Square Root And Cube Root - Study Mode

[#1] $$left( {frac{{sqrt {625} }}{{11}} imes frac{{14}}{{sqrt {25} }} imes frac{{11}}{{sqrt {196} }}}
ight){kern 1pt} $$ xa0 xa0 is equal to :
Correct Answer

(A) 5

Explanation

Solution: $$eqalign{
& { ext{Give}},{ ext{Expression}} cr
& = frac{{25}}{{11}} imes frac{{14}}{5} imes frac{{11}}{{14}} cr
& = 5 cr} $$

[#2] $$sqrt {0.0169 imes ?} = 1.3$$
Correct Answer

(B) 100

Explanation

Solution: $$eqalign{
& { ext{Let}}sqrt {0.0169 imes x} = 1.3 cr
& { ext{Then}},,0.0169x = {left( {1.3}
ight)^2} = 1.69 cr
& Rightarrow x = frac{{1.69}}{{0.0169}} = 100 cr} $$

[#3] $${left( {sqrt 3 - frac{1}{{sqrt 3 }}}
ight)^2},{ ext{simplifies}},{ ext{to:}}$$
Correct Answer

(C) $$frac{4}{3}$$

Explanation

Solution: $$eqalign{
& {left( {sqrt 3 - frac{1}{{sqrt 3 }}}
ight)^2} cr
& = {left( {sqrt 3 }
ight)^2} + {left( {frac{1}{{sqrt 3 }}}
ight)^2} - 2 imes sqrt 3 imes frac{1}{{sqrt 3 }} cr
& = 3 + frac{1}{3} - 2 cr
& = 1 + frac{1}{3} cr
& = frac{4}{3} cr} $$

[#4] How many two-digit numbers satisfy this property. : The last digit (unit's digit) of the square of the two-digit number is 8 ?
Correct Answer

(D) None of these

Explanation

Solution: A number ending in 8 can never be a perfect square.

[#5] The square root of 64009 is:
Correct Answer

(A) 253

Explanation

Solution: $$eqalign{
& ,,,,,,,,2|overline 6 overline {40} overline {09} ,(,,253 cr
& ,,,,,,,,,,,|4 cr
& ,,,,,,,,,,,| - - - - - cr
& ,,,,45,|240 cr
& ,,,,,,,,,,,|225 cr
& ,,,,,,,,,,,| - - - - - cr
& 503,,|,,,,,,1509 cr
& ,,,,,,,,,,,|,,,,,,1509 cr
& ,,,,,,,,,,,| - - - - - cr
& ,,,,,,,,,,,|,,,,,,,,,,,x cr
& ,,,,,,,,,,,| - - - - - cr
& herefore sqrt {64009} = 253 cr} $$