RRB NTPC Quant quiz for Stage I : 05/03/2019

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. Find the unit digit in 1! + 2! + 3! + 4! + ………… + 50!. 

A)  

B) 3 

C) 5 

D) 8 

Q2.  What would be the remainder when 10⁶ – 12 is divided by 9? 

A)  

B) 4 

C) 7

D) 3 

Q3. 

A) –2 

B) –1 

C) 0 

D) 1 

Q4.  What is the maximum number of common tangents which can be drawn to the circle, if the distance between the circles is equal to the sum of their radii? 

A) 0 

B) 1 

C) 3 

D)  

Q5.   If the two circles touch each other internally, then only _______________ common tangent can be drawn to them

A)  0

B) 1 

C) 2  

D)  3 

Q6. Rajiv scored 20% marks in an exam and failed by 11 marks. If he scored 50% marks, then he gets 4 marks more than passing marks. What is the passing marks for the exam? 

A) 21 

B) 33 

C) 50 

D) 45 

Q7. A bag contains 1 rupee, 50 paise and 25 paise coins. The value of 1 rupee, 50 paise and 25 paise coins are in the ratio of 5 : 3 : 1 respectively. If total number of coins is 180, then what will be the total value (in Rs) of all coins? 

A)  112 

B) 180 

C) 196 

D) 108 

Q8.   If the difference between the average of d, e and e, h is 19, then what will be the difference between d and h? 

A) 38 

B) 22 
C) 9 

D) 19

Q9.  The marked price of a shirt is Rs 10400. A shopkeeper allows a discount of 20% on shirt and gains 30%. If no discount is given, then what will be the profit percentage? 

A) 62.5 

B) 37.5 

C)  44 

D) 45 

Q10. ΔDEF is right angled at E. If m∠D = 60°, then find the value of (cotF + 1/√3). 

A) 3√3/2 

B)  4/√3 

C)  (2+2√3)/√3 

D 7/2√3 

Solutions

Q1. Ans(B)

Last digit after 4! is O 
So, we calculate last digit, only 1! To 4! 
Req. last digit =- 1 + 1 × 2 + 3 × 2 × 1 + 4 × 3 × 2 × 1 
= 1 + 2 + 6 + 24 
= 33 
= 3 

Q2. Ans(C)

Q3. Ans(B)

Q4. Ans(C)

Q5. Ans(B)

Q6. Ans(A)

Q7. Ans(D)

Let value of coins are = 5x, 3x, 1x 
Then total number of coins = 5x + 3x × 2 + x × 4 = 15x 
Atq, 
15x = 180 
X = 12 
Total value of all coins = 5x + 3x + 1x = 9x 
= 9 × 12 = 108 

Q8. Ans(A)

\

Q9. Ans(A)

Q10. Ans(B)

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