Grab The Knowledge aptitude Book Education knowledge Learning APTITUDE PREPARATION-5

APTITUDE PREPARATION-5

Unknown 18:39 aptitude Book Education knowledge Learning

Unknown

SQUARE ROOTS AND CUBE ROOTS

IMPORTANT FACTS AND FORMULAE

Square Root: If x² = y, we say that the square root of y is x and we write, √y = x.

Thus, √4 = 2, √9 = 3, √196 = 14.

Cube Root: The cube root of a given number x is the number whose cube is x. We denote the cube root of x by ³√x.

Thus, ³√8 = ³√2 x 2 x 2 = 2, ³√343 = ³√7 x 7 x 7 = 7 etc.

Note:

1.√xy = √x * √y 2. √(x/y) = √x / √y = (√x / √y) * (√y / √y) = √xy / y

SOLVED EXAMPLES

Ex. 1. Evaluate √6084 by factorization method .

Sol. Method: Express the given number as the product of prime factors. 2 6084

Now, take the product of these prime factors choosing one out of 2 3042

every pair of the same primes. This product gives the square root 3 1521

of the given number. 3 507

Thus, resolving 6084 into prime factors, we get: 13 169

6084 = 2² x 3² x 13² 13 \ √6084 = (2 x 3 x 13) = 78.

Ex. 2. Find the square root of 1471369.

Sol. Explanation: In the given number, mark off the digits 1 1471369 (1213

in pairs starting from the unit's digit. Each pair and 1

the remaining one digit is called a period. 22 47

Now, 1² = 1. On subtracting, we get 0 as remainder. 44

Now, bring down the next period i.e., 47. 241 313

Now, trial divisor is 1 x 2 = 2 and trial dividend is 47. 241

So, we take 22 as divisor and put 2 as quotient. 2423 7269

The remainder is 3. 7269

Next, we bring down the next period which is 13. x

Now, trial divisor is 12 x 2 = 24 and trial dividend is

313. So, we take 241 as dividend and 1 as quotient.

The remainder is 72.

Bring down the next period i.e., 69.

Now, the trial divisor is 121 x 2 = 242 and the trial

dividend is 7269. So, we take 3as quotient and 2423

as divisor. The remainder is then zero.

Hence, √1471369 = 1213.

Ex. 3. Evaluate: √248 + √51 + √ 169 .

Sol. Given expression = √248 + √51 + 13 = √248 + √64 = √ 248 + 8 = √256 = 16.

Ex. 4. If a * b * c = √(a + 2)(b + 3) / (c + 1), find the value of 6 * 15 * 3.

Sol. 6 * 15 * 3 = √(6 + 2)(15 + 3) / (3 + 1) = √8 * 18 / 4 = √144 / 4 = 12 / 4 = 3.

Ex. 5. Find the value of √25/16.

Sol. √ 25 / 16 = √ 25 / √ 16 = 5 / 4

Ex. 6. What is the square root of 0.0009?

Sol. √0.0009= √ 9 / 1000 = 3 / 100 = 0.03.

Ex. 7. Evaluate √175.2976.

Sol. Method: We make even number of decimal places 1 175.2976 (13.24

by affixing a zero, if necessary. Now, we mark off 1

periods and extract the square root as shown. 23 75

69

\√175.2976 = 13.24 262 629

524

2644 10576

10576

x

Ex. 8. What will come in place of question mark in each of the following questions?

(i) √32.4 / ? = 2 (ii) √86.49 + √ 5 + ( ? )² = 12.3.

Sol. (i) Let √32.4 / x = 2. Then, 32.4/x = 4 <=> 4x = 32.4 <=> x = 8.1.

(ii) Let √86.49 + √5 + x² = 12.3.

Then, 9.3 + √5+x² = 12.3 <=> √5+x²= 12.3 - 9.3 = 3

<=> 5 + x²= 9 <=> x²= 9 - 5= 4 <=> x = √4 = 2.

Ex.9. Find the value of √ 0.289 / 0.00121.

Sol. √0.289 / 0.00121 = √0.28900/0.00121 = √28900/121 = 170 / 11.

Ex.10. If √1 + (x / 144) = 13 / 12, the find the value of x.

Sol. √1 + (x / 144) = 13 / 12 Þ ( 1 + (x / 144)) = (13 / 12 )²= 169 / 144

Þx / 144 = (169 / 144) - 1

Þx / 144 = 25/144 Þ x = 25.

Ex. 11. Find the value of √3 up to three places of decimal.

Sol.

1 3.000000 (1.732

1

27 200

189

343 1100

1029

3462 7100

6924 \√3 = 1.732.

Ex. 12. If √3 = 1.732, find the value of √192 - 1 √48 - √75 correct to 3 places

2

of decimal. (S.S.C. 2004)

Sol. √192 - (1 / 2)√48 - √75 = √64 * 3 - (1/2) √ 16 * 3 - √ 25 * 3

=8√3 - (1/2) * 4√3 - 5√3

=3√3 - 2√3 = √3 = 1.732

Ex. 13. Evaluate: √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

Sol. Given exp. = √(9.5 * 0.0085 * 18.9) / (0.0017 * 1.9 * 0.021)

Now, since the sum of decimal places in the numerator and denominator under the radical sign is the same, we remove the decimal.

\ Given exp = √(95 * 85 * 18900) / (17 * 19 * 21) = √ 5 * 5 * 900 = 5 * 30 = 150.

Ex. 14. Simplify: √ [( 12.1 )² - (8.1)²] / [(0.25)² + (0.25)(19.95)]

Sol. Given exp. = √ [(12.1 + 8.1)(12.1 - 8.1)] / [(0.25)(0.25 + 19.95)]

=√ (20.2 * 4) /( 0.25 * 20.2) = √ 4 / 0.25 = √400 / 25 = √16 = 4.

Ex. 15. If x = 1 + √2 and y = 1 - √2, find the value of (x² + y²).

Sol. x² + y² = (1 + √2)²+ (1 - √2)² = 2[(1)²+ (√2)²] = 2 * 3 = 6.

Ex. 16. Evaluate: √0.9 up to 3 places of decimal.

Sol.

9 0.900000(0.948

81

184 900

736

1888 16400

15104 \√0.9 = 0.948

Ex.17. If √15 = 3.88, find the value of √ (5/3).

Sol.

√ (5/3) = √(5 * 3) / (3 * 3) = √15 / 3 = 3.88 / 3 = 1.2933…. = 1.293.

Ex. 18. Find the least square number which is exactly divisible by 10,12,15 and 18.

Sol. L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 2² * 3²* 5.

To make it a perfect square, it must be multiplied by 5.

\ Required number = (2² * 3² * 5²) = 900.

Ex. 19. Find the greatest number of five digits which is a perfect square.

(R.R.B. 1998)

Sol. Greatest number of 5 digits is 99999.

3 99999(316

9

61 99

61

626 3899

3756

143

\ Required number == (99999 - 143) = 99856.

Ex. 20. Find the smallest number that must be added to 1780 to make it a perfect

square.

Sol.

4 1780 (42

16

82 180

164

16

\ Number to be added = (43)² - 1780 = 1849 - 1780 = 69.

Ex. 21. √2 = 1.4142, find the value of √2 / (2 + √2).

Sol. √2 / (2 + √2) = √2 / (2 + √2) * (2 - √2) / (2 - √2) = (2√2 – 2) / (4 – 2)

= 2(√2 – 1) / 2 = √2 – 1 = 0.4142.

22. If x = (√5 + √3) / (√5 - √3) and y = (√5 - √3) / (√5 + √3), find the value of (x²+ y²).

Sol.

x = [(√5 + √3) / (√5 - √3)] * [(√5 + √3) / (√5 + √3)] = (√5 + √3)² / (5 - 3)

=(5 + 3 + 2√15) / 2 = 4 + √15.

y = [(√5 - √3) / (√5 + √3)] * [(√5 - √3) / (√5 - √3)] = (√5 - √3)² / (5 - 3)

=(5 + 3 - 2√15) / 2 = 4 - √15.

\ x²+ y² = (4 + √15)²+ (4 - √15)² = 2[(4)² + (√15)²] = 2 * 31 = 62.

Ex. 23. Find the cube root of 2744.

Sol. Method: Resolve the given number as the product 2 2744

of prime factors and take the product of prime 2 1372

factors, choosing one out of three of the same 2 686

prime factors. Resolving 2744 as the product of 7 343

prime factors, we get: 7 49

7

2744 = 2³ x 7³.

\ ³√2744= 2 x 7 = 14.

Ex. 24. By what least number 4320 be multiplied to obtain a number which is a perfect cube?

Sol. Clearly, 4320 = 2³ * 3³ * 2² * 5.

To make it a perfect cube, it must be multiplied by 2 * 5²i.e,50.

Author : Unknown

Share this

More

:)

:(

hihi

:-)

:D

=D

:-d

;(

;-(

@-)

:P

:o

:>)

(o)

:p

:-?

(p)

:-s

(m)

8-)

:-t

:-b

b-(

:-#

=p~

$-)

(y)

(f)

x-)

(k)

(h)

cheer

Subscribe to: Post Comments (Atom)