Simplification

• ’BODMAS’ Rule : This rule depicts the correct sequence in which the operations are to be executed,so as to find out the value of a given expression.




• Here, ‘B’ stands for ’bracket’ ,’O’ for ‘of’ , ‘D’ for’ division’ and ‘M’ for ‘multiplication’, ‘A’ for ‘addition’ and ‘S’ for ‘subtraction’. Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order (), {} and [].




• After removing the brackets, we must use the following operations strictly in the order: (1)division (2) multiplication (3)addition (4)subtraction.




• Modulus of a real number : Modulus of a real number is defined as |5| = 5 and |-5| = -(-5) = 5. So modulus always gives positive value.




• Virnaculum (or bar): When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

4505 

2 

4a 

17/5 

315 

78 

133/10 

(21/10)*(15/14) 

108.45 

0.46 

12 

2 

3.5 

61/11 

1/8 

2/3 

3/2 

6 

 75

7,11,8 

1/50 

2/5 

2ft. 7 inches 

200 

 1000

180,60 

40 

8cm 

11/18 

480 

288 

5 

121966 

13 

40 

40 

3500 

15 

34 

12 

5040 

20 

Chapter 3: LINEAR EQUATIONS

Introduction

•  (a + b)² = a² + 2 ab + b²        
• (a - b)² = a² - 2ab + b²     
  
• (a + b)³ = a³ + 3a²b + 3ab² + b³      
  
• (a - b)³ = a³ - 3a²b + 3 ab² - b³       
  
• a² - b² = (a - b) (a + b)        
  
• a³ - b³ = (a - b)³ + 3 a b (a - b)        

The above formulae can be used for solving giant multiplication problems.

1. 1605 * 1605 = (1605)^2 = (1600 + 5)^2 which is (a+b)^2.

Find units digits of large exponents

Sometimes we have to calculate the powers of very large numbers like 13^35 which looks impossible but the answer is quite simple .. Recognize the cycle.

Suppose we need to find last digit of 2^ ²⁰¹²

2^1 = 2

2^2= 4

2^3 = 8

2^4 = 16 = 6

2^ 5 = 32 = 2

2^ 6 = 64 =4

Hence there is a cycle of size 4 like 2-4-8-6-2-4-8-6... so all powers of 2 in multiples of 4 like 4,8,12,16... have last digit 6 since 2012 also satisfies this 2^2012  has last digit 6.

So generalizing for k > 4 if 2^^k = 2^⁽⁴ⁿ⁺¹⁾ then last digit is 2, if 2^⁽⁴ⁿ⁺²⁾ = 4, 2^⁽⁴ⁿ⁺³⁾ = 8 , 2^⁽⁴ⁿ⁾ = 6.

Similarly we can solve for all digits:

For 3 cycle is: 3-9-7-1-3-9-7-1... here too the cycle is 4 and  So generalizing for k > 4 if 3^^k = 3^⁽⁴ⁿ⁺¹⁾ then last digit is 2, if 3^⁽⁴ⁿ⁺²⁾ = 4, 3^⁽⁴ⁿ⁺³⁾ = 8 , 3^⁽⁴ⁿ⁾ = 6.

For 4 cycle is: 4-6-4-6-4-6-... hence cycle is 2 So generalizing for k > 3 if 4^^k = 4^⁽²ⁿ⁺¹⁾ then last digit is 6, if 4^⁽²ⁿ⁾ = 4.

Similarly we can obtain cycles for all digits till 9 any digit above this doesn't matter as 13^n is same as 3^n and 17^ⁿ is same as 7^ⁿ etc.

Division problems

Dividend - Remainder = Divisor * Quotient.

Q. A number when divided by 342 gives a remainder 47 when the same number is divided by 19 what would be the remainder?

A. 

Step 1: (number - 19) = 342 * k + 47 i.e.  number = 19*18k + 19*2+9 = 19(18k+2)+9

 if this number is divided by 19 then it gives 18k+2 as quotient and 9 as remainder.

Problems on ages  

Linear equations

Q. if f(x) = log (1+x/1-x) then f(x) + f(y) is

1. f(x + y)

2. $f\frac{x+y}{1+xy}$

3. $(x+y) f \frac{1}{1+xy}$

4. $\frac{f(x) + f(y)}{1+xy}$

Ans . B

1. $f(x) + f(y) = log \frac{1+x}{1-x}$ + $log \frac{1+y}{1-y}$

2. $log \frac{(1+x)(1+y)}{(1-x)(1-y)}$

3. $log \frac{(1+x+y+xy)}{(1-x-y+xy)}$

4. $log \frac{(1+x+y+xy)}{(1+xy-(x+y))}$

5. $log \frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}$

6. $f \frac{x+y}{1+xy}$

Q. Number S is obtained by squaring the sum of digits of a two-digit number D. If difference between S and D is 27, then the two-digit number D is

1. 24

2. 54

3. 34

4. 45

Ans . B

1. check choices and only choice B satisfies the condition.

2. S = (5+4)^2 = 81

3. D – S = 81 – 54 = 27. Hence, the number = 54

Q. The nth element of a series is represented as $X_n = (-1)^n * X_{n-1}$ . If $X_0$ = x and x > 0, then which of the following is always true?

1. $X_n$ is positive if n is even

2. $X_n$ is positive if n is odd

3. $X_n$ is negative if n is even

4. None of these

Ans . D

1. $x_0 = x, x_1 = -x, x_2 = -x$

2. $x_3 = x, x_4 = x$

3. $x_5 = -x, x_6 = -x$

4. Choices (1), (2), (3) are incorrect.

Q. If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

1. 5/3

2. √ 19

3. 13/3

4. none

Ans . C

1. $xy + yz + zx = 3$

2. $xy + (y + x)z = 3 $

3. $xy + (y + x)(5 + x + y)= 3$

4. $xy + 5y+xy+y^2+5x+x^2+xy = 3$

5. As it is given that y is a real number, the discriminant for above equation must be greater than or equal to zero. $(x - 5)^2 - 4(x^2 - 5x + 3) \geq 0$

6. $3x^2 - 10x + 13 \geq 0$

7. $x = 1, \frac{13}{3}$

8. Largest value that x can have is $\frac{13}{3}$

Q. Neeraj has agreed to mow a lawn, which is a 20 m × 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round before he has mowed half the lawn?

1. 2.5

2. 3.5

3. 3.8

4. 4

Ans . C

1. Area = 40 * 20 = 800 $m^2$

2. If 3 rounds are done, area = 34 × 14 = 476 $m^2$ ⇒ Area > 3 rounds

3. if 4 rounds ⇒ Area left = 32 × 12 = 347 $m^2$

4. Hence, area should be slightly less than 4 rounds.

Q. The owner of a local jewellery store hired three watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave 1/2 of the diamonds he had then, and 2 more besides. He escaped with one diamond. How many did he steal originally?

1. 40

2. 36

3. 25

4. None of these

Ans . B

1. Since thief escaped with 1 diamond

2. Before 3 rd watchman he had (1 + 2) × 2 = 6 diamonds.

3. Before 2 nd watchman he had (6 + 2) × 2 = 16 diamonds

4. Before 1 st watchman he had (16 + 2) × 2 = 36 diamonds.

Q. Mayank, Mirza, Little and Jaspal bought a motorbike for Rs.60. Mayank paid one-half of the sum of the amounts paid by the other boys. Mirza paid one-third of the sum of the amounts paid by the other boys. Little paid one-fourth of the sum of the amounts paid by the other boys. How much did Jaspal have to pay?

1. Rs.15

2. Rs.13

3. Rs.17

4. None of these

Ans . B

1. Mayank paid 1/2 of the sum paid by other three.

2. Mayank paid 1/3 rd of the total amount = Rs.20

3. Similarly, Mirza paid Rs.15 and Little paid Rs.12. Remaining amount of Rs.60 – Rs.20 – Rs.15 – Rs.12 = Rs.13 is paid by Jaspal

Q. A rich merchant had collected many gold coins. He did not want anybody to know about him. One day, his wife asked, " How many gold coins do we have?" After a brief pause, he replied, "Well! if I divide the coins into two unequal numbers, then 48 times the difference between the two numbers equals the difference between the squares of the two numbers." The wife looked puzzled. Can you help the merchant's wife in finding out how many gold coins the merchant has?

1. 96

2. 53

3. 43

4. None of these

Ans . D

1. Let the number of gold coins = x + y

2. 48(x – y) = x² - y²

3. 48(x – y) = (x – y)(x + y) ⇒ x + y = 48

4. Hence, the correct choice will be none of these.

Q. A car rental agency has the following terms. If a car is rented for 5 hr or less, then, the charge is Rs. 60 per hour or Rs. 12 per kilometre whichever is more. On the other hand, if the car is rented for more than 5 hr, the charge is Rs. 50 per hour or Rs. 7.50 per kilometre whichever is more. Akil rented a car from this agency, drove it for 30 km and ended up playing Rs. 300. For how many hours did he rent the car?

1. 4 hr

2. 5 hr

3. 6 hr

4. None of these

Ans . C

1. By trial and error:

2. 30 × 12 = 360 > 300

3. 30 × 7.5 = 225 < 300

4. 50 × 6 = 300. Hence, he rented the car for 6 hr.

Q. The number of non-negative real roots of 2^x – x – 1 = 0 equals

1. 0

2. 1

3. 2

4. 3

Ans . C

1. 2^x – x – 1 = 0 and so 2^x – 1 = x

2. If we put x = 0, then this is satisfied and if we put x = 1, then also this is satisfied.

3. Now, if we put x = 2, the equation this is not valid.

Q. When the curves y = log₁₀x , y=x^-1 are drawn in the x-y plane, how many times do they intersect for values x ≥ 1 ?

1. Never

2. Once

3. Twice

4. More than twice

Ans . B

1. For the curves to intersect, log x = x

2. Thus log₁₀x = 1/x or x^x=10

3. This is possible for only one value of x such that 2 < x < 3.

Q. Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r ≠ 0?
x+ 2y – 3z = p
2x + 6y – 11z = q
x – 2y + 7z = r

1. 5p –2q – r = 0

2. 5p + 2q + r = 0

3. 5p + 2q – r = 0

4. 5p – 2q + r = 0

Ans . A

1. It is given that p+q+r≠0 , if we consider the first option, and multiply the first equation by 5, second by –2 and third by –1, we see that the coefficients of x, y and z all add up-to zero.

2. Thus, 5p – 2q – r = 0 No other option satisfies this.

Q. Let g(x) = max(5 – x, x + 2). The smallest possible value of g(x) is

1. 4.0

2. 4.5

3. 1.5

4. None of the above

Ans . D

1. We can see that x + 2 is an increasing function and 5 – x is a decreasing function. This system of equation will have smallest value at the point of intersection of the two. i.e. 5 – x = x + 2 or x = 1.5.

2. Thus smallest value of g(x) = 3.5

Q. The function f(x) = |x – 2| + |2.5 – x| + |3.6 – x|, where x is a real number, attains a minimum at

1. x = 2.3

2. x = 2.5

3. x = 2.7

4. None of the above

Ans . B

1. At x = 2, f(x) = 2.1

2. At x = 2.5, f(x) = 1.6

3. At x = 3.6, f(x) = 2.7.Hence, at x = 2.5, f(x) will be minimum.

Q. How many three digit positive integers, with digits x, y and z in the hundred's, ten's and unit's place respectively exist such that x < y, z < y and x ≠ 0?

1. 245

2. 285

3. 240

4. 320

Ans . C

1. If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values.

2. Thus with y = 2, a total of 1 × 2 = 2 numbers can be formed. With y = 3, 2 × 3 = 6 numbers can be formed. Similarly checking for all values of y from 2 to 9 and adding up we get the answer as 240.

Quiz

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