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SSC CHSL Solved question Set 3, Number system 1

3rd SSC CHSL Solved Question Set, 1st on Number system

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This is the 3rd solved question set of 10 practice problem exercise for SSC CHSL exam and the 1st on topic Number system. It contains,

  1. Question set on Number system for SSC CHSL to be answered in 15 minutes (10 chosen questions). This question set can be used as a mini-mock test.
  2. Answers to the questions, and
  3. Detailed conceptual solutions to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be referred to. But more importantly, to absorb the concepts, techniques and reasoning explained in the solutions, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

3rd Question set - 10 problems for SSC CHSL exam: 1st on topic Number system - answering time 15 mins

Q1. When two numbers are separately divided by 33, the remainders are respectively 21 and 28. If the sum of the two numbers is divided by 33, the remainder will be,

  1. 16
  2. 12
  3. 14
  4. 10

Q2. Which of the following numbers will always divide a six-digit number of the form $xyxyxy$ (where $1\leq x \leq 9$)?

  1. 1010
  2. 11011
  3. 11010
  4. 10101

Q3. In a two-digit number, the digit at unit's place 1 less than twice the digit at the ten's place. If the digits at the unit's place and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is,

  1. 35
  2. 47
  3. 59
  4. 23

Q4. If sum of five consecutive integers is $S$, then the largest of those in terms of $S$ is,

  1. $\displaystyle\frac{S-10}{5}$
  2. $\displaystyle\frac{S+5}{4}$
  3. $\displaystyle\frac{S+10}{5}$
  4. $\displaystyle\frac{S+4}{4}$

Q5. If a number is as much greater than 31 as it is less than 75, then the number is,

  1. 74
  2. 106
  3. 44
  4. 53

Q6. Arrangement of the fractions, $\displaystyle\frac{4}{3}$, $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$ and $\displaystyle\frac{5}{12}$ into ascending order is,

  1. $-\displaystyle\frac{7}{8}$,  $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$
  2. $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$,  $\displaystyle\frac{4}{3}$, $\displaystyle\frac{5}{12}$
  3. $-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$,   $\displaystyle\frac{4}{3}$, $\displaystyle\frac{5}{12}$
  4. $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$

Q7. A number whose one-fifth part increased by 4 is equal to the one-fourth part diminished by 10 is,

  1. 280
  2. 260
  3. 240
  4. 270

Q8. The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction increases by $\displaystyle\frac{1}{24}$. The difference of numerator and denominator of the fraction is,

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

Q9. A girl as asked to multiply a number by $\displaystyle\frac{7}{8}$. Instead she divided the number by $\displaystyle\frac{7}{8}$. She got the result 15 more than the correct result. The sum of the digits of the number is,

  1. 4
  2. 8
  3. 11
  4. 6

Q10. In a test, 1 mark is awarded for each correct answer and 1 mark is deducted for each wrong answer. If a boy answers all 20 items of the test and gets 8 marks, the number of questions answered correctly by him was,

  1. 16
  2. 14
  3. 8
  4. 12

Answers to the questions

Q1. Answer: Option a: 16.

Q2. Answer: Option d : 10101.

Q3. Answer: Option b: 47.

Q4. Answer: Option c: $\displaystyle\frac{S+10}{5}$.

Q5. Answer: Option d: 53.

Q6. Answer: Option a : $-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$.

Q7. Answer: Option a: 280.

Q8. Answer: Option b: 1.

Q9. Answer: Option c: 11.

Q10. Answer: Option b: 14.


3rd solution set—10 problems for SSC CHSL exam: 1st on topic Number system: answering time 15 mins

Q1. When two numbers are separately divided by 33, the remainders are respectively 21 and 28. If the sum of the two numbers is divided by 33, the remainder will be,

  1. 16
  2. 12
  3. 14
  4. 10

Solution 1: Problem analysis and solving in mind by division, quotient and remainder relation

In the second division, effectively sum of remainders obtained by separate division, that is, $21+28=49$ is to be divided again by 33 to obtain the remainder as, $49-33=16$. Solved in 10 seconds.

Let us show you the mathematical steps.

Let the numbers be $a$ and $b$.

We'll use the relation of results of division formalized by Euclid.

$a=33q_1+21$, where $q_1$ is the quotient and 21 the remainder, and

$b=33q_2+28$, where $q_2$ is the quotient and 28 the remainder.

Add the two to get,

$a+b=33(q_1+q_2)+49=33(q_1+q_2+1)+16$.

This is the result of dividing $(a+b)$ by 33. The divisor is 33, quotient is $(q_1+q_2+1)$ and remainder 16.

Till the remainder remains more than or equal to the divisor 33, you have to divide (or repeatedly subtract) the remainder by 33.

Answer: Option a: 16.

Key concepts used: Division and remainder concepts  -- Euclid's division lemma -- Solving in mind.

Q2. Which of the following numbers will always divide a six-digit number of the form $xyxyxy$ (where $1\leq x \leq 9$)?

  1. 1010
  2. 11011
  3. 11010
  4. 10101

Solution 2: Problem solving by trial on choice values and mathematical reasoning

The sum of digits of the given form of six-digit number is,

$3(x+y)$.

The number must be divisible by 3.

Out of the four choice values only 11010 and 10101 have integer sum, (sum of digits) divisible by 3, and so only these two options are divisible by 3, and the first two choice values cannot be the answer.

At the second step, identify that for getting the six-digit number, each of two possible choice values can only be multiplied by a factor equal to or greater than 10 (target $xyxyxy$ is a six-digit number and both the possible choice values, 11010 and 10101 are five digit numbers with most significant digit as 1).

This is the first pattern identification.

Try first 10.

Multiply 11010 by 10. Result 110100 is not in the form of $xyxyxy$. Now multiply 10101 by 10 getting 101010. This is exactly in the form of $xyxyxy$.

As in at least one situation, choice value of option 3 fails but of option 4 succeeds, answer must be 10101.

We'll show you the formal solution now.

Solution 2: Problem solving by Place value mechanism

Break up $xyxyxy$ into two parts by place value mechanism,

$xyxyxy=x(100000+1000+10)+y(10000+100+1)$

$=x(101010)+y(10101)$

$=10101(10x+y)$

With value of $x$ at least 1, 10101 will then be always a factor of $xyxyxy$.

Answer: Option d: 10101.

Key concepts used: Choice value test --  Divisibility -- Mathematical reasoning -- Place value mechanism.

Q3. In a two-digit number, the digit at unit's place 1 less than twice the digit at the ten's place. If the digits at the unit's place and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is,

  1. 35
  2. 47
  3. 59
  4. 23

Solution 3: Problem analysis and solution by place value mechanism and linear equation solving

Let the original number be, $(10x+y)$, where $x$ and $y$ are the tens's and unit's digits.

By the first condition,

$2x-y=1$,

Or, $y=2x-1$.

By the second condition,

$(10y+x)-(10x+y)=(10x+y)-20$,

Or, $10y+x+20=2(10x+y)=20x+2y$.

Substitute expression for $y$,

$10(2x-1)+x+20=20x+4x-2$,

Or, $21x-10+20=24x-2$,

Or, $3x=12$,

Or, $x=4$.

So, $y=2x-1=7$.

Original number is 47.

Answer: Option b: 47.

We'll show you a second method of solution by trial on choice values.

Solution 3: Problem solving by trial on choice values and mathematical reasoning

When you interchange the unit's and ten's digits the difference between the two numbers is more than the original number by 20.

This means the result of the difference and the original number must have the same unit's digit (because of difference 20).

We'll subtract units digit from the ten's digit with 10 added for taking care of borrow mechanism for each choice value.

For 35, result is, $13-5=8$. This is an invalid choice.

For 47, result is, $14-7=7$, a valid choice value.

For 59, result is, $15-9=6$, an invalid choice.

For 23, result is, $12-3=9$, also an invalid choice.

A conceptually simpler way to test is to reverse the digits, subtract the original number from this new number and subtract this difference from the original number. It must be 20.

For the first choice value of 35, test result is, $53-35=18$, $35-18=17\neq 20$. Choice invalid.

For the second choice value of 47, test result is, $74-47=27$, $47-27=20$. It is the answer. No further test is necessary as only one choice must be the answer.

A lot of number system problems can be solved with this type of quick method customized for the specific problem but nevertheless based on mathematical reasoning and general number system concepts.

Answer: Option b: 47.

Key concepts used: Place value mechanism -- Solving a pair of linear equations -- Trial on choice values -- Mathematical reasoning -- Unit's digit analysis.

Q4. If sum of five consecutive integers is $S$, then the largest of those in terms of $S$ is,

  1. $\displaystyle\frac{S-10}{5}$
  2. $\displaystyle\frac{S+5}{4}$
  3. $\displaystyle\frac{S+10}{5}$
  4. $\displaystyle\frac{S+4}{4}$

Solution 4: Problem solving using the concept sum and average of consecutive numbers

The average of 5 consecutive numbers is the middle number, as the number of numbers is odd. This equals in this case to,

$\displaystyle\frac{S}{5}$, as sum is five times average.

The largest number is 2 more than this middle number, as the numbers are consecutive.

So it equals,

$\displaystyle\frac{S}{5}+2=\displaystyle\frac{S+10}{2}$.

Answer: Option c: $\displaystyle\frac{S+10}{5}$.

Key concepts used: Sum of consecutive numbers -- Average of consecutive numbers -- Property of consecutive numbers, the numbers increase by 1 at each step -- Solving in mind.

Q5. If a number is as much greater than 31 as it is less than 75, then the number is,

  1. 74
  2. 106
  3. 44
  4. 53

Solution 5: Problem analysis and solving by average concept

As the number is greater than 31 by the amount it is less than 75, it is equidistant from these two boundary numbers, and so is the average of the two numbers.

It is then,

$\displaystyle\frac{31+75}{2}=53$.

Difference between 53 and 31 is 22. and the difference between 75 and 53 is also 22.

Answer: Option d: 53.

Key concepts used: Average concept -- Solving in mind.

Q6. Arrangement of the fractions, $\displaystyle\frac{4}{3}$, $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$ and $\displaystyle\frac{5}{12}$ into ascending order is,

  1. $-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$
  2. $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$, $\displaystyle\frac{4}{3}$, $\displaystyle\frac{5}{12}$
  3. $-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{4}{3}$, $\displaystyle\frac{5}{12}$
  4. $-\displaystyle\frac{2}{9}$, $-\displaystyle\frac{7}{8}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$

Solution 6: Problem analysis and Solving by elementary fraction comparison and basic number comparison concepts

It is enough to compare the positive fraction $\displaystyle\frac{5}{12}$, with the other positive fraction $\displaystyle\frac{4}{3}$, and the negative fraction $\left(-\displaystyle\frac{7}{8}\right)$ with the other negative fraction $\left(-\displaystyle\frac{2}{9}\right)$ as positive fractions are always greater than any negative fraction.

Out of the two positive fractions it is easy to see that $\displaystyle\frac{4}{3}$ is much larger than $\displaystyle\frac{5}{12}$ as the former is larger than 1 and the latter lesser than 1.

Among the two negative fractions now the absolute value $\displaystyle\frac{7}{8}$ is much larger than that of $\displaystyle\frac{2}{9}$ as the numerator and denominator values are very close together for the former whereas these are far apart in the latter.

You can actually get the relative values by calculating their difference as,

$\displaystyle\frac{7}{8}-\displaystyle\frac{2}{9}=\displaystyle\frac{63-16}{72}=\frac{47}{72}$.

So when these two fractions are multiplied by $(-1)$, $\left(-\displaystyle\frac{7}{8}\right)$ becomes lesser than $\left(-\displaystyle\frac{2}{9}\right)$ (further away from origin because of larger megative value).

Finally then the ascending order arrangement of the four fractions is given by,

$-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$.

Answer: Option a : $-\displaystyle\frac{7}{8}$, $-\displaystyle\frac{2}{9}$, $\displaystyle\frac{5}{12}$, $\displaystyle\frac{4}{3}$.

Key concepts used: Basic number comparison, fraction comparison, and problem solving concepts of:

  1. negative numbers are smaller than positive numbers,
  2. problem partitioning—the four fractions are partitioned into two sets requiring only two comparisons in total. This is a general Problem solving technique,
  3. negative numbers with larger absolute values are smaller than negative numbers with smaller absolute values,
  4. Comparison of fractions by actual subtraction, and
  5. fractions larger than 1 are larger than fractions lesser than 1.

If you want to know how to compare fractions in more details you may refer to our article,

How to compare fractions easily in a few steps.

Q7. A number whose one-fifth part increased by 4 is equal to the one-fourth part diminished by 10 is,

  1. 280
  2. 260
  3. 240
  4. 270

Solution 7: Problem solving by Problem definition and Fraction arithmetic

By the problem statement,

$\displaystyle\frac{1}{5}N+4=\displaystyle\frac{1}{4}N-10$, where $N$ is the desired number,

Or, $N\left(\displaystyle\frac{1}{4}-\displaystyle\frac{1}{5}\right)=14$,

Or, $\displaystyle\frac{1}{20}N=14$.

So, $N=20\times{14}=280$.

Answer: Option a: 280.

Key concepts used:  Problem definition to form the variable relation from word description -- Fraction arithmetic -- Solving in mind.

Q8. The sum of the numerator and denominator of a positive fraction is 11. If 2 is added to both numerator and denominator, the fraction increases by $\displaystyle\frac{1}{24}$. The difference of numerator and denominator of the fraction is,

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

Solution 8: Problem analysis and formation of fraction expression in terms of numerator and denominator variables

Assume the numerator and denominator of the fraction to be $x$ and $y$ respectively.

By the first condition, $x+y=11$.

And by the second condition,

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

Add 1 to both sides to create $(x+y)$ in the numerator in both sides,

$\displaystyle\frac{x+y+4}{y+2}=\displaystyle\frac{x+y}{y}+\displaystyle\frac{1}{24}$,

Or, $\displaystyle\frac{15}{y+2}-\displaystyle\frac{11}{y}=\displaystyle\frac{1}{24}$.

Conventional method to solve for $y$ is to solve the resulting quadratic equation.

Solution 8: Conventional solution by solving a quadratic equation

The relation in $y$ is,

$\displaystyle\frac{15}{y+2}-\displaystyle\frac{11}{y}=\displaystyle\frac{1}{24}$,

Or, $y(y+2)=24(15y-11y-22)=96y-24\times{22}$,

Or, $y^2-94y+24\times{22}=0$, we keep the numeric term, $24\times{22}=528$ in the form of product of factors as it is easier to partition the numeric term into two factors sum of which is the coefficient of the middle term,

Or, $y^2-(6+88)y+6\times{88}$

$=y(y-6)-88(y-6)$

$=(y-6)(y-88)=0$.

So, either $(y-6)=0$ or $(y-88)=0$.

It follows, either $y=6$, or $y=88$.

As sum of $x$ and $y$ is 11, $y$ cannot be 88. So it must be 6.

And $x=11-y=5$.

Difference between denominator and numerator is 1.

Answer: Option b: 1.

Key concepts used: Solution of a quadratic equation by splitting the numeric term -- Conventional solution.

How do you break up such a large numeric term of 528 into two numbers, sum of which will be 94?

This is not easy unless you follow a method.

Factorize 528 as,

$528=24\times{22}=4\times{6}\times{2}\times{11}$.

Don't break up 528 into prime factors—that would be too many factors to deal with.

First test that sum of already known two parts, 22 and 24 is only 46 falling far short of 94.

This test is important as it gives you an idea about the relative values of the numbers involved and shows you a path to follow.

How to increase the sum of the two factor parts?

A reasonable way is to multiply 11, the largest factor, with as large a factor as possible with the result just falling short of 94 and then add the remaining factor.

With this approach 88 and 6 can be identified easily.

If you want to know in more details about factorizing quadratic equations, you may refer to our long and exhaustive article,

How to factorize quadratic equations easily by factor analysis.

Let us show you a second assured solution by analyzing choice value possibilities.

Solution 8: Quick solution by Choice value possibility analysis and mathematical reasoning

The sum of $x$ and $y$ being 11, the five possible pairs of values for $x$ and $y$ are,

$1+10=11$

$2+9=11$

$3+8=11$,

$4+7=11$, and

$5+6=11$.

Now consider the RHS denominator of 24 in the relation for $y$,

$\displaystyle\frac{15}{y+2}-\displaystyle\frac{11}{y}=\displaystyle\frac{1}{24}$.

For this relation to be satisfied, the factors of RHS and LHS denominators must match.

With $y=10$, the factor 5 cannot be eliminated from the denominator with the numerator result, and this corresponds to choice value of 9. So this choice and the possibility of $y=10$ is invalid.

$y=9$ and $x=2$ resulting $y-x=7$ is not among choice value possibilities of 5, 1 or 3, and so this possibility is invalid.

With $y=8$, again the factor 5 in the denominator cannot be eliminated by the numerator result. So this possibility is also invalid.

For the same reason possibility $y=7$ is invalid.

In the last possibility, with $y=6$ the relation in $y$ is at last satisfied,

$\displaystyle\frac{15}{8}-\displaystyle\frac{11}{6}=\displaystyle\frac{45-44}{24}=\frac{1}{24}$.

Difference between denominator and numerator of the fraction is then 1. 

Answer: Option b: 1.

Q9. A girl as asked to multiply a number by $\displaystyle\frac{7}{8}$. Instead she divided the number by $\displaystyle\frac{7}{8}$. She got the result 15 more than the correct result. The sum of the digits of the number is,

  1. 4
  2. 8
  3. 11
  4. 6

Solution 9: Problem analysis and Solving by precise problem definition and fraction arithmetic

Let the number be $x$.

By the first statement, the result obtained by multiplying $x$ with $\displaystyle\frac{7}{8}$ is,

$\displaystyle\frac{7}{8}x$

And, by the second statement, the result formed by dividing $x$ by $\displaystyle\frac{7}{8}$ is,

$\displaystyle\frac{8}{7}x$.

The second result is larger than the first. So,

$\displaystyle\frac{8}{7}x-\displaystyle\frac{7}{8}x=15$,

Or, $x\left(\displaystyle\frac{8}{7}-\displaystyle\frac{7}{8}\right)=x\left(\displaystyle\frac{15}{56}\right)=15$,

So,

$x=56$, and sum of the digits is 11.

Answer: Option c: 11.

Key concepts used: Problem definition -- Fraction arithmetic -- Solving in mind.

Q10. In a test, 1 mark is awarded for each correct answer and 1 mark is deducted for each wrong answer. If a boy answers all 20 items of the test and gets 8 marks, the number of questions answered correctly by him was,

  1. 16
  2. 14
  3. 8
  4. 12

Solution 10: Problem analysis and solving by problem definition and linear equation formation

Let $c$ be the number of correct answers so that, $w=(20-c)$ is the number of wrong answers as, $(c+w)=20$.

By the problem definition then,

$c\times{1}-(20-c)\times{1}=8$,

Or, $2c=28$,

Or, $c=14$.

So the number of questions answered correctly by the boy was 14.

Answer: Option b: 14.

Key concepts used: Problem definition -- Linear equation solving -- Solving in mind.


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