Fraction Questions for SSC-CGL

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Fraction Questions for SSC CGL
Fraction Questions for SSC CGL

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.

Download Fraction Questions for SSC CGL

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Question 1: If $2x – \frac{1}{2x} = 6$, then the value of $x^2 + \frac{1}{16x^2}$

a) $\frac{19}{2}$

b) $\frac{17}{2}$

c) $\frac{18}{3}$

d) $\frac{15}{2}$

Question 2: If x,y are positive acute angles, x + y < 90o and sin(2x -20o) = cos(2y + 20o), then the values of sec(x + y) is

a) $\sqrt{2}$

b) $\frac{1}{\sqrt{2}}$

c) $1$

d) $0$

Question 3: If $x^2=y + z, y^2=z + x, z^2=x+y$, then the value $\frac{1}{1+x} + \frac{1}{1+y} + \frac{1}{1+z}$

a) 1

b) 2

c) 0

d) -1

Question 4: The sum of a non-zero number and thrice its reciprocal is 13/2. Find the number.

a) 6

b) 7

c) 8

d) 9

Question 5: The reciprocal of the sum of the reciprocals of 8/7 and 5/6 is:

a) 83/40

b) 42/83

c) 83/42

d) 40/83

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

a) 81

b) 361

c) 729

d) 6561

Question 7: If A : B = 2 : 5, B : C = 4 : 3 and C : D = 2 : 1, then what is value of A : C : D?

a) 6 : 5 : 2

b) 7 : 20 : 10

c) 8 : 30 : 15

d) 16 : 30 : 15

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

a) $\frac{59}{81}$

b) $\frac{100}{27}$

c) $\frac{319}{81}$

d) $\frac{913}{81}$

Question 9: If 27x + 27[x-(1/3)] = 99, then what is the value of x?

a) 2

b) 3

c) 4

d) 5

Question 10: If $a^{3} + b^{3}$ = 19 and a + b = 1, then what is the value of ab?

a) 5

b) -6

c) 7

d) -9

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Answers & Solutions:

1) Answer (A)

$2x – \frac{1}{2x} = 6$
or $x – \frac{1}{4x} = 3$
squaring on both sides:
$x^2 + \frac{1}{16x^2} – \frac{1}{2}$ = 9
or $x^2 + \frac{1}{16x^2}$ = 19/2

2) Answer (A)

Given sin(2x-20) = cos(2y+20)
as x and y are positive acute angles
hence cos(90-2x+20) = cos(2y+20)
or 90-2x+20 = 2y+20
or x+y = 45
or sec(x+y) = $\sqrt2$

3) Answer (A)

$x^2 = y+z$
$y^2 = z+x$
on substracting above two eq. we will get
$x^2 – y^2 = y – x$
So either $x=y$
or $x+y= -1$ (it is not possible as $z^2$ can not be negative)
So $x=y=z=2$
So given eq. will reduce to a value 1

4) Answer (A)

Let the number be $x$

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

=> $\frac{x^2 + 3}{x} = \frac{13}{2}$

=> $2x^2 + 6 = 13x$

=> $2x^2 – 13x + 6 = 0$

=> $2x^2 – x – 12x + 6 = 0$

=> $x(2x – 1) – 6(2x – 1) = 0$

=> $(2x – 1)(x – 6) = 0$

=> $x = 6 , \frac{1}{2}$

=> Ans – (A)

5) Answer (D)

Sum of the reciprocals of 8/7 and 5/6

= $\frac{7}{8} + \frac{6}{5}$

= $\frac{7(5)+6(8)}{40} = \frac{35+48}{40}$

= $\frac{83}{40}$

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

=> Ans – (D)

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6) Answer (C)

Given : $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

=> $\frac{yz+zx+xy}{xyz}=0$

=> $xy+yz+zx=0$ ———–(i)

Also, $x+y+z=9$

Squaring both sides,

=> $(x+y+z)^2=(9)^2$

=> $x^2+y^2+z^2+2(xy+yz+zx)=81$

=> $x^2+y^2+z^2+2(0)=81$

=> $x^2+y^2+z^2=81$

To find : $x^3 + y^3 + z^3 – 3xyz$

= $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

= $(9)(81-0)=729$

=> Ans – (C)

7) Answer (D)

Given : $\frac{A}{B}=\frac{2}{5}$ ———–(i)

$\frac{B}{C}=\frac{4}{3}$ ———–(ii)

and $\frac{C}{D}=\frac{2}{1}$ ———–(iii)

Multiplying equation (ii) by 10 and (iii) by 15 to make ‘C’ equal in both

=> $\frac{B}{C}=\frac{40}{30}$ and $\frac{C}{D}=\frac{30}{15}$

Now, multiplying equation (i) by 8 to make ‘B’ equal in both

=> $\frac{A}{B}=\frac{16}{40}$

Thus, we get = $A:B:C:D=16:40:30:15$

$\therefore$ A : C : D = 16 : 30 : 15

=> Ans – (D)

8) Answer (C)

Given : $\frac{x}{y}=\frac{4}{9}$

Let $x=4$ and $y=9$

To find : $\frac{(7x^2-19xy+11y^2)}{y^2}$

= $\frac{7(4)^2-19(4)(9)+11(9)^2}{(9)^2}$

= $\frac{112-684+891}{81}$

= $\frac{319}{81}$

=> Ans – (C)

9) Answer (A)

Expression : 27x + 27[x-(1/3)] = 99

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

=> $27x+9(3x-1)=99$

=> $27x+27x-9=99$

=> $54x=99+9=108$

=> $x=\frac{108}{54}=2$

=> Ans – (A)

10) Answer (B)

Given :  $a^{3} + b^{3}$ = 19 ———–(i)

Also, $a+b=1$ ———–(ii)

Cubing both sides, we get :

=> $(a+b)^3=(1)^3$

=> $a^3+b^3+3ab(a+b)=1$

Substituting values from equation (i) and (ii)

=> $19+3ab(1)=1$

=> $3ab=1-19=-18$

=> $ab=\frac{-18}{3}=-6$

=> Ans – (B)

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