SSC CGL level Solution Set 28, Number system 5

28th SSC CGL level Solution Set, 5th on topic Number System

SSC CGL Solution set28 number system

This is the 28th solution set of 10 practice problem exercise for SSC CGL exam and 5th on topic Number System. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set. If you have not taken the test yet, you may refer to SSC CGL level Question set 28 on Number System, take the test and then come back to this solution set.

We repeat here the method of taking a 10 problem test if you have not gone through it already.

Method for taking this 10 problem test and get the best results from the test set:

  1. Before start, go through the tutorials Numbers, Number system and basic arithmetic operations, Factorizing or finding out factors, HCF and LCM, Basic concepts on fractions and decimals part 1, Ratio and proportion or any other short but good material to refresh your concepts if you so require.
  2. Answer the questions in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.
  3. When the time limit of 12 minutes is over, mark up to which you have answered, but go on to complete the set.
  4. At the end, refer to the answers given in this companion 28th SSC CGL solution set on Number System 5, to mark your score at 12 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.
  5. Identify and analyze the problems that you couldn't do to learn how to solve those problems.
  6. Identify and analyze the problems that you solved incorrectly. Identify the reasons behind the errors. If it is because of your shortcoming in topic knowledge improve it by referring to only that part of concept from the best source you get hold of. You might google it. If it is because of your method of answering, analyze and improve those aspects specifically.
  7. Identify and analyze the problems that posed difficulties for you and delayed you. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.
  8. Give a gap before you take a 10 problem practice test again.

Important: both practice tests and mock tests must be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.

Resources that should be useful for you

You may refer to:

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but generally for any hard MCQ test.

Tutorials that you should refer to

Numbers, Number system and basic arithmetic operations

Factorizing or finding out factors

HCF and LCM

Basic concepts on fractions and decimals part 1

Ratio and proportion

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28th solution set - 10 problems for SSC CGL exam: topic Number System - Answering time 12 mins

Problem 1.

When positive integer A is divided by 5 remainder is 2, and when a second positive integer B is divided by 4 the remainder 1. Which one among the following cannot be the sum of A and B?

  1. 13
  2. 14
  3. 12
  4. 16

Solution 1 - Problem analysis and solving

For both the division operations, let us use the very basic relationships that arise in a division,

$A = 5q_1 + 2$, where $q_1$ is the integer quotient of this division.

$B = 4q_2 + 1$, where $q_2$ is the integer quotient of this division.

So $A+B = 5q_1 + 4q_2 + 3$,

From only one equation mathematically deducing the value of $A+B$ is not possible and so we will resort to mathematical reasoning testing each choice value for feasibility of being the solution. It is understood that in this types of questions solution is always one.

Testing choice values with mathematical reasoning

If $q_1 = q_2 = 1$, 12 is the sum. So this choice value is out of consideration.

With $q_1 = 2$ and $q_2 = 0$, 13 is the sum. So this choice value is also out of consideration.

With $q_1 = 1$ and $q_2 = 2$ we get sum as 16, and so it is out of consideration.

But for the choice value of 14, no value pair for $q_1$ and $q_2$ can make the sum as 14. So this is the solution.

Answer: Option b: 14.

Key concepts used: In a basic division, relation between divisor, dividend, quotient and remainder -- testing choice values with mathematical reasoning.

Problem 2.

A positive integer $n$ divides a second positive integer $m$ with result 7.55. What can be the possible value of the remainder of the division among the following?

  1. 12
  2. 22
  3. 18
  4. 15

Solution 2 - Problem analysis and solving

The decimal part of the result indicates the existence of a remainder that is less than the divisor $n$.

Expressing the division operation formally in relation to quotient and remainder we have,

$m = n\times{q} + R$, where $R$ is the integer remainder and $q$ is the integer quotient,

Or, $\displaystyle\frac{m}{n} = q + \displaystyle\frac{R}{n}$, where $q=7$ is the quotient.

By this relationship we get decimal form of remainder as,

$\displaystyle\frac{R}{n} = 0.55 = \frac{55}{100} = \frac{11}{20}$.

It means, remainder will be a multiple of 11. Among the choice values then 22 at Option b can be the only possible value of the remainder

Answer: Option b : 22.

Key concepts used: Basic remainder concept in a division -- ratio concepts.

Problem 3.

Bablu scored a prize point at every round of a village fair game he played. The prize points in a round were, 5 points for 1st position, 4 points for second position and 3 points for third position. If at the end, the product of his points in each round were 2700, how many rounds did he play?

  1. 5
  2. 6
  3. 7
  4. 8

Solution 3 - Problem analysis and solving

By the nature of the problem statement the product value of 2700 will be a product of only the factors of 5, 4 and 3. Breaking up 2700 in terms of these three factors we get,

$2700 = 4\times{5}\times{5}\times{3}\times{3}\times{3}$.

So, Bablu played six rounds and scored points, 4, 5, 5, 3, 3, and 3 in the six rounds in any sequence.

Answer: b: 6.

Key concepts used: Problem modeling -- factorization.

Problem 4.

If $0 \lt a \lt 1$, which one of the following is of minimum value?

  1. $\displaystyle\frac{1}{a^2}$
  2. $\displaystyle\frac{1}{\sqrt{a + 1}}$
  3. $\displaystyle\frac{1}{a^2 + 1}$
  4. $\displaystyle\frac{1}{(a+1)^2}$

Solution 4.

By the given condition, $a$ is a proper fraction greater than 0 and lesser than 1.

All the choice values being on the inverse, let us find the largest among the denominators. This will give us the minimum value among the choices.

As $(a + 1)^2 = a^2 + 2a + 1$, it is greater than both $a^2 + 1$ and $a^2$, ($a$ being a positive number).

We will now have to compare only $\sqrt{a + 1}$ with $(a+1)^2$.

Squaring both of these we get,

$a + 1$ and $(a + 1)^4$.

As the term $a+1 \gt 1$, with $a$ as a positive number, when we raise $(a+1)$ to higher powers, the result will always be larger than its lower powers. By this mathematical reasoning we conclude,

$(a + 1)^4 \gt (a + 1)$

Or, $(a + 1)^2 \gt \sqrt{a + 1}$.

So then $(a+1)^2$ turns out to be the largest denominator among the four choices and thus its corresponding fraction value is the minimum.

Answer: Option d: $\displaystyle\frac{1}{(a+1)^2}$ .

Key concepts used: Concept of change of values of numbers and fractions when inverted and raised to higher powers..

Problem 5.

In a certain game, $m$ number of players scored 2 points each and $n$ number of players scored 5 points each. The total score was 50 points. What is the least possible difference between $m$ and $n$ (absolute value only)?

  1. 1
  2. 3
  3. 5
  4. 7

Solution 5.

By the problem statement,

$2m + 5n=50$.

We have to find minimum difference between $m$ and $n$ (absolute value) among all feasible pairs of values of $m$ and $n$ satisfying the equation.

Trying out $m=5$ and $n=8$, difference is 3.

As $n$ can only be even (otherwise sum can't be even), next possible choice of $n=6$ gives $m=10$, a difference of 4. There are no more feasible choices for further consideration.

So, 3 is the minimum difference.

Answer: Option b: 3.

Key concept used: Problem modeling  -- testing feasible values towards the solution using concepts on numbers.

Problem 6.

The denominator of a fraction is 3 more than its numerator. If the numerator is increased by 7 and the denominator is decreased by 2 we get 2. The sum of the numerator and denominator is,

  1. 17
  2. 13
  3. 5
  4. 19

Solution 6

Let the fraction be $\displaystyle\frac{x}{y}$.

By the first problem statement then,

$y=x+3$.

On change of numerator and denominator values,

$\displaystyle\frac{x+7}{y-2}=2$,

Or, $x + 7 = 2y - 4$.

Substituting the value of $y$,

$x+7 = 2x + 6 - 4 = 2x + 2$,

Or $x = 5$ and $y=8$.

So, $x+y=13$.

Answer: Option b : 13.

Key concepts used: Problem modeling -- fraction concepts -- simplification.

Problem 7.

The sum of numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction is increased by $\frac{1}{24}$. The difference of numerator and denominator of the fraction is,

  1. 1
  2. 3
  3. 5
  4. 9

Solution 7

Let the oroiginal fraction be $\displaystyle\frac{x}{y}$, where denominator $y \gt x$, the numerator as in a proper fraction.

From the first statement then,

$x + y = 11$ and from the second statement,

$\displaystyle\frac{x+2}{y+2} - \displaystyle\frac{x}{y} = \displaystyle\frac{1}{24}$,

Subtracting 1 from the first term and adding 1 to the second term we get,

$\displaystyle\frac{x - y}{y+2} - \displaystyle\frac{x - y}{y} = \displaystyle\frac{1}{24}$,

Or, $(y -x)\left(\displaystyle\frac{1}{y} - \displaystyle\frac{1}{y+2}\right) = \displaystyle\frac{1}{24}$, terms reversed in position as $y \gt x$.

As the denominator 24 on the RHS doesn't have any odd factor, the terms on the left must have only even denominator. By this constraint along with conditions, $x+y=11$, and $y \gt x$, possible values of $y$ are only three,

$x = 5$, $y = 6$,

$x = 3$, $y=8$, or,

$x = 1$, $y=10$ .

The second and third value of $y$ brings in a factor of 5 in one of the two denominators in the LHS which is not acceptable by the nature of the factors of the denominator on the RHS.

On the other hand, the value of $y=6$ satisfies all conditions and the equation.

So, $x=5$ and $y=6$ and $y-x=1$.

Answer: a: 1.

Key concepts used: Transformation of the equation formed by the second statement, taking out the factor of $y-x$ achieved by subtracting and adding 1 to the two terms on the LHS -- from the nature of the factors of the denominators in the RHS and LHS and testing the small number of possible values of $y$, getting the final feasible value of $y$ quickly.

Problem 8.

A tree increases annually by $\frac{1}{8}$th of its height. If it stands 81 feet high today, what was its height (in feet) two years ago?

  1. 68
  2. 64
  3. 66
  4. 72

Solution 8

If $h$ was the height two years ago, after first year, height,

$h_1 = h + \frac{1}{8}h = \frac{9}{8}h$, and after 2 years the height will be,

$h_2 = 81 = \frac{9}{8}h + \frac{9}{64}h = \frac{81}{64}h$,

Or, $h = 64$ feet.

Answer: Option b: 64.

Key concepts used: Compound reduction.

Problem 9.

The value of $\displaystyle\frac{1}{15} + \displaystyle\frac{1}{35} + \displaystyle\frac{1}{63} + \displaystyle\frac{1}{99} +  \displaystyle\frac{1}{143}$ is,

  1. $\displaystyle\frac{5}{39}$
  2. $\displaystyle\frac{7}{39}$
  3. $\displaystyle\frac{4}{39}$
  4. $\displaystyle\frac{2}{39}$

Solution 9 - Problem analysis and solving

To solve this sum of fractions we will adopt the strategy of taking up addition of only two suitable fractions in each step. A suitable pair of fractions will have common factors in their denominators such that the result of addition is simplified to the maximum possible extent. This result of addition of a suitable pair of fractions will then be added to the next suitable fraction. Thus at each step, we reduce the number of terms by 1 and also keep the calculations to the minimum.

Examining for common factors in the denominators for quick simplification with least amount of calculation, we detect common factor 11 between denominators of last two terms. The last denominator = $11\times{13}$ and the last but one denominator = $9\times{11}$. LCM of the two will be, $9\times{11}\times{13}$. When we add these two, the numerator will be = $13+9=22$ and will cancel out the 11 in the denominator leaving 2 in the numerator. Thus the addition of these two terms results in $\displaystyle\frac{2}{9\times{13}}$.

Now we will combine this term with $\displaystyle\frac{1}{63}$, as both the denominators have 9 as a common factor. The resultant denominator is then = $7\times{9}\times{13}$ and the numerator = $13 + 2\times{7} = 27$ so that 9 is canceled out from the denominator leaving 3 in the numerator and 7 and 13 as factors in the denominator. The result at this stage, $\displaystyle\frac{3}{7\times{13}}$.

Next when we combine this fraction with $\displaystyle\frac{1}{35}$, the denominator of the sum gets three factors, 5, 7 and 13 but the numerator this time = $13 + 15=28$ canceling out 7 in the denominator and leaving 4 in the numerator. The result at this stage is thus, $\displaystyle\frac{4}{5\times{13}}$

In the last stage when we combine this with $\displaystyle\frac{1}{15}$, the denominator gets three factors, 3, 5 and 13 and the numerator = $13 + 12=25$, canceling 5 in the denominator and also leaving 5 in the numerator so that the final answer will be $\displaystyle\frac{5}{39}$.

Answer: Option a: $\displaystyle\frac{5}{39}$.

Key concepts used: Carrying out the fraction addition taking two suitable terms at each step getting a simplified result at the step and combining this result with the next suitable fraction at the next step -- we could very well have proceeded from left to right of the expression -- all through we didn't have to write, the whole process was carried out mentally..

Problem 10.

Which one of the following is a factor of the sum of the first 25 natural numbers?

  1. 13
  2. 12
  3. 24
  4. 26

Solution 10.

The sum of first 25 natural numbers will be 25 times $\displaystyle\frac{25+1}{2}=13$. Among the choice values, only 13 is the possible factor of the sum.

Answer: a: 13.

Key concepts used: Sum of first $n$ natural numbers where $n$ is odd.


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