Quantitative Aptitude Questions Daily Quiz Day – 64

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1) If x + 4 is a factor of 3x2 + kx + 8 then what is the value of k?

(a) 4

(b) –4

(c) –14

(d) 14

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

(a) 1

(b) –1

(c) 1/2

(d) –1 / 2

3) If (a + 4)3 = a3 + 12a2 + ka + 64, then what is the value of k?

(a) 12

(b) 24

(c) 36

(d) 48

4) If(x + (1/x))2 =5 and x > 0, then what is the value of x3+(1/x3)?

(a) 2√5

(b) 3√5

(c) 4√5

(d) 5√5

5) If (x2 + 1 / x) = 4(1/4), then what is the value of x3+ (1/x3) ?

(a) 529/16

(b) 527/64

(c) 4913/64

(d) 4097/64

6) PQR is an isosceles triangle with sides PQ = PR = 45 cm and QR = 72 cm. PN is a median to base QR. What will be the length (in cm) of PN?

(a) 36

(b) 24

(c) 27

(d) 32

7) In the given figure, ∠BAC= 70o, ∠ACB  = 45o and ∠DEA = 140o. What is the value of ∠BDE?

(a) 10o

(b) 15 o

(c) 20 o

(d) 25 o

8) If the centroid of triangle ABC is G and BG = 9 cm, then what will be the length (in cm) of median BE?

(a) 12

(b) 14

(c) 15

(d) 13.5

9) Angles of a triangle are (y + 36)o, (2y – 14)o and (y – 22)o. What is the value (in degrees) of 2y?

(a) 45

(b) 70

(c) 90

(d) 100

10) In the given figure, PQRS is a parallelogram and U is the midpoint of QR. If PQ = 4 cm, then what is the value of PT (in cm)?

(a) 6

(b) 6.5

(c) 7.5

(d) 8

Answers:

1) Answer: D

X + 4, said to be a factor of the eqn. if x = –4, satisfy the given eqn.

So,

3 (–4)² + K (–4) + 8 = 0

48 – 4K + 8 = 0

4K = 56

K = 14

2) Answer: D

x + y + z = 0

x² + y² + z² + 2 (xy + yz + zx) = 0

x² + y² + z² = – 2(xy + yz + zx)

Put the value of x² + y² + z² in given eqn.

= (xy+yz+zx)/(–2(xy+yz+zx) ) = (-1/2)

3) Answer: D

(a + 4)³ = a³ + 12a² + ka + 64

a³ + 64 + 3 × 4 × a (a + 4) = a³ + 12a² + ka + 64

k = 48

4) Answer: A

(x+1/x)2=5,  x3+1/x3 =?

(x+1/x)=√5

(x+1/x) 3=(x3+1/x3 )+3(x+1/x)

5√5=x3+1/x3 +3√5

x3+1/x3 =2√5

5) Answer: D

(x2+1)/x=17/4

x+1/x=17/4

(x3+1/x3 )=(x+1/x)3–3(x+1/x)

=(17/4) 3–3×17/4

=4097/64

6) Answer: C

In isosceles triangle median is also the altitude

by Pythagoras theorem.

PN² = PQ² – QN²

= 45² – 36²

PN = 27

7) Answer: D

∠EBD = ∠BAC + ∠ACB    (by exterior angle theorem)

∠DEB = 180° – ∠AED (angle on one line)

= 40°

In ∆DBE,

∠BDE = 180° – (∠DEB + ∠BED)

= 180° –(40° + 115°)

= 25°

8) Answer: D

BE=3/2×BG

=3/2×9

= 13.5

9) Answer: C

y + 36° + 2y – 14 + y – 22° = 180°

4y = 180°

2y = 90°

10) Answer: D

SP = 2 × UQ

UQ = y, SP = 2y

In ∆TQU and ∆TPS

(TQ/TP) = (UQ/SP),

(x/(x+4)) = y/2y

2x=x+4

x=4

PT = 4+4 =8

 

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