• ’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.

Ans .

4505

1. Explanation :

 5005 - 5000 + 10 = 5005 - (5000/10) = 5005-500 = 4505.

Ans .

2

1. Explanation :

18800+470+20=(18800/470)+20=40/20=2.

Ans .

4a

1. Explanation :

Given expression=b-[b-(a+b)-{b-(b-a+b)}+2a]
=b-[b-a-b-{b-2b+a}+2a]
=b-[-a-{b-2b+a+2a}]
=b-[-a-{-b+3a}]=b-[-a+b-3a]
=b-[-4a+b]=b+4a-b=4a.

Ans .

17/5

1. Explanation :

9/2+19/6+x+7/3 = 67/5
Then x=(67/5)-(9/2+19/6+7/3)
x=(67/5)-((27+19+14)/6)=((67/5)-(60/6)
x=((67/5)-10)=17/5

Ans .

315

1. Explanation :

Let the number be x. then 4/15 of 5/7 of x-4/9 of 2/5 of x=8
So 4/21x-8/45x=8
Therefore (4/21-8/45)x=8 ; (60-56)/315x=8;
4/315x=8
x=(8*315)/4=630
1/2x=315
Hence required number = 315.

Ans .

78

1. Explanation :

Given exp. =[13/4 / {5/4-1/2(5/2-(3-2)/12)}]=[13/4 / {5/4-1/2(5/2-1/12)}]
=[13/4 / {5/4-1/2((30-1)/12)}]=[13/4 / {5/4-29/24}]
=[13/4 / (30-39)/24}]=[13/4 / 1/24]=[(13/4)*24]=78


Ans .

133/10

1. Explanation :

$$108 / 9 + \frac{2}{5} * \frac{13}{4} = \frac{108}{9} + \frac{13}{10} = 12 + \frac{13}{10} = \frac{133}{10} = 13\frac{3}{10}$$

Ans .

(21/10)*(15/14)

1. Explanation :

Given $$\frac{(7/2)*(2/5)*(3/2)}{(7/2)/(15/4)} / 5.25 = \frac{21}{10} / \frac{525}{100} = (21/10)*(15/14)$$

Ans .

108.45

1. Explanation :

Given exp. = 12.05*(5.4/0.6) = (12.05*9) = 108.45


Ans .

0.46

1. Explanation :

Given exp. = 0.6*0.6+(0.6*6) = 0.36+0.1 = 0.46


Ans .

12

1. Explanation :

(17.28/x) = 2*3.6*0.2
x = (17.28/1.44) = (1728/14) = 12

Ans .

2

1. Explanation :

(364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412
x = (364.824/182.412) =2


Ans .

3.5

1. Explanation :

8.5-{5.5-(7.5+(2.8/x))}*(4.25/0.04) = 306
8.5-{5.5-{(7.5x+2.8)/x)}*(425/4) = 306
8.5-{(5.5x-7.5x-2.8)/x}*(425/4) = 306
8.5-{(-2x-2.8)/x}*106.25  = 306
8.5-{(-212.5x-297.5)/x} = 306
(306-221)x = 297.5
x =(297.5/85) = 3.5


Ans .

61/11

1. Explanation :

$$[x^2 + y^2]/[x^2 - y^2] = [\frac{x^2}{y^2} + 1]/[\frac{x^2}{y^2} - 1] = = [(36/25)+1] / [(36/25)-1] = (61*25)/(25*11) = 61/11$$

Ans .

1/8

1. Explanation :

$$4 - \frac{5}{1 + \frac{1}{3 + \frac{1}{2 + \frac{1}{4}}}} = 4 - \frac{5}{1 + \frac{1}{3 + \frac{1}{9}}} = 4 - \frac{5}{1 + \frac{1}{31/9}} = 4 - \frac{5}{1 + 9/31} = 4 - \frac{5}{(40/31)} = 4 – (5*31)/ 40 = 4 – (31/8) = 1/8$$

Ans .

2/3

1. Explanation :

$$\frac{2x}{1 + \frac{1}{1 + \frac{x}{1 - x}}} = 1 ; \frac{2x}{1 + \frac{1}{1/(1- x)}} = 1; \frac{2x}{1+ (1 – x)} = 1 ; 2x = 2-x ; 3x = 2 ; x = (2/3).$$

Ans .

3/2

1. Explanation :

(a/b)=3/4 ;b=(4/3) a.
8a+5b=22 ; 8a+5*(4/3)a=22 ; 8a+(20/3); a=22;
44a = 66 ; a=(66/44)= 3/2


Ans .

6

1. Explanation :

(x /4)-((x-3)/6)=1; (3x-2(x-3) )/12 = 1 ; 3x-2x+6=12 ; x=6

Ans .

 75

1. Explanation :

The given equations are:
2x+3y=34 …(i) and, ((x + y) /y)=13/8 ; 8x+8y=13y ; 8x-5y=0 …(ii)
Multiplying (i) by 5,(ii) by 3 and adding, we get : 34x=170 or x=5.
Putting x=5 in (i), we get: y=8.
5y+7x=((5*8)+(7*5))=40+35=75


Ans .

7,11,8

1. Explanation :

The given equations are:
2x+3y+z=55 …(i); x + z - y=4 …(ii); y -x + z =12 …(iii)
Subtracting (ii) from (i), we get: x+4y=51  …(iv)
Subtracting (iii) from (i), we get: 3x+2y=43  …(v)
Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.
Putting x=7 in (iv), we get: 4y=44 or y=11.
Putting x=7,y=11 in (i), we get: z=8


Ans .

1/50

1. Explanation :

Given expression = (2/3)*(3/4)*(4/5) *…….* (99/100) = 2/100 = 1/50

Ans .

2/5

1. Explanation :

Given expression=((1/2)-(1/3))+((1/3)-(1/4))+((1/4)-(1/5))+
((1/5)-(1/6))+….+ ((1/9)-(1/10))
=((1/2)-(1/10))=4/10 = 2/5


Ans .

2ft. 7 inches

1. Explanation :

Length of board = 7ft. 9 inches=(7*12+9)inches=93 inches.
Length of each part = (93/3) inches = 31 inches = 2ft. 7 inches

Ans .

200

1. Explanation :

Let the share of each nephews be Rs.x.
Then,share of each daughter=rs4x;share of each son=Rs.5x;
So,5*5x+4*4x+2*x=8600
25x+16x+2x=8600
=43x=8600
x=200;


Ans .

 1000

1. Explanation :

Part of salary left=1-(2/5+3/10+1/8)
Let the monthly salary be Rs.x
Then, 7/40 of x=1400
X=(1400*40/7)
=8600
Expenditure on food=Rs.(3/10*800)=Rs.2400
Expenditure on conveyence=Rs.(1/8*8000)=Rs.1000


Ans .

180,60

1. Explanation :

Let Arun’s marks in mathematics and english be x and y
Then 1/3x-1/2y=30
2x-3y=180……>(1)
x+y=240…….>(2)
solving (1) and (2)
x=180
and y=60


Ans .

40

1. Explanation :

Suppose x bottles can fill the tin completely
Then4/5x-3/4x=6-4
X/20=2
X=40
Therefore required no of bottles =40

Ans .

8cm

1. Explanation :

Let the total length be xm
Then black part =x/8cm
The remaining part=(x-x/8)cm=7x/8cm
White part=(1/2 *7x/8)=7x/16 cm
Remaining part=(7x/8-7x/16)=7x/16cm
7x/16=7/2
x=8cm

Ans .

11/18

1. Explanation :

Let the total no of workers be x
No of women = x/3
No of men = x-(x/3)=2x/3
No of women having  children =1/3 of ½ ofx/3=x/18
No of men having children=2/3 of ¾ of2x/3=x/3
No of workers having children = x/8 +x/3=7x/18
Workers having no children=x-7x/18=11x/18=11/18 of all workers


Ans .

480

1. Explanation :

Let the total no of mangoes in the crate be x
Then the no of bruised mango = 1/30 x
Let the no of unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480

Ans .

288

1. Explanation :

Let no of passengers in the beginning be x
After first station no passengers=(x-x/3)+280=2x/3 +280
After second station no passengers =1/2(2x/3+280)+12
½(2x/3+280)+12=248
2x/3+280=2*236
2x/3=192
x=288

Ans .

5

1. Explanation :

$$a^2 + b^2 + 2ab = (a+b)^2; = 117+2*24 = 225 a+b=15; a^2 + b^2 - 2ab = (a-b)^2 = 117-2*54 ; a-b=3 ; a+b/a-b = 15/3 = 5$$

Ans .

121966

1. Explanation :

Given expression=(75983)2
-(45983)2/(75983-45983)
=(a-b)2/(a-b)
=(a+b)(a-b)/(a-b)
=(a+b)
=75983+45983
=121966


Ans .



1. Explanation :

given expression is $$\frac{a^3 - b^3}{a^2+ab+b^2} = (a-b) = (343-113) = 230$$

Ans .

13

1. Explanation :

Let the population of two villages be equal after  p years
Then,68000-1200p=42000+800p
2000p=26000
p=13

Ans .

40

1. Explanation :

Let at present there be x boys.
Then,no of girls at present=5x
Before the boys had left:no of boys=x+45
And no of girls=5x
X+45=2*5x
9x=45
x=5
no of girls in the beginning=25+15=40

Ans .

40

1. Explanation :

Suppose a worker remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40


Ans .

3500

1. Explanation :

Let the no of fifty rupee notes be x
Then,no of 100 rupee notes =(85-x)
50x+100(85-x)=5000
x+2(85-x)=100
x=70
so,,required amount=Rs.(50*70)= Rs.3500

Ans .

15

1. Explanation :

Let the number of keepers be x then,
Total number of heads =(50 + 45 + 8 + x)= (103 + x).
Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).
(312 + 2x)-(103 + x) =224; x =15.
Hence, number of keepers =15

Ans .

34

1. Explanation :

Suppose Arun has Rs. X and Sjal has Rs. Y. then,
2(x-30)= y+30  => 2x-y =90  …(i)
and  x +10 =3(y-10) => x-3y = - 40       …(ii)
Solving (i) and (ii), we get x =62 and y =34.
Arun has Rs. 62 and Sajal has Rs. 34

Ans .

12

1. Explanation :

Let the cost of a pen ands a pencil be Rs. X and Rs. Y respectively.
Then, 2x  + 3y = 86 ….(i) and 4x + y =112.
Solving (i) and (ii), we get: x = 25 and y = 12.
Cost of a pen =Rs. 25 and the cost of a pencil =Rs. 12

Ans .

5040

1. Explanation :

Let the total amount be Rs. X the, $$\frac{x}{14} - \frac{x}{18} = 80; \frac{2x}{126} = 80; x = 63 * 80 = 5040$$

Ans .

20

1. Explanation :

Suppose Mr. Bhaskar is on tour for x days. Then $$\frac{360}{x} - \frac{360}{x+4} = 3; \frac{1}{x} - \frac{1}{x-4} = \frac{1}{120} ; x(x+4) = 4 * 120 = 480 ; x(x+4) = 4 * 120 = 480$$. x2 + 4x – 480 = 0 ; (x+24) (x-20) = 0 ; x =20. Hence Mr. Bhaskar is on tour for 20 days.

## 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

Here ages of two people shall be given as a ratio. Then their ages before x years or after x years shall be given. This ratio would be equal to a value. The question would then be to calculate their present ages.

Methodology: Assume age of the lesser entity as 'x'. The older entity become 'kx'. Then their ages 'b' years ago would be 'x-b' and 'kx-b' respectively. This would be equal to some value say 10.

Obtaining the linear equation: (x-b) + (kx-b) = 10; When we solve this we get value of x.

Q1. One year ago Jaya was four times as old as her daughter.  Six years hence, Jaya's age will exceed her daughter's age by 9 years. The ratio of the present ages of Jaya and her daughter is :
Ans: daughter's age 1 yr ago= x , Jaya's age 1 year ago = 4x
6 years hence:  (4x+7) - (x+7) = 9 , we solve this to get x;

Q2. Five years ago, the total of the ages of a father and his son was 40 years. The ratio of their present ages is 4 : 1. What is the present age of the father ?
Ans.
Present age: son = x , father = 4x;
Five years ago : sons age = x-5 , fathers age = 4x-5;
linear equation: (x-5)+(4x-5)=40;

Q. One year ago, the ratio of Gaurav’s and Sachin’s age was 6: 7 respectively. Four years hence, this ratio would become 7: 8. How old is Sachin ?
A. let the age be 'x' and 'y' respectively. x : y = 6 : 7 so 7x - 6y = 0 and
(x+4) : (y+4) = 7 : 8 so 8x+32 = 7y+28 and we get 8x-7y=-4 solving both we get x = 24, y=28.

Q. Abhay’s age after six years will be three-seventh of his fathers age. Ten years ago the ratio of their ages was 1 : 5. What is Abhay’s father's age at present?
A. Abhay's age ten years ago = x, so his father is 5x. After 6 yrs from present Abhay shall be x+16 and his father shall be 5x+16 and we know that
(x+16) = 3/7 * (5x+16)

### Linear equations

Q. A man divides Rs. 8600 amongst 5 sons, 4 daughters and 2 nephews. If each daughter gets 4 times as much as nephews and each son gets 5 times as much as nephews. how much does each daughter get?
A. Let x be the share of each nephew so one daughter get 4x and son 5x.
5*(5x) + 4(4x) + 2x = 8600

Q. A man  spends 2/5 of his  salary on house rent, 3/10 of his  salary on  food and 1/8 of his salary  on  conveyance. if  he  has  Rs.1400  left  with  him, find  his  expenditure  on  food  and conveyance.
A. Salary = x ; x = 1400 + 0.4x + 0.3x+ o.125x

Q. A third of A marks in mathematics exceeds a half of his marks in English by 80. if he got 240 marks In two subjects together how many marks did he get in English?
A. x = marks in maths, y=marks in english. (x/3) - (y/2) = 80 and x + y = 240
Solving both equations gives the result.

Q. A  train  starts  full of passengers at  the  first  station  it drops 1/3 of  the passengers and takes 280 more at the second station it drops one half the new total and takes twelve more . On  arriving  at  the  third  station  it  is  found  to  have  248  passengers.  Find  the  no  of passengers in the beginning?
A. total capacity = x;
at first station : x- (x/3) + 280
at second station:  1/2{(2x/3)+280}+12
at third station: 248 = 1/2{(2x/3)+280}+12

Q. in a  office 1/3 of the workers are women ½ of the women are married and 1/3 of the married women have children if ¾ of the men are married and 2/3 of the married men have children what part of workers are without children?
A. Total workers be x. women = x/3, men = 2x/3, married women = x/3 * 1/2 = x/6;  women with children = x/6*1/3 = x/18; married men = 2x/3 * (3/4) = x/2; men with children = 2x/6 = x/3.
workers without children = x - (x/18+x/3)

Q. Village  X  has  a  population  of  68000,which  is  decreasing  at  the  rate  of  1200  per year. Village Y has a population  of 42000, which is increasing at  the  rate of 800 per year  .in how many years will  the population of  the  two villages be equal?
A. Assume 'x' as the number of years after which both populations shall be equal. So we get linear equation as: 68000 - 1200x = 42000 + 800x

Q. An employer pays Rs.20 for each day a worker works and for forfeits Rs.3 for each day is idle  at  the  end  of  sixty days  a worker  gets Rs.280  .  for how many days did  the worker remain ideal?
A. worker has 'x' working days and 'y' idle days so
20 * x - 3 * y = 280
x + y = 60

Q. When an amount was distributed among 14 boys, each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount?
A. Assume total amount be 'x'.
So when 'x' is given to 14 boys each gets 'x/14' and when divided amongst 18 boys each gets 'x/18' and we know that x/14 - x/18 = 80

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 2x – x – 1 = 0 equals

1. 0

2. 1

3. 2

4. 3

Ans . C

1. 2x – x – 1 = 0 and so 2x – 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 xx=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.

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