125876 + Interview Questions in General QA Page 6 InterviewSolution

251.	What is the ‘great adventure’ referred to in the letter?
Answer» The great adventure in contesting-in the elections for presidentship and campaign trail.

Discussion

252.	Identify the character/speaker.1. ‘I’m going to start this moment.2. ‘By tube or aero-train?’3. ‘By tube’4. ‘Then you’ll be here? At eleven fifty-nine this evening.’5. ‘Paris time?’
Answer» 1. Edith 2. Bennett 3. Edith 4. Bennett 5. Edith

252.

Identify the character/speaker.1. ‘I’m going to start this moment.2. ‘By tube or aero-train?’3. ‘By tube’4. ‘Then you’ll be here? At eleven fifty-nine this evening.’5. ‘Paris time?’

Answer»

1. Edith

2. Bennett

3. Edith

4. Bennett

5. Edith

Discussion

253.	What does he expect his daughters to do?
Answer» He expected his daughters to grow up in a world with no limits on their dreams and no achievements beyond their reach, and to grow into compassionate, committed women who will help build that world.

Discussion

254.	Use the set of words in your own sentences.1. approve – determined2. took place – incident3. follow – create4. desire – perform5. declare – support6. overjoyed – award7. effect – sound
Answer» 1. Though his parents did not approve of his interest in music, he was determined to become a musician. 2. The incident of theft took place last evening, when the family had gone out for a function. 3. If you do not follow traffic rules, it will only create more traffic problems. It was Sheetal’s desire to perform before her dance gurus. 4. All the students have declared their support to Raj, who is standing for the elections. 5. My parents were overjoyed when I got an award for the Best Student. 6. The sound of Lord Krishna’s flute had a magical effect on the people.

254.

Use the set of words in your own sentences.1. approve – determined2. took place – incident3. follow – create4. desire – perform5. declare – support6. overjoyed – award7. effect – sound

Answer»

1. Though his parents did not approve of his interest in music, he was determined to become a musician.

2. The incident of theft took place last evening, when the family had gone out for a function.

3. If you do not follow traffic rules, it will only create more traffic problems.

It was Sheetal’s desire to perform before her dance gurus.

4. All the students have declared their support to Raj, who is standing for the elections.

5. My parents were overjoyed when I got an award for the Best Student.

6. The sound of Lord Krishna’s flute had a magical effect on the people.

Discussion

255.	You’ll find the following contracted forms in the letter written by Obama to his daughters. Read the full forms given below. 1. You’ve 2. hasn’t 3. i’d4. shouldn’t 5. we’ve
Answer» 1. You’ve – you have 2. hasn’t – has not 3. i’d - I would 4. shouldn’t – should not 5. we’ve – we have

255.

You’ll find the following contracted forms in the letter written by Obama to his daughters. Read the full forms given below. 1. You’ve 2. hasn’t 3. i’d4. shouldn’t 5. we’ve

Answer»

1. You’ve – you have

2. hasn’t – has not

3. i’d - I would

4. shouldn’t – should not

5. we’ve – we have

Discussion

256.	Put the following statements in the correct order of occurrence:(1) “If you’ll unwrap that package you may see why you had me going a while at first.”(2) “It’s Christmas Eve, boy.”(3) “Don’t you like me just as well, anyhow?”(4) “It’s sold, I tell you.”
Answer» (3) “Don’t you like me just as well, anyhow?” (4) “It’s sold, I tell you.” (2) “It’s Christmas Eve, boy.” (1) “If you’ll unwrap that package you may see why you had me going a while at first.”

256.

Put the following statements in the correct order of occurrence:(1) “If you’ll unwrap that package you may see why you had me going a while at first.”(2) “It’s Christmas Eve, boy.”(3) “Don’t you like me just as well, anyhow?”(4) “It’s sold, I tell you.”

Answer»

(3) “Don’t you like me just as well, anyhow?”

(4) “It’s sold, I tell you.”

(2) “It’s Christmas Eve, boy.”

(1) “If you’ll unwrap that package you may see why you had me going a while at first.”

Discussion

257.	Data or formula can be copied by : (a) Through cut, copy, paste command on the edit menu (b) Through smart screen menu command (c) Through buttons of standard toolbar (d) All of the above
Answer» (d) All of the above

Discussion

258.	Verify the result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer» For the configuration of the two charges of the same magnitude and sign, the null point is the mid point of the line joining the two charges. If the test charge is displaced slightly from the null point along the line, it will return back due to the restoring force that comes into the day. But if the charge is displaced slightly from the null – point along normal to the line it will not return. This is because the resultant force due to the configuration of two charges will take it away from the null point. For the test charge to be in stable equilibrium restoring force must come into play, when it is displaced in any direction. Hence the test charge cannot be in stable equilibrium.

Discussion

259.	Magnification of lens is found to be +2. What type of lens is that?
Answer» Magnification +ve indicates the image is erect and virtual. Magnification 2 indicates it is magnified. Magnified virtual image is formed by only convex lens.

Discussion

260.	What is the nature of the image formed by a convex lens if the magnification produced by the lens is +3 ?
Answer» The image will be virtual and erect, since the magnification has positive value

Discussion

261.	What are the type of Colour blindness in human?
Answer» Colour blindness (colour vision deficiency) is a condition in which certain colours cannot be distinguished, and is most commonly due to an inherited condition. Cones (colour sensitive receptors) containing single visual pigments selective for red, green, and blue light, are present in the normal human eye. Disturbances of colour vision will occur if the amount of pigment per cone is reduced or if one or more of the three cone systems are absent. Protanomaly is referred to as "red-weakness", an apt description of this form of colour deficiency. Deuteranomalous person adjust your television and he would add more green and subtract red. He is considered "green weak". Nyctalopia (Greek for "night blindness") is a condition making it difficult or impossible to see in relatively low light. It is a symptom of several eye diseases. Night blindness may exist from birth, or be caused by injury or malnutrition (for example, a lack of vitamin A). The outer area of the retina is made up of more rods than cones. The rod cells are the cells that enable us to see in poor illumination. This is the reason why loss of side vision often results in night blindness.

261.

What are the type of Colour blindness in human?

Answer»

Colour blindness (colour vision deficiency) is a condition in which certain colours cannot be distinguished, and is most commonly due to an inherited condition.

Cones (colour sensitive receptors) containing single visual pigments selective for red, green, and blue light, are present in the normal human eye.

Disturbances of colour vision will occur if the amount of pigment per cone is reduced or if one or more of the three cone systems are absent.

Protanomaly is referred to as "red-weakness", an apt description of this form of colour deficiency.

Deuteranomalous person adjust your television and he would add more green and subtract red. He is considered "green weak".

Nyctalopia (Greek for "night blindness") is a condition making it difficult or impossible to see in relatively low light. It is a symptom of several eye diseases.

Night blindness may exist from birth, or be caused by injury or malnutrition (for example, a lack of vitamin A).

The outer area of the retina is made up of more rods than cones. The rod cells are the cells that enable us to see in poor illumination. This is the reason why loss of side vision often results in night blindness.

Discussion

262.	Viscosity is the property of liquids by virtue of which they …… (a) oppose the relative motion of its parts (b) push neighbouring molecules (c) attract other molecules (d) become conducting
Answer» (a) oppose the relative motion of its parts

Discussion

263.	Define relative density (or) specific gravity?
Answer» The relative density of a substance is defined as the ratio of the density of a substance to the density of water at 4°C. It is a dimensionless positive scalar quantity.

Discussion

264.	Explain the classification of longitudinal strain?
Answer» Longitudinal strain can be classified into two types: – 1. Tensile strain: If the length is increased from its natural length then it is known as tensile strain. 2. Compressive strain: If the length is decreased from its natural length then it is known as compressive strain

264.

Explain the classification of longitudinal strain?

Answer»

Longitudinal strain can be classified into two types: –

1. Tensile strain: If the length is increased from its natural length then it is known as tensile strain.

2. Compressive strain: If the length is decreased from its natural length then it is known as compressive strain

Discussion

265.	Define elastic limit?
Answer» Elastic limit: The maximum stress within which the body regains its original size and shape after the removal of deforming force is called the elastic limit.

Discussion

266.	What is meant by “Breaking stress or tensile strength”?
Answer» The maximum stress ulitimate stress point beyond which the wire breaks is called breaking stress or tensile strength.

Discussion

267.	What happens to the pressure inside a soap bubble when air is blown into it?
Answer» When air is blown into a soap bubble, the pressure inside a bubble is decreased P = \(\frac{4T}{R}\)

Discussion

268.	If a stream of air is blown under one of the pans of a physical balance in equilibrium, then the pan will ……..(a) go up (b) go down (c) not be affected (d) go up or down depending on the velocity of the stream
Answer» Correct answer is (b) go down

Discussion

269.	Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals?(a) a 18 kg adult dog(b) a 6 kg cat(c) a 18 kg pregnant dog
Answer» Correct answer is (a)1 tsp (b) 1 1/2 tsp (c) 2 tsp

269.

Animals: The label on a bottle of pet vitamins lists dosage guidelines. What dosage would you give to each of these animals?(a) a 18 kg adult dog(b) a 6 kg cat(c) a 18 kg pregnant dog

Answer»

Correct answer is

(a)1 tsp (b) 1 1/2 tsp (c) 2 tsp

Discussion

270.	"The collision between two hydrogen atoms is perfectly elastic, so the momentum is conserved." Do you agree with this statement?
Answer» No, because momentum is conserved even when the collision is inelastic.

Discussion

271.	A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.
Answer» No, because resistive force of air also acts on the body which is a non-conservative force. So the gain in KE would be smaller than the loss in PE.

Discussion

272.	Few children in your village have complained of swollen necks. More and more children have complained about swollen necks every month. A villager, Hari Singh is superstitious and he thinks that the village is cursed by a ‘devil’. Shyam, another villager, who thinks it as a disease which needs consultation with a doctor.(a) Do you agree with Hari Singh? (b) What can be the reason behind swollen necks of the young children in the village? How can the problem be overcome? (c) What social change is required in the village to make it free form blind faith? (d) What values are shown by Shyam?
Answer» (a) No. Since the case of swollen neck is slowly spreading every month, it seems more likely to be a disease and not any curse. (b) The reason behind swollen necks of the young children in the village is swelling of the lymph nodes which occurs when there is some infection in the lymph nodes. The condition is called Lymphadenitis and it occurs when the child is suffering rom some respiratory infection or has cold. The problem can be overcome by giving the children antibiotics to fight the infection (if caused by bacteria) and medicines to reduce the swelling. (c) The villagers need to be educated to make them free from blind faith. They need to understand that they cannot have blind faith in anything that does not have evidence. Education is the only weapon that can make people free from believing in blind faith and superstitions. (d) Shyam portrays the image of a sensible man who seems to be educated and does not believes in anything that does not have evidence.

272.

Few children in your village have complained of swollen necks. More and more children have complained about swollen necks every month. A villager, Hari Singh is superstitious and he thinks that the village is cursed by a ‘devil’. Shyam, another villager, who thinks it as a disease which needs consultation with a doctor.(a) Do you agree with Hari Singh? (b) What can be the reason behind swollen necks of the young children in the village? How can the problem be overcome? (c) What social change is required in the village to make it free form blind faith? (d) What values are shown by Shyam?

Answer»

(a) No. Since the case of swollen neck is slowly spreading every month, it seems more likely to be a disease and not any curse.

(b) The reason behind swollen necks of the young children in the village is swelling of the lymph nodes which occurs when there is some infection in the lymph nodes. The condition is called Lymphadenitis and it occurs when the child is suffering rom some respiratory infection or has cold. The problem can be overcome by giving the children antibiotics to fight the infection (if caused by bacteria) and medicines to reduce the swelling.

(c) The villagers need to be educated to make them free from blind faith. They need to understand that they cannot have blind faith in anything that does not have evidence. Education is the only weapon that can make people free from believing in blind faith and superstitions.

(d) Shyam portrays the image of a sensible man who seems to be educated and does not believes in anything that does not have evidence.

Discussion

273.	\( (a-b) x+(a+b) y=2 a^{2}-2 b^{2} \) \( (a+b)(x+y)=4 a b \)Solve:(a – b) x + (a + b) y = 2a2 – 2b2 and (a + b) (x + y) = 4ab
Answer» The given system of equations are: (a - b)x + (a + b)y = 2a² - 2b² So, (a - b)x + (a + b)y - 2a² - 2b² = 0 (a - b)x + (a + b)y - 2(a² - b²) = 0 ........(i) And (a + b)(x + y) = 4ab So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii) The given system of equation is in the form of a₁x + b₁y - c₁ = 0 and a₂x + b₂y - c₂ = 0 Compare (i) and (ii) , we get a₁ = a - b, b₁ = a + b, c₁ = -2(a² + b²) a₂ = a + b, b₂ = a + b, c₂ = -4ab By cross-multiplication method x2(a+b)(a2−b2+2ab) =−y2(a−b)(a2+b2) =1−2b(a+b) Now, x2(a+b)(a2−b2+2ab)=1−2b(a+b) ⇒x=2ab−a2+b2b And, −y2(a−b)(a2+b2)=1−2b(a+b) ⇒y=(a−b)(a2−b2)b(a+b) The solution of the system of equations are 2ab−a2+b2b and (a−b)(a2−b2)b(a+b) respectively. Given that, (a – b) x + (a + b) y = 2a² – 2b² (a + b) (x + y) = 4ab (a – b) x + (a + b) y – 2(a²– b²) = 0 (a + b)x + (a + b)y – 4ab = 0 On comparing both the equation with the general form we get a₁ = a – b, b₁ = a + b, c₁ = -2, a₂ = a + b, b₂ = a + b, c₂ = -4ab Now by using cross multiplication we get x/(b₁c₂– b₂c₁) = y/(c₁a₂– c₂a₁) = 1/(a₁b₂ – a₂b₁) ⇒ x/(-(a + b)4ab + 2(a + b) (a² – b²)) = y/(− 2(a² − b²)(a + b) + 4ab(a – b)) = 1/((a − b)(a + b) − (a + b)(a + b)) ⇒ x/(2(a + b)(a² – b² + 2ab)) = 1/-2b(a + b) x = (2ab – a² + b²)/b and, = -y/(2(a – b) (a² + b²) -2b (a + b)) = 1/ -2b(a + b) y = (a – b)(a² + b²)/ b(a + b) Hence, x = (2ab – a² + b²)/b and y = (a – b)(a² + b²)/ b(a + b)

273.

\( (a-b) x+(a+b) y=2 a^{2}-2 b^{2} \) \( (a+b)(x+y)=4 a b \)Solve:(a – b) x + (a + b) y = 2a2 – 2b2 and (a + b) (x + y) = 4ab

Answer»

The given system of equations are:
(a - b)x + (a + b)y = 2a² - 2b²
So, (a - b)x + (a + b)y - 2a² - 2b² = 0
(a - b)x + (a + b)y - 2(a² - b²) = 0 ........(i)
And (a + b)(x + y) = 4ab
So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)
The given system of equation is in the form of
a₁x + b₁y - c₁ = 0
and a₂x + b₂y - c₂ = 0
Compare (i) and (ii) , we get
a₁ = a - b, b₁ = a + b, c₁ = -2(a² + b²)
a₂ = a + b, b₂ = a + b, c₂ = -4ab
By cross-multiplication method
x2(a+b)(a2−b2+2ab) =−y2(a−b)(a2+b2) =1−2b(a+b)
Now, x2(a+b)(a2−b2+2ab)=1−2b(a+b)

⇒x=2ab−a2+b2b

And, −y2(a−b)(a2+b2)=1−2b(a+b)
⇒y=(a−b)(a2−b2)b(a+b)
The solution of the system of equations are 2ab−a2+b2b and (a−b)(a2−b2)b(a+b) respectively.

Given that,

(a – b) x + (a + b) y = 2a² – 2b²

(a + b) (x + y) = 4ab

(a – b) x + (a + b) y – 2(a²– b²) = 0

(a + b)x + (a + b)y – 4ab = 0

On comparing both the equation with the general form we get

a₁ = a – b, b₁ = a + b, c₁ = -2,

a₂ = a + b, b₂ = a + b, c₂ = -4ab

Now by using cross multiplication we get

x/(b₁c₂– b₂c₁) = y/(c₁a₂– c₂a₁) = 1/(a₁b₂ – a₂b₁)

⇒ x/(-(a + b)4ab + 2(a + b) (a² – b²)) = y/(− 2(a² − b²)(a + b) + 4ab(a – b))

= 1/((a − b)(a + b) − (a + b)(a + b))

⇒ x/(2(a + b)(a² – b² + 2ab)) = 1/-2b(a + b)

x = (2ab – a² + b²)/b

and,

= -y/(2(a – b) (a² + b²) -2b (a + b)) = 1/ -2b(a + b)

y = (a – b)(a² + b²)/ b(a + b)

Hence, x = (2ab – a² + b²)/b and y = (a – b)(a² + b²)/ b(a + b)

Discussion

274.	If x + 1 is a factor if the polynomial 2x2 + kx, then k = A. -2 B. -3 C. 4 D. 2
Answer» Let, f (x) = 2x² + kx Since, x + 1 is divided by f (x) so, f (-1) =0 2 (-1) + k (-1) = 0 k = 2

274.

If x + 1 is a factor if the polynomial 2x2 + kx, then k = A. -2 B. -3 C. 4 D. 2

Answer»

Let, f (x) = 2x² + kx

Since, x + 1 is divided by f (x)

so,

f (-1) =0

2 (-1) + k (-1) = 0

k = 2

Discussion

275.	Himi traveled 240 km. by bus and 360 km. by train then determine,(i) Ratio of travelling by train and travelling by bus.(ii) Ratio of travelling by bus and by train.(iii) Ratio of travelling by train and total travelling.(iv) Ratio of travelling by bus and total travelling.
Answer» (i) Journey by train = 360 km. Journey by bus = 240 km. ∴ Ratio of journey travelled by train to journey travelled by bus = 360 : 240 = 130/120 : 240/120 = 3 : 2 (ii) katio of journey travelled by bus to journey’ travelled by train = 240 : 360 = 2 : 3 (iii) Ratio of journey travelled by train to total distance = 360 : (240 + 360) = 360 : 600 = 360/120 : 600/120 (HCF (360,600) = 120) = 3 : 5 (iv) Ratio of journey travelled by bus to total distance = 240 : 600 = 240/120 : 600/120 (HCF (240,600) = 120) = 2 : 5

275.

Himi traveled 240 km. by bus and 360 km. by train then determine,(i) Ratio of travelling by train and travelling by bus.(ii) Ratio of travelling by bus and by train.(iii) Ratio of travelling by train and total travelling.(iv) Ratio of travelling by bus and total travelling.

Answer»

(i) Journey by train = 360 km.

Journey by bus = 240 km.

∴ Ratio of journey travelled by train to journey travelled by bus = 360 : 240

= 130/120 : 240/120

= 3 : 2

(ii) katio of journey travelled by bus to journey’ travelled by train

= 240 : 360

= 2 : 3

(iii) Ratio of journey travelled by train to total distance

= 360 : (240 + 360)

= 360 : 600

= 360/120 : 600/120 (HCF (360,600) = 120)

= 3 : 5

(iv) Ratio of journey travelled by bus to total distance

= 240 : 600

= 240/120 : 600/120 (HCF (240,600) = 120)

= 2 : 5

Discussion

276.	A businessman purchased goods for Rs. 14000. He paid Rs. 350 as auto rent and Rs. 150 as wages. For earning 5% profit, at what price he should sell goods.
Answer» Cost price of products = Rs 14,000 Transportation = Rs 350 Labor = Rs 150 ∴ Total cost price of products = 14,000 + 350 + 150 = Rs 14,500 Profit % = 5% ∴ Profit = 5% of 14,500 = 14,500 x 5/100 = Rs 725 ∴ Selling Price = Total Cost Price + Profit = 14,500 + 725 = Rs 15,225 Hence, he will sell his products for Rs 15,225.

276.

A businessman purchased goods for Rs. 14000. He paid Rs. 350 as auto rent and Rs. 150 as wages. For earning 5% profit, at what price he should sell goods.

Answer»

Cost price of products = Rs 14,000

Transportation = Rs 350

Labor = Rs 150

∴ Total cost price of products

= 14,000 + 350 + 150

= Rs 14,500

Profit % = 5%

∴ Profit = 5% of 14,500

= 14,500 x 5/100

= Rs 725

∴ Selling Price = Total Cost Price + Profit

= 14,500 + 725

= Rs 15,225

Hence, he will sell his products for Rs 15,225.

Discussion

277.	If the cost of one kg sugar is Rs 18, then cost of 3 kg sugar is Rs 54. Thus relation is called(a) Direct Proportion(b) Inverse Proportion(c) Ratio(d) None of the above
Answer» (a) Direct Proportion

Discussion

278.	The compound interest of ? 20,000 at the rate of 8% per year for 2 years is(a) Rs 3,200(b) Rs 1,728(c) Rs 1,600(d) Rs 3,328
Answer» (d) Rs 3,328

Discussion

279.	There are 1275 trees in total in the field of Mohan. There are 36% trees that have fruits. Determine number of trees having fruits in the field.
Answer» Total number of trees = 1275 Number of prolific trees = 36% of 1275 = 1275 x 36/100 = 459 Hence, the required number of prolific trees is 459.

279.

There are 1275 trees in total in the field of Mohan. There are 36% trees that have fruits. Determine number of trees having fruits in the field.

Answer»

Total number of trees = 1275

Number of prolific trees = 36% of 1275

= 1275 x 36/100

= 459

Hence, the required number of prolific trees is 459.

Discussion

280.	By selling 90 ball pens for Rs 160 a person loses 20%. How many ball pens should be sold for Rs 96 so as to have a profit of 20% ?
Answer» S.P. of 90 ball pens = Rs 160, Loss = 20% \(\therefore\) C.P. of 90 ball pens = Rs \(\frac{160\times100}{(100-20)}\) = Rs \(\frac{160\times100}{80}\) = Rs 200 \(\therefore\) C.P. of 1 ball pen = Rs \(\frac{20}{9}\) Suppose x ball pens are sold to earn a profit of 20%. Then, C.P. of x ball pens = Rs \(\frac{20}{9}X\) S.P. of x ball pens = Rs 96 \(\therefore\) Profit = Rs\(\Big(96-\frac{20}{9}X\Big)\) Given, profit % = 20% \(\therefore\) \(\frac{\Big(96-\frac{20}{9}X\Big)}{\frac{20X}{9}}\) x 100 = 20 \(\Rightarrow\) \(\big(96-\frac{20X}{9}\big)\)x 5 = \(\frac{20X}{9}\) \(\Rightarrow\) 96 - \(\frac{20X}{9}\) = \(\frac{4X}{9}\) \(\Rightarrow\) \(\frac{24X}{9}\) = 96 \(\Rightarrow\) x = \(\frac{96\times9}{24}\) = 36. \(\therefore\) 36 ball pens should be sold for Rs 96 to earn a profit of 20%

280.

By selling 90 ball pens for Rs 160 a person loses 20%. How many ball pens should be sold for Rs 96 so as to have a profit of 20% ?

Answer»

S.P. of 90 ball pens = Rs 160, Loss = 20%

\(\therefore\) C.P. of 90 ball pens = Rs \(\frac{160\times100}{(100-20)}\) = Rs \(\frac{160\times100}{80}\) = Rs 200

\(\therefore\) C.P. of 1 ball pen = Rs \(\frac{20}{9}\)

Suppose x ball pens are sold to earn a profit of 20%.

Then, C.P. of x ball pens = Rs \(\frac{20}{9}X\)

S.P. of x ball pens = Rs 96

\(\therefore\) Profit = Rs\(\Big(96-\frac{20}{9}X\Big)\)

Given, profit % = 20%

\(\therefore\) \(\frac{\Big(96-\frac{20}{9}X\Big)}{\frac{20X}{9}}\) x 100 = 20 \(\Rightarrow\) \(\big(96-\frac{20X}{9}\big)\)x 5 = \(\frac{20X}{9}\)

\(\Rightarrow\) 96 - \(\frac{20X}{9}\) = \(\frac{4X}{9}\) \(\Rightarrow\) \(\frac{24X}{9}\) = 96

\(\Rightarrow\) x = \(\frac{96\times9}{24}\) = 36.

\(\therefore\) 36 ball pens should be sold for Rs 96 to earn a profit of 20%

Discussion

281.	A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is(A) 21cm (B) 23cm (C) 25cm (D) 19cm
Answer» (A) 21cm As we know, Volume of cuboid = lbh Where, l = length, b = breadth and h = height For given cuboid, Length, l = 49 cm Breadth, b = 33 cm Height, h = 24 cm Volume of cube = 49× (33) × (24) cm³ Now, Let the radius of cube be r. As volume of sphere = 4/3 πr³ Where r = radius of sphere Also, Volume of cuboid = volume of sphere molded So, 49(33)(24) = 4/3 πr³ ⇒ πr³ = 29106 ⇒ r³ = 29106 × 22/7 ⇒ r³ = 9261 ⇒ r = ∛ 9261 cm = 21 cm Hence, radius of sphere is 21 cm

281.

A solid piece of iron in the form of a cuboid of dimensions 49cm × 33cm × 24cm, is moulded to form a solid sphere. The radius of the sphere is(A) 21cm (B) 23cm (C) 25cm (D) 19cm

Answer»

(A) 21cm

As we know,

Volume of cuboid = lbh

Where, l = length, b = breadth and h = height

For given cuboid,

Length, l = 49 cm

Breadth, b = 33 cm

Height, h = 24 cm

Volume of cube = 49× (33) × (24) cm³

Now,

Let the radius of cube be r.

As volume of sphere = 4/3 πr³

Where r = radius of sphere

Also,

Volume of cuboid = volume of sphere molded

So,

49(33)(24) = 4/3 πr³

⇒ πr³ = 29106

⇒ r³ = 29106 × 22/7

⇒ r³ = 9261

⇒ r = ∛ 9261 cm = 21 cm

Hence, radius of sphere is 21 cm

Discussion

282.	In a chemical factory two wheels of different radius were connected with a belt. Radius of the bigger wheel is 21cm and radius of the smaller wheel is 7cm. If the bigger wheel rotates completely 100 times, find out the number of times that the smaller wheel rotates.
Answer» Given the radius of the bigger wheel R = 21 cm Circumference of the bigger wheel C = 2πR = 2 x \(\frac{22}{7}\) x 21 = 132 cm If the bigger wheel completes 100 rotations, Distance covered by the bigger wheel = Number of rotates x circumference = 100 x 132 = 13,200 cm Now, radius of smaller wheel r = 7 cm. Circumference of the smaller wheel C = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44cm If the smaller wheel completes ‘n’ rotations, Distance covered by the smaller wheel = number of rotations x circumference = n x 44 = 44n cm Distance covered by the small wheel = Distance covered by the bigger wheel 44n = 13200 cm Divide with 44 as both sides, \(\frac{44n}{44}= \frac{13200}{44} = 300\) n = 300 ∴ Number of rotations made by the smaller wheel = 300.

282.

In a chemical factory two wheels of different radius were connected with a belt. Radius of the bigger wheel is 21cm and radius of the smaller wheel is 7cm. If the bigger wheel rotates completely 100 times, find out the number of times that the smaller wheel rotates.

Answer»

Given the radius of the bigger wheel R = 21 cm

Circumference of the bigger wheel C = 2πR

= 2 x \(\frac{22}{7}\) x 21 = 132 cm

If the bigger wheel completes 100 rotations,

Distance covered by the bigger wheel

= Number of rotates x circumference

= 100 x 132 = 13,200 cm

Now, radius of smaller wheel r = 7 cm.

Circumference of the smaller wheel C = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44cm

If the smaller wheel completes ‘n’ rotations,

Distance covered by the smaller wheel

= number of rotations x circumference

= n x 44 = 44n cm

Distance covered by the small wheel =

Distance covered by the bigger wheel 44n = 13200 cm

Divide with 44 as both sides,

\(\frac{44n}{44}= \frac{13200}{44} = 300\)

n = 300

∴ Number of rotations made by the smaller wheel = 300.

Discussion

283.	Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.
Answer» Given details are, Edge of three cubes are = 3cm, 4cm, 5cm Sum of volume of these cubes = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³ After these cubes are melted, a new cube is formed. Let edge length of this new cube be ‘a’ cm a³ = 216 a = √216 = 6cm Edge of new cube is = 6cm ∴ Surface area of new cube = 6 × a² = 6 × 6² = 6 × 36 = 216cm²

283.

Three cubes whose edges measure 3 cm, 4 cm, and 5 cm respectively are melted to form a new cube. Find the surface area of the new cube formed.

Answer»

Given details are,

Edge of three cubes are = 3cm, 4cm, 5cm

Sum of volume of these cubes = 3³ + 4³ + 5³

= 27 + 64 + 125

= 216 cm³

After these cubes are melted, a new cube is formed.

Let edge length of this new cube be ‘a’ cm

a³ = 216

a = √216

= 6cm

Edge of new cube is = 6cm

∴ Surface area of new cube = 6 × a²

= 6 × 6²

= 6 × 36

= 216cm²

Discussion

284.	The cost of preparing the walls of a room 12m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.
Answer» Given details are, Length of room = 12 m Let width of room be ‘b’ m Let height of room be ‘h’ metre Now, Area of floor = 12×b m² = 12b m² Cost of matting the floor at the rate of 85 paise per square metre = Rs 91.80 12b × 0.85 = 91.80 12b = 91.80/0.85 12b = 108 b = 108/12 = 9m Now, Breadth of room = 9m Area of 4 walls = 2 (l×h + b×h) = 2 (12×h + 9×h) = 2 (12h + 9h) = 2 (21h) = 42h m² Cost for preparing walls at the rate of Rs 1.35 per square metre = Rs 340.20 42h × 1.35 = 340.20 42h = 340.20/1.35 42h = 252 h = 252/42 = 6m ∴ Height of room is 6m.

284.

The cost of preparing the walls of a room 12m long at the rate of Rs 1.35 per square metre is Rs 340.20 and the cost of matting the floor at 85 paise per square metre is Rs 91.80. Find the height of the room.

Answer»

Given details are,

Length of room = 12 m

Let width of room be ‘b’ m

Let height of room be ‘h’ metre

Now,

Area of floor = 12×b m² = 12b m²

Cost of matting the floor at the rate of 85 paise per square metre = Rs 91.80

12b × 0.85 = 91.80

12b = 91.80/0.85

12b = 108

b = 108/12

= 9m

Now, Breadth of room = 9m

Area of 4 walls = 2 (l×h + b×h)

= 2 (12×h + 9×h)

= 2 (12h + 9h)

= 2 (21h)

= 42h m²

Cost for preparing walls at the rate of Rs 1.35 per square metre = Rs 340.20

42h × 1.35 = 340.20

42h = 340.20/1.35

42h = 252

h = 252/42

= 6m

∴ Height of room is 6m.

Discussion

285.	A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12m × 8m × 6m, find the cost of iron sheet at Rs. 17.50 per metre.
Answer» Given details are, Dimensions of tank = 12m × 8m × 6m Where, length = 12m Breadth = 8m Height = 6m Area of sheet required = total surface area of tank with one top open = l × b + 2 (l×h + b×h) = 12 × 8 + 2 (12×6 + 8×6) = 96 + 240 = 336 m² Let length be l₁ Breadth be b₁ Given, b₁ = 4m l₁ × b₁ = 336 l₁ = 336/b₁ = 336/4 = 84m ∴ Cost of iron sheet at the rate of Rs 17.50 per metre = 17.50 × 84 = Rs 1470

285.

A tank open at the top is made of iron sheet 4 m wide. If the dimensions of the tank are 12m × 8m × 6m, find the cost of iron sheet at Rs. 17.50 per metre.

Answer»

Given details are,

Dimensions of tank = 12m × 8m × 6m

Where, length = 12m

Breadth = 8m

Height = 6m

Area of sheet required = total surface area of tank with one top open

= l × b + 2 (l×h + b×h)

= 12 × 8 + 2 (12×6 + 8×6)

= 96 + 240

= 336 m²

Let length be l₁

Breadth be b₁

Given, b₁ = 4m

l₁ × b₁ = 336

l₁ = 336/b₁

= 336/4

= 84m

∴ Cost of iron sheet at the rate of Rs 17.50 per metre = 17.50 × 84 = Rs 1470

Discussion

286.	If a trench with measurements 5m × 6m × 3m is dug out then its volume is …………… m3 A) 915 B) 90 C) 165 D) 115
Answer» Correct option is (B) 90 Volume of trench \(=lbh\) \(=5\times6\times3\,m^3=90\,m^3\) Correct option is B) 90

286.

If a trench with measurements 5m × 6m × 3m is dug out then its volume is …………… m3 A) 915 B) 90 C) 165 D) 115

Answer»

Correct option is (B) 90

Volume of trench \(=lbh\)

\(=5\times6\times3\,m^3=90\,m^3\)

Correct option is B) 90

Discussion

287.	A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?
Answer» Match box is in the form of cuboid. Its length, l = 4 cm. breadth, b = 2.5 cm. height, h = 1.5 cm. ∴ Volume of Cuboid, V = l × b × h = 4 × 2.5 × 1.5 V = 15 cm³ . Volume of 1 match box is 15 cm³ . Volume of 12 match boxes …?… . = 15 × 12 = 180 cm³ .

287.

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes ?

Answer»

Match box is in the form of cuboid.

Its length, l = 4 cm.

breadth, b = 2.5 cm.

height, h = 1.5 cm.

∴ Volume of Cuboid, V = l × b × h

= 4 × 2.5 × 1.5

V = 15 cm³ .

Volume of 1 match box is 15 cm³ .

Volume of 12 match boxes …?… .

= 15 × 12

= 180 cm³ .

Discussion

288.	The surface area of a 10cm × 4cm × 6cm brick is(a) 84cm2 (b) 124cm2 (c) 164cm2 (d) 180cm2
Answer» (c) 164cm² Explanation: We know that total surface area of cuboid= 2(l b + b h + h l) = 2 (40 + 30 + 12) = 164

288.

The surface area of a 10cm × 4cm × 6cm brick is(a) 84cm2 (b) 124cm2 (c) 164cm2 (d) 180cm2

Answer»

(c) 164cm²

Explanation:

We know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (40 + 30 + 12)

= 164

Discussion

289.	How many bricks of measurements 25 cm × 11.25 cm × 6 cm are required to built a wall with measurements 8m × 6m × 22.5 cm? A) 6000 B) 6400 C) 2000 D) 8400
Answer» Correct option is (B) 6400 Number of bricks \(=\frac{\text{Volume of wall}}{\text{Volume of a brick}}\) \(=\frac{800\times600\times22.5\,cm^3}{25\times11.25\times6\,cm^3}\) \(=\frac{32}5\times1000\) \(=32\times200=6400\) Correct option is B) 6400

289.

How many bricks of measurements 25 cm × 11.25 cm × 6 cm are required to built a wall with measurements 8m × 6m × 22.5 cm? A) 6000 B) 6400 C) 2000 D) 8400

Answer»

Correct option is (B) 6400

Number of bricks \(=\frac{\text{Volume of wall}}{\text{Volume of a brick}}\)

\(=\frac{800\times600\times22.5\,cm^3}{25\times11.25\times6\,cm^3}\)

\(=\frac{32}5\times1000\)

\(=32\times200=6400\)

Correct option is B) 6400

Discussion

290.	The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm × 3 cm×0.75 cm can be put in this box?
Answer» Given, External dimensions of closed wooden box = 48cm × 36cm × 30cm Thickness of wood = 1.5 cm So inner dimensions of box would be = [(48 - 3)×(36 - 3)×(30 - 3)] = 45cm × 33cm × 27cm Volume of box = 45 × 33 × 27 = 38880 cm³ Volume of a brick = 6 × 3 × 0.75 = 13.5 cm³ So, number of bricks can put in box = \(\cfrac{38880}{13.5}\) = 2970

290.

The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm × 3 cm×0.75 cm can be put in this box?

Answer»

Given,

External dimensions of closed wooden box = 48cm × 36cm × 30cm

Thickness of wood = 1.5 cm

So inner dimensions of box would be = [(48 - 3)×(36 - 3)×(30 - 3)]

= 45cm × 33cm × 27cm

Volume of box = 45 × 33 × 27 = 38880 cm³

Volume of a brick = 6 × 3 × 0.75 = 13.5 cm³

So, number of bricks can put in box = \(\cfrac{38880}{13.5}\) = 2970

Discussion

291.	How many bricks, each of size 25 cm × 13.5cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33cm thick?
Answer» We know that volume of cuboid = length × breadth × height Volume of the brick = 25 × 13.5 × 6 = 2025 cm³ Volume of the wall = 800 × 540 × 33= 14256000 cm³ Therefore, total number of bricks required =Volume of the wall/ volume of the each brick = 1425000/2025 = 7040 bricks

291.

How many bricks, each of size 25 cm × 13.5cm × 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33cm thick?

Answer»

We know that volume of cuboid = length × breadth × height

Volume of the brick = 25 × 13.5 × 6 = 2025 cm³

Volume of the wall = 800 × 540 × 33= 14256000 cm³

Therefore, total number of bricks required

=Volume of the wall/ volume of the each brick

= 1425000/2025

= 7040 bricks

Discussion

292.	The size of matchbox is 4cm × 2.5 cm × 1.5 cm. What is the volume packet containing 144 matchboxes? How many such packets can be place in a carton of size 1.5cm × 84cm × 60 cm?
Answer» We know that volume of cuboid = length × breadth × height Volume occupied by single matchbox = 4 × 2.5 × 1.5 = 15 cm³ Volume of a packet containing 144 matchbox = 15 × 144 = 2160 cm³ Volume of carton is 1.5cm × 84cm × 60 cm = 756000 cm³ Therefore, total number of packets can be accommodated in a carton is Volume of the carton/ volume of the box = 75600/2160 = 350 packets

292.

The size of matchbox is 4cm × 2.5 cm × 1.5 cm. What is the volume packet containing 144 matchboxes? How many such packets can be place in a carton of size 1.5cm × 84cm × 60 cm?

Answer»

We know that volume of cuboid = length × breadth × height

Volume occupied by single matchbox = 4 × 2.5 × 1.5 = 15 cm³

Volume of a packet containing 144 matchbox = 15 × 144 = 2160 cm³

Volume of carton is 1.5cm × 84cm × 60 cm = 756000 cm³

Therefore, total number of packets can be accommodated in a carton is

Volume of the carton/ volume of the box

= 75600/2160

= 350 packets

Discussion

293.	How many bricks, each of size 25 cm x 13.5 cm x 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?
Answer» We know that, Volume of cuboid = Length × Breadth × Height Firstly, Volume of the brick = 25 × 13.5 × 6 = 2025 cm³ Now, Volume of the wall = 800 × 540 × 33 = 14256000 cm³ Hence, Total number of bricks required = \(\frac{Volume\,of\,the\,wall}{Volume\,of\,each\,brick}\) = \(\frac{14256000}{2025}\) = 7040 bricks

293.

How many bricks, each of size 25 cm x 13.5 cm x 6 cm, will be required to build a wall 8 m long, 5.4 m high and 33 cm thick?

Answer»

We know that,

Volume of cuboid = Length × Breadth × Height

Firstly,

Volume of the brick = 25 × 13.5 × 6

= 2025 cm³

Now,

Volume of the wall = 800 × 540 × 33

= 14256000 cm³

Hence,

Total number of bricks required = \(\frac{Volume\,of\,the\,wall}{Volume\,of\,each\,brick}\)

= \(\frac{14256000}{2025}\)

= 7040 bricks

Discussion

294.	Briefly discuss the classes and objects.
Answer» A class is a definition of an object. It is data type just like int. A class is a type, and an object of this class is just like a variable. In other words, class is a blue print and object can be considered as any real time entity (existing thing) that can perform a set of related activities.

Discussion

295.	What is encapsulation?
Answer» It is the method of combining the data, attributes and methods in the same entity is called encapsulation.

Discussion

296.	Define the term data hiding.
Answer» Data Hiding is the mechanism where the details of the class are hidden from the user.

Discussion

297.	Define the term data abstraction.
Answer» The process of hiding the representation of various data items and implementation details of a function from the user program is called data abstraction.

Discussion

298.	A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of liquid can be places in it > Also, find the volume of the wood used in it.
Answer» Given, External dimensions of box = 25cm × 18cm × 15cm Thickness of wood = 2 cm So, internal dimensions of box = (25 - 4) cm × (18 - 4) cm × (15 - 4) cm = 21cm × 14cm × 11cm Volume of liquid can place in it = 21 × 14 × 11 = 3234 cm³ Volume of wood use in it = external volume – internal volume. = 6750 - 3234 = 3516 cm³

298.

A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of liquid can be places in it > Also, find the volume of the wood used in it.

Answer»

Given,

External dimensions of box = 25cm × 18cm × 15cm

Thickness of wood = 2 cm

So, internal dimensions of box = (25 - 4) cm × (18 - 4) cm × (15 - 4) cm

= 21cm × 14cm × 11cm

Volume of liquid can place in it = 21 × 14 × 11 = 3234 cm³

Volume of wood use in it = external volume – internal volume.

= 6750 - 3234 = 3516 cm³

Discussion

299.	How many planks of size 2cm × 25cm × 8 cm can be prepared from a wooden block 5m long, 70cm broad and 32 cm thick, assuming that there is no wastage?
Answer» We know that volume of cuboid = length × breadth × height Volume of the block = 500 × 70 × 32 = 1120000 cm³ Volume of each plank = 2 × 25 × 8= 400 cm³ Therefore, total number of planks that can be made = Volume of the block/ volume of the each plank = 1120000/400 = 2800 planks

299.

How many planks of size 2cm × 25cm × 8 cm can be prepared from a wooden block 5m long, 70cm broad and 32 cm thick, assuming that there is no wastage?

Answer»

We know that volume of cuboid = length × breadth × height

Volume of the block = 500 × 70 × 32 = 1120000 cm³

Volume of each plank = 2 × 25 × 8= 400 cm³

Therefore, total number of planks that can be made

= Volume of the block/ volume of the each plank

= 1120000/400

= 2800 planks

Discussion

300.	The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?
Answer» Given, Dimension of cinema hall = 100m × 50m × 18m Volume of cinema hall = 100 × 50 × 18 m³ Number of persons can sit in cinema hall = \(\cfrac{volume\,of\,hall}{one\,person\,air\,volume}\) \(=\cfrac{100\times50\times18}{150}\) = 600

300.

The dimensions of a cinema hall are 100 m, 50 m and 18 m. How many persons can sit in the hall, if each person requires 150 m3 of air?

Answer»

Given,

Dimension of cinema hall = 100m × 50m × 18m

Volume of cinema hall = 100 × 50 × 18 m³

Number of persons can sit in cinema hall = \(\cfrac{volume\,of\,hall}{one\,person\,air\,volume}\)

\(=\cfrac{100\times50\times18}{150}\) = 600

Discussion

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