SSC Quantitative Aptitude Questions (Day-28)

Dear Aspirants, you can find the Quantitative Aptitude questions with detailed explanations for the SSC exams. Nowadays the competitive level of the exam has been increasing consistently. Due to the great demand for the government job, the level of the toughness reached greater. Candidates have to enhance the preparation process in order to drive in the right path. It doesn’t need to clear the prescribed cutoff. You must have to score good marks more than the cut off marks to get into the final provisional list. Here we have updating the Quantitative Aptitude questions with detailed explanations on a daily basis. You can practice with us and measure your level of preparation. According to that you can sculpt yourself in a proper way. SSC aspirants kindly make use of it and grab your success in your career.

Start Quiz

1) When a = 61, b = 63 and c = 65, then what is the value of a³ + b³ + c³ – 3abc?

(a) 1456

(b) 2268

(c) 4536

(d) 5460

2) If x = 3 – 2√2, then √x + (1/√x) is equal to ____.

(a) 0

(b) 1

(c) 2

(d) 2√2

3) If  then what is the value of ?

(a) 0

(b) 1

(c) 17/5

(d) 3

4) If x²+ 1/x² = 7/4 for x > 0, then what is the value of x+ 1/ x?

(a) 2

(b) √15/2

(c) √5

(d) √3

5) If x + y + z = 0, then what is the value of  ?

(a) 0

(b) 1/3

(c) 1

(d) 3

6) What is the ratio of in-radius and circum-radius of an equilateral triangle?

(a) 1 : 2

(b) 1 : 3

(c) 1 : 4

(d) 3 : 2

7) In ΔPQR, ∠P : ∠Q : ∠R = 1 : 3 : 5. What is the value (in degrees) of ∠R – ∠P?

(a) 30

(b) 80

(c) 45

(d) 60

8) A circular wire of length 168 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 5 : 7. What is the length (in cm) of the diagonal of the rectangle?

(a) √4127

(b) √3137

(c) √1813

(d) √3626

9) In ΔABC, a line parallel to side BC cuts the side AB and AC at points D and E respectively and also point D divide AB in the ratio of 1 : 4. If area of ΔABC is 200 cm², then what is the area (in cm²) of quadrilateral DECB?

(a) 192

(b) 50

(c) 120

(d) 96

10) In the given figure, PB is one-third of AB and BQ is one-third of BC. If the area of BPDQ is 20 cm², then what is the area (in cm²) of square ABCD?

(a) 45

(b) 30

(c) 40

(d) 60

Answers:

1) Answer: B

ATQ,

a = 61

b = 63

c = 65

= 2268

2) Answer: D

x = 3 – 2√2

x = (3 – 2√2)^1/2 = (√2² + 1² – 2√2)^1/2 = (√2 – 1)

√x + (1/√x)  = √2 – 1  + √2  + 1 = 2√2

3) Answer: C

ATQ,

Put x = 0, Z = 1331 & y = 3

4) Answer: B

x² + (1/x²) = 7/4

Add 2 both sides.

x² + (1/x²) + 2 = (7/4) + 2 = (7+18)/4 =15/4

x + (1/x) = √15/2

5) Answer: D

x+y+z=0

Put, x = 1, y = 1 & z = – 2

x²/yz+y²/xz+z²/xy=1/(-2)+1/(-2)+4/1

=4-1=3

6) Answer: A

We know that.

The ratio of in-radius & circum radius of an equilateral triangle divides 1 : 2 ratio.

7) Answer: B

∠P+∠Q+∠R=(1+3+5)x=180

So, 9x = 180 ⇒x=20

So, ∠R-∠P=100-20=80

8) Answer: D

We know that the perimeters of rectangle

2(l+b) = 168

2(5x+7x) = 168  x = 7

So, length of diagonal = √l² + b² =√35² + 49² = √3626

9) Answer: B

AD/AB=1/5

(Area of ADE)/(Area of ABC)=(∆ADE/5)^2=Area/200

So, 25 ratio = 200

1 ratio = 8

So, Area of DEBC = 200 – 8= 192 cm²

10) Answer: D

Given that,

ABCD is square.

Then, PB = (1/3)AB

BQ =  (1/3) BC

Area of BPDQ = 20cm²

Let, DC = 3cm, AB = 3 cm, AP = 2cm PB = 1 cm & CQ = 2cm ,BQ = 1cm

So, the area of ABCD = 3 × 3 (Let)

= 9 unit square.

& Area of ADP = (1/2)×3×2 = 3  unit square

Area of DQC =(1/2)×3×2= 3  unit square

So, Given that,

Area of D P B Q is given = 20cm²

So, that is the value of = 9 – (3 + 3) = 3 unit square.

So, 3 Ratio = 20cm²

9 Ratio = 60 cm²

So, area of square ABCD = 60 cm²