PSAT

PSAT

 

Be top of the class when you study PSAT online

PSAT tests are important for several reasons. However the most important ones are because study PSAT online is great practice for the SAT. Your results here will give students a very good indication of their own strengths and weaknesses. The second reason why PSAT is so important is because passing this test is the only way to qualify for a National Merit Scholarship. Due to these compelling reasons, AtHome Tuition offers a comprehensive study PSAT online course for its valued students.
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Going into a PSAT test, it’s invaluable that students know exactly what to expect. Study PSAT online with AtHome Tuition and you will be ready for anything.

Binomial Theorem

The Binomial Theorem 

 

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We know that any algebraic expression with two variables is called a binomial.  Observe the following powers of the binomial (x + y).      (x + y)0 = 1      (x + y)1 = x + y      (x + y)2 = x2 + 2xy + y2      (x + y)3 = x3 + 3x2y + 3xy2 + y3      (x + y)4 = (x + y)3(x + y)                  = (x3 + 3x2y + 3xy2 + y3) (x + y)                  = x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3+ y4      (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5

What do you notice in the above expansion of various powers of (x + y)?

The number of terms in the expansion is one more than the exponent. In each expansion: The exponent of the first term is same as the exponent of the binomial. The exponent of y in the first term is zero. Subsequently, in each successive term, the exponent of x decreases by 1 with a simultaneous increase of 1 in the exponent of y. The sum of the exponents of x and y in each term is equal to the exponent of the binomial. The exponent of x in the last term is zero and that of y is equal to the exponent of the binomial.

Binomial theorem for a positive integral exponent

Theorem:

If n is a natural number,

Example 1

Example 2

Example 3

Find the constant term or the term independent of x in the expansion of (3x – 5/x2 )9. Solution: Tr +1 = 9Cr (3x)9-r * (–5/x9)r         = 9Cr 39-r * x9-r * (–1)r * 5r/x2r         =9Cr 39-r* 5r (–1)r * x9-3r    ————————    (i) To get the constant term of the expansion we have to find r so that 9 – 3r = 0 ⇒ r = 3 ∴    4th term is the one independent of x, i.e., the constant term.

Try these questions

1. Expand (x + 1/y)7   Find the middle term of the following expansion 2. (x/a + y/b)6 3. (√a – b)8 4. (x2/y – y2/x)8 5. (xy – 1/x2y2)5 6. (a/x + x/a)5 7. (x – 3/y)5 8. Find the term containing x5 in the expansion of (x – 1/x)11 9. Write the 14th term in the expansion of (3 + x)15 10. Write the 10th term in the expansion of (3 + x)12

Answers

1. Expand (x + 1/y)7   Solution:     (x + 1/y)7 = 7C0 x7 + 7C1 x6(1/y) + 7C2 x5 (1/y)2 + 7C3 x4 (1/y)3 + 7C4 x3(1/y)4                   + 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + 7C7 (1/y)7                = x7 + 7x6(1/y) + 7C2 x5(1/y)2 + 7C3 x4(1/y)3 + 7C4 x4(1/y)4                      + 7C5 x2 (1/y)2 + 7C6 x(1/y)6 + (1/y))7 here                    7!  =  7 *6 *5 *4 *3 *2 *1                           7!                  7!           7 *6 *5 *4 *3 *2 *1           7C2 =  ————  =  ——  =  ——————————  =  21                       (7 – 2)!2!        5!2!           5 *4 *3 *2 *1 *2 *1                             7!               7!        7 * 6 * 5           7C3 =  ————  =  ——  =  ————  =  35                      (7 – 3)!3!        4!3!        3 * 2 *1           7C4  =  7C3 = 35 (since nCr = nCn-r )           7C5 = 7C2 = 21          7C6 = 7C1 = 7 ∴  (x + 1/y)7 = x7 + 7x6(1/y) + 21x5(1/y)2 + 35x4(1/y)3 + 35x3(1/y)4 + 21x2(1/y)5                                             + 7x(1/y)6 + (1/y)7   2. (x/a + y/b)6 Solution: Expansion contains 7 terms in it and 4th term is the middle term.   T4= T1+3 = 6C3 (x/a)6-3 (y/b)3                = 6C3 (x/a)3 (y/b)3   3. (√a – b)8 Solution: Expansion contains 9 terms in it and 5th term is the middle term. T5 = T1+4 = 8C4 (√a)8-4 (b)4               = 8C4 (√a)4 (b)4               = 8C4 a2b4   4. (x2/y – y2/x)8 Solution: Expansion contains 9 terms in it and 5th term is the middle term. T5 = T1+4 = 8C4 (x2/y)8–4 (–y2/x)4                = 8C4 (x2/y)4 (–y2/x)4                                        x8 *   y8                = 8C4  ————                           x4 * y4                = 8C4  x4 y4         5. (xy – 1/x2 y2)5 Solution: Expansion contains 6 terms in it and 3rd and 4th terms are the middle terms.       6. (a/x + x/a)5 Solution: Expansion contains 5 terms in it and 3rd term is the middle term.           7. (x – 3/y)5 Solution: Expansion contains 10 terms in it and 5th and 6th terms are the middle terms.           8. Find the term containing x5 in the expansion of (x – 1/x)11 Solution:           9. Write the 14th term in the expansion of (3 + x)15 Solution:           10. Write the 10th term in the expansion of (3 + x)12 Solution:

Cube and Cube Roots 

 

 

Cubes

You have learned that if x is a non-zero number then x* x* x* = x³ and is read as the 'cube of x' or simply 'x cubed'.

Therefore 27 = 3 * 3* 3 = 3³ or 27 is '3 cubed’.

8 = 2* 2* 2 = 23 or '2 cubed'

Consider the following table

Digit	Cube of the digit
1	13 = 1* 1* 1 = 1
2	23 = 2* 2* 2 = 8
3	33 = 3* 3* 3 = 27
4	43 = 4 * 4 * 4 = 64
5	53 = 5 * 5 * 5 = 125
6	63= 6 * 6 * 6 = 216
7	73 = 7 * 7 * 7 = 343
8	83 = 8 * 8 * 8 = 512
9	93 = 9 * 9 * 9 = 729
10	103 = 10 * 10 * 10 = 1000
11	113 = 11 * 11 * 11 = 1331
12	123 = 12 * 12 * 12 = 1728
13	133= 13 * 13 * 13 = 2197
14	143 = 14 * 14 * 14 = 2744
15	153 = 15 * 15 * 15  = 3375
16	163 = 16 * 16 * 16 = 4096
17	173 = 17 * 17 * 17 = 4913
18	183 = 18 * 18* 18 = 5832
19	193 = 19* 19* 19 = 6859
20	203 = 20* 20* 20 = 8000

These integers 1,8,27 . . . 8000 are called perfect cubes.

A non-zero number x is a perfect cube if there is an integer m such that x = m* m* m

Perfect Cubes

How do we find out whether a given number is a perfect cube?

If a prime p divides a number m then p³ will divide m³.

In the prime factorization of a perfect cube, every prime occurs 3 times of a multiple of three times.

For example

In order to check whether a number is a perfect cube or not, we find its prime factors and group together triplets of the prime factors. If no factor is left out then the number is a perfect cube. However if one of the prime factors is a single factor or a double factor then the number is not a perfect cube.

Example : 15

Examine if (i) 392 and (ii) 106480 are perfect cubes.

Solution:

(i)
	392=2*2*2*7*7 7 is a double factor, it is not a part of a triplet so it is not a perfect cube.
(ii)
	106480 One prime factor 2 and the prime factor 5 are not parts of a triplet so 106480 is not a perfect cube.
Example : 16
Is 19683 a perfect cube?
Solution:
19683
Since the prime factor 3 forms three triplets so 19683 is a perfect cube.
Properties of Cubes
If you look at the table in the first page, you will notice that numbers with their units digits 1, 4, 5, 6 or 9 have perfect cubes whose units digits are also 1, 4, 5, 6, 9 respectively. A number with a units digit of 2 has a cube whose units digit is 8 and vice-versa. A number with a units digit of 3 has a cube whose units digit is 7 and vice-versa.
The cube of a negative number is negative. (-1)3 = -1 -1 -1 = -1 = -13 (-2)3 = -2 -2 -2 = -8 = - (2)3 etc. So, negative numbers can be cubes.
Finding the Cubes of a Number
We can find the cube of a number by multiplying the number with itself three times. Another method which is based on the algebraic identity can also be used. Recall that to find the square of a two digit number you used (a + b)2 = a2 + 2ab + b2 and you formed three columns and worked out the square. Here we will use the algebraic identity (a + b)3 = a3 + 3a2b +3ab2 +b3 and form 4 columns and we will find the cube by a similar method as that of the squares.

Example : 17

Find the cube of 89, using the alternate method.

Solution:

We take a = 8 b = 9.

I a3	II 3a2b	III 3ab2	IV b3
83	3 *  82 *  9	3 *  8 *  92	93
512	3 *  64 *  9	24 *  81	729
512	1728	1944	729
+ 192	+ 201	+ 72
704	1929	2016

(89)³ = 704969

Reasoning:

Make 4 columns. In the first column write a³, second 3a²b, third 3ab², and in the fourth write b³.

Write the values of each.

In the first we get a³ = 8³ = 512
In the second, we get 3a2b = 3* 8²* 9 = 1728
In the third, we get 3ab2 = 3 * 8 * 9² = 1944
In the fourth, we get b³ = 9³ = 729

Underline the digit of b3 in this case 9. Add the digits 72 to the value of 3ab2 in this case 1944

Underline the units digit, in this case 6. Add the digits 201 to the value of 3a2b in this case 1728  we get

Underline the units digit, in this case 9. Add the digits 192 to the value of a3 in this case 512  we get

Underline all the digits. The required cube is 704969. We combine all the underlined digits to get the value of the number required.

Example :18

Examine if 53240 is a perfect cube. If not, find the smallest number by which it must be multiplied to form a perfect cube. Also find the smallest number by which it must be divided to form a perfect cube.

Solution:

Reason

Find the prime factors of 53240

53240 5 is not a part of a triplet. For 53240 to become a perfect cube we need to multiply it by 5 * 5 = 25. The smallest number by which 53240 must be multiplied to form a perfect cube is 25. If we divide 53240 by 5 then the resulting number will be a perfect cube. So 5 is the least number by which 53240 must be divided to obtain a perfect cube.

Cube Roots

If n = m³ the m is the cube root of n. We write this as m =  or n^1/3

From Table 2 we have

13 = 1	so
23 = 8	so
33 = 27
43 = 64
53 = 125
63 = 216
73 = 343
83 = 512
93 = 729
103 = 1000

Cube Root by Prime Factorization Method

You have already seen that in the prime factorization of a perfect cube, primes occur as triplets. We therefore can find using the following algorithm.

Step 1	Find the prime factorization of n.
Step 2	Group the factors in triplets such that all three factors in triplet are the same.
Step 3	If some prime factors are left ungrouped, the number n is not a perfect cube and the process stops.
Step 4	If no factor is left ungrouped, choose one factor from each group and take their product. The product is the cube root of n.

Example 19

Find the cube root of  a) 91125      b) 551368

Solution:

a)
	Reason
	Find the prime factorization of 91125 Group the prime factors as triplets such that all the factors in each triplet are the same Choose a factor from each triplet Multiply and get your answer
	91125
b)
	Reason
	Find the prime factors of 551368 Group the factors in triplets each factor of the triplet being the same. Select a factor from each triplet multiply these factors. Write your answer.

Cube Root using Units Digits

Perfect cubes which are six digit numbers can be obtained using the method of the units digits.

If a six digit perfect cube has a units digit of 0, 1, 4, 5, 6 or 9 then its cube root will have a units digit of 0, 1, 4, 5, 6 or 9.

If however the units digit of the cube is 8 then the units digit of the cube root will be 2.

If the units digit of the cube is 2 then the units digit of the cube root will be 8.

If the units digit of the cube is 7, the units digit of the cube root will be 3 and if the units digit of the cube is 3 the units digit of the cube root will be 7.

The cube root of a six digit perfect cube will have at the most, two digits, because the least seven digit number 1000000 = 1003 and its cube root 100 is a three digit number.

We determine the two digits of the cube root as follows:

Step 1	Look at the digit in the units place of the perfect cube and determine the digit in the units place of the cube root as discussed above.
Step 2	Strike out from the right, the last three digits, that is, the units the tens and the hundreds digits. If nothing is left we stop. The digit in step 1 is the cube root.
Step 3	Consider the digits left over from Step 2. Find the largest single digit number whose cube is less than or equal to those left over digits. This is the tens digit.

Example : 20

Find the cube roots of the following numbers

a) 512              b) 2197            c) 117649

Solution:

a)	512
	The digit in the units place is 2. Therefore the digit in the units place of its cube root is 8.       Strike out the 3 digits – the units, the tens and the hundreds. No number is left. The required cube root is 8. or
b)	2197
	Units digit of the cube root is 3 2 13 = 1 < 2 1 = tens digit = 13
	Reason
	Units digit of 2197 is 7 so the units digit of its cube root is 3. Strike out the units, tens and hundreds digitsDigit left is 2. Find the largest single digit number whose cube is less than or equal to this left over digit 2 In this case it is 1.
c)	117649
	Units digit of the cube root = 9 117 43 = 64 < 117 < 125 = 53 4 is the tens digit
	Reason
	Units digit of 117649 is 9. So the units digit of its cube root is 3. Strike out the units, tens and hundreds digits number left. The largest single digit number whose cube 64 is less than 117 is 4.

Try these problems

1. Find the cubes of 402 819 2. Using the alternative method to find the cubes of 56 87 3. Find the smallest number by which the following numbers must be multiplied to obtain a perfect cube. Also find the cube root of the resulting number. 137592 107811 35721 4. Find the smallest number by which the following numbers must be divided so that the products are perfect cubes. Also find the cube root of the resulting number. 7803 8192 26244 5. Fill in the blanks by observing the pattern. 9* 12 + 13 = 10 = 12 *  10 8 *  22 + 23 = 40 = 22 *  10 –* – + – = 90 = 32 *  10 6* 42 + 43 = 160 = ––* –– 5* 52 + 53 = –– = ––* –– 6. Find the cube roots of the following numbers by finding their units and tens digits. 389017 91125 46656 110592 Answers to Practice Problems 1. a. Cube of 402 = (402)3 = 402 *  402 *  402 = 64964808   b. Cube of 819 = (819)3 = 819 *  819 *  819 = 549353259 2. a. Consider 56 a = 5 and b = 6 I a3 II 3a2b III 3ab2 IV b3 Check examples for reasons. 53 125 +50 175 3 *  52 *  6  450 +56 506 3 *  5 *  62 540 +21 561 63 216 563 = 175616 b. Consider 87 a = 8 and  b = 7 I a3 II 3a2b III 3ab2 IV b3 Check examples for reasons. 83   512 +146   658 3 *  82 *  7 1344 +121 1465 3 *  8 *  72 1176   +34 1210 73 343 873 = 658503 3. a. 137592 = In order to obtain a perfect cube we need 7 to complete the triplet and 13 *  13 to complete the next unfinished triplet. The least number by which 137592 must be multiplied to form a perfect cube is 7 *  13 * 13 = 1183. b.   107811 = The extra factor 3 is not a part of a triplet. So we need 3 *  3, to get a perfect cube. The least number by which 107811 must be multiplied is 3 *  3 = 9 so that we get a perfect cube.     c. We need a 7 to complete the triplet. The least number by which 35721 must be multiplied to form a perfect triplet is 7. 4. a.   17 *17 is the least factor needed to divide the number 7803 to get a perfect cube. 289 is the least by which 7803 must be divided to obtain a perfect square. b.   2 is the least factor which is not a part of a triplet.  2 is the least number by which 8192 must be divided so that it forms a perfect cube. c.     2 *  2 and 3 *  3 are not parts of triplets  2 *  2 *  3 *  3 = 36 is the least number by which 26244 must be divided to obtain a perfect cube. 5. 9 * 12 + 13 = 10 = 12 *10 9 * 12 + 13 = 10 = 12 *10 8 * 22 + 23 = 40 = 22*10 7 * 32 + 33 = 90 = 32*10 6 * 42 + 43 = 160 = 42*10 5 * 52 + 53 = 250 = 52 *10 6. a. 389017 Units digit of the cube root = 3 Number left = 389 73 = 343 < 389 < 512 = 83 Tens digit of the cube root is 7. Reason Units digit of 389017 is 7   Units digit of the cube root is 3. Strike out the units, tens, and hundreds digits. The largest single digit number whose cube 343 is less than 389 is 7. b. 91125 Units digit of the cube root = 5 Number left = 91 43 = 5 < 91 < 125 = 53 Tens digit of the cube root is 4. Reason Units digit of 91125 is 5.   Units digit of the cube root is 5. Strike out the units, tens, and hundreds digits. The largest single digit number whose cube 64 is less than 91 is 4. c. 46656 Units digit of the cube root = 6 Number left = 46 33 = 27 < 46 < 64 = 43 Tens digit of the cube root = 3 Reason Units digit of 46656 is 6.  Units digit of the cube root is 6. Strike out the units, tens, and hundreds digit. The largest single digit number whose cube 27 is less than 46 is 3. d. 110592 Units digit of the cube root = 8 Number left = 110 43 = 64 < 110 < 125 = 53 Tens digit of the cube root = 4 Reason Units digit of 110592 is 2   The cube root's units digit is 8. Strike off the units, tens, and hundreds digits. The single largest digit whose cube 64 is less than 110 is 4.

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