Initial salary = 100
After two successive increments, final salary = 187.5
First increment = x%
Second increment =2x%

Going from the options, x=25%

The question is "After two successive increments, Gopal's salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was"

Hence, the answer is '25'

Single digit numbers where none of the digits is repeated = 1 to 9
⇒ n(single digit numbers)= 9
Double digit numbers = AB (where A is any number from 1 to 9 and B is any number except A)
⇒ n(double digit numbers) = 9 × 9 = 81
Triple digit numbers = ABC (where A is any number from 1 to 4, B is anything except A and C is anything except A, B)
⇒ n(triple digit numbers) = 4 × 9 × 8 = 288
Total number of numbers = 288 + 81 + 9 = 378

Total land area =20+30=50 hectares
In the total land area
Wheat part =1120×50=552
Mustard part =920×50=452
In Vimal's land area
Mustard
Wheat part =58×30=754
Mustard part =38×30=454
So, the land area for Rajesh
Wheat part =552−754=354
Mustard part =452−454=454
Ratio, Wheat : Mustard =354:454=7:9

A necessary condition for the system to have no solution for (𝑥,𝑦), is
The condition is a1a2=b1b2≠c1c2
i.e., p3=−4k≠2a
For p3=−4k;pk+12=0
For −4k≠2a;−4a≠2k⇒2a+k≠0
For p3≠2a;pa−6≠0

The question is "For some constant real numbers p,k and a, consider the following system of linear equations in x and y :

A necessary condition for the system to have no solution for (x,y), is "

Hence, the answer is '2a+k≠0'

Put x=1/x,

Solving (1) & (2),
(1) ⇒

Subtracting, we get

So,

Now,

Sum of the roots =−b/a
Answer =−3

The question is "For any non-zero real number x, let f(x)+2f(1x)=3x. Then, the sum of all possible values of x for which f(x)=3, is"

Hence, the answer is '-3'

We are thinking about two scenarios where x≥y and x≥−y. When x≥y,

So, the possibilities are ( 1,1 ), ( 1,0 ) and ( 1,−1 )
When x=0,
The possibilities are (0,1) and (0,−1)
When x=−1
The possibilities are ( −1,1 ), ( −1,0 ) and ( −1,−1 )
So, the number of distinct integer solutions =8

The question is "The number of distinct integer solutions (x,y) of the equation |x+y|+|x−y|=2, is"

Hence, the answer is '8'

Question:
A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is ?

Solution:

Interior angle of a regular octagon = (n−2)/n × 180°
= (8−2)/8 × 180° = 135°

Area of square ACEG = a²

In △BAC, using cosine rule:
a² = b² + c² − 2bc·cosC
a² = 6² + 6² − 2(6)(6)cos135°
a² = 72 − (72 × −√2/2)
a² = 72 + 36√2
a² = 36(2 + √2)

Therefore, Area = 36(2 + √2)

Answer: 36(2 + √2)  

Initial quantity of water in the container =x
So, initial quantity of milk in the container =300−x
Part of mixture taken out =2x300
Part of mixture that is retained =1−2x300
So, part of milk that is retained =(300−x)(1−2x300)
Also, quantity of milk that is retained after the iteration =72% of 300=216
Then,

The total content itself 300 litre. So, x=420 is inadmissible.
Thus, amount of water that was initially poured = 30 litre

Let, CP=x
Then, MP = 1.2 x ;

Also,
Savings = Rs. 15
Savings with respect to percentage =10% of MP=0.12x
Thus, 0.12x=15

Hence,

The question is "Gopi marks a price on a product in order to make 20% profit. Ravi gets 10% discount on this marked price, and thus saves Rs 15. Then, the profit, in rupees, made by Gopi by selling the product to Ravi, is"

Hence, the answer is '10'

Choice B is the correct answer.

Question:
If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8 and 8^f = 9, then the value of the product abcdef is ?

Solution:

From 3^a = 4  ⇒  a = log_3(4) = ln4 / ln3  
From 4^b = 5  ⇒  b = log_4(5) = ln5 / ln4  
From 5^c = 6  ⇒  c = log_5(6) = ln6 / ln5  
From 6^d = 7  ⇒  d = log_6(7) = ln7 / ln6  
From 7^e = 8  ⇒  e = log_7(8) = ln8 / ln7  
From 8^f = 9  ⇒  f = log_8(9) = ln9 / ln8

Multiply:
abcdef = (ln4/ln3)·(ln5/ln4)·(ln6/ln5)·(ln7/ln6)·(ln8/ln7)·(ln9/ln8)

All intermediate terms cancel telescopically, leaving:
abcdef = ln9 / ln3

But ln9 = ln(3^2) = 2 ln3, so
abcdef = (2 ln3) / ln3 = 2

Answer: 2

Distance = XT
From the table, we have:

Also,

Solving (1) & (2),

Adding the equations,

Distance =XT=720km

The question is "A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is "

Hence, the answer is '720'

Solution:

We are given the equation:
10^x + 4/(10^x) = 17/2

Step 1: Let 10^x = y
So, the equation becomes:
y + 4/y = 17/2

Multiply through by y:
2y^2 - 17y + 8 = 0

Step 2: Solve quadratic equation.
2y^2 - 17y + 8 = 0
Factorizing:
(2y - 1)(y - 8) = 0

So, y = 8 or y = 1/2

Step 3: Back-substitute.
10^x = 8  →  x = log10(8) = log10(2^3) = 3 log10(2)
10^x = 1/2  →  x = log10(1/2) = -log10(2)

Step 4: Sum of all possible values.
= 3 log10(2) + (-log10(2))
= 2 log10(2)

Final Answer:
2 log10(2)

Solving this, we get:

Now,

The question is "Aman invests Rs 4000 in a bank at a certain rate of interest, compounded annually. If the ratio of the value of the investment after 3 years to the value of the investment after 5 years is 25 : 36, then the minimum number of years required for the value of the investment to exceed Rs 20000 is"

Hence, the answer is '9'

Solution:

We are given:
t1 = 1, t2 = -1, and tn = ((n-3)/(n-1)) * t(n-2) for n ≥ 3.

Step 1: Compute first few terms.
t2 = -1
t4 = (4-3)/(4-1) × t2 = (1/3) × (-1) = -1/3
t6 = (6-3)/(6-1) × t4 = (3/5) × (-1/3) = -1/5
t8 = (8-3)/(8-1) × t6 = (5/7) × (-1/5) = -1/7

So, we see the pattern:
t2 = -1,  t4 = -1/3,  t6 = -1/5,  t8 = -1/7, ...

Step 2: General pattern.
t(2k) = -1/(2k-1)

Step 3: Inverse values.
1/t2 = -1
1/t4 = -3
1/t6 = -5
1/t8 = -7
...
1/t2024 = -2023

Step 4: Sum of series.
1/t2 + 1/t4 + 1/t6 + ... + 1/t2024
= -(1 + 3 + 5 + 7 + ... + 2023)

The series inside is the sum of first 1012 odd numbers.

Sum of first n odd numbers = n²
So, sum = (1012)² = 1024144

Final Answer:
= -1024144

Solution:

Given:
OA = OB = radius (r)
∠AOB = 120°
⇒ ∠OAB = ∠OBA = 30°

Consider OM ⟂ AB.

In right △OMB:
cos30° = MB / OB = (√3)/2 = (5√3)/OB
⇒ OB = 10

Also,
sin30° = OM / OB = 1/2 = OM / 10
⇒ OM = 5

Now, area of smaller region =
Area of sector − Area of △AOB

= (120/360) × π × (10)^2 − (1/2) × (10√3) × 5

= (1/3) × 100π − 25√3

= 25(4π/3 − √3)

Answer:
25(4π/3 − √3)

We are given that Sam completes a piece of work in 20 days. We are also given that Mohit is twice as fast, so
he should take only 10 days. Mohit is thrice as fast as Ayna, so he would take 30 days.
Let's take the total work to be 60 units; this would give the work done per day for Mohit, Sam, and Ayna to be 6,
3 and 2, respectively.
On the first day Sam and Mohit work: doing 9 units
On the second day Sam and Ayan work: doing 5 units
On the third day, Mohit and Ayan work: doing 8 units
Essentially doing 22 units in a 3 days cycle.
After two such cycles, there will be 60-44 = 16 units of work left
On day 7, Sam and Mohit would work 9 units, leaving 7 units
On day 8, Sam and Ayan would work 5 units, leaving 2 units
And on day 9, Ayan and Mohit would complete the remaining work.
So Sam worked for a total of 2+2+2= 6 days and on each day he did 3 units of work, completing 18 units of
work.
The ratio of work done by Sam would be 3/10
Therefore, Option B is the correct answer.

Question:
If 10^68 is divided by 13, what is the remainder?

Solution:

We calculate remainders of powers of 10 modulo 13:

10^1 mod 13 = 10  
10^2 mod 13 = 9  
10^3 mod 13 = 12  
10^4 mod 13 = 3  
10^5 mod 13 = 4  
10^6 mod 13 = 1  

So, after every power of 6, the remainders repeat in a cycle: {10, 9, 12, 3, 4, 1}.

Now, 68 ÷ 6 = 11 remainder 2.  
Thus, 10^68 mod 13 corresponds to the 2nd element in the cycle, which is 9.

Answer: 9

Let the numbers be a (smallest), b (middle), c (largest).
Given: (a + b + c)/3 = 28  ⇒  a + b + c = 84.         ...(1)

After changes:
 smallest → a + 7
 middle → b
 largest → c − 10

New mean:
[(a + 7) + b + (c − 10)] / 3 = (a + b + c − 3)/3.
But a + b + c = 84, so new mean = (84 − 3)/3 = 81/3 = 27.

Given: new mean = b + 2 ⇒ 27 = b + 2 ⇒ b = 25.       ...(2)

Given difference after changes:
(c − 10) − (a + 7) = 64
⇒ c − a − 17 = 64
⇒ c − a = 81.                                         ...(3)

From (1) and (2): a + c = 84 − b = 84 − 25 = 59.      ...(4)

Solve (3) and (4):
Add: (c − a) + (a + c) = 81 + 59 ⇒ 2c = 140 ⇒ c = 70.

Answer:
The largest number originally is 70.

Question:
If (a + b√3)^2 = 52 + 30√3, where a and b are natural numbers, find a + b.

Solution:

Expand the left side:
(a + b√3)^2 = a^2 + 2ab√3 + 3b^2

Equate real and irrational parts with the right side 52 + 30√3:

Real part:    a^2 + 3b^2 = 52
Irrational:   2ab = 30  ⇒  ab = 15

Find natural-number factor pairs of 15:
(ab = 15) ⇒ (a,b) ∈ {(1,15), (3,5), (5,3), (15,1)}

Check each pair against a^2 + 3b^2 = 52:

(1,15): 1^2 + 3·15^2 = 1 + 675 = 676 ≠ 52  
(3,5):  3^2 + 3·5^2  = 9 + 75  = 84  ≠ 52  
(5,3):  5^2 + 3·3^2  = 25 + 27 = 52  ✓  
(15,1): 15^2 + 3·1^2 = 225 + 3 = 228 ≠ 52

Only (a,b) = (5,3) satisfies both equations.

Therefore:
a + b = 5 + 3 = 8

Answer: 8

Here, with respect to boys
Swimming + Running = 50% + 70% = 120%
The extra 20% is meant for the boys who opted both swimming and running.
Similarly, with respect to girls
Swimming + Running = 80% + 60% = 140%
The extra 40% is meant for the boys who opted both swimming and running.
In pictorial representation:

So, percentage of people who opted for both swimming and running = 20% boys + 40% girls

The question is "In a group of 250 students, the percentage of girls was at least 44% and at most 60%. The rest of the students were boys. Each student opted for either swimming or running or both. If 50% of the boys and 80% of the girls opted for swimming while 70% of the boys and 60% of the girls opted for running, then the minimum and maximum possible number of students who opted for both swimming and running, are"

Hence, the answer is '72 and 80, respectively'

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

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

--------------------------------------------------
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

--------------------------------------------------
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  

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

--------------------------------------------------
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

--------------------------------------------------
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

--------------------------------------------------
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  

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

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

--------------------------------------------------
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.

We need to solve:

   1/(x + 5) ≤ 1/(2x − 3)

--------------------------------------------------
Step 1: Identify restrictions
Denominators ≠ 0
x ≠ −5,  x ≠ 3/2

--------------------------------------------------
Step 2: Bring to common denominator
1/(x + 5) − 1/(2x − 3) ≤ 0

⇒ [ (2x − 3) − (x + 5) ] / [ (x + 5)(2x − 3) ] ≤ 0

⇒ (x − 8) / [ (x + 5)(2x − 3) ] ≤ 0

--------------------------------------------------
Step 3: Critical points
x − 8 = 0   ⇒  x = 8  
x + 5 = 0   ⇒  x = −5  
2x − 3 = 0  ⇒  x = 3/2  

So, critical points: −5, 3/2, 8

--------------------------------------------------
Step 4: Check intervals
Intervals are: (−∞, −5), (−5, 3/2), (3/2, 8), (8, ∞)

• For (−∞, −5): numerator (x−8) < 0, denominator (x+5)(2x−3) > 0 → fraction < 0 (valid)  
• For (−5, 3/2): numerator (x−8) < 0, denominator (x+5)(2x−3) < 0 → fraction > 0 (not valid)  
• For (3/2, 8): numerator (x−8) < 0, denominator (x+5)(2x−3) > 0 → fraction < 0 (valid)  
• For (8, ∞): numerator (x−8) > 0, denominator (x+5)(2x−3) > 0 → fraction > 0 (not valid)

--------------------------------------------------
Step 5: Include equality
At x = 8 → numerator = 0 ⇒ inequality holds (included)

At x = −5, 3/2 → denominators = 0 ⇒ excluded

--------------------------------------------------
Final Solution:
x < −5   or   3/2 < x ≤ 8

--------------------------------------------------
Answer: Option A

1/A + 1/V= 1/ 150 ------------(1)
1/V + 1/S= 1/ 100 ------------(2)
75/A + 135/V+ 45/S= 1 ----(3)
Subtracting eqn (1) from eqn(2),
1/S – 1/A= 1/300 -----(4)
Multiplying eqn (1) with 135, we get 135/A + 135/V= 135/150= 9/10 --------(5)
Subracting eqn (5) from eqn (3), -60/A + 45/S= 1/10-----(6)
Solving, eqn (4) and eqn (6), we get 1/S= 1/150, 1/A= 1/300 . Substituting the value of 1/S in eqn (2) we get 1/V = 1/300.
Anil takes 300 days to finish the task.
Vimal takes 300 days to finish the task.
Sunil takes 150 days to finish the task.
Let total work= 300 units
Anil does 1 unit/day.
Vimal does 1 unit/day.
Sunil does 2 units/day.
If Anil works for 6k days, Vimal works for 3k days and Sunil works for 2k days.
If 6k= 138 days.
Anil would have finished 138 units, Vimal would have completed 69 units and Sunil would have finished 46*2= 92 units.
138+69+92= 299 units would be completed.
In 139th day, only Anil works and therefore (299+1= 300 units) work is finished.
Total work got completed in 139 days.

Let 'a' be sum of bonuses of 1 st 10 employees, ' b ' be the sum of bonuses of 2nd 10 employees, ' c ' be the sum of bonuses of 3rd 10 employees and ' d ' be the sum of bonuses of last 10 employees.

Adding the three equations we get,

In 2023, the average bonus of the first 10 employees increased by 100%.

In 2023, the average bonus of the last 10 employees increased by 200%.

Average in the year 2023 =(400k+1000k+2400k/40)=3800k/40=380k/4=95000

We can take AD to be x and CB to be 3-x; since the perimeter of ABCD is 6, and AB and CD are given as 2 and 1
cm, respectively, that leaves AD+BC as three only.
We can see that triangles AEB and DEC are similar, with the lengths of AB being double of CD, essentially
making D the mid-point of AE and C the mid-point of EB.
Through this, we get the length of DE and CE to be x and 3-x.
Give us the AE and BE lengths as 2x and 6-2x, respectively.
Giving the perimeter of AEB as 2x+6-2x+2 = 8 cm
Therefore, Option A is the correct answer.

Given: AB = BC = AC = a = 2r  
OB = 2r / √3  

Let x and y be the radii of the bigger and smaller circles.

We have:  
r + y = (2r / √3)  
2r + y = x  

So,  
2r + (2r / √3) − r = x  
r + (2r / √3) = x  

Also,  
(2r / √3) − r = y  

Thus,  
x / y = [ r + (2r / √3) ] / [ (2r / √3) − r ]  

= { (2/√3) + 1 } / { (2/√3) − 1 }  

= (2 + √3) / (2 − √3)  

= (2 + √3)(2 + √3) / (2 − √3)(2 + √3)  

= (4 + 3 + 4√3) / (4 − 3)  

= (7 + 4√3) : 1  

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The question is:  
"Three circles of equal radii touch (but not cross) each other externally.  
Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles.  
If the radius of X is more than that of Y, the ratio of the radii of X and Y is ?"

Final Answer:  
(7 + 4√3) : 1

Let's take Rajesh and Garima's ages to be R and G, respectively.

From the given ratio, we can see that Rajesh is older than Garima, 
so let's take R = G + x.

When Rajesh was of age G, which was x years ago, 
Garima was of (G − x) years old.

So the ratio is:  
(G + x − x) : (G − x) = G : (G − x)

This gives G = 3x, which in turn gives R = 4x.

We are asked the ratio when Garima becomes 4x years old.  
At that time, Rajesh will be 5x years old.

Therefore, their ratio is:  

5x : 4x = 5 : 4

The roots α, β of the equation 3x² + λx − 1 = 0.

Sum of the roots: α + β = −λ/3  
Product of the roots: αβ = −1/3  

Given: 1/α² + 1/β² = 15  

=> (α² + β²) / (αβ)² = 15  

Now,  
α² + β² = (α + β)² − 2αβ  
          = (λ² / 9) − 2(−1/3)  
          = (λ² / 9) + (2/3)  
          = (λ² + 6) / 9  

Substitute into the condition:  
(λ² + 6) / (9 × 1/9) = 15  
=> λ² + 6 = 15  
=> λ² = 9  
=> λ = +3 or −3  

Now,  
α³ + β³ = (α + β)³ − 3(αβ)(α + β)  
         = (−λ³ / 27) − 3(−1/3)(−λ/3)  
         = (−λ³ / 27) − (λ / 3)  
         = −(λ³ / 27 + λ / 3)  

Case 1: λ = +3  
α³ + β³ = −(27/27 + 3/3) = −(1 + 1) = −2  
(α³ + β³)² = (−2)² = 4  

Case 2: λ = −3  
α³ + β³ = −(−27/27 + (−3)/3) = −(−1 − 1) = 2  
(α³ + β³)² = (2)² = 4  

Final Answer:  
(α³ + β³)² = 4

Looking at the additional information about the prime numbers should make one realise 
that they are the key to solving the question.

f(16000) can be written as

Now, we can try to find these individual values:

For any prime p: f(p) = 1

This way, we can find the function output for any prime number raised to a power.

We can see that each new exponent is twice the previous output + 1, solving this way 
till prime raised to power 8.

f(p^2) = f(p) f(p) + f(p) + f(p) = 1 + 1 + 1 = 3  
f(p^3) = f(p^2) f(p) + f(p^2) + f(p) = 3 + 3 + 1 = 7  
f(p^4) = 7 + 7 + 1 = 15  
f(p^5) = 15 + 15 + 1 = 31  
f(p^6) = 31 + 31 + 1 = 63  
f(p^7) = 63 + 63 + 1 = 127  
f(p^8) = 127 + 127 + 1 = 255  

So, f(2^8 × 5^4) = (255 × 15) + 255 + 15 = 4095

Let the total number of fruits= 5x.
Therefore, the number of mangoes in the stock= (40/100)*5x= 2x.
Let the number of apples= 5y.
He sold x mangoes, 96 bananas and 2y apples which is equal to 50% of the total fruits.
x+ 96+ 2y= 2.5x
2x+192+4y= 5x
192+ 4y= 3x
The smallest value of y which will satisfy the equation is y= 3.
192+ 12= 204= 3x
x= 68
The smallest possible total number of fruits in the stock at the beginning of the day is 5x= 5*68= 340.

The question is "A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is"

Hence, the answer is '340'

In-radius of the triangle= (a+b-h)/2, where a and b are the lengths of two legs of the triangle and h is the hypotenuse.
In-radius= (8+6-10)/2= 4/2= 2 units

The question is "The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the incircle of the triangle is"

Hence, the answer is '2'

Let the cost price of the item be C
We are given that Bina sells this at 19% loss or at (1 - 0.19)C = 0.81C at 4860
This gives us the value of C at Rs. 6000
If Bina had sold this at 17% profit, the selling price would have been
So Shyam bought the product at 4860 and sold it to Hari at 7020
Giving the profit made by Shyam to be 7020-4860=2160
Therefore, 2160 is the correct answer.

Given: A closed rectangular box  

Step 1: Surface area of the box  
2(lb + lh + bh) = 846  

Step 2: Sum of the lengths of all edges  
4(l + b + h) = 144  
=> l + b + h = 144 / 4 = 36  …(1)  

Step 3: Square equation (1)  
(l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + lh + bh) = 36^2 = 1296  
l^2 + b^2 + h^2 = 1296 − 846 = 450  

Step 4: Diameter of the sphere (space diagonal of the box)  
d = √(l^2 + b^2 + h^2) = √450 = 15√2 = 2r  

Step 5: Radius of the sphere  
r = (15√2) / 2 = 15/√2  

Step 6: Volume of the sphere  
V = (4/3) π r^3 = (4/3) π (15√2 / 2)^3  
V = (4/3) π * 1125√2 / 8  
V = 1125 π √2 / 6  
V = 1125 π √2

Given: √(29 − 12√5) = a + b√n  

Step 1: Square both sides  
(a + b√n)^2 = 29 − 12√5  
a^2 + n b^2 + 2ab√n = 29 − 12√5  

Step 2: Equate rational and irrational parts  
Rational part: a^2 + n b^2 = 29  
Irrational part: 2ab√n = −12√5  
=> ab√n = −6√5  
=> n = 5, ab = −6  

Step 3: Solve a^2 + 5b^2 = 29 with ab = −6  
Possible values: (a = 3, b = −2) or (a = −3, b = 2)  

Step 4: Identify positive and negative roots  
Positive root: −3 + 2√5  
Negative root: 3 − 2√5  

Step 5: To maximize a + b + n, modify the expression:  
Take a = −3, b = 1, n = 20  

Step 6: Compute maximum possible value  
a + b + n = −3 + 1 + 20 = 18  

Answer: Maximum possible value of (a + b + n) = 18

Let the initial quantity of grains be x kg.  

Step 1: First customer  
Grains sold = (1/2)x + 3  
Remaining grains = x - [(1/2)x + 3] = (1/2)x - 3  

Step 2: Second customer  
Grains sold = (1/2)((1/2)x - 3) + 3 = (1/4)x - 3/2 + 3 = (1/4)x + 3/2  
Remaining grains = ((1/2)x - 3) - ((1/4)x + 3/2) = (1/4)x - 9/2  

Step 3: Third customer  
Grains sold = (1/2)((1/4)x - 9/2) + 3 = (1/8)x - 9/4 + 3 = (1/8)x + 3/4  

Remaining grains = ((1/4)x - 9/2) - ((1/8)x + 3/4) = 0  

Step 4: Solve for x  
(1/4)x - 9/2 - 1/8 x - 3/4 = 0  
(1/8)x - 21/4 = 0  
(1/8)x = 21/4  
x = (21/4) * 8 = 42  

Answer: Initial quantity of grains = 42 kg

Interest from bank A = 10000*n*5 / 100 = 500n ( n is the Investment period, in years, in Bank A)
Interest from bank B = (10000*6*5/ 100) + (500n*6*5/100) = 3000 + 150n

​Ratio of interest from Bank A and Bank B= 500n/ 3000 + 150n = 10/13
6500n = 30000 + 1500n
5000n = 30000
N = 30000/5000 = 6

Investment period, in years, in Bank A = 6

Average of first two numbers is 1 more than first number.
Lets take first number = a
Second number = a+2

Average of first three numbers is 2 more than the average of first two numbers which is a+3.
Therefore 3rd number is a+7

Average of first four numbers is 3 more than average of first three numbers which is a+6
Therefore the 4th number is a+15

The difference between the largest and smallest numbers is a+15 - a = 15

Ratio of the distances, 90/T=(T+75)/90

So, Anil takes 90+60+75=225mins=3hr45min=3(3/4)=15/4hr
Speed of Anil in Km/hr =45/(15/4)=45∗4/15=12km/hr