SSC CGL Tier II level Solution Set 29 Number system 1 | SureSolv

SSC CGL Tier II level Solution Set 29 Number system 1

29th SSC CGL Tier II level Solution Set, 1st on Number system problems

SSC-CGL-Tier-II-Solutions-29-Number-System-1

In this 29th solution set of 10 practice problem exercise for SSC CGL Tier II exam and 1st on topic Number System, a few of the problems should pose a bit of challenge to the student.

Solutions are all concept and method based and mostly in mind. The problem solving approach should prove valuable to the students.

Its paired question set can be used as a mini-mock test on number system, even for competitive tests other than SSC CGL.

We should mention that in an MCQ test, you need to deduce the answer in shortest possible time and select the right choice. You do not have to write the steps. Writing takes up valuable seconds that you can save by solving problems in mind and writing as little as possible. This is what we call Solving in mind.

All solutions are focused towards solving in mind and so should prove to be especially valuable to the student.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts in the topic area
  • is adequately fast in mental math calculation
  • should try to solve each problem using the basic and rich concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his or her head rather than on paper.

Actual problem solving is done in the fourth layer. You need to use your problem solving abilities to gain an edge in competition.

Before going through these solutions you should take the test by referring to SSC CGL Tier II level Question Set 29 on Number system 1.


29th solution set - 10 problems for SSC CGL Tier II exam: 1st on Number system problems - time 12 mins

Problem 1.

The smallest fraction that should be added to the sum of $2\frac{1}{2}$, $3\frac{1}{3}$, $4\frac{1}{4}$, and 5\frac{1}{5}$, to make the result a whole number is,

  1. $\frac{43}{60}$
  2. $\frac{13}{60}$
  3. $\frac{17}{60}$
  4. $\frac{1}{4}$

Solution 1: Problem analysis and solving in mind by mixed fraction breakup technique

If you break up each of the given 4 mixed fractions into an integer and a fraction part the problem is reduced to making the sum of the fraction parts a whole number. For example, the first mixed fraction broken up would be,

$2\frac{1}{2}=2+\displaystyle\frac{1}{2}$.

The sum of the integer parts, $(2+3+4+5=14)$ will automatically be a whole number. But the sum of the fraction parts would be,

$\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{5}=\displaystyle\frac{77}{60}=1\frac{17}{60}$.

You just have to add $(60-17)=43$ to the numerator of the fraction part to make the sum a whole number, that is, you have to add the smallest fraction, $\displaystyle\frac{43}{60}$ to the sum of the given four fractions to make the sum a whole number.

Answer: Option a: $\displaystyle\frac{43}{60}$.

The problem can easily and quickly solved in mind if you apply the mixed fraction breakup technique.

Key concepts used: Mixed fraction breakup technique  -- Fraction arithmetic -- Solving in mind.

Problem 2.

Three electronic devices make a beep after every 48 seconds, 72 seconds and 108 seconds respectively. They beeped together at 10 am. The time when the devices will next make a beep together at the earliest is,

  1. 10:07:48 hours
  2. 10:07:12 hours
  3. 10:07:36 hours
  4. 10:07:24 hours

Solution 2: Problem analysis and solution by LCM and clock time concept

After the three devices made a beep at 10 am, the next earliest time they will make the next beep will be AFTER a time duration that is the LCM of the beep frequencies. At that point of time all three beeps will be synchronized.

We'll find the LCM of 48, 72 and 108 by factorization method.

The factors of the three are,

$48=12\times{2}\times{2}$

$72=12\times{2}\times{3}$, and

$108=12\times{3}\times{3}$.

We have considered 12, the HCF of the three numbers, as a factor of each of the three numbers to reduce the total number of factors and speed up the process of finding the LCM.

In the LCM of the three, all factors taken together are to appear once. So it will be,

$LCM=12\times{2}\times{2}\times{3}\times{3}=432$.

432 seconds is 7 minutes 12 seconds.

So the earliest next beep will be at 10:07:12 hours.

Answer: Option b: 10:07:12 hours.

Key concepts used: Problem definition by converting the synchronization problem to an LCM finding problem -- LCM by factorization -- Solving in mind.

Problem 3.

Two baskets have 640 oranges. If $\displaystyle\frac{1}{5}$th of the oranges in the first basket be taken to the second basket then number of oranges in both baskets become equal. The number of oranges in the first basket was,

  1. 300
  2. 600
  3. 400
  4. 800

Solution 3: Problem analysis and quick solution by portions remaining in transfer concept

After the oranges have breen transferred from the first to the second basket, their number of oranges became equal. As the total number of oranges do not change and remains at 640, the number of oranges in each basket after the transfer is half of 640, that is, 320.

As $\displaystyle\frac{1}{5}$th of the oranges in the first basket have been transferred, $\displaystyle\frac{4}{5}$th of the original number remained after the transfer.

As this is equal to 320, the original number of oranges in the first basket was,

$\displaystyle\frac{5}{4}\times{320}=400$.

Answer: Option c: 400.

Key concepts used: Portions remaining concept -- Precise problem definition-- Event sequencing to clearly understand what happens because of the transfer -- Solving in mind.

Problem 4.

Which of the following statement(s) is/are true?

  1. The sum of first 20 odd numbers is 400.
  2. The total number of positive factors of 72 is 12.
  3. The largest two digit prime number is 97.
  1. Only II and III
  2. Only I and III
  3. Only I and II
  4. All are true

Solution 4 : Problem analysis and solving in mind

This is in fact a combination of three small problems.

You will adopt the strategy to decide the truth of the one that takes minimum time. Choosing 97 as the largest two digit prime takes just a second. So proposition III is true. You can eliminate the third choice by this result.

Now you take up the second mini-problem of finding number of factors of 72. Quick technique is to factorize 72 first into two close-by factors 9 and 8. Then it becomes easy.

9 has 2 factors—$3\times{3}=9$ and 8 has 3, $2\times{2}\times{2}=8$.

So 72 has 5 prime factors.

Proposition mentions just 'factors'. So you have to consider the other possible non-prime factors, 4, 6, 8, 9, 12, 18 and 36. So total number of possible factors of 72 is indeed 12—proposition II is true.

To decide between first and fourth choice you will now evaluate the sum of first 20 odd numbers 1 to 39.

Quick technique is to sum up the first 19 odd numbers as the sum of 1 to 37 with average as middle 10th number 19. Sum of these 19 numbers is,

$19\times{19}=361$.

Add 39 to complete the series and you will get the sum as 400.

Proposition I is also true.

Answer is the choice d—All are true.

Answer: Option d: All are true.

Key concepts used:Quick sum of first 20 odd numbers: sum of first 19 odd numbers as 19 times middle number average 19 with 39 added to the sum -- Prime number identification -- Finding all factors -- Solving in mind.

Problem 5.

N is the largest two digit number which when divided by 3, 4 and 6 leaves the remainders 1, 2 and 4 respectively. What is the remainder when N is divided by 5?

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

Solution 5: Problem analysis and solution by basic number system concepts and trial by mathematical reasoning

Any number on division by 4 when leaves a remainder of 2, must be an even number.

So the target number must be an even number, but not divisible by 3 or 4 considering in addition the first condition.

An even number that is not divisible by 3 or 4 when divided by 3 would leave a remainder of 1 and when divided by 4 would leave a remainder of 2.

As you need to find the largest such number which also satisfies the third condition, you would test out first the 90s.

In the 90s—90 to 98, you have two candidates that are even numbers not divisible by 3 or 4—98 and 94. Dividing 98 by 6 remainder is 2—it's not the number, but dividing 94 by 6, you get the remainder 4. So 94 is the number you were searching for.

If you divide 94 by 5, remainder will still be 4.

Answer: Option a: 4.

Key concepts used: Divisibility -- Remainder analysis -- Number system concepts -- Mathematical reasoning -- Solution by selected trial -- Solving in mind.

Problem 6.

Twenty one times a positive integer is less than its square by 100. The value of the positive integer is,

  1. 26
  2. 42
  3. 25
  4. 41

Solution 6 : Problem analysis and solving in mind by precise problem definition and quadratic equation solution

Imagine $x$ to be the number you are trying to find.

So, $x^2-21x=100$,

Or, $x^2-21x-100=0$,

Or, $(x-25)(x+4)=0$,

So $x=25$.

$25^2$ is 625 and $21\times{25}=525$. Difference is 100. This is result verification if you want.

Answer: Option c: 25.

Key concepts used: Problem definition by forming the quadratic equation -- Solving a quadratic equation -- Result verification -- Solving in mind.

You may also solve it very quickly by number estimation from the relation, $x(x-21)=100$.

Problem 7.

A General of an army wanted to form a square from 36562 armies. After the square formation, he found some armies still left. How many armies were left?

  1. 81
  2. 97
  3. 65
  4. 36

Solution 7: Problem analysis and quick solution by square root estimation

If you assume $36562=n^2 + x$, your job turns out to find $x$.

Without using the standard method of taking square root of a number we'll take the square root of 364, the first three digits from left and will multiply the result with 10.

Testing out squares close to 365 you note,

$19^2=361$, and $20^2=400$.

As 365 is very near to 365, assume the estimated square root of $36500$ to be 190. Actually the square of 190 will be,

$36100$.

Now just test square of 191 as,

$191^2=(190+1)(190+1)$

$=36100+2\times{190} +1$

$= 36100+381=36481$.

This is less than 36562 by,

$36562-36481=81$.

This separation of 81 is too small to accomodate the next square of 192 (just look back to note that 381 had to be added to $190^2$ to get $191^2$).

So 81 is the answer.

This is use of number estimation.

Answer: Option a: 81.

Key concepts used: Nearest square root by mathematical reasoning and number estimation.

A little writing was needed for this solution.

Alternately you can take the square root of 36562 by the standard method and take the integer part as the nearest square root. To find out the exact square root by the standard method, you may refer to the article,

How to take a square root.

Problem 8.

The sum of two positive integers is 80 and the difference between them is 20. What is the difference between squares of these numbers?

  1. 2000
  2. 1600
  3. 1400
  4. 1800

Solution 8: Problem analysis and instant solution by using algebraic product relation

If $a$ and $b$ are the two numbers, you have,

$a+b=80$, and

$a-b=20$.

Take the product of the two and you will get the desired difference between their squares as,

$a^2-b^2=1600$.

It is a case of pattern discovery and applying the corresponding method.

Answer: Option b: 1600.

Key concepts used: Pattern and method discovery -- Instant solution -- Solving in mind.

Problem 9.

Sum of three fractions is $2\frac{11}{24}$. On dividing the largest fraction by the smallest fraction $\displaystyle\frac{7}{6}$ is obtained which is greater than the middle fraction by $\displaystyle\frac{1}{3}$. The smallest fraction is,

  1. $\displaystyle\frac{5}{8}$
  2. $\displaystyle\frac{5}{6}$
  3. $\displaystyle\frac{3}{7}$
  4. $\displaystyle\frac{3}{4}$

Solution 9: Problem analysis and quick solution by target driven substitution and advanced fraction arithmetic

By the third and last condition, the middle fraction value is,

$\displaystyle\frac{7}{6}-\displaystyle\frac{1}{3}=\displaystyle\frac{5}{6}$.

Subtracting this from the sum of three fractions, you'll get the sum of first and third fractions as,

$2\frac{11}{24}-\displaystyle\frac{5}{6}=\displaystyle\frac{13}{8}$.

Assume smallest fraction as $x$.

So largest fraction is,

$\displaystyle\frac{13}{8}-x=\displaystyle\frac{7}{6}\times{x}$, by the second condition,

Or, $\displaystyle\frac{13}{8}=\frac{13}{6}x$,

Or, $x=\displaystyle\frac{6}{8}=\frac{3}{4}$.

Answer: Option d: $\displaystyle\frac{3}{4}$.

Key concepts used: Advanced fraction arithmetic -- Substitution technique -- Target driven substitution -- Mathematical reasoning.

Problem 10.

A certain number when successively divided by 8 and 11 leaves remainder 3 and 7 respectively. Find the remainder if the same number is divided by 88.

  1. 51
  2. 57
  3. 59
  4. 61

Solution 10: Problem analysis and quick solution by Euclid's division lemma and successive division concepts

Assume $q_1$ and $q_2$ to be the quotient of the first and second division and $x$ the number.

For the first division,

$x=8q_1+3$,

And for the second division,

$q_1=11q_2+7$.

Substitute the expression for $q_1$ into the first equation and get,

$x=8(11q_2+7)+3=88q_2+59$.

When the number is divided by 88 remainder will then be 59. In this case new quotient will also be $q_2$.

Answer: Option c: 59.

Key concepts used: Remainder theorem -- Euclid's division lemma -- Concept of successive division - in the second division, the first quotient is divided by the second divisor -- Solving in mind.

This problem can easily be solved in mind if you are clear about the relations between the dividend, divisor, quotient and remainder as well as the mechanism of successive division.


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7 steps for sure success in SSC CGL tier 1 and tier 2 Competitive tests

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Numbers and Number system and basic mathematical operations

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HCF and LCM

Fractions and decimals basic concepts part 1

Basic and Rich Algebra concepts for elegant solutions of SSC CGL problems - though on Algebra, it should be useful.

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