# SIMPLIFICATION

Simplification

Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly.

Rules of Simplification

V  Vinculum

B  Remove Brackets – in the order ( ) , { }, [ ]

O  Of
D  Division

M  Multiplication

S  Subtraction

Classification

 Types Description Natural Numbers: all counting numbers ( 1,2,3,4,5….∞) Whole Numbers: natural number + zero( 0,1,2,3,4,5…∞) Integers: All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞) Even & Odd Numbers : All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞) Prime Numbers: It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞) Composite Numbers: Natural numbers which are not prime Co-Prime: Two natural number a and b are said to be co-prime if their HCF is 1.

Divisibility

 Numbers IF A Number Examples Divisible by 2 End with 0,2,4,6,8 are divisible by 2 254,326,3546,4718 all are divisible by 2 Divisible by 3 Sum of its digits  is divisible by 3 375,4251,78123 all are divisible by 3.  [549=5+4+9][5+4+9=18]18 is divisible by 3  hence 549 is divisible by 3. Divisible by 4 Last two digit divisible by 4 5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4. Divisible by 5 Ends with 0 or 5 225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5. Divisible by 6 Divides by Both 2 & 3 4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6. Divisible by 8 Last 3 digit divide by 8 746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8. Divisible by 10 End with 0 220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10. Divisible by 11 [Sum of its digit in odd places-Sum of its digits in even places]= 0 or multiple of 11 Consider the number 39798847 (Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3) (23-12) 23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

hence ,represent it as:

dividend = ( divisorquotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

Rules

1. Modulus of a Real Number:

Modulus of a real number a is defined as

 |a| = a, if a > 0 –a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

1. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

Sum Rules

(1+2+3+………+n) = 1/n(n+1)

(12+22+32+………+n2) = 1/n (n+1) (2n+1)

(13+23+33+………+n3) = 1/4 n2 (n+1)2

Questions:

Level-I:

1. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
 A. 45 B. 60 C. 75 D. 90

2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
 A. 20 B. 80 C. 100 D. 200

3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
 A. Rs. 3500 B. Rs. 3750 C. Rs. 3840 D. Rs. 3900

4. If a – b = 3 and a2 + b2 = 29, find the value of ab.
 A. 10 B. 12 C. 15 D. 18

5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
 A. Rs. 1200 B. Rs. 2400 C. Rs. 4800 D. Cannot be determined E. None of these

6.

A sum of Rs. 1360 has been divided among A, B and C such that A gets  of what B gets and B gets  of what C gets. B’s share is:
 A. Rs. 120 B. Rs. 160 C. Rs. 240 D. Rs. 300

7. One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
 A. Rs. 30,000 B. Rs. 50,000 C. Rs. 60,000 D. Rs. 90,000

8. A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
 A. 30 birds B. 60 birds C. 72 birds D. 90 birds

9. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:
A.
 1 7
B.
 1 8
C.
 1 9
D.
 7 8

10. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?
 A. 10 B. 35 C. 62.5 D. Cannot be determined E. None of these

Level-II:

1. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
 A. 160 B. 175 C. 180 D. 195

12. Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
 A. 256 B. 432 C. 512 D. 640 E. None of these

13. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
 A. 22 B. 23 C. 24 D. 26

14.
 (469 + 174)2 – (469 – 174)2 = ? (469 x 174)
 A. 2 B. 4 C. 295 D. 643

15. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
 A. 19 B. 28 C. 30 D. 37

1. Find the value of 1/(3+1/(3+1/(3-1/3)))
 A.) 3/10 B.) 10/3 C.) 27/89 D.) 89/27

1. Find the value of
 A.) 3½ 99; B.) 34/99 C.) 2.131313 D.) 3.141414

18.Find the value of

((0.1)3 + (0.6)3 + (0.7)3 − (0.3)(0.6)(0.7))/((0.1)2 + (0.6)2 + (0.7)2 − 0.006 − 0.42 − 0.07)

 A.) 14/10 B.) 1.35 C.) 13/10 D.) 0

1. Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)
 A.) 0.56 B.) 0.65 C.) 0.54 D.) 0.45
1. Find the value of
 A.) 1.5 B.) -1.5 C.) 1 D.) 0

Level-I

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10      x – y = 20 …. (i)

and x + 20 = 2(y – 20)      x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

 Then, 10x = 4y   or   y = 5 x. 2

15x + 2y = 4000

 15x + 2 x 5 x = 4000 2

20x = 4000

x = 200.

 So, y = 5 x 200 = 500. 2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

Explanation:

2ab = (a2 + b2) – (a – b)2

= 29 – 9 = 20

ab = 10.

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

Divide equation (i) by 2, we get the below equation.

=> x +  2y =  800. — (iii)

Now subtract (iii) from (ii)

x +  6y = 1600  (-)

x +  2y =  800

—————-

4y =  800

—————-

Therefore, y = 200.

Now apply value of y in (iii)

=>  x + 2 x 200 = 800

=>  x + 400 = 800

Therefore x = 400

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

Explanation:

Let C’s share = Rs. x

 Then, B’s share = Rs. x ,   A’s share = Rs. 2 x x = Rs. x 4 3 4 6

 x + x + x = 1360 6 4

 17x = 1360 12

 x = 1360 x 12 = Rs. 960 17

 Hence, B’s share = Rs. 960 = Rs. 240.

Explanation:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

 1 x = 1 (150000 – x) 3 2

 x + x = 75000 3 2

 5x = 75000 6

 x = 75000 x 6 = 90000 5

Savings in Public Provident Fund = Rs. (150000 – 90000) = Rs. 60000

Explanation:

Let the total number of shots be x. Then,

 Shots fired by A = 5 x 8

 Shots fired by B = 3 x 8

 Killing shots by A = 1 of 5 x = 5 x 3 8 24

 Shots missed by B = 1 of 3 x = 3 x 2 8 16

 3x = 27 or x = 27 x 16 = 144. 16 3

 Birds killed by A = 5x = 5 x 144 = 30. 24 24

Explanation:

 Original share of 1 person = 1 8

 New share of 1 person = 1 7

 Increase = 1 – 1 = 1 7 8 56

 Required fraction = (1/56) = 1 x 8 = 1 (1/8) 56 1 7

Explanation:

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

 New capacity of bucket = 2 x 5

 Required number of buckets = 25x (2x/5)

 = 25x x 5 2x

 = 125 2

= 62.5

Level-II:

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

Explanation:

Let total number of children be x.

 Then, x x 1 x = x x 16     x = 64. 8 2

 Number of notebooks = 1 x2 = 1 x 64 x 64 = 512

Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140      x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

Explanation:

 Given exp. = (a + b)2 – (a – b)2 ab

 = 4ab ab

= 4 (where a = 469, b = 174.)

Explanation:

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x        120x = 40

 x = 1 3

 Number of floors covered by David in (1/3) min. = 1 x 57 = 19. 3

So, their paths cross at (11 +19) i.e., 30th floor.

Explanation:

1/[3 + (1/(3+1/(3 – 1/3)))]

=> 1/[3 + 1/(3 + 1/(8/3))]

=> 1/[3 + 1/(3 + 3/8)]

=> 1/[3 + 8/27]

=> 1/(89/27)

=> 27/89

Explanation:

6/9 + 7/9 + 9/9 + 69/99

2/3 + 7/9 + 1 + 69/99

(66 + 77 + 99 + 69)/99

311/99 => 3.141414

Explanation:

((0.1)3 + (0.6)3 + (0.7)3 − (0.3)(0.6)(0.7))/((0.1)2 + (0.6)2 + (0.7)2 − 0.006 − 0.42 − 0.07)

=> (0.1 + 0.6 + 0.7)3/(0.1 + 0.6 + 0.7)2

=> 0.1 + 0.6 + 0.7 => 1.4 = 14/10

11/30 − [1/6 + 1/5 + [7/12 − 7/12]]

11/30 − [1/6 + 1/5 + [0]]

11/30 − [(5 + 6)/30]

11/30 − 11/30 = 0.

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