Category: CBSE Sample Papers

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 8 with Solutions

Max. Marks: 35
Time: 2 Hours

General Instructions:
Read the following instructions carefully:

• There are 12 questions in this question paper with internal choice.
• Section A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• Section B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• Section C- Q. No. 12 is case-based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A

Question 1.
If the concentration is expressed in mol L^-1 units and time in seconds, what would be the unit of K^*
(a) For a zero-order reaction
(b) For a first order reaction
(c) Why molecularity of a reaction cannot be zero?
Answer:
(a) For zero order reaction unit of K is mol L^-1 sec^-1.
(b) For first order reaction unit of K is sec’^-1.
(c) Molecularity is the number of given reactant molecules or atoms that are colliding in the elementary reaction. Thus, a minimum of one reactant molecule, atom or ion is required to initiate a chemical
reaction. Hence, molecularity cannot be zero.

Question 2.
Complete the following equations:

Answer:

Question 3.
Predict the product of the following reactions:

Answer:

Section – B

Question 4.
(a) Name the isomerism shown by the complex [C0(NH₃)₅(NO₂)]Cl₂.
(b) What type of bond is present between metal and carbon in metal carbonyls?
(c) [FeF₆]^3- has five unpaired electrons while [Fe(CN)₆]^3-has only one unpaired electron. Explain.
OR
The crystal field theory assumes that the interaction between the metal ion and ligand is purely electrostatic. When ligands approach central metal atom/ion the five degenerated orbitals of the central metal atom become differential. In a complex, the central metal atom or ion is surrounded by various atoms or groups of atoms called ligands.
(a) What is the type of bond in metal complex according to crystal field theory?
(b) What is the number of ligands in Cu₂ [Fe(CN)₆].
(c) Which one is lower energy set between t_2g and eg, after splitting of five degenerate orbital or metal atom/ion on complexation in tetrahedral complex?
Answer:
(a) Linkage isomerism occurs in complex compounds with ambidentate ligands like NO₂^–, SCN^–, CN^–, S₂O₃^-2.
These ligands have two donor atoms but at a time only one atom is directly linked to central metal atom of the complex. Linkage isomerism is shown by [Co(NH₃)₅(NO₂)] Cl₂. The isomers are [Co(NH₃)₅NO₂]Cl and [Co(NH₃)₅ONO]Cl₂.

(b) In a metal carbonyl, the metal-carbon bond possesses both σ and π character. Here M – C π bond is formed by the donation of lone pair of electrons from carbonyl carbon to the vacant orbital of metal. M – C π bond is formed by the back donation of electrons from a filled d-orbital of metal into the vacant antibondirig n orbital of carbon monoxide (Carbonyl ligand).

(c) In [FeF₆]^3-Fe(III) has got 3d⁵ systgem. As F is a weak field ligand, according to crystal field theory (CFT) electronic arrangement will be as follows:

Hence, it has 5 unpaired electrons.
In [Fe(CH)₆]^3- Fe(III) has got 3d⁵ system. As CN^– is a strong field ligand, according to CFT electronic
the arrangement will be as follows:

Hence, it has only one unpaired electron.
OR
(a) Ionic bond or electrostatic bond.
(b) Number of ligands = 6.
(c) Lower energy set = eg.

Ionic Strength Calculator is a free online tool that displays the ionic strength for the given concentration and valency.

Question 5.
Following ions are given:
Cr²⁺, Cu²⁺, Cu⁺, Fe²⁺, Fe³⁺, Mn³⁺
Identify the ion which is
(a) A strong reducing agent.
(b) Unstable in aqueous solution.
(c) A strong oxiding agent.
Give suitable reason in each case.
OR
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
(a) Electronic configurations
(b) Oxidation states
(c) Ionisation enthalpies.
Answer:
(a) Cr²⁺’ is a stronger reducing agent because it can lose one of its electron to become more stable Cr³⁺
in which the t_2g level of d-orbital is half filled arid eg level is empty.
(b) Cu⁺ is unstable in aqueous solution because it disproportionates in water to form Cu²⁺ and Cu.
(c) Mn³⁺ is a stronger oxidising agent because it has 4 electrons in its valence shell and when it gains
one electron to form Mn²⁺, it results in the half-filled (d⁵) configuration that has extra stability.
OR
(a) In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbitais are respectively filled.
We know that elements in the same vertical column generally have similar electronic configurations.
In the first transition series, two elements show unusual electronic configurations:
Cr (24) = 3d⁵4s¹ (Because of extra stability of half filled and fully filled orbitais)
Cu (29) = 3d¹⁰ 4s¹
Similarly, there are exceptions in the second transition series. These are:
Mo(42) = 4d⁵5s¹
Te(43) = 4d⁶5s¹
Ru(44) = 4d⁷5s¹
Rh(45) = 4d⁸5s¹
Pd(46) = 4d¹⁰5s⁰
Ag(47) = 4d¹⁰5s¹
There are some exceptions in third transition series as well. These are:
W(74) = 5d⁴6s²
Pt(78) = 5d⁹6s¹
Au(79) =5d¹⁰6s¹

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.
(b) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and minimum at the extreme ends.

However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series.
All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states.
The stability of the +2 and +3 oxidation states decreases in the second and the third transition series,
wherein higher oxidation states are more common.

For example, chromate ion (CrO₄)^2- is strong oxidant while molybdate (MoO₄)^2- and tungstate (WO₄)^2-
are stable. For example: WCl₆, ReF₇, RuO₄, etc., are stable but no such compound of 1st transition series exists.

(c) In each of the three transition series, the first ionisation enthalpy increases from left to right. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. The removal of one electron alters the relative energies of 4s and 3d orbitals. Hence, there is marginal and irregular increase in ionisation enthalpies.

Question 6.
(a) How would you account for the fact that the transition elements have high enthalpies of atomisation?
(b) Name the catalyst used in Ostwald’s process for the manufacturing of nitric acid.
(c) Transition metals generally form coloured compounds. Explain why?
Answer:
(a) Transition elements have large number of unpaired electrons in their atoms due to which they have stronger interatomic interactions and hence stronger bonding between atoms, leading to high
atomisation enthalpies.

(b) Platinised asbestos is used as catalyst in the Ostwald’s process for the manufacturing of nitric acid.

(c) Transition metals generally form coloured compounds because they generally possess one or more unpaired electrons. When visible light falls on a transition metal compound or ion, the unpaired electrons present in the lower energy d-orbital get promoted to high energy d-orbitals, called d-d transition, due to the absorption of visible light. Since the energy involved in d-d transition is quantised, only a definite wavelength gets absorbed, remaining wavelengths present in the visible region got transmitted. Therefore, transmitted light shows some colour complementary to the absorbed colour.

Question 7.
Write one difference in each of the following:
(a) Multimolecular colloid and Associated colloid
(b) Coagulation and Peptization
(c) Lyophobic sol and Lyophilic sol
Answer:
(a) Multìmolecular colloids: It consists of aggregates of a large number of atoms or smaller molecules. Its size is in the range of 1-1000rim. Associated colloids: At low concentration, they act as strong electrolytes whereas at high concentrations form aggregates or micelles

(b) Coagulation is the process of aggregating together the colloidal particles so as to change them into large sized particle by the addition of excess of an electrolyte whereas peptisation is the process of converting a fresh precipitate into colloidal particles by adding small amount of suitable electrolyte.

(c) Lyophobic colloidal sols are not hydrated and have weak affinity with the dispersion medium whereas lyophilic colloidal sols are heavily hydrated and have strong affinity with the dispersion medium.

Question 8.
(a) A galvanic cell consists of metallic zinc plate dipped in 0.1 M zinc nitrate solution and a lead plate dipped in 0.02 M lead nitrate solution. Write chemical equation for the electrode reaction and calculate the e.m.f. of the cell at 25°C.
E⁰_Zn²⁺/Zn = – 0-76 V, E°_pb²⁺/Pb = – 0.13 V
(b) Why does the conductivity of a solution decrease with dilution?
(c) What is the use of salt bridge in an electrochemical cell?
Answer:
(a) The half cell reactions can be written as:
Pb²⁺+ 2e^– → Pb, E°_pb²⁺/Pb=-0.13V
Zn²⁺ + 2e^– → Zn, E°_{Zn²⁺ /Zn} =-0.76 V
So, E°cell = E°cathode — E°anode
=-0.13V-(-0.76V)
=-0.13+0.76 -0.63V
Now, overall reaction:
Zn+Pb²⁺ → Zn²⁺Pb;E_cell=?

E_cell = E°cell – \(\frac{0.0591}{2} \log \left[\frac{\mathrm{Zn}^{+2}}{\mathrm{~Pb}^{+2}}\right]\)
= 0.63 – \(\frac{0.06}{2} \log \left(\frac{0.1}{0.02}\right)\)
=0.63-0.03 log5
=0.63-0.03 x 0.6910
= 0.63 — 0.02 = 0.61V

(b) The conductivity of a solution is the conductance of ions present in a unit volume of the solution. The number of ions (responsible for carrying current) decreases when the solution is diluted. As a result, the conductivity of a solution decreases with dilution.

(c) The purpose of the salt bridge is to act as a source of spectator ions that can migrate Into each of the half cells to preserve electrical neutrality.

Question 9.
(a) An organic compound X has molecular formula C₅H₁₀O. It does not reduce Fehling’s solution but
forms a bisulphate compound. It also gives positive iodoform test. What are possible structures of X? Explain your reasoning relating structure.
(b) Write IUPAC name of the following:

OR
Predict the product:

Answer:
(a) As the compound gives adduct with bisulphite, i.e., bisulphate compound, it contains a carbonyl group. But it does not reduce Fehling’s solution so it must be a ketone. Now, as this gives a positive test with iodoform, it is methyl ketone. So, based on the above information the following are the possible structures of the compound:

Question 10.
Answer the following questions:
(a) What is Hinsberg reagent?
(b) Why do primary amines have higher boiling points than tertiary amines?
(c) Why are aliphatic amines stronger bases than aromatic amines?
Answer:
(a) Benzenesulphonyl chloride (C₆H₅ – SO₂Cl) is known as the Hinsberg reagent.

(b) Primary amines are involved in intermolecular association due to H-bonding. While 3° amines do not have intermolecular association due to lack of H-bonding. Thus, Molecular association of primary amine leads to the increase of its effective molar mass and thus, 1° amine has higher boiling point than 3° amine.

(c) Due to the -I effect of the benzene ring, the electrons on the N-atom are involved in resonance and are available as base. Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily. This is the reason that aliphatic amines are stronger bases than aromatic amines.

Question 11.
Predict the product:

Or
(a) Arrange the following in increasing acidic strength:

(b) Distinguish between:
(i) Acetone and formaldehyde
(ii) C₆H₅COCH₃ and C₆H₅CH₂ – CHO
Answer:

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
The term order of a reaction is the sum of exports of concentration terms appearing in the rate equation. It is an experimentally determinable quantity. It may be whole number, fraction, zero on even negative. Knowledge of order does not require knowledge of mechanism. Order of a ‘reaction’ may change with change in experimental condition namely pressure, temperature etc.

A 1st order reaction is a reaction whose rate-determining step (r.d.s) involves only one molecule. Thus the step is A → Product.
A 2nd order reaction may be of two types:
2 A→ Y Product
or A + B → Product.
(a) Give the unit of rate constant of a reaction.
(b) In what condition order and molecularity of a reaction becomes equal.
(c) In which type of reaction order and molecularity are different.
(d) Write expression for rate constant of first-order reaction. Mention unit of rate constant of a first-order reaction.
OR
(d) Show that half-life period of a first-order reaction is independent of initial concentration of the reactant.
Answer:
(a) Unit of rate constant = (Concentration)^1-n Time^-1
When n order of reaction
(b) In elementary reaction, order and molecularity are same.
(c) In complex multistep reaction, order and molecularity are different.
(d) Rate = K [R]¹, unit of k in first order reaction rate.
= \(\frac{\mathrm{mol} \cdot \mathrm{L}^{-1}}{s} \times \frac{1}{\left(\mathrm{~mol}, \mathrm{~L}^{-1}\right)} \) = s^-1
OR
(d) From First order reaction:
k= \(\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]} \)
At t_1/2‘[R] = \(\frac{[\mathrm{R}]_{0}}{2}\)

So, in a first-order reaction, t_1/2 is independent of initial concentration.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 6 with Solutions

Max. Marks: 35
Time: 2 Hours

General Instructions:
Read the following instructions carefully:

• There are 12 questions in this question paper with internal choice.
• Section A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• Section B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• Section C- Q. No. 12 is case-based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A

Question 1.
Arrange the following in the order of their property indicated (Any two):

decreasing values of Ka.
(b) Ethanol (I), acetic acid (II), Phenol (III), Benzoic acid (IV) decreasing values of acidic strength

increasing values of boiling point.
Answer:

(b) Ethanol (I)< Phenol (III)< acetic acid (II)< Benzoic acid (IV)

Question 2.
What pressure of H₂ would be required to make the e.m.f. of the hydrogen electrode zero in pure water at 25°C?
Answer:
The cell reaction is:
2H⁺(aq) + 2e ⇌ H₂(g)
According to Nemst equation, E° = E° – \(\frac{0.059}{2} \log \frac{p\left(\mathrm{H}_{2}\right)}{\left[10^{-7}\right]^{2}}\)
Since, E =0, [H⁺] = 10^-7 M for pure water
0=0- \(\frac{0.059}{2} \log \frac{p\left(\mathrm{H}_{2}\right)}{[\mathrm{H}]^{2}}\)
∴ 0 = \(\frac{0.059}{2} \log p\left(\mathrm{H}_{2}\right)+\frac{0.059}{2} \log 10^{-14}\)
∴ log p(H₂) = log 10^-14
∴ p(H₂) = 10 ^-14atm.

Enter the percent yield and theoretical yield in the below online actual yield calculator and then click calculate button to find the answer.

Question 3.
Answer the following questions:
(a) The boiling points of aldehydes and ketones are lower than that of the corresponding alcohols and acids? Explain why.
(b) Oxidation of toluene to benzaldehyde with CrO₃ is carried out in pressure of acetic anhydride. Why?
Answer:
(a) The boiling points of aldehydes and ketones are lower than that of corresponding alcohols and acids due to absence of intermolecular hydrogen bonding in aldehydes and ketones.

(b) Oxidation of toluene to benzaldehyde is carried out with CrO₃ in presence of acetic anhydride as it traps aldehyde as gem diacetate and further oxidation does not takes place to give carboxylic acid.

Section – B

Question 4.
Account for the following:
(a) Tertiary amines do not undergo acylation reaction.
(b) Silver chloride dissolves in methylamine solution.
(c) Amines are more basic than comparable alcohols.
OR
Convert the following:
(a) Aniline to chlorobenzene
(b) Nitrobenzene to phenol
(c) Aniline to benzoic acid
Answer:
(a) Due to the lack of acidic hydrogen, tertiary amines do not undergo acylation reaction.
(b) Methylamine can act as a base or a ligand, with silver chloride it acts as a ligand. A silver methylamine complex is formed:
AgCl + 2CH₃NH₂ [Ag(CH₃NH₂)₂]⁺ Cl^– Soluble complex
(c) Lone pair of ‘N’ is easily available for donation compared to oxygen as ‘N’ is less electronegative than ‘O’. Therefore amines are more basic than corresponding alcohols.

Question 5.
(a) Give an example of an organometallic compound having sandwich structure,
(b) Write IUPAC name of the following compounds—
(i) K₃[Fe(C₂O₄)₃]
(ii) [Co(NH₃)₆]Cl₃.
OR
(a) Cu(OH)₂ is soluble in NH₄OH but not is NaOH solution. Why?
(b) Draw the structure of Zeise’s salt.
(c) Name the central metal atom present in hemoglobin and chlorophyll.
Answer:
(a) Ferrocene
(b)
(i) Potassium trio xalatoferrate (III)
(ii) Hexaamine cobalt (III) chloride
OR
(a) Cu(OH)₂ is soluble is NH₄OH due to the formation of soluble complex [Cu(NH₃)₂](OH)₂.
But no such complex is formed by the reaction of Cu(OH)₂ with NaOH.

(c) Iron (Fe) and Magnesium (Mg).

Question 6.
Account for the following:
(a) Europium(II) is more stable than Cerium(II)
(b) Transition elements from interstitial compounds.
(c) Separation of Zr and Hf from a mixture is difficult.
Answer:
(a) Electronic configuration Eu²⁺ is [Xe] 4f⁷ 5d⁰ while Ce²⁺ is [Xe]
4f²5d⁰. Hence, Eu(II) has stable configuration as d-orbital is half filled whereas Ce²⁺ configuration has no such extra stability.

(b) In transition metals small size atoms like carbon, boron, nitrogen etc., occupy the interstices or holes present in the metal lattice. e.g., TiC, VH_0.6, Fe₃H. These compounds are more malleable, have a high melting point and are chemically inert.

(c) Separation of Zr and Hf from a mixture is difficult due to their similar atomic radii they have common physical properties due to which their separation is difficult.

Question 7.
Two molecules of organic compound ‘A’ on treatment with a strong base gives two compounds ‘B’ and ‘C’. compound ‘B’ on dehydration with Cu gives A’ while acidification of ‘C’ yields carboxylic acid ‘D’ having molecular formula of CH₂O₂. Identify the A, B, C, D.
Answer:

Question 8.
Depict the galvanic cell in which the given reaction
Zn(s)+ 2Ag⁺_(aq) → Zn²⁺_(aq) + 2Ag_(s) takes place. Further show:
(a) Which electrode is negatively charged?
(b) The carriers of the current in the cell.
(c) Individual reaction at each electrode.
Answer:
The galvanic cell in which the given reaction takes place is1 depicted as:
Zn_(s) | Zn²⁺_(aq) || Ag⁺_(aq) |Ag_(s)
(a) Zn electrode (anode) is negative charged.
(b) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.
Ions are Zn²⁺ and Ag⁺.
(c) The reaction takmg place at the anode is given by’
Zn_(s) → Zn²⁺_(aq) + 2e^–
The reaction taking place at the cathode is given by,
Ag⁺_(aq) + e ^– → Ag_(s).

Question 9.
What happens when:
(a) Amines react with carboxylic acid.
(b) Aniline reacts with bromine water.
(b) Aniline reacts with cone. HNOs and cone. H₂SO₄.
OR
(a) Write IUPAC name of (CH₃)₂ CH – NH₂
(b) Complete the following:

Answer:
(a) Amines react with carboxylic acid to form salt.

(b) Aniline reacts with bromine water to form 2, 4,. 6-tribromoaniline

(c) Aniline reacts with conc. HNO₃ and Conc. H₂SO₄ to form o, m, and p-nitro aniline in different
proportions.

OR
(a) Propan-2-amine.

Question 10.
Explain what happens when an opposing external voltage is applied on a galvanic cell when it is(a) smaller than (b) equal to (c) greater than cell potential.
Answer:
(a) Cell performs normally. Electrons flow from anode to cathode and conventional current in the opposite way.
(b) No current flow, The external voltage balances cell potential, hence equilibrium is reached.
(c) The cell now acts as an electrolytic cell, zinc acts as cathode and copper acts as anode. Flow of charge is from copper to zinc and conventional current flows from zinc to copper electrode.

Question 11.
Answer the following questions:
(a) Why is KMnO₄ solution used to clean surgical instruments in hospitals?
(b) Out of cobalt and zinc salts, which is attracted in a magnetic field. Explain with reasons.
(c) Name a transition element which does not exhibit variable oxidation states.
OR
Answer the following questions:
(a) The melting and boiling points of Zn, Cd, and Hg are low. Why?
(b) Write chemical equation for the disproportionation of Mn (IV) in acidic solution.
(c) Why do Zr and Hf exhibit similar properties?
Answer:
(a) KMnO₄ solution is used to clean surgical instruments because of the germicidal action of KMnO₄.
(b) In cobalt salts, Co has d⁷ electrons in the outer shell and hence it has three unpaired electrons.
Therefore it will be attracted in a magnetic field. But in case of Zr, it has got d¹⁰ configuration, and therefore, will not be attracted by magnetic field.

(c) Scandium (Z = 21) does not exhibit variable oxidation states.
OR
(a) The metallic bonds present in Zn, Cd and Hg are weak as the electrons in d-subshell are paired for them. Hence, they have low melting and boiling points.
(b) 2MnO₄^2- +4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O.
(c) Hf and Zn have almost similar size due to lanthanoid contraction. Hence, their properties are similar.

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
All chemical reactions proceed through one or more transition-state intermediates whose content of free energy is greater than that of either the reactants or the products. For the simple reaction
R (reactants) ⇌ P (products),
we can write , where S is the reaction intermediate with the highest free energy; K± is the equilibrium constant for the reaction R ⇌ S, the conversion of the reactant to the high-energy intermediate S; and v is the rate constant for conversion of S into the product P.

The energetic relation between the initial reactants and the products of a reaction can usually be depicted as shown in Figure given below. The free energy of activation ΔG± is equal to the difference in free energy between the transition-state intermediate S and the reactant R.

Because ΔG± generally has a very large positive value, only a small fraction of the reactant molecules will at any one time have acquired this free energy, and the overall rate of the reaction will be limited by the rate of formation of S.

(a) What is the main difference between a photosensitizer and a catalyst?
(b) Give reason, why the rate of a reaction generally increases with rise in temperature.
(c) A reaction is found to be zero order, will its molecularity be zero?
(d) A reaction is second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is:
(i) doubled and
(ii) reduced to half.
OR
(d) Why order of a reaction cannot be determined by looking at the balanced chemical reaction?
Answer:
(a) A catalyst can only change the speed of the reaction while a photosensitizer only initiates the reaction.
(b) Increase in temperature causes a total increase in the energy of the reacting species. Therefore, more and more reacting species are able to cross the activation energy required to form the product. The hence overall rate of reaction increases.
(c) No, molecularity of a reaction can not be zero.
(d) Rate = k[A]².
(i) When concentration of ‘A’ is doubled, the rate becomes 4 times.
(ii) When concentration of ‘A’ is reduced to half, the rate becomes ¼ times.
OR
(d) The sum of the stoichiometric coefficients of the reactants in a balanced chemical reaction displays the total number of moles involved in the reacting species but may or may not depict the correct order of the reaction. If the reaction is not an elementary reaction then only the slowest step reactants decides the order of the reaction.

Students can access the CBSE Sample Papers for Class 12 Computer Science with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Computer Science Term 2 Set 2 with Solutions

Time: 2 Hours
Maximum Marks: 35

General Instructions:

• The question paper is divided into 3 sections – A, B and C
• Section A, consists of 7 questions (1 – 7). Each question carries 2 marks.
• Section B, consists of 3 questions (8 – 10). Each question carries 3 marks.
• Section C, consists of 3 questions (11 – 13). Each question carries 4 marks.
• Internal choices have been given for question numbers 7, 8 and 12.

Section – A
(2 Marks Each)

Question 1.
Write the types of operations used in stack.
Answer:
There are two types of operations in Stack:

• Push- To add data into the stack.
• Pop- To remove data from the stack.

Question 2.
(a) Expand the following:
(i) GSM
(ii) GPRS
Answer:

• GSM: Global System for Mobile communication.
• GPRS: General Packet Radio Service.

(b) Which type of network out of LAN, PAN and MAN is formed, when you connect two mobiles using Bluetooth to transfer a video?
Answer:
PAN (Personal Area Network)

Question 3.
Differentiate between Cardinality and Degree of a table with the help of an example.
Answer:
Cardinality is defined as the number of rows in a table. Degree is the number of columns in a table.
e.g. consider the following table

Cno	Cname
A101	Rahul
A102	Riya
A103	Darsh

Cardinality: 3 Degree: 2

Question 4.
db = MySQLdb.connect(‘localhost’ , ‘LibMan’, ‘isspwd’, ‘Library’)
(a) What is “Library” in the statement?
Answer:
(a) Database name

(b) What is the use of connect()?
Answer:
connect() function is used to connect or establish a connection with My SQL database.

Question 5.
Write the output of the queries (a) to (d) based on the table, Grocery given below:
Table: Grocery

Pr_Code	Pr_Name	PrJPrice	Pr_Manufac
P101	Toothpaste	99	Dantkanti
PI 02	Soap	20	Lux
P103	Soap	25	Breeze
P104	Detergent	199	Surf Excel
PI 05	Shampoo	345	Loreal

(a) SELECT Pr_Name, Avg(Pr_Price) FROM Grocery Group BY Pr_Name;
Answer:

Pr_Code	Avg (Pr_Price)
P101	199.00
PI 02	345.00
P103	22.50
P104	99.00

(b) SELECT DISTINCT Pr_Manufac FROM Grocery;
Answer:

Pr_ Manfac

Dantikanti

Lux

Breeze

Surf Execel

Loreal

(c) SELECT COUNT(DISTINCT Pr_Name) FROM Grocery;
Answer:
4

(d) SELECT Pr_Name, MAX(Pr_Price) , MIN(Pr_Price) FROM Grocery GROUP BY Pr_ Name
Answer:

Pr_Name	MAX(Pr_Price)	MIN(Pr_Price)
Detergent	199	199
Shampoo	345	345
Soap	25	20
Toothpaste	99	99

Question 6.
(a) MySQL supports different character sets, which command is used to display all character sets?
Answer:
SHOW CHARACTER SET

(b) What will happen if the data being loaded into a text column exceeds the maximum size of that
type?
Answer:
Data will be truncated

Question 7.
Observe the following table and answer the parts (a) and (b):
Table: Store

Item Code	Items	Qty	Rate
10	Gel Pen classic	1150	25
11	Sharpener	1500	10
12	Ball Pen 0.5	1600	12
13	Eraser	1600	5
15	Ball Pen 0.25	800	20

(a) In the above table, can we have Qty as primary key. [Answer as Yes/No.]. Justify your answer.
Answer:
No, Because there is a duplication of values and primary key value cannot be duplicate.

(b) What is the cardinality and degree of the above table?
Answer:
Degree : 4 Cardinality : 5

OR

(a) Identify the candidate key(s) from the table store.
Answer:
Item Code and Items

(b) Consider the following table Company:

Comp Code	Cost TJ;	Manufacturer	Item Code
C101	15	ABC	10
C202	7	ABC	11
C303	10	XYZ	12
C404	2	PQR	13
C505	12	MNO	15

Which field will be considered as the foreign key if the tables Store and Company are related in a database?
Answer:
Item Code

Section – B
( 3 Marks Each)

Question 8.
Write a function in Python PUSH (Arr), where Arr is a list of numbers. From this list push all numbers divisible by 5 into a stack implemented by using a list. Display the stack if it has at least one element, otherwise display appropriate error message.
OR
Write a function in Python POP(Arr), where Arr is a stack implemented by a list of numbers. The function returns the value deleted from the stack.
Answer:
def PUSH (Arr, value):
s = [ ]
for x in range (0, len (Arr)): if Arr [x] % 5==0:
s.append (Arr [x]) if len (s) ==0:
print (“Empty stack”) else:
print (s)

OR

def popStack (st):
if len (st) == 0:# if stack is empty print (“Underflow”) else:
l=len (st) val=st[1-1] print (val) st.pop(1-1) return val

Question 9.
(a) A table FLIGHT has 4 rows and 2 columns and another table AIRHOSTESS has 3 rows and 4 columns. How many rows and columns will be there if we obtain the Cartesian product of these two tables?
Answer: Total number of rows will be 12 and total number of columns will be 6.

(b) There is a column Salary in a Table EMPLOYEE. The following two statements are giving different outputs. What may be the possible reason?
SELECT COUNT(*) FROM EMPLOYEE;
SELECT COUNT(Salary) FROM EMPLOYEE;
Answer:
SELECT COUNT (*)
FROM EMPLOYEE:
This statement returns the number of records in the table.
SELECT COUNT(Salary)
FROM EMPLOYEE;
This statement returns the number of values (NULL values will not be counted) of the specified column.

Question 10.
Hina has created database named employee in MySQL.
She needs create table employee – Info in the database to stored various employees across the country. The data employee – Info has the follwing structure:

FIELD NAME	DATA TYPE	REMARKS
Employee ID	Integer	Primary key
Employee Name	Varchar (255)
Emergency Contact Name	Varchar (255)
Phone Number	Integer
Address	Varchar (255)
City	Varchar (255)
Country	Varchar (255)

Help her to complete the task by suggesting appropriate SQL commands.
Answer:
CREATE DATABASE Employee;
CREATE TABLE Employee – Info
EmployeelD int primary key,
EmployeeName varchar (255),
EmergencyContactName varchar(255),
PhoneNumber int,
Address varchar (255),
City varchar (255),
Country varchar(255)
) ;

Section – C
(4 Marks Each)

Question 11.
Write SQL queries for (a) to (c) and find outputs for SQL queries (d), which are based on the tables.
Table: CUSTOMER

CNO	CNAME	ADDRESS
101	Richa Jain	Delhi
102	Surbhi Sinha	Chennai
103	Lisa Thomas	Bengalore
104	Imran Ali	Delhi
105	Roshan Singh	Chennai

Table: TRANSACTION

TRNO	CNO	AMOUNT	TYPE	DOT
T001	101	1500	Credit	2017-11-23
T002	103	2000	Debit	2017-05-12
T003	102	3000	Credit	2017-06-10
T004	103	12000	Credit	2017 – 09 – 12
T005	101	1000	Debit	2017 – 09 – 05

(a) To display details of all transactions of TYPE Credit from table TRANSACTION.
Answer:
SELECT * FROM TRANSACTION WHERE TYPE =”Credit”;

(b) To display the CNO and AMOUNT of all Transactions done in the month of September 2017 from table TRANSACTION.
Answer:
SELECT CNO,AMOUNT FROM TRANSACTION WHERE (MONTH(DOT)=”September” AND YEAR(DOT)=2017);

(c) To display the last date of transaction (DOT) from the table TRANSACTION and customer name from CUSTOMER for the customer having CNO as 103.
Answer:
SELECT MAX(DOT), CNAME FROM TRANSACTION, CUSTOMER WHERE CNO=”103″ AND CUSTOMER.CNO=TRANSACTION.CNO:

(d) SELECT CNO, COUNT(*), MAX (AMOUNT) FROM TRANSACTION GROUP BY CNO HAVING COUNT (*)> 1;
Answer:

CNO	MAX (AMOUNT)	COUNT(*)
101	1500	2
103	12,000	2

Question 12.
(a) Identify the type of topology from the following:
(i) Each node is connected with the help of a single cable.
Answer:
Bus topology

(ii) Each node is connected with central switching through independent cables.
Answer:
Star topology

OR

Server side scripting : ASP, JSP
Client side scripting : JavaScript, VBScript
OR
Name two server side scripting language and two client side scripting language.
(b) What is the difference between packet and message switching?
Answer:

Packet switching	Message switching
1. There is a tight upper limit on the block size. A fixed size of packet is specified.	1. In message switching there was no upper limit.
2. All the packets are stored in main memory in switching office.	2. In message switching packets are stored on disk. This increases the performance as access time is reduced.

Question 13.
Meerut school in Meerut is starting up the network between its different wings. There are four buildings named as S, J, A and H. The distance between various buildings is as follows:

A to S	200 m
A to J	150 m
A to H	50 m
S to J	250 m
StoH	350 m
J to H	350 m

Numbers of computers in each Buildings

S	130
J	80
A	160
H	50

(a) What type of topology is best suited for above network?
Answer:
Star topology

(b) Suggest the most suitable place (i.e. building) to house the server of this school.
Answer:
Building A

(c) The school is planning to link its head office situated in New Delhi with the offices in hilly areas. Suggest a way to connect it economically.
Answer:
Radio waves would be an economic way to connect it.

(d) The company wants internet accessibility in all the blocks. What is the suitable and cost-effective technology for that?
Answer:
Broadband 

Converter 350 Degrees F To C.

Students can access the CBSE Sample Papers for Class 12 Accountancy with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Accountancy Standard Term 2 Set 1 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions

• This question paper comprises two Parts A and B. There are 12 questions in the question paper. All questions are compulsory.
• Question nos. 1 to 3 and 10 are Short Answer Type I Questions carrying 2 marks each.
• Question nos. 4 to 6 and 11 are Short Answer Type II Questions carrying 3 marks each.
• Question nos. 7 to 9 and 12 are Long Answer Type Questions earning 5 marks each.
• There is no overall choice. However, an internal choice has been provided in 3 questions of three marks and 1 question of five marks.

Maximum Marks : 40
Time : 2 Hours

Part A
(Accounting for Not-for-Profit Organisations, Partnership Firms and Companies)

Question 1.
Following information has been provided by M/s Achyut Health Care. You are
required to calculate the amount of medicines consumed during the year 2020-21 (2)

Particulars	Amt (₹)
Stock of Medicines as on 1st April, 2020	15,00,000
Creditors for Medicines as on 1st April, 2020	3,50,000
Stock of Medicines as on 31st March, 2021	10,00,000
Creditors for Medicines as on 31st March, 2021	4,20,000
Cash Purchases Of Medicines During the Year 2020-21	2,00,000
Credit Purchases of Medicines During the Year 2020-21	6,00,000

Answer:
Calculation of amount of medicines consumed during the year 2020-21.

Question 2.
Distinguish between ‘dissolution of partnership’ and ‘dissolution of partnership firm’ based on
(i) Settlement of assets and liabilities
(ii) Economic relationship (2)
Answer:

Basis	Dissolution of Partnership	Dissolution of Partnership Firm
Settlement of Assets and Liabilities	Assets are revalued and liabilities are reassessed.	Assets are sold and liabilities are paid off.
Economic Relationship	Economic relationship between the partners continues, though in a changed form.	Economic relationship between the partners comes to an end.

Question 3.
Suresh, Ramesh and Tushar were partners of a firm sharing profits in the ratio of 6 : 5 : 4. Ramesh retired and his capital after making adjustments on account of reserves, revaluation of assets and reassessment of liabilities stood at ₹ 2,50,400. Suresh and Tushar agreed to pay him ₹ 2,90,000 in full settlement of his claim.
Pass necessary journal entry for the treatment of goodwill. Show workings clearly. (2)
Answer:

Working Notes

2. Gaining Ratio between Suresh and Tushar is 6 : 4

Question 4.
From the following information given by Modern Dance Academy, calculate the amount of subscription received during the year 2020-21.
(i) Subscription credited to income and expenditure account for the year ending 31st March, 2021 amounted to ₹ 3,00,000 and each member is required to pay an annual subscription of ₹ 3,000.
(ii) Subscription in arrears as on 1st April, 2020 amounted to ₹ 16,000.
(iii) During the year 2020-21,10 members made partial payment of ₹ 26,000 towards subscription, 8 members failed to pay the subscription amount and 5 members paid the subscription amount for the year 2021-22.
(iv) During the year 2019-20,12 members paid the subscription amount for the year 2020-21. (3)
Or
Following information is given by Alchemy Medical College, Library department for the year 2020-21.

Particulars	Amt (₹)
Books and Journals Fund as on 1st April, 2020	4,50,000
7% Books and Journals Fund Investments as on 1st April, 2020	4,00,000
Interest on Books and Journals Fund Investments	13,000
Donations for Books and Journals	20,000
Books Purchased	70,000
General Fund as on 1st April, 2020	10,00,000

Show the accounting treatment of the above mentioned items in the balance sheet qf the Alchemy Medical College as at 31st March, 2021.
Answer:
Calculation of amount of subscription received during the year 2020-21

Or

Working Note
Interest on Books and Journals Investments = 4,00,000 × 7 /100 = ₹ 28,000
Accrued Interest = 28,000 -13,100 = ₹ 15,000

Question 5.
Harihar, Hemang and Harit were partners with fixed capitals of ₹ 3,00,000, ₹ 2,00,000 and ₹ 1,00,000 respectively. They shared profits in the ratio of their fixed capitals. Harit died on 31st May, 2020, whereas the firm closes its books of accounts on 31st March every year. According to their partnership deed, Harit’s representatives would be entitled to get share in the interim profits of the firm on the basis of sales. Sales and profit for the year 2019-20 amounted to ₹ 8,00,000 and ₹ 2,40,000 respectively and sales from 1st April, 2020 to 31st May, 2020 amounted to X1,50,000. The rate of profit to sales remained constant during these two years. You are required to
(i) Calculate Harit’s share in profit.
(ii) Pass journal entry to record Harit’s share in profit. (3)
Answer:
(i) Ratio of Profit to Sales = \(\frac{2,40,000}{8,00,000}[latex] × 100 = 30%
8/10/100
Profit upto the date of death = 1,50/100 × 30% = ₹ 45,000
Profit sharing Ratio between Harihar, Hemang and Harit = 3 : 2 : 1
∴ Harit’s Share of Profit = 45,000 × 1/6= ₹ 7,500

Note As capitals are fixed, therefore profit is transferred to Harit’s current account.

Question 6.
Vedesh Ltd. purchased a running business of Vibhu Enterprises for a sum of ₹ 12,00,000. Vedesh Ltd. paid ₹ 60,000 by drawing a promissory note in favour of Vibhu Enterprises, ₹ 1,90,000 through bank draft and balance by issue of 8% debentures of ₹ 100 each at a discount of 5%. The assets and liabilities of Vibhu Enterprises consisted of fixed assets valued at ₹ 17,30,000 and trade payables at ₹ 3,20,000.
You are required to pass necessary journal entries in the books of Vedesh Ltd. (3)
Or
Youth Ltd. took a loan of ₹ 15,00,000 from State Bank of India against the security of tangible assets. In addition to principal security, it issued 10,000 11% debentures of ₹ 100 each as collateral security. Pass necessary journal entries for the above transactions, if the company decided to record the issue of 11% debentures as collateral security and show the presentation in the balance sheet of Youth Ltd.
Answer:


Working Note
Amount due to Vibhu Enterprises after drawing bills payable and bank overdraft
= 12,00,000 – 60,000 -1,90,000 = ₹ 9,50,000
Number of Debentures Issued = [latex]\frac{9,50,000}{95}\) = 10,000 Debentures
Or

Notes to Accounts

Question 7.
Madhav, Madhusudan and Mukund were partners in Jaganath Associates. They decided to dissolve the firm on 31st March, 2021.
Pass necessary journal entries for the following transactions after various assets (other than cash) and third-party liabilities have been transferred to realisation account
(i) Old machine fully written-off was sold for ₹ 42,000 while a payment of ₹ 6,000 is made to bank for a bill discounted being dishonoured.

(ii) Madhusudan accepted an unrecorded asset of ₹ 80,000 at ₹ 75,000 and the balance through cheque, against the payment of his loan to the firm of ₹ 1,00,000.

(iii) Stock of book value of ₹ 30,000 was taken by Madhav, Madhusudan and Mukund in their profit sharing ratio.

(iv) The firm had paid realisation expenses amounting to ₹ 5,000 on behalf of Mukund.

(v) There was a vehicle loan of ₹ 2,00,000 which was paid by surrender of asset to the
bank at an agreed value of ₹ 1,40,000 and the shortfall was met from firm’s bank account. (5)
Or
Gini, Bini and Mini were in partnership sharing profits and losses in the ratio of 5 : 2 : 2. Their balance sheet as at 31st March, 2021 was as follows

On 31st March, 2021, Gini retired from the firm. All the partners agreed to revalue the assets and liabilities on the following basis
(i) Bad debts amounted to ₹ 5,000. A provision for doubtful debts was to be maintained at 10% on debtors.
(ii) Partners have decided to write-off existing goodwill.
(iii) Goodwill of the firm was valued at ₹ 54,000 and be adjusted into the capital accounts of Bini and Mini, who will share profits in future in the ratio of 5:4.
(iv) The assets and liabilities valued as Inventories ₹ 1,30,000; Machinery ₹ 82,000; Furniture ₹1,95,000 and Building ₹ 6,00,000.
(v) Liability of ₹ 23,000 is to be created on account of claim for workmen compensation.
(vi) There was an unrecorded investment in shares of ₹ 25,000. It was decided to pay-off Gini by giving her unrecorded investment in full settlement of her part payment of ₹ 28,000 and remaining amount after two months.
Prepare Revaluation Account and Partners’ Capital Accounts as on 31st March, 2021.
Answer:

Or


Working Note
Calculation of Gaining Ratio
Bini’s Gain = = \(\frac{5}{9}-\frac{2}{9}=\frac{3}{9}\); Mini’s Gain = \(\frac{4}{9}-\frac{2}{9}=\frac{2}{9}\)
∴ Gaining Ratio between Bini and Mini = \(\frac{3}{9}: \frac{2}{9}\) = 3 : 2 Good will the Firm = ₹ 54,000
Gini’s Share of goodwill = 54,000 × \(\frac{5}{9}\) = ₹ 30,000

The pH calculator tool provides expected pH values for a variety of common laboratory and industrial chemicals.

Question 8.
Yogadatra Ltd. (pharmaceutical company) appointed marketing expert, Mr. Kartikay as the CEO of the company, with a target to penetrate their roots in the rural regions. Mr. Kartikay discussed the ways and means to achieve target of the company with financial, production and marketing departmental heads and asked the finance manager to prepare the budget. After reviewing the suggestions given by all the departmental heads, the finance manager proposed requirement of an additional fund of ₹ 52,50,000.

Yogadatra Ltd. is a zero-debt company. To avail the benefits of financial leverage, the finance manager proposed to include debt in the capital structure. After deliberations, on 1st April, 2020, the board of directors had decided to issue 6% debentures of ₹ 100 each to the public at a premium of 5%, redeemable after 5 years at ₹ 110 per share.

You are required to answer the following questions
(i) Calculate the number of debentures to be issued to raise additional funds.
(ii) Pass journal entry for the allotment of debentures.
(iii) Pass journal entry to write-off loss on issue of debentures.
(iv) Calculate the amount of annual fixed obligation associated with debentures.
(v) Prepare loss on issue of debentures account. (5)
Answer:
(i) Number of Debentures to be Issued = \(\frac{52,50,000}{105}\) = 50,000 Debentures

(iv) Interest on debentures is the annual fixed obligation associated debentures.
Thus, interest on 6% debentures = 50,00,000 × 6 /100 = ₹ 3,00,000

Question 9.
From the following receipts and payments account and additional information provided by Ramanath Club, prepare income and expenditure account for the year ending on 31st March, 2021.

Additional Information
(i) Subscription received during the year includes ₹ 25,000 as donation for building.
(ii) Telephone bill unpaid as on 31st March, 2020 was ₹ 4,000 and on 31st March, 2021 ₹ 2,600.
(iii) Value of 8% government securities on 31st March, 2020 was ₹ 80,000.
(iv) Additional government securities worth ₹ 30,000 were purchased on 31st March, 2021. (5)
Answer:

Note No interest will be provided on government securities worth ₹ 30,000 as they are purchased on the last day of financial year.

Part B
(Analysis of Financial Statements)

Question 10.
State whether the following transactions will result in inflow, outflow or no flow of cash while preparing cash flow statement
(i) Decrease in outstanding employees benefits by ₹ 3,000.
(ii) Increase in current investment by ₹ 6,000. (2)
Answer:
(i) Outflow Decrease in outstanding employee benefit will be deducted as it is a decrease in current
liability. Thus, it is an outflow of cash.
(ii) No Flow Increase in current investment (marketable) is included in cash and cash equivalents. Thus, no flow of cash.

Question 11.
From the following details provided by Kumud Ltd., prepare comparative statement of profit and loss for the year ended 31st March, 2021. (3)

Particulars	31st March, 2020	31st March, 2021
Revenue from Operations	₹ 30,00,000	₹ 35,00,000
Other Income	₹ 3,00,000	₹ 4,50,000
Cost of Materials Consumed	₹ 20,00,000	₹ 23,00,000
Other Expenses	₹ 1,00,000	₹ 1,20,000
Tax Rate	40%	40%

Or
From the following balance sheets of Vinayak Ltd. as at 31st March,2021, prepare a common size balance sheet


Answer:


Or

Question 12.
On the basis of information given by Aradhana Ltd., prepare cash flow statement for the year ending 31st March, 2021

Notes to Accounts

Additional Information
(i) Debentures were redeemed on 1st April, 2020.
(ii) Tax paid during the year ₹ 2,80,000. (5)
Answer:

Working Notes

Students can access the CBSE Sample Papers for Class 12 Applied Mathematics with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Informatics Practices Term 2 Set 2 with Solutions

Time: 2 Hours
Maximum Marks: 40

General Instructions:

• The question paper is divided into 3 sections -A, B and C
• Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
• Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
• Section C comprises of 4 questions of 4 marks each. It contains one case study based question. Internal choice has been provided in one question.

Section – A (2 Marks)

Question 1.
Evaluate: ∫\(\frac{e^{x}(x-3)}{(x-1)^{3}}\)dx
OR
The marginal cost of producing x units of a product is given by the marginal cost function as MC = \(\sqrt{x+1}\). The cost of producing 3 units is ₹ 7800. Find the cost function.
Answer:

Commonly Made Error:
Some candidates use product rule in ∫e^x {f(x) + f'(x)} in both parts which is wrong. Students should apply the rule that ∫e^x {f(x) + f'(x)} = e^xf(x) + C only.

Answering Tip: ∫e^x{f(x) + f'(x)} is specific situation in integration by parts. This needs to be sufficiently practiced by students.

OR

We know that,
Cost Function, C(x) = ∫MC(x)dx
where MC(x) is Marginal Cost, Function
C(x) = ∫\(\sqrt{x+1}\) dx ……… (1)
Let x + 1 = t dx = dt
So, C(t) = ∫\(\sqrt{t}\) dt

which is the required cost function

Question 2.
What is the present value of a sequence of payment of ₹ 1000 made at the end of every 6 months and continuing forever, if money is worth 8% per annum compounded semi-annually ?
Answer:
The given annuity is a perpetuity.
present value of perpetuity = \(\frac{\text { Cash flow }}{\text { Interest rate }}\)
Here, cash flow = ₹ 1000
interest rate = \(\frac{8 / 2}{100}\)
= \(\frac{4}{100}\) = 0.04
So, present value = \(\frac{1000}{0.04}\)
= ₹ 25,000

Question 3.
A Machine purchased two years ago. If its value depreciates at the annual rate of 10 %.The present value of the machine is ₹ 97200,then find the value of the machine after 3 years?
OR
If the present value of a perpetuity of ₹ 600 payable at end of every 6 months is ₹ 10,000, then find the rate of interest.
Answer:
Given, P = ₹ 97,200, i = 10% p.a.
⇒ i = \(\frac{10}{100}\) = 0.1
So, value after 3 years
= 97,200 × (1 – 0.1)³
= 97,200 × 0.729
= ₹ 70,858.80
= ₹ 70,859 (approx.)

OR

Let rate of interest be r% per annum,
then i = \(\frac{r}{200}\) …… (i)
Given, R = ₹ 600 and P = ₹ 10,000
Using P = \(\frac{R}{i}\)
⇒ i = \(\frac{R}{P}=\frac{600}{10,000}\) …. (ii)
From equations (i) & (ii)
⇒ \(\frac{r}{200}=\frac{600}{10,000}\) [∴ i = \(\frac{r}{200}\)]
⇒ r = \(\frac{120,000}{10,000}\)
= 12%
∴ Rate of interest is 12%

Question 4.
The standard deviation of a sample of size 50 is 6.3. Determine the standard error whose population’s standard deviation is 6.
Answer:
Sample size n = 50
Sample S.D. s = 6.3
Population S.D. σ = 6
The standard error for sample S.D. is given by

Thus standard error for samples S.D. = 0.6

Question 5.
From the following time series obtain trend value by 3 yearly moving averages.

Answer:
Calculation of trend values by three yearly moving average method.

Question 6.
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300. Formulate a LPP for finding how many of each should be produced daily to maximize the profit ? It is being given that at least one of each must be produced.
Answer:
Let x be the number of necklaces manufactured and y be the number of bracelets manufactured. The total number of necklaces and bracelets it can handle is at most 24.
x + y ≤ 24 …(i)
The x items takes x hours to manufacture and y items take \(\frac{y}{2}\) hours to manufacture and the y 2 maximum time available is 16 hours. So,
x + \(\frac{y}{2}\) ≤16
2x + y ≤ 32 …(ii)
The profit on one necklace is given as ₹ 100 and the profit on one bracelet is given as ₹ 300. Let the profit be Z. To maximize the profit,
Z = 100x + 300y
Therefore, the required LPP is
Max. Z = 100x + 300y
Subject to constraints,
x + y ≤ 24
2x + y ≤ 32
x, y ≥ 1
x > 0, y > 0

Section – B [3 marks each]

Question 7.
If the demand curve is given by D(x) = 50 – 0.06x². Find the surplus or profit of the consumers if the level of sale amounts to 20 units.
Answer:
As the number of units is 20, its price rises up = D (20)
= 50 – 0.06 × 20² = 26.
The profit of consumer’s:

= 50(20 – 0) – 0.06\(\left[\frac{20^{3}}{3}-0\right]\) – 520
= 1000 – 160 – 520
= 1000 – 680
= 320
The consumer’s gain is ₹ 320, if the level of sales is twenty units.

Question 8.
Fit a straight line trend by the method of least square to the following data on sales (₹ in lakhs) for the period 2011-2018.

OR
Calculate trend values from the following data assuming 5- yearly moving averages.

Answer:
x_i = t_i – A, Here A = 2014 + 2015/2 = 2014.5

Now,
a = \(\frac{\Sigma y_{i}}{n}\) = \(\frac{1052}{8}\) = 131.5
b = \(\frac{\Sigma x_{i} y_{i}}{\Sigma x_{i}^{2}}\) = \(\frac{1232}{168}\) = 7.33
Sp, tend equation is
y = 131.5 + 7.33x

OR

Calculation of Trend values by moving average method:

Commonly Made Error: Few students find the calculation tedious in such type of questions.

Answering Tip: Do not leave the question in the middle. Do calculation with little patience.

Question 9.
Find the student’s -t for the following variable values in a sample of eight:
– 4, – 2, – 2, 0, 2, 2, 3, 3 taking the mean of the universe to be zero.
Answer:

x	X – x̄	(x – x̄)2
-4	-4.25	18.0625
-2	-2.25	5.0625
-2	-2.25	5.0625
0	-0.25	0.0625
2	1.75	3.0625
2	1.75	3.0625
3	2.75	7.5625
3	2.75	7.5625
Σ  x =  2		Σ (x – x̄)2 =  49.5000

X̄ = mean
= \(\frac{\sum x}{n}\)
= \(\frac{2}{8}\)
= 0.25
Now, compute the standard deviation using formula as,

H₀ = The mean of universe, μ = 0, we get

Question 10.
A company borrowed ₹ 60,000 for renovation. The company plans to set up a sinking fund that will pay back the loan at the end of 5 years. Assuming a rate of 10% compounded semi-annually, Find the amount of sinking fund of the ordinary annuity. Calculate the amount of the sinking fund. [Given that (1.05)¹⁰ = 1.06288.]
Answer:
Given, P = ₹ 60,000, r = 10% or 0.10, No. of years, n = 5 years and No. of payments per year, m = 2 (Semi annually)
Sinking Fund is as:

which is the required amount

Section – C (4 marks each)

Question 11.
A manufacturing company makes two types of teaching aids A and B of Mathematics for class X. Each type of A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of ₹ 80 on each piece of type A and ₹ 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.
Answer:
Let the number of two types of teaching aids A and B be x and y, respectively.
Writing the given data in tabular as

The profit on type A is ₹ 80 and type B is ₹ 120.
Thus, the required L.EE is
Max. Z = 80x + 120y …….. (i)
Subject to constraints
9x + 12y ≤ 180
or 3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0, y ≥ 0
Consider the equation,
3x + 4y = 60

x	0	20
y	15	0

and
x + 3y = 30

x	0	30
y	10	0

On solving eq, (ii) and (iii), we get
x = 12 and y = 6
The graphical representation of the system of in equations is as shown below

From graph, feasible region is
OABCO, whose corner points are (0, 0), A(20, 0), B(12,6) and C(0,10).
The values of Z at corner points are as follows:

Corner Points	Value of Z = 80x + 120y
O(0,0).	Z = 80 × 0 + 120 × 0 = 0
A(20, 0)	Z = 80 × 20 + 120 × 0 = 1600
B(12, 6)	Z = 80 × 12 + 120 × 6 = 1680
C(0,10)	Z = 80 × 0 + 120 × 10 = 1200

From the table, the maximum value of Z is ₹ 1680.
Hence, 12 pieces of type A and 6 pieces of type B should be manufactured per week to get maximum profit of ₹ 1680 per week.

Commonly Made Error: A number of candidates could not form the correct inequalities subject to given constraints. Some noted incorrect corner points to find optimum value. A few candidates did not draw rough sketch of the feasible region.

Answering Tip: Do adequate practice to frame inequalities using given constraints and finding corner points by solving inequalities.

Question 12.
A bond has issued with the face (Par) value of ₹ 1,000 at 10% coupon for three years The required rate
of return is 8%. What is the value of the bond if the coupon amount is payable on half-yearly basis ? [Given (1.04)^-6 = 0.79031]
Answer:
Given, P = ₹ 1,000 Annual Coupon Payment
= ₹ 1,000 × 10% = ₹ 100
Semi-annual Coupon Payment,
C = ₹100 ÷ 2 = ₹ 50
r = 8% ÷ 2 = 4%
= 0.04
N = 3 years × 2 = 6 periods for semi-annual coupon payments

Bond value is ₹ 1052.42

Application for completing products and balance the equation calculator of chemical reactions.

Question 13.
Rohan purchased a laptop of worth ₹ 80,000. He paid ₹ 20,000 as cash down payment and remaining balance in equal monthly installments in 2 years. If bank charges 9.% p.a. compounded monthly. Calculate the EMI [Given (1.0075)²⁴ = 1.1964]
OR
A firm anticipates an expenditure of ₹ 5,00,000 for plant modernization at end of 10 years from now. How much should the company deposit at the end of year into a sinking fund earning interest 5% per annum. [Given log 1.05 = 0.0212, antilog (0.2120) = 1.629]
Answer:
Given, Cost of laptop = ₹ 80,000
Down payment = ₹ 20,000
∴ Balance = ₹ 60,000
So, p = ₹ 60,000,
i = \(\frac{9}{12 \times 100}\)
= 0.0075
and n = 2 × 12 = 24

The required value of EMI is ₹ 2741.24

OR

Given, A = ₹ 5,00,000, r = 5% and n = 10

Now, let x = (1.05)¹⁰ …(iii)
Taking log both sides, we get
log x = 10 log (1.05)
= 10 × 0.0212
= 0.2120
⇒ x = antilog (0,2120)
x = 1.629 ……… (iv)
Thus, (iii) (1.05)¹⁰ = 1.629
Now, From (ii) P = \(\frac{500000 \times 0.05}{1.629-1}\)
= \(\frac{25000}{0.629}\)
P = 39745.63
Hence, the company should deposit ₹ 39745.63 every year into the sinking fund.

CASE STUDY

Question 14.
A veterinary doctor was examining a sick cat brought by a pet lover. When it was brought to the hospital, it was already dead. The pet lover wanted to find its time of death. He took the temperature of the cat at 11.30 p.m. which was 94.6°F. He took the temperature again after one hour; the temperature was lower than the first observation, it was 93.4°F. The room in which the cat was put is always at 70°F. The normal temperature of the cat is taken as 98.6°F when it was alive. The doctor estimated the time of death using Newton’s law of cooling which is governed by the differential equation: \(\frac{d T}{d t}\) ∝ (T – 70), where 70°F is the room temperature and T is the temperature of the object at time t.
Substituting the two different observations of T and t made, in the solution of the differential equation \(\frac{d T}{d t}\) = k(T – 70) where k is a constant of proportion, time of death is calculated.

(i) What is the degree of the differential equation? Also, find the solution of the differential equation. (2)
Answer:
The given differential equation is
\(\frac{d T}{d t}\) = k(T – 70)
Since there is only one constant available in the given equation, therefore, degree of this differential equation is one.
The given equation can be written as:
\(\frac{d T}{(T-70)}\) = kdt
Integrating both sides we get
log|T – 70| = kt + C

(ii) Find the value of integration constant if it is given that t = 0, when T = 72. If the temperature was measured 2 hours after 11.30 p.m., will the time of death change ? (2)
Answer:
We have,
log |T – 70| = kt + C
Putting the value of t = 0, T = 72, we get
C = log 2
So the time of death is independent of all other factors.