Square Root & Cube Root
Step 1: First of all group the number in pairs of 2 starting from the right.
Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.
Step 3: To get the unit’s place digit of the square root
Remember the following
If number ends in | Unit’s place digit of the square root |
1 | 1 or 9(10-1) |
4 | 2 or 8(10-2) |
9 | 3 or 7(10-3) |
6 | 4or 6(10-4) |
5 | 5 |
0 | 0 |
Lets see the logic behind this for a better understanding
We know,
12=1
22=4
32=9
42=16
52=25
62=36
72=49
82=64
92=81
102=100
Now, observe the unit’s place digit of all the squares.
Do you find anything common?
We notice that,
Unit’s place digit of both 12 and 92 is 1.
Unit’s place digit of both 22 and 82 is 4
Unit’s place digit of both 32 and 72 is 9
Unit’s place digit of both 42 and 62 is 6.
Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.
Remember,
If first pair of the original number > Result obtained on multiplication then select the greater number out of the two numbers as the unit’s place digit of the square root.
If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.
Let us consider an example to get a better understanding of the method
Example 1: √784=?
Step 1: We start by grouping the numbers in pairs of two from right as follows
7 84
Step 2: To get the ten’s place digit,
We find that nearest square to first group (7) is 4 and √4=2
Therefore ten’s place digit=2
Step 3: To get the unit’s place digit,
We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).
Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,
2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8
Ans:√784=28
Cube roots of perfect cubes
It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.
We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.
Points to remember for speedy calculation of cube roots
1. To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.
13 | = | 1 |
23 | = | 8 |
33 | = | 27 |
43 | = | 64 |
53 | = | 125 |
63 | = | 216 |
73 | = | 343 |
83 | = | 512 |
93 | = | 729 |
103 | = | 1000 |
2. From the above cubes of 1 to 10, we need to remember an interesting property.
13 = 1 | => | If last digit of the perfect cube = 1, last digit of the cube root = 1 |
23 = 8 | => | If last digit of the perfect cube = 8, last digit of the cube root = 2 |
33 = 27 | => | If last digit of the perfect cube = 7, last digit of the cube root = 3 |
43 = 64 | => | If last digit of the perfect cube = 4, last digit of the cube root = 4 |
53 = 125 | => | If last digit of the perfect cube =5, last digit of the cube root = 5 |
63 = 216 | => | If last digit of the perfect cube = 6, last digit of the cube root = 6 |
73 = 343 | => | If last digit of the perfect cube = 3, last digit of the cube root = 7 |
83 = 512 | => | If last digit of the perfect cube = 2, last digit of the cube root = 8 |
93 = 729 | => | If last digit of the perfect cube = 9, last digit of the cube root = 9 |
103 = 1000 | => | If last digit of the perfect cube = 0, last digit of the cube root = 0 |
It’s very easy to remember the relations given above because
1 | -> | 1 | (Same numbers) |
8 | -> | 2 | (10’s complement of 8 is 2 and 8+2 = 10) |
7 | -> | 3 | (10’s complement of 7 is 3 and 7+3 = 10) |
4 | -> | 4 | (Same numbers) |
5 | -> | 5 | (Same numbers) |
6 | -> | 6 | (Same numbers) |
3 | -> | 7 | (10’s complement of 3 is 7 and 3+7 = 10) |
2 | -> | 8 | (10’s complement of 2 is 8 and 2+8 = 10) |
9 | -> | 9 | (Same numbers) |
0 | -> | 0 | (Same numbers) |
Also see
8 -> 2 and 2 -> 8
7 -> 3 and 3-> 7
Questions
Level-I
1. | The cube root of .000216 is: | |||||||
|
2. |
| |||||||
|
3. | The least perfect square, which is divisible by each of 21, 36 and 66 is: | |||||||
|
4. | 1.5625 = ? | |||||||
|
5. | If 35 + 125 = 17.88, then what will be the value of 80 + 65 ? | |||||||
| ||||||||
6. |
If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is: | |||||||
|
7. |
| |||||||
|
8. | A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is: | |||||||
|
9. | The square root of (7 + 35) (7 – 35) is | |||||||
|
10. |
| |||||||||||
Level-II
| ||||||||||||
11. |
| |||||||||||
|
12. | 0.0169 x ? = 1.3 | |||||||
|
13. |
| |||||||||||||
|
14. | How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ? | |||||||
|
15. | The square root of 64009 is: | |||||||
|
16. √29929 = ? | ||||||||
|
17. √106.09 = ? | ||
10.6 | ||
10.5 | ||
10.3 | ||
10.2 | ||
| ||
| ||
18. ?/√196 = 5 | ||
76 | ||
72 | ||
70 | ||
75 | ||
| ||
| ||
Answers
Level-I
Answer:1 Option B
Explanation:
(.000216)1/3 | = |
| 216 |
| 1/3 |
106 |
|
= |
| 6 x 6 x 6 |
| 1/3 |
102 x 102 x 102 |
|
= | 6 |
102 |
= | 6 |
100 |
= 0.06
Answer:2 Option A
Explanation:
Let | x | = | 162 |
128 | x |
Then x2 = 128 x 162
= 64 x 2 x 18 x 9
= 82 x 62 x 32
= 8 x 6 x 3
= 144.
x = 144 = 12.
Answer:3 Option A
Explanation:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444
Answer:4 Option B
Explanation:
1|1.5625( 1.25
|1
|——-
22| 56
| 44
|——-
245| 1225
| 1225
|——-
| X
|——-
1.5625 = 1.25.
Answer:5 Option D
Explanation:
35 + 125 = 17.88
35 + 25 x 5 = 17.88
35 + 55 = 17.88
85 = 17.88
5 = 2.235
80 + 65 = 16 x 5 + 65
= 45 + 65
= 105 = (10 x 2.235) = 22.35
Answer:6 Option C
Explanation:
4a2 – 4a + 1 + 3a = (1)2 + (2a)2 – 2 x 1 x 2a + 3a
= (1 – 2a)2 + 3a
= (1 – 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
Answer:7 Option C
Explanation:
x = | (3 + 1) | x | (3 + 1) | = | (3 + 1)2 | = | 3 + 1 + 23 | = 2 + 3. |
(3 – 1) | (3 + 1) | (3 – 1) | 2 |
y = | (3 – 1) | x | (3 – 1) | = | (3 – 1)2 | = | 3 + 1 – 23 | = 2 – 3. |
(3 + 1) | (3 – 1) | (3 – 1) | 2 |
x2 + y2 = (2 + 3)2 + (2 – 3)2
= 2(4 + 3)
= 14
Answer:8 Option C
Explanation:
Money collected = (59.29 x 100) paise = 5929 paise.
Number of members = 5929 = 77
Answer:9 Option B
Explanation:
(7 + 35)(7 – 35) | = | (7)2 – (35)2 | = 49 – 45 = 4 = 2 |
Answer:10 Option B
Explanation:
5 | – | 10 | + 125 | = | (5)2 – 20 + 25 x 55 |
2 | 5 | 25 |
= | 5 – 20 + 50 |
25 |
= | 35 | x | 5 |
25 | 5 |
= | 355 |
10 |
= | 7 x 2.236 |
2 |
= 7 x 1.118 |
= 7.826 |
Level-II
Answer:11 Option A
Explanation:
Given Expression = | 25 | x | 14 | x | 11 | = 5. |
11 | 5 | 14 |
Answer:12 Option B
Explanation:
Let 0.0169 x x = 1.3.
Then, 0.0169x = (1.3)2 = 1.69
| 1.69 | = 100 |
0.0169 |
Answer:13 Option C
Explanation:
| 3 – | 1 |
| 2 | = (3)2 + |
| 1 |
| 2 | – 2 x 3 x | 1 |
3 | 3 | 3 |
= 3 + | 1 | – 2 |
3 |
= 1 + | 1 |
3 |
= | 4 |
3 |
Answer:14 Option D
Explanation:
A number ending in 8 can never be a perfect square.
Answer:15 Option A
Explanation:
2 |64009( 253
|4
|———-
45 |240
|225
|———-
503| 1509
| 1509
|———-
| X
|———-
64009 = 253.
Answer:16 Option A
Explanation:
√29929 = So, √29929 = 173
Answer:17 Option C
Answer:18 Option C
- PPSC Mains Tests and Notes Program
- PPSC Prelims Exam 2024- Test Series and Notes Program
- PPSC Prelims and Mains Tests Series and Notes Program
- PPSC Detailed Complete Prelims Notes