of favourable outcomes = 1 P(E) = 1/4 Probability of getting no tails is 1/4 . (iv)Let E be an event of getting atmost one...

Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 tails. Favourable outcomes = TT Number of favourable...

Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When 2 coins are tossed, the possible outcomes are HH. HT, TH, TT. Number of possible outcomes = 4 (i)Let E be an event of getting 2 heads. Favourable outcomes = HH Number of favourable...

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(iii) a black card
(iv) a heart

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting a black card. There will be 23 black cards remaining since 3 spades are removed. Number of favourable outcomes = 23 P(E) = 23/49 Hence the probability of getting...

All the three face cards of spades are removed from a well-shuffled pack of 52 cards. A card is then drawn at random from the remaining pack. Find the probability of getting
(i) a black face card
(ii) a queen

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of cards = 52-3 = 49. [since 3 face cards of spade are removed] So number of possible outcomes = 49. (i) Let E be the event of getting black face card. There will be 3 black...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(xi) Let E be the event of getting a neither a spade nor a jack. There are 13 spades and 3 other jacks. So remaining cards = 52-13-3 = 36 There will be 36 cards which are neither a spade nor a jack....

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(ix) a non-ace
(x) non-face card of black colour

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(ix) Let E be the event of getting a non ace card. There will be 48 non ace cards. Number of favourable outcomes = 48 P(E) = 48/52 = 24/26 = 12/13 Hence the probability of getting a non ace card is...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(vii) a black face card
(viii) a black card

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(vii) Let E be the event of getting a black face card. There will be 6 black face cards. Number of favourable outcomes = 6 P(E) = 6/52 = 3/26 Hence the probability of getting a black face card is...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(v) a king or a queen
(vi) a non-face card

CBSE, Class 10, Exercise 1, Maths, ML Aggarwal, Probability

(v) Let E be the event of getting a king or a queen. There will be 4 cards of king and 4 cards of queen. Number of favourable outcomes = 4+4 = 8 P(E) = 8/52 = 2/13 Hence the probability of getting a...

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. Find the probability that it is
(i) white
(ii) not red.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of balls = 24 Number of red balls = x. Number of white balls = 2x. Number of blue balls = 3x. x+2x+3x = 24 6x = 24 x = 24/6 = 4 Number of red balls = x = 4 Number of white...

A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball is twice that of a red ball, find the number of balls in the bag.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of red balls = 6 Let number of blue balls be x. Total number of balls = 6+x Probability of drawing a red ball = 6/(6+x) Probability of drawing a blue ball = x/(6+x) Given the...

A bag contains 15 balls of which some are white and others are red. If the probability of drawing a red ball is twice that of a white ball, find the number of white balls in the bag.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of balls in the bag = 15. Let the number of white balls be x. Then number of red balls = 15-x. The probability of drawing a white ball = x/15. Probability of drawing a red...

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(iii) a number which is a perfect square
(iv) a prime number less than 30.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the card is a perfect square. Outcomes favourable to E are {4,9,16,25,36,49,64,81,100} Number of favourable outcomes = 9 P(E) = 9/100 Hence the...

Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn at random from this box. Find the probability that the number on the card is
(i) an even number
(ii) a number less than 14

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {2,3,…101} Number of possible outcomes = 100 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,…90} Number of possible outcomes = 90 (i) Let E be the event of getting the number on the disc is a two-digit number. Outcomes favourable to E are...

Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {3,5,7,9..…29} Number of possible outcomes = 14 (i) Let E be the event of getting the number on the ticket is a prime number. Outcomes favourable to E are...

Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(i) divisible by 5
(ii) a perfect square number.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {13,14,15,…60} Number of possible outcomes = 48 (i) Let E be the event of getting the number on the card is divisible by 5. Outcomes favourable to E are...

.A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random from the box. Find the probability that the number on the ball is :
(iii) neither divisible by 5 nor by 10
(iv) an even number.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the ball is neither divisible by 5 nor by 10. Outcomes favourable to E are {1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19} Number of favourable outcomes =...

.A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn at random from the box. Find the probability that the number on the ball is :
(i) a prime number
(ii) divisible by 3 or 5

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4…19} Number of possible outcomes = 19 (i) Let E be the event of getting the number on the ball is a prime number. Outcomes favourable to E are...

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(iii) divisible by 3
(iv) divisible by 3 and 2 both

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Let E be the event of getting the number on the card is divisible by 3. Outcomes favourable to E are {3,6,9,12,15} Number of favourable outcomes = 5 P(E) = 5/15 = 1/3 Hence the probability of...

A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed thoroughly. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) Odd
(ii) prime

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4…15} Number of possible outcomes = 15 (i) Let E be the event of getting the number on the card is odd. Outcomes favourable to E are {1,3,5,7,9,11,13,15}...

A box contains 25 cards numbered 1 to 25. A card is drawn from the box at random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,4 ….25} Number of possible outcomes = 25 (i) Let E be the event of getting the number on the card is an even number. Outcomes favourable to E are...

Cards marked with numbers 1, 2, 3, 4,…20 are well shuffled and a card is drawn at random. What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {1,2,3,….20} Number of possible outcomes = 20 (i) Let E be the event of getting the number on the card is a prime number. Outcomes favourable to E are...

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of integers between 0 and 100 = 99 Number of possible outcomes = 99 (i) Let E be the event of getting an integer divisible by 7. Outcomes favourable to E are...

Sixteen cards are labeled as a, b, c,…, m, n, o, p. They are put in a box and shuffled. A boy is asked to draw a card from the box. What is the probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes are {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p} Number of possible outcomes = 16 (i) Let E be the event of getting a vowel. Outcomes favourable to E are { a,e,i,o} Number of...

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the adjoining figure) and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: The possible outcomes of the game are {1,2,3,4,5,6,7,8} Number of possible outcomes = 8 (i) Let E be the event of arrow pointing 8. Outcomes favourable to E is 8. Number of favourable...

A die has 6 faces marked by the given numbers as shown below: The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When a die is thrown, the possible outcomes are {1,2,3,-1,-2,-3} Number of possible outcomes = 6 (i) Let E be the event of getting a positive integer. Outcomes favourable to E are {1,2,3}...

In a single throw of a die, find the probability of getting:
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(vii) Let E be the event of getting a number between 3 and 6. Outcomes favourable to E is 4,5. Number of favourable outcomes = 2 P(E) = 2/6 = 1/3 Hence the probability of getting a number between 3...

In a single throw of a die, find the probability of getting:
(v) a number less than 8
(vi) a number divisible by 3

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(v) Let E be the event of getting a number less than 8. Outcomes favourable to E is 1,2,3,4,5,6. Number of favourable outcomes = 6 P(E) = 6/6 = 1 Hence the probability of getting a number less than...

A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: When a die is thrown, the possible outcomes are 1,2,3,4,5,6. So Sample space = { 1,2,3,4,5,6} Number of possible outcomes = 6 Even numbers are (2,4,6). Number of favourable outcomes = 3...

A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Peter, a trader, will only accept the shirts which are good, but Salim, another trader, will only reject the shirts which have major defects. One shirts is drawn at random from the carton. What is the probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of shirts = 100 Number of good shirts = 88 Number of shirts with minor defects = 8 Number of shirts with major defects = 4 Peter accepts only good shirts. So number of shirts...

A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2 coins and ten Rs 5 coins. It is equally likely that one of the coins will fall down when the bank is turned upside down, what is the probability that the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of 50 paisa coins = 100 Number of 1 rupee coins = 50 Number of 2 rupee coins = 20 Number of 5 rupee coins = 10 Total number of coins = 100+50+20+10 = 180 (i) Probability of getting...

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(iii) not green
(iv) neither white nor black.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii)Probability of not green = Probability of getting red, white and black = (6+8+3)/22 = 17/22 Hence the probability of not green is 17/22. (iv) Probability of neither white nor black =...

A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls. One ball is drawn at random from the bag. Find the probability that the ball is :
(i) white
(ii) red or black

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of red balls = 6 Number of white balls = 8 Number of green balls = 5 Number of black balls = 3 Total number of marbles = 6+8+5+3 = 22 (i)Probability of white balls, = 8/22 = 4/11...

A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at random from the box, what is the probability that it will be
(iii) not black?
(iv) green?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

(iii) Probability of not black = Probability of white and blue = (7+8)/20 = 15/20 = 3/4 Hence the probability of not black is 3/4. (iv) Since there are no green marbles in the box, the probability...

A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of vowels in the word ‘TRIANGLE’ = 3 Total number of letters = 8 Probability that the letter chosen is a vowel , P(E) = 3/8 Hence the probability that the letter chosen is a vowel...

There are 40 students in Class X of a school of which 25 are girls and the others are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of students = 40 Number of girls = 25 Number of boys = 40-25 = 15 (i) Probability of getting a girl, P(E) = 25/40 = 5/8 Hence the probability of getting a girl is 5/8. (ii)...

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from a bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: (i) Number of red balls = 3 Number of black balls = 5 Total number of balls = 3+5 = 8 Probability that the ball drawn is red , P(E) = 3/8 Hence the probability that the ball drawn is red...

Two players, Sania and Sonali play a tennis match. It is known that the probability of Sania winning the match is 0.69. What is the probability of Sonali winning ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Probability of Sania winning the match, P(E) = 0.69 Probability of Sonali winning = Probability of Sania losing, = 1-0.69 = 0.31 Hence the probability of Sonali winning is...

If the probability of winning a game is 5/11, what is the probability of losing ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Given probability of winning the game, P(E) = 5/11 We know that, Probability of losing game, = 1-5/11 = (11-5)/11 = 6/11 Hence the probability of losing game is 6/11.

In a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Number of prized tickets = 5 Number of blank tickets = 995 Total number of tickets = 5+995 = 1000 The probability of winning a prize, P(E) P(E) = 5/1000 = 1/200 Hence the required...

A box contains 600 screws, one-tenth are rusted. One screw is taken out at random from this box. Find the probability that it is a good screw.

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Total number of screws = 600 Number of possible outcomes = 600 Number of rusted screws = one tenth of 600 = (1/10)×600 = 60 Number of remaining good screws = 600-60 = 540 Number of...

A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Anjali takes out a ball from the bag without looking into it. What is the probability that she takes out
(i) yellow ball ?
(ii) red ball ?
(iii) blue ball ?

CBSE, Class 10, Exercise 1, maths, Maths, ML Aggarwal, Probability

Solution: Anjali takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them. Let Y be the event ‘the ball taken out is yellow’, B be the event...

A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :
(i) the total surface area of the can in contact with water when the sphere is in it.
(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

(i)Given radius of the cylinder, r = 3.5 cm Diameter of the sphere = height of the cylinder = 3.5×2 = 7 cm So radius of sphere, r = 7/2 = 3.5 cm Height of cylinder, h = 7 cm Total surface area of...

Water is flowing at the rate of 15 km/h through a pipe of diameter 14 cm into a cuboid pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given speed of water flow = 15 km/h Diameter of pipe = 14 cm So radius of pipe, r = 14/2 = 7 cm = 0.07 m Dimensions of cuboidal pond = 50 m × 44 m Height of water in pond = 21 cm = 0.21 m...

The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate (i) the radius of the solid sphere. (ii) the number of cones recast. (Use π = 3.14).

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Given surface area of the solid metallic sphere = 1256 cm2 4R2 = 1256 4×3.14×R2 = 1256 R2 = 1256/4×3.14 R2 = 100 R = 10 Hence the radius of solid sphere is 10 cm. (ii)Volume of the...

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 616 cm2 4R2 = 616 4×(22/7)R2 = 616 R2 = 616×7/4×22 R2 = 49 R = 7 Volume of the solid metallic sphere = (4/3)R3 = (4/3)×73 = (1372/3) cm3 Diameter of...

A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water upto the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cone, h = 11 cm Radius of the cone, r = 2.5 cm Volume of the cone = (1/3)r2h = (1/3)×2.52×11 = (11/3)×6.25 cm3 When lead shots are dropped into vessel, (2/5) of water...

Solve the inequation and represent the solution set on the number line. $-3+x\le 8x/3+2\le 14/3+2x,x\in I$

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question,, Therefore, we get \[-3\text{ }+\text{ }x\text{ }\le \text{ }8x/3\text{ }+\text{ }2\] and \[8x/3\text{ }+\text{ }2\text{ }\le \text{ }14/3\text{ }+\text{ }2x\]...

A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of metallic cones, r = 2 cm Height of cone, h = 3 cm Volume of cone = (1/3)r2h = (1/3)×22×3 = 4 cm3 Radius of the solid sphere, R = 6 cm Volume of the solid sphere = (4/3)R3 =...

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the metallic sphere, R = 10.5 cm Volume of the sphere = (4/3)R3 = (4/3)×10.53 = 1543.5 cm3 Radius of cone, r = 3.5 cm Height of the cone, h = 3 cm Volume of the cone =...

Solve the given inequation, and graph the solution on the number line: 2y – 3 < y + 1 ≤ 4y + 7; y ∈ R.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[2y\text{ }\text{ }3\text{ }<\text{ }y\text{ }+\text{ }1\text{ }\le \text{ }4y\text{ }+\text{ }7\] Therefore, we get \[2y\text{ }\text{ }3\text{ }<\text{...

Solve the following inequation and represent the solution set on the number line:4x-19<3x

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question,, Therefore, we get \[4x\text{ }\text{ }19\text{ }<\text{ }3x/5\text{ }\text{ }2\] and \[3x/5\text{ }\text{ }2\text{ }\le \text{ }-2/5\text{ }+\text{ }x\] \[4x\text{...

A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the metal cylinder, r = 14 cm Height of the metal cylinder, h = 21 cm Radius of the sphere, R = 3.5 cm Volume of the metal cylinder = r2h = (22/7)×142×21 = 22×2×14×21 =...

Find the number of metallic circular discs with 1.5 cm base diameter and height 0.2 cm to be melted to form a circular cylinder of height 10 cm and diameter 4.5 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the circular cylinder, h = 10 cm Diameter of circular cylinder = 4.5 cm So radius, r = 4.5/2 = 2.25 cm Volume of circular cylinder = r2h = ×2.252×10 = 50.625 cm3 Base...

Given 20 – 5 x < 5 (x + 8), find the smallest value of x, when (i) x ∈ I (ii) x ∈ W (iii) x ∈ N.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[20\text{ }\text{ }5\text{ }x\text{ }<\text{ }5\text{ }\left( x\text{ }+\text{ }8 \right)\] \[20\text{ }\text{ }5x\text{ }<\text{ }5x\text{ }+\text{ }40\] \[-5x\text{...

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm?

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Edge of the cube, a = 44 cm Volume of cube = a3 = 443 = 85184 cm3 Diameter of shot = 4 cm So radius of shot, r = 4/2 = 2 cm Volume of a shot = (4/3)r3 = (4/3)×(22/7)×23 = 704/21 cm3 Number...

If x ∈ I, find the smallest value of x which satisfies the inequation 2x+5/2>5x/3+2

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[\left( 4x\text{ }+\text{ }5 \right)/2\text{ }>\text{ }\left( 5x\text{ }+\text{ }6 \right)/3\] [Taking L.C.M] \[3\text{ }\left( 4x\text{ }+\text{ }5 \right)\text{...

How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the cuboidal solid = 9 cm× 11 cm× 12 cm Volume of the cuboidal solid = 9×11×12 = 1188 cm3 Diameter of shot = 3 cm So radius of shot, r = 3/2 = 1.5 cm Volume of shot =...

Given: P {x : 5 < 2x – 1 ≤ 11, x ∈ R} Q {x : – 1 ≤ 3 + 4x < 23, x ∈ I} where R = (real numbers), I = (integers) Represent P and Q on number line. Write down the elements of P ∩ Q.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to ques, \[~P\text{ }\{x\text{ }:\text{ }5\text{ }<\text{ }2x\text{ }\text{ }1\text{ }\le \text{ }11,\text{ }x\in R\}\] And \[~Q\text{ }\{x\text{ }:\text{ }\text{ }1\text{ }\le \text{...

A = {x : 11x – 5 > 7x + 3, x ∈ R} and B = {x : 18x – 9 ≥ 15 + 12x, x ∈ R} Find the range of set A ∩ B and represent it on a number line

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to ques, \[~A\text{ }=\text{ }\{x\text{ }:\text{ }11x\text{ }\text{ }5\text{ }>\text{ }7x\text{ }+\text{ }3,\text{ }x\in R\}\] and \[B\text{ }=\text{ }\{x\text{ }:\text{ }18x\text{...

A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Radius of the solid circular cylinder, r = 14 cm Height, h = 12 cm Volume of the cylinder = r2h = ×142×12 = ×196×12 = 2352 = 2352×22/7 = 7392 cm3 Edge of the cube, a = 2 cm Volume of cube...

If P is the solution set of -3x + 4 < 2x – 3, x ∈ N and Q is the solution set of 4x – 5 < 12, x ∈ W, find (i) P ∩ Q (ii) Q – P.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question,s, \[-3x\text{ }+\text{ }4\text{ }<\text{ }2x\text{ }\text{ }3\] where x ∈ N and \[4x\text{ }\text{ }5\text{ }<\text{ }12\] where x ∈ W Therefore, solving...

If x ∈ I, A is the solution set of 2 (x – 1) < 3x – 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩ B.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question,s, \[2\text{ }\left( x\text{ }\text{ }1 \right)\text{ }<\text{ }3x\text{ }\text{ }1\] and \[4x\text{ }\text{ }3\text{ }\le \text{ }8\text{ }+\text{ }x\text{ }for\text{ }x\in...

There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given radius of the glass jar, R = 8 cm Diameter of the sphere = 12 cm Radius of the sphere, r = 12/2 = 6 cm When the sphere is removed from the jar, volume of water decreases. Let h be...

Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line.

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[2x\text{ }\text{ }5\text{ }\le \text{ }5x\text{ }+\text{ }4\text{ }<\text{ }11\] Therefore, we get \[2x\text{ }\text{ }5\text{ }\le \text{ }5x\text{ }+\text{ }4\text{...

A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given internal diameter of cylindrical can = 21 cm Radius of the cylindrical can, R = 21/2 cm Diameter of sphere = 10.5 cm Radius of the sphere, r = 10.5/2 = 21/4 cm Let the rise in water...

Solve the following inequation, write down the solution set and represent it on the real number line: -2 + 10x ≤ 13x + 10 ≤ 24 + 10x, x ∈ Z

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[-2\text{ }+\text{ }10x\text{ }\le \text{ }13x\text{ }+\text{ }10\text{ }\le \text{ }24\text{ }+\text{ }10x\] Therefore, we get \[-2\text{ }+\text{ }10x\text{...

Solving the following inequation, write the solution set and represent it on the number line -3(x – 7) ≥ 15 – 7x >(x+1)/3 , x ∈ R

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

According to question, \[3\left( x\text{ }\text{ }7 \right)\text{ }\ge \text{ }15\text{ }\text{ }7x\text{ }>\] \[3x\text{ }+\text{ }21\text{ }\ge \text{ }15\text{ }\text{ }7x\text{...

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone. (2002)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of hollow sphere = 4 cm Internal radius, r = 4/2 = 2 cm External diameter = 8 cm External radius, R = 8/2 = 4 cm Volume of the hollow sphere, V = (4/3)(R3-r3) V =...

Solve the following inequation and represent the solution set on the number line: $-3<-1/2-2x/3\le 5/6,x\in R$

Class 10, Exercise 1, ICSE, Linear Inequations, Maths, ML Aggarwal

Given in equation, \[-3\text{ }<\text{ }-\left( 3\text{ }+\text{ }4x \right)/6\text{ }\le \text{ }5/6\] [Taking L.C.M] \[-18\text{ }<\text{ }-3\text{ }\text{ }4x\text{ }\le \text{ }5\]...

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal of the tube is ½ cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone correct to one decimal place.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal radius of the tube, r = 3 cm Thickness of the tube = ½ cm = 0.5 cm External radius of tube = 3+0.5 = 3.5 cm Height of the tube, h = 21 cm Volume of the tube = (R2-r2)h =...

A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm and height is 72 cm, find the uniform thickness of the cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the sphere, r = 6 cm Volume of the sphere = (4/3)r3 = (4/3)×63 = 288 cm3 Let r be the internal radius of the hollow cylinder. External radius of the hollow cylinder, R = 4...

A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given inner diameter of the pipe = 6 cm So inner radius, r = 6/2 = 3 cm Outer diameter = 10 cm Outer radius, R = 10/2 = 5 cm Let h be the height of the pipe. Volume of pipe = (R2-r2)h =...

Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m1 = 1 kg Mass of the bigger sphere, m2 = 7 kg The spheres are melted to form a new sphere. So...

A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 2.5 mm = 0.25 cm Radius of the cylinder, r = 12 cm Volume of the cylinder = r2h = ×122×0.25 = ×144×0.25 = 36 cm3 Let R be the radius of the sphere. Volume...

Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of each sphere, r = 2 cm Volume of a sphere = (4/3)r3 = (4/3)×23 = (4/3)×8 = (32/3) cm3 Volume of 8 spheres = 8×(32/3) = (256/3) cm3 Let R be radius of new sphere. Volume of...

The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 9 cm Diameter of the cylinder = 40 cm Radius of the cylinder, r = 40/2= 20 cm Volume of the cylinder = r2h = ×202×9 = ×400×9 = 3600 cm3 Height of the...

A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the cylindrical vessel = 22 m × 20 m Let the rainfall be x cm. Volume of water = (22×20×x)m3 Diameter of the cylindrical base = 2 m So radius of the cylindrical base =...

A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemisphere, r = 8 cm Volume of the hemisphere, V = (2/3)r3 = (2/3)×83 = (1024/3) cm3 Radius of cone, R = 6 cm Since hemisphere is melted and recasted into a cone, the...

The radius of a sphere is 9 cm. It is melted and drawn into a wire of diameter 2 mm. Find the length of the wire in metres.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Radius of the sphere, r = 9 cm Volume of the sphere, V = (4/3)r3 = (4/3)××93 = 12××81 = 972 cm3 Diameter of the wire = 2 mm So radius of the wire = 2/2 = 1 mm = 0.1 cm Since the sphere is...

The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the metallic sphere = 6 cm Radius of the sphere, r = 6/2 = 3 cm Volume of the sphere, V = (4/3)r3 = (4/3)××33 = 4××9 = 36 cm3 Length of the wire, h = 36 m = 3600 cm Since...

The adjoining figure shows a model of a solid consisting of a cylinder surmounted by a hemisphere at one end. If the model is drawn to a scale of 1 : 200, find
(i) the total surface area of the solid in π m².
(ii) the volume of the solid in π litres.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 8 cm Radius of the cylinder, r = 3 cm Radius of hemisphere , r = 3 cm Scale = 1:200 Hence actual radius, r = 200×3 = 600 Actual height, h = 200×8 = 1600...

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given height of the cylinder, H = 13 cm Radius of the cylinder, r = 5 cm Radius of the hemisphere, r = 5 cm Height of the cone, h = 12 cm Radius of the cone, r = 5 cm Slant height of the cone, l =...

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. Their common diameter is 3.5 cm and the height of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the volume of the solid. (Take π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given height of the cylinder, H = 10 cm Height of the cone, h = 6 cm Common diameter = 3.5 cm Common radius, r = 3.5/2 = 1.75 cm Volume of the solid = Volume of the cone + Volume of the...

The adjoining figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and the cone are each of 4 cm. Find the volume of the solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given common radius, r = 7 cm Height of the cone, h = 4 cm Height of the cylinder, H = 4 cm Volume of the solid = Volume of the cone + Volume of the cylinder + Volume of the hemisphere =...

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and the height of the cylinder are 6 cm and 12 cm respectively. If the slant height of the conical portion is 5 cm, find the total surface area and the volume of the rocket. (Use π = 3.14).

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given diameter of the cylinder = 6 cm Radius of the cylinder, r = 6/2 = 3 cm Height of the cylinder, H = 12 cm Slant height of the cone, l = 5 cm Radius of the cone, r = 3 cm Height of the...

A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let the radius of the dome be r. Internal diameter = 2r Given internal diameter is equal to total height. Total height of the building = 2r Height of the hemispherical area = r So height...

A circular hall (big room) has a hemispherical roof. The greatest height is equal to the inner diameter, find the area of the floor, given that the capacity of the hall is 48510 m³.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Let the radius of the hemisphere be r. Inner diameter = 2r Given greatest height equal to inner diameter. So total height of the hall = 2r Height of the hemispherical part = r Height of cylindrical...

A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is 2/3 of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm = 7/2 cm Volume of hemisphere = (2/3)r3 = (2/3)×(22/7)×(7/2)3 = (2/3)×(22/7)×(7/2)×(7/2)×(7/2) = (22/3)×(7/2)×(7/2) =...

The adjoining figure shows a hemisphere of radius 5 cm surmounted by a right circular cone of base radius 5 cm. Find the volume of the solid if the height of the cone is 7 cm. Give your answer correct to two places of decimal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemisphere, r = 5 cm Radius of cone, r = 5 cm Height of the cone, h = 7 cm Volume of the solid = Volume of the hemisphere + Volume of the cone = (2/3)r3 + (1/3)r2h =...

The adjoining figure shows a wooden toy rocket which is in the shape of a circular cone mounted on a circular cylinder. The total height of the rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted green and the cylindrical portion red, find the area of the rocket painted with each of these colours. Also, find the volume of the wood in the rocket. Use π = 3.14 and give answers correct to 2 decimal places.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; (i) Given height of the rocket = 26 cm Height of the cone, H = 6 cm Height of the cylinder, h = 26-6 = 20 cm Diameter of the cone = 5 cm Radius of the cone, R = 5/2 = 2.5 cm Diameter of...

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and of base radius 7 cm is drilled out. Find the volume and the total surface of the remaining solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution; Given height of the cylinder, H = 30 cm Radius of the cylinder, r = 7 cm Height of cone, h = 24 cm Radius of cone, r = 7 cm Slant height of the cone, l = √(h2+r2) l = √(242+72) l =...

An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50 m. If the diameter of the base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds and stitching. Give your answer to the nearest m².

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

solution; Given height of the tent above the ground = 85 m Height of the cylindrical part, H = 50 m height of the cone, h = 85-50 h = 35 m Diameter of the base, d = 168 m Radius of the base of...

A circus tent is in the shape of a cylinder surmounted by a cone. The diameter of the cylindrical portion is 24 m and its height is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas used to make the tent.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given diameter of the cylindrical part of tent, d = 24 m Radius, r = d/2 = 24/2 = 12 m Height of the cylindrical part, H = 11 m Since vertex of cone is 16 m above the ground, height of cone, h =...

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. If the total height of the toy is 15.5 cm, find the total surface area of the toy.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given radius of the cone, r = 3.5 cm Radius of hemisphere, r = 3.5 cm Total height of the toy = 15.5 cm Height of the cone = 15.5 – 3.5 = 12 cm Slant height of the cone, l = √(h2+r2) l = √(122+3.52)...

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder (as shown in the given figure). If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cylinder, h = 10 cm Radius of the cylinder, r = 3.5 cm Radius of the hemisphere = 3.5 cm Total surface area of the article = curved surface area of the cylinder +...

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given edge of the cube, a = 7 cm Diameter of the hemisphere, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Surface area of the hemisphere = 2r2 = 2×(22/7)×3.52 = 44×12.25/7 = 539/7 = 77 cm2 Surface area...

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of the wood in the entire stand, correct to 2 decimal places.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Dimensions of the cuboid = 15 cm× 10 cm × 3.5 cm Volume of the cuboid = 15×10×3.5 = 525 cm3 Radius of each depression, r = 0.5 cm Depth, h = 1.4 cm Volume of conical depression = (1/3)r2h...

16 glass spheres each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm x 8 cm x 8 cm and then the box is filled with water. Find the volume of the water filled in the box.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given dimensions of the box = 16 cm ×8 cm ×8 cm So volume of the box = lbh = 16×8×8 = 1024 cm3 Radius of the glass sphere, r = 2 cm Volume of the sphere = (4/3)r3 = (4/3)×(22/7)×23 =...

A cone of maximum volume is curved out of a block of wood of size 20 cm x 10 cm x 10 cm. Find the volume of the remaining wood.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Given dimensions of the block of wood = 20 cm × 10 cm× 10 cm Volume of the block of wood = 20×10×10 = 2000 cm3 [Volume = lbh] Diameter of the cone, d = 10 cm Radius of the cone , r = d/2 = 10/2 = 5...

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the remaining material.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Given edge of the cube, a = 14 cm Radius of the cone, r = 14/2 = 7 cm Height of the cone, h = 14 cm Volume of the cube = a3 = 143 = 14×14×14 = 2744 cm3 Volume of the cone = (1/3)r2h =...

The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs Rs 0.75, find the cost of the glass for making the trophy

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given side of the cube, a = 28 cm Radius of the cylinder, r = 28/2 = 14 cm Height of the cylinder, h = 28 cm Volume of the cube = a3 = 283 = 28×28×28 = 21952 cm3 Volume of the cylinder =...

Write whether the following statements are true or false. Justify your answer :
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is 2r equals the volume of a hemisphere of radius r.
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is 1 : 2 : 3.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Let the radius of sphere be r. Then height of the cylinder, h = 2r Radius of cylinder = r Volume of cylinder = r2h = ×r2×2r = 2r3 Volume of sphere = (4/3)r3 = (2/3)× 2r3 = (2/3)× Volume...

The surface area of a solid sphere is 1256 cm². It is cut into two hemispheres. Find the total surface area and the volume of a hemisphere. Take π = 3.14.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 1256 cm2 4r2 = 1256 4×3.14×r2 = 1256 r2 = 1256×/3.14×4 r2 = 100 r = 10 cm Total surface area of the hemisphere = 3r2 = 3×3.14×102 = 3×3.14×100 = 942 cm2...

The water for a factory is stored in a hemispherical tank whose internal diameter is 14 m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its capacity. Find the volume of water pumped into the tank.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of the hemispherical tank, d = 14 m So radius, r = 14/2 = 7 m Volume of the tank = (2/3)r3 = (2/3)×(22/7)×(7)3 = 718.667 m3 = 718.67 m3 (approx) = 718.67 kilolitre...

A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the hemispherical bowl, r = 3.5 cm = 7/2 cm Volume of the hemisphere = (2/3)r3 = (2/3)×(22/7)×(7/2)3 = 11×49/6 = 539/6 Hence the volume of the hemispherical bowl is...

Find the volume of a sphere whose surface area is 154 cm².

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 154 cm2 4r2 = 154 4×(22/7)×r2 = 154 r2 = (154×7)/(4×22) r2 = 49/4 r = 7/2 Volume of the sphere = (4/3)r3 = (4/3)×(22/7)×(7/2)3 = 539/3 = 179.666 = 179.67...

A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given side of the cube, a = 4 cm Volume of the cube = a3 = 43 = 4×4×4 = 64 cm3 Diameter of the sphere = 4 cm So radius of the sphere, r = d/2 = 4/2 = 2 cm Volume of the sphere = (4/3)r3 =...

(a) If the ratio of the radii of two sphere is 3 : 7, find :
(i) the ratio of their volumes.
(ii) the ratio of their surface areas.
(b) If the ratio of the volumes of the two sphere is 125 : 64, find the ratio of their surface areas.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Let the radii of two spheres be r1 and r2. Given ratio of their radii = 3:7 Volume of sphere = (4/3)r3 Ratio of the volumes = (4/3)r13/(4/3)r23 = r13/ r23 = 33/73 = 27/343 Hence the...

A sphere and a cube have the same surface area. Show that the ratio of the volume of the sphere to that of the cube is √6 :√π

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let r be the radius of the sphere and a be the side of the cube. Surface area of sphere = 4r2 Surface area of cube = 6a2 Given sphere and cube has same surface area. 4r2 = 6a2 r2/a2 = 6/4...

The radius of a spherical balloon increases from 7 cm to 14 cm as air is jumped into it. Find the ratio of the surface areas of the balloon in two cases.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius of the spherical balloon, r = 7 cm Radius of the spherical balloon after air is pumped, R = 14 cm Surface area of the sphere = 4r2 Ratio of surface areas of the balloons =...

A hemispherical brass bowl has inner- diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given inner diameter of the brass bowl, d = 10.5 cm Radius, r = d/2 = 10.5/2 = 5.25 cm Curved surface area of the bowl = 2r2 = 2×(22/7)×5.252 = 173.25 cm2 Rate of tin plating = Rs.16 per...

Find:
(i) the curved surface area.
(ii) the total surface area of a hemisphere of radius 21 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the hemisphere, r = 21 cm Curved surface area of the hemisphere = 2r2 = 2×(22/7)×212 = 2×22×3×21 = 2772 cm2 Hence the curved surface area of the hemisphere is 2772 cm2....

Find the diameter of a sphere whose surface area is 154 cm2 .

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given surface area of the sphere = 154 cm2 Surface area of the sphere = 4r2 4×(22/7)×r2 = 154 r2 = 154 ×7/(22×4) = 49/4 r = √49/2 Diameter = 2×r = 2×√49/2 = √49 = 7 Hence the diameter of...

Find the surface area of a sphere of diameter:
(i) 21 cm
(ii) 3.5 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given diameter of the sphere, d = 21 cm Radius, r = d/2 = 21/2 = 10.5 Surface area of the sphere = 4r2 = 4×(22/7)×10.52 = 1386 cm2 Hence the surface area of the sphere is 1386 cm2....

Find the volume of a sphere of radius :
(i) 0.63 m
(ii) 11.2 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the sphere, r = 0.63 m Volume of the sphere, V = (4/3)r3 = (4/3)×(22/7)×0.633 = 1.047 m3 = 1.05 m3 (approx) Hence the volume of the sphere is 1.05 m3. (ii) Given radius...

Find the surface area of a sphere of radius : (i) 14 cm (ii) 10.5 cm

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: (i) Given radius of the sphere, r = 14 cm Surface area of the sphere = 4r2 = 4×(22/7)×142 = 4×22×14×2 = 2464 cm3 Hence the surface area of the sphere is 2464 cm2. (ii) Given radius of the...

A semi-circular lamina of radius 35 cm is folded so that the two bounding radii are joined together to form a cone. Find
(i) the radius of the cone.
(ii) the (lateral) surface area of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

(i)Given radius of the semi circular lamina, r = 35 cm A cone is formed by folding it. So the slant height of the cone, l = 35 cm Let r1 be radius of cone. Semicircular perimeter of lamina becomes...

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

So the height of the resulting cone, h = 8 cm Radius, r = 6 cm Slant height, l = 10 cm Volume of the cone, V = (1/3)r2h V = (1/3)×3.14×62×8 V = (1/3)×3.14×36×8 V = 3.14×12×8 V = 301.44 cm3 Hence the...

The volume of a right circular cone is 9856 cm3 and the area of its base is 616 cm2 . Find
(i) the slant height of the cone.
(ii) total surface area of the cone.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given base area of the cone = 616 cm2 r2 = 616 (22/7)×r2 = 616 r2 = 616×7/22 r2 = 196 r = 14 Given volume of the cone = 9856 cm3 (1/3)r2h = 9856 (1/3)×(22/7)×142 ×h = 9856 h =...

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given perimeter of the base of a cone = 44 cm 2r = 44 2×22/7×r = 44 r = 44×7/(2×22) r = 7 cm Slant height, l = 25 height, h = √(l2-r2) h = √(252-72) h = √(625-49) h = √576 h = 24 cm Volume...

The perimeter of the base of a cone is 44 cm and the slant height is 25 cm. Find the volume and the curved surface of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Find what length of canvas 2 m in width is required to make a conical tent 20 m in diameter and 42 m in slant height allowing 10% for folds and the stitching. Also find the cost of the canvas at the rate of Rs 80 per metre.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the conical tent, d = 20 m radius, r = d/2 = 20/2 = 10 m Slant height, l = 42 m Curved surface area of the conical tent = rl = (22/7)×10×42 = 22×10×6 = 1320 m2 So the...

(a) The ratio of the base radii of two right circular cones of the same height is 3 : 4. Find the ratio of their volumes.
(b) The ratio of the heights of two right circular cones is 5 : 2 and that of their base radii is 2 : 5. Find the ratio of their volumes.
(c) The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone. Find: (i) the ratio of their volumes.
(ii) the ratio of their lateral surface areas.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (a) Let r1 and r2 be the radius of the given cones and h be their height. Ratio of radii, r1:r2 = 3:4 Volume of cone, V1 = (1/3)r12h Volume of cone, V2 = (1/3)r22h V1 /V1 = (1/3)r12h/...

A Jocker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the cloth required to make 10 such caps.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the cone, h = 24 cm Radius, r = 7 cm We know that l2 = h2+r2 l2 = 242+72 l2 = 576+49 l2 = 625 l = √625 l = 25 Curved surface area = rl = (22/7)×7×25 = 22×25 = 550 cm2 So...

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given slant height of conical tomb, l = 25 m Base diameter, d = 14 m So radius, r = 14/2 = 7 m Curved surface area = rl = (22/7)×7×25 = 550 m2 Hence the curved surface area of the cone is...

The height of a cone is 15 cm. If its volume is 1570 cm2 , find the radius of the base. (Use π = 3.14)

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of a cone, h = 15 cm Volume of the cone = 1570 cm3 (1/3)r2h = 1570 (1/3)3.14 ×r2×15 = 1570 5 ×3.14×r2 = 1570 r2 = 1570/5×3.14 = 314/3.14 = 100 r = 10 Hence the radius of the...

If the volume of a right circular cone of height 9 cm is 48π cm3 , find the diameter of its base.

CBSE, Class 10, Exercise 1, Maths, Mensuration, ML Aggarwal

Solution: Given height of a cone, h = 9 cm Volume of the cone = 48 (1/3)r2h = 48 (1/3)r2×9 = 48 3r2 = 48 r2 = 48/3 = 16 r = 4 So diameter = 2×radius = 2×4 = 8 cm Hence the diameter of the cone is 8...

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters ?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter, d = 3.5 m So radius, r = 3.5/2 = 1.75 Depth, h = 12 m Volume of the cone = (1/3)r2h =(1/3)×(22/7)×1.752×12 = (22/7)× 1.752×4 = 38.5 m3 = 38.5 kilolitres [1 kilolitre = 1m3]...

Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given radius, r = 7 cm Slant height, l = 25 cm We know that l2 = h2+r2 Height of the conical vessel, h = √(l2-r2) = √(252-72) = √(625-49) = √576 = 24 cm Volume of the cone = (1/3)r2h...

Find the volume of the right circular cone with
(i) radius 6 cm and height 7 cm
(ii) radius 3.5 cm and height 12 cm.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Given radius, r = 6 cm Height, h = 7 cm Volume of the cone = (1/3)r2h =(1/3)×(22/7)×62×7 = 22×12 = 264 cm3 Hence the volume of the cone is 264 cm3. (ii) Given radius, r = 3.5 cm Height,...

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find , (i)radius of the base (ii)total surface area of the cone.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Given curved surface area of the cone = 308 cm2 Slant height of the cone, l = 14 cm rl = 308 (22/7)×r×14 = 308 r = 308×7/(22×14) = 7 Hence the radius of the cone is 7 cm. (ii)Total...

Diameter of the base of a cone is 10.5 cm and slant height is 10 cm. Find its curved surface area.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given diameter of the cone = 10.5 cm Radius, r = d/2 = 10.5/2 = 5.25 cm Slant height of the cone, l = 10 cm Curved surface area of the cone = rl = (22/7)×5.25×10 = 165 cm2 Hence the curved...

Write whether the following statements are true or false. Justify your answer.
(i) If the radius of a right circular cone is halved and its height is doubled, the volume will remain unchanged.
(ii) A cylinder and a right circular cone are having the same base radius and same height. The volume of the cylinder is three times the volume of the cone.
(iii) In a right circular cone, height, radius and slant height are always the sides of a right triangle.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i)Volume of cone = (1/3)r2h If radius is halved and height is doubled, then volume = (1/3)(r/2)2 2h = (1/3)r2h/2 If the radius of a right circular cone is halved and its height is...

A soft drink is available in two packs
(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm
(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Length of the can, l = 5 cm Width, b = 4 cm Height, h = 15 cm Volume of the can = lbh = 5×4×15 = 300 cm3 (ii) Diameter of the cylinder, d = 7 cm Radius, r = d/2 = 7/2 = 3.5 cm Height,...

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given length of the pencil, h = 14 cm Diameter of the pencil = 7 mm radius, R = 7/2 mm = 7/20 cm Diameter of the graphite = 1 mm Radius of graphite, r = ½ mm = 1/20cm Volume of graphite =...

The given figure shows a metal pipe 77 cm long. The inner diameter of a cross-section is 4 cm and the outer one is 4.4 cm. Find its
(i) inner curved surface area
(ii) outer curved surface area
(iii) total surface area.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given height of the metal pipe = 77 cm Inner diameter = 4 cm Inner radius, r = d/2 = 4/2 = 2 cm Outer diameter = 4.4 cm Outer radius, R = d/2 = 4.4/2 = 2.2 cm (i)Inner curved surface area...

A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 11.2 cm and its length is 21 cm. The metal thickness is 0.4 cm. Calculate the volume of the metal.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given internal diameter of the tube = 11.2 cm Internal radius, r = d/2 = 11.2/2 = 5.6 cm Length of the tube, h = 21 cm Thickness = 0.4 cm Outer radius, R= 5.6+0.4 = 6 cm Volume of the...

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let r1 and r2 be the radius of the two cylinders and h1 and h2 be their heights. Given ratio of the diameter = 3:4 Then the ratio of radius r1:r2 = 3:4 Given volume of both jars are same....

The ratio between the curved surface and the total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm2.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Given the ratio of curved surface area and the total surface area = 1:2 Total surface area = 616 cm2 Curved surface area = 616/2 = 308 cm2 2rh = 308 rh = 308/2 = 308×7/2×22 rh = 49 …(i)...

(i) The sum of the radius and the height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm2 . Find the height and the volume of the cylinder.
(ii) The total surface area of a cylinder is 352 cm2 . If its height is 10 cm, then find the diameter of the base.

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: (i) Let r be the radius and h be the height of the cylinder. Given the sum of radius and height of the cylinder, r+h = 37 cm Total surface area of the cylinder = 1628 cm2 2r(r+h) = 1628...

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?

CBSE, Class 10, Exercise 1, maths, Maths, Mensuration, ML Aggarwal

Solution: Let the radius of the base of a right circular cylinder be r and height be h. Volume of the cylinder, V1 = r2h The radius of the base of a right circular cylinder is halved and the height...