Problems On Numbers - Study Mode

[#1] The sum of two numbers is 40 and their product is 375. What will be the sum of their reciprocals ?
Correct Answer

(B) $$frac{8}{75}$$

Explanation

Solution: Let the numbers be x and y Then, x + y = 40 and xy = 375 $$eqalign{
& herefore frac{1}{x} + frac{1}{y} = frac{{x + y}}{{xy}} cr
& ,,,,,,,,,,,,,,,,,,,,,,, = frac{{40}}{{275}} cr
& ,,,,,,,,,,,,,,,,,,,,,,, = frac{8}{{75}} cr} $$

[#2] A two-digit number is 7 times the sum of its two digits. The number that is formed by reversing its digits is 18 less than the original number. What is the number ?
Correct Answer

(A) 42

Explanation

Solution: Let the ten's digit be x and the unit's digit be y Then, number = 10x + y $$eqalign{
& herefore 10x + y = 7left( {x + y}
ight) cr
& Leftrightarrow 3x = 6y cr
& Leftrightarrow x = 2y cr} $$ Number formed by reversing the digits = 10y + x $$eqalign{
& herefore left( {10x + y}
ight) - left( {10y + x}
ight) = 18 cr
& Leftrightarrow 9x - 9y = 18 cr
& Leftrightarrow x - y = 2 cr
& Leftrightarrow 2y - y = 2 cr
& Leftrightarrow y = 2 cr
& { ext{So, }}x = 2y = 4 cr} $$ Hence, ∴ Required number = 10x + y = 40 + 2 = 42

[#3] If the difference between the reciprocal of a positive proper fraction and the fraction itself be $$frac{9}{20}$$, then the fraction is :
Correct Answer

(B) $$frac{4}{5}$$

Explanation

Solution: Let the fraction be $$frac{a}{1}$$ Then, $$eqalign{
& Leftrightarrow frac{1}{a} - a = frac{9}{{20}} cr
& Leftrightarrow frac{{1 - {a^2}}}{a} = frac{9}{{20}} cr
& Leftrightarrow 20 - 20{a^2} = 9a cr
& Leftrightarrow 20{a^2} + 9a - 20 = 0 cr
& Leftrightarrow 20{a^2} + 25a - 16a - 20 = 0 cr
& Leftrightarrow 5aleft( {4a + 5}
ight) - 4left( {4a + 5}
ight) = 0 cr
& Leftrightarrow left( {4a + 5}
ight)left( {5a - 4}
ight) = 0 cr
& Leftrightarrow a = frac{4}{5},,,,,,,,left[ {x08ecause a
e - frac{5}{4}}
ight] cr} $$

[#4] The sum of two numbers is 75 and their difference is 25. The product of the two numbers is :
Correct Answer

(B) 1250

Explanation

Solution: Let the numbers be a and b According to the question, $$eqalign{
& a + b = 75 cr
& a - b = 25 cr
& x08ecause {left( {a + b}
ight)^2} - {left( {a - b}
ight)^2} = 4ab cr
& Rightarrow {75^2} - {25^2} = 4ab cr
& Rightarrow 4ab = left( {75 + 25}
ight)left( {75 - 25}
ight) cr
& left[ {x08ecause {a^2} - {b^2} = left( {a + b}
ight)left( {a - b}
ight)}
ight] cr
& Rightarrow 4ab = 100 imes 50 cr
& Rightarrow ab = frac{{100 imes 50}}{4} cr
& Rightarrow ab = 1250 cr} $$

[#5] Three-forth of a number is 60 more than its one-third. The number is :
Correct Answer

(C) 144

Explanation

Solution: Let the number be x Then, $$eqalign{
& Leftrightarrow frac{3}{4}x - frac{1}{3}x = 60 cr
& Leftrightarrow frac{{5x}}{{12}} = 60 cr
& Leftrightarrow x = left( {frac{{60 imes 12}}{5}}
ight) cr
& Leftrightarrow x = 144 cr} $$