100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Observe the diagram 1. Which step of cardiac cycle is shown here? 2. Which sound is produced during the step and how is it produced?
Answer» 1. Ventricular systole 2. The ‘lub’ sound. This is produced by the closing of AV valve in order to prevent backward flow of blood to auricle.

1.

Observe the diagram 1. Which step of cardiac cycle is shown here? 2. Which sound is produced during the step and how is it produced?

Answer»

1. Ventricular systole

2. The ‘lub’ sound. This is produced by the closing of AV valve in order to prevent backward flow of blood to auricle.

Discussion

2.	First aid treatment of chocking is1. Try to remove object with your finger2. Give five sharp back blows between shoulder blades3. Call of ambulance4. Give something to drink
Answer» Correct Answer - Option 2 : Give five sharp back blows between shoulder blades Explanation: Choking It occurs when a foreign object like a piece of food or small objects lodged in the throat or windpipe, blocking the flow of air. Choking cuts off oxygen to the brain, give first aid as quickly as possible. The universal sign for choking is hands clutched to the throat. If the person doesn't give the signal, look for these indications: Inability to talk Difficulty breathing or noisy breathing Squeaky sounds when trying to breathe Cough, which may either be weak or forceful Skin, lips and nails turning blue or dusky Skin that is flushed then turns pale or bluish in colour Loss of consciousness The American Red Cross recommends a "five-and-five" approach to delivering first aid for chocking: Give 5 back blows: Stand to the side and just behind a choking adult. For a child, kneel down behind. Place one arm across the person's chest for support. Bend the person over at the waist so that the upper body is parallel with the ground. Deliver five separate back blows between the person's shoulder blades with the heel of your hand. Give 5 abdominal thrusts: Perform five abdominal thrusts (also known as the Heimlich manoeuvre). Alternate between 5 blows and 5 thrusts until the blockage is dislodged.

2.

First aid treatment of chocking is1. Try to remove object with your finger2. Give five sharp back blows between shoulder blades3. Call of ambulance4. Give something to drink

Answer» Correct Answer - Option 2 : Give five sharp back blows between shoulder blades

Explanation:

Choking

It occurs when a foreign object like a piece of food or small objects lodged in the throat or windpipe, blocking the flow of air.

Choking cuts off oxygen to the brain, give first aid as quickly as possible.

The universal sign for choking is hands clutched to the throat. If the person doesn't give the signal, look for these indications:

Inability to talk
Difficulty breathing or noisy breathing
Squeaky sounds when trying to breathe
Cough, which may either be weak or forceful
Skin, lips and nails turning blue or dusky
Skin that is flushed then turns pale or bluish in colour
Loss of consciousness

The American Red Cross recommends a "five-and-five" approach to delivering first aid for chocking:

Give 5 back blows:

Stand to the side and just behind a choking adult. For a child, kneel down behind.
Place one arm across the person's chest for support.
Bend the person over at the waist so that the upper body is parallel with the ground.
Deliver five separate back blows between the person's shoulder blades with the heel of your hand.

Give 5 abdominal thrusts:

Perform five abdominal thrusts (also known as the Heimlich manoeuvre).
Alternate between 5 blows and 5 thrusts until the blockage is dislodged.

Discussion

3.	The gross motor development of 8 - 10 months old infant is1. Holds head erect2. Sits without support3. Crawling4. Stands with support
Answer» Correct Answer - Option 2 : Sits without support Explanation: Gross Motor Skill Milestones for Infants 0 - 2 months: Kicks aimlessly, stretches hand and feet, closed fist. 2 - 4 months: Follows a moving person with eyes, stares at a bright object, lifts chest short distance when placed on the abdomen, holds head, rolls from back to the side, begins to grasp toy in hand. 4 - 6 months: Holds head steady when carried, holds toys and reaches for objects, raises hands to be lifted, lifts head and shoulder and rolls over. Lifts self by hands or forearms when lying on the stomach. The child has free and active movement and can respond to music. 6 - 8 months: Can hold the head and sit erect without support. Bangs spoons or pats floors. Can pick up objects from floor, table. Can hold a toy in each hand. Has learnt to use the thumb to grip things. 8 - 10 months: Begins to crawl, can pull self up to stand by holding on to furniture. 10 - 12 months: Stands with slight support, can push light objects. Picks up small and big objects and examines them. 1 - 2 years: Walks without support, drinks from a cup, can handle a spoon, plays with push and pull toys, can climbing stairs, jump, run and come down as well, scribbles, eats by self.

3.

The gross motor development of 8 - 10 months old infant is1. Holds head erect2. Sits without support3. Crawling4. Stands with support

Answer» Correct Answer - Option 2 : Sits without support

Explanation:

Gross Motor Skill Milestones for Infants

0 - 2 months: Kicks aimlessly, stretches hand and feet, closed fist.
2 - 4 months: Follows a moving person with eyes, stares at a bright object, lifts chest short distance when placed on the abdomen, holds head, rolls from back to the side, begins to grasp toy in hand.
4 - 6 months: Holds head steady when carried, holds toys and reaches for objects, raises hands to be lifted, lifts head and shoulder and rolls over. Lifts self by hands or forearms when lying on the stomach. The child has free and active movement and can respond to music.
6 - 8 months: Can hold the head and sit erect without support. Bangs spoons or pats floors. Can pick up objects from floor, table. Can hold a toy in each hand. Has learnt to use the thumb to grip things.
8 - 10 months: Begins to crawl, can pull self up to stand by holding on to furniture.
10 - 12 months: Stands with slight support, can push light objects. Picks up small and big objects and examines them.
1 - 2 years: Walks without support, drinks from a cup, can handle a spoon, plays with push and pull toys, can climbing stairs, jump, run and come down as well, scribbles, eats by self.

Discussion

4.	What is the role of decomposers in the ecosystem? Write the aquatic organisms in order of who eats whom and form a chain of at least 3 steps.
Answer» Various roles played by decomposers in the ecosystem are: i. They decompose biodegradable substances into useful substances. Decomposers feed on dead things. ii. They release nutrients into soil by decomposing dead and decaying matter, thus making the soil fertile. iii. They maintain the nutrient pool by returning back the nutrients in the pool. The names of aquatic organisms in order to form a food chain of at least 3-steps are as follows: Phytoplankton → Zooplankton → small fish→ big fish Phytoplanktons are the producers, Zooplanktons are the primary consumers which are mainly herbivores, small fishes are the secondary consumers and big fishes are the tertiary consumers which are mainly omnivores.

4.

What is the role of decomposers in the ecosystem? Write the aquatic organisms in order of who eats whom and form a chain of at least 3 steps.

Answer»

Various roles played by decomposers in the ecosystem are:

i. They decompose biodegradable substances into useful substances. Decomposers feed on dead things.

ii. They release nutrients into soil by decomposing dead and decaying matter, thus making the soil fertile.

iii. They maintain the nutrient pool by returning back the nutrients in the pool.

The names of aquatic organisms in order to form a food chain of at least 3-steps are as follows:

Phytoplankton → Zooplankton → small fish→ big fish

Phytoplanktons are the producers, Zooplanktons are the primary consumers which are mainly herbivores, small fishes are the secondary consumers and big fishes are the tertiary consumers which are mainly omnivores.

Discussion

5.	What are trophic levels? Give an example of food chain and state the different trophic levels in it.
Answer» Trophic levels: Hierarchical levels in an ecosystem, comprising organisms that share the same function in the food chain and the same nutritional relationship to primary sources of energy. An example of food chain: grass ⟶ grasshopper ⟶ frog ⟶ snake • Plants are the producers and represents the first trophic level. • Grasshoppers are the primary consumers, and represent the 2nd trophic level. • Frogs are the secondary consumers, and represent the third trophic level • Snakes are the tertiary consumers and represent the fourth trophic level.

5.

What are trophic levels? Give an example of food chain and state the different trophic levels in it.

Answer»

Trophic levels:

Hierarchical levels in an ecosystem, comprising organisms that share the same function in the food chain and the same nutritional relationship to primary sources of energy.

An example of food chain:

grass ⟶ grasshopper ⟶ frog ⟶ snake

• Plants are the producers and represents the first trophic level.

• Grasshoppers are the primary consumers, and represent the 2nd trophic level.

• Frogs are the secondary consumers, and represent the third trophic level

• Snakes are the tertiary consumers and represent the fourth trophic level.

Discussion

6.	What are the methods used by plants to get rid of excretory products?
Answer» Plants also excrete metabolic wastes products just like any other organism. Plants excrete the excretory products by the following process: i. Plants consume water for their life processes; however, they do not utilize it completely. The excess water is expelled through the process of transpiration. ii. Other excretory wastes get accumulated in the cell vacuoles, eventually forming resin and gum. This is usually observed in old xylem tissues. Some of these wastes are stored in the leaves which later droop and wither away. iii. CO₂ or gaseous wastes are eliminated through stomata. iv. Some other forms of wastes can be excreted into the soil. Excretory products can also be stored as oils and latex in the barks of trees The main waste materials of plants are Oxygen , Carbon dioxide , water and other harmful substances . Oxygen is the waste product of photosynthesis and carbon dioxide is the waste product of respiration . • Plant gets rid of oxygen and carbon dioxide and other gases through stomata present on leaves and lenticles present on stems . • Access water is removed from the plant body through stomata by the process of 'Transpiration'. • Metabolic waste products are stored in the leaves and the bark of the plants that gets removed whn leaves fall off . • Other waste products like rubber , resin , gums are removed through bark . And some waste products are also removed into soil through roots.

6.

What are the methods used by plants to get rid of excretory products?

Answer»

Plants also excrete metabolic wastes products just like any other organism.

Plants excrete the excretory products by the following process:

i. Plants consume water for their life processes; however, they do not utilize it completely. The excess water is expelled through the process of transpiration.

ii. Other excretory wastes get accumulated in the cell vacuoles, eventually forming resin and gum. This is usually observed in old xylem tissues. Some of these wastes are stored in the leaves which later droop and wither away.

iii. CO₂ or gaseous wastes are eliminated through stomata.

iv. Some other forms of wastes can be excreted into the soil. Excretory products can also be stored as oils and latex in the barks of trees

The main waste materials of plants are Oxygen , Carbon dioxide , water and other harmful substances . Oxygen is the waste product of photosynthesis and carbon dioxide is the waste product of respiration .

• Plant gets rid of oxygen and carbon dioxide and other gases through stomata present on leaves and lenticles present on stems .

• Access water is removed from the plant body through stomata by the process of 'Transpiration'.

• Metabolic waste products are stored in the leaves and the bark of the plants that gets removed whn leaves fall off .

• Other waste products like rubber , resin , gums are removed through bark . And some waste products are also removed into soil through roots.

Discussion

7.	Find the number of solid spheres, each of diameter 6 cm, that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm?
Answer» Given that the diameter of the sphere is 2R = 6 cm. ∴ R = \(\frac{6}{2}\) = 3 cm. Hence, the radius of the sphere is R = 3 cm. Height and diameter of cylinder are h = 45 cm and 2r = 4 cm. Hence, the radius of the cylinder is r = \(\frac{4}{2}\) = 2 cm. Therefore, the volume of the cylinder is v = \(\pi r\) ²ℎ = \(\pi\) × 2 × 2 × 45 = 180\(\pi\) cm³. And volume of the sphere V = \(\frac{4}{3}\) \(\pi r\) ³ = \(\frac{4}{3}\) \(\pi\) × 3 × 3 × 3 = 4 × 9\(\pi\) = 36\(\pi\) cm³ . Let n solid spheres are required to making a solid metal cylinder. Therefore, the volume of cylinder = n × The volume of sphere. ∴ n× 36\(\pi\) = 180\(\pi\) ⇒ n = \(\frac{180 \pi}{36\pi}\) = \(\frac{180}{36}\) = \(\frac{30}{6}\) = 5. Hence, total 5 solid spheres are required to melting for making a solid metal cylinder of given dimensions.

7.

Find the number of solid spheres, each of diameter 6 cm, that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm?

Answer»

Given that the diameter of the sphere is 2R = 6 cm.

∴ R = \(\frac{6}{2}\) = 3 cm.

Hence, the radius of the sphere is R = 3 cm.

Height and diameter of cylinder are h = 45 cm and 2r = 4 cm.

Hence, the radius of the cylinder is r = \(\frac{4}{2}\) = 2 cm.

Therefore, the volume of the cylinder is v = \(\pi r\) ²ℎ = \(\pi\) × 2 × 2 × 45 = 180\(\pi\) cm³.

And volume of the sphere V = \(\frac{4}{3}\) \(\pi r\) ³ = \(\frac{4}{3}\) \(\pi\) × 3 × 3 × 3 = 4 × 9\(\pi\) = 36\(\pi\) cm³ .

Let n solid spheres are required to making a solid metal cylinder.

Therefore, the volume of cylinder = n × The volume of sphere.

∴ n× 36\(\pi\) = 180\(\pi\)

⇒ n = \(\frac{180 \pi}{36\pi}\) = \(\frac{180}{36}\)

= \(\frac{30}{6}\) = 5.

Hence, total 5 solid spheres are required to melting for making a solid metal cylinder of given dimensions.

Discussion

8.	Solve \( : x^{2} d y+\left(x y+y^{2}\right) d x=0 \), given that \( y-1 \) when \( x=1 \)
Answer» x²dy + (xy + y²)dx = 0 \(\frac{dy}{dx} = \frac{-(xy + y^2)}{x^2}\) = \(-\frac{y}{x}-(\frac{y}{x})^2\) Let y = vu \(\frac{dy}{dx} = v + x \frac{dv}{dx}\) \(v + x \frac{dv}{dx}\) = -v -v² \(x \frac{dv}{dx}\) = -(v² + 2v) \(\frac{dv}{v^2 + 2v} = \frac{-dx}{x}\) \(\Rightarrow \int \frac{dv}{(v + 1)^2 - 1} = -\int \frac{dx}{x}\) \(\Rightarrow \frac{1}{2} log\left\|\frac{v + 1 - 1}{v + 1 + 1}\right\| = -log\,x + log\, c\) \(\Rightarrow\) \(log \left(\frac{v}{v + 2}\right) = 2\,log\,\frac{c}{x}\) \(\Rightarrow\) \(log \left(\frac{y/x}{\frac{y}{x} + 2}\right) = log\,\left(\frac{c}{x}\right)^2\) \(\Rightarrow\) \(\frac{y}{y + 2x} = \frac{c^2}{x^2}\) \(\therefore\) when x = 1, y = 1 \(\therefore\) c² = \(\frac{1}{1 + 2} = \frac{1}{3}\) \(\therefore\) solution of given differential equation is \(\frac{y}{y + 2x} = \frac{1}{3x^2}\) \(\Rightarrow\) y + 2x = 3x²y \(\Rightarrow\) y + 2x - 3x²y = 0.

8.

Solve \( : x^{2} d y+\left(x y+y^{2}\right) d x=0 \), given that \( y-1 \) when \( x=1 \)

Answer»

x²dy + (xy + y²)dx = 0

\(\frac{dy}{dx} = \frac{-(xy + y^2)}{x^2}\)

= \(-\frac{y}{x}-(\frac{y}{x})^2\)

Let y = vu

\(\frac{dy}{dx} = v + x \frac{dv}{dx}\)

\(v + x \frac{dv}{dx}\) = -v -v²

\(x \frac{dv}{dx}\) = -(v² + 2v)

\(\frac{dv}{v^2 + 2v} = \frac{-dx}{x}\)

\(\Rightarrow \int \frac{dv}{(v + 1)^2 - 1} = -\int \frac{dx}{x}\)

\(\Rightarrow \frac{1}{2} log\left|\frac{v + 1 - 1}{v + 1 + 1}\right| = -log\,x + log\, c\)

\(\Rightarrow\) \(log \left(\frac{v}{v + 2}\right) = 2\,log\,\frac{c}{x}\)

\(\Rightarrow\) \(log \left(\frac{y/x}{\frac{y}{x} + 2}\right) = log\,\left(\frac{c}{x}\right)^2\)

\(\Rightarrow\) \(\frac{y}{y + 2x} = \frac{c^2}{x^2}\)

\(\therefore\) when x = 1, y = 1

\(\therefore\) c² = \(\frac{1}{1 + 2} = \frac{1}{3}\)

\(\therefore\) solution of given differential equation is

\(\frac{y}{y + 2x} = \frac{1}{3x^2}\)

\(\Rightarrow\) y + 2x = 3x²y

\(\Rightarrow\) y + 2x - 3x²y = 0.

Discussion

9.	If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations ?
Answer» Length of the road roller, h = 1.4 m Radius of the road roller, r = 0.92/2 = 0.45 m Now, Area of the field pressed in one rotation = Curved surface area of the road roller = 2πrh = 2 x 22/7 x 0.45 x 1.4 = 3.96 m² ∴ Area of the field pressed in 500 rotations = 500 × Area of the field pressed in one rotation = 500 × 3.96 = 1980 m² Thus, the area of the field pressed in 500 rotations is 1980 m².

9.

If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations ?

Answer»

Length of the road roller, h = 1.4 m

Radius of the road roller, r = 0.92/2 = 0.45 m

Now,

Area of the field pressed in one rotation = Curved surface area of the road roller

= 2πrh

= 2 x 22/7 x 0.45 x 1.4

= 3.96 m²

∴ Area of the field pressed in 500 rotations = 500 × Area of the field pressed in one rotation = 500 × 3.96 = 1980 m²

Thus, the area of the field pressed in 500 rotations is 1980 m².

Discussion

10.

If a ∶ b = 5 ∶ 8 and b ∶ c = 11 ∶ 4 then find the a ∶ b ∶ c.1. 55 ∶ 88 ∶ 322. 55 ∶ 35 ∶ 443. 44 ∶ 25 ∶ 374. 65 ∶ 55 ∶ 63

Answer» Correct Answer - Option 1 : 55 ∶ 88 ∶ 32

Given:

a ∶ b = 5 ∶ 8

b ∶ c = 11 ∶ 4

Calculation

a	b	c
5 × 11 = 55	8 × 11 = 88
	11 × 8 = 88	4 × 8 = 32

⇒ a ∶ b ∶ c = 55 ∶ 88 ∶ 32

∴ The value of a ∶ b ∶ c is 55 ∶ 88 ∶ 32

Discussion

11.	If A/B = 13/8, then find the value of (A2 + B2)/(A2 – B2).1. 163/1352. 246/1333. 233/1054. 225/105
Answer» Correct Answer - Option 3 : 233/105 Calculation (A² + B²)/(A²– B²) ⇒ (169 + 64)/(169 - 64) ⇒ 233/105 ∴ The correct answer is 233/105

11.

If A/B = 13/8, then find the value of (A2 + B2)/(A2 – B2).1. 163/1352. 246/1333. 233/1054. 225/105

Answer» Correct Answer - Option 3 : 233/105

Calculation

(A² + B²)/(A²– B²)

⇒ (169 + 64)/(169 - 64)

⇒ 233/105

∴ The correct answer is 233/105

Discussion

12.	If 3 × 4 = 21; 4 × 5 = 36; 5 × 6 = 55, then find 6 × 7.1. 912. 673. 424. 78
Answer» Correct Answer - Option 4 : 78 3 × 4 = 21; 4 × 5 = 36; 5 × 6 = 55 See the pattern 3 × (3 + 4) = 21 ⇒ 3 × 7 = 21 ⇒ 21 = 21 (satisfied) Similarly, 4 × 5 = 36 ⇒ 4 × (4 + 5) = 36 ⇒ 4 × 9 = 36 ⇒ 36 = 36 (satisfied) Similarly, 5 × 6 = 55 ⇒ 5 × (5 + 6) = 55 ⇒ 5 × 11 = 55 ⇒ 55 = 55 (satisfied) Similarly, 6 × 7 ⇒ 6 × (6 + 7) ⇒ 6 × 13 ∴ Required value is 78

12.

If 3 × 4 = 21; 4 × 5 = 36; 5 × 6 = 55, then find 6 × 7.1. 912. 673. 424. 78

Answer» Correct Answer - Option 4 : 78

3 × 4 = 21; 4 × 5 = 36; 5 × 6 = 55

See the pattern

3 × (3 + 4) = 21

⇒ 3 × 7 = 21

⇒ 21 = 21 (satisfied)

Similarly,

4 × 5 = 36

⇒ 4 × (4 + 5) = 36

⇒ 4 × 9 = 36

⇒ 36 = 36 (satisfied)

Similarly,

5 × 6 = 55

⇒ 5 × (5 + 6) = 55

⇒ 5 × 11 = 55

⇒ 55 = 55 (satisfied)

Similarly,

6 × 7

⇒ 6 × (6 + 7)

⇒ 6 × 13

∴ Required value is 78

Discussion

13.	If a noiseless channel bandlimited to 5 kHz is sampled every 1msec, what will be the value of sampling frequency? a. 250 samples/sec b. 500 samples/sec c. 800 samples/sec d. 1000 samples/sec
Answer» d. 1000 samples/sec

Discussion

14.	Information rate basically gives an idea about the generated informationper _ by source. a. Second b. Minute c. Hour d. None of the above
Answer» Correct option is a. Second

Discussion

15.	Which lossy method for audio compression is responsible for encoding the difference between two consecutive samples? a. Silence Compression b. Linear Predictive Coding (LPC) c. Adaptive Differential Pulse Code modulation (ADPCM) d. Code Excited Linear Predictor (CELP)
Answer» c. Adaptive Differential Pulse Code modulation (ADPCM)

Discussion

16.	Which among the following is used to construct the binary code that satisfies the prefixcondition? a. Information Rate b. Noiseless Channel c. Channel Coding Theorem d. Kraft Inequality
Answer» d. Kraft Inequality

Discussion

17.	Which coding technique/s exhibit/s the usability of fixed length codes? a. Lempel Ziv b. Huffman c. Both a and b d. None of the above
Answer» a. Lempel Ziv

Discussion

18.	Explain briefly elements of business opportunities. With examples.
Answer» The basic elements of business opportunities are (A) Market for the idea B) Availability of Funds. For an idea to an opportunity, it is essential to have a reasonable market base. Though initially there might not be sufficient customers or clients, however over time there must be demand and market for the opportunity. Second, for a successful opportunity to be profitable, the availability of funds is essential. Easy and cheap availability of capital determines the success of a business. I think the key elements of any business opportunities are: Business concept. You should effectively describe your business concept and ideas, its product, and the market it will serve. You should point out what exactly you will sell through your business and what are the competitions in your business. What is the market size of the business you are planning. Financial features. Highlights the important financial points of the business including sales, profits, cash flows and return on investment. Financial requirements. Clearly states the capital needed to start the business and to expand. It should detail how the capital will be used, and the equity, if any, that will be provided for funding. If the loan for initial capital will be based on security instead of equity, you should also specify the source of collateral. Current business position. Furnishes relevant information about the company, its legal form of operation, when it was formed, the principal owners, and key personnel. Major achievements. Details any developments within the company that are essential to the success of the business. Major achievements include items like patents, prototypes, location of a facility, any crucial contracts that need to be in place for product development, or results from any test marketing that has been conducted.

18.

Explain briefly elements of business opportunities. With examples.

Answer»

The basic elements of business opportunities are

(A) Market for the idea

B) Availability of Funds.

For an idea to an opportunity, it is essential to have a reasonable market base.
Though initially there might not be sufficient customers or clients, however over time there must be demand and market for the opportunity.
Second, for a successful opportunity to be profitable, the availability of funds is essential.
Easy and cheap availability of capital determines the success of a business.

I think the key elements of any business opportunities are:

Business concept. You should effectively describe your business concept and ideas, its product, and the market it will serve. You should point out what exactly you will sell through your business and what are the competitions in your business. What is the market size of the business you are planning.
Financial features. Highlights the important financial points of the business including sales, profits, cash flows and return on investment.
Financial requirements. Clearly states the capital needed to start the business and to expand. It should detail how the capital will be used, and the equity, if any, that will be provided for funding. If the loan for initial capital will be based on security instead of equity, you should also specify the source of collateral.
Current business position. Furnishes relevant information about the company, its legal form of operation, when it was formed, the principal owners, and key personnel.
Major achievements. Details any developments within the company that are essential to the success of the business. Major achievements include items like patents, prototypes, location of a facility, any crucial contracts that need to be in place for product development, or results from any test marketing that has been conducted.

Discussion

19.	Agile Limited has launched a new range of air conditioners in order to add value to the usability of the product. The new range of air conditioners have an inbuilt air purifier and are available in attractive colours. Identify the type of marketing philosophy being described inthe above lines. (a) Product concept (b) Production concept (c) Marketing concept (d) Societal marketing concept
Answer» Correct option is (a) Product concept

Discussion

20.	This technique of scientific management aims to determine the amount and frequencyof rest intervals that should be provided to the employees during working hours. (a) Time study (b) Method study (c) Motion study (d) None of the above
Answer» Correct option is (d) None of the above

Discussion

21.	What are the challenges you are facing during the out break of COVID-19? How it will influence the GDP the world?
Answer» The problems we are facing during COVID-19 are: Minimum supply of food due to high demand. Economic crisis is the major problem as we can not go for work due to which there is a huge number of job cuts in every industry. Not able to start our work. Effect of COVID-19 on GDP of world: Well Coronavirus is going to affect more on GDP of world. Some of the reasons I'm pointing out here: There is no cashflow between countries which is effecting more on GDP. Manufaturing industry is affected more due to COVID-19 which play major role in powering the GDP. Travel and Hospitality industry is measurely affected which also affect GDP of any country. Tourism is going to be affected more for a year which also play important role in GDP.

21.

What are the challenges you are facing during the out break of COVID-19? How it will influence the GDP the world?

Answer»

The problems we are facing during COVID-19 are:

Minimum supply of food due to high demand.
Economic crisis is the major problem as we can not go for work due to which there is a huge number of job cuts in every industry.
Not able to start our work.

Effect of COVID-19 on GDP of world:

Well Coronavirus is going to affect more on GDP of world. Some of the reasons I'm pointing out here:

There is no cashflow between countries which is effecting more on GDP.
Manufaturing industry is affected more due to COVID-19 which play major role in powering the GDP.
Travel and Hospitality industry is measurely affected which also affect GDP of any country.
Tourism is going to be affected more for a year which also play important role in GDP.

Discussion

22.	‘Twinkle Stars’ is a well-known resort for organising parties, especially for children. However, in past 6 months its popularity has reduced considerably as a new resort with better ambience and facilities has opened within its vicinity. Name the related feature of business environment which has influenced the business of ‘Twinkle Stars’ adversely. (a) Totality of external forces (b) Dynamic nature (c) Interrelatedness (d) Uncertainty
Answer» Correct option is (b) Dynamic nature

Discussion

23.	Why do stars twinkle?
Answer» The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Discussion

24.	Botulism means a. Food adultration b. Food poisioning by streptococcus bacteria c. Chemical contamination of food d. Food processing
Answer» c. Chemical contamination of food

Discussion

25.	Penicillin is the drug of choice for a. Scarlet fever b. Whooping cough c. Brucellosis d. Cholera
Answer» a. Scarlet fever

Discussion

26.	A jar contains 20 marbles: 4 red, 6 white and 10 blue. If you remove marbles one at a time, randomly, what is the minimum number that must be removed to be certain that you have at least 2 marbles of each color?
Answer» Here we have 10 Blue 6 White and 4 Red. The worst case scenario you pull all white and blue before you pull 2 red. Hence minimum number of marbles you need the pull is all 10 white, 6 White and 2 Red. Number of marbles to be removed = 10+6+210+6+2 = 18

26.

A jar contains 20 marbles: 4 red, 6 white and 10 blue. If you remove marbles one at a time, randomly, what is the minimum number that must be removed to be certain that you have at least 2 marbles of each color?

Answer»

Here we have 10 Blue 6 White and 4 Red. The worst case scenario you pull all white and blue before you pull 2 red.

Hence minimum number of marbles you need the pull is all 10 white, 6 White and 2 Red.

Number of marbles to be removed = 10+6+210+6+2 = 18

Discussion

27.	The drug of choice for dermal, oral and vaginal candidiasis is a. Griseofulvin b. Amphoterein B c. Gentian violet d. Nystatin
Answer» The drug of choice for dermal, oral and vaginal candidiasis is Nystatin .

Discussion

28.	candidiasis is caused by
Answer» Candidiasis is a fungal infection caused by a yeast (a type of fungus) called Candida.

Discussion

29.	A and B can complete a work in 8 days together. B alone can do in12 days. after working 4days B left the work. how many days will A take to complete the work
Answer» ( A + B) 's 1 days work = 1/8 B's 1 days work = 1/12 ∴ A's 1 day's work = (1/8) - (1/12) = (3 - 2)/24 = 1/24 ∴ A can complete the work in 24 days Now, B's 4 days work = 4/12 = 1/3 Remaining work = 1 - 1/3 = 2/3 As, time taken by A to complete the work is 24 days. ∴ Time taken by A to do 2/3 of the work = 2/3 x 24 = 16 days

Discussion

30.	APTITUDE61. According to Wikipedia, 'monologue' is presented by a single character most often to express their mental thoughts aloud, though sometimes also to directly address another character or the audience. Which of the following does NOT represent a monologue?(A) A poet reciting a poem on the stage(B) Anshuman hosting an award show(C) A mother narrating a story to her child(D) A teacher taking up student queries in the class
Answer» (D) A teacher taking up student queries in the class

Discussion

31.	\( x=1, y-c,-5 \quad\left\|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right\| \) is positive then (a) \( a b c>-11 \) (b) \( a b c<-8 \) (c) \( a b c>-8 \) (d) None of these
Answer» \(\begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix}>0\) ⇒ a\(\begin{vmatrix}b&1\\1&c\end{vmatrix}-1\begin{vmatrix}b&1\\1&c\end{vmatrix}\) + \(1\begin{vmatrix}1&b\\1&1\end{vmatrix}>0\) ⇒ a(bc - 1) - 1(c - 1) + 1(1 - b) > 0 ⇒ abc - a - c + 1+ 1 - b >0 ⇒ abc - (a + b + c) + 2 >0 ⇒ abc >(a + b + c) - 2

31.

\( x=1, y-c,-5 \quad\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right| \) is positive then (a) \( a b c>-11 \) (b) \( a b c<-8 \) (c) \( a b c>-8 \) (d) None of these

Answer»

\(\begin{vmatrix}a&1&1\\1&b&1\\1&1&c\end{vmatrix}>0\)

⇒ a\(\begin{vmatrix}b&1\\1&c\end{vmatrix}-1\begin{vmatrix}b&1\\1&c\end{vmatrix}\) + \(1\begin{vmatrix}1&b\\1&1\end{vmatrix}>0\)

⇒ a(bc - 1) - 1(c - 1) + 1(1 - b) > 0

⇒ abc - a - c + 1+ 1 - b >0

⇒ abc - (a + b + c) + 2 >0

⇒ abc >(a + b + c) - 2

Discussion

32.	\( f(x)=\left\{\begin{array}{cc}x^{2}\left\|\cos \frac{\pi}{x}\right\|, & x \neq 0 \\ 0, & x=0\end{array}, x \in R\right. \) then \( f \) isDifferentiable both at \( x=0 \) and at \( x=2 \)8) Differentiable at \( x=0 \) but not differentiable at \( x=2 \)C) Not differentiable at \( x=0 \) but differentiable at \( x=2 \)Differentiable neither at \( x=0 \) nor at \( x=2 \)
Answer» Option(B) is correct. f(x) = \(\begin{cases}x^2\|cos\frac{\pi}x\|&;x\neq 0\\0&;x=0\end{cases}\) Left hand derivative of f(x) at x = 0 is Df(0^-) = \(\lim\limits_{h\to 0}\frac{f(0-h)-f(0)}{-h}\) = \(\lim\limits_{h\to 0}\frac{(-h)^2\|cos(\frac{\pi}{-h})\|-0}{-h}\) = \(\lim\limits_{h\to 0}-h\|cos(\frac{\pi}h)\|\) = 0 (\(\because \lim\limits_{h\to 0}cos(\frac{\pi}h)\)) gives a bounded value) Right hand derivative of f(x) at x = 0 is Df(0⁺) = \(\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}\) = \(\lim\limits_{h\to 0}\frac{(+h)^2\|cos(\frac{\pi}{-h})\|-0}{h}\) = \(\lim\limits_{h\to 0}h\|cos(\frac{\pi}h)\|\) = 0 (\(\because\) \(\lim\limits_{h\to 0}cos\frac{\pi}4\) gives a bounded value) \(\because\) Df (0^-) = Df(0⁺) \(\therefore\) f is differentiable at x = 0 Left hand derivative of f(x) at x = 2 is Df(2^-) = \(\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{-h}\) = \(\lim\limits_{h\to 0}\frac{(2-h)^2\|cos(\frac{\pi}{2-h})\|-0}{-h}\) (\(\because f(2)=4\times\|cos\frac{\pi}2\|= 4\times0 = 0\)) = \(4\lim\limits_{h\to 0}\cfrac{-\|cos(\frac{\pi}{2-h})\|}{-h}\) (\(\because\frac{\pi}{2-h}>\frac{\pi}2\)) (⇒ cos (\(\frac{\pi}{2-h}\)) lies in 2^nd quadrant so negative.) = \(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})\times\frac{-\pi}{(2-h)^2}\times{-1}}{1}\) = -4π\(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})}{(2-h)^2}\) = -4π\(\cfrac{sin\frac{\pi}2}4\) = \(\frac{-4\pi}4=-\pi\) Right hand derivative of f(x) at x = 2 is Df(2⁺) = \(\lim\limits_{h\to 0}\frac{f(2+h)-f(2)}{h}\) = \(\lim\limits_{h\to 0}\frac{(2+h)^2\|cos(\frac{\pi}{2+h})\|-0}{h}\) (\(\because\) f(2) = 0) = \(4\lim\limits_{h\to 0}\cfrac{cos(\frac{\pi}{2+h})}{h}\) (\(\because \frac{\pi}{2+h}<\frac{\pi}2\)) (⇒\|cos(\(\frac{\pi}{2+h}\))\|) = cos(\(\frac{\pi}{2+h}\)) = \(4\lim\limits_{h\to 0}\cfrac{-sin(\frac{\pi}{2+h})\times\frac{-\pi}{(2+h)^2}}{1}\) = \(4\pi\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2+h})}{(2+h)^2}\) = \(4\pi\times\cfrac{sin\frac{\pi}2}4\) = 4π x \(\cfrac{sin\frac{\pi}2}4\) = 4π x 1/4 = π \(\because\) Df(2^-) \(\neq\) Df(2⁺) \(\therefore\) Function f(x) is not differentiable at x = 2. Hence, function f(x) is differentiable at x = 0 but not differentiable at x = 2.

32.

\( f(x)=\left\{\begin{array}{cc}x^{2}\left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x=0\end{array}, x \in R\right. \) then \( f \) isDifferentiable both at \( x=0 \) and at \( x=2 \)8) Differentiable at \( x=0 \) but not differentiable at \( x=2 \)C) Not differentiable at \( x=0 \) but differentiable at \( x=2 \)Differentiable neither at \( x=0 \) nor at \( x=2 \)

Answer»

Option(B) is correct.

f(x) = \(\begin{cases}x^2|cos\frac{\pi}x|&;x\neq 0\\0&;x=0\end{cases}\)

Left hand derivative of f(x) at x = 0 is

Df(0^-) = \(\lim\limits_{h\to 0}\frac{f(0-h)-f(0)}{-h}\)

= \(\lim\limits_{h\to 0}\frac{(-h)^2|cos(\frac{\pi}{-h})|-0}{-h}\)

= \(\lim\limits_{h\to 0}-h|cos(\frac{\pi}h)|\)

= 0

(\(\because \lim\limits_{h\to 0}cos(\frac{\pi}h)\)) gives a bounded value)

Right hand derivative of f(x) at x = 0 is

Df(0⁺) = \(\lim\limits_{h\to 0}\frac{f(0+h)-f(0)}{h}\)

= \(\lim\limits_{h\to 0}\frac{(+h)^2|cos(\frac{\pi}{-h})|-0}{h}\)

= \(\lim\limits_{h\to 0}h|cos(\frac{\pi}h)|\)

= 0 (\(\because\) \(\lim\limits_{h\to 0}cos\frac{\pi}4\) gives a bounded value)

\(\because\) Df (0^-) = Df(0⁺)

\(\therefore\) f is differentiable at x = 0

Left hand derivative of f(x) at x = 2 is

Df(2^-) = \(\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{-h}\)

= \(\lim\limits_{h\to 0}\frac{(2-h)^2|cos(\frac{\pi}{2-h})|-0}{-h}\) (\(\because f(2)=4\times|cos\frac{\pi}2|= 4\times0 = 0\))

= \(4\lim\limits_{h\to 0}\cfrac{-|cos(\frac{\pi}{2-h})|}{-h}\)

(\(\because\frac{\pi}{2-h}>\frac{\pi}2\))

(⇒ cos (\(\frac{\pi}{2-h}\)) lies in 2^nd quadrant so negative.)

= \(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})\times\frac{-\pi}{(2-h)^2}\times{-1}}{1}\)

= -4π\(\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2-h})}{(2-h)^2}\)

= -4π\(\cfrac{sin\frac{\pi}2}4\) = \(\frac{-4\pi}4=-\pi\)

Right hand derivative of f(x) at x = 2 is

Df(2⁺) = \(\lim\limits_{h\to 0}\frac{f(2+h)-f(2)}{h}\)

= \(\lim\limits_{h\to 0}\frac{(2+h)^2|cos(\frac{\pi}{2+h})|-0}{h}\) (\(\because\) f(2) = 0)

= \(4\lim\limits_{h\to 0}\cfrac{cos(\frac{\pi}{2+h})}{h}\)

(\(\because \frac{\pi}{2+h}<\frac{\pi}2\))

(⇒|cos(\(\frac{\pi}{2+h}\))|) = cos(\(\frac{\pi}{2+h}\))

= \(4\lim\limits_{h\to 0}\cfrac{-sin(\frac{\pi}{2+h})\times\frac{-\pi}{(2+h)^2}}{1}\)

= \(4\pi\lim\limits_{h\to 0}\cfrac{sin(\frac{\pi}{2+h})}{(2+h)^2}\)

= \(4\pi\times\cfrac{sin\frac{\pi}2}4\)

= 4π x \(\cfrac{sin\frac{\pi}2}4\)

= 4π x 1/4 = π

\(\because\) Df(2^-) \(\neq\) Df(2⁺)

\(\therefore\) Function f(x) is not differentiable at x = 2.

Hence, function f(x) is differentiable at x = 0 but not differentiable at x = 2.

Discussion

33.	If \( A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right] \), and \( B=\operatorname{adj} A \quad \& \quad C=3 A \), then \( \mid \frac{\|\operatorname{adj} B\|}{\|C\|}= \) (a) 72 (b) 2 (c) 8 (d) 16
Answer» The correct option is (c) 8. \|A\| = \(\begin{vmatrix}1&1&2\\1&3&4\\1&-1&3\end{vmatrix}\) = \(=1\begin{vmatrix}3&4\\-1&3\end{vmatrix}-1\begin{vmatrix}1&4\\1&3\end{vmatrix}+2\begin{vmatrix}1&3\\1&-1\end{vmatrix}\) = 1(9 + 4) - 1(3 - 4) + 2(-1 - 3) = 13 + 1 - 8 = 6 \|B\| = \|adj A\| = \|A\|^3-1 = \|A\|² = 6² = 36 \|adj B\| = \|B\|^3-1 = \|B\|² = 36² = 6² x 6² \|C\| = \|3A\| = 3³\|A\| = 3³ x 6 \(\therefore\) \(\frac{\|adj B\|}{\|C\|} = \frac{6^2\times6^2}{3^3\times6}\) = \(\frac{6\times2^2}3\) = 2³ = 8

33.

If \( A=\left[\begin{array}{lll}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3\end{array}\right] \), and \( B=\operatorname{adj} A \quad \& \quad C=3 A \), then \( \mid \frac{|\operatorname{adj} B|}{|C|}= \) (a) 72 (b) 2 (c) 8 (d) 16

Answer»

The correct option is (c) 8.

|A| = \(\begin{vmatrix}1&1&2\\1&3&4\\1&-1&3\end{vmatrix}\) = \(=1\begin{vmatrix}3&4\\-1&3\end{vmatrix}-1\begin{vmatrix}1&4\\1&3\end{vmatrix}+2\begin{vmatrix}1&3\\1&-1\end{vmatrix}\)

= 1(9 + 4) - 1(3 - 4) + 2(-1 - 3)

= 13 + 1 - 8

= 6

|B| = |adj A| = |A|^3-1 = |A|² = 6² = 36

|adj B| = |B|^3-1 = |B|² = 36² = 6² x 6²

|C| = |3A| = 3³|A| = 3³ x 6

\(\therefore\) \(\frac{|adj B|}{|C|} = \frac{6^2\times6^2}{3^3\times6}\) = \(\frac{6\times2^2}3\) = 2³ = 8

Discussion

34.	\( 2 A+B+X=0 \), where \( A=\left[\begin{array}{cc}-1 & 2 \\ 3 & 4\end{array}\right] \) and \( B=\left[\begin{array}{cc}3 & -2 \\ 1 & 5\end{array}\right] \) the matrix is
Answer» 2A + B + x = 0 ⇒ x = -2A - B = \(-2\begin{bmatrix} -1&2\\3&4\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}\) \(=\begin{bmatrix}2&-4\\-6&-8\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}\) \(=\begin{bmatrix}2-3&-4+2\\-6-1&-8-5\end{bmatrix}\) \(=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}\)

34.

\( 2 A+B+X=0 \), where \( A=\left[\begin{array}{cc}-1 & 2 \\ 3 & 4\end{array}\right] \) and \( B=\left[\begin{array}{cc}3 & -2 \\ 1 & 5\end{array}\right] \) the matrix is

Answer»

2A + B + x = 0

⇒ x = -2A - B = \(-2\begin{bmatrix} -1&2\\3&4\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}\)

\(=\begin{bmatrix}2&-4\\-6&-8\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}\)

\(=\begin{bmatrix}2-3&-4+2\\-6-1&-8-5\end{bmatrix}\)

\(=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}\)

Discussion

35.	\(\begin{bmatrix}-3 & 9 \\-13 & -1 \\\end{bmatrix}\)\(\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}=\)[-3, 9][-13, -1][1, 0][0, 1] =(A) \(\begin{bmatrix}-3 & 0 \\0 & -1 \\\end{bmatrix}\)(B) \(\begin{bmatrix}-2 & 9 \\-13 & -1 \\\end{bmatrix}\)(C) \(\begin{bmatrix}-3 & 9 \\-13 & -1 \\\end{bmatrix}\)(D) \(\begin{bmatrix}1 & 0 \\0 & 1 \\\end{bmatrix}\)
Answer» Correct answer is (C) \(\begin{bmatrix}-3 & 9 \\-13 & -1 \\\end{bmatrix}\)

Discussion

36.	\(\begin{bmatrix}1\\0\end{bmatrix}\)[2, 3] =[1, 0][2, 3] =(A) [2, 3](B) \(\begin{bmatrix}2\\3\end{bmatrix}\)(C) \(\begin{bmatrix}2 & 0\\0 & 3\end{bmatrix}\)(D) \(\begin{bmatrix}2 & 3\\0 & 0\end{bmatrix}\)
Answer» Correct answer is (D) \(\begin{bmatrix}2 & 3\\0 & 0\end{bmatrix}\)

Discussion

37.	[x-1 y + 2] = [3 5] ⇒ (x, y) =(A) (2, 1) (B) (4, 3) (C) (3, 4) (D) (1, 2)
Answer» Correct answer is (B) (4, 3)

Discussion

38.	For two independent events A and B, if P(A) = x, P(B) = 1/7 and P(A∪B) = 3/7 then the value of x is(A) 1/3(B) 2/3(C) 3/5(D) None of these
Answer» Correct answer is (A) 1/3

Discussion

39.	Find the angle of elevation of the sum (sun's altitude) when the length shadow of a vertical pole is equal to its height.
Answer» The angle of elevation of the sun = 45°.

Discussion

40.	Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?
Answer» Let sandhya invest Rs. x. ∴ When r = 10% total amount = x + 10x/100 = x + x/10 = 11x/10 When r = 12\(\frac12\)% = \(\frac{25}2\)% Total amount = x + \(\cfrac{\frac{25}2x}{100}\) = x + \(\frac x8=\frac{9x}8\) According to given condition, we have \(\frac{9x}8-\frac{11x}{10}=1250\) ⇒ \(\frac{45x-44x}{40}=1250\) ⇒ x = 1250 x 40 = 50000 Hence, sandhya invest Rs.50,000.

40.

Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?

Answer»

Let sandhya invest Rs. x.

∴ When r = 10%

total amount = x + 10x/100 = x + x/10 = 11x/10

When r = 12\(\frac12\)% = \(\frac{25}2\)%

Total amount = x + \(\cfrac{\frac{25}2x}{100}\) = x + \(\frac x8=\frac{9x}8\)

According to given condition, we have

\(\frac{9x}8-\frac{11x}{10}=1250\)

⇒ \(\frac{45x-44x}{40}=1250\)

⇒ x = 1250 x 40

= 50000

Hence, sandhya invest Rs.50,000.

Discussion

41.	solve \( x-2 y=0,3 x+4 y=20 \) (By using eliminationmethod)
Answer» Given the equations x−2y=0 or, x=2y.......(1). And 3x+4y=20......(2). Now, using values of x in (2) we get, 10y=20 or, y=2. Then x=4. So, solutions is x=4,y=2.

Discussion

42.	The first term of an infinite geometric series is 21. The seconds term and the sum of the series are both positive integers. The possible value(s) of the second term can not beA. 14B. 20C. 12D. 18
Answer» Correct Answer - A

Discussion

43.	Let `a_(1),a_(2),a_(3)`, be three positive numbers which are `G.P.` with common ratio r. The inequality `a_(3) gt a_(2) + 2a_(1)` do not holds if r is equal toA. 2B. 1.5C. 0.5D. 2.5
Answer» Correct Answer - A

Discussion

44.	The sequence form by common terms of `17,21,25,"….."` & `16,21,26,"…."` hasA. common difference 20B. sum of first 100 terms is 100 100C. sum of first 1000 terms is 101100D. tenth term is 201.
Answer» Correct Answer - A

Discussion

45.	If `log_(x)a, a^(x//2), log_(b)x` are in G.P. then x is equal toA. `log_(a) (log_(b)a)`B. `(log(log a) - log(log b))/(log a)`C. `log_(b) (log_(a)b)`D. none of these
Answer» Correct Answer - A

Discussion

46.	If `a ,b ,c`are first three terms of a G.P. If the harmonic mean of `aa n db`is 20 and arithmetic mean of `b & c`is 5, thenno term of this G.P.is square of an integerarithmetic mean of `a ,b ,c`is 5c. `b=+-6`d. common ratio of this G.P. is 2A. no term of this G.P. is square of an integerB. arithmetic mean of `a,b,c,` is `5`C. `b = +- 6`D. common ratio of this G.P. is 2
Answer» Correct Answer - A

Discussion

47.	If `S_n` denotes the sum of first n terms of an arithmetic progression and an denotes the `n^(th)` term of the same A.P. given `S_n = n^2p` ; where `p,n in N`, thenA. `a_(1) = p`B. common difference `= 2p`C. `S_(p) = p^(3)`D. `a_(p) = 2p^(3) - p`
Answer» Correct Answer - A

Discussion

48.	a, b, c are the first three terms of geometric series. If the H.M. of a and b is 12 and that of b and c is 36 then which of the following hold(s) good?A. `n(n+1)`B. `n^(2)`C. `n(4-n)`D. `n(6-n)`
Answer» Correct Answer - A

Discussion

49.	a, b, c are the first three terms of geometric series. If the H.M. of a and b is 12 and that of b and c is 36 then which of the following hold(s) good?A. Sum of the first term and common ratio of the G.P. Is 11.B. Sum of the first five terms of the G.P. is 948.C. If the value of the first term and common ratio of the given G.P. is taken as the first term and common difference of an A.P. then its `8^(th)` term is 29`.D. The number `648` is one of the term of the G.P.
Answer» Correct Answer - A

Discussion

50.	The number of terms of an A.P. is even; the sum of the odd terms is310; the sum of the even terms is 340; the last term exceeds the first by 57.Find the number of terms and the first terms of series.A. Number of terms of the series `= 20`B. Number of terms of the series `= 10`C. The first term of series = 4D. The first tem of series = 5
Answer» Correct Answer - A

Discussion

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Find the angle of elevation of the sum (sun's altitude) when the length shadow of a vertical pole is equal to its height.

Sandhya invested a certain amount in a bank. When the rate of interest changed from \( 10 \% \) to \( 121 / 2 \% \) her annual income increased by \( ₹ 1250 \). How much did Sandhya invest?

solve \( x-2 y=0,3 x+4 y=20 \) (By using eliminationmethod)

The first term of an infinite geometric series is 21. The seconds term and the sum of the series are both positive integers. The possible value(s) of the second term can not beA. 14B. 20C. 12D. 18

Let `a_(1),a_(2),a_(3)`, be three positive numbers which are `G.P.` with common ratio r. The inequality `a_(3) gt a_(2) + 2a_(1)` do not holds if r is equal toA. 2B. 1.5C. 0.5D. 2.5

The sequence form by common terms of `17,21,25,"….."` & `16,21,26,"…."` hasA. common difference 20B. sum of first 100 terms is 100 100C. sum of first 1000 terms is 101100D. tenth term is 201.

If `log_(x)a, a^(x//2), log_(b)x` are in G.P. then x is equal toA. `log_(a) (log_(b)a)`B. `(log(log a) - log(log b))/(log a)`C. `log_(b) (log_(a)b)`D. none of these

If `S_n` denotes the sum of first n terms of an arithmetic progression and an denotes the `n^(th)` term of the same A.P. given `S_n = n^2p` ; where `p,n in N`, thenA. `a_(1) = p`B. common difference `= 2p`C. `S_(p) = p^(3)`D. `a_(p) = 2p^(3) - p`

a, b, c are the first three terms of geometric series. If the H.M. of a and b is 12 and that of b and c is 36 then which of the following hold(s) good?A. `n(n+1)`B. `n^(2)`C. `n(4-n)`D. `n(6-n)`

\( x=1, y-c,-5 \quad\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right| \) is positive then (a) \( a b c>-11 \) (b) \( a b c<-8 \) (c) \( a b c>-8 \) (d) None of these

\( 2 A+B+X=0 \), where \( A=\left[\begin{array}{cc}-1 & 2 \\ 3 & 4\end{array}\right] \) and \( B=\left[\begin{array}{cc}3 & -2 \\ 1 & 5\end{array}\right] \) the matrix is

\(\begin{bmatrix}1\\0\end{bmatrix}\)[2, 3] =[1, 0][2, 3] =(A) [2, 3](B) \(\begin{bmatrix}2\\3\end{bmatrix}\)(C) \(\begin{bmatrix}2 & 0\\0 & 3\end{bmatrix}\)(D) \(\begin{bmatrix}2 & 3\\0 & 0\end{bmatrix}\)