SSC CPO Solved question Set 1, Algebra 1 | SureSolv

SSC CPO Solved question Set 1, Algebra 1

1st SSC CPO Solved Question Set, 1st on Algebra

ssc-cpo-solved-question-set-1-algebra-1

This is the 1st solved question set of 10 practice problem exercise for SSC CPO exam and the 1st on topic Algebra. It contains,

  1. Question set on Algebra for SSC CPO to be answered in 15 minutes (10 chosen questions)
  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 read.

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.

1st Question set - 10 problems for SSC CPO exam: 1st on topic Algebra - answering time 15 mins

Q1. What is the value of $\displaystyle\frac{(a^2+b^2)(a-b)-(a-b)^3}{a^2 b-ab^2}$?

  1. $-1$
  2. $0$
  3. $1$
  4. $2$

Q2. What is the value of $\left(\displaystyle\frac{x^2-x-6}{x^2+x-12}\right) \div \left(\displaystyle\frac{x^2+5x+6}{x^2+7x+12}\right)$?

  1. $1$
  2. $\displaystyle\frac{x-3}{x+4}$
  3. $\displaystyle\frac{x-3}{x+3}$
  4. $\displaystyle\frac{x+4}{x-3}$

Q3. If $\displaystyle\frac{1}{x+2}=\displaystyle\frac{3}{y+3}=\displaystyle\frac{1331}{z+1331}=\displaystyle\frac{1}{3}$, then what is the value of $\displaystyle\frac{x}{x+1}+\displaystyle\frac{4}{y+2}+\displaystyle\frac{z}{z+2662}$?

  1. $0$
  2. $1$
  3. $3$
  4. $\displaystyle\frac{3}{2}$

Q4. If $p=\displaystyle\frac{5}{18}$, then $27p^3-\displaystyle\frac{1}{216}-\displaystyle\frac{9}{2}p^2+\displaystyle\frac{1}{4}p$ is equal to,

  1. $\displaystyle\frac{10}{27}$
  2. $\displaystyle\frac{8}{27}$
  3. $\displaystyle\frac{5}{27}$
  4. $\displaystyle\frac{4}{27}$

Q5. If $p(x+y)^2=5$ and $q(x-y)^2=3$, then the simplified value of $p^2(x+y)^2+4pqxy-q^2(x-y)^2$ is,

  1. $-(p+q)$
  2. $-2(p+q)$
  3. $2(p+q)$
  4. $(p+q)$

Q6. What is the simplified value of $\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^8+\displaystyle\frac{1}{x^8}\right)\left(x-\displaystyle\frac{1}{x}\right)$

$\hspace{10mm}\times{\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)\left(x+\displaystyle\frac{1}{x}\right)\left(x^4+\displaystyle\frac{1}{x^4}\right)}$?

  1. $\displaystyle\frac{x^{32}-\displaystyle\frac{1}{x^{32}}}{x+\displaystyle\frac{1}{x}}$
  2. $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x+\displaystyle\frac{1}{x}}$
  3. $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x^2+\displaystyle\frac{1}{x^2}}$
  4. $x^{64}+\displaystyle\frac{1}{x^{64}}$

Q7. If $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$ and $x+y+z=11$, then what is  the value of $x^3+y^3+z^3-3xyz$?

  1. $2662$
  2. $14641$
  3. $1331$
  4. $3993$

Q8. If $x=\sqrt[3]{7}+3$, then the value of $x^3-9x^2+27x-34$ is,

  1. $1$
  2. $0$
  3. $-1$
  4. $2$

Q9. If $\left(x+\displaystyle\frac{1}{x}\right)=3\sqrt{2}$, then what is the value of $\left(x^5+\displaystyle\frac{1}{x^5}\right)$?

  1. $717\sqrt{2}$
  2. $789\sqrt{2}$
  3. $1581\sqrt{2}$
  4. $178\sqrt{3}$

Q10. If $\displaystyle\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\displaystyle\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=62$, then what is the value of $x$ $(x \lt 0)$?

  1. $3$
  2. $-4$
  3. $16$
  4. $0$

Answers to the questions

Q1. Answer: Option d: $2$.

Q2. Answer: Option a: $1$.

Q3. Answer: Option d: $\displaystyle\frac{3}{2}$.

Q4. Answer: Option b: $\displaystyle\frac{8}{27}$.

Q5. Answer: Option c: $2(p+q)$.

Q6. Answer: Option c: $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x^2+\displaystyle\frac{1}{x^2}}$.

Q7. Answer: Option c: $1331$.

Q8. Answer: Option b: $0$.

Q9. Answer: Option a: $717\sqrt{2}$.

Q10. Answer: Option b: $-4$.


1st solution set - 10 problems for SSC CPO exam: 1st on topic Algebra - answering time 15 mins

Q1. What is the value of $\displaystyle\frac{(a^2+b^2)(a-b)-(a-b)^3}{a^2 b-ab^2}$?

  1. $-1$
  2. $0$
  3. $1$
  4. $2$

Solution 1: Quick solution by key pattern identification of common factor between numerator and denominator

First identify the common factor of $(a-b)$ between numerator and denominator and eliminate it. Further simplification is straightforward.

The target expression is,

$\displaystyle\frac{(a^2+b^2)(a-b)-(a-b)^3}{a^2b-ab^2}$

$=\displaystyle\frac{(a-b)[(a^2+b^2)-(a-b)^2]}{ab(a-b)}$

$=\displaystyle\frac{2ab}{ab}=2$.

Answer: Option d: $2$.

Key concepts used: Key pattern identification -- Factorization and common factor elimination -- Solving in mind.

Q2. What is the value of $\left(\displaystyle\frac{x^2-x-6}{x^2+x-12}\right) \div \left(\displaystyle\frac{x^2+5x+6}{x^2+7x+12}\right)$?

  1. $1$
  2. $\displaystyle\frac{x-3}{x+4}$
  3. $\displaystyle\frac{x-3}{x+3}$
  4. $\displaystyle\frac{x+4}{x-3}$

Solution 2: Factorization of quadratic equation and common factor identification and elimination

Identify the key pattern that all four quadratic equations in the target expression can easily be factorized. Next step is just cancellation of common factors between numerator and denominator.  

The target expression,

$\left(\displaystyle\frac{x^2-x-6}{x^2+x-12}\right) \div \left(\displaystyle\frac{x^2+5x+6}{x^2+7x+12}\right)$

$=\displaystyle\frac{(x-3)(x+2)}{(x+4)(x-3)} \times{ \displaystyle\frac{(x+4)(x+3)}{(x+2)(x+3)}}$, inverting the second term which is the dividend

$=1$, all four pairs of factors cancel out between numerator and denominator.

Answer: Option a: $1$.

Key concepts used: Quadratic equation factorization -- Key pattern identification -- Solving in mind.

Q3. If $\displaystyle\frac{1}{x+2}=\displaystyle\frac{3}{y+3}=\displaystyle\frac{1331}{z+1331}=\displaystyle\frac{1}{3}$, then what is the value of $\displaystyle\frac{x}{x+1}+\displaystyle\frac{4}{y+2}+\displaystyle\frac{z}{z+2662}$?

  1. $0$
  2. $1$
  3. $3$
  4. $\displaystyle\frac{3}{2}$

Solution 3: Quick solution by splitting chained equation into three independent equations, evaluation of variable values and substitution

The given is a chained equation that is to be first split into three standalone equations and $x$, $y$ and $z$ evaluated.

The given equation is,

$\displaystyle\frac{1}{x+2}=\displaystyle\frac{3}{y+3}=\displaystyle\frac{1331}{z+1331}=\displaystyle\frac{1}{3}$.

First split the chained equation into three independent equations equating each of the first three LHS with the numeric RHS, the fourth term from left. This would enable you to get the values of $x$, $y$ and $z$ directly.

The results are,

$\displaystyle\frac{1}{x+2}=\displaystyle\frac{1}{3}$,

Or, $x+2=3$,

Or, $x=1$.

$\displaystyle\frac{3}{y+3}=\displaystyle\frac{1}{3}$,

Or, $y+3=9$,

Or, $y=6$.

$\displaystyle\frac{1331}{z+1331}=\displaystyle\frac{1}{3}$,

Or, $z+1331=3\times{1331}$,

Or, $z=2662$.

Substitute these variable values in the target expression,

$\displaystyle\frac{x}{x+1}+\displaystyle\frac{4}{y+2}+\displaystyle\frac{z}{z+2662}$

$=\displaystyle\frac{1}{2}+\displaystyle\frac{4}{6+2}+\displaystyle\frac{2\times{1331}}{4\times{1331}}$

$=\displaystyle\frac{3}{2}$.

Answer: Option d: $\displaystyle\frac{3}{2}$.

Key concepts used: Splitting of chained equation into three independent equations to evaluate $x$, $y$ and $z$ --  Chained equation treatment technique -- Substitution -- Solving in mind.

Q4. If $p=\displaystyle\frac{5}{18}$, then $27p^3-\displaystyle\frac{1}{216}-\displaystyle\frac{9}{2}p^2+\displaystyle\frac{1}{4}p$ is equal to,

  1. $\displaystyle\frac{10}{27}$
  2. $\displaystyle\frac{8}{27}$
  3. $\displaystyle\frac{5}{27}$
  4. $\displaystyle\frac{4}{27}$

Solution 4: Solving in mind by key pattern identification of target expression as an expanded cube of sum and then getting the value of cubed expression from the given equation

Identify that the target expression is the expanded form of a cube of sum by rearranging the terms,

$27p^3-\displaystyle\frac{1}{216}-\displaystyle\frac{9}{2}p^2+\displaystyle\frac{1}{4}p$

$=27p^3-\displaystyle\frac{9}{2}p^2+\displaystyle\frac{1}{4}p-\displaystyle\frac{1}{216}$

$=(3p)^3-3.(3p)^2.\left(\displaystyle\frac{1}{6}\right)+3.(3p).\left(\displaystyle\frac{1}{6}\right)^2-\left(\displaystyle\frac{1}{6}\right)^3$

$=\left(3p-\displaystyle\frac{1}{6}\right)^3$.

From the given equation get the value of $\left(3p-\displaystyle\frac{1}{6}\right)$ as,

$p=\displaystyle\frac{5}{18}$,

Or, $3p=\displaystyle\frac{5}{6}$,

Or, $\left(3p-\displaystyle\frac{1}{6}\right)=\displaystyle\frac{4}{6}=\frac{2}{3}$.

Substitute this value of cubed expression in the transformed target expression,

$\left(3p-\displaystyle\frac{1}{6}\right)^3=\left(\displaystyle\frac{2}{3}\right)^3=\displaystyle\frac{8}{27}$.

Answer: Option b: $\displaystyle\frac{8}{27}$.

Key concepts used: Key pattern identification -- Cube of sum expansion -- Substitution-- Solving in mind.

Q5. If $p(x+y)^2=5$ and $q(x-y)^2=3$, then the simplified value of $p^2(x+y)^2+4pqxy-q^2(x-y)^2$ is,

  1. $-(p+q)$
  2. $-2(p+q)$
  3. $2(p+q)$
  4. $(p+q)$

Solution 5: Solve quickly by key pattern identification, base equalization and two stage substitution

It is apparent that first and third term of the target expression are easily simplified by direct substitution of the RHSs from the given equations,

$p^2(x+y)^2+4pqxy-q^2(x-y)^2$

$=5p+4pqxy-3q$.

Question is, how to transform the middle term $4pqxy$ in terms  of $p$ and $q$.

Easiest way to do this is to first equalize the factors $p$ and $q$ in the LHSs of the two given equations to $pq$ by multiplying the first equation by $q$ and the second by $p$. This is one form of application of base equalization technique, where bases are the factors, $p$ and $q$.

Now if you subtract the second result from the first, the difference of $(x+y)^2$ and $(x-y)^2$ becomes simply $4xy$ giving you $4pqxy$ in the LHS of the subtraction result. The RHS is in terms of only $p$ and $q$.

Let us show you the deductions. 

Multiplying the first given equation by $q$, the second given equation by $p$ and subtracting the second result from the first,

$pq(x+y)^2-pq(x-y)^2=5q-3p$,

Or, $5q-3p=pq\left[(x+y)^2-(x-y)^2\right]$,

Or, $5q-3p=pq(4xy)=4pqxy$.

Substitute this result in the transformed target expression,

$5p+4pqxy-3q$

$=5p+(5q-3p)-3q=2(p+q)$.

Answer: Option c. $2(p+q)$.

Key concepts used: Key pattern identification --  Base equalization technique -- Two stage given expression transformation and substitution -- Solving in mind.

Q6. What is the simplified value of $\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^8+\displaystyle\frac{1}{x^8}\right)\left(x-\displaystyle\frac{1}{x}\right)$

$\hspace{10mm}\times{\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)\left(x+\displaystyle\frac{1}{x}\right)\left(x^4+\displaystyle\frac{1}{x^4}\right)}$?

  1. $\displaystyle\frac{x^{32}-\displaystyle\frac{1}{x^{32}}}{x+\displaystyle\frac{1}{x}}$
  2. $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x+\displaystyle\frac{1}{x}}$
  3. $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x^2+\displaystyle\frac{1}{x^2}}$
  4. $x^{64}+\displaystyle\frac{1}{x^{64}}$

Solution 6: Solving in mind by missing element identification and combining the like factors

Identify that if you multiply the third and fifth factors you would get a promising result,

$\left(x-\displaystyle\frac{1}{x}\right)\left(x+\displaystyle\frac{1}{x}\right)=\left(x^2-\displaystyle\frac{1}{x^2}\right)$.

But, now among the four other remaining factors, you don't have $\left(x^2+\displaystyle\frac{1}{x^2}\right)$.

If you had it, you could have multiplied your current result with this sum of inverses of squares in $x$ getting another promising result of subtractive sum of inverses in 4th power of $x$, that is, $\left(x^4-\displaystyle\frac{1}{x^4}\right)$.

This is your missing element. You don't have it in the problem expression. No problem, introduce it by multiplying and dividing the last resultant expression by $\left(x^2+\displaystyle\frac{1}{x^2}\right)$.

This is the technique of missing element identification and introduction.

Let us show you the result,

$\displaystyle\frac{1}{\left(x^2+\displaystyle\frac{1}{x^2}\right)}\times{\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^8+\displaystyle\frac{1}{x^8}\right)\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)}$

$\hspace{8mm}\times{\left(x^2-\displaystyle\frac{1}{x^2}\right)\left(x^2+\displaystyle\frac{1}{x^2}\right)\left(x^4+\displaystyle\frac{1}{x^4}\right)}$.

Now it is straightforward combining like factors (or in general, like terms).

Combine 4th and 5th factors to get the promising factor of $\left(x^4-\displaystyle\frac{1}{x^4}\right)$. Next combine it with $\left(x^4+\displaystyle\frac{1}{x^4}\right)$ to get, $\left(x^8-\displaystyle\frac{1}{x^8}\right)$. And continue to combine this way.

Following are the results,

$\displaystyle\frac{1}{\left(x^2+\displaystyle\frac{1}{x^2}\right)}\times{\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)\left(x^8+\displaystyle\frac{1}{x^8}\right)}$

$\hspace{8mm}\times{\left(x^4-\displaystyle\frac{1}{x^4}\right)\left(x^4+\displaystyle\frac{1}{x^4}\right)}$

$=\displaystyle\frac{1}{\left(x^2+\displaystyle\frac{1}{x^2}\right)}\times{\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)\left(x^8+\displaystyle\frac{1}{x^8}\right)}$

$\hspace{8mm}\times{\left(x^8-\displaystyle\frac{1}{x^8}\right)}$

$=\displaystyle\frac{1}{\left(x^2+\displaystyle\frac{1}{x^2}\right)}\times{\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^{16}+\displaystyle\frac{1}{x^{16}}\right)\left(x^{16}-\displaystyle\frac{1}{x^{16}}\right)}$

$=\displaystyle\frac{1}{\left(x^2+\displaystyle\frac{1}{x^2}\right)}\times{\left(x^{32}+\displaystyle\frac{1}{x^{32}}\right)\left(x^{32}-\displaystyle\frac{1}{x^{32}}\right)}$

$=\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x^2+\displaystyle\frac{1}{x^2}}$.

Answer: Option c: $\displaystyle\frac{x^{64}-\displaystyle\frac{1}{x^{64}}}{x^2+\displaystyle\frac{1}{x^2}}$.

Key concepts used: Key pattern identification -- Technique of missing element identification and introduction -- Combining like factors (terms) -- Principle of collection of like terms -- Solving in mind.

If you can identify the missing element and are able to introduce it, solution should take a few tens of seconds. Explaining and writing the deductive steps take large space and time, you know.

Q7. If $\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=0$ and $x+y+z=11$, then what is the value of $x^3+y^3+z^3-3xyz$?

  1. $2662$
  2. $14641$
  3. $1331$
  4. $3993$

Solution 7: Quick solution by key pattern identification, square of three variable sum and expanded form of three variable sum of cubes

The first key pattern identified is by evaluation of $xy+yz+zx$ from given equation,

$\displaystyle\frac{1}{x}+\displaystyle\frac{1}{y}+\displaystyle\frac{1}{z}=\displaystyle\frac{xy+yz+zx}{xyz}=0$,

Or, $xy+yz+zx=0$.

Now let us use the expanded form of three variable sum of cubes to determine what more are required to evaluate the target expression,

$x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$,

So target expression,

$E=x^3+y^3+z^3-3xyz=11(x^2+y^2+z^2)$.

It remains only to evaluate $x^2+y^2+z^2$.

Value of this expression we would get easily by the three variable square of sum,

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$,

Or, $x^2+y^2+z^2=(x+y+z)^2=(11)^2=121$.

So value of target expression is,

$E=11(x^2+y^2+z^2)=11\times{121}=1331$.

Answer: Option: c: $1331$.

Key concepts used: Key pattern identification -- Three variable square of sum -- Three variable sum of cubes -- Solving in mind.

If you know the expanded relations well, you should be able to solve the problem mentally.

Q8. If $x=\sqrt[3]{7}+3$, then the value of $x^3-9x^2+27x-34$ is,

  1. $1$
  2. $0$
  3. $-1$
  4. $2$

Solution 8: Quick solution by key pattern identification of similarity of target expression with expanded $(x-3)^3$ from given expression

The key pattern identified from the given expression is the similarity of the expanded $(x-3)^3$ with the target expression,

$(x-3)^3=x^3-9x^2+27x-27$.

The first three terms of the target expression are same as the expanded form of $(x-3)^3$. This is use of End state analysis.

So we would transform the given expression to,

$x=\sqrt[3]{7}+3$,

Or, $x-3=\sqrt[3]{7}$.

And then raise this transformed equation to its cube to get,

$(x-3)^3=7$,

Or, $x^3-9x^2+27x-27=7$,

Or, $x^3-9x^2+27x-34=0$.

Answer: Option b: $0$.

Key concepts used: Key pattern identification of maximum similarity of cube of modified given expression with target expression -- End state analysis approach -- Cube of sum expansion -- Solving in mind.

Q9. If $\left(x+\displaystyle\frac{1}{x}\right)=3\sqrt{2}$, then what is the value of $\left(x^5+\displaystyle\frac{1}{x^5}\right)$?

  1. $717\sqrt{2}$
  2. $789\sqrt{2}$
  3. $1581\sqrt{2}$
  4. $178\sqrt{3}$

Solution 9: Quick solution by raising power of $x$ in sum of inverses using principle of inetraction of inverses

If you raise the given sum of inverses to its square, the mutually inverse variables in the middle term cancel out to leave just a numeric value,

$\left(x+\displaystyle\frac{1}{x}\right)=3\sqrt{2}$,

Or, $\left(x+\displaystyle\frac{1}{x}\right)^2=18$,

Or, $x^2+\displaystyle\frac{1}{x^2}=16$.

Now we will use the two factor expanded form of $\left(x^3+\displaystyle\frac{1}{x^3}\right)$,

$x^3+\displaystyle\frac{1}{x^3}=\left(x+\displaystyle\frac{1}{x}\right)\left(x^2-1+\displaystyle\frac{1}{x^2}\right)$

$=3\sqrt{2}\times{(16-1)}=45\sqrt{2}$.

In the last step, we would multiply $\left(x^2+\displaystyle\frac{1}{x^2}\right)$ with $\left(x^3+\displaystyle\frac{1}{x^3}\right)$ to get the value of $\left(x^5+\displaystyle\frac{1}{x^5}\right)$,

$\left(x^2+\displaystyle\frac{1}{x^2}\right)\left(x^3+\displaystyle\frac{1}{x^3}\right)$

$=\left(x^5+\displaystyle\frac{1}{x^5}\right)+\left(x+\displaystyle\frac{1}{x}\right)$.

So,

$\left(x^5+\displaystyle\frac{1}{x^5}\right)=16\times{45\sqrt{2}}-3\sqrt{2}=717\sqrt{2}$.

Answer: Option a: $717\sqrt{2}$.

Key concepts used: Key pattern identification of similarity of target expression with product of sum of inverses of squares with sum of inverses of cubes -- Two factor expansion of sum of cubes -- Principle of interaction of inverses -- Solving in mind.

With clear concepts, quick decision making based on key pattern identification and reasonably accurate mental math skill, it should easily be possible to solve this problem in mind.

Q10. If $\displaystyle\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\displaystyle\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=62$, then what is the value of $x$ $(x \lt 0)$?

  1. $3$
  2. $-4$
  3. $16$
  4. $0$

Solution 10: Quick solution by identifying simple numerator and denominator of two LHS terms combined

Assume dummy variable $p$ for $\sqrt{x^2-1}$,

$p=\sqrt{x^2-1}$.

This dummy variable substitution is not necessary but it makes mental visualization of the solution steps comfortably easy.

By this substitution the given equation is simplified to,

$\displaystyle\frac{x+p}{x-p}+\displaystyle\frac{x-p}{x+p}=62$,

Or, $\displaystyle\frac{(x+p)^2+(x-p)^2}{x^2-p^2}=62$,

Or, $4x^2-2=62$, as denominator $x^2-p^2=1$,

Or, $x^2=16$,

Or, $x=-4$, as by given condition $x$ must be negative.

Answer: Option b: $-4$.

Key concepts used: Dummy variable substitution -- Simplified result of sum of $(x+p)^2$ and $(x-p)^2$ as $2(x^2+p^2)$ -- Solving in mind.

End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of mental steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

Concept based pattern and method formation, and,

Identification of the key pattern and use of the method associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the concept based pattern identification and use of quick problem solving method.


SSC CPO level Question and Solution sets

Algebra

SSC CPO level Solved Question set 1 on Algebra 1

SSC CPO level Solved Question set 2 on Algebra 2

Trigonometry

SSC CPO level Solved Question set 3 on Trigonometry 1

SSC CPO level Solved Question set 4 on Trigonometry 2

Number system

SSC CPO level Solved Question set 5 on Number system 1

Surds and Indices

SSC CPO level Solved Question set 6 on Surds and indices 1