Quantitative Aptitude – Number Sequences, Series, Averages, Number Systems

A sequence is a list of numbers written in a special order such as (1, 2, 3, 4…), which typically follows a pattern. Sequences are usually set in brackets ( ) to notate the sequence, and each element (also known as a “member” or “term”) of the sequence is separated by a comma, like this:

(8, 4, 2, 1)

In the Arithmetic Progression there are two cases when the number of terms is odd and second one is when number of terms is even.

  • So when the number of terms is odd the average will be the middle term.
  • when the number of terms is even then the average will be the average of two middle terms.

A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, thesequence 2, 6, 18, 54, … is a geometric progression with common ratio 3.

Average

 

What is Average?

The result obtained by adding several quantities together and then dividing this total by the number of quantities is called Average.

The main term of average is equal distribution of a value among all which may distribute persons or things. We obtain the average of a number using formula that is sum of observations divided by Number of observations.

Here is average based some fact and formula and some average shortcut tricks examples. The problem is given in Quantitative Aptitude which is a very essential paper in ssc exam. Given below are some more example for practicing.

Formula:

  • Average: = (Sum of observations / Number of observations).

 

Find the Average Speed

  • If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr then the average speed during the whole journey is given by-
  • If a person covers A km at x km/hr and B km at y km/hr and C km at z km/hr, then the average speed in covering the whole distance is-

When a person leaves the group and another person joins the group in place of that person then-

  • If the average age is increased,
    Age of new person = Age of separated person + (Increase in average × total number of persons)
  • If the average age is decreased,
    Age of new person
     = Age of separated person – (Decrease in average × total number of persons)

 

        When a person joins the group-In case of increase in average

  • Age of new member = Previous average + (Increase in average × Number of members including new member)

       

        In case of decrease in average

  • Age of new member = Previous average – (Decrease in average × Number of members including new member)

 

In the Arithmetic Progression there are two cases when the number of terms is odd and second one is when number of terms is even.

  • So when the number of terms is odd the average will be the middle term.
  • when the number of terms is even then the average will be the average of two middle terms.

 

Some Important Examples

 

Examples 1: what will be the average of 13, 14, 15, 16, 17?

Solution: Average is the middle term when the number of terms is odd, but before that let’s checks whether it is in A.P or not, since the common difference is same so the series is in A.P. So the middle term is 15 which is our average of the series.

 

Example 2: What will be the average of 13, 14, 15, 16, 17, 18?

Solution: We have discussed that when the number of terms are even then the average will be the average of two middle terms.

Now the two middle terms are 15 and 16, but before that the average we must check that the series should be A.P. Since the common difference is same for each of the term we can say that the series is in A.P. and the average is (16+15)/2 = 15.5

Example 3:The average of five numbers is 29. If one number is excluded the average becomes 27. What is the excluded number ?

Answer :

let the excluded number is

= (29 x 5) – ( 27 x 4 )

= 145 – 108

= 37 .

Example 4: Find the average of first 20 natural numbers?

Answer:

Sum of first n natural numbers = n ( n + 1 ) /2

So, we can find easily average of first 20 natural numbers 20 x 21 / 2 = 210

So, then Required average is = 210 / 20 = 10.5.

 

 

Example 5

Find the average of first 20 multiplies of 5 .

Answer:

Required average = 5 ( 1 + 2 + 3 +……………….. + 20) /20

= ( 5 x 20 x 21 / 20 x 2) = 2100 / 40 = 52.5 .

So the Required average is 52.5.

 

Questions:

Level-I:

1.In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A.6.25
B.6.5
C.6.75
D.7

 

2.A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
A.
284years
7
B.
315years
7
C.
321years
7
D.None of these

 

3.A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A.Rs. 4991
B.Rs. 5991
C.Rs. 6001
D.Rs. 6991

 

4.The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
A.0
B.1
C.10
D.19

 

5.The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
A.76 kg
B.76.5 kg
C.85 kg
D.Data inadequate
E.None of these
 

 

6.

 

 

The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?

A.23 years
B.24 years
C.25 years
D.None of these

 

7.The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A.3500
B.4000
C.4050
D.5000

 

8.The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
A.35 years
B.40 years
C.50 years
D.None of these

 

9.A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
A.Rs. 7.98
B.Rs. 8
C.Rs. 8.50
D.Rs. 9

 

10.In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother’s view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
A.67 kg.
B.68 kg.
C.69 kg.
D.Data inadequate
E.None of these
 

 

 

 

 

11.

 

 

Level-II:

 

 

The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

A.17 kg
B.20 kg
C.26 kg
D.31 kg

 

12.The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
A.47.55 kg
B.48 kg
C.48.55 kg
D.49.25 kg

 

13.A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
A.250
B.276
C.280
D.285

 

14.If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
A.53.33
B.54.68
C.55
D.None of these

 

15.A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:
A.10
B.20
C.40
D.73
 

 

 

16.

 

 

The average age of P, Q, R and S is 30 years. How old is R?

I.The sum of ages of P and R is 60 years.
 II.S is 10 years younger than R.
A.I alone sufficient while II alone not sufficient to answer
B.II alone sufficient while I alone not sufficient to answer
C.Either I or II alone sufficient to answer
D.Both I and II are not sufficient to answer
E.Both I and II are necessary to answer

 

17.
How many candidates were interviewed everyday by the panel A out of the three panels A, B and C?
I.The three panels on average interview 15 candidates every day.
 II.Out of a total of 45 candidates interviewed everyday by the three panels, the number of candidates interviewed by panel A is more by 2 than the candidates interviewed by panel c and is more by 1 than the candidates interviewed by panel B.
A.I alone sufficient while II alone not sufficient to answer
B.II alone sufficient while I alone not sufficient to answer
C.Either I or II alone sufficient to answer
D.Both I and II are not sufficient to answer
E.Both I and II are necessary to answer

 

 

18.

 

What is the average age of children in the class?
I.The age of the teacher is as many years as the number of children.
 II.Average age is increased by 1 year if the teacher’s age is also included.
A.I alone sufficient while II alone not sufficient to answer
B.II alone sufficient while I alone not sufficient to answer
C.Either I or II alone sufficient to answer
D.Both I and II are not sufficient to answer
E.Both I and II are necessary to answer

 

 

Answers:

Level-I:

Answer:1 Option A

 

Explanation:

Required run rate =282 – (3.2 x 10)=250   = 6.25
4040

 

 

Answer:2 Option B

 

Explanation:

Required average
=67 x 2 + 35 x 2 + 6 x 3
2 + 2 + 3
=134 + 70 + 18
7
=222
7
= 315years.
7

Answer:3 Option A

 

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs. [ (6500 x 6) – 34009 ]

= Rs. (39000 – 34009)

= Rs. 4991.

 

 

Answer:4 Option D

 

Explanation:

Average of 20 numbers = 0.

Sum of 20 numbers (0 x 20) = 0.

It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).

 

 

Answer:5 Option C

 

Explanation:

Total weight increased = (8 x 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

 

Answer:6 Option A

 

Explanation:

Let the average age of the whole team by x years.

11x – (26 + 29) = 9(x -1)

11x – 9x = 46

2x = 46

x = 23.

So, average age of the team is 23 years.

 

Answer:7 Option B

 

Explanation:

Let P, Q and R represent their respective monthly incomes. Then, we have:

P + Q = (5050 x 2) = 10100 …. (i)

Q + R = (6250 x 2) = 12500 …. (ii)

P + R = (5200 x 2) = 10400 …. (iii)

Adding (i), (ii) and (iii), we get:  2(P + Q + R) = 33000  or   P + Q + R = 16500 …. (iv)

Subtracting (ii) from (iv), we get P = 4000.

  • P’s monthly income = Rs. 4000.

 

 

 

Answer:8 Option B

 

Explanation:

Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.

Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.

Husband’s present age = (90 – 50) years = 40 years.

 

Answer:9 Option A

 

Explanation:

Total quantity of petrol
consumed in 3 years
=4000+4000+4000 litres
7.5088.50
= 40002+1+2 litres
15817
=76700 litres
51

Total amount spent = Rs. (3 x 4000) = Rs. 12000.

 Average cost = Rs.12000 x 51= Rs.6120   = Rs. 7.98
76700767

 

Answer:10 Option A

 

Explanation:

Let Arun’s weight by X kg.

According to Arun, 65 < X < 72

According to Arun’s brother, 60 < X < 70.

According to Arun’s mother, X <= 68

The values satisfying all the above conditions are 66, 67 and 68.

 Required average =66 + 67 + 68=201= 67 kg.
33

 

 

Level-II:

 

Answer:11 Option D

Explanation:

Let A, B, C represent their respective weights. Then, we have:

A + B + C = (45 x 3) = 135 …. (i)

A + B = (40 x 2) = 80 …. (ii)

B + C = (43 x 2) = 86 ….(iii)

Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv)

Subtracting (i) from (iv), we get : B = 31.

B’s weight = 31 kg.

 

Answer:12 Option C

 

Explanation:

Required average
=50.25 x 16 + 45.15 x 8
16 + 8
=804 + 361.20
24
=1165.20
24
= 48.55

 

Answer:13 Option D

 

Explanation:

Since the month begins with a Sunday, to there will be five Sundays in the month.

Required average
=510 x 5 + 240 x 25
30
=8550
30
= 285

 

Answer:14 Option B

 

Explanation:

Required average
=55 x 50 + 60 x 55 + 45 x 60
55 + 60 + 45
=2750 + 3300 + 2700
160
=8750
160
= 54.68

Answer:15 Option C

 

Explanation:

Let there be x pupils in the class.

Total increase in marks =x x1=x
22

 

x= (83 – 63)x= 20      x= 40.
22

 

 

Answer:16 Option D

 

Explanation:

P + Q + R + S = (30 x 4)      P + Q + R + S = 120 …. (i)

  1. P + R = 60 …. (ii)
  2. S = (R – 10) …. (iii)

From (i), (ii) and (iii), we cannot find R.

Correct answer is (D)

 

Answer:17 Option B

 

Explanation:

  1. Total candidates interviewed by 3 panels = (15 x 3) = 45.
  2. Let xcandidates be interviewed by C.

Number of candidates interviewed by A = (x + 2).

Number of candidates interviewed by B = (x + 1).

x + (x + 2) + (x + 1) = 45

3x = 42

x = 14

Hence, the correct answer is (B).

 

 

 

Answer:18 Option D

 

Explanation:

Let there be x children.

I gives, age of teacher = x years.

II gives, average age of (x + 1) persons = (x + 1) years.

Teacher’s age = (x + 1) (x + 1) – x2 = (x2 + 1 + 2x) – x2 = (1 + 2x)

Thus, teacher’s age cannot be obtained.

Correct answer is (D)

 

Data analysis

Data analysis is a primary component of data mining and Business Intelligence (BI) and is key to gaining the insight that drives business decisions. Organizations and enterprises analyze data from a multitude of sources using Big Data management solutions and customer experience management solutions that utilize data analysis to transform data into actionable insights.

Basic numeracy

Number system

The Natural Numbers

The natural (or counting) numbers are 1,2,3,4,5, etc. There are infinitely many natural numbers. The set of natural numbers, {1,2,3,4,5,…}, is sometimes written N for short.

The whole numbers are the natural numbers together with 0.

(Note: a few textbooks disagree and say the natural numbers include 0.)

The sum of any two natural numbers is also a natural number (for example, 4+2000=2004), and the product of any two natural numbers is a natural number (4×2000=8000). This is not true for subtraction and division, though.

The integers are the set of real numbers consisting of the natural numbers, their additive inverses and zero.

{…,−5,−4,−3,−2,−1,0,1,2,3,4,5,…}{…,−5,−4,−3,−2,−1,0,1,2,3,4,5,…}

The set of integers is sometimes written JJ or ZZ for short.

The sum, product, and difference of any two integers is also an integer. But this is not true for division… just try 1÷21÷2.

 

Rational Numbers

The Rational Numbers The rational numbers are those numbers which can be expressed as a ratio between two integers. For example, the fractions 1 / 3 and

−1111 / 8 are both rational numbers. All the integers are included in the rational numbers, since any integer z can be written as the ratio z /1.

 

All decimals which terminate are rational numbers (since 8.27 can be written as 827 /100.) Decimals which have a repeating pattern after some point are also rationals:

for example

 

0.0833333… = 1 /12

The set of rational numbers is closed under all four basic operations, that is, given any two rational numbers, their sum, difference, product, and quotient is also a rational number (as long as we don’t divide by 0).

 

Irrational Numbers

The Irrational Numbers An irrational number is a number that cannot be written as a ratio (or fraction).  In decimal form, it never ends or repeats. The ancient Greeks discovered that not all numbers are rational; there are equations that cannot be solved using ratios of integers.

 

The Real Numbers

The real numbers is the set of numbers containing all of the rational numbers and all of the irrational numbers.  The real numbers are “all the numbers” on the number line.  There are infinitely many real numbers just as there are infinitely many numbers in each of the other sets of numbers.  But, it can be proved that the infinity of the real numbers is a bigger infinity.

The Complex Numbers

The complex numbers are the set {a+bia+bi | aa and bb are real numbers}, where ii is the imaginary unit, −1−−−√−1.

 

 

 

 

Numbers and their relations

 

Fractions

 

 

 

 1/4                                 1/2                       3/8

 

The top number says how many slices we have.

 

 

Equivalent Fractions Some fractions may look different, but are really the same, for example:

 

 

4/8  =                           2/4    =                        1/2

 

Numerator / Denominator

We call the top number the Numerator, it is the number of parts we have.
We call the bottom number the Denominator, it is the number of parts the whole is divided into.

Numerator

____________

Denominator

Adding fractions

 

                      

 

       1/4              +             1/4                        =          1/2

 

 

 

Subtracting Fractions

Step 1. Make sure the bottom numbers (the denominators) are the same

Step 2. Subtract the top numbers (the numerators). Put the answer over the same denominator.

Step 3. Simplify the fraction

 

 

Example      : ¾  −  ¼ = ?

The bottom numbers are already the same

Subtract the top numbers and put the answer over the same denominator.

3/4 – ¼  = 3-1 /4 = 2/4

Simplify the fraction.

2/4= ½

Number system

  1. Basic Formulae
  2. (a + b)(a – b) = (a2 – b2)
  3. (ab)2 = (a2 + b2 + 2ab)
  • (a– b)2 = (a2 + b2 – 2ab)
  1. (ab + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
  2. (a3b3) = (a + b)(a2 – ab + b2)
  3. (a3– b3) = (a – b)(a2 + ab + b2)
  • (a3b3 + c3 – 3abc) = (a + b + c)(a2 + b2 + c2 – ab – bc – ac)
  • When ab + c = 0, then a3 + b3 + c3 = 3abc

 

  1. Types of Numbers
  2. Natural Numbers

Counting numbers 1,2,3,4,5,…1,2,3,4,5,… are called natural numbers

 

  1. Whole Numbers

All counting numbers together with zero form the set of whole numbers.

Thus,

(i) 0 is the only whole number which is not a natural number.

(ii) Every natural number is a whole number.

III. Integers

All natural numbers, 0 and negatives of counting numbers i.e., …,−3,−2,−1,0,1,2,3,……..,−3,−2,−1,0,1,2,3,….. together form the set of integers.

(i) Positive Integers: 1,2,3,4,…..1,2,3,4,….. is the set of all positive integers.

(ii) Negative Integers: −1,−2,−3,…..−1,−2,−3,….. is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers: 0 is neither positive nor negative.

So, 0,1,2,3,….0,1,2,3,…. represents the set of non-negative integers,

while 0,−1,−2,−3,…..0,−1,−2,−3,….. represents the set of non-positive integers.

 

  1. Even Numbers

A number divisible by 2 is called an even number, e.g.,2,4,6,82,4,6,8, etc.

 

  1. Odd Numbers

A number not divisible by 2 is called an odd number. e.g.,1,3,5,7,9,11,1,3,5,7,9,11, etc.

 

  1. Prime Numbers

A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.

  • Prime numbers up to 100 are :2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.:2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.
  • Prime numbers Greater than 100: Let pp be a given number greater than 100. To find out whether it is prime or not, we use the following method:

 

Find a whole number nearly greater than the square root of pp. Let k>∗jpk>∗jp. Test whether pp is divisible by any prime number less than kk. If yes, then pp is not prime. Otherwise, pp is prime.

Example: We have to find whether 191 is a prime number or not. Now, 14>V19114>V191.

Prime numbers less than 14 are 2,3,5,7,11,13.2,3,5,7,11,13.

191 is not divisible by any of them. So, 191 is a prime number.

 

VII. Composite Numbers

Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4,6,8,9,10,12.4,6,8,9,10,12.

 

Note:

(i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

  1. 3. Remainder and Quotient

“The remainder is rr when pp is divided by k” means p=kq+rp=kq+r the integer qq is called the quotient.

For instance, “The remainder is 1 when 7 is divided by 3” means 7=3∗2+17=3∗2+1. Dividing both sides of p=kq+rp=kq+r by k gives the following alternative form pk=q+rkpk=q+rk

Example:

The remainder is 57 when a number is divided by 10,000. What is the remainder when the same number is divided by 1,000?

(A) 5 (B) 7 (C) 43 (D) 57 (E) 570

Solution:

Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as 10,000n+5710,000n+57, where nn is an integer.

Rewriting 10,000 as 1,000∗101,000∗10 yields 10,000n+57=1,000(10n)+5710,000n+57=1,000(10n)+57

Now, since nn is an integer, 10n10n is an integer. Letting 10n=q10n=q , we get

10,000n+57=1,000∗q+5710,000n+57=1,000∗q+57

Hence, the remainder is still 57 (by the p=kq+rp=kq+r form) when the number is divided by 1,000. The answer is (D).

Method II (Alternative form):

Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as 10,000n+5710,000n+57. Dividing this number by 1,000 yields

10,000n+57100010,000n+571000 =10,000n1000+571000=10,000n1000+571000 =10n+571000=10n+571000

Hence, the remainder is 57 (by the alternative form pk=q+rkpk=q+rk ), and the answer is (D).

 

  1. Even, Odd Numbers

A number n is even if the remainder is zero when nn is divided by 2:n=2z+02:n=2z+0, or n=2zn=2z.

A number nn is odd if the remainder is one when nn is divided by 2:n=2z+12:n=2z+1.

The following properties for odd and even numbers are very useful – you should memorize them:

even * evenodd * oddeven * oddeven + evenodd + oddeven + odd=even=odd=even=even=even=oddeven * even=evenodd * odd=oddeven * odd=eveneven + even=evenodd + odd=eveneven + odd=odd

 

Example:

If nn is a positive integer and (n+1)(n+3)(n+1)(n+3) is odd, then (n+2)(n+4)(n+2)(n+4) must be a multiple of which one of the following?

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

Solution:

(n+1)(n+3)(n+1)(n+3) is odd only when both (n+1)(n+1) and (n+3)(n+3) are odd. This is possible only when nn is even.

Hence, n=2mn=2m, where mm is a positive integer. Then,

(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)(n+2)(n+4)=(2m+2)(2m+4)=2(m+1)2(m+2)=4(m+1)(m+2)

=4 * (product of two consecutive positive integers, one which must be even)=4 * (product of two consecutive positive integers, one which must be even) =4 * (an even number), and this equals a number that is at least a multiple of 8=4 * (an even number), and this equals a number that is at least a multiple of 8

Hence, the answer is (D).

Questions

Level-I

1.If one-third of one-fourth of a number is 15, then three-tenth of that number is:
A.35
B.36
C.45
D.54

 

2.Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:
A.9
B.11
C.13
D.15

 

3.The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?
A.3
B.4
C.9
D.Cannot be determined
E.None of these

 

4.The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
A.4
B.8
C.16
D.None of these

 

5.A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:
A.18
B.24
C.42
D.81

 

 6.The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?
A.69
B.78
C.96
D.Cannot be determined
E.None of these

 

7.The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
A.20
B.30
C.40
D.None of these

 

8.A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:
A.3
B.5
C.9
D.11

 

9.In a two-digit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:
A.24
B.26
C.42
D.46

 

10.Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
A.3
B.10
C.17
D.20

 

Level-II

 

11.The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:
A.380
B.395
C.400
D.425

 

12.The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:
A.20
B.23
C.169
D.None of these

 

13.A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:
A.145
B.253
C.370
D.352

 

14.The sum of two number is 25 and their difference is 13. Find their product.
A.104
B.114
C.315
D.325

 

15.What is the sum of two consecutive even numbers, the difference of whose squares is 84?
A.34
B.38
C.42
D.46

 

 

  1. 16. If both 112and 33are factors of the number a * 43 * 62 * 1311, what is the smallest possible value of ‘a’?
  2. 121
  3. 3267
  4. 363
  5. 33
  6. 37

 

  1. 17. Find the largest five digit number that is divisible by 7, 10, 15, 21 and 28.
  2. 99840
  3. 99900
  4. 99990
  5. 99960
  6. 99970

 

  1. 18. Anita had to multiply two positive integers. Instead of taking 35 as one of the multipliers, she incorrectly took 53. As a result, the product went up by 540. What is the new product?
  2. 1050
  3. 540
  4. 1440
  5. 1520
  6. 1590

 

 

 

Answers:

 

Level-I

 

 

Answer:1 Option D

 

Explanation:

Let the number be x.

Then,1of1of x = 15      x = 15 x 12 = 180.
34

 

So, required number =3x 180= 54.
10

 

Answer:2 Option D

 

Explanation:

Let the three integers be xx + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

Third integer = x + 4 = 15.

 

Answer:3 Option B

 

Explanation:

Let the ten’s digit be x and unit’s digit be y.

Then, (10x + y) – (10y + x) = 36

9(x – y) = 36

x – y = 4.

 

Answer:4 Option B

 

Explanation:

Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.

Let ten’s and unit’s digits be 2x and x respectively.

Then, (10 x 2x + x) – (10x + 2x) = 36

9x = 36

x = 4.

Required difference = (2x + x) – (2x – x) = 2x = 8.

 

 

Answer:5 Option B

 

Explanation:

Let the ten’s and unit digit be x and8respectively.
x

 

Then,10x +8+ 18 = 10 x8x
xx

10x2 + 8 + 18x = 80 + x2

9x2 + 18x – 72 = 0

x2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2.

 

Answer:6 Option D

 

Explanation:

Let the ten’s digit be x and unit’s digit be y.

Then, x + y = 15 and x – y = 3   or   y – x = 3.

Solving x + y = 15   and   x – y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y – x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

 

Answer:7 Option A

 

Explanation:

Let the numbers be ab and c.

Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

(a + b + c) = 400 = 20.

 

Answer:8 Option D

 

Explanation:

Let the ten’s digit be x and unit’s digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

 

Answer:9 Option A

 

Explanation:

Let the ten’s digit be x.

Then, unit’s digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

(11x + 2)(2x + 2) = 144

22x2 + 26x – 140 = 0

11x2 + 13x – 70 = 0

(x – 2)(11x + 35) = 0

x = 2.

Hence, required number = 11x + 2 = 24.

 

Answer:10 Option A

 

Explanation:

Let the number be x.

Then, x + 17 =60
x

x2 + 17x – 60 = 0

(x + 20)(x – 3) = 0

x = 3.

 

Answer:11 Option C

 

Explanation:

Let the numbers be x and y.

Then, xy = 9375 andx= 15.
y

 

xy=9375
(x/y)15

y2 = 625.

y = 25.

x = 15y = (15 x 25) = 375.

Sum of the numbers = x + y = 375 + 25 = 400.

 

Answer:12 Option B

 

Explanation:

Let the numbers be x and y.

Then, xy = 120 and x2 + y2 = 289.

(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529

x + y = 529 = 23.

 

Answer:13 Option B

 

Explanation:

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred’s digits is smaller than the unit’s digit.

Hence, required number = 253.

 

Answer:14 Option B

 

Explanation:

Let the numbers be x and y.

Then, x + y = 25 and x – y = 13.

4xy = (x + y)2 – (x– y)2

= (25)2 – (13)2

= (625 – 169)

= 456

xy = 114.

 

Answer:15 Option C

 

Explanation:

Let the numbers be x and x + 2.

Then, (x + 2)2 – x2 = 84

4x + 4 = 84

4x = 80

x = 20.

The required sum = x + (x + 2) = 2x + 2 = 42.

 

Answer:16  Option C

Explanation:

112 is a factor of the given number.
In the given expression, a * 43 * 62 * 1311 none of the other factors, viz., 4, 6 or 13 is either a power or multiple of 11.
Hence, “a” should include 112
The question states that 33 is a factor of the given number. 62 is a part of the number.
62 can be expressed as 32 * 22.
Therefore, if 33 has to be a factor of the given number a * 43 * 62 * 1311, then we will need another 3 as part of the number.
Therefore, “a” should be at least 112 * 3 = 363 if the given number has to have 112 and 33 as its factors.
Correct answer choice (C)

 

Answer:17  Option C

Explanation:

The number should be divisible by 10 (2, 5), 15 (3, 5), 21 (3, 7), and 28 (4, 7).

Hence, it is enough to check whether the number is divisibile by 3, 4, 5 and 7.

Test of divisibility by 3: Sum of the digits will be divisible by 3 if a number is divisible by 3.

Test of divisibility by 4: The righmost two digits viz., the units and tens digits will be divisible by 4 if a number is divisible by 4. For e.g, 1232 is divisible by 4 because 32 is divisible by 4.

Test of divisibility by 5: The units digit is either 5 or 0.

99960 is the only number which is divisible by 3, 4, 5 and 7.

Correct answer choice (C)

 

Answer:18  Option E

Explanation:

Let the number that Anita wanted to multiply be ‘X’.
She was expected to find the value of 35X.
Instead, she found the value of 53X.

The difference between the value that she got (53X) and what she was expected to get (35X) is 540.

i.e., 53X – 35X = 540
or (53 – 35) * X = 540
X = 30

Therefore, the correct product = 53 * 30 = 1590

Correct answer choice (E)

 

 

 

 

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