Number System - Study Mode
[#201] In which set of numbers every pair is coprime to each other?
Correct Answer
(D) 21, 32, 43
Explanation
Solution: Given, 35, 48, 55 24, 35, 49 42, 55, 69 21, 32, 43 Concept used: Co-prime numbers = Co-prime numbers are ones with just one common element, which is 1. Calculations: According to the question, Factors of (35, 48, 55) = 5, 1 ⇒ 35 = 5 × 7 × 1 ⇒ 48 = 2 × 2 × 2 × 2 × 3 × 1 ⇒ 55 = 5 × 11 × 1 Factors of (24, 35, 49) = 7, 1 ⇒ 24 = 2 × 2 × 2 × 3 × 1 ⇒ 35 = 5 × 7 × 1 ⇒ 43 = 7 × 7 × 1 Factors of (42, 55, 69) = 3, 1 ⇒ 42 = 2 × 3 × 7 × 1 ⇒ 55 = 5 × 11 × 1 ⇒ 69 = 3 × 23 × 1 Factors of (21, 32, 43) = 1 ⇒ 21 = 3 × 7 × 1 ⇒ 32 = 2 × 2 × 2 × 2 × 2 × 1 ⇒ 43 = 43 × 1 ∴ The set of number which are co-prime number are (21, 32, 43)
[#202] Which of the following given value is greater than $$
oot 3 of {12} ,?$$
Correct Answer
(D) $$
oot {12} of {33214} $$
Explanation
Solution: $$eqalign{
& { ext{Given, }}
oot 3 of {12} cr
& {x08f{Concept ,used:}} cr
& { ext{If }}a > b,{ ext{ then:}} cr
& {a^{frac{1}{3}}} > {b^{frac{1}{3}}} cr
& {x08f{Calculation:}} cr
& { ext{Option}}left( { ext{A}}
ight):
oot 9 of {1500} = {left( {1500}
ight)^{frac{1}{9}}} cr
& Rightarrow
oot 3 of {12} = {12^{frac{1}{3}}} cr
& Rightarrow { ext{LCM}}left( {9,,3}
ight) = 9 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3} imes 9}}{ ext{ and }}{left( {1500}
ight)^{frac{1}{9} imes 9}} cr
& i.e.,{left( {12}
ight)^3}{ ext{ and }}{left( {1500}
ight)^1} cr
& i.e.,left( {1728}
ight){ ext{ and }}left( {1500}
ight) cr
& { ext{Since, }}1728 > 1500 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3}}} > {left( {1500}
ight)^{frac{1}{9}}} cr
& { ext{Option}}left( { ext{B}}
ight):
oot 5 of {60} = {left( {60}
ight)^{frac{1}{5}}} cr
& Rightarrow
oot 3 of {12} = {12^{frac{1}{3}}} cr
& Rightarrow { ext{LCM}}left( {5,,3}
ight) = 15 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3} imes 15}}{ ext{ and }}{left( {60}
ight)^{frac{1}{5} imes 15}} cr
& i.e.,{left( {12}
ight)^5}{ ext{ and }}{left( {60}
ight)^3} cr
& {12^2} imes {12^3}{ ext{ and }}{12^3} imes {5^3} cr
& 144 imes {12^3}{ ext{ and }}125 imes {12^3} cr
& { ext{1}}{{ ext{2}}^3}{ ext{ is common both but }}114 > 125 cr
& Rightarrow { ext{Hence }}{12^5} > {60^3}{ ext{ or }}{12^{frac{1}{3}}} > {60^{frac{1}{5}}} cr
& { ext{Option}}left( { ext{C}}
ight):
oot 6 of {121} = {left( {121}
ight)^{frac{1}{6}}} cr
& Rightarrow
oot 3 of {12} = {12^{frac{1}{3}}} cr
& Rightarrow { ext{LCM}}left( {6,,3}
ight) = 6 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3} imes 6}}{ ext{ and }}{left( {121}
ight)^{frac{1}{6} imes 6}} cr
& i.e.,{left( {12}
ight)^2}{ ext{ and }}{left( {121}
ight)^1} cr
& i.e.,{left( {12}
ight)^2}{ ext{ and }}{left( {11}
ight)^2} cr
& { ext{Since, }}12 > 11 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3}}} > {left( {121}
ight)^{frac{1}{6}}} cr
& { ext{Option}}left( { ext{D}}
ight):
oot {12} of {33214} = {left( {33214}
ight)^{frac{1}{{12}}}} cr
& Rightarrow
oot 3 of {12} = {12^{frac{1}{3}}} cr
& Rightarrow { ext{LCM}}left( {12,,3}
ight) = 12 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3} imes 12}}{ ext{ and }}{left( {33214}
ight)^{frac{1}{{12}} imes 12}} cr
& i.e.,{left( {12}
ight)^4}{ ext{ and }}{left( {33214}
ight)^1} cr
& i.e.,{left( {20736}
ight)^1}{ ext{ and }}{left( {33214}
ight)^1} cr
& { ext{Since, }}33214 > 20736 cr
& Rightarrow {left( {12}
ight)^{frac{1}{3}}} > {left( {33214}
ight)^{frac{1}{{12}}}} cr
& herefore { ext{ The required greatest value is }}
oot {12} of {33214} . cr} $$
[#203] Sum of three fractions is $$2frac{{11}}{{24}}.$$ xa0On dividing the largest fraction by the smallest fraction, $$frac{7}{6}$$ is obtained which is $$frac{1}{3}$$ greater than the middle fraction. The smallest fraction is
Correct Answer
(B) $$frac{3}{4}$$
Explanation
Solution: Let the three fractions be p, q and r, where p < q < r. According to the question, $$eqalign{
& frac{r}{p} = frac{7}{6} cr
& Rightarrow r = frac{7}{6}p cr
& { ext{Again, middle fraction}} cr
& = q = frac{7}{6} - frac{1}{3} = frac{{7 - 2}}{6} = frac{5}{6} cr
& herefore p + q + r = 2frac{{11}}{{24}} cr
& Rightarrow p + frac{5}{6} + frac{7}{6}p = frac{{59}}{{24}} cr
& Rightarrow p + frac{{7p}}{6} = frac{{59}}{{24}} - frac{5}{6} cr
& Rightarrow frac{{6p + 7p}}{6} = frac{{59 - 20}}{{24}} = frac{{39}}{{24}} cr
& Rightarrow 13p = frac{{39}}{{24}} imes 6 = frac{{39}}{4} cr
& Rightarrow p = frac{{39}}{{4 imes 13}} = frac{3}{4} cr} $$
[#204] Each prime number has . . . . . . . . factor/factors.
Correct Answer
(D) 2
Explanation
Solution: Each prime number have to distinct factor number itself and 1.
[#205] The greatest perfect square number of digits is?
Correct Answer
(B) 998001
Explanation
Solution: Six digit largest number 999999 Sis digit largest square number = 999999 - 1998 = 998001