Answer: Option C. -> 66.67
:
C
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).

Answer: Option D. ->  10%
:
D
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 $10\%.$

Answer: Option D. -> 16
:
D
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).

Answer: Option A. -> 25
:
A
Total Cost Price = Rs 5000 . Total profit = $10\%$ . Total Selling Price $=1.1\times 5000=5500$
Now go from answer options
If the number of footballs sold at profit is 25, then revenue for these 25 balls $=1.3\times 25\times 100=3250$
And the number of footballs sold at loss will be 25; revenue for these balls $=0.9\times 25\times 100=2250$
Sum= 5500 = $10\%$ profit on 5000
Thus, the assumption is right

Answer: Option C. -> 12
:
C
Let total work is LCM of (15, 20) = 60
Work done by Anil in a day = 4 units
Work done by Sunil in a day = 3 units
Anil worked for (t – 6) days and Sunil worked for t days.
4(t – 6) + 3t = 60
t = 12; Option (c).
2nd method:- During last 6 days Sunil alone will work i.e $30\%$ of work will be done by Sunil alone. Remaining $70\%$of work will be done together. In one day both will do $\left(\frac{100}{15}\right)+\left(\frac{100}{20}\right)=11.67\%$ , then$70\%$will complete in 6 days. Total 6+6= 12 days.Option (c).

Answer: Option A. -> 509
:
A
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).

Answer: Option A. -> 10,000
:
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.

:
1st car:
C.P = 150000
S.P = 195000
Profit = 45000
2nd car:
C.P = 175000
S.P = 157500
loss = 17500
Overall profit = 45000 - 17500 = 27500

Answer: Option D. -> 53
:
D
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)

Answer: Option C. -> 33.33%
:
C
Approach 1:
$\%$Nita scored more than Gita $=\frac{25}{100-25}\times 100\%=33.33\%$
Approach 2:
Rita’s score is lesser by $25\%$$\left(\frac{1}{4}\right)$. Nita’s score will be greater by $\frac{1}{x-1}\%=\left(\frac{1}{3}\right)=33.33\%$