In what time Rs. 8.000 at 3% simple interest per annum will produce some earning as Rs. 6.000 does in 5 years at 4% simple interest per annum ?

Solution
Let the required time
= t years
According to the question,
\(\frac{8000\times 3\times t}{100}=\frac{6000\times 5\times 4}{100}\)
⇒24t= 120
⇒t=\(\frac{120}{24}=5\: years\)
Kanti started a business investing Rs. 9.000. After five months Sudhakar joined in that business investing Rs. 8.000. If in the end of year they earn profit of Rs. 6.970. then what will be the share of Sudhakar in profit?

Solution
Ratio of equivalent capital of Kanti and Sudhakar for I month
= 9000 × 12 : 7 × 8000
= 9 × 12 : 7 × 8
= 27: 14
Sum of the ratios = 27 + 14 = 41
∴ Share of Sudhakar
\(\left ( \frac{14}{41}\times 6970 \right )= Rs\: 2380\)
A wheel makes 1000 revolutions to cover a distance of 88 km. The diameter of the wheel is

Solution
Distance covered by wheel in
1 revolution = π × diameter
∴ Distance covered in 1000 revolutions
= 1000 π × diameter
∴ 1000 π × diameter = 88 km
= 88 × 1000m
⇒Diameter
\(\left ( \frac{88\times 1000\times 7}{1000\times 22} \right )m = 28 m\)
If the side of a square is increased by 25%. its area will increase by

Solution
Required increase
=\(\left ( 25+25+\frac{25\times 25}{100} \right )\)%
= 56.25%
After 5 years. the age of father will be thrice to that of his son. while his age was seven times to that of
his son 5 years ago. The present age of the father is

Solution
Let the present ages of father and son be x and y years respectively.
According to the question,
x+ 5 = 3 (y + 5)
⇒ x 3y = 10 (1)
and, x  5 = 7 (y  5)
⇒x5 = 7y35
⇒x 7y = 30 (ii)
By equation (i) (ii),
x 3y  (x 7y) = 10  (30)
⇒ x  3y  x + 7y = 40
⇒4y=40 ⇒ u=^{40}⁄_{4}=10
By equation (i),
x3x1O=1O
⇒ x = 30 + 10 = 40 years
When 1 is added to numerator and denominator both of a fraction it becomes 4. When 1 is subtracted from numerator and denominat both. it becomes 7. The numerator of the original fraction is

Solution
Let the original fraction =^{x}⁄_{y}
According to the question,
\(\frac{x+1}{y+1}=4\)
⇒ x+ 1 = 4y + 4
⇒ x4y = 3......(i)
and,\(\frac{x1}{y+1}=7\)
⇒ x 1 = 7y7
⇒x7y =  6
⇒x+ 6 = 7y
⇒ \(y=\frac{x+6}{7}.....(ii)\)
∴ x4y=3⇒x\(\frac{4(x+6)}{7}\)=3[From equation (ii),]
\(\Rightarrow \frac{7x4x24}{7}=3\)
⇒3x= 21 + 24 = 45
\(\Rightarrow x=\frac{45}{3}=15\)
Two trains are running on parallel lines at the speeds of 50 kmph and 30 kmph respectively in the same direction. The faster train crosses a man sitting in the slower train in 18 seconds. The length of the faster traln is

Solution
Relative speed oftrain
= (50 30) kmph = 20 kmph
=20×^{5}⁄_{18}=^{50}⁄_{9}gm/sec.
∴ Required length of train
=^{50}⁄_{9}×18 = 100 metre
The ratio of the area of a square of side a units and an equilateral triangle of side a units, is

Solution
Required ratio =\(a^{2}:\frac{\sqrt{3}}{4}a^{2}\)
\(=4:\sqrt{3}\)
A 270m long train is running at the speed of 25 kmph. In what time will it cross a man coming from the opposite direction at the speed of 2kmph?

Solution
Relative speed of train
= (25 + 2) kmph = 27 kmph
Distance coveredby train in crossing the man = 270 m
Now, 27kmph
= 27 × ^{5}⁄_{18} m/sec = ^{15}⁄_{2}m/sec
∴ Required time =\(\frac{270}{\frac{15}{2}}=\frac{270\times 2}{15}\)
= 36 secords
A railway train completes the distance between P and Q station in 48 minutes. If its speed is increased
by 5 km/hr, then it will complete the same distance in 45 minutes. What was the initial speed of that train in km/hr ?

Solution
Let the original speed of train =x kmph.
∴ Distance covered in 48 minutes
or ^{48}⁄_{60} =^{4}⁄_{5} hour =^{4x}⁄_{5}km
After increasing speed by 5kmph,
\(\frac{\frac{4x}{5}}{x+5}=\frac{45}{60}=\frac{3}{4}\Rightarrow \frac{4x}{5(x+5)}=\frac{3}{4}\)
⇒ 16x= 15x+ 75
⇒ x= 75 kmph