If ax = by, then:
If ax = by, then:
[A].
| log | a | = | x | 
| b | y | 
[B].
| log a | = | x | 
| log b | y | 
[C].
| log a | = | y | 
| log b | x | 
[D].
Answer: Option C
Explanation:
ax = by
log ax = log by
x log a = y log b
| log a | = | y | . | |
| log b | x | 
[B].
[C].
[D].
Answer: Option C
Explanation:
Let the principal be P and rate of interest be R%.
| Required ratio = | 
 | = | 6PR | = | 6 | = 2 : 3. | ||||
| 
 | 9PR | 9 | 
Video Explanation: https://youtu.be/GaaEDwTWc6w
[B].
[C].
[D].
Answer: Option C
Explanation:
| (A + B + C)’s 1 day’s work = | 1 | ; | 
| 6 | 
| (A + B)’s 1 day’s work = | 1 | ; | 
| 8 | 
| (B + C)’s 1 day’s work = | 1 | . | 
| 12 | 
| (A + C)’s 1 day’s work | 
 | ||||||||||||||
| 
 | |||||||||||||||
| 
 | |||||||||||||||
| 
 | 
So, A and C together will do the work in 8 days.
[B].
[C].
[D].
Answer: Option D
Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
| Required number of ways | = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
| 
 | ||||||||||||||||||||
| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. | 
[B].
[C].
[D].
Answer: Option C
Explanation:
Let the speed of the train be x km/hr and that of the car be y km/hr.
| Then, | 120 | + | 480 | = 8 | 1 | + | 4 | = | 1 | ….(i) | 
| x | y | x | y | 15 | 
| And, | 200 | + | 400 | = | 25 | 1 | + | 2 | = | 1 | ….(ii) | |
| x | y | 3 | x | y | 24 | 
Solving (i) and (ii), we get: x = 60 and y = 80.
Ratio of speeds = 60 : 80 = 3 : 4.
[B].
[C].
[D].
Answer: Option D
Explanation:
One of AB, AD and CD must have given.
So, the data is inadequate.
[B].
[C].
[D].
Answer: Option B
Explanation:
| Perimeter = Distance covered in 8 min. = | 12000 | x 8 | m = 1600 m. | |
| 60 | 
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
Length = 480 m and Breadth = 320 m.
Area = (480 x 320) m2 = 153600 m2.