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.