Four terms are in A.P. such that their sum is 100 and greatest of them is 4 times the least. Find ‘the second term.

Solution

Let terms be a-3d, a-d, a + d, a + 3d

∴ Adding all terms 4a = 100 ∴ a = 25

Smallest term = 25-3d

Greatest term = 25 + 3d

∴ 4(25-3d) = (25 + 3d) ∴ 100-12d = 25 + 3d

∴ 15d = 75 ∴ d = 5

Second term = a - d = 25 - 5 = 20

A father’s age was 5 times his son’s age 5 years ago and will be 3 times son’s age after 2 years. What is father’s present age?

Solution

Let the ages 5 years ago be x for son and 5x for father.

∴ Today the age of son is x+5 and father is 5x + 5. After two years age of son will be x + 7 and father will be 5x + 7.

After two years equation of their age will be 5x + 7 = 3 (x + 7) ∴ x = 7

∴ present age of son is x + 5 = 7 + 5=12 and father is 5x + 5 = 40

∴ Ratio of present ages is 40 : 12 = 10 : 3

A fraction is multiplied by itself and product is divided by reciprocal. The fraction thus obtained was 18\(\frac{26}{27}\). The original fraction IS

Solution

Let fraction be x

\(∴\frac{x^{2}}{1/x}=x^{3}=18\frac{26}{27}=\frac{512}{27}\)

∴ x³ = (⁸⁄₃)³ ∴ x = ⁸⁄₃ = 2²⁄₃

Ann, Mark, Dave and Paula line up at a ticket window. In how many ways can they arrange themselves so that Dave is third in line from the window?

Solution

Fixing the position of Dave as third from window, Ann, Mark, Paula can arrange themselves in 3! = 6 number of Ways.

^x⁄₁ + ¹⁄_x + ¹⁄_x = \(\frac{10}{x}\),x > 0, x = ?

Solution

Substitute the values.

^x⁄₁ + ¹⁄_x + ¹⁄_x = \(\frac{10}{x}\)

\(\frac{x^{2}+1+x^{2}+1}{x}=\frac{10}{x}\)

∴ 2x² + 2 = 10

∴ x² = 4

∴ x = 2

A team won 40% of 15 games it has already played. If team were to win 75% of its remaining games, it will have won 60% of all its games. How many remaining games are there?

Solution

Let x be the number of remaining games. The team has won 40% of 15 games i.e. 6 games.

The remaining games are x and they won 75% games i.e. 0.75x.

He has won 60% of the total games. Therefore equation will be 6 + 0.75x = 0.60 (15 + x)

6 + 0.75x = 9 + 0.60x

0.15x= 3 ∴ x= 20

If 12 printing presses take 6 days, working 16 hours per day to print 4800 newspapers, how long should it take 6 printing presses working 18 hours a day to print 3600 newspapers?

Solution

According to principles of chain rule -\(\frac{12\times 6\times 16}{4800}=\frac{6\times 18\times D}{3600}\)= man hours required for 1 newspaper

∴ D = \(\frac{12\times 6\times 16}{6\times 18}=\frac{3600}{4800}\)

∴ D = 8 days.

What is the farthest distance between 2 points on a cylinder of radius 4.5 and height 40?

Solution

The figure shows a cylinder of radius 4.5 and height 40.

Clearly A and B are farthest points on the cylinder. The distance AB can be calculated by using Pythagoras theorem as ∠ACB = 90°

(AB)² = (AC)² + (BC)² ∴(AB)² = (40)² + (9)²

9,40,41 is a Pythagoras triplet. AB = 41

Plane is flying from A to B at 500 mph. another plane flying from B to A travels 50 mph faster than the first plane. A and B are 4200 miles apart. If both planes depart at the same time, how far are they from A when they pass each other?

Solution

Combined speed of both the planes in one hour is 500 + 550 = 1050 mph.

∴ They will cover distance of 4200 miles in 4 hours and meet each other.

In 4 hours plane flying from A will cover 2000 miles.

The ratio of arithmetic mean of two numbers to one of numbers is 3:5. What is the ratio of smaller number to larger?

Solution

Let numbers be x and y

∴ \(\frac{\frac{x+y}{2}}{x}=\frac{3}{5}\) ∴ \(\frac{5x + 5y}{2}=3x\)

∴ 5x+ 5y = 6x

∴ x = 5y

y/x = 1/5

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