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Pulmonary vein brings oxygenated blood from the lungs and open into left auricle. If the pulmonary veins are tied with a thread the oxygenated blood will not supply to the heart and body parts from the lungs. Hence the person will die because of lack of oxygen.

Diagonal Relationship of Lithium with Magnesium

The main points of similarity :

(1) Both have almost similar electronegativities.

(2) Both Li and Mg are quite hard. They are harder and lighter than other elements in their respective groups.

(3) Both LiOH and Mg(OH)₂ are weak bases.

(4) Both form ionic nitrides when heated in atmosphere of Nitrogen.

6 Li + N₂ → 2 Li₃N

3 Mg + N₂ → 2 Mg₃N₂

(5) The hydroxides of both lithium and magnesium decompose upon heating.

2 LiOH → Li₂O + H₂O

Mg(OH)₂ → MgO + H₂O

(6) Both lithium and magnesium combine with oxygen to form monooxide while other members of their respective groups form peroxide and superoxide.

4 Li + O₂ → 2 Li₂O

2 Mg + O₂ → 2 MgO

The Li₂O and MgO thus formed do not combine with excess O₂ to form peroxide and superoxide.

(7) The carbonates of these metals decompose on heating to the corresponding oxides with the evolution of carbon dioxide.

Li₂CO₃ → Li₂O + CO₂

MgCO₃ → MgO + CO₂

(8) Both Lithium and magnesium do not form solid bicarbonates.

(9) Both Lithium and magnesium nitrate decompose on heating producing nitrogen dioxide.

4 LiNO₃ → 2 Li₂O + 4 NO₂ + O₂

2 Mg(NO₃)₂ → MgO + 4 NO₂ + O₂

(10) The hydroxides, carbonate, oxalates, phosphates and fluorides of both Lithium and magnesium are sparingly soluble in water.

(11) Because of the covalent character, LiCl and MgCl₂ are soluble in ethanol.

(12) Both Lithium perchlorate and magnesium perchlorate are highly soluble in ethanol.

(13) LiCl and MgCl₂ are deliquescent and crystallise from aqueous solution as hydrate LiCl·2 H₂O and MgCl₂ ·6 H₂O.

Hello, Galen Today I received a postcard from Pushpa. Remember I had told you that She has gone to Sri Lanka on a holiday? Well, she has written from Colombo. She has written that she had visited Pinnawala Elephant Orphanage. It had 84 elephants. She said that it is the biggest herd of elephants in the world that is living under human supervision. She also added that she was glad that they had come there because she was learning a lot. The Elephant Orphanage was truly worth visiting. She said that next day, they are going to the national park. She would be returning next week and added that she was looking forward to meeting me then.

1. The teacher said to the boys, “Leave this place”.

2. I said to him, “Please give me a glass of water”.

3. The captain said, “Stand at ease”.

4. My brother said, “I may go to Kolkata”.

5. He said, “The earth moves around the sun”.

6.  Ravi said to Ganesh, “When are you going to the library?”

7.The policeman said to the boy, “Show me your licence”.

8. She said to him, “I am going to the market now”.

9. He said, “I am buying a cell phone today”.

10. He said, “Who are you?”

(a). Sharun asked me if I am coming to school, the next day.

(b). Ashok said that they had to visit the historical building of Delhi, since they were there.

(c). run Amutha asked whether he had read ‘The Wind in the Willows’.

(d). The teacher told us that we had to conduct the experiment very carefully

(d). Tejaswar exclaimed happily that it was a great news.

1. Ramu says that he is busy.

2. Suresh said that he liked dancing.

3. He said that he was going to the cinema.

4. Ravi said that he had bought a cycle.

5. Ram said that Visu had come at night. I

6. He said that honesty is the best policy.

7. She said that she had done her homework.

8. He says that he is happy.

9. The teacher said that Hari would definitely pass.

10. Raju said that he would be there that evening.

Failure of pollen to fertilized the same flower or other flower in the plants or it is due to failure to penetrate the full length of style effect fertilization e.g. Mustard.

● When our body encounters an antigenic protein or a pathogen for the first time it produces a response which is of low intensity and our body retains memory of the first encounter. 

● The subsequent encounter with the same pathogen elicits a highly intensified response carried out with the help of two special types of lymphocytes present in our blood, B- lymphocytes, and T-lymphocytes. 

 ● The B-lymphocytes produce an army of proteins in response to these pathogens into our blood to fight with them. These proteins are called antibodies. The T-cells themselves do not secrete antibodies but help B-cells produce them.

1. False – It is prominent protein digesting enzyme present in pancreatic juice. 

2. False – It is the process of swallowing of food . from the pharynx to oesophagus. 

3. True – The collective secretion of intestinal gland is known as intestinal juice or succus entericus. 

4. False – Enterokinase present in succus entericus activates the inactive trypsinogen into active trypsin.

Correct option is A) Despite

Correct option: (c) Geographical

Correct option: (d) rider

Correct option: (a) after

Given

Ramkishan sells an article at a profit of 6%

Profit when sold for Rs. 200 more = 10%

Formula used

SP = CP + (Profit %/100) × CP

Calculation

Profit %= 6%

SP= CP+ (6/100) × CP = 106CP/100

Now, if he had sold it for RS.200 more,

profit % = 10%

⇒ SP + 200 = CP + (10/100) × CP

⇒ SP + 200 = 11CP/10

Substituting SP = 106CP/100 in the above equation we get,

⇒ 106CP/100 + 200 = 11CP/10

⇒ 4CP/100 = 200

⇒ CP = Rs. 5000

He sells goods at loss of 25%.

He used weight of 600 grams instead of 1000 grams at the time of selling.

Formula used :

Profit % = {(Selling price – Cost price )/cost price} × 100

Calculations:

Let the Cost price of 1000 grams of goods be  ₹ 100      ----(1)

As he promises to sell his goods at a loss of 25 %

The selling price of 1000 grams = ₹ 75

But, as he weighs only 600 grams,

⇒ he takes ₹ 75 for 600 grams

From equation (1)

We can get a cost price of 600 grams = ₹ 60

⇒ {(75 –  60) / 60} × 100

⇒ Profit % = 25

∴ The actual profit percentage of the shopkeeper is 25%.

Given:

Successive discount in shop I = 12.5%

Successive discount in shop II = 20% and 5%

Marked price of pant = Rs. 640

Formula used:

Successive decrease = x + y – [(xy)/100]

Calculation:

Overall Discount% offered by shopkeeper I = 12.5% + 12.5% - [(12.5 × 12.5)/100]

= 25 - 1.5625

= 23.4375%

∴ Discount price = (23.4375/100) × 640

= Rs. 150

∴ Selling price of shopkeeper II = 640 - 150

= Rs. 490 

Overall Discount% offered by shopkeeper II = 20 + 5 – [(20 × 5)/100]

= 25 – 1

= 24%

∴ Discount price = (24/100) × 640

= Rs. 153.6

∴ Selling price of shopkeeper II = 640 - 153.6

= Rs. 486.4

Given:

A shopkeeper sells an item that is marked as Rs. 3685 for Rs. 2845

Formula used:

%discount = [(Mark price – Selling price)/Mark price] × 100

Calculation:

Mark price (M.P.) = 3685

Selling price (S.P.) = 2845

%discount = [(M.P. –S.P.)/M.P.] × 100

⇒ [(3685 – 2845)/3685] × 100

⇒ (840/3685) × 100

⇒ 840 × 20/737 = 22.795%

∴ Discount of offering is 22.795%

GIVEN:

Three numbers 2800, 2401 and 1470.

CALCULATION:

Let same remainder = ‘x’

Required maximum number = HCF of (2800 – x), (2401 – x) and (1470 – x)

We know that if two numbers be divisible by a certain number, then their difference is also divisible by that number.

HCF of (2800 – x), (2401 – x) and (1470 – x) = HCF of [(2800 – x) – (2401 – x)], [(2800 – x) – (1470 – x)] and [(2401 – x) – (1470 – x)] = HCF of 399, 1330 and 931 = 133

Hence, required greatest number = 133

Given:

The given numbers are 12.25 and 2/3.

Formula used/Concept Used:

HCF of fraction = (HCF of numerators)/(LCM of denominators)

Calculation:

Fraction conversion of 12.25 = 49/4

HCF of numerator = HCF of (49, 2) = 1

And, LCM of denominator = LCM of (4, 3) = 12

Calculation:

Successive discounts of 10%, 20% and 30% = 100 – [100 × 90% × 80% × 70%]

⇒ 100 – 50.4

⇒ 49.6%

∴ The successive discount of 10%, 20% and 30% is 49.6%

Given:

The given divisors are 415 and 450 and remainders are 5 and 0 respectively.

Concept Used:

When we need to find the greatest number in that case we calculate the HCF of numbers.

Here we have to find the greatest number which when divides x and y leaves the remainder a and b and the number can be found by formula

∴ Required number = HCF of [(x - a), (y - b)]

Calculation:

Required number = HCF of [(415 - 5) and (450 - 0)] = HCF of (410 and 450)

Given:

Total words to type, (work) = 300 words

Time taken by Kaushik to write 3 words = Time taken by Kumar to write 5 words

Concept used:

Work = time × efficiency

Calculation:

The ratio of efficiency of Kaushik to Kumar = 3 ∶ 5

From the table we can see that Kumar finishes his essay in 60 mins.

Number of words by Kaushik in 60 mins = 60 × 3 = 180 word.

Percentage of the essay completed by Kaushik = 180/300 × 100 = 60%

∴ The percentage of the essay completed by Kaushik is 60%.

Given:

The ratios of numbers = 1 ∶ 2.

HCF of number = 7.

Calculation:

Let the numbers are in ratio x ∶ 2x.

∴ HCF of (x and 2x) = x

Now, HCF of numbers is 7

∴ x = 7

Given:

CP = Rs. 3200

Spent on repair = Rs. 600

SP = Rs. 4280

Where CP = Cost price, SP = Selling price

Formula used:

\({\rm{Profit\% \;}} = {\rm{\;}}\frac{{{\rm{SP\;}} - {\rm{\;CP}}}}{{{\rm{CP}}}}{\rm{\;}} × {\rm{\;}}100\)

Calculation:

New CP = Rs. (3200 + 600) = Rs. 3800

\({\rm{Profit\% \;}} = {\rm{\;}}\frac{{4280{\rm{\;}} - {\rm{\;}}3800}}{{3800}}{\rm{\;}} × {\rm{\;}}100\)

⇒ 480/38 × 100

⇒ 12.63%

∴ profit percentage is 12.63%

The least possible number which when divided by 11, 12 and 13 leaves the remainder 7, 8 and 9

Concept used:

Basic concept of number system and LCM

In this type of question when a, b and c divided and leaves remainder d, e and f

If the difference of terms is same

Let (a - d) = (b - e) = (c - f) = K

∴ Required number is LCM of (a, b, c) -K

Calculation:

By using the identity

⇒ (11 - 7) = (12 - 8) = (13 - 9) = 4

Here the difference between the division and the respective remainder is same

∴ Required number = LCM of (11, 12 and 13) – 4 = 1712

Hence, option (2) is correct

Given:

Marked up % = 70% 

If the final selling price after the discount results in the shopkeeper making no profit and no loss

Concept used:

D = M.P – S.P

D% = (D/M.P) × 100

Where,

S.P → Selling price 

M.P → Marked price 

D → Discount 

Calculations:

Let the cost price of the article be 100x 

So, M.P of the article = 100x × (170/100) = 170x 

If the final selling price after the discount results in the shopkeeper making no profit and no loss

So, Selling price of the article = 100x 

Discount = 170x – 100x = 70x 

Discount% = (70x/170x) × 100 = 41.17% 

∴ Required discount% is 41.17% 

GIVEN:

Time taken by outlet pipe to empty = 24 hours

Time taken by both inlet and outlet pipes to fill = 16 hours

Rate of filling of inlet pipe = 2 ltr/min

EXPLANATION:

Time taken by outlet pipe to empty = 24 hours

Time taken by both inlet and outlet pipes to fill = 16 hours

LCM of 24 and 16 is 48 so, let the capacity of the tank be 48 units

Per hour efficiency of outlet pipe = \(\frac{48}{24}= -2\) units/hour (here '-' sign indicates outlet)

Per hour efficiency of both pipes together = \(\frac{48}{16}=3\) units/hour

Per hour efficiency of inlet pipe = Per hour efficiency of both pipes together - Per hour efficiency of the outlet pipe

\(\Rightarrow 3-(-2)=3+2=5 units/hour\)

5 units/hour = \(\frac1{12}\) units/min

 \(\frac1{12}\) units/min = 2 ltr/min

\(\frac1{12}\) units = 2 ltr 

1 unit = 24 ltr

Capacity of tank 48 units = 48 \(\times\) 24 = 1152 ltr

Solution:
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for
some integer q ≥ 0 and r = 0,1,2,3,4,5 because
0 ≤ r ≤ 6.
So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Here, a cannot be 6q or 6q + 2 or 6q + 4, as they are divisible by 2.
6q + 1
6 is divisible by 2 but 1 is not divisible by 2.
6q + 3
6 is divisible by 2 but 3 is not divisible by 2.
6q + 5
6 is divisible by 2 but 5 is not divisible by 2.
Since, 6q + 1, 6q + 3, 6q + 5 are not divisible by 2, they are odd numbers.
Therefore, any odd integer is of the form
6q + 1, or 6q + 3, or 6q + 5.

Solution:
By Euclid’s division algorithm,
92690 = 7378 x 12 + 4154
7378 = 4154 x 1+ 3224
4154 = 3224 x 1 + 930
3224 = 930 x 3 + 434
930 = 434 x 2 + 62
434 = 62 x 7 + 0
Therefore, HCF (92690, 7378) = 62
7161 = 62 x 115 + 31
62 = 31 x 2 + 0
Therefore, HCF(7161, 62) = 31
Hence, HCF (92690, 7378, 7161) = 31

Given:

The ratios of numbers are 7 ∶ 11 and HCF of number is 5

Concept Used:

Basic concept of ratio and HCF

Calculation:

Let the numbers are in the ratio 7x ∶ 11x and we know that HCF is the common factors in both the numbers

∴ HCF of (7x and 11x) = x

Now, HCF of numbers is 5

∴ x = 5

Now, the numbers are in the ratio 7x ∶ 11x

So, the numbers are 35 and 55

∴ Difference of numbers will be = 55 - 35 = 20

Hence, option (1) is correct

Given:

Cost price of old typewriter = Rs 19200

Selling price of old typewriter = Rs 24150

Profit = 15%

Formula Used:

Selling price = (Cost price + amount spend on repairing) × (100 + profit %)/100

Calculation:

Let the amount spend on repairing of typewriter be x.

Selling price = (Cost price + amount spend on repairing) × (100 + profit %)/100

⇒ 24150 = (19200 + x) × (100 + 15)/100

⇒ 24150 = (19200 + x) × 115/100

⇒ 19200 + x = 24150 × 100/115

⇒ 19200 + x = 21000

⇒ x = 21000 – 19200

⇒ x = 1800

Given:

Numbers are 270, 675, 1215

Calculation:

Let us assume that the remainder left is m

So, numbers that are completely divisible are

⇒ 270 – m, 675 – m, 1215 – m

Now subtract the pair

⇒ 675 – m – 270 + m = 405      ----1

⇒ 1215 – m – 675 + m = 540      ----2

Now find the HCF of 405 and 540

⇒ 405 = 135 × 3

⇒ 540 = 135 × 4

HCF of 405 and 540 = 135

∴ The largest number by which 270, 675 and 1215 are divided, leave the equal remainder in each case is 135