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.

Quadratic Equations

Short Answer Type

251. By reduction of Re. 1 per kg in the price of sugar. Ritika can buy one kg sugar more for Rs. 56. Find the original price of sugar per kilograms.

89 Views

252. A piece of cloth costs Rs. 200. If the piece were 5 m longer and each metre of cloth costed Rs. 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre.

97 Views

Long Answer Type

253. Two very good friends ‘Fox’ and ‘Eagle’ lived at the top of a cliff of height 6 m whose base was at a distance of 10 m from a point A on the ground. The fox descends the cliff and went straight to the point A. The eagle flew vertically up to height x and then flew in a straight line to point A. The distance travelled by each being the same. Find the value of x.

98 Views

Short Answer Type

254. On a pillar 9 metres high is perched a peacock. From a distance of 27 metres, a beautiful snake is coming to its hole at the bottom of a pillar. Seeing a beautiful snake, the peacock flying in an oblique direction, pounces upon it. If their speeds are equal of what distance away the hole is the beautiful snake caught.

104 Views

Long Answer Type

255. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Let the speed of the stream be x km/hr.
∴ Speed of the boat upstream = (18 – x) km/hr.
Speed of the boat downstream = (18 + x) km/hr.
Time taken for going 24 km upstream = $fraction numerator 24 over denominator 18 minus straight x end fraction space hours$

Time taken for going 24 km downstream = $<pre>uncaught exception: mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56 #0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>$
According to given condition,
∴ $space fraction numerator 24 over denominator 18 minus straight x end fraction minus fraction numerator 24 over denominator 18 plus straight x end fraction equals 1$

$rightwards double arrow space space 24 left parenthesis 18 plus straight x right parenthesis space minus space 24 left parenthesis 18 minus straight x right parenthesis space equals space left parenthesis 18 minus straight x right parenthesis left parenthesis 18 plus straight x right parenthesis rightwards double arrow space space space space space space space space space space space space space space straight x squared plus 48 straight x minus 324 space equals space 0$
Using the quadratic formula, we get
$straight x equals fraction numerator negative 48 plus-or-minus square root of 48 squared plus 1926 end root over denominator 2 end fraction equals fraction numerator negative 48 plus-or-minus square root of 3600 over denominator 2 end fraction equals space fraction numerator negative 48 plus-or-minus 60 over denominator 2 end fraction equals 6 space or space minus 54$
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = –54. Therfore, x = 6 gives the speed of the stream as 6km/h.
$rightwards double arrow space space space space space space space 21 over 4 cross times fraction numerator 2 straight x over denominator 25 minus straight x squared end fraction equals 1 rightwards double arrow space space space space space space space space space 21 over 2 cross times fraction numerator straight x over denominator 25 minus straight x squared end fraction equals 1 rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 21 straight x space equals space 50 minus 2 straight x squared$
$rightwards double arrow space space space space space space space space space space space space space 2 straight x squared plus 21 straight x minus 50 space equals space 0 rightwards double arrow space space space space 2 straight x squared plus 25 straight x minus 4 straight x minus 50 space equals space 0 rightwards double arrow space space space space straight x left parenthesis 2 straight x plus 25 right parenthesis minus 2 left parenthesis 2 straight x plus 25 right parenthesis space equals space 0 rightwards double arrow space space space space space space space space space space space space space left parenthesis 2 straight x plus 25 right parenthesis space left parenthesis straight x minus 2 right parenthesis space equals space 0 rightwards double arrow space space 2 straight x space plus 25 space equals 0 space space space space space space straight x space minus space 2 space equals 0 rightwards double arrow space space space space space 2 straight x space equals space minus 25 space space space space space rightwards double arrow space space straight x space equals space 2 rightwards double arrow space space space space space straight x space equals space fraction numerator negative 25 over denominator 2 end fraction$
$straight x equals fraction numerator negative 25 over denominator 2 end fraction$ is rejected.
Hence, the speed of the stream is 2 km/hr.

257 Views

Short Answer Type

256. If the equation (1 + m²)x² + 2mcx + (c² – a²) = 0 has equal roots. Prove that c² = a² (1 + m²).

96 Views

257. If the roots of the equation (b – c)x² + (c - a)x + (a b) = 0 are equal then prove that 2b = a + c.

84 Views

258. Find the value of ‘p’ for which the given quadratic equation has equal roots.
5x² – 4x + 2 + p(4x² – 2x – 1) = 0.

77 Views

Long Answer Type

259.

Solve for 'x': $fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis space left parenthesis straight x minus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x minus 3 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x minus 4 right parenthesis end fraction equals 1 over 6$

90 Views

Short Answer Type

260. For what value of p (4 – p)x² – (2p + 4)x + (8p + 1) = 0 is a perfect square.

246 Views