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Given:

Terms differ by 40.

The measure is 2/7

Calculation:

Let the terms be x and x + 40

⇒ x/(x + 40) = 2/7

⇒ 7x = 2x + 80

⇒ 5x = 80

⇒ x = 80/5 = 16

⇒ x + 40 = 56

∴ The ratio is 16 : 56.

(f) The narrator was much impressed with Bill. He first meets him at the station. He finds him to be friendly and cheerful. Bill is very helpful in his eyes as he offers to take him around in search of Lutkins. The narrator admires him when he goes looking for Lutkins on his behalf. Bill is full of a wonderful village charm. The narrator finds Bill to have a unique country wisdom. He admires him as a story teller. He appreciates him a lot when Bill even goes to Lutkins’ mother’s place to find him. For the narrator, Bill is a friendly man who helps others generously. He is so impressed by Bill that he decides to settle down in the village.

Given:

The efficiency of A = 2 × efficiency of B

Time taken by A = time taken by B - 20

Concept used:

Time = (Total work)/(Efficiency)

1.) When work is the same the ratio of time and efficiency is inversely proportional. eg - if time = 3 : 4 then, efficiency = 4 : 3 

2.) The time required to complete the work is the reciprocal of the fraction of work done in one time period (one day).

Calculation:

Efficiency → A : B = 2 : 1

Time taken → A : B = 1 : 2

Difference in time = 2 - 1 = 1 ratio

∴ 1 ratio = 20 days 

A's time = 20 days 

B's time = 40 days

(A + B)'s one day work = \(\frac{1}{20}+\frac{1}{40}=\frac{3}{40}\)

Time taken to complete the work = \(\frac{40}{3}=13\frac13days\)

∴ Time taken to complete the work is \(13\frac13days\) 

Concept used:-

Trick to find square root of a number.

Method and Calculation:-

 • Find the unit digit of square root by looking at unit digit of given number.

For ✓2209, unit digit must be 3 or 7

 • Find tens digit by looking at number formed by first 2 digit.

16 < 22 < 25 (16 and 25 are squares of 4 and 5, so tens digit of square root will be 4.

 • Find the square root from the narrowed down options.

✓2209 = 43 or 47. To find this, we can find the square of number between these numbers ending in 5 i.e. 45.

45² = 2025 (for left part multiply tens digit with one more than tens digit, right part is 25)

Clearly, ✓2209 = 47 (2209 is more than 2025)

Given:

Number of days taken, if Heena and Karan work together = 12 days

Heena alone completes the work in 18 days.

Calculation:

Let the number of days taken by Karan to complete work be x.

Work done in 1 days by Heena = (1/18)

Work done in 1 day by Karan = (1/x)

Work done in 1 day when both work together = (1/12)

According to question:

(1/18) + (1/x) = 1/12

⇒ 1/x = (3 - 2)/36 = 1/36

⇒ x = 36 days

∴ Karan completes the work in 36 days.

√ 3249 = 57 and √ 2209 = 47 and √ 961 = 31

√ 3249 - √ 2209 + √ 961 

= 57 - 47 + 31

= 41

√ 225 = 15 and √ 256 = 16 and √ 961 = 31

(√ 225 + √ 256) ÷ √ 961 

= (15 + 16) ÷ 31

= 31 ÷ 31

= 1

\(\left| \frac{z_1 + \bar {z_2}}{\bar {z_1} + z_2}\right| = \left|\frac{(z_1 + \bar{z_2})}{(\bar{z_1} + z_2)} \times \frac{z_1 + \bar{z_2}}{z_1 + \bar{z_2}} \right|\)

\(= \left|\frac{(z_1 + \bar{z_2})^2}{(\bar{z_1}+z_2)(\overline{\bar{z_1} + z_2})}\right|\)

\(\)\(= \left|\frac{(z_1 + \bar{z_2})^2}{(\bar{z_1}+z_2)^2}\right|\)      \((\because z\;\bar z = |z|^2)\) 

\(= \frac{|(z_1 + \bar{z_2})^2|}{|\bar {z_1} + z_2|^2}\)

\(= \frac{|z_1 + \bar{z_2}|^2}{|\overline{z_1 + \bar{z_2}}|^2}\)

\(= \frac{|z_1 + \bar{z_2}|^2}{|{z_1 + \bar{z_2}}|^2}\)       \((\because |\bar z| = |z|)\)

\(= 1\)

Hence, \(\left| \frac{z_1 + \bar {z_2}}{\bar {z_1} + z_2}\right| = 1\)

The smallest odd number is 1.

Calculation:

The series follows the following pattern

3

3 × 0.5 + 0.5 = 2

2 × 1 + 1 = 3

3 × 1.5 + 1.5 = 6

6 × 2 + 2 = 14

14 × 2.5 + 2.5 = 37.5

∴ The wrong number in the given series is 4

Given:

The profit is 84000

Calculation:

⇒ The ratio of their profit = 2 : 3 : 7

⇒ The profit of Rajesh = (3/12) × 84000 = 21,000

∴ The required result will be 21,000.

Given:

CP of X = 3 × CP of Y

SP of Y = Rs. 18,000

Loss% = 10%

Formula used:

CP = SP × 100/(100 – L%)

Where

CP → Cost price

SP → Selling Price

L% → Loss percent

Calculation:

CP of Y = CP_Y = (SP of Y) × 100/(100 – L%)

⇒ CP_Y = 18000 × 100/(100 – 10)

⇒ CP_Y = 18000 × (100/90)

⇒ CP_Y = 20,000

CP_X = 3 × CP_Y

⇒ CP_X = 3 × 20,000

∴ Cost price of X is Rs. 60,000

Given:

Sathi and Rathin invested some money in a business in the ratio 6 ∶ 5

Profit was shared between Sathi and Rathin in the ratio 7 ∶ 10

Concept used:

(Invest of money A)/(invest of money B)× (invest of time A)/(invest of time B)= (Profit of A)/(profit of B)

Calculation:

Suppose Sathi withdrew her money after p months

So,

⇒ (6 × p)/(5 × 12) = 7/10

⇒ 6p/60 = 7/10

⇒ p = 7

Rathin invest alone 12 – 7 = 5 month

∴ Rathin alone invest for 5 month.

Given:

(i) Amit’s investment = Rs.20,000

(ii) Trisha’s investment = Rs.25,000

(iii) Profit after one year = Rs.83,000

Common Mistake:

(i) If A invested after 4 months then it means B invested for (12 - 4) months or 8 months.

Calculations:

Amit’s Share : Trisha’s Share

⇒ 20000 × 12 : 25000 × (12 - 5)

⇒ 240 : 175

⇒ 48 : 35

⇒ Now let Amit’s share be 48x.

⇒ Trisha’s share be 35x.

According to the question,

⇒ 48x + 35x = 83000

⇒ 83x = 83000

⇒ X = 1000

Given:

Time taken to dig a pond by 63 people = 126 days

Formula used:

Total work = Number of persons × Number of days

Calculation:

Total work = 63 × 126

Time taken by 98 people to dig it = (63 × 126)/98 days

⇒ 81 days

∴ 98 people can dig it in 81 days

Given:

Krishna starts a business with 45,000.

Three months later Arjun joins him with 30,000.

The time period of Krishna is 12 months.

Concept Used:

L₁: L₂ = I₁ ×  T₁ : I₂ ×  T₂ 

Here, L = Loss, I = Investment, T = Time

Calculation:

Three months later Arjun joins him with 30, 000.

The time period of Arjun = (12 - 3) = 9 months

Then, L1: L2 = 45000 × 12 : 30000 × 9

⇒ L1: L2 = 2 : 1

∴ The ratio of the loss is 2 : 1.

Given:

Number of days taken by 45 men to dig a pond = 25 days

Concept used:

Total work = Number of men × Time taken

Calculation:

Total work = (45 × 25) units

Number of men required to dig it in 15 days = (45 × 25)/15 men

⇒ 75 men

∴ The number of men to be employed is 75

Given:

3 friends A, B and C invested Rs. 5000, 6000 and Rs. 7000 for 8 months, 4 months and 2 months respectively. Each of them invested Rs. 10,000 for the remaining part of the year.

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Given, A, B and C invested 5000, 6000 and 7000 for 8, 4 and 2 months respectively and each of them invested Rs. 10,000 for the remaining part of the year.

A                                     :       B                                        :        C

5000 × 8 + 10,000 × 4    :       6000 × 4 + 10,000 × 8       :        7000 × 2 + 10,000 × 10

80,000                            :       104000                               :        114000

80                                   :       104                                     :        114

40                                   :        52                                      :         57

Given:

A start with investment Rs. 21,000 

B Joins with Rs. 36,000

Formula used:

Profit = investment × time

Calculation:

Let B invested for x months in the business.

A invest for a year that means 12 months.

A's profit = 21000 × 12

B's profit = 36000 × x

According to question 

A's profit = B's profit

⇒ 21000 × 12 = 36000 × x

⇒ x = (21000 × 12)/36000

⇒ x = 7 months

B invest after 12 – 7 = 5 months

∴ B invest after 5 months of business started by A

Given:

Investment of Suresh = ₹ 7000

Investment of Ramesh = ₹ 5000

Investment of Garish = ₹ 3000

Total loss at the end of 1 year = ₹ 11800

Concept used:

Loss = Investment × time

Calculation:

Suresh invested for 6 months

Ramesh invested for 8 months

Garish invested for 12 months

Loss ratio = (7000 × 6) : (5000 × 8) : (3000 × 12)

⇒ Loss ratio = 21 : 20 : 18

Total loss in terms of ratio = 21 + 20 + 18 = 59

⇒ Loss share of Suresh = (11800/59) × 21 = 4200

∴ The loss share of Suresh is ₹ 4200.

Given :

Ramesh purchased the car at Rs x

Ramesh sold the car to Suresh at 90% profit at a selling price of Rs y

Suresh purchased the car at Rs y

Suresh sold the car to Ganesh at a 20% profit. at Rs 1140000

Ganesh purchased the car at Rs 1140000

Formula used :

Profit% = (selling price – cost price)/cost price) × 100

Calculations :

Suresh sold the car Rs 1140000 at 20% profit

20 = (1140000 – y/y) × 100 (using the above-given formula, here y is the cost price and 1140000 is the selling price for Suresh)

20/100 = (1140000 – y)/y

1/5 = (1140000 – y)/y

6y = 5 × 1140000

⇒ y = 950000

Similarly for Ramesh who sold the car at 15% profit at Rs y (y = 950000)

90 = (950000 – x)/x) × 100     (using above given formula, where x is the cost price and 950000 is the selling price for Ramesh)

90/100 = (950000 – x)/x

9/10 = (950000 – x)/x

19x = 950000 × 10

⇒ x = 500000 

Now

y – x = 950000 – 500000

⇒ 450000

∴ The value of y – x is Rs 450000.

Given:

Total work = 60 units

Work done by Ramesh in 10 days = 30% of total work

Work done by Suresh in 10 days = 40% of total work

Calculations:

Work done by Ramesh in 10 days = 60 × 30% = 18 units

Work done by Ramesh in 1 day = 18/10 = 1.8 units

Work done by Ramesh in 3 days = 1.8 × 3 = 5.4 units

Work done by Suresh in 10 days = 60 × 40% = 24 units

Work done by Suresh in 1 day = 24/10 = 2.4 units

Work done by Suresh in 3 days = 2.4 × 3 = 7.2 units

Difference = 7.2 - 5.4 = 1.8 units

∴ Difference of work done by Ramesh and Suresh in 3 days is 1.8 units.

Given:

M and N together can do a piece of work = 4 days.

N takes twice as much days alone as taken by M to complete the same piece of work.

Concepts used:

Total work = LCM

Efficiency = Work/Time

Calculation:

Let the number of days taken by M to complete the piece of work alone be 'a' days.

So, time taken by N alone to complete the work = 2a days

Total work = LCM of 'a' and '2a' = 2a

Time = Total work/Efficiency

⇒ 4 = 2a/3

⇒ a = 12/2 days

⇒ a = 6 days

Time taken by M to complete the work alone = a days = 6 days.  

∴ M takes 6 days alone to complete a piece of work. 

Let the price of article be Rs. Y

If the price of an article is increased by 25%

The new price = Y + (Y × 25)/100 = 5y/4

To restore its former value the new price must be decreased by = \(\frac{\frac{5y}{4}-y}{\frac{5y}{4}}\) \(\times 100\) =20%

T.S. Eliot wrote the above line.

Correct answer is 6

B₂H₆ + 6MeOH → 2B (OMe)₃ + 6H₂

1 mole of B₂H₆ reacts with 6 mole of MeOH to give 2 moles of B(OMe)₃. 

3 mole of B₂H₆ will react with 18 mole of MeOH to give 6 moles of B(OMe)₃

In conversion of \(^{238}_{92}U\) to \(^{206}_{82}Pb\) , 8α - particles and 6β particles are ejected. The number of gaseous moles initially = 1 mol

The number of gaseous moles finally = 1 + 8 mol; (1 mol from air and 8 mol of ₂He⁴) 

So the ratio = 9/1 = 9

Volume of larger sphere = (4/3)π*6*6*6) cm³ = 288π cm³. 

Volume of 1 small lead ball = ((4/3)π*(1/2)*(1/2)*(1/2)) cm³ = π/6 cm³ . 

Number of lead balls = (288π*(6/π)) = 1728.

Volume of the liquid in the cylindrical vessel = Volume of the conical vessel 

= ((1/3)* (22/7)* 12 * 12 * 50) ) cm³ = (22 *4 *12 * 50)/7 cm³ . 

Let the height of the liquid in the vessel be h. 

Then (22/7)*10*10*h =(22*4*12*50)/7 or h = (4*12*50)/100 = 24 cm

Volume = πr²h = ((22/7) x (7/2) x (7/2) x 40) = 1540 cm³ 

Curved surface area = 2πrh = (2x(22/7)x(7/2)x40)= 880 cm². 

Total surface area = 2πrh + 2πr² = 2πr (h + r) 

= (2 x (22/7) x (7/2) x (40+3.5)) cm² 

= 957 cm² 

Here, r = 21 cm and h = 28 cm. 

Slant height, l = √r² + h² = √(21)² + (28)² = √1225 = 35 cm

Inner radius = (3/2) cm = 1.5 cm, Outer radius = (1.5 + 1) = 2.5 cm. 

Volume of iron = [π x (2.5)² x 100 - π x (1.5)² x 100] cm³ 

= (22/7) x 100 x [(2.5)² - (1.5)² ] cm³ 

= (8800/7) cm³ 

Weight of the pipe = ((8800/7) x (21/1000))kg = 26.4 kg.

Correct option (4) p² – 4q – 12 = 0

Explanation:

If one root of equation x² + px + q = 0 is 2 – √3 

then other root will be 2+ √3

∴ equation x² – 4x + 1 = 0

⇒ p = –4 and q = 1

⇒ p² – 4q – 12 = 0 

Solution:

Given diameter of copper rod, d = 1 cm
Hence its radius, R= (1/2) cm
Height of rod, H = length of rod = 8 cm
∴ Volume of rod = πR²H cubic units
= π x (1/2)² x 8 cubic cm
= 2π cubic cm
Given that cylindrical rod is drawn into a cylindrical wire of length, h = 18 m
That is, h = 1800 cm
∴ Volume of cylindrical rod = Volume of cylindrical wire
Let r be the radius of the wire
Hence 2π = πr²h
That is, 2 = r² x 1800
r² = (1/900)
Therefore, r = (1/30) cm = 1/3 mm
Diameter or thickness = 2(1/3) = (2/3) mm

If and are relatively prime, then their HCF must be 1.

Equation of the parabola is y² = 6x 

i.e., S ≡ y² – 6x 

S₁₁ = (–6)² – 6.6 = 36 – 36 = 0 

∴ (6, – 6) lies on the parabola.

(d) Feldspar is not aluminosilicate

(c) (i) and (iii)

(a) VO₂⁺ < Cr2O₇^2- < MnO^-₄

VO₂⁺ < Cr2O₇^2- < MnO^-₄ greater the oxidation state, higher is the oxidising power.

(c) Octahedral

Option: (b) -cotx+c

Correct option  (C) a = 1 or 23/3

Explanation :

By hypothesis

6 - 26 =  ± 20

⇒ a = 46/6 or 6/6

⇒ a = 1 or 23/3

To verify : sin3 = 3sin – 4 sin3 

Given that = 30° 

L.H.S = sin 3 = sin (3× 30°) = sin(90°) = 1. (∵ sin90° = 1) 

R.H.S = 3 sin – 4sin³ = 3sin30° – 4 sin³30°

3 x \(\frac{1}{2} - 4(\frac{1}{2})^3\)   (\(\because sin 30° = 1\))

\(\frac{3}{2} - \frac{4}{8} = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1.\)

Hence, L.H.S = R.H.S. 

Therefore, for =30°, sin3 = 3sin – 4sin³. 

Hence Proved

Solution: 
65 + 76.8 = 141.8
141.8 + 38.4 = 180.2
180.2 + 19.2 = 199.4
199.4 + 9.6 = 209 
209 + 4.8 = 213.8

Given:

M₁ = 25

D₁ = 10

H₁ = 6

M₂ = 18

H₂ = 10

W₂ = 3 × W₁

Formula Used:

(M₁ × D₁ × H₁)/W₁ = (M₂ × D₂ × H₂)/W₂

Where M₁ → Initial Men

D₁ → Initial day

H₁ → Initial hour

M₂ → Final men

D₂ → Final day

H₂ → Final hour

Calculation:

(M₁ × D₁ × H₁)/W₁ = (M₂ × D₂ × H₂)/W₂

⇒ (25 × 10 × 6)/W₁ = (18 × D₂ × 10)/(3 × W₁)

⇒ (25 × 10 × 6 × 3 × W₁)/(18 × 10 × W₁)= D₂

⇒ D₂ = 25

Amines are stronger base than ammonia because +I effect of alkyl groups increases electron density on nitrogen atom.

Artificial Sweetner.