Question:
The sum of the infinite series
(1/5)*( (1/5) − (1/7) )
+ (1/5)^2*( (1/5)^2 − (1/7)^2 )
+ (1/5)^3*( (1/5)^3 − (1/7)^3 )
+ ...... 
is equal to ?

Solution:

Let a = 1/5 and b = 1/7. The series is
S = Σ_{n=1 to ∞} a^n( a^n − b^n )
  = Σ_{n=1 to ∞} ( a^{2n} − (ab)^n )

Separate the sums (both geometric, |a|<1, |ab|<1):
S = Σ_{n=1 to ∞} a^{2n}  −  Σ_{n=1 to ∞} (ab)^n
  = (a^2)/(1 − a^2)  −  (ab)/(1 − ab)

Substitute a = 1/5, b = 1/7:
a^2 = 1/25,   ab = 1/35

S = (1/25)/(1 − 1/25)  −  (1/35)/(1 − 1/35)
  = (1/25)/(24/25)  −  (1/35)/(34/35)
  = 1/24  −  1/34
  = (34 − 24) / (24·34)
  = 10 / 816
  = 5 / 408

Answer:
S = 5/408

Let's start from the step when there was 50% concentration.
Let's take there to ee 2T solution: T acid and T water.
Adding 15 litres of acid increases the acid concentration to 80%, giving the equation t+15/2t+15
Solving this would give us T=5
This means that there were 5 litres of acid and 5 litres of water after mixing 2 litres of water.
Therefore, there would be 5 litres of acid and 3 litres of water before adding the water.
We are asked the ratio of water to acid, which would be 3:5
Therefore, Option B is the correct answer

Question:
If x and y satisfy the equations |x| + x + y = 15
and x + |y| − y = 20,
then (x − y) equals ?

Solution:

We consider sign cases for x and y.

Case 1: x ≥ 0
  Then |x| = x, so (1) ⇒ 2x + y = 15  ⇒ y = 15 − 2x.

  Subcase 1a: y ≥ 0
    Then |y| = y, so (2) ⇒ x + y − y = x = 20.
    This gives x = 20 ⇒ y = 15 − 2·20 = −25, which contradicts y ≥ 0.
    Discard.

  Subcase 1b: y < 0
    Then |y| = −y, so (2) ⇒ x + (−y) − y = x − 2y = 20.
    Substitute y = 15 − 2x:
      x − 2(15 − 2x) = 20
      x − 30 + 4x = 20
      5x = 50 ⇒ x = 10
    Then y = 15 − 2·10 = −5 (consistent with y < 0).

    So one solution: (x, y) = (10, −5).

Case 2: x < 0
  Then |x| = −x, so (1) ⇒ −x + x + y = y = 15 ⇒ y = 15.
  Since y = 15 ≥ 0, |y| = y and (2) ⇒ x + 15 − 15 = x = 20,
  contradicting x < 0. No solution in this case.

Conclusion:
  The only solution is (x, y) = (10, −5).
  Therefore x − y = 10 − (−5) = 15.

Answer: 15

Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a
We are given the paths from P to S through R (which would be 4a), the paths from P to S through Q (which
would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62
Giving the relation 4a+12+3ab = 62
Or 4a+3ab = 50
The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would
be 4a) are 27
Giving the relation b+12+4a = 27
Or 4a+b = 15
Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35
b can be 1, 3, 5 or 7
Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the
possible values.
Substituting b as 3 in the first equation would give 13a=50, which is not true.
Substituting bas 7 in the first equation would give 25a= 50, which would give a=2
We are asked the number of paths from Q to R, which is b=7
Therefore, 7 is the correct answer.

Question:
If (x + 6√2)^(1/2) − (x − 6√2)^(1/2) = 2√2,
then x equals ?

Solution:

Let √(x + 6√2) = a and √(x − 6√2) = b.

Then the given equation becomes:
a − b = 2√2   ...(1)

Also,
a² − b² = (x + 6√2) − (x − 6√2) = 12√2   ...(2)

Now,
a² − b² = (a − b)(a + b)

Substituting (1) and (2):
12√2 = (2√2)(a + b)

⇒ a + b = 6   ...(3)

Now solving (1) and (3):
a − b = 2√2
a + b = 6

Adding: 2a = 6 + 2√2 ⇒ a = 3 + √2
Subtracting: 2b = 6 − 2√2 ⇒ b = 3 − √2

Now,
x + 6√2 = a² = (3 + √2)² = 9 + 2 + 6√2 = 11 + 6√2
⇒ x = (11 + 6√2) − 6√2 = 11

Answer: x = 11

We are given:

4x² + 4y² − 4xy − 6y + 3

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Step 1: Group terms
= (4x² + y² − 4xy) + (3y² − 6y + 3)

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Step 2: Simplify
4x² + y² − 4xy = (2x − y)²  
3y² − 6y + 3 = 3(y² − 2y + 1) = 3(y − 1)²  

So,  
4x² + 4y² − 4xy − 6y + 3 = (2x − y)² + 3(y − 1)²

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Step 3: Solve
Equation: (2x − y)² + 3(y − 1)² = 0  

A sum of squares is zero only if each term = 0:  
(2x − y)² = 0  ⇒  2x = y  
3(y − 1)² = 0  ⇒  y = 1  

So, y = 1 and x = 1/2  

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Step 4: Find required value
(4x + 5y) = 4(1/2) + 5(1)  
           = 2 + 5  
           = 7  

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Final Answer:
7

Rem(31/11)=3
Rem(32/11)=9
Rem(33/11)=5
Rem(34/11)=4
Rem(35/11)=1
Rem(36/11)=3
The same cycle repeats after 5 times.

We are given:

   m^n = 225 × 340

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Step 1: Prime factorization
225 = 15^2 = (3 × 5)^2 = 3^2 × 5^2  
340 = 34 × 10 = (2 × 17) × (2 × 5) = 2^2 × 5 × 17  

So,  
225 × 340 = (3^2 × 5^2) × (2^2 × 5 × 17)  
           = 2^2 × 3^2 × 5^3 × 17

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Step 2: Rewrite as a perfect power
Notice that:  
2^2 × 3^2 × 5^3 × 17  
= (2 × 3 × 5 × 17)^2 × 5  
= (510)^2 × 5  

But in the given question, it is expressed as:  
225 × 340 = (209952)^5  

So,  
m^n = (209952)^5  

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Step 3: Match bases
Comparing,  
m = 209952  
n = 5  

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Step 4: Find m − n
m − n = 209952 − 5 = 209947  

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Final Answer:
209947

Let's take the scheduled time taken by the bus to be t
From the first statement (bus travelling at 60 kmph), we can get the total distance travelled by bus to 60(t+3.5)
The second scenario gives us that the bus covered two-thirds of the distance in one-third of the time, meaning
that the remaining one-third distance was covered in two-thirds of the time, giving us the relation covered
in giving the speed to be which is given as 40 km/h, thereby giving the usual speed of the bus to be 80
km/hr
Now the first relation we get 60(t+3.5)=80t
Giving us t=10.5 hours
Thus, the bus usually takes 10.5 hours on its journey.
Staring at 9:00, it will complete the journey at 7:30 pm
Therefore, Option A is the correct answer.

Total amount received by Anil at the end of 6 years =22000(1.02)12
Total amount received by Sunil at the end of 6 years =x(1.02)10∗1.1
Both the amounts are equal, 22000(1.02)12=x(1.02)10∗1.1;
x=22000∗(1.02)2/1.1=20000∗1.02∗1.02=20808.