Question 1:

a2 = 2a1
a3 = 4a1
...
an = 2^(n-1) * a1

We are given a5 - a2 = 12

16a1 - 2a1 = 12 and so a1 = 6/7

Hence E

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Question 2:

3^x - 3^(x-1) = 162
3^(x-1) [ 3 - 1] = 162
3^(x-1) = 81 = 3^4
Hence x-1 = 4

So, x(x-1) = 20

Hence C

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Question 3:

You can do backsolving here to find the correct answer (in the interest
of time).

Basically, we are asked to find which one of the values under the
root sign yields a negative number. It can be either root (2- root(x))
or the outer root.

If you start with 4, you can see that 2- root(4) = 0. So, a number
greater than 4 would immediately satisfy the condition

2 - root(5) is negative and so root (2 - root(5)) will not be a real
number.

Hence E

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Question 4:

Only 2^6 and 2^5*3 fit the bill. All other numbers are > 100.
Hence answer is 2

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Question 5:

Interest is compounded annually.

First year interest = 120
2nd year interest = 10% (1200 + 120) = 132
3rd year interest = 10% (1320 + 132) = 145.2

Sum = 397.2

Hence closest = 400

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Question 6:

We can get a min value for B when A operates for the max time i.e. 8
hours.

8 * 9000 + B * 7000 = 100000
B = 4 hours

Thank you very much Jayhawk.

Q2) 3^x – 3^(x-1) = 162
X(x-1) =?
Another alternative approach besides Jay hawk’s approach:-
A very simple logic here is that the two numbers must be consecutive (x, x-1) are consecutive. Hence amongst the choices let’s find the number which is the product of 2 consecutive numbers
12=2*2*3=4*3
16=2*2*2*2=4*4(Hence discard)
20=2*5*2=5*4
30=2*5*3=5*6
81=3*3*3*3= (9*9) (Hence discard)
If you substitute the pair 5&4 will give the right answer i.e. 162
Hence it should be 5, 4 wherein product will be 20

Q1) Here a2=2a1
A3=2a2 = 4a1
A4=2a3=8a1
A5=2a4=16a1
A5-A2=16a1-2a1=14a1=12
Therefore a1 = 12/14=6/7

A ---- 1 hour ---- 9000 pencils

B -------- 1 hour ------- 7000 pencils

If A operates for 8 hours --------- 8*9000 = 72,000 pencils

So B has to produce ----- 28,000 pencils

B can do so in 28,000/7000 = 4 hours

Hence B has to work for a minimum of 4 hours

5)

$1200, 10% annual interest, for 3 years

P [1 + (R/100)] ^3 – P
=


= 1200 [1+ (10/100)] ^3
= 397.2
= Approx 400

4)

There are only 2 numbers

2*2*2*2*2*2 = 64
2*2*2*2*2*3=96
If you substitute 3 for 2 any further the number will become a 3 digit number