Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Pair of Linear Equations in Two Variables

Short Answer Type

131. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

105 Views

132. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

175 Views

133. Solve the following pairs of equations by reducing them to a pair of linear equations :

fraction numerator 1 over denominator 2 straight x end fraction plus fraction numerator 1 over denominator 3 straight y end fraction equals 2 fraction numerator 1 over denominator 3 straight x end fraction plus fraction numerator 1 over denominator 2 straight y end fraction equals 13 over 6

232 Views

134. Solve the following pairs of equations by reducing them to a pair of linear equations :

<pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>

fraction numerator 4 over denominator square root of straight x end fraction minus fraction numerator 9 over denominator square root of 9 end fraction equals negative 1

133 Views

135. Solve the following pairs of equations by reducing them to a pair of linear equations :

4 over straight x plus 3 straight y equals 14 3 over straight x minus 4 straight y equals 23

113 Views

136.
Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 5 over denominator x minus 1 end fraction plus fraction numerator 1 over denominator y minus 2 end fraction equals 2$

$fraction numerator 6 over denominator straight x minus 1 end fraction minus fraction numerator 3 over denominator straight y minus 2 end fraction equals 1$

Let space fraction numerator 1 over denominator straight x minus 1 end fraction equals straight u space and space fraction numerator 1 over denominator straight y minus 2 end fraction equals straight v

Thus, the given equations becomes
5u + v = 2 ...(i)
6u - 3v = 1 ...(ii)
For making the coefficient of v in (i) and (ii) equal, we multiply (i) by 3 and adding, we get

"<pre

Putting the value of u in (i), we get
5u + v = 2

$rightwards double arrow space space space 5 open parentheses 1 third close parentheses plus v space equals space 2$

Thus, the given equations becomes5u + v = 2 ...(i)6u - 3v = 1 ?

70 Views

137.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$space space space space space space space space space fraction numerator 7 straight x minus 2 straight y over denominator xy end fraction equals 5 space space space space space space space fraction numerator 8 straight x plus 7 straight x over denominator x y end fraction equals 15$

83 Views

138.

Solve the following pairs of equations by reducing them to a pair of linear equations:

6x + 3y = 6xy
2x + 4y = 5x

381 Views

139.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 10 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 4$

$fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 5 over denominator straight x minus straight y end fraction equals negative 2$

158 Views

140.

Solve the following pairs of equations by reducing them to a pair of linear equations:

$fraction numerator 1 over denominator 3 x plus y end fraction plus fraction numerator 1 over denominator 3 x minus y end fraction space equals space 3 over 4 fraction numerator 1 over denominator 2 left parenthesis 3 x plus y right parenthesis end fraction plus fraction numerator 1 over denominator 2 left parenthesis 3 x minus y right parenthesis end fraction space equals space fraction numerator negative 1 over denominator space space 8 end fraction$

85 Views

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

136. Solve the following pairs of equations by reducing them to a pair of linear equations: