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SSC CGL Exam Quantitative Aptitude Study Material

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Quadratic Equations

If a=√x+2+√x-2/√x+2√x-2 then the value of (a²-ax)is

a) 1
b) 2
c) -1
d) 0

✔Answer

Correct Answer Is : -1

Solution Is :

The HCF of X8 - 1 and X4 + 2X3 - 2X - 1

a) X²+1
b) X+1
c) X²-1
d) X-1

✔Answer

✔Solution

Correct Answer Is : X²-1

Solution Is :

The least number which when divided by 6,9,12,15,18 leaves the same remainder 2 in each case is:

a) 178
b) 182
c) 176
d) 180

✔Answer

✔Solution

Correct Answer Is : 182

Solution Is :

If X=X24 + 1/X12 = 7then the X72 + 1/X36 value of

a) 432
b) 433
c) 343
d) 322

✔Answer

✔Solution

Correct Answer Is : 322

Solution Is :

If 5x + 9y = 5 and 125x + 729y = 120 then the value of the product of x and y is

a) 135
b) 1/135
c) 1/9
d) 45

✔Answer

✔Solution

Correct Answer Is : 1/135

Solution Is : 5x +9y = 5 ...(i)
125x³+ 729y³=120 ...(ii)
Now cube both sides of equation (i), we get
⇒ (5x + 9y)³= (5)³
⇒125x³ + 729y³+135xy(5x +9y)=125
⇒ 125x³ + 729y³ +135xy 5 =125
Now, put the value of equation (i)
120 + 135 xy × 5 = 125
⇒ 135 × 5xy = 125 - 120
⇒135 × 5xy = 5 ⇒
xy=(5/(1355))=1/135

If p = 99 then, the value of p(p² + 3p + 3) is :

a) 999999
b) 988899
c) 989898
d) 998889

✔Answer

✔Solution

Correct Answer Is : 999999

Solution Is : Given P = 99
then P(P² + 3P + 3)
= P((P + 1)²+ P + 2)
= 99 ((99 + 1)² + 101)
= 99 × (10000 + 101)
= 999999

The least number which when divided by 48, 64.90, 120 will leave the remainders 38, 54, 80, 110 respectively, is

a) 2870
b) 2860
c) 2890
d) 2880

✔Answer

✔Solution

Correct Answer Is : 2870

Solution Is :

a)
b)
c)
d)

✔Answer

✔Solution

Correct Answer Is :

Solution Is :

a) 535
b) 1030
c) 790
d) 970

✔Answer

✔Solution

Correct Answer Is : 970

Solution Is :

a) 19-Feb
b) 17-Feb
c) 18-Mar
d) 15-Feb

✔Answer

✔Solution

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SSC CGL Exam Quantitative Aptitude Study Material

Quadratic Equations

If a=√x+2+√x-2/√x+2√x-2 then the value of (a²-ax)is

Correct Answer Is : -1

Solution Is :

The HCF of X8 - 1 and X4 + 2X3 - 2X - 1

Correct Answer Is : X²-1

Solution Is :

The least number which when divided by 6,9,12,15,18 leaves the same remainder 2 in each case is:

Correct Answer Is : 182

Solution Is :

If X=X24 + 1/X12 = 7then the X72 + 1/X36 value of

Correct Answer Is : 322

Solution Is :

If 5x + 9y = 5 and 125x + 729y = 120 then the value of the product of x and y is

Correct Answer Is : 1/135

If p = 99 then, the value of p(p² + 3p + 3) is :

Correct Answer Is : 999999

Solution Is : Given P = 99
then P(P² + 3P + 3)
= P((P + 1)²+ P + 2)
= 99 ((99 + 1)² + 101)
= 99 × (10000 + 101)
= 999999

The least number which when divided by 48, 64.90, 120 will leave the remainders 38, 54, 80, 110 respectively, is

Correct Answer Is : 2870

Solution Is :

Correct Answer Is :

Solution Is :

Correct Answer Is : 970

Solution Is :

Correct Answer Is : 19-Feb

Solution Is :

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SSC CGL Exam Quantitative Aptitude Study Material

Quadratic Equations

If a=√x+2+√x-2/√x+2√x-2 then the value of (a2-ax)is

Correct Answer Is : -1

Solution Is :

The HCF of X8 - 1 and X4 + 2X3 - 2X - 1

Correct Answer Is : X2-1

Solution Is :

The least number which when divided by 6,9,12,15,18 leaves the same remainder 2 in each case is:

Correct Answer Is : 182

Solution Is :

If X=X24 + 1/X12 = 7then the X72 + 1/X36 value of

Correct Answer Is : 322

Solution Is :

If 5x + 9y = 5 and 125x + 729y = 120 then the value of the product of x and y is

Correct Answer Is : 1/135

If p = 99 then, the value of p(p2 + 3p + 3) is :

Correct Answer Is : 999999

Solution Is : Given P = 99 then P(P2 + 3P + 3) = P((P + 1)2+ P + 2) = 99 ((99 + 1)2 + 101) = 99 × (10000 + 101) = 999999

The least number which when divided by 48, 64.90, 120 will leave the remainders 38, 54, 80, 110 respectively, is

Correct Answer Is : 2870

Solution Is :

Correct Answer Is :

Solution Is :

Correct Answer Is : 970

Solution Is :

Correct Answer Is : 19-Feb

Solution Is :

Kids Learning

Academics

Competitive Exams

Govt. Jobs Exams

If a=√x+2+√x-2/√x+2√x-2 then the value of (a²-ax)is

Correct Answer Is : X²-1

If p = 99 then, the value of p(p² + 3p + 3) is :

Solution Is : Given P = 99
then P(P² + 3P + 3)
= P((P + 1)²+ P + 2)
= 99 ((99 + 1)² + 101)
= 99 × (10000 + 101)
= 999999