A man goes from A to B at the speed of 5 kmph and returns from B to A at 4 kmph. His average speed during the whole journey is

Solution
Average speed
= \(\frac{2 \times 4 \times 5}{4 + 5} = \frac{40}{9}\)kmph
Note: If two equal distances are covered at two different speeds of x kmph and y kmph, then average speed = \(\frac{2xy}{x + y}\)kmph.
Two numbers are less than the third number by 30% and 37% respectively. By what per cent is the second number less than the first number?

Solution
Let the third number be x.
Then
First number = 70% of x = ^{7x}⁄_{10}
Second number = 63% of x
= \(\frac{63x}{100}\)
Difference = \(\frac{7x}{10}  \frac{63x}{100}\)
= \(= \frac{70x  63x}{100} = \frac{7x}{100}\)
Required percentage
= \(\left ( \frac{7x}{100} \times \frac{10}{7x} \times 100 \right )\% = 10\%\)
In an election a candidate gets 40% of votes polled and is defeated by the winning candidate by 298 votes. Find the total number of votes polled.

Solution
Let the total number of votes polled be x:
⇒ votes of winning = 100  40 = 60%
20% of X = 298
⇒ ^{x}⁄_{5} = 298
⇒ x = 298 × 5 = 1490
Ram bought 1600 eggs at the rate of Rs. 3.75 per dozen. He sold 900 of them at 2 eggs for Re 1 and the remaining at 5 eggs for Rs. 2. Find his loss or gain percent.

Solution
CP of 1600 eggs
= \(Rs.\left ( \frac{3.75 \times 1600}{12} \right ) = Rs.500\)
SP of 1600 eggs
= \(Rs.\left ( \frac{900}{2} + \frac{2}{5} \times 700 \right )\)
= Rs. (450 + 280) = Rs. 730
Profit = Rs. (730  500)
= Rs. 230
∴ Gain% = \(\frac{230}{500} \times 100 = 46\%\)
A shopkeeper earns 20% by allowing a discount of 10% on the marked price of an article. An article is marked at Rs.280, its actual price is

Solution
Let the cost price = x Rupees.
⇒ As per question
Price offer Allowed 10% on Marked Price
280 = C.P. x 20% Profit
⇒ \(280 \times \frac{100  10}{100} = x \times \frac{120}{100}\)
⇒ \(x = 280 \times \frac{90}{100} \times \frac{100}{120}\)
= \(280 \times \frac{90}{120}\)
= 7 × ^{90}⁄_{3} = Rs.210
If the interest is compounded annually and the compound interest after 3 years at 10% per annum on a sum is Rs. 331, the principal is

Solution
\(C.I. = P\left [ \left ( 1 + \frac{R}{100} \right )^{T}  1 \right ]\)
\(331 = P\left [ \left ( 1 + \frac{10}{100} \right )^{3}  1 \right ]\)
\(331 = P\left [ \left (\frac{11}{100} \right )^{3}  1 \right ]\)
\(331 = P\left [ \frac{1331  1000}{100} \right ]\)
⇒ \(\frac{P \times 331}{1000} = 331\)
⇒ P = Rs. 1000
A particle is thrown upwards at the velocity of 39.2 m/sec. Time to reach the highest point is

Solution
u = 39.2 m/sec
v = O
g = 9.8 m/sec^{2}
∴ v = u + gt
0 = 39.2  9.8 t
⇒ 39.2 = 9.8 t
⇒ t = ^{39.2}⁄_{9.8} = 4 second
In an examination,a student gets 20% of total marks and fails by 30 marks. Another student gets 32% of total marks which is more than the minimum pass marks by 42 marks. The pass percentage is

Solution
Let the maximum marks be x.
∴ 20% of x + 30
= 32% of x  42
⇒ \(\frac{x}{5} + 30 = \frac{x \times 32}{100}  42\)
⇒ \(\frac{x}{5} + 30 = \frac{8x}{25}  42\)
⇒ \(\frac{8x}{25}  \frac{x}{5} = 30 + 42\)
⇒ \(\frac{8x  5x}{25} = 72\)
⇒ \(\frac{8x  5x}{25} = 72\)
⇒ \(\frac{3x}{25} = 72\)
⇒ \(x = \frac{72 \times 25}{3} = 600\)
∴ Pass marks
= 20% of 600 + 30
= \(\frac{600 \times 20}{100} + 30 = 150\)
∴ Pass percentage
= \(\frac{150}{600} \times 100 = 25\%\)
At what rate of compound interest a sum will be ^{25}⁄_{16} times of itself in 2 years?

Solution
Let the principal be Rs. x.
\(A = P\left ( 1 + \frac{R}{100} \right )^{T}\)
\(\frac{25}{16}x = x\left ( 1 + \frac{R}{100} \right )^{2}\)
⇒ \(\left ( 1 + \frac{R}{100} \right )^{2} = \left ( \frac{5}{4} \right )^{2}\)
⇒ \( 1 + \frac{R}{100} = \frac{5}{4}\)
⇒ \( \frac{R}{100} = \frac{5}{4}  1 = \frac{1}{4}\)
⇒ R = ^{1}⁄_{4} × 100 = 25%
The simple interest on a certain sum at the rate of 4% per annum in 4 years is Rs. 80 more than the simple interest on the same sum at 5% per annum after 3 years. The sum is

Solution
Let the principal be Rs. x.
According to the question,
\(\frac{x \times 4 \times 4}{100}  \frac{x \times 5 \times 3}{100} = 80\)
⇒ 16x  15x = 80 × 100
⇒ x = Rs. 8000