Algebra Questions for SSC CPO, Answer and Solution 1 | Suresolv

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SSC CPO Solved question Set 1, Algebra 1

Algebra questions for SSC CPO with answers and solutions 1

Algebra questions for SSC CPO with answers and quick easy solutions Set 1

Algebra questions for SSC CPO with answers and easy solutions first set. These are previous year SSC CPO questions. Solutions by algebra techniques.

The set contains,

  1. Algebra questions for SSC CPO to be answered in 15 minutes (10 chosen questions)
  2. Answers to the questions, and
  3. Quick conceptual solutions to the questions.

Take the timed test, verify answers and know how to solve questions quick from solutions.

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.

Algebra Questions for SSC CPO 1st set - 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 Algebra questions for SSC CPO 1st set

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: $-4$.


Solutions to the Algebra questions for SSC CPO 1st set - answering time was 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,

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

Split it into three independent equations in $x$, $y$ and $z$ respectively,

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

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: Quick solution by key pattern identification and expanded form of cube of sum

Identify that the target expression is the expanded form of a cube of sum.

Rearrange the the terms of the target expression,

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

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

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

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

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 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

Multiply first given equation by $p$ to form first term of target expression,

$p^2(x+y)^2=5p$.

Same way multiply the second given equation by $-q$ to get the third term of the target expression,

$-q^2(x-y)^2=-3q$.

Substitute in target expression,

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

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

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

Objective is to eliminate $x^2$ and $y^2$ from the two given equations.

Way forward is easy.

Multiply first given equation by $q$ and the second by $p$ and subtract the second result from the first,

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

Or, $4pqxy=5q-3p$.

Substitute in the intermediate result of the target expression,

$E=5p-3q+4pqxy=(5p-3q)+(5q-3p)=2(p+q)$.

At the second stage, the factors of $(x+y)^2$ and $(x-y)^2$ are made equal so that the sum of squares can be interacted against each other. This is a form of base equalization technique.

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.


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