Let's consider the one-day earnings of Neeta, Geeta, Sita to be n, g, s respectively.
The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days
n + g = 6(s) → eq(1)
The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days
s + n = 2(g) → eq(2)
By solving eq (1) & (2) you will get the ratio n : g : s = 11 : 7 : 3
The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is 11:3

Keeping these roots on the number line and checking for the range where f(x) becomes negative.
For the numbers below -5, f(x) is positive.
For the numbers between -5 and -2, the value f(x) is negative.
For the numbers between -2 and 3, the value of f(x) is positive.
For the numbers between 3 and 9, the value of f(x) is negative.
For the numbers above 9, the value of f(x) is positive.
Hence the range should be -5 < x < -2 and 3 < x < 9.

Given that, there are two 800cc bottles of indigo solution with strengths of 33% and 17% respectively.

Some amount of the solution from bottle 1 is replaced by the same amount of solution from bottle 2.

In 800cc of a new solution of strength 21% has two solutions in the ratio 1:3.
200cc of 33% solution and 600cc of 17% solution combined to give 21% solution of indigo
Hence, the remaining solution in the bottle 2 is 200cc.

The are of the rhombus is given by,
Area = ½ × d₁ × d₂
Where d₁ & d₂ are the diagonals of the rhombus.

Area = ½ × d₁ × d₂
96 = ½ × d₁ × d₂
96 × 4 = 2 × d₁ × d₂
Also,

d₁² + d₂² = 400
(d₁ + d₂ )² = d₁² + d₂² + 2 × d₁ × d₂
(d₁ + d₂ )² = 400 + 4(96)
(d₁ + d₂ )² = 4(100 + 96)
(d₁ + d₂ )² = 4(196)
(d₁ + d₂) = 2(14)
d₁ + d₂ = 28
The cost of laying electric wires along the diagonals at the rate of ₹125 per meter
= 28 × 125
= ₹3500

|1 + mn| < |m + n| < 5
For two numbers ‘a’ and ‘b’,
|a| < |b| is equivalent to a² < b²


So, we can say that:
(1 + mn)² < (m + n)²
1 + 2mn +m²n² < m² + n² + 2mn
1 - n² - m² + m²n² < 0
(1 - n²) - m²(1 - n²) < 0
(1 - m²)(1 - n²) < 0


For the product to be negative, either one of the two terms has to be negative.
But they cannot simultaneously be 0.
The only possibility for either of the two terms to be positive is when
n = 0 and |m| > 1, or |n| > 1 and m = 0


Now for the case when m = 0 and |n| > 1
|m + n| < 5
|0 + n| < 5
So n can be ±±2, ±±3, ±±4
Which are 6 cases

Similarly for the case when n = 1 and |m| > 1
|m + n| < 5
|0 + m| < 5
So m can be ±±2, ±±3, ±±4
Again we have 6 cases.
Hence the answer is 12.

Revising the Cosine rule and the area of the triangle using the Sine rule…

We draw the described parallelogram ABCD.

From the triangle ABC,
∠A + ∠B + ∠C = 180⁰
∠A + ∠B + 50⁰ = 180⁰
∠A + ∠B = 130⁰
In the quadrilateral CFDE,
∠C + ∠F + ∠D + ∠E = 360⁰
50⁰ + 180⁰ - ∠A + ∠x + 180⁰ - ∠B = 360⁰
50⁰ + ∠x = ∠A + ∠B
50⁰ + ∠x = 130⁰
∠x = 80⁰
∠FDE = 80⁰

We will select the 4 digits first and arrange them later.
Out of the 4 digits, one of them should be 2 and one of them should be 3.
2, 3, , .
So, we just need to select the other two digits…
The two digits could be (1,1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).
So, the selection of numbers could be…
2, 3, 1, 1
2, 3, 2, 2
2, 3, 3, 3
2, 3, 1, 2
2, 3, 1, 3
2, 3, 2, 3
Each of these selections could be re-arranged in a number of ways.

So total number of possibilities = (12 + 4 + 4 + 12 + 12 + 6) = 50 ways.
Alternate method:
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
Think why we need to add (Arrangements with only 1)!

_, _, _, _
(All possible arrangements) = 3⁴
Each blank could be any one of 1, 2 or 3.
(Arrangements with either 1 or 2) = 2⁴
Each blank could be any one of 1 or 2.
(Arrangements with either 1 or 3) = 2⁴
Each blank could be any one of 1 or 3.
(Arrangements with only 1) = 1
Each blank is filled with 1.
(Arrangements with at least one 2 and one 3) = (All possible arrangements) - (Arrangements with either 1 or 2) - (Arrangements with either 1 or 3) + (Arrangements with only 1)
(Arrangements with at least one 2 and one 3) = 3⁴ - 2⁴ - 2⁴ + 1
(Arrangements with at least one 2 and one 3) = 81 - 16 - 16 + 1
(Arrangements with at least one 2 and one 3) = 82 - 32 = 50

Since the prices of the small and medium cups are in the ratio 2 : 5,
Let us assume the prices of the small, medium and large cups to be 2x, 5x and y.
We are given that the product of the three prices is 800.
Therefore, (2x)(5x)(y) = 800 —- (1)
If the price of the smallest and the medium cups are increased by 6, then the product becomes 3200
(2x + 6) (5x + 6) (y) = 3200 —- (2)

(2x + 6) (5x + 6) = 40 x²
10 x² + 42x + 36 = 40 x²
30 x² - 42x - 36 = 0
10 x² - 14x - 12 = 0
10 x² - 20x + 6x - 12 = 0
10 x(x - 2) + 6(x - 2) = 0
(10x + 6) (x - 2) = 0
x = 2 or x = -0.6
x can’t be negative and hence x = 2
WKT, (2x)(5x)(y) = 800
(4)(10)(y) = 800
y = 20
Therefore, the sum of the prices = 2x + 5x + y
= 2(2) + 5(2) + 20
= 4 + 10 + 20
= 34

Let R be the fraction of work done by Rahul in 1 hour.
and G be the fraction of work done by Gautam in 1 hour.
Initially Rahul works from 9AM to 5PM (8 hours) and Gautam works for 2 hours less.
8R + 6G = 1 whole unit of work
If they start together they finish 30 minutes earlier or if they start at 9 AM, they finish at 4:30PM (7.5 hours)
7.5R + 7.5G = 1 whole unit of work
8R + 6G = 7.5R + 7.5G
0.5R = 1.5G
R = 3G
This means Rahul is thrice as efficient as Gautam.
8R + 6G = 1 whole unit of work
8R + 2R = 1 whole unit of work
10R = 1 whole unit of work
R is the fraction of work done by Rahul in 1 hour.
Since 10R = 1
R alone takes 10 hours to finish the job.

Let the weight of Silver in the initial Alloy be Ag kgs
and the weight of Copper in the initial Alloy be Cu kgs
So, the total weight of the initial Alloy = (Ag + Cu) kgs
This Alloy is mixed with 3kgs of Pure Silver to form a new mixture.
Total weight of the mixture = (Ag + Cu) + 3 kgs
Weight of Silver in the mixture = (Ag + 3) kgs
Since this mixture contains 90% Silver,

The initial Alloy is mixed with 2kgs of another alloy having 90% Silver by weight.
Total weight of the mixture = (Ag + Cu) + 2 kgs
Weight of Silver in the mixture = (Ag + 90% of 2) kgs = (Ag + 1.8) kgs
Since this mixture contains 84% Silver,

g(x) = x + 3
f(x) = x² - 7x
f(g(x)) - 3x
= f(x + 3) - 3x
= (x + 3)² - 7(x + 3) - 3x
= x² + 9 + 6x - 7x - 21 - 3x
= x² - 4x - 12
= x² - 4x + 4 - 4 - 12
= x² - 4x + 4 - 16
= (x - 2)² - 16
f(g(x)) - 3x is minimum when (x - 2)² - 16 is minimum.
(x - 2)² - 16 is minimum when (x - 2)² is minimum.
Since (x - 2)² is non-negative, the minimum value it can take is 0.
Hence the minimum value of (x - 2)² - 16 = 0 - 16 = -16
Therefore, the minimum value of f(g(x)) - 3x is -16

Bank A has a rate of interest of 6% and compounds half yearly.
This is the same as having a 3% interest rate per half-year.
So, if a Principal, P is invested for an year in bank A, at the end of the year it becomes P(1.03)(1.03) = P(1.0609)
Therefore the interest rate when viewed as a Simple interest scheme is 6.09% per annum.
Rupa invested in Bank C, which has twice the interest rate as Bank B and the quantum for which the investment is made is also double, hence Rupa effectively gets 4 times the interest that Raju gets for the same investment in Bank A.
Let’s say Raju invested ₹ 10,000 in Bank B.
Since this is the same as investing in Bank A for 1 year, his interest would be 6.09% of 10,000 = ₹ 609.
Now, for the same investment, Rupa must earn 4 times that of ₹ 609
So, Rupa earns ₹ 2,436

he arithmetic mean of the scores of 25 students = 50
Sum of scores of these students = 25 × 50 = 1250
For the scores of the top 5 students to be as high as possible, the score of the bottom 20 students should be as low as possible.
The minimum score is 30, and the scores of the bottom 20 students are distinct integers.
In order for the bottom 20 scores to be as low as possible, they must be,
30, 31, 32, … 49
Sum of the bottom 20 scores
= 30 + 31 + 32 + … + 49
= (30 + 0) + (30 + 1) + (30 + 2) + … + (30 + 19)
= 20 × 30 + (0 + 1 + 2 + 3 + … + 19)
= 600 +

Let the number of small shirts be ‘x’
then the number of large shirts becomes 64 - x.
Let the price of a small shirt be ‘y’
then the price of a large shirt becomes y + 50
Money spent on small shirts = xy = 1800
Money spent on large shirts = (64 - x) (y + 50) = 5000
(64 - x) (y + 50) = 5000
64y + 3200 - xy - 50x = 5000
64y + 3200 - 1800 - 50x = 5000
64y + 1400 - 50x = 5000
64y - 50x = 3600
32y - 25x = 1800
32y - 25(1800/y) = 1800
32y² - 1800y - 25(1800) = 0
4y² - 9(25)y - 25(9)(25) = 0
y = 75
Price of a small shirt = ‘y’ = 75
Price of a small shirt = ‘y + 50’ = 125
The price of a large shirt and a small shirt together, in INR = 75 + 125 = 200

Let the fixed cost be ₹ F and the variable cost be ₹ V.
Since the profit per border is ₹200 when there are 50 borders
The expenses of the Hostel is,
F + 50(V) = 50 (1600 - 200)
F + 50(V) = 50 (1400) — (1)
Since the profit per border is ₹250 when there are 75 borders
The expenses of the Hostel is,
F + 75(V) = 75 (1600 - 250)
F + 75(V) = 75 (1350) — (2)
(2) - (1)
25(V) = 75(1350) - 50(1400)
25(V) = 25( 3(1350) - 2(1400) )
V = 3(1350) - 2(1400)
V = 4050 - 2800
V = 1250
F + 75(V) = 75 (1350)
F + 75(1250) = 75 (1350)
F = 75(100) = 7500
The Expenditure for 80 borders will be,
= F + 80(V)
= 7500 + 80(1250)
The revenue collected from 80 students is,
= 80(1600)
Hence, the profit is,
= 80(1600) - (7500 + 80(1250))
= 80(1600 - 1250) - 7500
= 80(350) - 7500
= 100(8×35 - 75)
= 20500
Hence the total profit when there are 80 borders is ₹20500.

3x + 2|y| + y = 7 
x + |x| + 3y = 1 
lets take x and y to be positive 
3x+3y=7 
2x+3y=1 
x=6, but y is negative so this case is invalid 
Lets take x is +ve and y is -ve 
3x-y=7 
2x+3y=1

à x=2 and y=-1 
Case is valid 
x+2y = 2 + 2(-1) = 0 

Let the number of matches to be played be ‘x’
We are given that 40 matches are already played and 30% of them are won.
If 60% of the remaining matches are won, then the overall win percentage is 50%.
30% of 40 + 60% of x = 50% of (40 + x)
0.3 (40) + 0.6 (x) = 0.5 (40 + x)
12 + 0.6 (x) = 20 + 0.5 (x)
0.1 (x) = 8
x = 80
The number of matches to be played is 80.
If the team wins 90% of the remaining matches, it would win 90% of 80 = 72 matches.
Total matches won by the team
= 30% of 40 + 72
= 12 + 72 = 84

From 1970 to 1980,
the male population increased by 40%
the female population increased by 20%
the overall population increased by 25%

1.4M + 1.2F = 1.25(M + F)
1.4M + 1.2F = 1.25M + 1.25F
1.4M - 1.25M = 1.25F - 1.2F
0.15M = 0.05F
F = 3M
From 1980 to 1990,
Female population increased by 25%
Therefore, the female population in 1990 = 1.25 × 1.2F = 1.5F
Since, the female population in 1990 is twice the male population,
Male population in 1990 = 1.5F ÷ 2 = 0.75F
Since F = 3M,
Male population in 1990 = 2.25M
Total population in 1970 = M + F = M + 3M = 4M
Total population in 1990 = 2.25M + 1.5F = 2.25M + 4.5M = 6.75M

Let the length be x and the breadth be y,
Then the area of the region will be
x×y = 60000
Then the cost of fencing the region will be
200x + 100x + 100y + 100y
300x + 200y

Now we know that the Arithmetic mean ≥ Geometric mean.