RS Aggarwal Quantitative Aptitude PDF Free download: SQUARE ROOTS AND CUBE ROOTS
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
- 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.
- 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 * 52 i.e,50.