You are here

SSC CGL level Solution Set 54, Number System 8

How to Solve Number System Questions: SSC CGL Set 54

Solution to number system questions for SSC CGL

How to solve number system questions for SSC CGL explains basic to advanced techniques of solving number system questions easy and quick.

If you have not taken this test yet, take it first at,

54th SSC CGL level Question Set on Number System 8.

How to solve number system questions for SSC CGL Solution set 54 - time to solve was 12 mins

Problem 1.

Sum of squares of two positive integers is 100 and the difference of their squares is 28. The sum of the numbers is,

  1. 15
  2. 14
  3. 13
  4. 12

Solution 1 : Problem analysis and execution:

Here, the value of $a^2+b^2$ is given as well as the value of $a^2-b^2$. Summing up the two, $b^2$ is eliminated and we get,


Or, $a^2=64$,

Or, $a=8$.

As $a^2-b^2=28$,


Or, $b=6$.

So, the sum of the numbers is,


Answer: b: 14.

Key concepts used: Basic algebraic concepts -- solving two linear equations.

Problem 2.

A number when divided by 6 leaves a remainder of 3. When the square of the number is divided by 6 the remainder is,

  1. 0
  2. 1
  3. 3
  4. 2

Solution 2 : Problem analysis and execution

If $n$ be the number,

$n=6\times{q_1}+3$, where $q_1$ is the quotient.

When $n$ is squared we have,

$n^2=a^2+2ab+b^2$, where, $a=6\times{q_1}$ and $b=3$,

Or, $n^2=a^2+2ab+9$

Then if $n^2$ is divided by 6, the first two terms on the RHS are divisible by 6 with remainders 0 (as both contain 6 as a factor), but dividing the numeric term 9 by 6 leaves a remainder of 3.

So when the square of the number is divided by 6 a remainder of 3 is left.

Answer: Option c : 3.

Key concepts used: Basic remainder concepts -- Euclid's division relation -- Basic algebraic concept.

Problem 3.

The average of 5 consecutive positive even numbers is 86. What is the product of the third and the fourth numbers?

  1. 7704
  2. 4077
  3. 4707
  4. 7568

Solution 3 : Problem analysis and execution:

If the first number is $n$, the second number is, $n+2$, the third, $n+4$, the fourth $n+6$ and the fifth $n+8$.

The average of the three numbers is equal to the middle third number $n+4$ which equals 86.

So the fourth number is,

$n+6=88$, and their product,


Without calculating the product we will test the choice values using free resource use principle, for factors, 2 and then 11 (out of 88).

Out of the two even numbers 7704 and 7568, only the number 7568 satisfies the divisibility test of 11 (difference of sums of alternate digits 0). So 7568 must be the desired product.

Second method of testing is to just multiply the unit's digits of the two numbers 86 and 88 which is 8 canceling out 7704.

This is use of Unit's digit product concept as well as application of many ways technique.

Without calculating the time consuming product we have instead used the choice values as free resources.

Answer: Option d: 7568.

Key concepts used: Even number series concept -- rich average concept -- free resource use principle -- unit's digit behavior analysis -- many ways technique.

Note: Average of odd number of uniformly increasing integers is always the middle number. This is a rich average concept.

Problem 4.

A four digit number is formed by repeating a 2 digit number like 1212, 3434 etc. Any number of this form is always divisible by,

  1. 13
  2. 11
  3. 7
  4. Smallest 3 digit prime number

Solution 4 : Problem analysis and execution:

Taking recourse to place value mechanism and number formation concept we can express such a number $p$ with given properties of the form $mnmn$ as,

$p=10^3m+10^2n+10m+n$, where the non-zero digits $m$ and $n$ are repeated,



So whatever be the non-zero values of $m$ and $n$, the number as described will always be divisible by 101, which is the smallest three digit prime number.

Answer: d: Smallest 3 digit prime number.

Key concepts used: Basic number system concepts -- place value mechanism -- number formation concepts -- number breakdown technique -- basic algebra concepts -- basic prime number concept.

Problem 5.

The product of two consecutive odd numbers is 28899. The square root of the smaller number is,

  1. 13
  2. 7
  3. 11
  4. 3

Solution 5 : Problem analysis and execution:

A quick divisiblity test with 13 results in 13 as a factor of 28899 and the quotient, 2223, which again turns out to be divisible by 13 with quotient 171, and 171 has two factors 9 and 19.

So the prime factors of 28899 are,

3, 3, 13, 13, and 19.

By combining the five factors into two consecutive odd numbers we can have only $13\times{13}=169$ and $9\times{19}=171$ as the two consecutive odd factors of 28899.

Square root of the smaller number is thus 13.

Alternate logic: There are only two squares in the factor set, $3^3=9$ and $13^2=169$. As the former is too small it cannot independently be one of the two consecutive odd factors of 28899 (it has to combine with the fifth prime factor 19). This is an instance of application of Many ways technique.

Answer: Option a: 13.

Key concepts used: Divisibility rules -- Basic factors and multiple concept -- free resource use -- rich number system concepts -- factor combining technique -- Basic number domain concepts -- Many ways technique

Problem 6.

The largest number that exactly divides any number of the form $p^3-p$, where $p$ is a natural number, is,

  1. 6
  2. 12
  3. 2
  4. 3

Solution 6 : Problem analysis and execution:


By factors in consecutive natural numbers concepts, three consecutive natural numbers will always have 2 and 3 as factors.

Reason: If $p$ is even in the form $2m$, the two numbers adjacent to $2m$ will be $2m-1$ and $2m+1$. If $2m-1$ is not divisible by 3, it must be 1 or 2 short of being a multiple of 3. So either $2m$ or $2m+1$ must be a multiple of 3, even if $2m-1$ is not a multiple of 3.

So the given expression will always have two factors 2 and 3, and so 6 as a factor.

6 being larger than the choice values of 2 and 3, it is the right choice value.

Answer: Option a : 6.

Key concepts used: Factors in consecutive natural numbers concepts -- Basic algebraic concepts -- Odd even number concept.

Problem 7.

How many numbers less than 1000 are multiples of both 13 and 10?

  1. 7
  2. 6
  3. 8
  4. 9

Solution 7 : Problem analysis and execution:

The first and smallest such number with both 13 and 10 as factors is 130 and seven times 130 is 910 which is the largest number smaller than 1000.

So in total there are 7 such numbers smaller than 1000 that are divisible by both 13 and 10.

Answer: Option a: 7.

Key concepts used: Basic factor concept -- number system concept.

Problem 8.

A number when divided by 5 leaves a remainder of 3. What will be the remainder when the square of the same number is divided by 5?

  1. 1
  2. 4
  3. 2
  4. 3

Solution 8 : Problem analysis and execution:

If $n$ is the number, and $q_1$ the quotient of first division,


When $n$ is squared it generates two terms each involving 5 as a factor leaving only the third term 9 to be considered for finding the desired remainder. So when the square of the number is divided by 5, the remainder is the remainder generated by dividing 9 by 5, which is 4.

Answer: Option b: 4.

Key concepts used: Euclid's division relation -- remainder concept -- basic algebra concepts -- divisibility concepts -- factor concepts.

Problem 9.

The largest number which when divides 398, 436 and 542 leaves remainders as 7, 11 and 15 respectively, is,

  1. 7
  2. 13
  3. 17
  4. 23

Solution 9 : Problem analysis and execution:

We resort to testing the choice values by dividing the three number with the choice values one by one.

The first choice 7 is invalid as the first division remainder is 7, violating basic remainder concepts.

13 when divides 398 generates remainder 8, and violates even the first remainder condition.

17 generates remainder 7 when dividing 398, remainder of 11 when dividing 436, and remainder 15 when dividing 542. So this is the valid choice. We don't test 23.

Note: If we get one choice value that satisfies all the conditions of the problem, we don't consider any other choice value for evaluation. The logic behind this is, there can be only one valid choice value in an MCQ choice set.

Answer: Option c: 17.

Key concepts used: Basic remainder concepts -- free resource use principle -- enumeration technique.

Problem 10.

In a series of odd numbers, the first number is 1, the second 3, the third 5 and so on. What will be the 200th odd number in the series?

  1. 357
  2. 421
  3. 599
  4. 399

Solution 10 : Problem analysis and execution:

The $n$th term of the odd number series can be represented by the expression,


where $n$ starts from 1 and increases monotonously by 1.

We conclude this by testing the expression on the first few terms with $n=1$, $n=2$ and $n=3$.

The 200th term is then,


Answer: Option d: 399.

Key concepts used: General expression of odd number -- Expression fitting with given series -- number series concept.

Note: In almost all the problems a clear understanding of the very basic concepts was called for.

Guided help on Number system, HCF LCM in Suresolv

To get the best results out of the extensive range of articles of tutorials, questions and solutions on Number system and HCF LCM in Suresolv, follow the guide,

Suresolv Number system, HCF LCM Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II, Bank PO and Other Competitive exams.

The guide list of articles is up-to-date.