SSC CGL 20191)If \( {x^8-1442x^4+1}=0\), then a possible value of (x - \( {1\over x}\)) is :
6
\(x^8 - 1442 x^4 + 1 = 0; x^4 - 1442 + \frac{1}{x^4} = 0; x^4 + \frac{1}{x^4} = 1442 ; x^4 + \frac{1}{x^4} + 2 = 1442 + 2; (x^2 + \frac{1}{x^2})^2 = (38)^2 ; x^2 + \frac{1}{x^2} = 38 ; x^2 + \frac{1}{x^2} - 2 = 38 - 2 ; (x - \frac{1}{x})^2 = 6^2 ; x - \frac{1}{x} = 6; (\because (a - b)^2 = a^2 - 2ab + b^2);\)
SSC CGL 20192)If \({ \sqrt{86-60\sqrt2}}=a-b\sqrt2\), then what will be the value of \( {\sqrt{a^2+b^2} }\) , correct to one decimal place?
7.8
\(\sqrt{86 - 60\sqrt{2}} = a - b\sqrt{2}; \sqrt{36 + 50 - 2 \times 30\sqrt{2}} = a - b\sqrt{2}; \sqrt{6^2 + (5\sqrt{2})^2 - 2 \times 6 \times 5\sqrt{2}} = a - b\sqrt{2}; \sqrt{(6 - 5\sqrt{2})^2 } = a - b\sqrt{2}; 6 - 5\sqrt{2} = a - b\sqrt{2}; \)
solve a = 6, b = 5, ans is 7.8
SSC CGL 20193)If \({a^2+b^2+c^2+96}=8(a+b-2c)\), then \({\sqrt {ab-bc+ca}}\) is equal to :
4
\(a^2 + b^2 + c^2 + 96 = 8(a + b - 2c)\);
\(a^2 + b^2 + c^2 + 96 - 8a - 8b + 16c = 0;\)
\((a^2 - 8a + 16) + (b^2 - 8b + 16) + (c^2 + 16c + 64) = 0; \) \((a - 4)^2 + (b - 4)^2 + (c + 8)^2 = 0;\)
\((a - 4)^2 = 0, (b - 4)^2 = 0, (c + 8)^2 = 0;\)
a = 4,
b = 4,
c = -8;
\(\sqrt{ab - bc + ca} = \sqrt{4 \times 4 + 4 \times 8 - 4 \times 8 } = \sqrt{16 + 32 - 32 } = 4\)
SSC CGL 20194)In a school, \({4 \over 9}\) of the number of students are girls and the rest are boys. \({3 \over 5}\) of the number of boys are below 12 years of age and \({5 \over 12}\) of the number of girls are 12 years or above 12 years of age.
If the number of students below 12 years of age is 480, then \({5 \over 18}\) of the total number of students in the school will be equal to :
225
Let the total student be x.
Number of girls = 4x/9;
Number of boys = x - 4x/9 = 5x/9;
Number of boys below 12 years = \(5x/9 \times 3/5 \)= x/3;
Number of girls are 12 years or above 12 years of age =\( 4x/9 \times 5/12 \)= 5x/27;
Number of girls below 12 years =\( \frac{4x}{9} - \frac{5x}{27}\) = 7x/27;
The number of students below 12 years of age = 480;
Number of boys below 12 years + number of girls below 12 years = 480;
\(\frac{x}{3} +\frac{7x}{27} \)= 480;
16x/27 = 480;
x = 810;
\(\frac{5}{18} \)of the total number of students in the school =\( 810 \times \frac{5}{18}\) = 225
SSC CGL 20195)If x + y + z =11, \({x^2 + y^2 + z^2 = 133}\) and \({x^3 + y^3 + z^3 = 881}\), then the value of \(\sqrt[3]{xyz}\) is :
-6
\((x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz);\)
\((11)^2 = 133 + 2(xy + yz + xz);\)
2(xy + yz + xz) = -12;
xy + yz + xz = -6;
\(x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + xz)); \)
881 - 3(xyz) = 11(133 + 6);
3xyz = 648;
xyz = -216;
\(\sqrt[3]{xyz} \) = -6
SSC CGL 20196)The value of \({(253)^3+(247)^3}\over {25.3\times25.3-624.91+24.7\times24.7}\) is \({50\times10^k}\), then the value of k is :
3
\(\frac{(253)^3 + (247)^3}{25.3 \times 25.3 - 624.91 + 24.7 \times 24.7} = 50 \times 10^k;\)
\(\frac{(253 + 247)(253)^2 + (247)^2 - 253 \times 247)}{\frac{1}{100}[(253)^2 - 253 \times 247 + (247)^2] } = 50 \times 10^k;\)
\(50000 = 50 \times 10^k;\)
\(50 \times 10^3 = 50 \times 10^k;\)
k = 3
SSC CGL 20197)If \({(\sqrt2+\sqrt5-\sqrt3)\times k=-12}\), then what will be the value of k ?
\({(\sqrt2+\sqrt5+\sqrt3)(2-\sqrt10)}\)
\(\left(\sqrt{2} + \sqrt{5} - \sqrt{3}\right) \times k\) = -12;
\((1.414 + 2.236 - 1.732) \times k \)= -12;
By the option B,
\((1.414 + 2.236 - 1.732) \times \left(\sqrt{2} + \sqrt{5} + \sqrt{3}\right)\left(2 - \sqrt{10}\right) \)= -12;
(1.414 + 2.236 - 1.732) x (1.414 + 2.236 + 1.732)(2 - 3.16) = -12;
\(1.9 \times 5.3 \times (-1.16) \)= -12;
-12 = -12
SSC CGL 20198)If x = \(\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}}\), then the value of\( \frac{\sqrt{2}-x}{\sqrt{2}+x}\) will be closest to:
0.17
\(x=\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}}; x^2 =(\sqrt{1+\frac{\sqrt{3}}{2}-}\sqrt{1-\frac{\sqrt{3}}{2}})^2; x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1+\frac{\sqrt{3}}{2}} \sqrt{1-\frac{\sqrt{3}}{2}};\)
\( x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1+\frac{\sqrt{3}}{2}} \sqrt{1-\frac{\sqrt{3}}{2}}; x^2 =1+\frac{\sqrt{3}}{2} + 1-\frac{\sqrt{3}}{2} - 2\sqrt{1- \frac{3}{4}} ; x^2 = 2 - 2\sqrt{ \frac{1}{4}} ; x^2 = 1;\)
x = 1 or x = -1;
Now,;\(\frac{\sqrt{2}-x}{\sqrt{2}+x} = \frac{\sqrt{2}-x}{\sqrt{2}+x} \times \frac{\sqrt{2}-x}{\sqrt{2}-x}; = \frac{2 - 2\sqrt{2}x + x^2}{2 - x^2}; \)Put the value of x = 1;\(= \frac{2 - 2\sqrt{2} + 1}{2 - 1}; = 3 - 2\sqrt{2} = 0.17 \)
SSC CGL 20199)If \( a^3 + b^3 = 218 \) and a + b = 2, then the value of ab is:
-35
\((a + b)^3 = a^3 + b^3 + 3ab(a + b)\);
\((2)^3 = 218 + 3ab(2)\);
-6ab = 218 - 8 = 210;
ab = -210/6 = -35
SSC CGL 201910)If \(2\sqrt{2}x^3-3\sqrt{3}y^3=\)\(\left(\sqrt{2}x-\sqrt{3}y\right)\)\(\left(Ax^2+By^2+Cxy\right)\), then the value of \(( A^2 + B^2 - C^2 )\) is:
7
\(2\sqrt{2}x^3-3\sqrt{3}y^3=\left(\sqrt{2}x-\sqrt{3}y\right)\left(Ax^2+By^2+Cxy\right);\)
(because\( a^3 - b^3 = (a - b)(a^2 + ab + b^2)\));
On compression,
A =\( (\sqrt{2})^2 \)= 2;
A = \((-\sqrt{3})^2\) = 3;
C = \(\sqrt{2}\sqrt{3}\) =\( \sqrt{6}\);
Now,
\(A^2 + B^2 - C^2 \)= \(2^2 + 3^2 - (\sqrt{6})^2\)
= 4 + 9 - 6
= 7