Dear Aspirants, you can find the Quantitative Aptitude questions with detailed explanations for the SSC exams. Nowadays the competitive level of the exam has been increasing consistently. Due to the great demand for the government job, the level of the toughness reached greater. Candidates have to enhance the preparation process in order to drive in the right path. It doesn’t need to clear the prescribed cutoff. You must have to score good marks more than the cut off marks to get into the final provisional list. Here we have updating the Quantitative Aptitude questions with detailed explanations on a daily basis. You can practice with us and measure your level of preparation. According to that you can sculpt yourself in a proper way. SSC aspirants kindly make use of it and grab your success in your career.
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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This post was last modified on March 11, 2021 3:36 pm