Explanation:

Let the speed of the water in still water = $x$

Given that speed of the stream = 3 kmph

Speed downstream $=\left(x+3\right)$ kmph

Speed upstream $=\left(x-3\right)$ kmph

He travels a certain distance downstream in 4 hour and come back in 6 hour.

ie, distance travelled downstream in 4 hour = distance travelled upstream in 6 hour

since distance = speed × time, we have

$\left(x + 3\right)4 = \left(x - 3\right)6$

=>$\left(x + 3\right)2 = \left(x - 3\right)3$

=>$2x + 6 = 3x - 9$

=>$x = 6+9 = 15\text{ kmph}$

Explanation:

Given that, time taken to travel upstream = 2 × time taken to travel downstream

When distance is constant, speed is inversely proportional to the time

Hence, 2 × speed upstream = speed downstream

Let speed upstream $=x$

Then speed downstream $=2x$

we have, $\dfrac{1}{2}(x + 2x)$ = speed in still water

$\Rightarrow \dfrac{1}{2}(3x) = 7.5$

$\Rightarrow 3x = 15$

$\Rightarrow x = 5$

i.e., speed upstream = 5 km/hr

Rate of stream $=\dfrac{1}{2}(2x-x) = \dfrac{x}{2}= \dfrac{5}{2} = 2.5\text{ km/hr}$

Explanation:

BD = Rs.100

TD = Rs.80

R = 10%

F = $\dfrac{BD \times TD}{BD-TD}$

$= \dfrac{100 \times 80}{(100-80)} = \dfrac{100 \times 80}{20} =Rs.400$

BD = Simple Interest on the face value of the bill for unexpired time $=\dfrac{FTR}{100}$

$\Rightarrow 100 = \dfrac{400 \times \text{T} \times 10}{100}$

$\Rightarrow 100 = 4 \times {T} \times 10$

$\Rightarrow 10 = 4 \times {T}$

$\Rightarrow {T} $= $\dfrac{10}{4} $= 2.5 years

Explanation:

Given that 1.12.91 is the first Sunday

Hence we can assume that 3.12.91 is the first Tuesday

If we add 7 days to 3.12.91, we will get second Tuesday

If we add 14 days to 3.12.91, we will get third Tuesday

If we add 21 days to 3.12.91, we will get fourth Tuesday

=> fourth Tuesday = [3.12.91 + 21 days ]= 24.12.91

Explanation:

Given that January 1, 2004 was Thursday.

Odd days in 2004 = 2 [because 2004 is a leap year]

Also note that we have taken the complete year 2004 because we need to find out the odd days from 01-Jan-2004 to 31-Dec-2004,

that is the whole year 2004

Hence January 1, 2005 = Thursday + 2 odd days = Saturday

Explanation:

A can give C 18 points in 90

=> While A scores 90 points, C scores $\left(90-18\right)$=72 points

$\text{=> While A scores }\dfrac{90}{72}\text{ points, C scores 1 point.}$

$\text{=> While A scores }\dfrac{90}{72} \times 120 =\dfrac{90}{6} \times 10\text{ = 150 points, C scores 120 point.}$

A can give B 20 points in 60

=> While A scores 60 points, B scores $\left(60-20\right)$=40 points

$\text{=> While A scores 1 point, B scores }\dfrac{40}{60} = \dfrac{2}{3}\text{ points}$

$\text{=> While A scores 150 points, B scores }\dfrac{2}{3} \times 150 = 100\text{ points}$

i.e., while C scores 120 points, B scores 100 points

Hence, in a 120 race, C can give B $\left(120-100\right)$=20 points

Explanation:

Let the required number days be $ x $.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

$\begin{cases}1 : 7 \\7 : 1 \end{cases}::7:x$

$\therefore$ 1 x 7 x $ x $ = 7 x 1 x 7

$\Rightarrow x $ = 7.

Explanation:
S.P. = 102% of Rs. 600 =$(\dfrac{102}{100}\times 600)$=Rs.612

Now, P.W. = Rs. 612 and sum = Rs. 688.50.

T.D. = Rs. (688.50 - 612) = Rs. 76.50.

Thus, S.I. on Rs. 612 for 9 months is Rs. 76.50.

Rate=$[\dfrac{100 \times 76.50 }{612 \times \dfrac{3}{4}}]$%=16$\dfrac{2}{3}$%

Explanation:

Speed of A = 6 km/hr

=$6\times\dfrac{5}{18}$

=$\dfrac{5}{3}m/s$

Time taken by A = $\dfrac{distance}{speed}$

=$\dfrac{100}{\left(\dfrac{5}{3}\right)}$

=60sec

A gives B a start of 4 m

i.e., B has to cover (100-4)

=96 metre

A beats B by 4 sec

=> Time taken by B to run the race = 60 + 4

= 64 sec

Speed of B =$ \dfrac{Distance}{Time}$

=$\dfrac{96}{64}m/s$

=$\dfrac{96}{64}\times\dfrac{18}{5}$

=5.4 km/hr

Explanation:

B.G. = S.I. on T.D.
on Rs. ($140 \times 20 \times \dfrac{1}{2} \times \dfrac{ 1}{100}$)
= Rs. 14
Therefore, B.G. – T.D. = Rs. 14
B.G. = Rs. (140 + 14)
=Rs. 154.