ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test

More Exercises

Solve the following equations (1 to 4) by factorisation :

Question 1.
(i) x² + 6x – 16 = 0
(ii) 3x² + 11x + 10 = 0
Solution:
x² + 6x – 16 = 0
⇒ x² + 8x – 2x – 16 = 0
⇒ x (x + 8) – 2 (x + 8) = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q1.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q1.2

Question 2.
(i) 2x² + ax – a² = 0
(ii) √3x² + 10x + 7√3 = 0
Solution:
(i) 2x² + ax – a² = 0
⇒ 2x² + 2ax – ax – a² = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q2.2

Question 3.
(i) x(x + 1) + (x + 2)(x + 3) = 42
(ii) \(\frac { 6 }{ x } -\frac { 2 }{ x-1 } =\frac { 1 }{ x-2 } \)
Solution:
(i) x(x + 1) + (x + 2)(x + 3) = 42
⇒ 2x² + 6x + 6 – 42 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q3.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q3.2

Question 4.
(i)\(\sqrt { x+15 } =x+3 \)
(ii)\(\sqrt { { 3x }^{ 2 }-2x-1 } =2x-2\)
Solution:
(i) \(\sqrt { x+15 } =x+3 \)
Squaring on both sides
x + 15 = (x + 3)²
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q4.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q4.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q4.3

Solve the following equations (5 to 8) by using formula :

Question 5.
(i) 2x² – 3x – 1 = 0
(ii) \(x\left( 3x+\frac { 1 }{ 2 } \right) =6\)
Solution:
(i) 2x² – 3x – 1 = 0
Here a = 2, b = -3, c = -1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q5.2

Question 6.
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(ii) \(\frac { 2 }{ x+2 } -\frac { 1 }{ x+1 } =\frac { 4 }{ x+4 } -\frac { 3 }{ x+3 } \)
Solution:
(i) \(\frac { 2x+5 }{ 3x+4 } =\frac { x+1 }{ x+3 } \)
(2x + 5)(x + 3) = (x + 1)(3x + 4)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q6.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q6.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q6.3

Question 7.
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
(ii) \(\frac { 4 }{ x } -3=\frac { 5 }{ 2x+3 } ,x\neq 0,-\frac { 3 }{ 2 } \)
Solution:
(i) \(\frac { 3x-4 }{ 7 } +\frac { 7 }{ 3x-4 } =\frac { 5 }{ 2 } ,x\neq \frac { 4 }{ 3 } \)
let \(\frac { 3x-4 }{ 7 } \) = y,then
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q7.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q7.3

Question 8.
(i)x² + (4 – 3a)x – 12a = 0
(ii)10ax² – 6x + 15ax – 9 = 0,a≠0
Solution:
(i)x² + (4 – 3a)x – 12a = 0
Here a = 1, b = 4 – 3a, c = -12a
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q8.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q8.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q8.3

Question 9.
Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)
Solution:
(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q9.1

Question 10.
Discuss the nature of the roots of the following equations:
(i) 3x² – 7x + 8 = 0
(ii) x² – \(\\ \frac { 1 }{ 2 } x\) – 4 = 0
(iii) 5x² – 6√5x + 9 = 0
(iv) √3x² – 2x – √3 = 0
Solution:
(i) 3x² – 7x + 8 = 0
Here a = 3, b = -7, c = 8
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q10.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q10.2

Question 11.
Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.
Solution:
(4 – k) x² + 2 (k + 2) x + (8k + 1) = 0
Here a = (4 – k), b = 2 (k + 2), c = 8k + 1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q11.1
or k – 3 = 0, then k= 3
k = 0, 3 Ans.

Question 12.
Find the values of m so that the quadratic equation 3x² – 5x – 2m = 0 has two distinct real roots.
Solution:
3x² – 5x – 2m = 0
Here a = 3, b = -5, c = -2m
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q12.1

Question 13.
Find the value(s) of k for which each of the following quadratic equation has equal roots:
(i)3kx² = 4(kx – 1)
(ii)(k + 4)x² + (k + 1)x + 1 =0
Also, find the roots for that value (s) of k in each case.
Solution:
(i)3kx² = 4(kx – 1)
⇒ 3kx² = 4kx – 4
⇒ 3kx² – 4kx + 4 = 0
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q13.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q13.2

Question 14.
Find two natural numbers which differ by 3 and whose squares have the sum 117.
Solution:
Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q14.1

Question 15.
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:
Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q15.1

Question 16.
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
Solution:
Ratio in two natural numbers = 3 : 4
Let the numbers be 3x and 4x
According to the condition,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q16.1

Question 17.
Two squares have sides A cm and (x + 4) cm. The sum of their areas is 656 sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.
Solution:
Side of first square = x cm .
and side of second square = (x + 4) cm
Now according to the condition,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q17.1
or x – 16 = 0 then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 – 4 = 20 cm

Question 18.
The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.
Solution:
Let breadth = x m
then length = (x + 12) m
Area = l × b = x (x + 12) m²
and perimeter = 2 (l + b) = 2(x + 12 + x) = 2 (2x + 12) m
According to the condition.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q18.1

Question 19.
A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.
Solution:
Area of rectangular garden = 100 cm²
Length of barbed wire = 30 m
Let the length of the side opposite to wall = x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q19.1

Question 20.
The hypotenuse of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.
Solution:
Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q20.1

Question 21.
A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.
Solution:
Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q21.1

Question 22.
Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.
(i) Write down the number of litres of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 litres of petrol more than car B in covering 400 km. write down an equation, in A and solve it to determine the number of litres of petrol used by car B for the journey.
Solution:
Distance travelled by car A in one litre = x km
and distance travelled by car B in one litre = (x + 5) km
(i) Consumption of car A in covering 400 km
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q22.1

Question 23.
The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream
Solution:
Speed of a boat in still water = 11 km/hr
Let the speed of stream = x km/hr.
Distance covered = 12 km.
Time taken = 2 hours 45 minutes
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q23.1

Question 24.
By selling an article for Rs. 21, a trader loses as much per cent as the cost price of the article. Find the cost price.
Solution:
S.P. of an article = Rs. 21
Let cost price = Rs. x
Then loss = x%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q24.1

Question 25.
A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.
Solution:
Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q25.1

Question 26.
Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.
Solution:
Let Partap’s present age = x years
40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test Q26.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 5 Quadratic Equations in One Variable Chapter Test are helpful to complete your math homework.

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Practical Based Questions for Class 10 Science Chemistry

Practical Based Questions for Class 10 Science Chemistry

Question 1.
A student dipped a strip of pH paper in distilled water taken in a tube. As expected, the pH paper acquired green colour. He then dissolved a pinch of common salt in the same tube. What will be the expected change in colour of the pH paper ?
Answer:
There will not be any change in the colour of the pH paper. It will remain green which indicates that the solution has maintained neutral character. Actually, sodium chloride is a salt prepared from strong base (NaOH) and strong acid (HCl). It is a neutral salt and is not expected to bring any change in the pH of distilled water.

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Question 2.
(a) Five solutions A, B, C, D and E when tested with universal indicator showed pW as 4, 1, 11,7 and 9 respectively. Which solutions is :

  1. neutral
  2. strongly alkaline
  3. strongly acidic
  4. weakly acidic acid
  5. weakly alkaline.

Arrange the solutions in increasing order of H-ion concentration.
Answer:

  1. D
  2. C
  3. B
  4. A
  5. E.

The increasing order of H-ion concentration :C<E<D<A<B.

Question 3.
Name the acid and base from which the following salts have been formed :

  1. Sodium acetate
  2. Ammonium chloride
  3. Calcium nitrate
  4. Sodium carbonate.

Answer:

  1. NaOH and CH3COOH
  2. NH4OH and HCl
  3. Ca(OH)2 and HNO3
  4. NaOH and H3CO3.

Question 4.
When water is added gradually to a white solid X, a hissing sound is heard and a lot of heat is produced forming the product Y. A suspension of Y in water is applied to the walls of a house during white washing.
(a) What could be the solid X ? Write its chemical formula.
(b) What could be the product Y ? Write its chemical formula.
Answer:
(a) The white solid X is quick lime. Its chemical formula is CaO.
(b) When water is added to the white solid X, product Y formed is calcium hydroxide. Its chemical formula is Ca(OH)r

Question 5.
Two drinks P and Q gave acidic and alkaline reactions respectively. One has a pH value of 9 and the other has pH value of 3. Which drink has the pH value of 9 ?
Answer:
The solution Q is of alkaline nature. It has a pH value of 9.

Question 6.
Take five test tubes and label them as A, B, C, D and E. Add 5 mL of five unknown solutions in them. Put a small strip of universal indicator in each of them. Following colours appear in these :
Solution A — Orange;
Solution B — Green;
Solution  C – Red;
Solution D – Blue;
Solution E – Violet.
Predict the nature of the solutions in these from the pH character.
Answer:
Solution-A (Weakly acidic);
Solution-B (Neutral);
Solution-C (Strongly acidic);
Solution-D (Weakly basic);
Solution-E (Strongly basic).

Question 7.
A metal A’ gives a compound ‘B’ (molecular mass 40) when it reacts with water. Compound ‘B’ gives a soluble compound ‘C’ on treatment with aluminium oxide. Identify A’, ‘B’ and ‘C’. Also give the reactions involved.
(CBSE 2013)
Answer:
The available data suggests that the metal A sodium (Na). It reacts with water to form sodium hydroxide ‘B’ with formula NaOH. On reacting with aluminium oxide, the compound ‘B’ forms sodium metaaluminate ‘C.
The chemical equations for the reactions are :
Practical Based Questions for Class 10 Science Chemistry 1

Question 8.
You want to study a decomposition reaction by taking ferrous sulphate crystals in a boiling tube. List two steps you would follow while doing the experiment. (CBSE 2014)
Answer:

  1. Take a test tube and dry it completely.
  2. Take a small amount of the given sample in the test tube. Hold it with a clamp and heat the tube over a burner. Crystals will first become dirty white and then change to brown.

Question 9.
On keeping iron nails in a blue coloured copper sulphate solution, it is observed that the colour of the solution turns light green after sometime. Give reasons for this colour change. Name the type of reaction.
(CBSE 2014)
Answer:
It is an example of metal displacement reaction. Iron has displaced copper from copper sulphate solution and has changed to iron sulphate which is light green in colour.
Practical Based Questions for Class 10 Science Chemistry 2

Question 10.
While studying the double displacement reaction, the solutions of barium chloride and sodium sulphate are mixed together.
(i) What do you observe as soon as the two solutions are mixed together ?
(ii) What will happen in the above observation made by you after ten minutes ? (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.
Practical Based Questions for Class 10 Science Chemistry 3
(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 11.
You want to perform an experiment to study a double displacement reaction in your school laboratory. Name two aqueous solutions required for the experiment. State the colour change you are likly to observe on mixing the two solutions. (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.
Practical Based Questions for Class 10 Science Chemistry 3
(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 12.
You are given two colourless solutions present in two test tubes. One out of these is ethyl alcohol and the other is acetic acid. Give three tests to identify these.
Answer:
Smell. The tube in which the liquid has a vinegar like or fruity smell is acetic acid while the other with a very little pungent smell is ethyl alcohol.
Litmus Test. Add a strip of blue litmus separately in both the tubes. The tube which turns blue litmus red contains acetic acid while the other which does not change its colour is that of ethyl alcohol.
Sodium hydrogen carbonate test. Add a small amount of solid sodium hydrogen carbonate (NaHCO3) in both these tubes. The one which gives a brisk effervescence contains acetic acid while the other in which no effervescence is noticed contains ethyl alcohol.

Question 13.
You are provided with two samples of hard water; one containing temporary hardness and the other permanent hardness. Without the help of any chemical, how will you identify the nature of the sample.
Answer:
Boil the two samples separately in beakers for sometime. If precipitate appears, filter it out. Now, add a strip of soap in each. The one which produces lather contained temporary hardness while the other in which no lather is formed, contained permanent hardness.

Question 14.
How will you distinguish between ethane and ethene with the help of a chemical test ?
Answer:
Pass the two gases separately through bromine dissolved in carbon tetrachloride taken in two test tubes. If the yellow colour of bromine gets discharged, the gas is ethene. If the colour remains intact, then the gas is ethane.

Question 15.
An unknown organic liquid does not turn blue litmus red and gives no effervescence with sodium hydrogen carbonate. However, when a dry piece of sodium pellet is added to the liquid, a gas is evolved with brisk effervescence. Identify the liquid.
Answer:
The given liquid is probably an alcohol (e.g., ethyl alcohol) which has no reaction with blue litmus or sodium hydrogen carbonate. But it evolves hydrogen gas on reacting with sodium metal
2C2H5OH + 2Na ————> 2C2H5ONa + H2

Question 16.
How will you distinguish between hydrochloric acid and ethanoic acid with the strip of a-universal pH paper ?
Answer:
The strip of universal pH paper will turn red in hydrochloric acid while its colour will change to orange in ethanoic acid. In fact, hydrochloric acid is a stronger acid than acetic acid and they have different pH values.

Question 17.
Give a simple test to distinguish soaps from detergents.
Answer:
Soaps donot give lather with hard water. However, detergents form lather with hard water.

Question 18.
Write the name of apparatus/chemicals required to study the following properties of ethanoic acid in the laboratory.
Nature, odour, solubility and action are sodium hydrogen carbonate. (CBSE 2015)
Answer:
Litmus paper, test tube, test tube, lime water.

Question 19.
A student added sodium hydrogen carbonate solution in ethenoic acid taken in a test tube and the gas evolved was tested with a burning splinter. Write the chemical equation for the evolution of this gas and its effect on burning splinter. (CBSE 2015, Sample Paper 2017)
Answer:
CH3COOH + NaHCO3 ———–> CH3COONa + H2O + CO2
Since CO2 is not a supporter of combustion, the burning splinter will be extinguished.

Question 20.
A student is studying the properties of acetic acid in his school laboratory. List two physical and two chemical properties which he must observe and note in his record book. (CBSE 2016)
Answer:
Complete oxidation: When ethanol is warmed with dilute alkaline solution of potassium permanganate (5% solution) called Baeyer’s reagent, ethanoic acid or acetic acid is formed as the product.
CH3CH2OH + 2(O) ————–> CH3COOH +H2O
Partial oxidation: Upon oxidation with chromic anhydride (CrO3) dissolved in acetic acid (CH3COOH), ethanol forms ethanal, also called acetaldehyde.
CH3CH2OH +  (O)  —————– > CH3CHO + H2O

Question 21.
A student adds a spoon full of powdered sodium hydrogen carbonate to a flask containing ethanoic acid. List two main observations, he must note in his note book, about the reaction that takes place. Also write chemical equation for the reaction. (CBSE 2016)
Answer:

  1. A colourless and odourless gas is evolved accompanied by brisk effervescence.
  2. When the gas is bubbled through lime water, it becomes milky.

Question 22.
A gas is liberated immediately with brisk effervescence when you add acetic acid to sodium hydrogen carbonate powder in a test tube. Name the gas and describe a test to confirm the identity of the gas.
(CBSE 2017)
Answer:
The gas evolved is CO2
CH3COOH + NaHCO3 ———–> CH3COONa + H2O + CO2
It turns lime water milky when bubbled in small amount. On bubbling the gas in excess, the milkiness disappears.

Question 23.
If you are asked to report two observations about the following two properties of acetic acid, what would you report (t) Odour (it) Effect on litmus. (CBSE 2017)
Answer:

  1. Acetic acid has vinegar smell.
  2. On adding a few drops of blue litmus solution to acetic acid, it acquires red colour.

Question 24.
Mention the essential materials (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of a saponification reaction.
(CBSE 2017)
Answer:
The essential materials are edible oil (from animal or vegetable origin), caustic alkali (NaOH or KOH) and common salt. The reaction mixture in the saponification reaction is of alkaline nature since it contains in it a caustic alkali. The nature can be tested by adding a few drops of red litmus to the mixture. Its colour will change to blue.

Question 25.
What do you observe when you add a few drops of acetic acid to a test tube containing.
(a) phenolphthalein
(b) distilled water
(c) universal indicator
(d) sodium hydrogen carbonate.
(CBSE 2017)
Answer:
(a) Acetic acid will remain colourless in phenolphthalein.
(b) Acetic acid will mix with distilled water to form a clear solution.
(c) A universal indicator will impart orange colour to acetic acid.
(d) A brisk effervescence will be noticed due to the evolution of CO2 gas.

 

Hope given Practical Based Questions for Class 10 Science Chemistry helpful to you.

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Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds

Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds

Question 1.
Kamla and Reema are best friends. On one evening Kamla went to the house of Reema and found her working in the kitchen. The gas burner was emitting yellow flame instead of blue flame. Kamla immediately asked Reema to put off the gas. She helped her in cleaning the fine holes of the gas burner with a needle. The entire operation took about fifteen minutes. The gas was now ignited and there was a blue flame. Please read the above narration and answer the following questions :

  1. Why was burner emitting yellow flame ?
  2. What was the purpose of cleaning the holes ?
  3. In what way Kamla helped Reema ?
  4. What lesson can we learn from this ?

Answer:

  1. The holes of the burner were blocked due to soot or some oily material released during cooking. The combustion was incomplete and the gas burnt with yellow flame.
  2. By cleaning the holes, the deposits were removed and the combustion was now complete.
  3. By doing this operation, there was a saving of fuel. This also helped in checking pollution.
  4. We should always keep gas burners and engines of scooters or cars clean so that the combustion is complete and there is no wastage of any fuel. This also checks pollution problem.

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Question 2.
Mohan and Sohan were working in a factory and were good friends. On one evening Sohan received a phone . call from Mohan that his eye sight had become very dim all of a sudden. Sohan immediately rushed to his place and came to know that this had occurred after consuming some alcohol. Sohan took him to an eye specialist. He checked his vision properly and gave some antibiotic drops. The specialist kept him under observation and by next morning, he was fully cured.

  1. Why did eye sight of Mohan become dim ?
  2. How was he cured ?
  3. What is the value associated with episode ?

Answer:

  1. Mohan had consumed adultrated liquor poisoned either with methyl alcohol or copper sulphate.
  2. Antibiotic given by the specialist neutralised the effect of poisoning. As a result, Mohan regained his sight.
  3. Timely help by Sohan saved Mohan from getting completely blind. This was need a service rendered by one friend to the other.

Question 3.
Teacher asked Hema to perform test for unsaturation in the laboratory for ethylene gas. She took some chlorine water in a tube and passed the vapours of the gas. Nothing happened Teacher asked her to pass the vapours of the gas into bromine water. The yellow colour of the gas immediately discharged.

  1. What was the mistake committed by Hema ?
  2. How did teacher help her ?
  3. Write chemical equation for the reaction.

Answer:

  1. Chlorine water has no colour. Therefore, on passing ethylene gas the colour of chlorine water did not discharge.
  2. Bromine water is yellow in colour. When ethylene gas was passed through bromine water, its colour got discharged. This is the test for unsaturation.
  3. The chemical equation for the reaction is :
    Value Based Questions in Science for Class 10 Chapter 4 Carbon and Its Compounds image - 1

Question 4.
A patient was suffering from high blood pressure and high cholesterol.He went to the specialist. He enquired about his eating habits. The patient told the doctor that he consumes both dalda ghee and desi ghee and also drinks milk with full fat. The doctor asked the patient to immediately stop these and instead use vegetable oil and also drink fat free milk.

  1. What was wrong with the eating habits of the patient ?
  2. How did doctor help him ?

Answer:

  1. Both dalda/desi ghee and milk full of fats are very harmful to our body. The fats get deposited and lead to high chrolestrol level and high blood pressure. This may ultimately result in either paralysis or death.
  2. Doctor gave the patient correct advice. The diet should be free from fats as much as possible particularly in the present set up when there is lack of exercise.

Question 5.
Ethanol, commonly called alcohol is an excellent solvent, is used in medicines and is an important chemical compound involved in synthesis of many chemical compounds. However in spite of its benefits to man, its impact on social behaviour has always been questioned. Media has often shown abnormal behaviour of people while drunk. It is considered as a curse in the lives of those who are addicted to alcohol – Alcoholic’ people are not only lowering their metabolism and affecting Central Nervous System, they are also a threat to the lives of others. Anger and rude behaviour are some of its ill effects.

  1. Comment on the statement – ‘Should production of alcohol be banned’ give three valid reasons to justify.
  2. As a student what initiative would you take in the common concern of‘Save Life, Do not Drink’. Give two suggestions.

Answer:
In favour of negative response :

  1. Ethanol is used as a solvent in the laboratory.
  2. Ethanol is used in the synthesis of large number of compounds.
  3. Ethanol is a constituent of many drugs.

In favour of positive response :

  1. Ethanol is habit forming.
  2. Excessive intake of ethanol is highly injurious to the body.
  3. Excessive consumption of ethanol is the major cause of crimes in our society.
  4. It is a big strain on the family budget. Many families are ruined.

Initiatives :

  1. Students must take a pledge not to drink alcohol.
  2. Students must create awareness in the society by organising skits/seminars and also through charts.

Question 6.
School going children generally bring tiffins and they eat food during the break. One category of students marked ‘A’ carry paranthas, butter and pickels while the other category of students marked ‘B’ bring chapatties, vegetables, salad and fruits .Whereas students A’ donot like to share their food, students ‘B’ would like to share it.

  1. Which acid is present in pickels ?
  2. Which group of students bring healthy food and why ?
  3. Which group of students bring unhealthy food and why ?
  4. Which group of students have better value system and why ?

Answer:

  1. Acetic acid is present in pickels
  2. Students marked ‘B’ bring healthy food since it has more neutrition values and it is easy to digest.
  3. Students marked A’ bring unhealthy food. Since it is quite difficulty to digest it and leads to obacity.
  4. Students marked ‘B’ have better value system. Sharing of food at school level will make them better citizens at a later stage in their life.

Question 7.
Cough syrups, generally contain alcohol. Some people are habitual of drinking ‘alcohol’. Instead of drinking ‘alcohol’, they have started using cough syrups which contains alcohol and cause addiction. To solve this problem, government is thinking to ban cough syrups.

  1. What is an alcohol ?
  2. Should production of cough syrups be banned ?
  3. As a student ‘what initiative would you take to make people aware of harmful effects of taking cough syrups unnecessarily’. Give two suggestions.

Answer:

  1. Alcohol is chemically ethyl alcohol and its formula is C2H5OH.
  2. No, production of all cough syrups cannot be banned since they given relief from cough and cold. However, they should be sold strictly if prescribed by the doctor on a proper prescription.
  3. We students must acquaint people around us the harmful effects of alcohols by holding seminars and by door to door convassing.

 

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Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals

Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals

These Solutions are part of Value Based Questions in Science for Class 10. Here we have given Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals

Question 1.
A customer approached a newly trained Goldsmith and asked him to make some ornaments. For this he gave advance money also. The Goldsmith bought a bar of 24-carat gold from the market and tried to make ornaments from it. He was not successful because the gold was too soft to handle. He approached a trained Goldsmith who asked him to use 22-carat gold for this purpose. He was now quite successful and could make the ornaments ?

  1. What is the difference between 24-carat gold and 22-carat gold ?
  2. What was wrong with 24-carat gold ?
  3. How did trained Goldsmith help the freshly trained Goldsmith ?
  4. What is the value associated with this episode ?

Answer:

  1. 24-carat gold is pure gold while 22-carat gold is comparatively less pure.
  2. Pure or 24-carat gold is very soft, little malleable and ductile. It is quite difficult to work on it. But 22 carat gold (alloy) is comparatively hard, more malleable and ductile
  3. Freshly trained Goldsmith realised his mistake and could complete the order. Otherwise he might have been in problem.
  4. The trained Goldsmith had a big heart and did not bother about the professional jealousy. He was very sincere in his approach and helped the junior colleage.

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Question 2.
A student was asked to purify a sample of copper extracted by a suitable method. He constructed a-cell in which a rod of impure copper was made cathode while that of pure copper as anode. The electrolyte was aqueous CuSO4 solution. On passing electric current nothing happened. He realised his mistake and connected the electrodes in the reverse order. He was now quite successful in his mission.

  1. What was his mistake ?
  2. How did he rectify it ?
  3. Write a chemical equation for the reaction.

Answer:

  1. Impure copper cannot be made cathode and pure copper as the anode.
  2. He changed over. Pure copper was made cathode while impure copper as the anode.
  3. The chemical equations for the process are :
    Value Based Questions in Science for Class 10 Chapter 3 Metals and Non-metals image - 1

 

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HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals

HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals

These Solutions are part of HOTS Questions for Class 10 Science. Here we have given HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals

Question 1.
A student has been collecting silver coins and copper coins. One day she observed a black coating on silver coins and a green coating on copper coins. Which chemical phenomenon is responsible for these coatings ? Write the chemical names of black and green coatings ?
Answer:
The phenomenon is known as corrosion. Air contains traces of hydrogen sulphide gas which reacts with silver metal present in the coin to form silver sulphide. It is black in colour.
HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals image - 1
Similarly, copper present in the coin reacts with oxygen and traces of both carbon dioxide and water vapours present in air to form a green mass. It is chemically basic copper carbonate :
HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals image - 2

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Question 2.
You are provided with three metals : Sodium, magnesium and copper. Using only water as the reactant, how will you identify them ?
Answer:
The metal which reacts violently with cold water and catches fire is sodium.

  1. The metal which evolves hydrogen gas upon heating with water is magnesium.
  2. The metal which does not react with water even on strong heating is copper.

Question 3.
E is an element amongst copper, zinc, aluminium and iron. It shows following properties :
(a) One of its ores is rich in E2O3
(b) E2O3 is not attacked by water,
(c) It forms two chlorides ECl2 and ECl3. Name the element and justify your answer.
Answer:
The clue for the correct answer is the formation of ECl2 and ECl3. This shows that the element E has variable valencies of 2 and 3. Out of the elements listed, only iron exists in divalent and trivalent forms.
(a) The ore rich in Fe2O3 is haematite.
(b) Haematite (Fe2O3) is not attacked by water.
(c) The two chlorides are : iron (II) chloride or FeCl2 and iron (III) chloride or FeCl3.

Question 4.
An element reacts with oxygen to form an oxide which dissolves in dilute hydrochloric acid. The oxide formed also turns a solution of red litmus blue. Is the element a metal or non-metal ? Explain with the help of a suitable example.
Answer:
The oxide of the element is basic as it turns red litmus solution blue. This means that the element is a metal (M). Let the metal be sodium (Na). The chemical equations that are involved are given as follows :
HOTS Questions for Class 10 Science Chapter 3 Metals and Non-metals image - 3

Question 5.
An element E combines with oxygen to form an oxide E2O which is a good conductor of electricity.
Give the following information:

  1. How many electrons will be present in the valence shell of the element E ?
  2. Write the formula of the compound formed when the element E combines with chlorine.

Answer:

  1. From the formula E2O of the oxide, it is clear that the valency of the element E is one. This means that it has only one electron in the valence shell.
  2. We know that chlorine is monovalent. Sine the valency of the element E is a also one, the formula of the chloride of the element is ECl.

Question 6.
An element A’ catches fire in water and burns with golden yellow flame in air. It reacts with another element ‘B’, present in group 17 to give a product ‘C’. An aqueous solution of product ‘C’ on electrolysis gives a compound ‘D’ and liberates hydrogen. Identify A, B, C and D.
Answer:
Since the element A’ catches fire in water and burns with golden yellow flame, it is sodium (Na). The element ‘B’ with atomic number 17 is chlorine (Cl). Both these combine to form sodium chloride (NaCl) which is designated as ‘C’. Upon electrolysis, sodium chloride gives sodium hydroxide (D) and evolves hydrogen along with chlorine.

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HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds

HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds

Question 1.
An organic compound A of molecular formula C2H4 on reduction gives another compound B of molecular formula C2H6. B on reaction with chlorine in the presence of sunlight gives C of molecular formula C2H5Cl.
(a) Name the compounds A, B and C.
(b) Write chemical equation for the conversion of A to B and name the type of reaction.
Answer:
The compound A of molecular formula C2H4 is an alkene. Upon reduction with hydrogen, it gives ‘B’ of molecular formula C2H6. The compound ‘B’ upon chlorinadon gives ‘C’ of molecular formula C2H5Cl.
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 1
The reaction to called addition reaction.

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Question 2.
Name the functional group of organic compounds that can be hydrogenated. With the help of suitable example, explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. Name any one natural source of organic compounds that are hydrogenated.
Answer:
The functional group which can be easily hydrogenated isHOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 2
The family is known as alkenes. The hydrogenation reaction can be carried by heating a member of the family (e.g. ethene) with hydrogen in the presence of a catalyst like nickel (Ni) For example.
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 3
Ethene is an unsaturated hydrocarbon while ethane is of saturated nature.
Edible oils such as coconut oil, olive oil, peanut oil etc. contain atleast one HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 2
in their molecules. These are regarded as unsaturated compounds. Upon hydrogenation, these get converted into fats which no longer contain any HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 2

Question 3.
An organic compound A’ of molecular formula C2H6O on oxidation with dilute alkaline KMnO4 solution gives an acid ‘B’ with the same number of carbon atoms. Compound A’ is often used for sterilization of skin by doctors. Name the compound. Write the chemical equation involved in the formation of ‘B’ from A.
Answer:
The compound ‘B’ should contain a —COOH group as it is an acid. Since it has only two carbon atoms, the other carbon atom must represent CH3 group. Thus, compound ‘B’ is ethanoic acid (CH3COOH). The compound A used for the sterilization of skin by doctors is ethanol C2H5OH (C2H6O). The chemical reaction involved in the oxidation by dilute alkaline KMnO4 solution also called Baeyers reagent is :
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 6

Question 4.
A to F are the structural formulae of some organic compounds :
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 7
(i) Give the letters which represent the same family.
(ii) Give the letters which do not represent hydrocarbons.
(iii) Flow can ‘C’ be converted into A ?
Answer:
(i) Letters ‘B’ and ‘D’ represent the family of alkynes.
(ii) Letters ‘E’ and ‘F’ donot represent any hydrocarbon.
(iii) ‘C’ can be converted into ‘A’ by passing hydrogen (H2) in the presence of Ni at 473 K.
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 8

Question 5.
(a) A test tube contains a brown liquid in it. The colour of the liquid remains the same when methane is passed through it but it disappears when ethene is passed. Suggest the name of the liquid brown in colour. Give the chemical equation involved.
(b) The formula of an ester is C3H7COOC2H5. Write the formulae of the acid and alcohol from which the ester is prepared.
Answer:
(a) The brown liquid seems to be bromine dissolved in water. Methane (CH4) is a saturated hydrocarbon and does not react with bromine. Ethene (C2H4) being unsaturated in nature, decolourises bromine and its colour therefore, disappears.
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 9
(b) In an ester, the left side in the molecular formula containing C3H7CO is derived from the acid while the right side having OC2H5 is from the alcohol. This means that the acid and alcohol participating in the ester are C3H7COOH and C2H5OH respectively. The formation of ester may be shown as follows :
HOTS Questions for Class 10 Science Chapter 4 Carbon and Its Compounds image - 10

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3

More Exercises

Question 1.
Find the dividend received on 60 shares of Rs, 20 each if 9% dividend is declared.
Solution:
Value of one share = Rs. 20
Value of 60 shares = Rs. 20 x 60
= Rs. 1200
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q1.1

Question 2.
A company declares 8 percent dividend to the share holders. If a man receives Rs. 2840 as his dividend, find the nominal value of his shares.
Solution:
Rate of dividend = 8%
Amount of dividend = Rs. 2840
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q2.1

Question 3.
A man buys 200 ten-rupee shares at Rs 12.50 each and receives a dividend of 8%. Find the amount invested by him and the dividend received by him in cash.
Solution:
Face value of 200 shares = Rs. 10 x 200
= Rs. 2000
(i) Amount invested for the purchase of 200 shares at the rate of Rs. 12.50 each
= Rs. 12.50 x 200
= Rs. 2500
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q3.1

NCVT MIS

Question 4.
Find the market price of 5% share when a person gets a dividend of Rs 65 by investing Rs. 1430.
Solution:
Amount of dividend = Rs. 65
Rate of dividend = 5%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q4.1

Question 5.
Salman buys 50 shares of face value Rs 100 available at Rs 132.
(i) What is his investment ?
(ii) If the dividend is 7.5% p.a., what will be his annual income ?
(iii) If he wants to increase his annual income by Rs 150, how many extra shares should he
Solution:
Face Value = Rs 100
(i) Market Value = Rs 132
No. of shares = 50
Investment = no. of shares x Market value
= 50 x 132 = Rs 6600
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q5.1

Question 6.
A lady holds 1800, Rs. 100 shares of a company that pays 15% dividend annually. Calculate her annual dividend. If she had bought these shares at 40% premium, what percentage return does she get on her investment ? Give your answer to the nearest integer.
Solution:
Total number of shares = 1800
Nominal value of each share = Rs. 100
Rate of dividend = 15%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q6.1

Question 7.
What sum should a person invest in Rs 25 shares, selling at Rs 36, to obtain an income of Rs 720, if the dividend declared is 12%? Also find the percentage return on his income.
(i) The number of shares bought by him.
(ii) The percentage return on his income.
Solution:
Nominal value of each share = Rs. 25
Market value of each share = Rs. 36
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q7.1

Question 8.
Ashok invests Rs 26400 on 12% Rs 25 shares of a company. If he receives a dividend of Rs 2475, find:
(i) the number of shares he bought.
(ii) the market value of each share. (2016)
Solution:
Investment = Rs 26400
Face value of each share = Rs 25
Rate of dividend = 12%
and total dividend = Rs 2475
We know,
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q8.1

Question 9.
Amit Kumar invests Rs 36,000 in buying Rs 100 shares at Rs 20 premium. The dividend is 15% per annum. Find :
(i) The number of shares he buys
(ii) His yearly dividend
(iii) The percentage return on his investment.
Give your answer correct to the nearest whole number.
Solution:
Investment = Rs 36000
Face value = Rs 100
Premium = Rs 20, dividend = 15%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q9.1

Question 10.
Mr. Tiwari invested Rs 29,040 in 15% Rs 100 shares at a premium of 20%. Calculate:
(i) The number of shares bought by Mr. Tiwari.
(ii) Mr. Tiwari’s income from the investment.
(iii) The percentage return on his investment.
Solution:
(i) M.V. of one share = \(\left[ \frac { 20 }{ 100 } \times 100+100 \right] \)
= Rs 120
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q10.1

Question 11.
A man buys shares at the par value of Rs 10 yielding 8% dividend at the end of a year. Find the number of shares bought if he receives a dividend of Rs 300.
Solution:
Face value of each share = Rs 10
Rate of dividend = 8% p.a.
Total dividend = Rs 300
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q11.1

Question 12.
A man invests Rs 8800 on buying shares of face value of rupees hundred each at a premium of 10%. If he earns Rs 1200 at the end of year as dividend, find :
(i) the number of shares he has in the company.
(ii) the dividend percentage per share.
Solution:
Investment = Rs 8800
Face value of each share = Rs 100
and market value of each share
= Rs 100 + Rs 10 = Rs 110
Total income = Rs 1200
Total face value
= Rs \(\\ \frac { 8800\times 100 }{ 110 } \)
= Rs 8000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q12.1

Question 13.
A man invested Rs. 45000 in 15% Rs. 100 shares quoted at Rs. 125. When the market value of these shares rose to Rs. 140, he sold some shares, just enough to raise Rs. 8400. Calculate :
(i) the number of shares he still holds.(2004)
(ii) the dividend due to him on these shares.
Solution:
Investment on shares = Rs. 45000
Face value of each share = Rs. 125
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q13.1

Question 14.
A company pays a dividend of 15% on its ten-rupee shares from which it deducts tax at the rate of 22%. Find the annual income of a man, who owns one thousand shares of this company.
Solution:
No. of shares = 1000
Face value of each are = Rs. 10
Rate of dividend = 15%,
Rate of tax deducted = 22%
Total face value of 1000 shares = Rs. 10 x 1000 = Rs. 10000
Total dividend = Rs 10000 x \(\\ \frac { 15 }{ 100 } \)
= Rs 1500
Tax deducted at the rate of 22 %
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q14.1

Question 15.
Ajay owns 560 shares of a company. The face value of each share is Rs. 25. The company declares a dividend of 9%. Calculate.
(i) the dividend that Ajay will get.
(ii) the rate of interest, on his investment if Ajay has paid Rs. 30 for each share. (2007)
Solution:
No. of shares = 560
Face value of each share = Rs. 25
Rate of dividend = 9% p.a.
Total face value of 560 shares = Rs. 25 x 560
= Rs. 14000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q15.1

Question 16.
A company with 10000 shares of nominal value of Rs. 100 declares an annual dividend of 8% to the share holders.
(i) Calculate the total amount of dividend paid by the company.
(ii) Ramesh bought 90 shares of the company at Rs. 150 per share.
Calculate the dividend he received and the percentage return on his investment. (1994)
Solution:
(i) Number of shares = 10000
Nominal value of each share = Rs. 100
Rate of annual dividend = 8%
Total face value of 10000 shares
= Rs. 100 x 10000
= Rs. 1000000
and amount dividend = Rs \(\\ \frac { 1000000\times 8 }{ 100 } \)
= Rs 80000
(ii) Number of shares = 90
Face value of each share = Rs. 150
Total face value of 90 shades
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q16.1

Question 17.
A company with 4000 shares of nominal value of Rs. 110 declares annual dividend of 15%. Calculate :
(i) the total amount of dividend paid by the company,
(ii) the annual income of Shah Rukh who holds 88 shares in the company,
(iii) if he received only 10% on his investment, find the price Shah Rukh paid for each share. (2008)
Solution:
Number of shares = 4000
Nominal (face) value of each share = Rs. 110
Total face value of 4000 shares = Rs. 110 x 4000
= Rs, 440000
Rate of annual dividend = 15%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q17.1

Question 18.
By investing Rs. 7500 in a company paying 10 percent dividend, an income of Rs. 500 is received. What price is paid for each Rs. 100 share
Solution:
Investment = Rs. 7500
Rate of dividend = 10%,
Total income = Rs. 500.
Face value of each share = Rs. 100
Total face value = \(\\ \frac { 100\times 500 }{ 10 } \) = Rs. 5000
If face value is Rs. 5000, then investment = Rs. 7500
and if face value is Rs. 100 then market value of
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q18.1

Question 19.
A man invests Rs. 8000 in a company paying 8% dividend when a share of face value of Rs. 100 is selling at Rs. 60 premium,
(i) What is his annual income,
(ii) What percent does he get on his money ?
Solution:
Investment = Rs. 8000
Face value of each share = Rs. 100
Market value = Rs. 100 + Rs. 60
= Rs. 160
Rate of dividend = 8% p.a.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q19.1

Question 20.
A man buys 400 ten-rupee shares at a premium of Rs. 2.50 on each share. If the rate of dividend is 8%, Find,
(i) his investment
(ii) dividend received
(iii) yield.
Solution:
No. of shares = 400
Face value of each share = Rs. 10
Market value of each share
= Rs. 10 + Rs. 2.50
= Rs. 12.50
Rate of dividend = 8%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q20.1

Question 21.
A man invests Rs. 10400 in 6% shares at Rs. 104 and Rs. 11440 in 10.4% shares at Rs. 143. How much income would he get in all ?
Solution:
In first case; Total investment = Rs. 10400
Rate of dividend = 6%
Market value of each share = Rs. 104
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q21.1
Total dividend from both cases = Rs. 600 + Rs. 832
= Rs. 1432 Ans.

Question 22.
Two companies have shares of 7% at Rs. 116 and 9% at Rs. 145 respectively. In which of the shares would the investment be more profitable ?
Solution:
Let the investment in each case = Rs. 116 x 145
Dividend in first case
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q22.1

Question 23.
Which is better investment : 6% Rs. 100 shares at Rs. 120 or 8% Rs. 10 shares at Rs. 15
Solution:
Let the investment in each case = Rs. 120 In the fist case,
Dividend on Rs. 120 = Rs. 6
In second case, Dividend on Rs. 10
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q23.1

Question 24.
A man invests Rs -10080 in 6% hundred- rupee shares at Rs. 112. Find his annual income. When the shares fall to Rs. 96 he sells out the shares and invests the proceeds in 10% ten-rupee shares at Rs. 8. Find the change in his annual income.
Solution:
Investment = Rs. 10080
Face value of each share = Rs. 100
Market value of each share = Rs. 112
Rate of dividend = 6%
Total income for the year
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q24.1

Question 25.
A man bought 360 ten-rupee shares paying 12% per annum. He sold them when the price rose to Rs. 21 and invested the proceeds in five-rupee shares paying \(4 \frac { 1 }{ 2 } \) % per annum at Rs. 3.5 per share. Find the annual change in his income.
Solution:
No. of shares bought = 360
Face value of each share = Rs. 10
Rate of dividend = 12%
Total face value of 360 shares
= Rs. 10 x 360
= Rs. 3600
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q25.1

Question 26.
A person invests Rs. 4368 and buys certain hundred-rupee shares at 91. He sells out shares worth Rs. 2400 when they have t risen to 95 and the remainder when they have fallen to 85. Find the gain or loss on the total transaction,
Solution:
Investment = Rs. 4368
Market value of each share = Rs. 91
Face value of each share = Rs. 100
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q26.1

Question 27.
By purchasing Rs. 50 gas shares for Rs. 80 each, a man gets 4% profit on his investment. What rate percent is company paying ? What is his dividend if he buys 200 shares ?
Solution:
Market value of each share = Rs 80
Face value of each share = Rs. 50
Interest on investment = 4%
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q27.1
Dividend = Rs 10000 x \(\\ \frac { 6.4 }{ 100 } \)
= Rs 640

Question 28.
Rs. 100 shares of a company are sold at a discount of Rs. 20. If the return on the investment is 15%. Find the rate of dividend declared
Solution:
Market value of each shares = 100 – 20
= Rs.80
Interest on investment of Rs. 80
= 15% x 80
= \(\\ \frac { 15 }{ 100 } \) x 80
= Rs 12
Dividend on face value of Rs. 100 = Rs. 12
Rate of dividend = 12%. Ans.

Question 29.
A company declared a dividend of 14%. Find tire market value of Rs. 50 shares if the return on the investment was 10%.
Solution:
Rate of dividend = 14%
Dividend on Rs. 50 = \(\\ \frac { 14\times 50 }{ 100 } \) = Rs 7
Now Rs. 10 is interest on the investment of
= Rs. 100
Rs. 7 will be the interest on
= \(\\ \frac { 100\times 7 }{ 10 } \) = Rs. 70
Hence Market value of Rs. 50 shares = Rs. 70Ans.

Question 30.
At what price should a 6.25% Rs. 100 shares be quoted when the money is worth 5%?
Solution:
Interest on Rs. 100 worth = Rs. 5
If interest is Rs. 5, then market value = Rs. 100
and if interest is Rs. 6.25, then market value
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q30.1
Market value of each share = Rs. 125 Ans.

Question 31.
At what price should a 6.25% Rs. 50 share be quoted when the money is worth 10%?
Solution:
Interest on Rs. 100
worth = Rs. 10
If the interest is Rs. 10, then market value = Rs. 100
and if interest is Rs. 6.25, then market
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q31.1

Question 32.
A company with 10000 shares of Rs. 100 each, declares an annual dividend of 5%.
(i) What is the total amount of dividend paid by the company ?
(ii) What would be the annual income of a man, who has 72 shares, in the company ?
(iii) If he received only 4% on his investment, find the price he paid for each share. (1998)
Solution:
No. of shares = 10000
Face value of each share = Rs. 100
Rate of dividend = 5%
(i) Total face value of 10000 shares
= Rs. 100 x 10000
= Rs. 1000000
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q32.1

Question 33.
A man sold some Rs. 100 shares paying 10% dividend at a discount of 25% and invested the proceeds in Rs. 100 shares paying 16% dividend quoted at Rs. 80 and thus increased his income by Rs. 2000. Find the number of shares sold by him.
Solution:
Face value of each share = Rs. 100
Market value of each share
= Rs. 100 – Rs.25
= Rs. 75
Rate of dividend = 10%
Let no. of shares = x
Selling price = x × 75 =Rs. 75x
Face value of x share = 100 x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q33.1

Question 34.
By selling at Rs. 77, some \(2 \frac { 1 }{ 4 } \) % shares of face value Rs. 100, and investing the proceeds in 6% shares of face value Rs. 100, selling at 110, a person increased his income by Rs, 117 per annum. How many shares did he sell ?
Solution:
Let the number of shares = x
On selling at Rs.77, the amt received x × 77 = Rs. 77 x
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q34.1

Question 35.
A man invests Rs. 6750, partly in shares of 6% at Rs. 140 and partly in shares of 5% at Rs. 125. If his total income is Rs. 280, how much has he invested in each ?
Solution:
Let the investment in first case = x
Then investment in second case = (6750 – x)
In first case, the dividend
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q35.1

Question 36.
Divide Rs. 20304 into two parts such that if one part is invested in 9% Rs. 50 shares at 8% premium and the other part is invested in 8% Rs. 25 shares at 8% discount, then the annual incomes from both the investment are equal
Solution:
Total amount = Rs 20304
Let amount invested in 9% Rs 50 at 8%
premium = x
Then amount invested in 8% Rs 25 at 8%
Discount = 20304 – x
Income from both investments are equal Now income from first type of shares
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q36.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 Q36.2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 3 Shares and Dividends Ex 3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Multiple Choice Questions

Question 1.
Upto which element, the Law of Octaves was found to be applicable ?
(a) Oxygen
(b) Calcium
(c) Cobalt
(d) Potassium.
Answer:
(b). Law was found to be applicable upto the element calcium.

More Resources

Question 2.
According to Mendeleev’s Periodic Law, the elements were arranged in the periodic table in the order of
(a) increasing atomic number
(b) decreasing atomic number
(c) increasing atomic masses
(d) decreasing atomic masses.
Answer:
(c).

Question 3.
In Mendeleev’s Periodic Table, gaps were left for the elements to be discovered later. Which of the following elements found a place in the periodic table later ?
(a) Germanium
(b) Chlorine
(c) Oxygen
(d) Silicon.
Answer:
(a). The element was called Eka-silicon.

Question 4.
Which of the following statement (s) about the Modern Periodic Table are incorrect ?
(i) The elements in the Modern Periodic Table are arranged on the basis of their decreasing atomic numbers
(ii) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic masses.
(iii) Isotopes are placed in adjoining group (s) in the Periodic Table
(iv) The elements in the Modern Periodic Table are arranged on the basis of their increasing atomic numbers
(a) (i) only
(b) (i), (ii) and (iii)
(c) (i), (ii) and (iv)
(d) (iv) only.
Answer:
(b). Statements (i), (ii) and (iii) are all in correct.

Question 5.
Which of the following statements about the Modern Periodic Table is correct ?
(a) It has 18 horizontal rows known as Periods
(b) It has 7 vertical columns known as Periods
(c) It has 18 vertical columns known as Groups
(d) It has 7 horizontal rows known as Groups.
Answer:
(c).

Question 6.
Which of the given elements A, B, C, D and E with atomic numbers 2, 3, 7, 10 and 30 respectively belong to the same period ?
(a) A, B, C
(b) B, C, D
(c) A, D, E
(d) B, D, E.
Answer:
(b). The elements B (Z = 3), C (Z = 7) and D(Z = 10) belong to the same period. It is second period.

Question 7.
The elements A, B, C, D and E have atomic numbers 9, 11, 17, 12 and 13 respectively. Which pair of elements belong to the same group ?
(a) A and B
(b) B and D
(c) A and C
(d) D and E.
Answer:
(c). Elements A (Z = 9) and C (Z = 17) belong to the same group. It is a halogen family (group 17).

Question 8.
Where would you locate the element with electronic configuration 2, 8 in the Modern Periodic Table ?
(a) Group 8
(b) Group 2
(c) Group 18
(d) Group 10.
Answer:
(c). It is a noble gas element Neon (Ne) present in group 18.

Question 9.
An element which is an essential constituent of all organic compounds belongs to
(a) group 1
(b) group 14
(c) group 15
(d) group 16.
Answer:
(b). The element is carbon. It belongs to group 14.

Question 10.
Which of the following is the outermost shell for elements of period 2 ?
(a) K shell
(b) L shell
(c) M shell
(d) N shell
Answer:
(b). In second (2) period, the electrons are filled in second shell also known as L-shell.

Question 11.
Which one of the following elements exhibits maximum number of valence electrons ?
(a) Na
(b) Al
(c) Si
(d) P
Answer:
(d). The element phosphorus (P) has five electrons (2, 5) in the valence shell.

Question 12.
Which of the following gives the correct increasing order of the atomic radii of O, F and N ?
(a) O, F, N
(b) N, F, O
(c) O, N, F
(d) F, O, N.
Answer:
(d). It is the correct order. These elements are present in second period.

Question 13.
Which among the following elements has the largest atomic radius ?
(a) Na
(b) Mg
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) present in group 1 has the largest atomic radius.

Question 14.
Which of the following elements would lose an electron easily ?
(a) Mg
(b) Na
(c) K
(d) Ca.
Answer:
(c). The element potassium (K) with maximum size would lose electron easily.

Question 15.
Which of the following elements does not lose an electron easily ?
(a) Na
(b) F
(c) Mg
(d) Al.
Answer:
(b). The element fluorine (F) with smallest size does not lose electron easily.

Question 16.
Which of the following are the characteristics of isotopes of an element ?
(i) Isotopes of an element have same atomic mass.
(ii) Isotopes of an element have same atomic number.
(iii) Isotopes of an element show same physical properties.
(iv) Isotopes of an element show same chemical properties.
(a) (i), (iii) and (iv)
(b) (ii), (iii) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Answer:
(d).

Question 17.
Arrange the following elements in the order of their decreasing metallic character Na, Si, Cl, Mg, Al
(a) Cl > Si > Al > Mg > Na
(b) Na > Mg > Al > Si > Cl
(c) Na > Al > Mg > Cl > Si
(d) Al > Na > Si > Ca > Mg.
Answer:
(b). The metallic character of the elements decreases along a period. These elements are present in third period.

Question 18.
Arrange the following elements in the order of their increasing non-metallic character Li, O, C, Be, F
(a) F < O < C < Be < Li
(b) Li < Be < C < O < F
(c) F < O < C < Be < Li
(d) F < O < Be < C < Li.
Answer:
(b). The non-metallic character of the elements increases along a period. These elements are present in second period in the order Li, Be, C, O, F.

Question 19.
What type of oxide would Eka-aluminium form ?
(a) EO3
(b) E3O2
(c) E2O3
(d) EO.
Answer:
(c). Is the correct answer.

Question 20.
Three elements B, Si and Ge are
(a) metals
(b) non-metals
(c) metalloids
(d) metal, non-metal and metalloid respectively.
Answer:
(c). is the correct answer. These are also called semi-metals and possess the characteristics of both metals and non-metals.

Question 21.
Which of the following elements will form an acidic oxide ?
(a) An element with atomic number 7
(b) An element with atomic number 3
(c) An element with atomic number 12
(d) An element with atomic number 19.
Answer:
(a). The element with atomic number (Z) = 7 is nitrogen. It forms acidic oxides such as N2O3, N2O5 etc.
All other elements are metals and they form basic oxides.

Question 22.
The element with atomic number 14 is hard and forms acidic oxide and a covalent halide. To which of the following categories does the element belong ?
(a) Metal
(b) Metalloid
(c) Non-metal
(d) Left-hand side element.
Answer:
(b). The element with atomic number (Z) = 14 is silicon (Si). It is a metalloid.

Question 23.
Which one of the following depicts the correct representation of atomic radius (r) of an atom ?
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 1
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(b). The atom (ii) has only one shell (K-shell). No electrons are present in the other shells. Therefore, the arrow represents correct atomic radius.
The arrow in atom (iii) also represents the correct atomic radius of the element.

Question 24.
Which one of the following does not increase while moving down the group of the periodic table ?
(a) Atomic radius
(b) Metallic character
(c) Valence
(d) Number of shells in an element.
Answer:
(c). In a group, the valence or valency does not change since all the elements present have same valence shell configuration.

Question 25.
On moving from left to right in a period in the periodic table, the size of the atom
(a) increases
(b) decreases
(c) does not change appreciably
(d) first decreases and then increases
Answer:
(b). Atomic size decreases along a period. However, the noble gas atom is an exception. It has very large size.

Question 26.
Which of the following set of elements is written in order of their increasing metallic character ?
(a) Be Mg Ca
(b) Na Li K
(c) Mg Al Si
(d) C O N
Answer:
(a). These elements belong to group 2. These are written in increasing order of their size. Since the metallic character increases down the group, the order is the correct.

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Short Answer Questions

Question 27.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as ? Give one example of such a set of elements.
Answer:
The arrangement is known as Dobereiner’s triad. For example, Calcium (Ca), Strontium (Sr) and Barium (Ba).

Question 28.
Elements have been arranged in the following sequence on the basis of their increasing atomic masses.
F, Na, Mg, Al, Si, P, S, Cl, Ar, K
(a) Pick two sets of elements which have similar properties.
(b) The given sequence represents which law of classification of elements ?
Answer:
(a) The elements that have similar properties belong to the same group. From the list of elements available, elements which belong to same group are : .
Na, K (Alkali metals) ; F, Cl (Halogens)
(b) The sequence is according to Newland’s law of octaves.

Question 29.
Can the following groups of elements be classified as Dobereiner’s triad ?
(a) Na, Si, Cl
(b) Be, Mg, Ca
Answer:
Atomic mass of Be = 9; Na = 23; Mg = 24; Si = 28; Cl = 35; Ca = 40 Explain by giving reason.
(a) No, these elements cannot be classified as triads because these do not have same properties. However, the atomic mass of Si (28) is almost the mean of the atomic masses of elements Na (23) and Cl (35).
Note : The Dobereiner’s triad is meaningful only if the elements present in the triad have almost identical properties.
(b) Yes, these elements can be classified as triads. These belong to the same group (2) and have almost identical properties. The atomic mass of Mg (24) is almost the mean of the atomic masses of the elements Be (9) and Ca(40) i.e., 9 + 40 = 49/2 = 24.5.

Question 30.
In Mendeleev’s Periodic Table, the elements were arranged in the increasing order of their atomic masses. However, cobalt with atomic mass of 58.93 amu was placed before nickel having an atomic mass of 58.71 amu. Give reason for the same.
Answer:
This is regarded as a defect in the Mendeleev’s periodic table. However, the main reason for making the arrangement was that the elements with similar characteristics must be grouped together. In this case,
• Cobalt (Co) should be in the company of the elements Rhodium (Rh) and Iridium (Ir).
• Nickel (Ni) should be in the company of the elements Palladium (Pd) and Platinum (Pt).

Question 31.
“Hydrogen occupies a unique position in Modern Periodic Table”. Justify the statement.
Answer:
The position of the element hydrogen is still not clear even in the Modern Periodic Table.
• In electronic configuration, it resembles alkali metals of group 1. All of them have only one electron in the valence shell. Actually, hydrogen has only one shell (K-shell) which has one electron. However, it is not a metal whereas alkali-metals are metallic in nature.
• In characteristics, it resembles halogens of group 17. For example, like halogens hydrogen is a non-metal and diatomic as well.
It has been therefore, decided to assign hydrogen a unique position in the Modern or Long Form. Periodic Table. It is placed at the top in group 1 of alkali metals. However, it is not a member of that group.*

Question 32.
Write the formulae of chlorides of Eka-silicon and Eka-aluminium, the ^elements predicted by Mendeleev.
Answer:
Eka-aluminium represents gallium (Ga) with valency three and Eka-silicon is for germanium (Ge) with valency four. The formulae of their respective chlorides are GaCl3 and GeCl4.

Question 33.
Three elements A, B and C have 3, 4 and 2 electrons respectively in their outermost shell. Give the group number to which they belong in the Modern Periodic Table. Also, give their valencies.
Answer:
The electrons present in the outermost shell are also known as valence electrons. The desired information is given :
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 2

Question 34.
If an element X is placed in group 14, what will be the formula and the nature of bonding of its chloride ?
Answer:
The element X present in group 14 has four valence electrons in its atom. It can complete its octet by sharing four valence electrons with the electrons of other atoms. Therefore, it will form covalent bonds with the four atoms of chlorine. The formula of the chloride of the element X is.
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 3

Question 35.
Compare the radii of two species X and Y. Give reasons for your answer.
(a) X has 12 protons and 12 electrons
(b) Y has 12 protons and 10 electrons
Answer:
The available information makes it clear that :
(a) Species X with equal number of protons (12) and electrons (12) is an atom.
(b) Species Y with two electrons less (10) than the number of protons (12), is a divalent cation y2+ of species X.
Therefore, the radius of species X is more as compared to that of y2+.
Note : The radius of cation is always less than that of atom while that of anion is more.

Question 36.
Arrange the following elements in increasing order of their atomic radii.
Answer:
(a) All the elements belong to the same period (second). The atomic size or radius decreases along a period. Therefore, increasing order of atomic radii is : F < N < Be < Li
(b) The listed elements are present in group 17 (halogen family). The atomic size or radius increases down
a group. Therefore, the correct increasing order of atomic radii is : Cl < Br < I < At.

Question 37.
Identify and name the metals out of the following elements whose electronic configurations are given below :
(a) 2, 8, 2
(b) 2, 8, 1
(c) 2, 8, 7
(d) 2, 1.
Answer:
(a) Electronic configuration : 2, 8, 2 (Mg is a metal)
(b) Electronic configuration : 2, 8, 1 (Na is a metal)
(c) Electronic configuration : 2, 8, 7 (Cl is a non-metal)
(d) Electronic configuration : 2, 1 (Li is a metal)

Question 38.
Write the formula of the compound formed when the element A (atomic number 19) combines with the element B (atomic number 17). Draw its electronic dot structure. ^What is the nature of the bond formed ? (CBSE 2013)
Answer:
Electronic configuration of element A (Z = 19) is 2, 8, 8, 1
Electronic configuration of element B(Z = 17) is 2, 8, 7.
The element A has one valence electron while the element B has seven electrons in its valence shell. One electron gets transferred from the atom of element A to the atom of element B. As a result of electron transfer, an ionic bond is formed.
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 4

Question 39.
Arrange the following elements in the increasing order of their metallic character. ,
Mg, Ca, K, Ge, Ga
Answer:
In general, the metallic character increases down the group and decreases along a period. From the relative positions of the elements in the periodic table, the increasing order of metallic character is :
Ge < Ga < Mg < Ca < K.

Question 40.
Identify the elements with the following property and arrange them in increasing order of their reactivity
(a) An metal which is soft and reactive
(b) The metal which is an important constituent of limestone
(c) The metal which exists in liquid state at room temperature
Answer:
(a) The metal may be either Na or K.
(b) The metal is Ca.
(c) The metal is Hg.

Question 41.
The increasing order of reactivity of the metals is :
Hg < Ca < Na < K.
Properties of the elements are given below. Where would you locate the following elements in the periodic table ?
(a) A soft metal stored under kerosene
(b) An element with variable (more than one) valency stored under water
(c) An element which is tetravalent and forms the basis of organic chemistry
(d) An element which is an inert gas with atomic number 2
(e) An element whose thin oxide layer is used to make other elements corrosion resistant by the process of “anodising.”
Answer:
(a) Sodium (Group 1 and Period 3) or Potassium (Group 1 and Period 4)
(b) Phosphorus (Group 15 and Period 3). It shows variable valencies 3 and 5 and is stored under water.
(c) Carbon (Group 14 and Period 2)
(d) Helium (Group 18 and Period 1)
(e) Aluminium (Group 13 and Period 3)

NCERT Exemplar Solutions for Class 10 Science Chapter 5 Long Answer Questions

Question 42.
An element placed in 2nd group and 3rd Period of the Periodic Table, burns in the presence of oxygen to form a basic oxide.
(a) Identify the element
(b) Write the electronic configuration
(c) Write the balanced equation when it burns in the presence of air
(d) Write a balanced equation when this oxide is dissolved in water
(e) Draw the electron dot structure for the formation of this oxide
Answer:
(a) The element is magnesium (Mg).
(b) The electronic configuration is 2, 8, 2.
(c) Magnesium burns in oxygen (air) to form magnesium oxide which is of basic nature.
2Mg(s) + O2(g) ————-> 2MgO(s)
(d) Magnesium hydroxide is formed.
MgO(s) + H2O(aq) ————> Mg(OH)2(s)
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 5

Question 43.
An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide.
(a) Where in the Periodic Table are elements X and Y placed ?
(b) Classify X and Y as metal(s), non-metal(s) or metalloid(s).
(c) What will be the nature of oxide of element Y ? Identify the nature of bonding in the compound formed
(d) Draw the electron dot structure of the divalent halide.
Answer:
Element X (Z = 17) is chlorine while element Y (Z = 20) is calcium.
(a) Chlorine (Cl) is a member of group 17 and period 3. Calcium (Ca) is present in group 2 and period 4.
(b) Chlorine is a typical non-metal while calcium is a metal.
(c) The oxide of calcium is calcium oxide (CaO). It is a basic oxide.
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 6

Question 44.
Atomic number of a few elements are : 10, 20, 7, 14
(a) Identify the elements
(b) Identify the Group number of these elements in the Periodic Table
(c) Identify the Periods of these elements in the Periodic Table
(d) What would be the electronic configuration for each of these elements ?
(e) Determine the valency of these elements.
Answer:
(a) The elements are : Neon (Z = 10), Calcium (Z = 20), Nitrogen (Z = 7) and Silicon (Z = 14)
(b) Group numbers : Neon (18), Calcium (2), Nitrogen (15), Silicon (14).
(c) Periods : Neon (2), Calcium (4), Fluorine (2), Silicon (3).
(d) Electronic Configuration : Neon (2, 8) ; Calcium (2, 8, 8, 2) Nitrogen (2, 5) ; Silicon (2, 8, 4)
(e) Valency : Neon (zero) ; Calcium (2) ; Nitrogen (3) ; Silicon (4)

Question 45.
(a) In this ladder symbols of elements are jumbled up. Rearrange these symbols of elements in the increasing order of their atomic number in the Periodic Table.
(b) Arrange them in the order of their group also.
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 7
Answer:
(a) H, He, Li, Be, B, C, N, O, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Br.
(b) Group 1 — H, Li, Na, K
Group 2 — Be, Mg, Ca
Group 13 — B, A1
Group 14 — C, Si
Group 15 — N, P
Group 16 — O, S
Group 17 — Cl, Br
Group 18 — He, Ne, Ar

Question 46.
Complete the following cross word puzzle
Across :
(1) An element with atomic number 12.
(3) Metal used in making cans and member of Group 14.
(4) A lustrous non-metal which has 7 electrons in its outermost shell.
Down :
(2) Highly reactive and soft metal which imparts yellow colour when subjected to flame and is kept in kerosene.
(5) The first element of second period
(6) An element which is used in making fluorescent bulbs and is second member of group 18 in the Modern Periodic Table.
(7) A radioactive element which is the last member of the halogen family.
(8) Metal which is an important constituent of steel and forms rust when exposed to moist air.
(9) The first metalloid in Modern Periodic Table whose fibres are used in making bullet proof vests.

1

7 2

3

8 9 5

4

6

                             (Cross-Puzzle)
Answer:

1M

7A

G

N

E

2s

I

U

M

S

O

3T

8I N D 9B

5L

A

R 4I o D I 6n

E

T

O U R T

E

I N M O H

O

N N I

N

E

U

M

                               (Puzzle Solved)

Question 47.
Mendeleev predicted the existence of certain elements not known at that time and named two of them as Eka-silicon and Eka-aluminium.
(a) Name the elements which have taken the place of these elements.
(b) Mention the group and the period of these elements in the Modern Periodic Table.
(c) Classify these elements as metals, non-metals or metalloids.
(d) How many valence electrons are present in each one of them ?
Answer:
(a) Germanium (Ge) for Eka-silicon and gallium (Ga) for Eka-aluminium
(b) Germanium (Group 14 and Period 4)
Gallium (Group 13 and Period 4)
(c) Germanium (Metalloid) ; Gallium (Metal)
(d) Germanium (Z = 32), Four valence electrons (2, 8, 18, 4)
Gallium (Z = 31) ; Three valence electrons (2, 8, 18, 3).

Question 48.
(a) Electropositive nature of the element(s) increases down the group and decreases across the period
(b) Electronegativity of the element decreases down the group and increases across the period
(c) Atomic size increases down the group and decreases across a period (left to right)
(d) Metallic character increases down the group and decreases across a period.
On the basis of the above trends of the Periodic Table, answer the following about the elements with atomic numbers 3 to 9.
(a) Name the most electropositive element among them.
(b) Name the most electronegative element.
(c) Name the element with smallest atomic size.
(d) Name the element which is a metalloid.
(e) Name the element which shows maximum valency.
Answer:
(a) Lithium (Z = 3) is the most electropositive element,
(b) Fluorine (Z = 9) is the most electronegative element.
(c) Fluorine (Z = 9) has the smallest atomic size.
(d) Boron (Z = 5) is a metalloid.
(e) Carbon (Z = 6) shows the maximum valency (4). However, the element nitrogen (Z = 7) can show valency 5 in some compounds (e.g., N2O5).

Question 49.
An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants.
(a) Identify the element X.
(b) Write the electronic configuration of X.
(c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphate crystals.
(d) What would be the nature (acidic/basic) of oxides formed ?
(e) Locate the position of the element in the Modern Periodic Table.
Answer:
(a) The available information suggests that the element X is sulphur.
(b) Electronic configuration of S(Z = 16) 2, 8, 6
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 8
(d) Fe2O3 (basic oxide), SO2(acidic oxide), SO3(acidic oxide)
(e) Sulphur is a member of group 16 and period 3.

Question 50.
An element X of group 15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell.
(a) Identify the element X. How many valence electrons does it have ?
(b) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it ?
(c) Draw the electron dot structure for ammonia. What type of bonds is formed in it ?
Answer:
(a) The available information suggests that the element X is nitrogen (N) and exists in diatomic form as N,
Electronic configuration of N(Z = 7) ; 2, 5. It has five valence electrons.
NCERT Exemplar Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements image - 9

Question 51.
Which group of elements could be placed in Mendeleev’s Periodic Table without disturbing the original order ? Give reason.
Answer:
Group of noble gases (called zero group) could be placed in the Mendeleev’s Periodic Table with out disturbing the original order.
Reasons : Elements present in a group have same valency in the Mendeleev’s Periodic Table. Based on their electronic configuration, the members of noble gas family have zero valency. That is why they are called inert gases. They could be easily placed in the Mendeleev’s Periodic Table as a separate group without disturbing the arrangement of other elements.

Question 52.
Give an account of the process adopted by Mendeleev for the classification of elements. How did he arrive at “Periodic Law” ?
Answer:

  1. The basis of classification of elements adopted was atomic masses of the elements. The elements were arranged in order of increasing atomic masses.
  2. Elements with similar properties were kept in a particular group in order of increasing atomic masses.
  3. Elements placed in a particular group (or sub group) were having same valency.

Note : Please note that when Mendeleev arranged the elements in the Periodic Table only 63 elements were known. Even noble gas or inert gas elements were adjusted at a later stage. Moreover, many groups were left in the table.

 

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test

More Exercises

Question 1.
Mr. Dhruv deposits Rs 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.
Solution:
Deposit per month = Rs 600
Rate of interest = 10% p.a.
Period (n) = 5 years 60 months.
Total principal for one month
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q1.1

Question 2.
Ankita started paying Rs 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs 500 per month in a \(2 \frac { 1 }{ 2 } \) years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?
Solution:
In case of Ankita,
Deposit per month = Rs 400
Period (n) = 3 years = 36 months
Rate of interest = 10%
Total principal for one month
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q2.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q2.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q2.3

Question 3.
Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits Rs 800 per month. If she gets Rs 48200 at the time of maturity, find
(i) the rate of simple interest,
(ii) the total interest earned by Shilpa
Solution:
Deposit per month (P) = Rs 800
Amount of maturity = Rs 48200
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q3.1

Question 4.
Mr. Chaturvedi has a recurring deposit account in Grindlay’s Bank for \(4 \frac { 1 }{ 2 } \) years at 11% p.a. (simple interest). If he gets Rs 101418.75 at the time of maturity, find the monthly instalment.
Solution:
Let each monthly instalment = Rs x
Rate of interest = 11 %
Period (n) = \(4 \frac { 1 }{ 2 } \) years or 54 months,
Total principal for one month
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q4.1

Question 5.
Rajiv Bhardwaj has a recurring deposit account in a bank of Rs 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs 15450 as maturity amount, find the total time for which the account was held.
Solution:
Deposit during the month (P) = Rs 600
Rate of interest = 7% p.a.
Amount of maturity = Rs 15450
Let time = n months
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q5.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test Q5.2

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking Chapter Test are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test

More Exercises

Question 1.
Solve the inequation : 5x – 2 ≤ 3(3 – x) where x ∈ { – 2, – 1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.
Solution:
5x – 2 < 3(3 – x)
⇒ 5x – 2 ≤ 9 – 3x
⇒ 5x + 3x ≤ 9 + 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q1.1

Question 2.
Solve the inequations :
6x – 5 < 3x + 4, x ∈ I.
Solution:
6x – 5 < 3x + 4
6x – 3x < 4 + 5
⇒ 3x <9
⇒ x < 3
x ∈ I
Solution Set = { -1, -2, 2, 1, 0….. }

Question 3.
Find the solution set of the inequation
x + 5 < 2 x + 3 ; x ∈ R
Graph the solution set on the number line.
Solution:
x + 5 ≤ 2x + 3
x – 2x ≤ 3 – 5
⇒ -x ≤ -2
⇒ x ≥ 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q3.1

Question 4.
If x ∈ R (real numbers) and – 1 < 3 – 2x ≤ 7, find solution set and represent it on a number line.
Solution:
-1 < 3 – 2x ≤ 7
-1 < 3 – 2x and 3 – 2x ≤ 7
⇒ 2x < 3 + 1 and – 2x ≤ 7 – 3
⇒ 2x < 4 and -2x ≤ 4
⇒ x < 2 and -x ≤ 2
and x ≥ -2 or -2 ≤ x
x ∈ R
Solution set -2 ≤ x < 2
Solution set on number line
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q4.1

Question 5.
Solve the inequation :
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } ,x\in R\)
Solution:
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le 1\frac { 3 }{ 5 } +\frac { 3x-1 }{ 7 } \)
\(\frac { 5x+1 }{ 7 } -4\left( \frac { x }{ 7 } +\frac { 2 }{ 5 } \right) \le \frac { 8 }{ 5 } +\frac { 3x-1 }{ 7 } \)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q5.1

Question 6.
Find the range of values of a, which satisfy 7 ≤ – 4x + 2 < 12, x ∈ R. Graph these values of a on the real number line.
Solution:
7 < – 4x + 2 < 12
7 < – 4x + 2 and – 4x + 2 < 12
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q6.1

Question 7.
If x∈R, solve \(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
Solution:
\(2x-3\ge x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
\(2x-3\ge x+\frac { 1-x }{ 3 } \) and \(x+\frac { 1-x }{ 3 } >\frac { 2 }{ 5 } x\)
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q7.1

Question 8.
Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.
Solution:
Let the positive integer = x
According to the problem,
5a – 6 < 4x
⇒ 5a – 4x < 6
⇒ x < 6
Solution set = {x : x < 6}
= { 1, 2, 3, 4, 5, 6}

Question 9.
Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.
Solution:
Let first least natural number = x
then second number = x + 1
and third number = x + 2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test Q9.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 4 Linear Inequations Chapter Test are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS

More Exercises

Choose the correct answer from the given four options (1 to 4) :

Question 1.
If Sharukh opened a recurring deposit account in a bank and deposited Rs 800 per month for \(1 \frac { 1 }{ 2 } \) years, then the total money
deposited in the account is
(a) Rs 11400
(b) Rs 14400
(c) Rs 13680
(d) none of these
Solution:
Monthly deposit = Rs800
Period (n) = \(1 \frac { 1 }{ 2 } \) years = 18 months
.’. Total money deposit = Rs 800 x 18
= Rs 14400 (b)

Question 2.
Mrs. Asha Mehta deposit Rs 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is
(a) Rs 65
(b) Rs 120
(c) Rs 130
(d) Rs 260
Solution:
Deposit per month (P) = Rs 250
Period (n) = 1 year = 12 months
Rate (r) = 8% p.a.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS Q2.1

Question 3.
Mr. Sharma deposited Rs 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is
(a) Rs 875
(b) Rs 6875
(c) Rs 10875
(d) Rs 12875
Solution:
Deposit (P) = Rs 500 per month
Period (n) = 2 years = 24 months
Rate (r) = 7% p.a.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS Q3.1

Question 4.
John deposited Rs 400 every month in a bank’s recurring deposit account for \(2 \frac { 1 }{ 2 } \) years. If he gets Rs 1085 as interest at the time of maturity, then the rate of interest per annum is
(a) 6%
(b) 7%
(c) 8%
(d) 9%
Solution:
Deposit (P) = Rs 400 per month
Period (n) = \(2 \frac { 1 }{ 2 } \) years = 3 months
Interest = Rs 1085
Let r% be the rate of interest
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS Q4.1

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 2 Banking MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.