MHTCET Archives - Noon Academy

In the figure given below, ∠ABC = ∠DAC and AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that ∆ACD is similar to ∆BCA
(ii) Find BC and CD
(iii) Find the area of ∆ACD : area of ∆ABC.

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

Solution:- From the question it is given that, ∠ABC = ∠DAC AB = 8 cm, AC = 4 cm, AD = 5 cm (i) Now, consider ∆ACD and ∆BCA ∠C = ∠C … [common angle for both triangles] ∠ABC = ∠CAD … [from the...

In the figure given below, ABCD is a trapezium in which DC is parallel to AB. If AB = 9 cm, DC = 6 cm and BB = 12 cm., find
(i) BP
(ii) the ratio of areas of ∆APB and ∆DPC.

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

Solution:- From the question it is given that, DC is parallel to AB AB = 9 cm, DC = 6 cm and BB = 12 cm (i) Consider the ∆APB and ∆CPD ∠APB = ∠CPD … [because vertically opposite angles are equal]...

In the figure (ii) given below, AB || DC and AB = 2 DC. If AD = 3 cm, BC = 4 cm and AD, BC produced meet at E, find
(i) ED
(ii) BE
(iii) area of ∆EDC : area of trapezium ABCD.

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

Solution:- From the question it is given that, AB || DC AB = 2 DC, AD = 3 cm, BC = 4 cm Now consider ∆EAB, EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1 (i) EA = 2, DA = 2 × 3 = 6 cm Then, ED = EA – DA = 6 –...

The number of solutions of $\sin x+\sin 3 x+\sin 5 x=0$ in the interval $\left[\frac{\pi}{2}, 3 \frac{\pi}{2}\right]$ is
(A) 2
(B) 3
(C) 4
(D) 5

maths, Maths

CORRECT OPTION: B $ \begin{array}{l} (\sin x+\sin 5 x)+\sin 3 x=0 \\ 2 \sin 3 x \cos 2 x+\sin 3 x=0 \\ \sin 3 x(2 \cos 2 x+1)=0 \\ \sin 3 x=0 \text { and } 2 \cos 2 x+1=0 \\ \cos 2 x=-\frac{1}{2}...

In the figure (i) given below, DE || BC and the ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

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

Solution:- From the question it is given that, DE || BC The ratio of the areas of ∆ADE and trapezium DBCE is 4 : 5 Now, consider the ∆ABC and ∆ADE ∠A = ∠A … [common angle for both triangles] ∠D = ∠B...

If $X=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)$ then $\frac{d y}{d x}$ at $\theta=\frac{\pi}{4}$ is
(A) 1
(B) 0
(C) $\frac{1}{\sqrt{2}}$
(D) $\sqrt{2}$

maths, Maths

CORRECT OPTION: A $ \begin{array}{l} \frac{d x}{d \theta}=e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) e^{\theta}=2 e^{\theta} \sin \theta \\ \frac{d y}{d...

$\quad$ In $\Delta A B C$ , with usual notations, if $a, b, c$ are in A.P. then $a \cos ^{2}\left(\frac{C}{2}\right)+c \cos ^{2}\left(\frac{A}{2}\right)=$
(A) $3 \frac{a}{2}$
(B) $3 \frac{c}{2}$
(C) $3 \frac{b}{2}$
(D) $\frac{3 a b c}{2}$

maths, Maths

CORRECT OPTION: C $ \begin{array}{l} 2 b=a+c \\ \therefore \frac{a}{2}\left(2 \cos ^{2}\left(\frac{C}{2}\right)\right)+\frac{C}{2}\left(2 \cos ^{2}\left(\frac{A}{2}\right)\right) \\...

In the adjoining figure, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1 : 2 and DP produced meets AB produced at Q. If area of ∆CPQ = 20 cm², find
(i) area of ∆BPQ.
(ii) area ∆CDP.
(iii) area of parallelogram ABCD.

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

Solution:- From the question it is given that, ABCD is a parallelogram. BP: PC = 1: 2 area of ∆CPQ = 20 cm² Construction: draw QN perpendicular CB and Join BN. Then, area of ∆BPQ/area of ∆CPQ =...

In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find
(i) EF : AD
(ii) area of ∆BEF : area of ∆ABD In the figure
(iii) given below, ABCD is a parallelogram
(iv) area of ∆FEO : area of ∆OBC.

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

Solution:- From the question it is given that, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O. AE : EB = 2 : 3 (i) We have to find EF : AD So, AB/BE = AD/EF EF/AD =...

In the figure (ii) given below, ABCD is a parallelogram. AM ⊥ DC and AN ⊥ CB. If AM = 6 cm, AN = 10 cm and the area of parallelogram ABCD is 45 cm², find
(i) AB
(ii) BC
(iii) area of ∆ADM : area of ∆ANB.

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

Solution:- From the question it is given that, ABCD is a parallelogram, AM ⊥ DC and AN ⊥ CB AM = 6 cm AN = 10 cm The area of parallelogram ABCD is 45 cm² Then, area of parallelogram ABCD = DC × AM =...

In the figure (i) given below, ABCD is a trapezium in which AB || DC and AB = 2 CD. Determine the ratio of the areas of ∆AOB and ∆COD.

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

Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and AB = 2 CD, Then, ∠OAB = ∠OCD … [because alternate angles are equal] ∠OBA = ∠ODC Then, ∆AOB ~ ∆COD So, area of...

In ∆ABC, AP : PB = 2 : 3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA. Find:
(i) area ∆APO : area ∆ABC.
(ii) area ∆APO : area ∆CQO.

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

Solution:- From the question it is given that, PB = 2: 3 PO is parallel to BC and is extended to Q so that CQ is parallel to BA. (i) we have to find the area ∆APO: area ∆ABC, Then, ∠A = ∠A … [common...

In the adjoining figure, ABC is a triangle. DE is parallel to BC and AD/DB = 3/2,
(i) Determine the ratios AD/AB, DE/BC0
(ii) Prove that ∆DEF is similar to ∆CBF. Hence, find EF/FB.
(iii) What is the ratio of the areas of ∆DEF and ∆CBF?

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

Solution:- (i) We have to find the ratios AD/AB, DE/BC, From the question it is given that, AD/DB = 3/2 Then, DB/AD = 2/3 Now add 1 for both LHS and RHS we get, (DB/AD) + 1 = (2/3) + 1 (DB + AD)/AD...

In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm, calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

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

Solution:- (i) Consider the ∆ABC and ∆DEC, ∠ABC = ∠DEC … [both angles are equal to 90o] ∠C = ∠C … [common angle for both triangles] Therefore, ∆ABC ~ ∆DEC … [by AA axiom] (ii) AC/CD = AB/DE...

In the given figure, DE || BC.
(i) Prove that ∆ADE and ∆ABC are similar.
(ii) Given that AD = ½ BD, calculate DE if BC = 4.5 cm.
(iii) If area of ∆ABC = 18cm2, find the area of trapezium DBCE

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

Solution:- (i) From the question it is given that, DE || BC We have to prove that, ∆ADE and ∆ABC are similar ∠A = ∠A … [common angle for both triangles] ∠ADE = ∠ABC … [because corresponding angles...

In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE.

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

Solution:- From the question it is given that, DE || BC and AD : DB = 1 : 2, ∠D = ∠B, ∠E = ∠C … [corresponding angles are equal] Consider the ∆ADE and ∆ABC, ∠A = ∠A … [common angles for both...

In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm, find the area of ∆ABC.

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

Solution:- From the question it is given that, DE || BC, DE = 6 cm, BC = 9 cm and area of ∆ADE = 28 sq. cm From the fig, ∠D = ∠B and ∠E = ∠C … [corresponding angles are equal] Now consider the ∆ADE...

In the figure (ii) given below, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm. (i) Prove that ∆OAB ~ ∆OCD. (ii) Find CD and OB. (iii) Find the ratio of areas of ∆OAB and ∆OCD.

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

Solution:- From the question it is given that, AB || DC. AO = 10 cm, OC = 5cm, AB = 6.5 cm and OD = 2.8 cm (i) We have to prove that, ∆OAB ~ ∆OCD So, consider the ∆OAB and ∆OCD ∠AOB = ∠COD …...

In the figure, (i) given below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm², find the area of ∆QOA.

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

Solution:- From the question it is given that, PO = 6 cm, QO = 9 cm and the area of ∆POB = 120 cm² From the figure, Consider the ∆AOQ and ∆BOP, ∠OAQ = ∠OBP … [both angles are equal to 90o] ∠AOQ =...

The area of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

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

Solution:- From the question it is given that, The area of two similar triangles are 36 cm² and 25 cm². Let us assume ∆PQR ~ ∆XYZ, PM and XN are their altitudes. So, area of ∆PQR = 36 cm2 Area of...

∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and area of ∆ABC = 54 sq. cm. Determine the area of ∆DEF.

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

Solution:- From the question it is given that, ∆ABC ~ ∆DEF BC = 3 cm, EF = 4 cm Area of ∆ABC = 54 sq. cm. We know that, Area of ∆ABC/ area of ∆DEF = BC2/EF2 54/area of ∆DEF = 32/42 54/area of ∆DEF =...

∆ABC ~ DEF. If area of ∆ABC = 9 sq. cm., area of ∆DEF =16 sq. cm and BC = 2.1 cm., find the length of EF.

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

Solution:- From the question it is given that, ∆ABC ~ DEF Area of ∆ABC = 9 sq. cm Area of ∆DEF =16 sq. cm We know that, area of ∆ABC/area of ∆DEF = BC2/EF2 area of ∆ABC/area of ∆DEF = BC2/EF2 9/16 =...

In the figure (2) given below AD is bisector of ∠BAC. If AB = 6 cm, AC = 4 cm and BD = 3cm, find BC

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

Solution:- From the question it is given that, AD is bisector of ∠BAC AB = 6 cm, AC = 4 cm and BD = 3cm Construction, from C draw a straight line CE parallel to DA and join AE ∠1 = ∠2 … [equation...

In the figure (1) given below, AB || CR and LM || QR.
(i) Prove that BM/MC = AL/LQ
(ii) Calculate LM : QR, given that BM : MC = 1 : 2.

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

Solution:- From the question it is given that, AB || CR and LM || QR (i) We have to prove that, BM/MC = AL/LQ Consider the ∆ARQ LM || QR … [from the question] So, AM/MR = AL/LQ … [equation (i)] Now,...

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O. Using Basic Proportionality theorem, prove that AO/BO = CO/DO

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

Solution:- From the question it is given that, ABCD is a trapezium in which AB || DC and its diagonals intersect each other at O Now consider the ∆OAB and ∆OCD, ∠AOB = ∠COD [because vertically...

In the adjoining given below, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. show that BC || QR.

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

Solution:- Consider the ∆POQ AB || PQ … [given] So, OA/AP = OB/BQ … [equation (i)] Then, consider the ∆OPR AC || PR OA/AP = OC/CR … [equation (ii)] Now by comparing both equation (i) and equation...

In the give figure, ∠D = ∠E and AD/BD = AE/EC. Prove that BAC is an isosceles triangle.

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

Solution:- From the given figure, ∠D = ∠E and AD/BD = AE/EC, We have to prove that, BAC is an isosceles triangle So, consider the ∆ADE ∠D = ∠E … [from the question] AD = AE … [sides opposite to...

(a) In the figure (i) given below, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm.

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

Solution:- From the given figure, CD || LA and DE || AC, Consider the ∆BCA, BE/BC = BD/BA By using the corollary of basic proportionality theorem, BE/(BE + EC) = BD/AB 4/(4 + 2) = BD/AB … [equation...

In figure (ii) given below, AB || DE and BD || EF. Prove that DC² = CF x AC.

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

Solution:- From the figure it is given that, AB || DE and BD || EF. We have to prove that, DC² = CF x AC Consider the ∆ABC, DC/CA = CE/CB … [equation (i)] Now, consider ∆CDE CF/CD = CE/CB …...

In figure (i) given below, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

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

Solution:- From the question it is given that, DE || BC and BD = CE So, we have to prove that ABC is an isosceles triangle. Consider the triangle ABC, AD/DB = AE/EC Given, DB = EC … [equation (i)]...

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: of a ∆PQR. For each of the following cases, state whether EF || QR:PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

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

Solution:- From the dimensions given in the question, Consider the ∆PQR So, PQ/PE = 1.28/0.18 = 128/18 = 64/9 Then, PR/PF = 2.56/0.36 = 256/36 = 64/9 By comparing both the results, 64/9 = 64/9...

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...

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 =...

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 =...

The Air Prevention and Control of Pollution Act came into force in
Option A 1975
Option B 1981
Option C 1985
Option D 1990

biology, Biology

The correct answer is Option B 1981. This is because - The Air Pollution Prevention and Control Act was passed in India in 1981, but it was revised in 1987 to include noise as an air contaminant.

Global warming can be controlled by
Option A Reducing deforestation, cutting down the use of fossil fuel.
Option B Reducing reforestation, increasing the use of fossil fuel.
Option C Increasing deforestation, slowing down the growth of human population.
Option D Increasing deforestation, reducing efficiency of energy usage.

biology, Biology

The correct answer is Option A Reducing deforestation, cutting down the use of fossil fuel. This is because - Cutting down on the use of fossil fuels, improving energy efficiency, reducing...

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,...

A sedentary sea anemone gets attached to the shell lining of hermit crab. The association is
Option A Ecto-parasitism
Option B Symbiosis
Option C Commensalism
Option D Amensalism

biology, Biology

The correct answer is Option B Symbiosis. This is because - A stationary sea anemone attaching itself to the hermit crab's shell lining is an example of symbiosis, in which the sea anemone grows on...

Which one of the following processes during decomposition is correctly described?
Option A Fragmentation – carried out by organisms such as earthworms.
Option B Humification – leads to an accumulation of a dark coloured substance humus which undergoes microbial action at a very fast rate.
Option C Catabolism – last step in the decomposition under fully anaerobic condition.
Option D Leaching – water soluble inorganic nutrients rise to the top layers of soil.

biology, Biology

The correct answer is Option A Fragmentation – carried out by organisms such as earthworms. This is because - Detritivores, such as earthworms, carry out the fragmentation process. Humification...

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...

A biologist studied the population of rats in a barn. He found that the average natality was 250, average mortality 240, immigration 20 and emigration 30. The net increase in population is
Option A 10
Option B 15
Option C 05
Option D Zero

biology, Biology

The correct answer is Option D Zero. This is because - The density of a population in a specific environment fluctuates over time due to changes in four basic processes, two of which (natality and...

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,...

During sewage treatment, biogas is produced which includes
Option A Methane, hydrogen sulphide, carbon dioxide
Option B Methane, oxygen, hydrogen sulphide
Option C Hydrogen sulphide, methane, sulphur dioxide
Option D Hydrogen sulphide, nitrogen, methane

biology, Biology

The correct answer is Option A Methane, hydrogen sulphide, carbon dioxide. This is because - Biogas is a mixture of gases formed during the breakdown of biomass in the absence of oxygen, primarily...

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...

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,...

In plant breeding programmes, the entire collection (of plants/seeds) having all the diverse alleles for all genes in a given crop is called
Option A Selection of superior recombinants
Option B Cross-hybridisation among the selected parents
Option C Evaluation and selection of parents.
Option D Germplasm collection

biology, Biology

The correct answer is Option D Germplasm collection. This is because - A prerequisite for optimal use of natural genes accessible in populations is the collection and preservation of all distinct...

(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...

The cell-mediated immunity inside the human body is carried out by
Option A T- lymphocytes
Option B B-lymphocytes
Option C Thrombocytes
Option D Erythrocytes

biology, Biology

The correct answer is Option A T- lymphocytes. This is because - T- lymphocytes, highly specialised cells that recognise 'non-self' and'self' cells, are responsible for cell-mediated immunity in the...

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...

Infection of Ascaris usually occurs by
Option A Drinking water containing eggs of Ascaris
Option B Eating imperfectly cooked pork
Option C Tse-tse fly

biology, Biology

The correct answer is Option A Drinking water containing eggs of Ascaris. This is because - Ascaris infection is spread through the consumption of polluted water, vegetables, and fruits that contain...

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...

The eye of octopus and eye of cat show different patterns of structure, yet they perform similar function. This is an example of
Option A Homologous organs that have evolved due to convergent evolution.
Option B Homologous organs that have evolved due to divergent evolution.
Option C Analogous organs that have evolved due to convergent evolution.
Option D Analogous organs that have evolved due to divergent evolution.

biology, Biology

The correct answer is Option C Analogous organs that have evolved due to convergent evolution. This is because - Organs that perform the same job but differ in origin and structure are referred to...

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...

According to Darwin, the organic evolution is due to
Option A Intraspecific competition
Option B Intraspecific competition
Option C Competition within closely related species.
Option D Reduced feeding efficiency in one species due to the presence of interfering species.

biology, Biology

The correct option is Option B Intraspecific competition. This is because - According to Darwin, organic evolution is due to intraspecific competition which is an interaction in which members of the...

Which enzyme/s will be produced in a cell in which there is a non-sense mutation in the lac Y gene?
Option A β- galactosidase
Option B Lactose permease
Option C Transacetylase
Option D Lactose permease and transacetylase

biology, Biology

The correct option is Option A β- galactosidase. This is because - Non-functional permease will result from a non-sense mutation in the lac Y gene, and a stop codon will be added to the sequence...

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...

The diagram shows an important concept in the genetic implication of DNA. Fill in the blanks A to C.
Option A A – Transcription, B – Transcription, C – James Watson
Option B A – Translation, B – Transcription, C – Erwin Chargaff
Option C A – Transcription, B – Translation, C – Francis Crick
Option D A – Translation, B – Extension, C – Rosalind Franklin

biology, Biology

The correct answer is Option C A – Transcription, B - Translation, C - Francis Crick. This is because - Francis Crick proposed the central dogma in molecular biology, which states that the genetic...

If both parents are carriers for thalassemia which is an autosomal recessive disorder, what are the chances of pregnancy resulting in an affected child?
Option A No chance
Option B 50 %
Option C 25 %
Option D 100 %

biology, Biology

The correct answer is Option C 25 %. This is because - The genotype of husband for thalassemia is XTX and the genotype for carrier women is XTX.

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...

The incorrect statement with regards to haemophilia is
Option A It is a sex-linked disease.
Option B It is a recessive disease.
Option C It is a dominant disease.
Option D A single protein involved in the clotting of blood is affected.

biology, Biology

The correct answer is Option C It is a dominant disease. This is because - Haemophilia is a sex-linked recessive disorder caused by a mutation in a single protein that is part of a cascade of...

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 …...

Which Mendelian idea is depicted by a cross in which the F1 generation resembles both the parents?
Option A Incomplete dominance
Option B Law of dominance
Option C Inheritance of one gene
Option D Co-dominance

biology, Biology

The correct answer is Option D Co-dominance. This is because - When both alleles of a pair are completely expressed in a heterozygote, it is said that the genes and traits are codominant. As a...

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...

Artificial insemination means
Option A Transfer of sperms of a healthy donor to a test tube containing ova
Option B Transfer of sperms of husband to a test tube containing ova
Option C Artificial introduction of sperms of a healthy donor into the vagina
Option D Introduction of sperms of a healthy donor directly into the ovary

biology, Biology

The correct answer is Option C Artificial introduction of sperms of a healthy donor into the vagina. This is because - Artificial insemination is the process of introducing sperm from the husband or...

Which of the following cannot be detected in a developing foetus by amniocentesis?
Option A Klinefelter’s syndrome
Option B Sex of the foetus
Option C Down’s syndrome
Option D Jaundice

biology, Biology

The correct option is Option D Jaundice. This is because - Amniocentesis is a foetal sex determination technique that uses the chromosomal pattern in the amniotic fluid around the growing embryo to...

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...

One of the legal methods of birth control is
Option A Abortion by taking an appropriate medicine
Option B By abstaining from coitus from day 10 to 17 of the menstrual cycle
Option C By having coitus at the time of day break
Option D By a premature ejaculation during coitus

biology, Biology

The correct option is Option B By abstaining from coitus from day 10 to 17 of the menstrual cycle. This is because - Periodic abstinence is a practise in which couples avoid or abstain from coitus...

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 –...

Which one of the following is not the function of placenta?
It Option A Facilities supply of oxygen and nutrients to the embryo
Option B Secretes estrogen
Option C Facilities the removal of carbon dioxide and waste material from the embryo
Option D Secretes oxytocin during parturition

biology, Biology

The correct option is Option D Secretes oxytocin during parturition. This is because - The placenta serves the following purposes: i. Allows the embryo to receive oxygen and nutrition; ii. Secretes...

(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 ∆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 =...

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

State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):

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

Solution:- (i) From the ΔABC and ΔPQR AB/PQ = 3.2/4 = 32/40 Divide both numerator and denominator by 8 we get, = 4/5 AC/PR = 3.6/4.5 = 36/45 Divide both numerator and denominator by 9 we get, = 4/5...

Menstrual flow occurs due to lack of
Option A Progesterone
Option B FSH
Option C Oxytocin
Option D Vasopressin

biology, Biology

The correct option is Option A Progesterone. This is because -If fertilisation fails, the ovary's corpus luteum degenerates, and progesterone and oestrogen release is stopped, resulting in a new...

What is the correct sequence of sperm formation?
Option A Spermatid, Spermatocyte, Spermatogonia, Spermatozoa
Option B Spermatogonia, Spermatocyte, Spermatozoa, Spermatid
Option C Spermatogonia, Spermatozoa, Spermatocyte, Spermatid
Option D Spermatogonia, Spermatocyte, Spermatid, Spermatozoa

biology, Biology

Correct Option D. The sperm mother cells divide frequently through mitotic divisions to generate spermatogonia during the multiplicative phase of spermatogenesis. The spermatogonia develop in size...

Select the answer which correctly matches the endocrine gland with the hormone it secretes and its function / deficiency symptom.

biology, Biology

The correct answer is Option C. This is because - Anterior pituitary produces growth hormone (GH). Posterior pituitary stores and releases two hormones called oxytocin and vasopressin. Corpus luteum...

Which of the following statements is correct in relation to the endocrine system?
Option A Adenohypophysis is under direct neural regulation of the hypothalamus.
Option B Organs in the body like gastrointestinal tract, heart, kidney and liver do not produce any hormones.
Option C Non-nutrient chemicals produced by the body in trace amounts that act as intercellular messengers are known as hormones.
Option D Releasing and inhibitory hormones are produced by the pituitary gland.

biology, Biology

The correct answer is Option C Non-nutrient chemicals produced by the body in trace amounts that act as intercellular messengers are known as hormones. This is because - Hormones are non-nutrient...

Parts A, B, C and D of the human eye are shown in the diagram. Select the option which gives correct identification along with its functions/ characteristics.
Option A A – Retina – contains photo receptors rods and cones.
Option B B – Blind spot – has only a few rods and cones.
Option C C – Aqueous chamber- reflects the light which does not pass through the lens.
Option D D – Choroid – its anterior part forms ciliary body.

biology, Biology

The correct answer is Option A A - Retina – contains photo receptors rods and cones. This is because - Retina contains two types of photoreceptor cells, rods and cones which contain the...

A diagram showing axon terminal and synapse is given. Identify correctly at least two of A-D.

biology, Biology

The correct answer is Option A A – Receptor, C – Synaptic vesicles This is because - When an impulse reaches the axon terminal, it causes synaptic vesicles to migrate towards the membrane, where...

The characteristic and an example of a synovial joint human is

biology, Biology

The correct answer is Option C. This is because - The presence of a fluid-filled synovial space between the articulating surfaces of two bones distinguishes synovial joints. Between the atlas and...

Select the correct statement with respect to locomotion in humans.
Option A A decreased level of progesterone causes osteoporosis in old people.
Option B Accumulation of uric acid crystals in joints causes their inflammation.
Option C The vertebral column has 10 thoracic vertebrae.
Option D The joint between adjacent vertebrae is a fibrous joint.

biology, Biology

The correct answer is Option B Accumulation of uric acid crystals in joints causes their inflammation. This is because - Inflammation of joints due to accumulation of uric acid crystals is called...

Figure shows human urinary system with structures labelled A to D. Select option which correctly identifies them and gives their characteristics and/ or functions.

biology, Biology

Correct Option A The front part of each kidney houses the adrenal gland. The adrenal medulla secretes two hormones, adrenaline or epinephrine, and nor-adrenaline or nor-epinephrine, which are both...

The diagram given here is the standard ECG of a normal person. The P-wave represents the

biology, Biology

Correct Option A. The P-wave depicts the atria's electrical stimulation (or depolarization), which causes both atria to constrict.

Figure shows schematic plan of blood circulation in humans with labels A to D. Identify the label and give its function/s.
Option A A – Pulmonary vein – takes impure blood from body parts, PO2 = 60 mm Hg
Option B B – Pulmonary artery – takes blood from heart to lungs, PO2 = 90 mm Hg
Option C C – Vena cava – takes blood from body parts to right auricle, PCO2 = 45 mm Hg
Option D D – Dorsal aorta – takes blood from heart to body parts, PO2 = 95 mm Hg

biology, Biology

The correct answer is Option C. This is because - Blood from the pulmonary veins and vena cava flows into the left and right ventricles via the left and right atria as the tricuspid and bicuspid...

The figure shows a diagrammatic view of human respiratory system with labels A, B, C and D. Select the option which gives correct identification and main function and/or characteristic.

biology, Biology

The correct answer is Option C. This is because - A - The tracheae is a lengthy tube with incomplete cartilaginous rings supporting it. B - The thoracic lining is in close touch with the outer...

A pregnant female delivers a baby who suffers from stunted growth, mental retardation, low intelligence quotient and abnormal skin. This is the result of
Option A Deficiency of iodine in diet
Option B Low secretion of growth hormone
Option C Cancer of the thyroid gland
Option D Over secretion of pars distalis

biology, Biology

The correct answer is Option A Deficiency of iodine in diet. This is because - Deficiency of iodine in our diet results in hypothyroidism and enlargement of the thyroid gland, commonly called goitre...

Select the correct match of the digested products in humans given in column I with their absorption site and mechanism in column II.

biology, Biology

The correct option is A. This is because - Small intestine is the principal organ for absorption of nutrients where the final products of digestion such as glucose, fructose, fatty acids, glycerol...

Two different dice are thrown at the same time. Find the probability of getting :
(iii) sum divisible by 5
(iv) sum of at least 11.

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

(iii)Let E be an event of getting a sum divisible by 5. Favourable outcomes = {(1,4),(2,3), (3,2), (4,1),(4,6), (5,5), (6,4)} Number of favourable outcomes = 7 P(E) = 7/36 Probability of getting a...

Two different dice are thrown at the same time. Find the probability of getting :
(i) a doublet
(ii) a sum of 8

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

Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...

Two different dice are thrown simultaneously. Find the probability of getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.

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

Solution: When two dice are thrown simultaneously, the sample space of the experiment is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1),(3,2), (3,3),...

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...

From a pack of 52 cards, a blackjack, a red queen and two black kings fell down. A card was then drawn from the remaining pack at random. Find the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.

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

Solution: Total number of cards = 52-4 = 48 [∵4 cards fell down] So number of possible outcomes = 48 (i) Let E be the event of getting black card. There will be 23 black cards since a black jack and...

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
(v) a spade
(vi) ‘9’ of black colour.

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

(v) Let E be the event of getting a spade. There will be 10 spades. Number of favourable outcomes = 10 P(E) = 10/49 Hence the probability of getting a spade is 10/49. (vi) Let E be the event of...

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:
(iii) a king of red colour
(iv) a card of diamond

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

(iii) Let E be the event of getting a king of red colour. There will be 2 cards of king of red colour. Number of favourable outcomes = 2 P(E) = 2/52 = 1/26 Hence the probability of getting a king of...

A card is drawn from a well-shuffled pack of 52 cards. Find the probability of getting:
(i) ‘2’ of spades
(ii) a jack .

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

Solution: Total number of cards = 52. So number of possible outcomes = 52. (i) Let E be the event of getting ‘2’ of spades. There will be only one card of ‘2’ spades. Number of favourable outcomes =...

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 :
(v) divisible by 3 or 2
(vi) a perfect square number.

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

(v) Let E be the event of getting the number on the card is divisible by 3 or 2 Outcomes favourable to E are {2,3,4,6,8,9,10,12,14,15} Number of favourable outcomes = 10 P(E) = 10/15 = 2/3 Hence the...

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...

Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.

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

Solution: There are 7 ways in which the month of February can occur, each starting with a different day of the week. (i)Only 1 way is possible for 5 Wednesdays to occur in February with 29 days....

Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.

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

Solution: For a leap year there are 366 days. Number of days in January = 31 Total number of January month types = 7 Number of January months with 5 Mondays = 3 (i)Probability that the month 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
(iii) a number greater than 2?
(iv) a number less than 9?

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

(iii) Let E be the event of arrow pointing a number greater than 2. Outcomes favourable to E are {3,4,5,6,7,8} Number of favourable outcomes = 6 P(E) = 6/8 = 3/4 Hence the probability of arrow...

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...

A stage in cell division is shown in the figure. Select the answer which gives correct identification of the stage with its characteristics.

biology, Biology

The correct answer is Option A Telophase Nuclear envelope reforms, Golgi complex reforms. This is because- Telophase shows the following key events: i. Chromosomes cluster at opposite spindle poles...

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...

The essential chemical components of many coenzymes are
Option A Proteins
Option B Nucleic acids
Option C Carbohydrates
Option D Vitamins

biology, Biology

The correct answer is Option D Vitamins. This is because- The essential chemical components of many coenzymes are vitamins, e.g., coenzyme nicotinamide adenine dinucleotide (NAD) and NADP contain...

Macromolecule chitin is
Option A Nitrogen containing polysaccharide
Option B Phosphorus containing polysaccharide
Option C Sulphur containing polysaccharide
Option D Simple polysaccharide

biology, Biology

The correct answer is Option A Nitrogen containing polysaccharide. This is because - Chitin is a polysaccharide composed of a nitrogen containing glucose derivative known as N-acetyl glucosamine.

In a single throw of a die, find the probability of getting:
(iii) a number greater than 5
(iv) a prime number

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

(iii)Let E be the event of getting a number greater than 5. Outcomes favourable to E is 6. Number of favourable outcomes = 1 P(E) = 1/6 Hence the probability of getting a number greater than 5 is...

Which one of the following organelles in the figure correctly matches with its function?
Option A Rough endoplasmic reticulum, formation of glycoproteins
Option B Golgi apparatus, proteins synthesis
Option C Golgi apparatus, formation of glycolipids
Option D Rough endoplasmic reticulum, protein synthesis

biology, Biology

The correct option is Option D Rough endoplasmic reticulum, protein synthesis. This is because - Rough endoplasmic reticulum proteins and smooth endoplasmic reticulum lipids both reach the Golgi...

In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5

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. Number of possible outcomes = 6 (i) Let E be the event of getting an odd number. Outcomes favourable to E are 1,3,5. Number of...

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 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
(i) black?
(ii) blue or black?

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

Solution: Number of blue marbles = 7 Number of white marbles = 8 Number of black marbles = 5 Total number of marbles = 7+8+5 = 20 (i) Probability of getting black , = 5/20 = 1/4 Hence the...

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random from the bag, find the probability that the ball drawn is: (i) red (ii) black or white (iii) not black.

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

Solution: Number of black balls = 5 Number of red balls = 7 Number of white balls = 3 Total number of balls = 5+7+3 = 15 (i)Probability that the ball drawn is red, = 7/15 (ii) Probability of black...

A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.

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

Solution: Total number of alphabets = 26 Number of vowels = 5 Total number of consonants = 26-5 = 21 Probability that the letter chosen is a consonant , = 21/26 Hence the required probability is...

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.

12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

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

Solution: Number of defective pens = 12 Number of good pens = 132. Total number of pens = 132+12 = 144 Probability of getting a good pen, P(E) P(E) = 132/144 = 11/12 Hence the required probability...

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...

What external changes are visible ‘after the last moult of a cockroach nymph?
Option A Mandibles become harder
Option B Anal cerci develop
Option C Both forewings and hindwings develop
Option D Labium develops

biology, Biology

The correct answer is Option C Both forewings and hindwings develop. This is because - The nymph develops into an adult after moulting 13 times. Wing pads are present in the next to last nymphal...

The H-zone in the skeletal muscle fibres is due to
Option A The absence of myofibrils in the central portion of A-band
Option B The central gap between myosin filaments in the A-band
Option C The central gap between actin filaments extending through myosin filaments in the A-band
Option D Extension of myosin filaments in the central portion of the A-band

biology, Biology

The correct answer is Option C The central gap between actin filaments extending through myosin filaments in the A-band. This is because - A considerably less dark zone known as the H-zone exists in...

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...

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 =...

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...

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 =...

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...

A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged? (2002)

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

Given height of the cone, h = 20 cm Diameter of the cone = 16.8 cm Radius of the cone, r = 16.8/2 = 8.4 cm Volume of water in the vessel = (1/3)r2h = (1/3)×8.42 ×20 = (1/3)×(22/7)×8.4 ×8.4 ×20 =...

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...

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...

A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.

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

Solution; Given inner diameter of the well = 6 m Radius of the well, r = 6/2 = 3 m Depth of the well, H = 22 m Volume of the soil dug out of well = r2H = ×32×22 = 198 m3 Width of the embankment = 5...

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 =...

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 =...

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 =...

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...

One of representatives of Phylum Arthropoda is
Option A Cuttlefish
Option B Silverfish
Option C Pufferfish
Option D Flying fish

biology, Biology

The correct answer is Option B Silverfish. This is because - Cuttlefish belongs to Phylum Mollusca. Silverfish belongs to Phylum Arthropoda. Pufferfish and flying fish belong to Phylum Chordata.

Which group of animals belong to the same phylum?
Option A Malarial parasite, Amoeba, Mosquito
Option B Earthworm, Pinworm, Tapeworm
Option C Prawn, Scorpion, Locusta
Option D Sponge, sea anemone, Starfish

biology, Biology

The correct answer is Option C Prawn, Scorpion, Locusta. This is because - Prawn, Scorpion, Locusta are examples of phylum Arthropoda.

Which of the following are correctly matched with respect to their taxonomic classification?
Option A Flying fish, cuttlefish, silverfish, -Pisces
Option B Centipede, millipede, spider, scorpion-Insecta
Option C House fly, butterfly, tsetsefly, silverfish-Insecta
Option D Spiny anteater, sea urchin, sea cucumber-Echinodermata

biology, Biology

The correct answer is Option C House fly, butterfly, tsetsefly, silverfish-Insecta. This is because - House fly, butterfly, tsetsefly, silverfish belongs to Phylum Insecta/Arthropoda which includes...

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 =...

Match the name of the animal (Column I) with one characteristic (Column II) and the phylum/class (column III) to which it belongs.

biology, Biology

The correct answer is This is because - Petromyzon is an ectoparasite on some fishes and belongs to class Cyclostomata. Hence, option A is the correct option.

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) =...

Which of the following represent maximum number of species among global biodiversity?
Option A Algae
Option B Lichens
Option C Fungi
Option D Mosses and ferns

biology, Biology

The correct answer is Option C Fungi. This is because - Fungi represents maximum number of species among global biodiversity.

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 =...

Kyoto protocol was endorsed at
Option A CoP-3
Option B CoP-5
Option C CoP-6
Option D CoP-4

biology, Biology

The correct option is Option A CoP-3. This is because - The Kyoto Protocol was adopted at the third session of the Conference of Parties to the UNFCCC, COP-3 in 1997 in Kyoto, Japan.

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...

Which one of the following is not used for ex situ plant conservation?
Option A Field gene banks
Option B Seed banks
Option C Shifting cultivation
Option D Botanical gardens

biology, Biology

The correct answer is Option C Shifting cultivation. This is because - Ex-situ conservation includes field gene banks, seed banks and botanical gardens. Shifting cultivation is not used for ex situ...