**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) If x + 4 is a factor of 3x ^{2} + 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} = a^{3} + 12a^{2} + 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 x^{3}+(1/x^{3})?**

(a) 2√5

(b) 3√5

(c) 4√5

(d) 5√5

**5) If (x ^{2} + 1 / x) = 4(1/4), then what is the value of x^{3}+ (1/x^{3}) ?**

(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= 70 ^{o}, ∠ACB **

**= 45**

^{o}and ∠**DEA = 140**

^{o}. What is the value of ∠**BDE?**

(a) 10^{o}

(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, x^{3}+1/x^{3} =?

(x+1/x)=√5

(x+1/x)^{ 3}=(x^{3}+1/x^{3} )+3(x+1/x)

5√5=x^{3}+1/x^{3} +3√5

x^{3}+1/x^{3} =2√5

**5) ****Answer: D**

(x^{2}+1)/x=17/4

x+1/x=17/4

(x^{3}+1/x^{3} )=(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