SSC Quantitative Aptitude Questions (Day-28)

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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

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