’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
5005 - 5000 + 10 = 5005 - (5000/10) = 5005-500 = 4505.
Ans .
2
18800+470+20=(18800/470)+20=40/20=2.
Ans .
4a
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
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
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
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
\( 108 / 9 + \frac{2}{5} * \frac{13}{4} = \frac{108}{9} + \frac{13}{10} = 12 + \frac{13}{10} = \frac{133}{10} = 13\frac{3}{10}\)
Ans .
(21/10)*(15/14)
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
Given exp. = 12.05*(5.4/0.6) = (12.05*9) = 108.45
Ans .
0.46
Given exp. = 0.6*0.6+(0.6*6) = 0.36+0.1 = 0.46
Ans .
12
(17.28/x) = 2*3.6*0.2 x = (17.28/1.44) = (1728/14) = 12
Ans .
2
(364.824/x) = (3794.1696 + 36.4824) – 3648.24 = 3830.652 – 3648.24 = 182.412 x = (364.824/182.412) =2
Ans .
3.5
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
\( [x^2 + y^2]/[x^2 - y^2] = [\frac{x^2}{y^2} + 1]/[\frac{x^2}{y^2} - 1] = = [(36/25)+1] / [(36/25)-1] = (61*25)/(25*11) = 61/11 \)
Ans .
1/8
\( 4 - \frac{5}{1 + \frac{1}{3 + \frac{1}{2 + \frac{1}{4}}}} = 4 - \frac{5}{1 + \frac{1}{3 + \frac{1}{9}}} = 4 - \frac{5}{1 + \frac{1}{31/9}} = 4 - \frac{5}{1 + 9/31} = 4 - \frac{5}{(40/31)} = 4 – (5*31)/ 40 = 4 – (31/8) = 1/8 \)
Ans .
2/3
\( \frac{2x}{1 + \frac{1}{1 + \frac{x}{1 - x}}} = 1 ; \frac{2x}{1 + \frac{1}{1/(1- x)}} = 1; \frac{2x}{1+ (1 – x)} = 1 ; 2x = 2-x ; 3x = 2 ; x = (2/3). \)
Ans .
3/2
(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
(x /4)-((x-3)/6)=1; (3x-2(x-3) )/12 = 1 ; 3x-2x+6=12 ; x=6
Ans .
75
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
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
Given expression = (2/3)*(3/4)*(4/5) *…….* (99/100) = 2/100 = 1/50
Ans .
2/5
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
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
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
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
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
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
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
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
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
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
\( a^2 + b^2 + 2ab = (a+b)^2; = 117+2*24 = 225 a+b=15; a^2 + b^2 - 2ab = (a-b)^2 = 117-2*54 ; a-b=3 ; a+b/a-b = 15/3 = 5 \)
Ans .
121966
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 .
given expression is \( \frac{a^3 - b^3}{a^2+ab+b^2} = (a-b) = (343-113) = 230 \)
Ans .
13
Let the population of two villages be equal after p years Then,68000-1200p=42000+800p 2000p=26000 p=13
Ans .
40
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
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
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
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
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
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
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
Suppose Mr. Bhaskar is on tour for x days. Then \( \frac{360}{x} - \frac{360}{x+4} = 3; \frac{1}{x} - \frac{1}{x-4} = \frac{1}{120} ; x(x+4) = 4 * 120 = 480 ; x(x+4) = 4 * 120 = 480\). x^{2} + 4x – 480 = 0 ; (x+24) (x-20) = 0 ; x =20. Hence Mr. Bhaskar is on tour for 20 days.
The above formulae can be used for solving giant multiplication problems.
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.
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.
Q. if f(x) = log (1+x/1-x) then f(x) + f(y) is
f(x + y)
$f\frac{x+y}{1+xy}$
$(x+y) f \frac{1}{1+xy}$
$\frac{f(x) + f(y)}{1+xy}$
Ans . B
$f(x) + f(y) = log \frac{1+x}{1-x}$ + $log \frac{1+y}{1-y}$
$log \frac{(1+x)(1+y)}{(1-x)(1-y)}$
$log \frac{(1+x+y+xy)}{(1-x-y+xy)}$
$log \frac{(1+x+y+xy)}{(1+xy-(x+y))}$
$log \frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}$
$f \frac{x+y}{1+xy}$
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
24
54
34
45
Ans . B
check choices and only choice B satisfies the condition.
S = (5+4)^2 = 81
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?
$X_n$ is positive if n is even
$X_n$ is positive if n is odd
$X_n$ is negative if n is even
None of these
Ans . D
$x_0 = x, x_1 = -x, x_2 = -x$
$x_3 = x, x_4 = x$
$x_5 = -x, x_6 = -x$
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?
5/3
√ 19
13/3
none
Ans . C
$xy + yz + zx = 3$
$xy + (y + x)z = 3 $
$xy + (y + x)(5 + x + y)= 3$
$xy + 5y+xy+y^2+5x+x^2+xy = 3$
As it is given that y is a real number, the discriminant for above equation must be greater than or equal to zero. $(x - 5)^2 - 4(x^2 - 5x + 3) \geq 0$
$3x^2 - 10x + 13 \geq 0$
$x = 1, \frac{13}{3}$
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?
2.5
3.5
3.8
4
Ans . C
Area = 40 * 20 = 800 $m^2$
If 3 rounds are done, area = 34 × 14 = 476 $m^2$ ⇒ Area > 3 rounds
if 4 rounds ⇒ Area left = 32 × 12 = 347 $m^2$
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?
40
36
25
None of these
Ans . B
Since thief escaped with 1 diamond
Before 3 rd watchman he had (1 + 2) × 2 = 6 diamonds.
Before 2 nd watchman he had (6 + 2) × 2 = 16 diamonds
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?
Rs.15
Rs.13
Rs.17
None of these
Ans . B
Mayank paid 1/2 of the sum paid by other three.
Mayank paid 1/3 rd of the total amount = Rs.20
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?
96
53
43
None of these
Ans . D
Let the number of gold coins = x + y
48(x – y) = x^{2} - y^{2}
48(x – y) = (x – y)(x + y) ⇒ x + y = 48
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?
4 hr
5 hr
6 hr
None of these
Ans . C
By trial and error:
30 × 12 = 360 > 300
30 × 7.5 = 225 < 300
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
0
1
2
3
Ans . C
2^{x} – x – 1 = 0 and so 2^{x} – 1 = x
If we put x = 0, then this is satisfied and if we put x = 1, then also this is satisfied.
Now, if we put x = 2, the equation this is not valid.
Q. When the curves y = log_{10}x , y=x^{-1} are drawn in the x-y plane, how many times do they intersect for values x ≥ 1 ?
Never
Once
Twice
More than twice
Ans . B
For the curves to intersect, log x = x
Thus log_{10}x = 1/x or x^{x}=10
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
5p –2q – r = 0
5p + 2q + r = 0
5p + 2q – r = 0
5p – 2q + r = 0
Ans . A
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.
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
4.0
4.5
1.5
None of the above
Ans . D
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.
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
x = 2.3
x = 2.5
x = 2.7
None of the above
Ans . B
At x = 2, f(x) = 2.1
At x = 2.5, f(x) = 1.6
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?
245
285
240
320
Ans . C
If y = 2 (it cannot be 0 or 1), then x can take 1 value and z can take 2 values.
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.
Score more than 80% marks and move ahead else stay back and read again!