{"id":40504,"date":"2020-02-12T18:45:21","date_gmt":"2020-02-12T13:15:21","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=40504"},"modified":"2020-02-12T18:45:21","modified_gmt":"2020-02-12T13:15:21","slug":"fraction-questions-for-ssc-cgl","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/fraction-questions-for-ssc-cgl\/","title":{"rendered":"Fraction Questions for SSC-CGL"},"content":{"rendered":"

Fraction Questions for SSC-CGL<\/strong><\/span><\/h2>\n

Download SSC CGL Questions on Fraction PDF based on previous papers very useful for SSC CGL exams. Fraction Questions for SSC exams.<\/p>\n

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Question 1:\u00a0<\/b>If $2x – \\frac{1}{2x} = 6$, then the value of $x^2 + \\frac{1}{16x^2}$<\/p>\n

a)\u00a0$\\frac{19}{2}$<\/p>\n

b)\u00a0$\\frac{17}{2}$<\/p>\n

c)\u00a0$\\frac{18}{3}$<\/p>\n

d)\u00a0$\\frac{15}{2}$<\/p>\n

Question 2:\u00a0<\/b>If x,y are positive acute angles, x + y < 90o <\/sup>and sin(2x -20o<\/sup>) = cos(2y + 20o<\/sup>), then the values of sec(x + y) is<\/p>\n

a)\u00a0$\\sqrt{2}$<\/p>\n

b)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n

c)\u00a0$1$<\/p>\n

d)\u00a0$0$<\/p>\n

Question 3:\u00a0<\/b>If $x^2=y + z, y^2=z + x, z^2=x+y$, then the value $\\frac{1}{1+x} + \\frac{1}{1+y} + \\frac{1}{1+z}$<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a00<\/p>\n

d)\u00a0-1<\/p>\n

Question 4:\u00a0<\/b>The sum of a non-zero number and thrice its reciprocal is 13\/2. Find the number.<\/p>\n

a)\u00a06<\/p>\n

b)\u00a07<\/p>\n

c)\u00a08<\/p>\n

d)\u00a09<\/p>\n

Question 5:\u00a0<\/b>The reciprocal of the sum of the reciprocals of 8\/7 and 5\/6 is:<\/p>\n

a)\u00a083\/40<\/p>\n

b)\u00a042\/83<\/p>\n

c)\u00a083\/42<\/p>\n

d)\u00a040\/83<\/p>\n

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Question 6:\u00a0<\/b>If $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$ and x + y + z = 9, then what is the value of $x^3 + y^3 + z^3 – 3xyz$ ?<\/p>\n

a)\u00a081<\/p>\n

b)\u00a0361<\/p>\n

c)\u00a0729<\/p>\n

d)\u00a06561<\/p>\n

Question 7:\u00a0<\/b>If A : B = 2 : 5, B : C = 4 : 3 and C : D = 2 : 1, then what is value of A : C : D?<\/p>\n

a)\u00a06 : 5 : 2<\/p>\n

b)\u00a07 : 20 : 10<\/p>\n

c)\u00a08 : 30 : 15<\/p>\n

d)\u00a016 : 30 : 15<\/p>\n

Question 8:\u00a0<\/b>If $\\frac{x}{y}=\\frac{4}{9}$, then what is the value of $\\frac{(7x^2-19xy+11y^2)}{y^2}$ ?<\/p>\n

a)\u00a0$\\frac{59}{81}$<\/p>\n

b)\u00a0$\\frac{100}{27}$<\/p>\n

c)\u00a0$\\frac{319}{81}$<\/p>\n

d)\u00a0$\\frac{913}{81}$<\/p>\n

Question 9:\u00a0<\/b>If 27x + 27[x-(1\/3)] = 99, then what is the value of x?<\/p>\n

a)\u00a02<\/p>\n

b)\u00a03<\/p>\n

c)\u00a04<\/p>\n

d)\u00a05<\/p>\n

Question 10:\u00a0<\/b>If $a^{3} + b^{3}$ = 19 and a + b = 1, then what is the value of ab?<\/p>\n

a)\u00a05<\/p>\n

b)\u00a0-6<\/p>\n

c)\u00a07<\/p>\n

d)\u00a0-9<\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$2x – \\frac{1}{2x} = 6$
\nor\u00a0$x – \\frac{1}{4x} = 3$
\nsquaring on both sides:
\n$x^2 + \\frac{1}{16x^2} – \\frac{1}{2}$ = 9
\nor\u00a0$x^2 + \\frac{1}{16x^2}$ = 19\/2<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Given sin(2x-20) = cos(2y+20)
\nas x and y are positive acute angles
\nhence cos(90-2x+20) = cos(2y+20)
\nor 90-2x+20 = 2y+20
\nor x+y = 45
\nor sec(x+y) = $\\sqrt2$<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$x^2 = y+z$
\n$y^2 = z+x$
\non substracting above two eq. we will get
\n$x^2 – y^2 = y – x$
\nSo either $x=y$
\nor $x+y= -1$ (it is not possible as $z^2$ can not be negative)
\nSo $x=y=z=2$
\nSo given eq. will reduce to a value 1<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the number be $x$<\/p>\n

According to ques, => $x + \\frac{3}{x} = \\frac{13}{2}$<\/p>\n

=> $\\frac{x^2 + 3}{x} = \\frac{13}{2}$<\/p>\n

=> $2x^2 + 6 = 13x$<\/p>\n

=> $2x^2 – 13x + 6 = 0$<\/p>\n

=> $2x^2 – x – 12x + 6 = 0$<\/p>\n

=> $x(2x – 1) – 6(2x – 1) = 0$<\/p>\n

=> $(2x – 1)(x – 6) = 0$<\/p>\n

=> $x = 6 , \\frac{1}{2}$<\/p>\n

=> Ans – (A)<\/p>\n

5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Sum of the reciprocals of 8\/7 and 5\/6<\/p>\n

= $\\frac{7}{8} + \\frac{6}{5}$<\/p>\n

= $\\frac{7(5)+6(8)}{40} = \\frac{35+48}{40}$<\/p>\n

= $\\frac{83}{40}$<\/p>\n

=> Reciprocal of 83\/40 = $\\frac{40}{83}$<\/p>\n

=> Ans – (D)<\/p>\n

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6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Given\u00a0:\u00a0$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$<\/p>\n

=> $\\frac{yz+zx+xy}{xyz}=0$<\/p>\n

=> $xy+yz+zx=0$ ———–(i)<\/p>\n

Also, $x+y+z=9$<\/p>\n

Squaring both sides,<\/p>\n

=> $(x+y+z)^2=(9)^2$<\/p>\n

=> $x^2+y^2+z^2+2(xy+yz+zx)=81$<\/p>\n

=> $x^2+y^2+z^2+2(0)=81$<\/p>\n

=> $x^2+y^2+z^2=81$<\/p>\n

To find\u00a0:\u00a0$x^3 + y^3 + z^3 – 3xyz$<\/p>\n

= $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$<\/p>\n

= $(9)(81-0)=729$<\/p>\n

=> Ans – (C)<\/p>\n

7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Given\u00a0: $\\frac{A}{B}=\\frac{2}{5}$ ———–(i)<\/p>\n

$\\frac{B}{C}=\\frac{4}{3}$ ———–(ii)<\/p>\n

and $\\frac{C}{D}=\\frac{2}{1}$ ———–(iii)<\/p>\n

Multiplying equation (ii) by 10 and (iii) by 15 to make ‘C’ equal in both<\/p>\n

=>\u00a0$\\frac{B}{C}=\\frac{40}{30}$ and $\\frac{C}{D}=\\frac{30}{15}$<\/p>\n

Now, multiplying equation (i) by 8 to make ‘B’ equal in both<\/p>\n

=> $\\frac{A}{B}=\\frac{16}{40}$<\/p>\n

Thus, we get\u00a0= $A:B:C:D=16:40:30:15$<\/p>\n

$\\therefore$\u00a0A : C : D =\u00a016 : 30 : 15<\/p>\n

=> Ans – (D)<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Given\u00a0:\u00a0$\\frac{x}{y}=\\frac{4}{9}$<\/p>\n

Let $x=4$ and $y=9$<\/p>\n

To find\u00a0:\u00a0$\\frac{(7x^2-19xy+11y^2)}{y^2}$<\/p>\n

= $\\frac{7(4)^2-19(4)(9)+11(9)^2}{(9)^2}$<\/p>\n

= $\\frac{112-684+891}{81}$<\/p>\n

= $\\frac{319}{81}$<\/p>\n

=> Ans – (C)<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Expression\u00a0:\u00a027x + 27[x-(1\/3)] = 99<\/p>\n

=> $27x+27(\\frac{3x-1}{3})=99$<\/p>\n

=> $27x+9(3x-1)=99$<\/p>\n

=> $27x+27x-9=99$<\/p>\n

=> $54x=99+9=108$<\/p>\n

=> $x=\\frac{108}{54}=2$<\/p>\n

=> Ans – (A)<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given\u00a0:\u00a0\u00a0$a^{3} + b^{3}$ = 19 ———–(i)<\/p>\n

Also, $a+b=1$ ———–(ii)<\/p>\n

Cubing both sides, we get\u00a0:<\/p>\n

=> $(a+b)^3=(1)^3$<\/p>\n

=> $a^3+b^3+3ab(a+b)=1$<\/p>\n

Substituting values from equation (i) and (ii)<\/p>\n

=> $19+3ab(1)=1$<\/p>\n

=> $3ab=1-19=-18$<\/p>\n

=> $ab=\\frac{-18}{3}=-6$<\/p>\n

=> Ans – (B)<\/p>\n

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We hope this Fraction questions for SSC CGL Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Fraction Questions for SSC-CGL Download SSC CGL Questions on Fraction PDF based on previous papers very useful for SSC CGL exams. Fraction Questions for SSC exams. Question 1:\u00a0If $2x – \\frac{1}{2x} = 6$, then the value of $x^2 + \\frac{1}{16x^2}$ a)\u00a0$\\frac{19}{2}$ b)\u00a0$\\frac{17}{2}$ c)\u00a0$\\frac{18}{3}$ d)\u00a0$\\frac{15}{2}$ Question 2:\u00a0If x,y are positive acute angles, x + y < […]<\/p>\n","protected":false},"author":42,"featured_media":40508,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,125,9,504,378,1493,1459,1611,1741,1268,1441],"tags":[3631,462],"class_list":{"0":"post-40504","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-ssc-stenographer","17":"category-stenographer","18":"tag-fraction-questions-for-ssc-cgl","19":"tag-ssc-cgl"},"better_featured_image":{"id":40508,"alt_text":"Fraction Questions for SSC CGL","caption":"Fraction Questions for SSC CGL\n","description":"Fraction Questions for SSC 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