# Quantitative Aptitude Questions Daily Quiz Day – 61

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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

ATQ,

a = 61

b = 63

c = 65 = 2268

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

ATQ,

Put x = 0, Z = 1331 & y = 3  x2 + (1/x2) = 7/4

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

x + (1/x) = √15/2

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

We know that.

The ratio of in-radius & circum radius of an equilateral triangle divides 1 : 2 ratio. ∠P+∠Q+∠R=(1+3+5)x=180

So, 9x = 180 ⇒x=20

So, ∠R-∠P=100-20=80

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 So, 25 ratio = 200

1 ratio = 8

So, Area of DEBC = 200 – 8= 192 cm2 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