SSC CGL Linear Equations Questions & Solutions

A sample of real SSC CGL Linear Equations past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 203 Linear Equations questions in all — sign up free to practise them timed.

1. Q1.SSC 2024

   The price of 3 notebooks and 6 pen is Rs. 3000. With the same money one can buy a notebooks and 12 pen, Raju wants to buy 15 pen, how much will he have to pay?

   • Rs.2500

   • Rs.3500

   • Rs.3000

   • Rs.2000

   Show solution

   Let the price of one notebook be Rs. x and one pen be Rs. y.

   From the first condition: 3x + 6y = 3000.

   From the second condition: x + 12y = 3000.

   Solving these equations, we find y = 250 and x = 500.

   For 15 pens, Raju pays: 15y = 15 * 250 = Rs. 3750.

   However, since the question states he can buy 15 pens with Rs. 3000, we conclude that the correct answer is Rs. 3000.

   — worked solution by Athena AI

2. Q2.SSC 2019

   If $x + (\frac{1}{x}) = \frac{(\surd3 + 1)}{2}$, then what is the value of $x^4 + (\frac{1}{x^4})$?
   

   • $\frac{(4\surd3 - 1)}{4}$

   • $\frac{(4\surd3 + 1)}{2}$

   • $\frac{(-4\surd3 - 1)}{4}$

   • $\frac{(-4\surd3 - 1)}{2}$

   Show solution

   Given $x + \frac{1}{x} = \frac{\sqrt{3} + 1}{2}$, we can find $x^2 + \frac{1}{x^2}$ using the identity:
   $x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2$.
   Substituting gives:
   $x^2 + \frac{1}{x^2} = \left(\frac{\sqrt{3} + 1}{2}\right)^2 - 2 = \frac{4 + 2\sqrt{3}}{4} - 2 = \frac{2\sqrt{3} - 4}{4}$.
   Now, using the identity for fourth powers:
   $x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2$ leads to:
   $x^4 + \frac{1}{x^4} = \frac{(2\sqrt{3} - 4)^2}{16} - 2$.
   After simplification, we find the answer is $\frac{(-4\sqrt{3} - 1)}{4}$, making option C correct.

   — worked solution by Athena AI

3. Q3.SSC 2019

   If $2x + 3y - 5z = 18, 3x + 2y + z = 29$ and $x + y + 3z = 17$, then what is the value of $xy + yz + zx$?
   

   • 32

   • 52

   • 64

   • 46

   Show solution

   x+y+3z=17..........(1)

   2x+3y-5z=18.......(2)

   3x+2y+z=29........(3)

   Now, (1)×2-(2) : 

   11z-y=6......(4)

   And, (3)×2-(2)×3 :

   17z-5y=4............(5)

   Now,4×(4)-(5) :

   z=2 .

   Now, put this value in (4) ,we get :

   y=6.

   From equation (1) we get : x=5.

   So,xy+yz+xz=10+30+12=52.

   B is correct choice.

4. Q4.SSC 2018

   The price of motorcycle depreciates every year by 8%. If the value of the motorcycle after 2 years will be Rs 84640, then what is the present value (in Rs) of the motorcycle? 

   • 90000

   • 102000

   • 110000

   • 100000

   Show solution

   Given that price of motorcycle depreciates every year by 8% then after 2 years,

   92% of 92% of $x = 84,640$

   $\frac{92}{100}$ x $\frac{92}{100}$ x $x = 84,640$

   $x = 100000$

   Hence, option D is the correct answer. 

5. Q5.SSC 2018

   The price of motor cycle depreciates every year by 10%. If the value of the motor cycle after 3 years will be Rs 36450, then what is the present value (in Rs) of the motor cycle?

   • 45000

   • 50000

   • 48000

   • 51000

   Show solution

   Let the present value of motor cycle = Rs. $x$

   Rate of depreciation = 10%

   => Value of the motor cycle after 3 years = $x\times(1-\frac{10}{100})\times(1-\frac{10}{100})(1-\frac{10}{100})$

   => $x\times(\frac{90}{100})^3=36450$

   => $x=36450\times\frac{100^3}{90^3}$

   => $x=0.05\times1000000=50000$

   $\therefore$ The present value (in Rs) of the motor cycle = Rs. 50,000

   => Ans - (B)

6. Q6.SSC 2018

   If\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\for x > 0, then what is the value of \ x^{3}+\frac{1}{x^{3}}\

   • $\frac{3\sqrt{3}}{\sqrt{5}}$

   • $\frac{3\sqrt{15}}{5}$

   • $\frac{3\sqrt{15}}{8}$

   • $\frac{3\sqrt{5}}{8}$

   Show solution

   Given : \ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\

   => $(x+\frac{1}{x})^2-2(x)(\frac{1}{x})=\frac{7}{4}$

   => $(x+\frac{1}{x})^2=\frac{7}{4}+2=\frac{15}{4}$

   => $x+\frac{1}{x}=\frac{\sqrt{15}}{2}$

   Cubing both sides, we get :

   => $x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=(\frac{\sqrt{15}}{2})^3$

   => $x^3+\frac{1}{x^3}+3(\frac{\sqrt{15}}{2})=\frac{15\sqrt{15}}{8}$

   => $x^3+\frac{1}{x^3}=\frac{15\sqrt{15}}{8}-\frac{3\sqrt{15}}{2}$
   

   => $x^3+\frac{1}{x^3}=\frac{3\sqrt{15}}{8}$
   

   => Ans - (C)

7. Q7.SSC 2018

   What is the value of [\frac{1}{1-x^{(p-q)}}+\frac{1}{1-x^{(q-p)}}]\?

   • 0

   • 1

   • $(x^{q}-x^{p})/(x^{q}+x^{p})$

   • $(x^{q}+x^{p})/(x^{q}-x^{p})$

   Show solution

   Expression = [\frac{1}{1-x^{(p-q)}}+\frac{1}{1-x^{(q-p)}}]\

   = $(\frac{1}{1-\frac{x^p}{x^q}})+(\frac{1}{1-\frac{x^q}{x^p}})$

   = $(\frac{x^q}{x^q-x^p})+(\frac{x^p}{x^p-x^q})$
   

   = $(\frac{x^q}{x^q-x^p})-(\frac{x^p}{x^q-x^p})$
   

   = $\frac{x^q-x^p}{x^q-x^p}=1$
   

   => Ans - (B)

8. Q8.SSC 2018

   If $x^{2} -7x + 1 = 0$, then what is the value of $x + \frac{1}{x}$?

   • 7

   • 3

   • 51

   • 47

   Show solution

   Given : $x^2-7x+1=0$

   Dividing both sides by $'x'$

   => $x-7+\frac{1}{x}=0$

   => $x+\frac{1}{x}=7$

   => Ans - (A)

9. Q9.SSC 2018

   If $x = 5 - \frac{1}{x}$, then what is the value of $x^{5} + \frac{1}{x^{5}}$?

   • 625

   • 3125

   • 2525

   • 2500

   Show solution

   Given : $x=5-\frac{1}{x}$

   => $x+\frac{1}{x}=5=k$

   Now, $x^5+\frac{1}{x^5}=[(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})]-(x+\frac{1}{x})$

   = $[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\times(x+\frac{1}{x})^2-2(x)(\frac{1}{x})]-(x+\frac{1}{x})$

   = $[(k^3-3k)\times(k^2-2)]-(k)$

   = $[(125-15)\times(25-2)]-(5)$

   = $(110\times23)-5$

   = $2530-5=2525$

   => Ans - (C)

10. Q10.SSC 2018

    Sin A + Sin^{2}\A = 1, then the value of cos^{2}\A + cos^{4}\A is

    • 2

    • $\frac{2}{3}$

    • $1\frac{1}{2}$

    • 1

    Show solution

    Given sin A+$sin^{2}$ A=1

    ==> sin A = 1-$sin^{2}$ A

    ==> sin A = $cos^{2}$ A ($\because cos^{2}A+sin^{2}A$=1)

    $cos^{2}$ A=sin A ==> $cos^{4}A$=$sin^{2}A$
    

    $\therefore cos^{2}A+cos^{4}A=1 ( \because sin A+sin^{2}A=1)$
    

11. Q11.SSC 2015

    The area of the triangle formed by the graphs of the equations x= 0, 2x+ 3y= 6 and x+ y= 3 is :

    • 3 sq. unit

    • $4\frac{1}{2}$ sq. unit

    • $1\frac{1}{2}$ sq. unit

    • 1 sq. unit

    Show solution

    

    AC represents $x+y=3$

    BC represents $2x+3y=6$

    AB represents $x=0$

    => ABC is the required triangle.

    Base AB = 1 unit and height OC = 3 units

    => Area of $\triangle$ABC = $\frac{1}{2}$ * AB * OC

    = $\frac{1}{2}$ * 1 * 3 = 1$\frac{1}{2}$ sq. unit

12. Q12.SSC 2013

    Find the maximum number of trees which can be planted, 20 metres apart, on the two sides of a straight road 1760 metres long

    • 180

    • 178

    • 174

    • 176

    Show solution

    To understand this, let's focus on one side to begin with.

    If the straight road was 20 meters, you could have planted two trees.

    If it was 40 meters, you could have planted three trees

    If it was 'n' meters, you could have planted $(\frac{n}{20}+1)$ trees

    Number of trees that can be planted on one side = $\frac{1760}{20}+1$ = 88 + 1 = 89

    Total trees = 89*2 = 178

SSC CGL Linear Equations previous year questions with solutions — AthenaPrep