Sample Paper

The main purpose of this test is to measure the candidate’s basic competencies in Arithmetic. All the Twenty questions of this test will be based on the following 15 topics.

• Number and numeric system.
• Four fundamental operations on the whole number.
• Fractional numbers and four fundamental operations on them.
• Factors and multiple including their properties.
• LCM and HCF of numbers.
• Decimals and fundamental operations on them.
• Conversion of fractions to decimals and vice versa.
• Applications of numbers measure length, mass, capacity, time, money, etc.
• Distance, time, and speed.
• Approximation of expressions.
• Simplification of Numerical Expressions.
• Percentage and its applications.
• Profit and loss.
• Simple interest.
• Perimeter, area, and volume.

 

Sample Questions: Number and Numeric System

Directions: For every question, four probable answers as (a), (b), (c), and (d) are given. Only one out of these is correct. Choose the correct answer. 

Q1. What is the difference between the greatest 7-digit number and the smallest 4-digit number? [JNV 2020]

(a) 9990999                   

(b) 9993999                   

(c) 9996999                   

(d) 9998999

Answer: d (Explanation: The greatest 7-digit number = 9999999, The smallest 4-digit number = 1000. Difference = 9999999 − 1000 = 9998999)

Q2. What will be the difference between the greatest 6-digit number and the greatest 5-digit number? [JNV 2020]

(a) 100000                   

(b) 100001                   

(c) 99999                   

(d) 900000

Answer: d (Explanation: The greatest 6-digit number = 999999, The greatest 5-digit number = 99999. Difference = 999999 − 99999 = 900000)

 

Sample Questions: Four fundamental operations on whole number

Q3. Which of the following is not equal to 25? [JNV 2020]

(a) 50 − (100 ÷ 4)                   

(b) 20 + (20 ÷ 4)                   

(c) 10 + (5 × 2) + (10 − 5)                   

(d) 24 + 2 × 1

Answer: d

Q4. If 15 − 15 ÷ 15 × 6  = x, then the value of x is [JNV 2020]

(a) 6                   

(b) 0                   

(c) 9                   

(d) 84

Answer: c

Sample Questions: Fractional number and four fundamental operations on them.

Q5. The value of 09 ÷ (03 × 03) is [JNV 2020]

(a) 0.01                   

(b) 0.1                   

(c) 1                   

(d) 10

Answer: d

Q6. Find the sum of 7.7 + 7.77 + 7.777 + 7.777. [JNV 2018]

(a) 28. 2828                   

(b) 28. 2847                   

(c) 30.0247                     

(d) 31.0247

Answer: d

Sample Questions: Factors and multiple including their properties

Q7. A common multiple of both 9 and 7 is A. This number is between 1200 and 1300. What is number A? [JNV 2018]

(a) 1197                     

(b) 1260                   

(c) 1206                     

(d) 1266

Answer: b (Explanation: The multiples of 7 between the range 1200 and 1300 is given by: 1204, 1211, 1218, 1225, 1232, 1239, 1246, 1253, 1260, 1267, 1274, 1281, 1288 and 1295. The multiples of 9 between the range 1200 and 1300 are given by 1206, 1215, 1224, 1233, 1242, 1251, 1260, 1269, 1278, 1287, and 1296. It is obvious from the multiples of 7 and 9, the common multiple is 1260.)

Q8. Which of the following numbers is divisible by 3, 4, 5 and 6? [JNV 2019]

(a) 36                   

(b) 60                   

(c) 80                   

(d) 90

Answer: b (Explanation: Multiples of 60 = ́2 × 2 × 3 × 5 or 4 × 3 × 5 ́or 6 ×10 . Hence, number 60 is divisible by 3, 4, 5, and 6.)

Sample Questions: LCM and HCF of numbers

Q9. The sum of HCF and LCM of 45, 60, and 75 is [JNV 2020]

(a) 330

(b) 960

(c) 915

(d) 630

Answer: c (Explanation: Prime factors of 45, 60 and 75; 45 = 3 × 3 × 5; 60 = 2 × 2 × 3 × 5; 75 =  3 × 5 × 5; Therefore  HCF = 3 × 5 [Common factors of 45, 60, and 75] = 15; LCM = 2 × 2 × 3 × 3 × 5 × 5 (to take the higest power of prime factor) = 900; Therefore, Sum of HCF and LCM = 15 + 900 = 915.)

Q10. The number of numbers which are multiples of both 3 and 5 in the first 100 natural numbers is [JNV 2019]

(a) 10

(b) 9

(c) 7

(d) 6

Answer: d (LCM of 3 and 5 = 15. The numbers which are multiples of both 3 and 5 are 15, 30, 45, 60, 75, 90. Total numbers = 6)

Sample Questions: Decimals and fundamental operations on them.

Q11. 140.75 × 0.01 is equal to [JNV 2020]

(a) 140.75

(b) 14000.75

(c) 1.4075

(d) 0.14075

Answer: c

Q12. Find the value of 3× 0.3 × 0.03 × 0 × 30 [JNV 2018]

(a) 81

(b) 8.1

(c) 0.81

(d) 0

Answer: d (Any number multiplied by 0 is equal to 0.)

Sample Questions: Conversion of fractions to decimals and vice versa.

Q13. The decimal equivalent of is [JNV 2015]

(a) 1.870

(b) 18.70

(c) 187

(d) 1870

Answer: a

Q14.  is equal to [JNV 2014]

(a) 101/10

(b) 1101/100

(c) 11/10

(d) 1001/100

Answer: a

Sample Questions: Applications of number in measure length, mass, capacity, time, money, etc.

Q15. 5045 grams is equal to [JNV 2019]

(a) 50 kg, 45 gm

(b) 5 kg, 45 gm

(c) 5 kg, 450 gm

(d) 50 kg, 450 gm

Answer: b

Q16. 5 minutes past 3, in the afternoon, is written as

(a) 5 : 30 am

(b) 5 : 30 pm

(c) 3 : 50 pm

(d) 3 : 05 pm

Answer: d

Q17. An article is sold for Rs. 500 and hence a loss is incurred. Had the article been sold for Rs. 700, the shopkeeper would have gained three times the former loss. What is the cost price of the article? [JNV 2019]

(a) Rs. 525

(b) Rs. 550

(c) Rs. 600

(d) Rs. 650

Answer: b (Explanation: Let the cost price (C.P) of the article be x, Selling price (S.P) = Rs.500; Since there is a loss, so C.P will be greater than S.P. Therefore, loss = C.P – S.P = x – 500; In Second case, since C.P remains the same and S.P = Rs. 700. Since there is a gain this time, gain = S.P – C.P = 700 – x; According to the given condition: Gain= 3 × Former loss; So, 700-x = 3( x-500); Solving for x we get x = Rs. 550.)

Sample Questions: Distance, time, and speed

Q18. A park is 1500 m long and 750 m wide. A cyclist has to take four rounds of this park. How much time will he take at the speed of 4.5 km/hr? [JNV-2020]

(a) 40 hr

(b) 20 hr

(c) 10 hr

(d) 4 hr

Answer: d (Explanation: First must first calculate the perimeter of the park. The park is in form of a rectangle. Perimeter of the rectangle = (Length +  Width) × 2 = (1500 + 750) × 2 = 4500 meters. The Cyclist takes four rounds. Four rounds is equivalent to; 4500 × 4 = 18000 meters. So, the cyclist covers a distance of 18000 meters. Since the speed is given in km/h, we convert this distance to kilometers. 1km = 1000m; so, 18000 meters = 18 km. Given speed and distance, the formula for getting time is: Time = Distance/Speed = 18/4.5 = 4 hours.) 

Q19. If a man travels at a speed of 30 km/hr, he reaches his destination 10 min late and if he travels at a speed of 42 km/hr, he reaches his destination 10min early. The distance traveled is [JNV 2019]

(a) 36 km

(b) 35 km

(c) 40 km

(d) 42 km

Answer: b (Explanation:  Let the distance be x km. Then, correct time at a speed of  30 km/hr = x/30 – 10/60 (reaches 10 minutes late). Correct time at a speed of 42 km /hr = x/42 + 10/60 (reaches 10 minutes early). Solving for x using both equation, we get x/30 – 1/6 = x/42 + 1/6; x = 35 Km.)

Q20. A passenger train, running at a speed of 80 km/hr leaves a railway station 6hr after a goods train leaves and overtake it in 4 hr. What is the speed of the goods train?

(a) 32 km/hr

(b) 48 km/hr

(c) 60 km/hr

(d) 50 km/hr

Answer: a (Explanation: Distance covered by the passenger train in 4 Hrs is 80 × 4 = 320 Km. Let, speed of goods train = V. Time taken by goods train = 10 hr. Since the distance covered by both the trains is same.Now, by using, Speed = D/t = 320/10 = 32 Km/hr.)

Sample Questions: Approximation of expressions

Q21. Find the approximate result of the following expression (in whole numbers).
49.6 × 10.2 −7.1 × 29.7 − 5.1 × 20.1 [JNV 2020]

(a) 390                   

(b) 290                   

(c) 209                   

(d) 190

Answer: d (Explanation: (d) 49.6 × 10.2 − 7.1 × 29.7 − 5.1 × 20.1 = 50 × 10 − 7 × 30 − 5 × 20 (take the nearest integer value) = 190.)

Q22. What is the approximate value of 275.0003 × 3.005? [JNV 2010]

(a) 825

(b) 830

(c) 810

(4) 835

Answer: a (Explanation: 275.0003 × 3.005 = 826.3759 ≈ 825)

Sample Questions: Simplification of Numerical Expressions

Q23. What is the product of 9680 × 10 × 14 × 0 × 8? [JNV 2016]

(a) 561260

(b) 642976

(c) 912040

(d) 0

Answer: d

Q24. The simplification of 10 + 4 ÷ 2− 3 × 2 + 4 ÷ 2 × 2 − 4 gives [JNV 2004, 1995]

(a) 0

(b) 1

(c) 6

(d) 8

Answer: c (Explanation: Division and Multiplication (Execute in Left to right sequence) have higher precedence than addition and subtraction.)

Sample Questions: Percentage and its applications

Q25. A student scored 18 marks out of 25 marks in the first test of Math. In the second test, he scored 22 marks. In the second test, the student marks exceed his first test by [JNV 2017] 

(a) 4%

(b) 8%

(c) 16%

(d) 80%

Answer: c (Explanation: More marks in second test = 22 − 18 = 4; Exceed in marks percentage = (4 × 100) / 25 = 16%)

Q26. A boy gets Rs. 20 per month and spends 50% of it. How much does he save in 1 yr? [JNV 2012]

(a) Rs. 100

(b) Rs. 50

(c) Rs. 120

(d) Rs. 40

Answer: c (Explanation: Monthly saving amount by the boy 20 × (50 / 100) = Rs. 10. Annually saving amount = 12 × 10 = Rs. 120.)

Sample Questions: Profit and loss

Q27. A shopkeeper bought 2 dozen brushes at the rate of Rs. 10 per dozen. If he sells them at Rs. 1 per brush, what profit will he earn? [ JNV 2002]

(a) Rs. 9

(b) Rs. 7

(c) Rs. 6

(d) Rs. 4

Answer: d (Explanation: The Cost Price of 2 dozen brushes = 2 × 10 = Rs. 20; Sale Price of 1 brush = Rs. 1; SP of 2 dozen or 24 brushes = 24 × 1 = Rs. 24. Therefore, Profit = 24 − 20 = Rs. 4)

Q28. An old table was purchased for Rs. 180 and Rs. 20 was spent on its repairs. If it was sold at a profit of 20%, the selling price of the table was [ JNV 2001]

(a) Rs. 200

(b) Rs. 216

(c) Rs. 240

(d) Rs. 250

Answer: c (Explanation: Total CP of the table = 180 +20 = Rs. 200. Profit = 20%. Therefore, SP = CP × (100+Profit %)/100 = 200 ×(100+20)/100 = Rs.240.)

Sample Questions: Simple Interest

Q29. Find the simple interest on Rs. 600 for 6 yr at 10% per annum. [JNV 1999]

(a) Rs. 300

(b) Rs. 350

(c) Rs. 360

(d) Rs. 380

Answer: c (Explanation: SI = (Principal × Rate × Time)/100 = (600 × 10 × 6)/100 = Rs. 360)

Q30. If SI on Rs. 5000 in 2 yr is Rs. 500, the amount is [JNV 1997]

(a) Rs. 4500

(b) Rs. 5500

(c) Rs. 5575

(d) Rs. 6000

Answer: b (Explanation: Rs. 5000 + Rs. 500 = Rs. 5500)

Sample Questions: Perimeter, area and volume

Q31. The area of square, whose perimeter is 48 m, is [JNV 2017, 2009, 2004]

(a) 48 m²

(b) 144 m²

(c) 1152 m²

(d) 2304 m²

Answer: b (Explanation: Side of the square = Perimeter / 4 = 48/4 = 12m. Area of the square = Side × Side = 12 × 12 = 144 square meters)

Q32. What is the volume of a box whose each edge measures 3 m in length? [JNV 2017, 2009]

(a) 54 cu m

(b) 27 cu m

(c) 18 cu m

(d) 9 cu m

Answer: b (Explanation: Volume of the box = 3 × 3 × 3 = 27 cu m)

Q33. The number of 15 cm square tiles required to lay a floor of size 3.6 m × 4.5 m is [JNV 2004] 

(a) 720

(b) 360

(c) 10800

(d) 5400

Answer: a (Explanation: Number of tiles required = Area of floor/Area of 1 tile = (3.6 × 4.5) / (0.15 x 0.15) = 720.)

Q34. A rectangle is formed by 100 cm wire. Find out the maximum area of this rectangle. [JNV 2010]

(a) 100 sq cm

(b) 400 sq cm

(c) 625 sq cm

(d) 10000 sq cm

Answer: c (Explanation: Maximum a square is formed in a rectangle. Perimeter of square = 100 cm/4 = 25 cm. Area of rectangle = Area of square = 25 × 25 = 625 sq cm)

Q35. The volumes of a cube and a cuboid are equal. If the dimensions of the cuboid are 18 cm, 12 cm and 8 cm the edge of the cube is [JNV 2001]

(a) 8 cm

(b) 10 cm

(c) 12 cm

(d) 16 cm

Answer: c (Explanation: Volume of cuboid = l × b × h = 18 × 12 × 8 = 1728 cu cm. Volume of cube = Volume of cuboid
= 18 × 12 × 8 cu cm. Therefore, Edge of cube = Cube root (18 × 12 × 8) = 12 cm.)