SSC CGL level Solution Set 76, Percentage 4 | SureSolv

SSC CGL level Solution Set 76, Percentage 4

76th SSC CGL level Solution Set, 4th on topic Percentage

ssc cgl solution set 76 percentage 4

This is the 76th solution set of 10 practice problem exercise for SSC CGL exam and 4th on topic Percentage. In most of the solutions special techniques and methods based on fundamental concepts have been used for solving the problems as quickly as possible, generally in mind.

If you have not taken the corresponding test yet, go through the test first by referring to, SSC CGL Question set 76 Percentage 4 and then only go through this solution set for gaining maximum benefits from this resource.

76th solution set - 10 problems for SSC CGL exam: 4th on topic Percentage - time 12 mins

Problem 1.

If $p\text{% of } p$ is 36 and $q\text{% of } q$ is 48, the ratio, $p:q$ is

  1. $3:4$
  2. $4:3$
  3. $3:8$
  4. $1:4$

Solution 1: Problem analysis and solving execution

By percentage conversion to decimal (divide the percentage figure by 100), the first statement translates to,

$p\times{0.01}=36$

Or, $.01p=36$, and the second statement to,

$q\times{0.01}=48$,

Or, $0.01q=48$.

Taking the ratio and simplifying,

$p:q=36:48=3:4$.

We have used just the very basic concepts and percentage conversions,

Percentage with a value of $x$ equals $\displaystyle\frac{1}{100}x=0.01x$.

Secondly, as the target requirement is ratio of $p$ and $q$, actual value calculation was not required, just  simple division of two equations gave us the result.

Answer: Option a: $3:4$.

Key concepts used: Basic percentage concepts -- Percentage to decimal conversion -- Efficient simplification, we didn't calculate the values of $p$ and $q$ as just their ratio was needed.

Problem 2.

In a school 40% of the students play football and 50% of the students play cricket. If 18% of the students play neither football nor cricket, percentage of students playing both is,

  1. 22%
  2. 40%
  3. 8%
  4. 32%

Solution 2 : Problem analysis and execution

Among the 40% of the students who play football there is a portion of students who play cricket also. These are the students who play both football and cricket.

When we say 50% of the students play cricket, in the same way, a portion of students play both football and cricket. This portion appears in both the 40% and 50% portions. This is common to both 40% and 50%.

If we assume this portion of the students who play both football and cricket as $x$% then, when we add portion of the students who play football and the portion of the students who play cricket, the total of $40+50=90$% contains $x$% twice, that is, $2x$%.

So the exact portion of students who play either football or cricket or both is,

$(90-x)$%, we have rectified the overcounting of $x$% twice by deducting it once from the total.

This can be shown pictorially in the Venn diagram as below,

venn-diagram-set-union-exclusion-percentage-4.jpg

This is set union concept.

Our target is to find portion of students who play neither football nor cricket.

Knowing that total portion of the students is 100% and portion of students who play either football or cricket or both is, $(90-x)$%, the portion of the students who play neither football nor cricket is the difference of the two, that is,

$100-(90-x)=18$, given is $18$%,

Or, $x=8$%.

The portion subtraction is set exclusion concept similar to arithmetic subtraction. Sets are the portions here and the concepts are the part of basic set theoretic concepts.

Answer: Option c : 8%.

Key concepts used: Basic percentage concepts -- Base Set theoretic concepts -- Set union and Set exclusion -- Percentage as actual values, all values were percentages, we never needed actual values.

Effectively there are four distinct portions or subsets of students.

First: Portion of students who play only football but not cricket. This is, $(40-8)=32$% of total students.

Second: Portion of students who play only cricket but not football. This is, $(50-8)=42$% of total students.

Third: Portion of students who play both football and cricket. This is, $8$% of students, and lastly,

Fourth: Portion of students who play neither football nor cricket. This is, $18$% of the total students.

These four portions together make up all the students, that is, 100% of the students.

When we say 40% of the students play football, this portion actually is a sum of first and third portions.

Problem 3.

If 40% of $P$ equals 25% of $Q$, then 60% of $P$ is $x$% of $Q$. Find $x$.

  1. 30
  2. 37.5
  3. 6.25
  4. 0.625

Solution 3 : Problem analysis, pattern identification and execution

Identifying the pattern that 40% of $P$ is given and 60% of $P$ is desired, we decide to use the ever dependable three-step unitary method without going into percentage conversion at all.

First step: 40% of $P$ equals 25% of $Q$,

Second step: So, 10% of $P$ equals $\displaystyle\frac{25}{4}$% of $Q$,

Third step: So, 60% of $P$ would equal $\displaystyle\frac{25}{4}\times{6}=37.5$% of $Q$.

Answer: Option b: $37.5$%.

Key concepts used: Key pattern identification -- Unitary method, bypassing percentage treatment altogether.

Problem 4.

The sum of two positive numbers is 20% of sum of their squares and 25% of the difference of their squares. If the numbers are $x$ and $y$, then $\displaystyle\frac{x+y}{x^2}$ is,

  1. $\displaystyle\frac{3}{8}$
  2. $\displaystyle\frac{1}{3}$
  3. $\displaystyle\frac{1}{4}$
  4. $\displaystyle\frac{2}{9}$

Solution 4 : Problem analysis and strategy decision

By the given compound statement we have the chained equality,

$x+y=0.2(x^2 + y^2)=0.25(x^2-y^2)$.

This is effectively three equations between the three parts. Analyzing the target expression with respect to the given equations we decide to eliminate $(x^2+y^2)$ by dividing two ratios with denominators as $(x^2+y^2)$. One of the numerators would be $x+y$ and the other $x^2$.

We would get the first ratio directly from the first part of the continued equality,

$x+y=0.2(x^2+y^2)$,

Or, $\displaystyle\frac{x+y}{x^2+y^2}=0.2=\displaystyle\frac{1}{5}$.

The second ratio we will get from the second part of the given equality. Let us see how.

Solution 4: Problem solving execution

By the continued equality statement we have the relation,

$0.2(x^2+y^2)=0.25(x^2-y^2)$.

Simplifying,

$\displaystyle\frac{x^2-y^2}{x^2+y^2}=\displaystyle\frac{4}{5}$

Adding 1 to both sides and simplifying,

$\displaystyle\frac{x^2}{x^2+y^2}=\frac{9}{10}$.

Dividing the earlier ratio, $(x^2+y^2)$ cancels out,

$\displaystyle\frac{x+y}{x^2}=\displaystyle\frac{2}{9}$.

The problem turns out to be more of algebraic manipulation rather than percentage treatments.

Answer: Option d: $\displaystyle\frac{2}{9}$.

Key concepts used: Basic percentage concepts -- Percentage to decimal conversion -- Algebraic manipulation with the objective of producing the desired expression.

Problem 5.

If $15\text{% of }(A+B)=25\text{% of }(A-B)$, then what percent of B is A?

  1. 10%
  2. 400%
  3. 60%
  4. 200%

Solution 5 : Problem analysis and execution

Using percentage to decimal conversion,

$0.15(A+B)=0.25(A-B)$,

Or, $0.1A=0.4B$,

Or, $A =4B= 400\text{ % of } B$.

Using integer to percent conversion, $A=4B$ means $A$ is 4 times of $B$ or, $400\text{% of }B$.

Answer: Option b: 400%.

Key concepts used:Percentage to decimal conversion -- Integer to percentage conversion -- Percentage concept.

Problem 6.

A number is divided into two parts in such a way that 80% of the 1st part is 3 more than 60% of the 2nd part and 80% of the 2nd part is 6 more than 90% of the first part. Then the number is,

  1. 135
  2. 125
  3. 145
  4. 130

Solution 6 : Problem analysis and pattern identification

By the first statement,

$0.8x=0.6y +3$, and by the second statement,

$0.8y=0.9x+6$, where $x$ and $y$ are the first and second parts of the number desired.

By solving the two linear equations we would get the values of $x$ and $y$ easily. This is the conventional approach.

Instead, with the objective of finding the answer in shortest possible time we examine the two equations closely and discover the key pattern,

As the coefficients of the RHSs of both equations are multiples of 3 and that of LHSs are not multiples of 3, both $x$ and $y$ must be multiples of 3. So their sum, the target number, must also be a multiple of 3.

We will use this result against the choice values to arrive at the result in a few seconds.

Solution 6: Lightning fast solution using factors multiples concept: Mathematical reasoning

As only 135 of the four choice values is a multiple of 3, that is the answer.

Note: If there were more than one choice values as multiple of 3, we would have gone the conventional way wasting only a few seconds. But we would in any case go along with this test as in case of success, time gain would be significant.

The conventional solution follows.

Solution 6: Conventional solution by solving two linear equations in two variables

Let us repeat the two equations,

$0.8x=0.6y+3$, and,

$0.8y=0.9x+6$.

To solve two linear equations usual approach is to eliminate one variable by equalizing its coefficients in the two equations by multiplying the two equations by suitable factors.

Instead, we multiply only the first equation by the fraction $\displaystyle\frac{4}{3}$ to equalize the coefficient of $y$ here, getting,

$0.8y+4=\displaystyle\frac{3.2}{3}x$.

Subtracting the second equation from this result,

$4=\displaystyle\frac{0.5}{3}x-6$,

Or, $x=60$.

Using this value of $x$ in equation 1, we get,

$48=0.6y+3$,

Or, $y=75$.

The desired number is then,

$60+75=135$.

It is up to you which method to choose. Generally though, if you can use factors multiples concept, answer is nearly instantaneous.

Answer: Option a : $135$.

Key concepts used: Basic percentage concepts  -- percentage to decimal conversion -- linear equations in two varibales  -- factors multiples concept -- Quick solution -- Choice value test.

Problem 7.

A reduction of 20% in the price of sugar enables a purchaser to obtain 8 kgs more for Rs.160. Then the price per kg before reduction was,

  1. Rs.4
  2. Rs.6
  3. Rs.5
  4. Rs.10

Solution 7 : Problem analysis and execution

If $P$ were the original sugar price per kg and $Q$ were the original quantity of sugar purchased before price fall,

$Q=\displaystyle\frac{160}{P}$, and after price fall,

$Q+8=\displaystyle\frac{160}{0.8P}$.

Subtracting the first from the second,

$8=\displaystyle\frac{160}{P}\left(\displaystyle\frac{5}{4}-1\right)$,

Or, $32P=160$,

Or, $P=5$.

Answer: Option c: Rs.5.

Key concepts used: Basic percentage concepts -- Percentage to decimal conversion -- Decimal to fraction conversion -- Change analysis technique.

Note: By equating the change in weight of 8 kgs to 160 multiplied by,

$\text{Change in inverse of }P=\left(\displaystyle\frac{5}{4}-1\right)=\displaystyle\frac{1}{4}$, as coefficient of inverse of $P$,

answer could been obtained a bit faster, but the concept would have been a bit more difficult to understand and apply.

Essentially this is adapted change analysis technique.

Problem 8.

The strength of a school increases and decreases in every alternate year by 10%. It started with increase in 2000. Then the strength of the school in 2003 as compared to that in 2000 was,

  1. increased by 9.8%
  2. increased by 8.9%
  3. decreased by 9.8%
  4. decreased by 8.9%

Solution 8 : Problem analysis and execution

At the beginning of 2001 the strength increased from that of 2000 by 10%. So,

$S_{2001}=1.1S_{2000}$, where $S_{2000}$ and $S_{2001}$ are the strengths at the start of 2000 and 2001.

Similarly at the end of 2001 because of reduction of strength of 2001 by 10%,

$S_{2002}=0.9S_{2001}$, and likewise,

$S_{2003}=1.1S_{2002}$.

Using the earlier values,

$S_{2003}=1.1\times{0.9}\times{1.1}S_{2000}$

$=1.089S_{2000}$

$=S_{2000}+0.089S_{2000}$.

Multiplying the increment decimal coefficient $0.089$ by 100, we get the increase as, $8.9$%.

Thus the strength in 2003 increased by 8.9% from the strength in 2000.

Mark that the percentage is on the strength of 2000.

Answer: Option b: increased by 8.9%.

Key concepts used: Basic percentage concepts -- Series of percentage increments and decrements --  Percentage reference concept -- Percentage to decimal conversion -- Decimal to percentage conversion.

Problem 9.

The price of an article was first increased by 10% and then again by 20%. If the last increased price was Rs.33, the original price was,

  1. Rs.25
  2. Rs.27.50
  3. Rs.26.50
  4. Rs.30

Solution 9 : Problem analysis and execution

If $P$ were the original price, after the first increase the price is,

$=1.1P$, and after the second increase,

$=1.2(1.1P)=1.32P=33$

So,

$P=25$.

For the second increase of 20%, the percentage reference is changed from $P$ to $(1.1P)$. This is application of Percentage reference concept.

Answer: Option a: Rs. 25.

Key concepts used: Basic percentage concepts -- Consecutive percentage increments -- Percentage reference concept -- Percentage to decimal conversion.

Problem 10.

Fresh fruit contains 68% of water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits?

  1. 52 kgs
  2. 32 kgs
  3. 40 kgs
  4. 80 kgs

Solution 10 : Problem analysis and execution

To solve this typical problem you have to understand the concept of fruit as fruit matter plus water and the drying process. By drying process the fruit matter weight remains unchanged while the weight of water gets reduced when the fruit is dried.

The second important point you must take care of is the percentage reference of water in fruit—it is always the percentage of the total weight of the fresh fruit or the dry fruit that is being referred to.

For example, water in fresh fruit is 68% and so its weight will be 68 kgs in 100 kgs of fresh fruit. What will be the weight of fruit matter in 100 kgs of fresh fruit? It will simply be, $(100-68)=32$ kgs. This fruit matter won't change in weight as the fresh fruit gets dry.

As the water in dry fruit is 20%, the fruit matter must be the rest, that is 80% of the total weight of the dry fruit.

Now we apply unitary method,

Step 1: 80% of the dry fruit is 32 kgs,

Step 2 and 3: 100% of the dry fruit is $32\times{\displaystyle\frac{5}{4}}=40$ kgs.

Answer: Option c: 40 kgs.

Key concepts used: Basic percentage concepts -- Fruit drying concept -- Event visualization, fruit matter remains unchanged in weight when fruit is dried, only the water gets reduced -- Percentage reference concept, the percentage weight of 20% of water in dry fruit is percentage of total weight of dry fruit, and not of fresh fruit.


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