Practice Squareroots and cuberoots - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The smallest number by which 3888 must be divided so that the resulting number is a perfect square is
(a)
(b)
(c)
(d)
Resolving 3888 into its prime factors, we find that
3888=2×2×2×2×3×3×3×3×3
3888 = (2×2) × (2×2) × (3×3) × (3×3) ×3
Here we find that prime factor 3 is appearing alone.
So, if we divide 3888 by 3, we will get a perfect square number
Q-2) $(13)^2 - (4)^3$ - $√{676}$ + 2 = $(?)^2$
(a)
(b)
(c)
(d)
(e)
169 – 64 – $√{676} + 2 = (?)^2$
= 169 – 64 – 26 + 2 = $(?)^2$ = 171 – 90 = 81
∴ ? = 9
Q-3) The square of a natural number when subtracted from its cube results in 48. The number is
(a)
(b)
(c)
(d)
Let the natural number be 'x'.
∴ $x^3-x^2$=48
⇒$x^2$(x-1)=48
⇒$4^2$(4-1)=48
∴ x= 4
Q-4) $√{450 +890 + 685}$ = ?
(a)
(b)
(c)
(d)
(e)
$√{450 + 890 + 685} = √{2025}$ = 45
Q-5) $(13)^2 - (5)^2 - √676 +7 = (?)^2$
(a)
(b)
(c)
(d)
(e)
169 – 25 – 26 + 7 = $(?)^2 = 125 = ?^2⇒? = √125 = 5 √5$
Q-6) $(656 ÷ 164)^2 = √{?}$
(a)
(b)
(c)
(d)
(e)
$√{?}=4^2$ =16
∴? = 256
Q-7) You have a rectangular frame that is 40 cm by 60 cm. Can you put a square picture that has an area of 800 $cm^2$ completely inside the frame?
(a)
(b)
(c)
(d)
Q-8) What is the least number to be added to 2000 to make it a perfect square?
(a)
(b)
(c)
(d)
(e)
| 4 4 | 2000 16 | 45 |
| 85 5 | 400 425 | |
| - 25 |
Clearly, the required least number is 25.
Q-9) The smallest number by which 136 must be multiplied so that it becomes a perfect square is
(a)
(b)
(c)
(d)
Resolve 136 into prime factors and make group of two of each prime factor
136=2×2×2×17
136=(2×2)×2×17
We find that 2 and 17 doesn't appear in group of two. So, 136 has to be multiplied with 34 to make it a perfect square.
Q-10) ($^3$$√{795657}$ × 7) ÷ (3.8 × 5.5) = ?
(a)
(b)
(c)
(d)
(e)
Q-11) $√54 × √2120 ÷ √460$ = ?
(a)
(b)
(c)
(d)
(e)
73.86 × 46.04 ÷ 21.44 = ?
? = 74 × 46 ÷ 22
? = 154.7 ≈ 160
Q-12) 1190 ÷ $√7225$ × ? = 3094 =
(a)
(b)
(c)
(d)
(e)
$1190/√{7225}$ ×? = 3094 = or, ${1190 × ?}/85$ = 3094
or, ? = ${3094 × 85}/1190$ = 221
Q-13) The least number to be subtracted from 24136 to make it a perfect square
(a)
(b)
(c)
(d)
Let us extract the square root from 24136.
∴ 24136, is 111 more than $(155)^2$ .
So if we subtract 111 from 24136, we will get a perfect sq. number.
Q-14) $√3100 × √567 ÷ √250$ = ? ÷ 8
(a)
(b)
(c)
(d)
(e)
$√3100 × √567 ÷ √250 = ? ÷ 8$
⇒56 × 24 ÷ 16 ≈ ? ÷ 8 ⇒${56 × 24}/16 ≈ ?/8$
⇒? = ${56 × 24 × 8}/16$ ≈ 672
∴required answer = 670
Q-15) What must be added to 24136 to make it a perfect square?
(a)
(b)
(c)
(d)
∴ 24136 < $(156)^2$
24136 < 24336
∴ we add 24336 – 24136 = 200
so that it becomes a perfect square
Q-16) 38% of 295 + 62% of 445 = ?
(a)
(b)
(c)
(d)
(e)
$√{{444/37} + 15 + 11 × ?}$ = 7
⇒12+15+11× ? = 49⇒11 × ? = 49 - 27 = 22
∴ ? = ${22}/11$ = 2
Q-17) $(16)^2 - 5^3 + √169 = (?)^2$
(a)
(b)
(c)
(d)
(e)
$(16)^2 - 5^3 + √{169} = (?)^2$
256 – 125 + 13 = $(?)^2$
144 = $(?)^2$
? = ± 12
Q-18) $√225 + √2304 = ? - (12)^2$
(a)
(b)
(c)
(d)
(e)
? = $√{225} + √{2304} + (12)^2 $ = 15 + 48 + 144 = 207
Q-19) The hypotenuse of an isosceles right angled triangular field has a length of $30√2$ m, the length of other side is
(a)
(b)
(c)
(d)
$(30√2)^2=a^2+a^2$
1800=$2a^2$
$a^2$=900
a = 30m
Q-20) The area of a circular play ground is ${3168}/{7}m^2$ . The diameter of the ground is
(a)
(b)
(c)
(d)
Area = $πr^2=3168/7$
$r^2 = {3168}/7×7/{22}$=144
$r =√144$ = 12m
Diameter = 24 m