RRB NTPC Quant quiz for Stage I : 17/03/2019

RRB NTPC Quant quiz for Stage I

Railway NTPC Quant quiz for (Phase I), RRB (JE), RRB ALP, SSC various exams and other competitive exams.

Q1. Six times of a positive integer is equal to 8 less than twice the square of that number. Find the number.

A) 4

B) 12

C) 24

D) 32

Q2.    In a workshop 18 men and 14 children can complete a work in 30 days. 15 men can complete the same work in 60 days. In how many days will 14 men and 7 children complete that work?

A) 45

B) 36

C) 40

D) 48

Q3.  If P. cos θ + Q. sin θ = 8 and P. sin θ – Q. cos θ = 5, what is the value of (P2 + Q2)?

A) 89

B) 97

C) 101

D) 99

Q4.   P, Q, and R are the midpoints of the sides AB, BC, and CA respectively of the ΔABC. If the area of ΔPQR is 8 cm², find the area of ΔABC?

A) 40 sqcm

B) 32 sqcm

C) 16 sqcm

D) 24 sqcm

Q5. If x + y = 5 and x2 + y2 = 25, which one of the following is the value of x3 + y3?

A) 450

B) 361

C) 243

D) 125

Q6. AB is a diameter and AC is a chord of a circle such that ∠ BAC =30°. The tangent at C intersects AB produced in D. Then,

A) BC  < BD

B) BC > BD

C) BC = BD

D) None of these

Q7.  3 years ago the age of a father was 1 year more than 4 times his son’s age. After 5 years the sum of their ages will be 57 years. Find the present age of the father.

A) 25 years

B) 36 years

C) 48 years

D) 40 years

Q8.  Four pipes of lengths 75 cm, 100 cm, 125 cm, and 200 cm are to be cut into parts of equal lengths. Each part must be as long as possible. What is the maximum number of pieces that can be cut?

A) 15

B) 20

C) 30

D) 25

Q9. If 3x + 3x+1 = 36, then the value of xx  is : 

A) 64

B) 2125

C) 81

D) 4

Q10. Find the values of k for which 64x2 + kx + 1 = 0 has equal roots.

A) ± 16

B) ± 14

C) ± 12

D) ± 10

Solutions

Q1. Ans(A)

Traditional Approach :

Let the number be x.

Then, according to the question,

6x = 2x2 – 8

⇒  2x2 – 6x – 8 = 0

⇒  (x – 4)(2x + 2) = 0

⇒  x = 4, –1

Elimination Approach :

Let’s take option A ; 4

6 times of 4 = 24

Square of 4 = 16

∴  Twice of square of 4 = 32

We can clearly observe that 6 times of 4 is 8 less than twice the square of it.

Hence, it gets confirmed that 4 is the answer.

Option A, is hence the correct answer.

Q2. Ans(A)

Given, 15 men can complete a work in 60 days and the same work is done by 18 men and 14 children in 30 days.
So, (18M + 14C) × 30 = 15M × 60
On solving we get, efficiency of  6 Men = efficiency of 7 Children ……….(1)
Also it is asked that in how many days will 14 men and 7 children complete that work.
As, 7 children = 6 men ( from 1)
So 14 men + 7 children = 14 Men + 6 Men = 20 men
Let the no. of days in which 14 men and 7 children will complete that work be ‘a’
Therefore, 20 Men will also complate the same work in ‘a’ days.
According to the question:
15M × 60 = 20M × a

a = 45 days

So, 14 men and 7 children will complete that work in 45 days.

Option A, is hence the correct answer.

Q3. Ans(A)
Given, P. cos θ + Q. sin θ = 8
⇒ P2. cos2 θ + Q2. sin2 θ + 2PQ. sin θ.cos θ = 64     …… (i)
Also given, P. sin θ – Q. cos θ = 5
 
⇒ P2. sin2 θ + Q2. cos2 θ – 2PQ. sin θ.cos θ = 25     ……… (ii)
Adding equations (i) and (ii)

P2 + Q2 = 89

Option A, is hence the correct answer

Q4. Ans(B)
Given, the area of ΔPQR is 8 cm²
By the property of triangle (joining mid point of the sides)

⇒ Area of ΔABC = 4 × area of ΔPQR = 4 × 8 = 32 cm²

Option B, is hence the correct answer.

 

Q5. Ans(D)
As we know, (x + y)2 = x2 + y2 + 2xy

⇒ 52 = 25 + 2xy          [Given x + y = 5 and x² + y² = 25]

So,  
 

⇒ 2xy = 0
⇒ xy = 0
We know, x3 + y3 = (x + y) (x2 + y2 – xy)
⇒ x3 + y3 = 5 × (25 – 0)

⇒ x3 + y3 = 125

Option C, is hence the correct answer.

 

Q6. Ans(C)

∠ ACB = 90° ( ∠ in a semi-circle).

So, ∠ ABC = 60° & ∠ CBD = 120°

Now, ∠ BCD = ∠ BAC = 30°(∠ s in alt. segments)

∴ ∠ BDC = 180° – ( 30°+120°) = 30°

Thus, ∠ BCD = ∠ BDC . So, BC = BD.

 

Q7. Ans(B)

Let present age of father be F and son be S.

F – 3 = 4(S – 3) + 1

⇒ F – 3 = 4S – 12 + 1

⇒ F – 3 = 4S – 11

⇒ F = 4 S – 11 + 3 = 4 S – 8 

⇒F = 4S – 8   …(i)

Also,

(F + 5) + (S + 5) = 57

⇒ F + S + 10 = 57

⇒ F + S = 47

⇒ F = 47 – S    …(2)

Putting value of F from eqn.(i)  in eqn. (ii), we get:

 4 S – 8 = 47 – S 

⇒ 5S = 47 + 8 = 55

⇒ S = 11

Age of son = 11

Putting the age of son in eqn. (2), Age of father (F) = 47 – 11 = 36 years

Option B, is hence the correct answer.

 

Q8. Ans(B)

Given, lengths of four pipes are 75, 100, 125, and 150 cm.

Now,

H.C.F of 75, 100, 125 and 200 is 25.

As,

75 = 25 × 3
100 = 25 × 4
125 = 25 × 5

200 = 25 × 8

So, the pipe, length of which is 75cm is cut into three equal parts of 25cm each.and similarly pipe of 100cm length cut into 4 equal parts and so on.

∴ Number of pieces = 3 + 4 + 5 + 8 = 20.

Length of the piece of pipe that can be cut so that length of all pieces are equal = HCF of the given lengths= 25 cm
Option B, is hence the correct answer.
Q9. Ans(D)

3x + 3x+1 = 36 ⇒ 3x + 3x × 31 = 36

⇒ 3x (1 + 3) = 36 ⇒ 3x = 9 = 32

⇒ x = 2 ∴ xx = 22 = 4

Option D, is hence the correct answer.

Q10. Ans(A)
Given equation, 64x2 + kx + 1 = 0
Condition for real and equal roots, b2 – 4ac = 0
⇒ k2 – 4 × 64 × 1 = 0
⇒ k2 = 256

⇒ k = ± 16

Option A, is hence the correct answer.

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