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SSC CGL level Solution Set 71 on number system 10

Solution to number system questions for SSC Set 71

Solutions to number system questions for SSC Set 71

Learn to solve 10 number system questions for SSC Set 71 in 15 minutes using basic concepts and advance problem solving techniques. But take the test first.

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Number systems questions for SSC Set 71.

10 problems number system problems for SSC Set 71 Solutions - time to solve was 15 mins

Problem 1.

If 1 is added to both the numerator and denominator of a fraction it become $\displaystyle\frac{1}{4}$. If 2 is added to both the numerator and denominator it becomes $\displaystyle\frac{1}{3}$. Sum of the numerator and denominator is,

  1. 13
  2. 8
  3. 27
  4. 22

Solution 1: Problem analysis and execution:

As the operations in both cases on the numerator and denominator are uniform, we decide to use Componendo dividendo.

Assuming the fraction to be, $\displaystyle\frac{x}{y}$, in the first case,


Adding 1 to both sides,


Subtracting both sides from 1,


Dividing second by first,


Doing similar operations in the second case we get,


Dividing first by second,

$\displaystyle\frac{x+y+2 +2}{x+y+2}=\frac{6}{5}$,

Or, $1 + \displaystyle\frac{2}{x+y+2}=1+\frac{1}{5}$,

Or, $x+y+2=10$,

Or, $x+y=8$.

Answer: Option b: 8.

Key concepts used: Pattern recognition and use -- Componendo dividendo -- efficient simplification.

The process of componendo dividendo is so easy and systematic that it can be carried out in mind. We do not need to write the steps as above.

Problem 2.

The difference between two positive numbers is 3. If sum of their squares is 369, then the sum of the numbers is,

  1. 25
  2. 81
  3. 27
  4. 33

Solution 2: Problem analysis and execution:

As far as our idea of squares of small two digit numbers goes, $15^2 + 12^2 = 369$.

Method: $16^2=256$ is too near to 369 because $13^2=169$ does not add up to 369. Next lower pair would be 15 and 12.

Answer: Option c: 27.

Key concepts used: Sense of squares of small two digit numbers -- Number estimation skill.

Alternate straightforward algebraic solution

If $a$ and $b$ are the numbers, given, 



We will use the second after squaring the first to get $2ab$ and add $2ab$ to $a^2+b^2$ again to get $(a+b)^2$.


Or, $a^2-2ab+b^2=9$,

Or, $2ab=369-9=360$,

Or, $a^2 +b^2+2ab=(a+b)^2=360+369=729=27^2$.

So $a+b=27$

This is a simple enough method and will work even if you can't use your number estimation senses.

Problem 3.

If 2 is subtracted from the sum of digits of a two digit number whose unit's digit is twice the ten's digit, the result is one-sixth of the number. The number is,

  1. 25
  2. 36
  3. 24
  4. 48

Solution 3: Problem analysis and solving

Testing the second condition, we find choice value 25 invalid. Testing the first condition on the rest of the three choice value we find 24 satisfies this first condition also.

Answer: Option c: 24.

Alternate solution using place value mechanism

Assuming the ten's digit as $x$, the unit's digit is $2x$, and by first condition,


Or, $x=2$,

The number, 24.

Problem 4.

A number when divided by 361 gives a remainder 47. If the same number is divided by 19 what will be the remainder?

  1. 1
  2. 3
  3. 8
  4. 9

Solution 4: Problem analysis and execution:

We identify the pattern, $361=19^2$.

So the remainder of dividing the number by 19 produces remainder equal to that produced by dividing 47 by 19, which is 9. The rest portion of the number as fully divisible by 361 which is a multiple of 19, produces 0 remainder when divided by 19.

Answer: Option d: 9.

Key concepts used: Remainder concept -- Pattern identification.

Problem 5.

If total head count of cows and hens in a farm is 180 and the total leg count of same cows and hens is 420, the number of cows is,

  1. 130
  2. 150
  3. 30
  4. 50

Solution 5: Problem analysis and execution:

Legs of each cow is 4 whereas that of a hen is 2. So dividing the total leg count by 2 reduces legs per cow from 4 to 2 and that of hen from 2 to 1 which is equivalent of the sum of twice the head count of cows and the head count of hens.

By this halving operation, relationship of legs of cows and hens is transformed to relationship of heads of cows and hens.

Subtracting total head count of 180 from 210, the result of dividing the total leg count by 2, eliminates head count of hens, leaving only the head count of cows, which is 30.

Answer: Option c: 30.

Key concepts used: Mathematical reasoning -- variable domain transformation.

This solution is conceptual. In case of difficulty you can always get the answer solving two linear equations.

Problem 6.

In a three digit number, the digit at the hundred's place is two times the digit at the unit's place and sum of the three digits is 18. If the digits are reversed the number is reduced by 396. The difference of hundred's and ten's digits of the number is,

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

Solution 6: Problem analysis and execution:

From the first condition, the possible pairs of hundred's and unit's digits are,

(2, 1), (4, 2), (6, 3) and (8, 4).

As the sum of the three digits is 18, the first two pairs turn out to be infeasible and three digit possibilities are reduced to,

(6, 9, 3) and (8, 6, 4).

By the third condition, 693 is eliminated because of its odd unit's digit (we haven't calculated, we have analyzed the behavior of the unit's digit).

864 satisfies all conditions.

Answer: Option b: 2.

Key concepts used: Mathematical reasoning in number domain -- formation of possible digit combinations by limiting number of possible solutions -- unit's digit behavior analysis technique.

Alternate simpler solution

When you reverse a three digit number $xyz$ to $zyx$ and subtract one from the other, mentally expanding each number by place value mechanism, the result can be expressed as,

$100(x-z)+(z-x)=99(x-z)=396$. as the middle digit doen't change position, its effect cancels out. Here $x$ is hundred's digit and $z$ unit's digit.

So, $x-z=4$.

As $x$ is twice $z$, $x=8$, $z=4$ and $y=18-12=6$.

So, $x-y=2$.

Concepts used: Place value mechanism -- effect of reversing and subtraction and simple algebra.

Problem 7.

If the sum of two numbers are multiplied by each number separately, the products so obtained are 247 and 114. The sum of the numbers is,

  1. 20
  2. 19
  3. 21
  4. 23

Solution 7: Problem analysis and execution:

We will use factorization and detect the common factor between the two. That will be the sum of the two numbers.

$247 = 13\times{19}$, and

$114 = 6\times{19}$.

So the common factor is 19 and it is the desired sum of the two numbers which are 6 and 13.

Answer: Option b: 19.

Key concepts used: Mathematical reasoning in number domain -- factorization.

Alternate simpler solution

$x(x+y)=147$, ad $y(x+y)=114$.

Adding the two,


A much simpler solution.

Problem 8.

Find the maximum value of $F$ in the following equation, $5E9+2F8+3G7=1114$ where $E$, $F$ and $G$ stands for any integer.

  1. 5
  2. 9
  3. 7
  4. 8

Solution 8: Problem analysis and execution:

Adding the unit's digits we get, $9+8+7=24$. In the result, the uint's digit is 4, matching with the result. But the ten's digit of the result is 1 whereas adding the three unit's digits resulted in a carry of 2.

So, the sum of ten's position variables $E+F+G=9$, creating another carry to hundred's position and 1 as the ten's digit of the final sum. In line with this result, the carry of 1 added to sum of hundred's digits, $5+2+3=10$ produces desired result of 11 in hundred's position.

Answer: Option b: 9.

Key concepts used: Place value mechanism -- process of addition.

Problem 9.

The least number that should be added to 2055 so that the sum is exactly divisible by 27 is,

  1. 24
  2. 28
  3. 27
  4. 31

Solution 9: Problem solving execution

Noticing the pattern that 27 is divisible by 9 with sum of digits 9 and 2055 is divisible by 3 with sum of digits 3, to make 2055 divisible by 27 we must at least add a number the sum of digits of which is 6. Such a number to be added is divisible by 3, and out of two such choices, 24 is smaller and must be the answer. 

Answer: Option a: 24.

Key concepts used: Pattern identification -- rules of divisibility by 3 and 9.

Conventional solution

Dividing 2055 by 27 and forming quotient mentally as, 76 with remainder 3, the least number to add becomes 24. But we have to carry out the division.

Problem 10.

A fraction having denominator 30 and lying between $\frac{5}{8}$ and $\frac{7}{11}$ is,

  1. $\frac{19}{30}$
  2. $\frac{18}{30}$
  3. $\frac{20}{30}$
  4. $\frac{21}{30}$

Solution 10: First solution to inclusion in a range problem

The problem involves fraction comparisons between the lower and the upper range boundaries with the choice values. Without a proper strategy, this process is cumbersome and time consuming.

We will employ Range comparison strategy along with easy fraction comparison possibilities.

While comparing the larger fraction of the range $\displaystyle\frac{7}{11}$ with the first choice $\displaystyle\frac{21}{30}$, we equalize the numerators by converting the first fraction to $\displaystyle\frac{21}{33}$.

Fraction comparison by numerator equalization

If one numerator out of the two fractions to be compared is a factor of the second numerator, multiply the first numerator (and also the first denominator) by a suitable factor to make it equal to the second numerator. Now with equal numerators, the fraction with lower denominator will be the larger.

For example, if we have to compare fractions $\displaystyle\frac{7}{9}$ with the fraction $\displaystyle\frac{35}{43}$, finding first numerator 7 as a factor of second numerator 35, we multiply the first numerator and first denominator both by $\displaystyle\frac{35}{7}=5$ to transform the fraction to,


With the two numerator now equal, as the second denominator 43 is smaller than the first 45, the second fraction will be larger than the first (smaller number dividing same number produces larger result).

In general,

In two fractions,

$\displaystyle\frac{a}{b}$ and $\displaystyle\frac{a}{c}$, numerators are equal, and

if $c \lt b$, second fraction will be larger,

if $b \lt c$, first fraction will be larger.

This is fraction comparison by numerator equalization, which is infrequently used but is as effective as the more frequently used fraction comparison by denominator equalization.

Fraction comparison by denominator equalization

If two fractions to be compared are $\displaystyle\frac{3}{4}$ and $\displaystyle\frac{19}{24}$, multiply both numerator and denominator of first fraction by $\displaystyle\frac{24}{4}=6$ to transform it to,


With both denominators now 24, as the first numerator 18 is smaller than the second 19, the second fraction is larger.

In general, in two fractions,

$\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{b}$, denominators are equal;

if $c \gt a$, second fraction will be larger,

if $a \gt c$, first fraction will be larger.

This is fraction comparison by denominator equalization.

Direct method of fraction comparison

When you can't use the faster methods of numerator or denominator equalization, you have to go by the conventional direct method.

When we are to compare the two fractions,

$\displaystyle\frac{13}{18}$ and $\displaystyle\frac{17}{24}$,

identify the unique factor in a denominator that is not present in the other denominator and cross-multiply the numerators with these unique factors. Whichever result is larger will result in the larger fraction.

In this case, the unique factor of 18 not present in 24 is 3, and unique factor of of 24 not present in 18 is 4.

As cross-multiplying the numerators with these unique fctors,

$13\times{4}=52$ is larger than $17\times{3}=51$,

the first fraction is larger.

This is a derived method from method of addition and subtraction of fractions where we find the LCM of the two denominators.

Coming back to our problem, in this case, the first choice is not only the largest of the four choices (and that's why to be compaed with the upper range boundary first), it can also be compared easily by numerator equalization.

With equal numerator and smaller denominator then the first choice is larger than the given range and so cannot be our choice.

Next we take up comparing the next smaller choice, $\displaystyle\frac{20}{30}$, or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$. This time by direct comparison we find the fourth choice also larger than the upper range boundary $\displaystyle\frac{7}{11}$.

Leaving the upper range boundary now we will take the lower range boundary $\displaystyle\frac{5}{8}$ and compare it with the lowest of the four choices, $\displaystyle\frac{18}{30}$, or $\displaystyle\frac{9}{15}$. This second choice being smaller than the lower range boundary, this is also out of consideration leaving the third choice $\displaystyle\frac{19}{30}$ as the only remaining choice possible and so the answer.

Answer: Option a: $\displaystyle\frac{19}{30}$.

Key concepts used: Range comparison strategy -- Fraction comparison by numerator equalization -- Fraction comparison by direct method -- pattern identification -- efficient simplification -- strategic choice of comparison fraction pairs to minimize calculations -- Range comparison strategy - Analytical approach example.

Though all processing was done mentally, without the strategic selection of fraction pairs for comparison, the time taken would have been longer.

Range inclusion strategy

When upper and lower values of a range are given and we are to find which one of the four given choice values is within the range, we will try to find outside range choices by comparing the largest choice with the upper range value and the smallest choice with the lower range boundary.

Solution 10: Recommended quickest solution

First step: Sorted choice values

First thing we notice is the perfectly sorted four choice values. In this range inclusion type problem where we have to place one of four given values within a given range of two values, we must first mentally arrange the four choice values in sorted (ascending or descending) order. In this case the choice values are very conveniently sorted, and we identify the pattern.

Second step: first comparison: Compare the higher range boundary value with the second largest choice value (and not the largest choice value)

Effectively we halve the set of choice values to be compared with this modified step. This concept is taken from Binary search algorithm.

If result of comparison gives choice value as smaller, in this special case, this particular choice value becomes the answer, as in an MCQ problem there can't be more than one answer, that is, more than one choice value smaller than the higher range but larger than the lower range.

In reality, second largest choice value $\displaystyle\frac{20}{30}$ is larger than the higher range boundary $\displaystyle\frac{7}{11}$, and so is out of range. Only two possible choices are left.

At the outset, the equivalent action would have been to compare the second lowest choice value with the lower range boundary, but convenience of comparing $\displaystyle\frac{20}{30}$ or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$ is much more than comparing $\displaystyle\frac{19}{30}$ with $\displaystyle\frac{5}{8}$.

Convenience factor of action

When there are two possible courses of action, we take the action with higher convenience factor that results in quicker solution.

So we compare $\displaystyle\frac{20}{30}$ or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$ in the first comparison step.

Third step: second comparison

Now we can either compare the higher boundary value with the next lower choice value or use the lower boundary value in comparison. We take the second course of action as convenience factor of comparing $\displaystyle\frac{18}{30}$ or $\displaystyle\frac{9}{15}$ with $\displaystyle\frac{5}{8}$ is more.

This is a systematic process and will give you the result quickest in just maximum of two comparisons.

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