| Join Telegram |  |
| Join Whatsapp Groups |  |
NRA CET Quantitative Aptitude Questions and Answers PDF Download: Are you among those candidates who applied for the NRA CET Exam? and we very well know that now you will be searching all over the internet for the model questions and answers, aren’t we correct! well, this article helps you guys to find all the information regarding the NRA CET Quantitative Aptitude Questions & Answers PDF. Before starting the preparation, you guys should have a basic idea regarding what sort of questions will be asked and how and where to start the preparation.
We as a team tried to accommodate all sorts of model questions and answers with explanations for every specific topic in quantitative aptitude for NRA CET Exam. For better convenience and easy access, we have provided a link for the NRA CET Quantitative Aptitude Questions and Answers PDF at the end of this article. You can click on that link and download the model question paper at any point of time. We hope this will help you to attempt well.
NRA CET Quantitative Aptitude Questions and Answers
From this section, candidates can get the complete and detailed information regarding the NRA CET Quantitative Aptitude Questions and Answers. In the NRA CET Exam, the Quantitative Aptitude section CONTAINS only 25 Questions and the maximum number of marks is 50 marks. With the help of the model question paper provided at the end of this article, you can be able to test your knowledge and speed up your preparation. In the below section, we have provided the NRA CET Quantitative Aptitude Questions and Answers and their explanations. Kindly go through them.
NRA CET Quantitative Aptitude Questions and Answers, Explanation
- How many terms are there in the G.P 3, 6, 12, … , 768?
A. 7
B. 8
C. 10
D. 9
Answer: D
Explanation:
a = 3, r = 6⁄2 = 2
Let number of terms = n
-> tn = 768
-> 768 = arn-1
-> 768 = 3 x 2n-1
-> 768⁄3 = 2n-1
-> 256 = 2n-1
-> 28 = 2n-1
-> n = 9
Therefore the number of terms = 9
2. The difference between the two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number?
A. 295
B. 360
C. 270
D. 240
Answer: C
Explanation:
Let the smaller number be x. Then a larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270.
3. 3889 + 12.952 – ? = 3854.002
A. 47.752
B. 47.95
C. 47.095
D. 47.932
Answer: B
Explanation:
Let 3889 + 12.952 – x = 3854.002.
Then x = (3889 + 12.952) – 3854.002
= 3901.952 – 3854.002
= 47.95
4. What is the product of all the numbers in the dial of a telephone?
A. 1,59,480
B. 1,58,480
C. 1,59,450
D. None of the above
Answer: D
Explanation: Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
5. Find the solution and simplify. 5x−12−11x=3x+12+3x
A. -2
B. 0
C. -3
D. -1
Answer: A
Explanation:
6. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
A. 26
B. 22
C. 24
D. 23
Answer: A
Explanation:
Let the number of hens be x and the number of cows be y.
Then, x + y = 48 …. (i)
and 2x + 4y = 140 x + 2y = 70 …. (ii)
Solving (i) and (ii) we get: x = 26, y = 22.
The required answer = 26.
7. (0.25 x 0.25 – 0.24 x 0.24) / 0.49 = ?
A. 0.49
B. 0.01
C. 0.0006
D. 0.1
Answer: B
Explanation:
As a^2 – b^2 = (a+b)(a-b)
∴ (0.25 x 0.25 – 0.24 x 0.24) / 0.49 = (0.25 + 0.24)(0.25 – 0.24) / 0.49
= (0.49 x 0.01) / 0.49 = 0.01
8. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
A. 20
B. 10
C. 17
D. 3
Answer: D
Explanation:
Let the number be x.
Then, x + 17 = 60/x
x^2 + 17x – 60 = 0
(x + 20)(x – 3) = 0
x = 3.
9. 7X2 is a three-digit number and X is the missing digit. If the number is divisible by 6, the missing digit is
A. 5
B. 3
C. 4
D. 2
Answer: B
Explanation:
If the given number is divisible by 6 so it would be divisible by 2 and 3. As the last digit is 2 whatever be the value of X, it would be divisible by 2.
Now, 7+X+2 = 9 + X, must be divisible by 3.
∴X = 3 makes the number divisible by 3, so 3 is the required digit.
10. If 40% of an amount is 250, what will be 60% of that amount?
A. 375
B. 400
C. 300
D. 320
Answer: A
Explanation:
Let the amount be x.
As per question: 40% of x = 250
(40/100)*x = 250
x= 25000/40
i.e x= 625
Now 60% of 625 = (60/100)*625 = 375
11. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m^3) is:
A. 6420
B. 4830
C. 8960
D. 5120
Answer: D
Explanation:
Clearly, l = (48 – 16)m = 32 m,
b = (36 -16)m = 20 m,
h = 8 m.
Volume of the box = (32 x 20 x 8) m^3 = 5120 m^3.
12. Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
A. 12 kmph
B. 14 kmph
C. 8 kmph
D. 11 kmph
Answer: A
Explanation:
Let the distance travelled by x km.
Then, x/10 – x/15 = 2
3x-2x = 60
x=60km
Time taken to travel 60 km at 10 km/hr =60/10hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = 60/5 kmph.= 12 kmph.
13. A blend contains drain and water in the proportion 5:1. Including 5 liters of water, the proportion of milk to the water gets to be 5:2. The amount of milk in the first blend is
A. 32.5 liters
B. 25 liters
C. 16 liters
D. 22.75 liters
Answer: B
Explanation:
In the given mixture, Let milk = 5x liter and water =x liters.
Then, 5x/x+5= 5/2
=> 10x= 5x+25
=>5x =25
=>x=5
∴ milk in original mixture= (5*5) =25 liters
14. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is?
A. 5000
B. 3500
C. 4050
D. 4000
Answer: D
Explanation:
P + Q = (5050 x 2) = 10100 …. (i)
Q + R = (6250 x 2) = 12500 …. (ii)
P + R = (5200 x 2) = 10400 …. (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 …. (iv)
Subtracting (ii) from (iv), we get P = 4000.
i.e P’s monthly income = Rs. 4000.
15. A sum of money at simple interest becomes Rs. 3000 in 2 years and Rs. 3540 in 5 years. Find the rate of interest.
A. 4.5 %
B. 6%
C. 4 %
D. 5%
Answer: B
Explanation:
After two years the amount is 3000 and in the next three years, it becomes Rs. 3540.
So, we have:
P = Rs. 3000
Amount = Rs. 3540
Interest = 3540 – 3000 = 540
Number of years = 3 years
Rate of interest =?
Formula => Rate of Interest= (I*100)/(P*T)
R= (540*100)/ (3000*3)
= 6%
16. At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?
A. 7.5%
B. 6.5%
C. 6%
D. 7%
Answer: C
Explanation:
Let the rate be R% p.a.
17. The bar graph given below shows the percentage distribution of the total expenditures of a company under various expense heads during 2003.
I) The total amount of expenditures of the company is how many times of expenditure on research and development?
A. 18
B. 20
C. 27
D. 8
Answer: B
Explanation:
II) If the expenditure on advertisement is 2.10 crores then the difference between the expenditure on transport and taxes is?
A. Rs. 35 lakhs
B. Rs. 1.25 crores
C. Rs. 95 lakhs
D. Rs. 65 lakhs
Answer: A
Explanation:
Let the total expenditure be Rs. x crores.
Total expenditure = Rs. 14 crores
and so, the difference between the expenditures on transport and taxes
= Rs. [(12.5 – 10)% of 14] crores
= Rs. [2.5% of 14] crores
= Rs. 0.35 crores
= Rs. 35 lakhs
III) What is the ratio of the total expenditure on infrastructure and transport to the total expenditure on taxes and interest on loans?
A. 13:11
B. 8:7
C. 5:4
D. 9:7
Answer: A
Explanation:
Let the total amount of expenditures be Rs. x.
Then, the total expenditure on infrastructure and transport
IV) If the interest on loans amounted to Rs. 2.45 crores then the total amount of expenditure on an advertisement, taxes, and research and development is?
A. Rs. 4.2 crores
B. Rs. 7 crores
C. Rs. 3 crores
D. Rs. 5.4 crores
Answer: A
Explanation:
Let the total expenditure be Rs. x crores.
Then, 17.5% of x = 2.45 => x = 14.
Therefore Total expenditure = Rs. 14 crores.
and so, the total expenditure on advertisement, taxes and Research and Development
= Rs. [(15 + 10 + 5)% of 14] crores
= Rs. [30% of 14] crores
= Rs. 4.2 crores.
V) The expenditure on the interest on loans is by what percent more than the expenditure on transport?
A. 20%
B. 5%
C. 10%
D. 40%
Answer: D
Explanation:
18. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
A. 11760
B. 266
C. 86400
D. 5040
Answer: A
Explanation:
19. The owner of the cell phone shop charges 23% more than the cost price. If a customer paid 7011 for a cell phone, find the cost price of the cell phone.
A. 6750
B. 5860
C. 5220
D. 5700
Answer: D
Explanation:
Let a number x compare the assumed selling price and actual selling price.
i.e., 123 * x =7011
x = 57
That means if we multiply the assumed value by 57, we get the original selling price.
So, multiply the cost price by 57 to get the original cost price,
i.e., 100 * 57 = 5700.
20. There is a 10% discount on a dozen pairs of trousers marked at Rs. 8000. How many pairs of trousers can be bought with Rs. 2400?
A. 2
B. 8
C. 4
D. 7
Answer: B
Explanation:
21. A group of workers undertakes a task. They can complete the task in 30 days. If 5 of them did not turn in for the work and the remaining workers complete the task in 40 days, find the original number of workers.
A. 25 days
B. 23 days
C. 20 days
D. 21 days
Answer: C
Explanation:
Let the original number of workers = X
X workers can complete the work in 30 days. And (X – 50) complete the same task in 40 days.
Apply formula: M1D1W2 =M2D2W1
W1=W2 as the task is the same in both cases.
Therefore, X * 30 = (X – 5) * 40
30 X = 40X – 200
200 = 40X -30X
200 = 10 X
X = 200/10 = 20days
22. The ratio of the total amount distributed in all the males and females as salary is 6: 5. The ratio of the salary of each male and female is 2: 3. Find the ratio of the no. of males and females.
A. 5:9
B. 9:5
C. 5:7
D. 7:5
Answer: B
Explanation:
The total salary of males: the total salary of females = 6:5
The salary of each male: salary of each female = 2:3
To find the number of men and women, divide the total salary of males and females by the salary of each male and female.
i.e., 6/2: 5/3
Or, 18: 10 = 9: 5
So, the ratio of the number of males and females = 9:5
23. The least number which when divided by 5, 6, 7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is
A. 3363
B. 1677
C. 2523
D. 1683
Answer: D
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
The required number is of form 840k + 3
The least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.
24. In The adjoining figure, ABCD is a rhombus whose diagonals intersect at O. IF ∠OAB =40⁰ and ∠ABO =x⁰, then X= ?
A. 35 degree
B. 40 degree
C. 50 degree
D. 45 degree
Answer: C
Explanation:
We know that the diagonals of a rhombus
bisect each other at right angle. So ,∠ AOB = 90⁰.
Now ,∠ OAB + ∠ABO + ∠AOB = 180⁰
⇒ 40 +x + 90 = 180 ⇒ x=50.
25. In the following table find, the average score of the students
A. 39
B. 41
C. 40
D. 42
Answer: B
Explanation:
NRA CET Quantitative Aptitude Questions and Answers PDF
| NRA CET Quantitative Aptitude Questions and Answers – Important links | |
| To download NRA CET Quantitative Aptitude Questions and Answers PDF | Click Here |
For more updates about the NRA CET Quantitative Aptitude Questions, Answers, follow our page @ freshersnow.com.