**Dear Aspirants,** **Our SSC Crackers team** is providing a new series of Quantitative Aptitude Questions for **Upcoming Exam **so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

**1) When a = 61, b = 63 and c = 65, then what is the value of a ^{3} + b^{3} + c^{3} – 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 th****en what is the value of ****?**

(a) 0

(b) 1

(c) 17/5

(d) 3

**4) If x ^{2}+ 1/x^{2} = 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 cm ^{2}, then what is the area (in cm^{2}) 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 cm ^{2}, then what is the area (in cm^{2}) 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} = (√2^{2} + 1^{2} – 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**

x^{2} + (1/x^{2}) = 7/4

Add 2 both sides.

x^{2} + (1/x^{2}) + 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

x^{2}/yz+y^{2}/xz+z^{2}/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 = √l^{2} + b^{2} =√35^{2} + 49^{2} = √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 cm^{2}

**10) Answer: D**

Given that,

ABCD is square.

Then, PB = (1/3)AB

BQ = (1/3) BC

Area of BPDQ = 20cm^{2}

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 = 20cm^{2}

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

So, 3 Ratio = 20cm^{2}

9 Ratio = 60 cm^{2}

So, area of square ABCD = 60 cm^{2}