Simplification - Study Mode
[#76] The value of $$frac{3}{{{1^2} imes {2^2}}} + $$ xa0 $$frac{5}{{{2^2} imes {3^2}}} + $$ xa0 $$frac{7}{{{3^2} imes {4^2}}} + $$ xa0 $$frac{9}{{{4^2} imes {5^2}}} + $$ xa0 $$frac{{11}}{{{5^2} imes {6^2}}} + $$ xa0 $$frac{{13}}{{{6^2} imes {7^2}}} + $$ xa0 $$frac{{15}}{{{7^2} imes {8^2}}} + $$ xa0 $$frac{{17}}{{{8^2} imes {9^2}}} + $$ xa0 $$frac{{19}}{{{9^2} imes {{10}^2}}}$$ xa0 is = ?
Correct Answer
(B) $$frac{{99}}{{100}}$$
Explanation
Solution: $$eqalign{
& { ext{Given expression,}} cr
& left( {frac{1}{{{1^2}}} - frac{1}{{{2^2}}}}
ight) + left( {frac{1}{{{2^2}}} - frac{1}{{{3^2}}}}
ight) + left( {frac{1}{{{3^2}}} - frac{1}{{{4^2}}}}
ight) + left( {frac{1}{{{4^2}}} - frac{1}{{{5^2}}}}
ight) . . . . + left( {frac{1}{{{9^2}}} - frac{1}{{{{10}^2}}}}
ight) cr
& = left( {frac{1}{{{1^2}}} - frac{1}{{{2^2}}}}
ight) cr
& = left( {1 - frac{1}{{100}}}
ight) cr
& = frac{{99}}{{100}} cr} $$
[#77] The sum of the first 99 terms of the series $$frac{3}{4} + frac{5}{{16}} + frac{7}{{144}} + frac{9}{{400}} + .....$$
Correct Answer
(C) $$frac{{9999}}{{10000}}$$
Explanation
Solution: $$eqalign{
& { ext{Given expression,}} cr
& { ext{ = }}frac{{4 - 1}}{{4 imes 1}} + frac{{9 - 4}}{{9 imes 4}} + frac{{16 - 9}}{{16 imes 9}} + ..... cr
& = left( {1 - frac{1}{4}}
ight) + left( {frac{1}{4} - frac{1}{9}}
ight) + left( {frac{1}{9} - frac{1}{{16}}}
ight) + ..... cr} $$ $$ = left( {frac{1}{{{1^2}}} - frac{1}{{{2^2}}}}
ight) + $$ xa0 $$left( {frac{1}{{{2^2}}} - frac{1}{{{3^2}}}}
ight) + $$ xa0 $$left( {frac{1}{{{3^2}}} - frac{1}{{{4^2}}}}
ight) + $$ xa0 $$.....$$ $$eqalign{
& herefore ,{99^{{ ext{th}}}}{ ext{ term of the series}} cr
& { ext{ = }}left( {frac{1}{{{{99}^2}}} - frac{1}{{{{100}^2}}}}
ight) cr
& herefore ,,,{ ext{Given expression,}} cr} $$ $$left( {frac{1}{{{1^2}}} - frac{1}{{{2^2}}}}
ight) + $$ xa0 $$left( {frac{1}{{{2^2}}} - frac{1}{{{3^2}}}}
ight) + $$ xa0 $$left( {frac{1}{{{3^2}}} - frac{1}{{{4^2}}}}
ight) + $$ xa0 . . . . . $$ + $$ $$left( {frac{1}{{{{98}^2}}} - frac{1}{{{{99}^2}}}}
ight) + $$ xa0 $$left( {frac{1}{{{{99}^2}}} - frac{1}{{{{100}^2}}}}
ight)$$ $$eqalign{
& = left( {1 - frac{1}{{{{100}^2}}}}
ight) cr
& = left( {1 - frac{1}{{10000}}}
ight) cr
& = frac{{9999}}{{10000}} cr} $$
[#78] The smallest fraction which should be subtracted from the sum of $${ ext{1}}frac{3}{4}$$,xa0 $${ ext{2}}frac{1}{2}$$,xa0 $$5frac{7}{{12}}$$,xa0 $${ ext{3}}frac{1}{3}$$ xa0andxa0 $${ ext{2}}frac{1}{4}$$ xa0to make the result a whole number is = ?
Correct Answer
(A) $$frac{5}{{12}}$$
Explanation
Solution: $$eqalign{
& { ext{Sum of given fractions}} cr
& { ext{ = }}frac{7}{4} + frac{5}{2} + frac{{67}}{{12}} + frac{{10}}{3} + frac{9}{4} cr
& = left( {frac{{21 + 30 + 67 + 40 + 24}}{{12}}}
ight) cr
& = frac{{185}}{{12}} cr
& { ext{The whole number just }} cr
& { ext{less than }}frac{{185}}{{12}}{ ext{ is 15}} cr
& { ext{let }}frac{{185}}{{12}} - x = 15 cr
& { ext{Then,}} cr
& x = left( {frac{{185}}{{12}} - 15}
ight) cr
& ,,,,,, = frac{5}{{12}} cr} $$
[#79] $$sqrt {110 + frac{1}{4}} $$ xa0 is equal to = ?
Correct Answer
(D) 10.5
Explanation
Solution: $$eqalign{
& { ext{According to question,}} cr
& sqrt {110 + frac{1}{4}} , cr
& Rightarrow sqrt {frac{{441}}{4}} cr
& Rightarrow frac{{21}}{2} cr
& Rightarrow 10.5 cr} $$
[#80] By what least number should 675 be multiplied so as to obtain a perfect cube number ?
Correct Answer
(B) 5
Explanation
Solution: $$eqalign{
& { ext{According to question,}} cr
& { ext{5}},,{ ext{| 675}} cr
& - - - - - - cr
& 5,,|,,135 cr
& - - - - - - cr
& 5,,|,,27 cr
& - - - - - - cr
& 3,,|,,9 cr
& - - - - - - cr
& 3,,|,,3quad cr
& - - - - - - cr
& ,,,,,|,,1 cr
& { ext{Factors are}} cr
& = ,5 imes 5 imes 3 imes 3 imes 3 cr} $$ ∴ It must be multiplied by 5 to make a perfect cube.