Let total work be 100 units.

Ghulam’s one day work = $\frac{100}{25}$ = 4 units.

Pranab’s one day work = 12 units.

Krishna’s one day work = 2 units.

Combined work of the trio = 4 + 12 + 2 = 18 units.

Hence, They will take  $\frac{100}{18}=\frac{50}{9}$ days. Option(a).  

2nd method:- In place of unit, we can take in percentage form also.In one day, $G=4\%,P=12\%,K=2\%$

 Total in one day= $18\%$ , Then  $100\%$ in $\frac{100}{8}$ days i.e $\frac{50}{9}$ days. Option(a).

Let amount borrowed be Rs. X

Amount to be paid back = Rs. 1.04X

Amount received due to lending at 6% compound interest payable half -yearly = $x(1+\frac{6}{200}{)}^2=1.0609x$

Profit = 1.0609X – 1.04X = 209

0.0209X = 209, X = 10,000; ‘A’ borrowed Rs.10, 000 initially. Option(a)

Alternate Solution

We see that the answer options are far apart and hence the exact calculations are not required. There is an increase of 2% and he is gaining 209 rupees in one year.Hence the amount has to be close to 10000 and hence the answer.

Let the average score after ${18}^{th}$  Mock CSAT be x.
$1^{st}$  Method:- Weighted Average:  $\frac{18x+98}{19}=x+4;x=22$ answer is  $x+4=26$, option (c).
$2^{nd}$  Method:- Using Alligation

Use the reverse gear approach now.

Put X = 22; we will get 18: 1; answer is x + 4 = 26, option (c).

Let speed of son be s,and speed of father be f,then
$\frac{s}{f}=\frac{3}{2}$ (inverse ratio of the time) ( If D is the distance ,then  $\frac{D}{f}=30$ and $\frac{D}{s}=20$ which implies $\frac{s}{f}=\frac{3}{2}$)

Let speed of dad be 20 and son be 30m/s

Now in 5 mins, the father covers 100 m

And son covers 150 m

The son covers 100 m in 3.33 min,

Hence the son will be able to catch up with the father after 8.3 mins from the start. Option (b).

The LCM of 18, 24 and 36 is 72. Let total work is be 72 units.

Sushilkumar Shinde and, Dr. Farooq’s 1 day work  $=\frac{1}{18}\times 72=4units$

Dr. Farooq and S. Jaipal Reddy’s 1 day work   $=\frac{1}{24}\times 72=3units$

S. Jaipal Reddy and Sushilkumar Shinde’s 1 day work   $=\frac{1}{36}\times 72=2units$

Combined work of Ministers  $=\frac{4+3+2}{2}=\frac{9}{2}$  units

Hence, they will take  $\frac{72}{9}\times 2=16$  days;
Option (d).

Average Speed  $=\frac{2\times 50\times 100}{50+100}=\frac{10,000}{150}=66.67$ . Distance is kept constant, hence Speed and Time are inversely proportional, hence, we can use Harmonic Mean.  $H.M=\left(\frac{2ab}{a+b}\right)$   Option(c).

Alternative,   Average  speed= total distance / total time= 66.67. Option (c).

Option (d)

Conventional approach :

$FinalValue=InitialValue{(1+\frac{r}{100})}^t$

Thus, rate  $={\left(\frac{FV}{IV}\right)}^{\frac{1}{t}}-1$

$\Rightarrow \frac{FV}{IV}=\frac{48}{36}=1.33$

Going from answer options

If 10 is the answer

${1.1}^3={11}^2\times 11=12\times 11\approx 1.33$

Therefore r is approximately  $10\%.$

Option (c )

Total = 65 pages

Go from answer options

Option $\left(a\right)\to F=6$

A= 4

Together  $\frac{10}{24}$  in one hour . This multiplied by (65-4=61),

should be an integer which is not true. Hence, answer a is eliminated Similarly, options (b),(d)are eliminated.

Option (c)  $=\frac{35}{25}\times 60,$which is an integer

Reverse gear method 2

If Francis writes 35 pages, then Amit writes 30 pages. Use answer options, to find each person’s individual time the time difference should be one hour. This happens only with option (c)

Let the larger number and smaller number is L and S respectively.

$0.2L=0.3S-2.3\&2L-3S=-23$

$L-S=10\&2L-2S=20$

Solving the above two equations we get L = 53. Hence option (e)