Number System Questions for SSC CGL Set 12 with Solutions

Submitted by Atanu Chaudhuri on Wed, 13/01/2021 - 02:39

Number system questions for SSC CGL set 12 with answers and solutions

12th Solved Number sytem questions for SSC CGL

Practice on 10 selected Number system questions for SSC CGL Set 12, verify your performance from answers and learn to solve quickly from solutions.

A few of these 10 are difficult number system questions that are solved quickly by advanced problem solving techniques.

This set contains,

Questions on Number system for for SSC CGL to be answered in 15 minutes (10 chosen questions)
Answers to the questions, and
Detailed conceptual solutions to the questions.

Special attention is given for quick solution of all the questions. Use the question set for timed mock test and clarify the doubts from the solutions for getting best results.

IMPORTANT: To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Algebra quickly, 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.

12th set of 10 Number system Questions for SSC CGL - answering time 15 mins

Q1. Two baskets together have 640 oranges. If one-fifth of the oranges in the first basket be taken to the second basket, the number of oranges in both baskets become equal. The number of oranges in the first basket is,

$400$
$300$
$600$
$800$

Q2. In a fraction when 3 is added to both the numerator and the denominator, it becomes $\frac{4}{5}$ and it becomes $\frac{1}{2}$ when 2 is subtracted from both the numerator and denominator. Find the fraction.

$\displaystyle\frac{15}{16}$
$\displaystyle\frac{9}{16}$
$\displaystyle\frac{14}{16}$
$\displaystyle\frac{11}{16}$

Q3. The sum of three numbers is 2, the first number is $\frac{1}{2}$ times the second number and the 3rd number is $\frac{1}{4}$ times the second number. The second number is,

$\displaystyle\frac{10}{9}$
$\displaystyle\frac{7}{6}$
$\displaystyle\frac{8}{7}$
$\displaystyle\frac{9}{8}$

Q4. A, B, C and D purchase a gift worth Rs.60. A pays $\frac{1}{2}$ of what others are paying, B pays $\frac{1}{3}$rd of what others are paying, and C pays $\frac{1}{4}$th of what others are paying. What is the amount paid by D?

Rs.15
Rs.14
Rs.16
Rs.13

Q5. The square of sum of two successive natural numbers exceeds the sum of their squares by 112. Find the lower of the two.

Q6. The number of composite numbers lying between 67 and 101 is,

$24$
$26$
$27$
$23$

Q7. Three bells chime at intervals of 5 secs, 6 secs and 7 secs. If the bells toll together at 6 am, how many times would the three toll together in the next 1 hour?

$16$
$17$
$18$
$15$

Q8. Ratio of a number and its sum of digits is 8 and the quotient of the product of its digits and that of the sum is $\frac{14}{9}$. Find the number.

$27$
$45$
$72$
$54$

Q9. A number $x$ when divided by 289 leaves 18 as remainder. The same number when divided by 17 leaves $y$ as remainder. The value of $y$ is,

Q10. A number $ab$ divides a second number $xy$ where $xy < ab$. If the result of division is $0.xyxyxy.......$, then the value of $ab$ is,

$11$
$66$
$99$
$33$

Answers to the Number system questions for SSC CGL Set 12

Q1. Answer: Option a: $400$.

Q2. Answer: Option d: $\displaystyle\frac{11}{16}$.

Q3. Answer: Option c: $\displaystyle\frac{8}{7}$.

Q4. Answer: Option d: Rs.13.

Q5. Answer: Option b: $7$.

Q6. Answer: Option c: $27$.

Q7. Answer: Option b: $17$.

Q8. Answer: Option c: $72$.

Q9. Answer: Option a: $1$.

Q10. Answer: Option c: $99$.

Solutions to the 12th set of 10 Number system Questions for SSC CGL - answering time was 15 mins

$400$
$300$
$600$
$800$

Solution 1: Quick solution by equating half of total oranges to known portion of first basket oranges

What happened (being clear about EVENTS),

After $\frac{1}{5}$th of first basket oranges are transferred to the second basket, number of oranges became equal in both baskets.
After the transfer, this equal number of oranges in the two baskets totaling 640, must be $\displaystyle\frac{640}{2}=320$, and,
After the transfer, $\displaystyle\frac{4}{5}$th of the original number of oranges remained in the first basket.

Conclusion,

$\displaystyle\frac{4}{5}\text{ of original number in 1st basket}=320$,

So, $\text{Original number of oranges in 1st basket}=\frac{5}{4}\times{320}=400$.

In problem solving, especially for solving problems with transactions, taking stock of events and the effect of events make the situation clear.

What remained in the first basket is a known portion of the whole. And it equals half of total. Simple isn't it?

Answer: Option a: $400$.

Key concepts used: Event sequencing -- Leftover portion analysis -- Solving in mind.

Q2. n a fraction when 3 is added to both the numerator and the denominator, it becomes $\frac{4}{5}$ and it becomes $\frac{1}{2}$ when 2 is subtracted from both the numerator and denominator. Find the fraction.

$\displaystyle\frac{15}{16}$
$\displaystyle\frac{9}{16}$
$\displaystyle\frac{14}{16}$
$\displaystyle\frac{11}{16}$

Solution 2: Quick solution by substituting value of $\displaystyle\frac{x}{y}$ from the proportion to the given equation bypassing solving a pair of linear equations

Let the fraction be,

$\displaystyle\frac{x}{y}$, where $x$ and $y$ are the numerator and the denominator respectively.

By the 1st problem statement,

$\displaystyle\frac{x+3}{y+3}=\displaystyle\frac{4}{5}$................................(1)

By the second statement,

$\displaystyle\frac{x-2}{y-2}=\frac{1}{2}$................(2)

You can cross-multiply both the equations and solve the two resulting linear equations.

Instead, we will first subtract each equation from 1 and take the ratio of the two results to eliminate $x$ at one stroke. This is application of Dividendo concept of the powerful method of Componendo dividendo.

Let's show you.

Subtracting equation (1) and equation (2) respectively from 1 you get,

$\displaystyle\frac{y-x}{y+3}=\displaystyle\frac{1}{5}$, and,

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

Take the ratio of the two results to get a linear equation in only $y$,

$\displaystyle\frac{y-2}{y+3}=\displaystyle\frac{2}{5}$

$\Rightarrow 5y-10=2y+6$

$\Rightarrow 3y=16$

$\Rightarrow y=\displaystyle\frac{16}{3}$.

Substitute in equation (2) as it is simpler,

$2x-4=y-2=\displaystyle\frac{10}{3}$,

$\Rightarrow x=\displaystyle\frac{11}{3}$.

$\Rightarrow \text{Desired fraction}=\displaystyle\frac{x}{y}=\frac{11}{16}$.

Answer: Option d: $\displaystyle\frac{11}{16}$.

Key concepts used: Instead of solving a pair of linear equation by cross-multilying and eliminating one variable, by Dividendo technique one variable is quickly eliminated to get the value of the second variable in no time -- Numerator equalization -- Solving in mind.

Q3. The sum of three numbers is 2, the first number is $\frac{1}{2}$ times the second number and the 3rd number is $\frac{1}{4}$ times the second number. The second number is,

$\displaystyle\frac{10}{9}$
$\displaystyle\frac{7}{6}$
$\displaystyle\frac{8}{7}$
$\displaystyle\frac{9}{8}$

Solution 3: Quick solution by converting two numbers in terms of the third and substitution in their sum acheving quick elimination

Let $a$, $b$ and $c$ by the first, second and the third numbers. Their sum is 2. So,

$a+b+c=2$.................................................(1)

First number $a$ is $\frac{1}{2}$ times the second number $b$. So,

$a=\frac{1}{2}b$.

3rd number $c$ is $\frac{1}{4}$ times the second number $b$. So,

$c=\frac{1}{4}b$.

Substitute values of $a$ and $c$ in terms of $b$ in equation (1),

$\frac{1}{2}b+b+\frac{1}{4}b=2$,

$\Rightarrow b=\displaystyle\frac{8}{7}$.

Answer: Option c: $\displaystyle\frac{8}{7}$.

Key concepts used: Conversion of two numbers in terms of the third number and substitution in their sum to eliminate the two numbers at one stroke -- Solving in mind.

Rs.15
Rs.14
Rs.16
Rs.13

Solution 4: Quick solution by Key pattern identification and missing element introduction

Let the payments of A, B, C and D are represented by the variables, $a$, $b$, $c$ and $d$ respectively.

By the first statement,

$a+b+c+d=60$..............................(1)

And by the second statement,

$a=\frac{1}{2}(b+c+d)$,

$\Rightarrow 2a=b+c+d$.

Identify the key pattern that introducing the missing element $a$ in both sides of the equation, you can transform the RHS to a known given value,

$3a=a+b+c+d=60$, Or, $a=20$.

By the 3rd statement,

$b=\frac{1}{3}(a+c+d)$,

$\Rightarrow 3b=a+c+d$.

Add $b$ to both sides as before,

$4b=60$, Or, $b=15$.

From the 4th statement,

$c=\frac{1}{4}(a+b+d)$,

$\Rightarrow 4c=a+b+d$.

Add $c$ to both sides,

$5c=60$, Or, $c=12$.

So value of $d$ from equation (1) is,

$d=60-(a+b+c)=60-47=13$.

Answer: Option d: Rs.13.

Key concepts used: Key pattern identification of the uniformly missing element in the RHS of the three given equations -- Missing element introduction technique --- Solving in mind.

Q5. The square of sum of two successive natural numbers exceeds the sum of their squares by 112. Find the lower of the two.

Solution 5: Quick solution by factor analysis of the product of two successive natural numbers

Let the two successive numbers by $a$, $a+1$. Being successive, variable needed is only 1.

By the problem statement,

$(a+b)^2=a^2+b^2+112$,

$\Rightarrow 2ab=112$,

$\Rightarrow ab=56$.

By using intuition and factor analysis of 56, you can spot the two numbers from their product of 56 as, 7 and 8. Lower of the two is 7, your answer.

Answer: Option b: $7$.

Key concepts used: Factor analysis of the product of two successive natural numbers-- Solving in mind.

Q6. The number of composite numbers lying between 67 and 101 is,

$24$
$26$
$27$
$23$

Solution 6: Quick solution by enumerating prime numbers in the range and reducing all numbers in the range by number of prime numbers in the range to get number of composite non-prime numbers

A composite number is a number having at least two prime factors and so it is not a prime number.

To use this concept first identify the prime numbers in the range $67 < x < 101$ as,

71, 73, 79, 83, 89, and 97, a total of only 6 prime numbers.

Between 67 and 101, you have $101-67-1=33$ numbers.

Reducing this by 6 primes, you find the number of composite numbers between 67 and 101 to be 27.

Answer: Option c : $27$.

Key concepts used: Enumerating prime numbers in the range and reducing all numbers by number of prime numbers to get number of composite non-prime numbers in the range -- Solving in mind.

Q7. Three bells chime at intervals of 5 secs, 6 secs and 7 secs. If the bells toll together at 6 am, how many times would the three toll together in the next 1 hour?

$16$
$17$
$18$
$15$

Solution 7: Quick solution by key pattern identification, concept of bells chiming together at LCM of chiming durations of the bells and dividing a given duration by LCM duration to get number of synchronized ringing together in the given duration

We know hy the concept of bells ringing together,

The three bells will toll together next when the time elapsed from now is exactly divisible by 5, 6 and 7, and also that time elapsed is the shortest such duration.

In other words, if $d$ secs is the shortest duration after which the three bells will toll together next, it is by definition the LCM of 5, 6, and 7 which is 210 secs.

So after every 210 secs the three bells will together.

To get the number of times they will toll together then, you just have to divide the interval of 1 hour by 210 secs.

To bring the two durations, the dividend and the divisor to the same unit, we will represent 1 hour as 60 minutes and 210 secs as 3 minutes 30 secs or $3\frac{1}{2}=\frac{7}{2}$ mins.

Divide 60 minutes by $\frac{7}{2}$ mins to split the duration into whole number of LCM durations,

$\displaystyle\frac{60}{\frac{7}{2}}=\frac{120}{7}=17\frac{1}{7}$.

After 17 synchronized rings time elapsed is $59\frac{1}{2}$ mins and in the remaining half a minute to the end of 1 hour duration, the three bells cannot ring together one more time.

So in next 1 hour from 6 am, the three bells will toll together a total of just 17 times.

Answer: Option b: $17$.

Key concepts used: Bells chiming together at LCM of their ringing duration -- LCM concept -- Dividing a duration by the LCM of component durations to get number of synchronized rings -- Solving in mind.

Q8. Ratio of a number and its sum of digits is 8 and the quotient of the product of its digits and that of the sum is $\frac{14}{9}$. Find the number.

$27$
$45$
$72$
$54$

Solution 8: Very quick solution by applying factors multiples concept on the equation obtained by place value mechanism

Let the number be comprising of two digits $(ab)$ where $a$ is the tens digit and $b$ is the units digit.

By place value concept,

$(ab)=10a+b$, $(ab)$ indicates just the number and is not the product of its two digits.

By problem description,

$\displaystyle\frac{10a+b}{a+b}=8$,

$\Rightarrow 10a+b=8a+8b$,

$\Rightarrow 2a=7b$.

From this equation it follows,

$a$ and $b$ must be multiples of 7 and 2 respectively with same multiplying factor. For example if $a$ is the $n$th multiple of 7, $b$ must also be the $n$th multiple of $2$ for the equation to be satidfied.

Without going into the second part of the problem description we conclude,

As $a$ is a digit of a two-digit number it must be a single digit integer, and so it must be 7 itself (not 14 or 21 or any other higher multiple of 7), and so $b$ is 2.

The number is then, 72.

Answer: Option c: $72$.

Key concepts used: Ratio simplification - Place value mechanism -- Factors multiples concept -- Mathematical reasoning using the available properties of the variables less than 10 -- Solving in mind.

Q9. A number $x$ when divided by 289 leaves 18 as remainder. The same number when divided by 17 leaves $y$ as remainder. The value of $y$ is,

Solution 9: Quick solution by Euclid's division lemma and remainder concept

From the first statement by Euclid's division lemma,

$x=p\times{289}+18$, where $p$ is the quotient,

$\Rightarrow x=p\times{17^2}+(17+1)$...........(1)

Identified the pattern of $289=17^2$ and $18=(17+1)$ for exploiting the patterns in the imminent second division with 17 as the divisor.

In the second division,

$x=q\times{17}+y=p\times{17^2}+(17+1)$, from equation (1); $q$ is the quotien this time,

$\Rightarrow x=17q+y=17(17p+1)+1$...............(2)

The equation of the first division has been transformed so that the RHS and LHS both represent division of $x$ by 17 in equation (2). So the remainders must be equal in equation (2).

That is, $y=1$.

Answer: Option a: $1$.

Key concepts used: Euclid's division lemma -- Remainder concept -- Key pattern identification -- Solving in mind.

Intuitive solution

If you are clear about the concepts and are able to identify the key patterns, solution will be in tens of seconds.

Q10. A number $ab$ divides a second number $xy$ where $xy < ab$. If the result of division is $0.xyxyxy.......$, then the value of $ab$ is,

$11$
$66$
$99$
$33$

Solution 10: Quick solution by Division concept and the concept of rational number equivalent to a repeating non-terminating decimal

By the concept of rational number equivalent to a non-terminating repeating decimal,

$0.xyxyxy.......=\displaystyle\frac{xy}{99}=\frac{xy}{ab}$, as given,

Or, $ab=99$.

Instant solution by using the right concepts. We'll show the derivation of this equivalence soon.

Answer: Option c: $99$.

Key concepts used: Rational number equivalent to repeating non-terminating decimal -- Division concept -- solving in mind.

Rational number equivalent to a non-terminating repeating decimal

Let the repeating non-terminating decimal be,

$0.xyzxyzxyz.................$.

which has the sequence $xyz$ repeating indefinitiely in the decimal portion.

Let $p=0.xyzxyzxyz.....$.

Multiply both sides by 1000,

$1000p=xyz.xyzxyzxyz.......$.

The decimal part remains unchanged as it is non-terminating and continues repeating the sequence indefinitely. Reducing one number of sequence $xyz$ doesn't change the decimal.

Break up the RHS,

$1000p=xyz+0.xyzxyzxyz...=xyz+p$

$\Rightarrow 999p=xyz$

$\Rightarrow p=\displaystyle\frac{xyz}{999}$ which is a rational number.

In general, if the repeating pattern in the non-terminating decimal $p$ has $n$ digits, the rational number equivalent to $p$ is,

$p=\displaystyle\frac{\text{repeating digits}}{10^n-1}$.

If $n=2$ as in our problem,

$p=\displaystyle\frac{xy}{10^2-1}=\frac{xy}{99}$, $xy$ form the repeating digits.

Concepts and techniques summary

The concepts and techniques used for quick solution of the questions are,

Event sequencing, Leftover portion analysis, Numerator equalization, Dividendo technique, Solving a pair of linear equations, Missing element introduction technique, Key pattern identification, Numerator equalization, Factor analysis, Prime numbers, Composite numbers, Mathematical reasoning, Euclid's division lemma, Remainder concept, Conversion of a non-terminating repeating decimal to rational fraction.

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SSC CGL Solved question Set 99, Number system 12