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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Which is the smallest whole number? |
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Answer» ‘0’ (zero) is the smallest whole number. |
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| 152. |
Are the following statements true: (a) 40 persons : 200 persons = Rs.15 : Rs.75 (b) 7.5 liters : 15 liters = 5 kg = 10 kg (c) 99 kg : 45 kg = Rs.44 : Rs.20 (d) 32 m : 64 m = 6 sec. : 12 sec. (e) 45 km : 60 km = 12 hours : 15 hours |
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Answer» (a) 40 persons : 200 persons = 40/200 =1/5 = 1 : 5 Rs.15 : Rs.75 = 15/75 = 1/5 = 1 : 5 Since, 40 persons : 200 persons = Rs.15 : Rs.75 Hence, the statement is true. (b) 7.5 liters : 15 liters = 7.5/15 = 75/150 = 1/2 = 1 : 2 5 kg : 10 kg = 5/10 = 1/2 = 1 : 2 Since, 7.5 liters : 15 liters = 5 kg : 10 kg Hence, the statement is true. (c) 99 kg : 45 kg = 99/45 = 11/5 = 11 : 5 Rs.44 : Rs.20 = 44/20 =11/5 = 11 : 5 Since, 99 kg : 45 kg = Rs.44 : Rs.20 Hence, the statement is true. (d) 32 m : 64 m = 32/64 = 1/2 = 1 : 2 6 sec : 12 sec = 6 /12 = 1 /2 = 1 : 2 Since, 32 m : 64 m = 6 sec : 12 sec Hence, the statement is true. (e) 45 km : 60 km = 45 /60 =3/4 = 3 : 4 12 hours : 15 hours = 12/15 = 4/5 = 4 : 5 Since, 45 km : 60 km ¹ 12 hours : 15 hours Hence, the statement is true |
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| 153. |
The distance between the school and the house of a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days. |
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Answer» Distance between school and home = 1.875 km Distance between home and school = 1.875 km Total distance covered in one day = 1.875+1.875 = 3.750 km Distance covered in six days = 3.750 x 6 = 22.500 km Therefore, 22 km 500 m distance covered in six days. |
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| 154. |
Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.(a) 5m = 60 (10, 5, 12, 15)(b) n + 12 = 20 (12, 8, 20, 0)(c) p - 5 = 5 (0, 10, 5, –5)(d) q/2 = 7 (7, 2, 10, 14)(e) r - 4 = 0 (4, –4, 8, 0)(f) x + 4 = 2 (–2, 0, 2, 4) |
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Answer» (a) 5m = 60 Putting the given values in L.H.S., 5x10=50 ∵ L.H.S. ¹ R.H.S. ∴ m = 10 is not the solution. 5x12=60 ∵L.H.S. = R.H.S. ∴ m = 12 is a solution 5x5=25 ∵ L.H.S. ¹ R.H.S. ∴ m = 5 is not the solution. 5x15=75 ∵ L.H.S. ¹ R.H.S. ∴ m = 15 is not the solution. (b) n + 12 = 20 Putting the given values in L.H.S. 12+12=24 ∵ L.H.S. ¹ R.H.S ∴ n = 12 is not the solution. 20+12=32 ∵ L.H.S. ¹ R.H.S ∴ n = 20 is not the solution. 8+12=20 ∵ L.H.S. = R.H.S. ∴ n = 8 is a solution. 0+12=12 ∵ L.H.S. ¹ R.H.S ∴ n = 0 is not the solution. (c) p - 5 = 5 Putting the given values in L.H.S., 0-5=-5 ∵ L.H.S. ¹ R.H.S ∴ p = 0 is not the solution. 5-5=0 ∵ L.H.S. ¹ R.H.S ∴ p = 5 is not the solution. 10-5=5 ∵ L.H.S. = R.H.S ∴ p = 10 is a solution. -5-5= -10 ∵ L.H.S. ¹ R.H.S ∴ p = -5 is not the solution. (d) q/2=7 Putting the given values in L.H.S., 7/2 ∵ L.H.S. ¹ R.H.S ∴ q = 7 is not the solution. 10/2=5 ∵ L.H.S. ¹ R.H.S ∴ q = 10 is not the solution. 2/2=1 ∵ L.H.S. ¹ R.H.S ∴ q = 2 is not the solution. 14/2=7 ∵ L.H.S.= R.H.S ∴ q = 14 is a solution. (e) r-4=0 putting the given values in L.H.S., 4-4=0 ∵ L.H.S.= R.H.S ∴ r=4 is a solution. 8-4=4 ∵ L.H.S. ¹ R.H.S ∴ r=8 is not the solution. -4-4= -4 ∵ L.H.S. ¹ R.H.S ∴ r = -4 is not the solution. 0-4= -4 ∵ L.H.S. ¹ R.H.S ∴ r = 0 is not the solution. (f) x + 4 = 2 Putting the given values in L.H.S., -2+4=2 ∵ L.H.S. = R.H.S ∴ x = -2 is a solution. 2+4=6 ∵ L.H.S. ¹ R.H.S ∴ x = 2 is not the solution. 0+4=4 ∵ L.H.S. ¹ R.H.S ∴ x = 0 is not the solution. 4+4=8 ∵ L.H.S. ¹ R.H.S ∴ x = 4 is not the solution. |
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| 155. |
A taxi-driver, filled his car petrol tank with 40 liters of petrol on Monday. The next day, he filled the tank with 50 liters of petrol. If the petrol costs Rs.44 per liter, how much did he spend in all on petrol? |
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Answer» Petrol filled on Monday = 40 liters Petrol filled on next day = 50 liters Total petrol filled = 90 liters = 44 x (100 – 10) = 44 x 100 – 44 x 10 = 4400 – 440 = Rs.3960 Therefore, he spent Rs.3960 on petrol. |
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| 156. |
State whether the following statements are true or false. Give reasons for your answers.(i) Every natural number is a whole number. |
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Answer» (i) True, ( Consider the whole numbers and natural numbers separately. We know that whole number series is 0,1, 2, 3, 4,5..... . We know that natural number series is1, 2, 3, 4,5..... . So, we can conclude that every number of the natural number series lie in the whole number series. Therefore, we conclude that, yes every natural number is a whole number. |
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| 157. |
Write the three whole numbers occurring just before 10001. |
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Answer» 10,001 – 1 = 10,000 10,000 – 1 = 9,999 9,999 – 1 = 9,998 |
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| 158. |
Write the next three natural numbers after 10999. |
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Answer» 10,999 + 1 = 11,000 11,000 + 1 = 11,001 11,001 + 1 = 11,002 |
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| 159. |
Estimate each of the following using general rule: (a) 730 + 998 (b) 796 – 314 (c) 12,904 + 2,888 (d) 28,292 – 21,496 |
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Answer» (a) 730 round off to 700 998 round off to 1000 Estimated sum = 700+1000 = 1700 (b) 796 round off to 800 314 round off to 300 Estimated difference = 800-300 = 500 (c) 12904 round off to 13000 2888 round off to 3000 Estimated sum = 13000+3000 = 16000 (d) 28292 round off to 28000 21496 round off to 21000 Estimated difference= 28000-21000 = 7000 |
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| 160. |
Estimate the following products using general rule: (a) 578 x 161 (b) 5281 x 3491 (c) 1291 x 592 (d) 9250 x 29 |
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Answer» (a) 578 x 161 578 round off to 600 161 round off to 200 The estimated product = 600 x 200 = 1,20,000 (b) 5281 x 3491 5281 round of to 5,000 3491 round off to 3,500 The estimated product = 5,000 x 3,500 = 1,75,00,000 (c) 1291 x 592 1291 round off to 1300 592 round off to 600 The estimated product = 1300 x 600 = 7,80,000 (d) 9250 x 29 9250 round off to 10,000 229 round off to 30 The estimated product = 10,000 x 30 = 3,00,000 |
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| 161. |
Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens): (a) 439 + 334 + 4317 (b) 1,08,737 – 47,599 (c) 8325 – 491 (d) 4,89,348 – 48,365 |
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Answer» (a) 439 round off to 400 334 round off to 300 4317 round off to 4300 Estimated sum = 400+300+4300 = 5000 (b) 108734 round off to 108700 47599 round off to 47600 Estimated difference = 108700-47600 = 61100 (c) 8325 round off to 8300 491 round off to 500 Estimated difference = 8300-500 = 7800 (d) 489348 round off to = 489300 48365 round off to = 48400 Estimated difference = 489300-48400 = 440900 |
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| 162. |
Write opposite of the following: (a) Increase in weight (b) 30 km north (c) 326 BC (d) Loss of Rs.700 (e) 100 m above sea level |
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Answer» (a) Decrease in weight (b) 30 km south (c) 326 AD (d) Profit of Rs.700 (e) 100 m below sea level |
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| 163. |
See the figure and find the ratio of: (a) Number of triangles to the number of circles inside the rectangle. (b) Number of squares to all the figures inside the rectangle. (c) Number of circles to all the figures inside the rectangle. |
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Answer» (a) Ratio of number of triangle to that of circles = 3/2 = 3 : 2 (b) Ratio of number of squares to all figures = 2/7 = 2 : 7 (c) Ratio of number of circles to all figures = 2/7 = 2:7 |
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| 164. |
Use the figure to name: (a) Five points (b) A line (c) Four rays (d) Five line segments |
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Answer» (a) Five points are: O, B, C, D, E (b) A line: DE bar, DB bar, OE bar, OB bar (c) Four rays: OD, OE, OC, OB (d) Four line segments: DE bar, OE bar, OC bar, OB bar, OD bar. |
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| 165. |
Between which two numbers in tenths place on the number line does each of the given number lie? (a) 0.06 (b) 0.45 (c) 0.19 (d) 0.66 (e) 0.92 (f) 0.57 |
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Answer» All the numbers lie between 0 and 1. (a) 0.06 is nearer to 0.1. (b) 0.45 is nearer to 0.5. (c) 0.19 is nearer to 0.2. (d) 0.66 is nearer to 0.7. (e) 0.92 is nearer to 0.9. (f) 0.57 is nearer to 0.6. |
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| 166. |
Write each of the following as decimals: (a) seven-tenths (b) Two tens and nine-tenths (c) Fourteen point six (d) One hundred and two-ones (e) Six hundred point eight |
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Answer» (a) seven-tenths = 7 tenths = 7/10 = 0.7 (b) 2 tens and 9-tenths = 2 x 10 + 9/10 = 20 + 0.9 =20.9 (c) Fourteen point six = 14.6 (d) One hundred and 2-ones = 100 + 2 x 1 = 100 + 2 = 102 (e) Six hundred point eight = 600.8 |
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| 167. |
(a) Identify three triangles in the figure: (b) Write the names of seven angles. (c) Write the names of sic line segments. (d) Which two triangles have ∠B as common? |
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Answer» (a) The three triangles are: △ABC, △ABD, △ADC (b) Angles are: ∠ADB, ∠ADC, ∠ABD, ∠ACD, ∠BAD, ∠CAD, ∠BAC (c) Line segments are: AB bar, AC bar, AD bar, BD bar, DC bar, BC bar (d) Triangles having common ∠B: △ABC, △ABD, |
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| 168. |
Raju bought a book of Rs.35.65. He gave Rs.50 to the shopkeeper. How much money did he get back from the shopkeeper? |
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Answer» Total amount given to shopkeeper = Rs.50 Cost of book = Rs.35.65 Amount left = Rs.50.00 = Rs.35.65 = Rs.14.35 Therefore, Raju got back Rs.14.35 from the shopkeeper. |
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| 169. |
Subtract:23 - (-12) |
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Answer» 23– (–12) = 23 + 12 = 35 |
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| 170. |
Subtract:(-32) - (-40) |
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Answer» (–32) – (–40) = –32 + 40 = 8 |
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| 171. |
Write seven consecutive composite numbers less than 100 so that there is no prime number between them. |
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Answer» 90, 91, 92, 93, 94, 95, 96 |
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| 172. |
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students) |
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Answer» Number of students = s Number of pencils to each student = 5 Therefore, total number of pencils needed are = 5s |
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| 173. |
Write down separately the prime and composite numbers less than 20. |
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Answer» Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 Composite numbers: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 |
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| 174. |
Give three pairs of prime numbers whose difference is 2.[Remark: Two prime numbers whose difference is 2 are called twin primes.] |
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Answer» 3 and 5; 5 and 7; 11 and 13 |
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| 175. |
What is the greatest prime number between 1 and 10? |
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Answer» The greatest prime number between 1 and 10 is ‘7’. |
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| 176. |
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3.Find such pairs of prime numbers up to 100. |
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Answer» 17 and 71; 37 and 73; 79 and 97 |
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| 177. |
Express the following as the sum of two odd numbers:(a) 44 (b) 36 (c) 24 (d) 18 |
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Answer» (a) 3 + 41 = 44 (b) 5 + 31 = 36 (c) 7 + 17 = 24 (d) 7 + 11 = 18 |
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| 178. |
Find the areas of the rectangles whose sides are:2 km and 3 km |
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Answer» Area of rectangle = length x breadth = 2 km x 3 km = 6 km2 |
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| 179. |
A room us 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room? |
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Answer» Length of room = 4 m and breadth of room = 3 m 50 cm = 3.50 m Area of carpet = length x breadth = 4 x 3.50 = 14m2 |
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| 180. |
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples. |
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Answer» Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6. Example: (i) 2 x 3 x 4 = 24 (ii) 4 x 5 x 6 = 120 |
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| 181. |
State whether the following statements are true or false:(a) The sum of three odd numbers is even.(b) The sum of two odd numbers and one even number is even.(c) The product of three odd numbers is odd.(d) If an even number is divided by 2, the quotient is always odd.(e) All prime numbers are odd.(f) Prime numbers do not have any factors.(g) Sum of two prime numbers is always even.(h) 2 is the only even prime number.(i) All even numbers are composite numbers.(j) The product of two even numbers is always even. |
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Answer» (a) False (b) True (c) True (d) False (e) False (f) False (g) False (h) True (i) False (j) True |
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| 182. |
What is the sum of any two:(a) Odd numbers.(b) Even numbers |
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Answer» (a) The sum of any two odd numbers is an even number. Example: 1 + 3 = 4, 3 + 5 = 8 (b) The sum of any two even numbers is an even number. Example: 2 + 4 = 6, 6 + 8 = 14 |
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| 183. |
Find the areas of the rectangles whose sides are:2 m and 70 cm |
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Answer» Area of rectangle = length x breadth = 2 m x 70 cm = 2 m x 0.7 m = 2.7 m2 |
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| 184. |
The sum of three consecutive numbers is always divisible by 4. Verify this statement with the help of some examples. |
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Answer» 3 + 5 = 8 and 8 is divisible by 4. 5 + 7 = 12 and 12 is divisible by 4. 7 + 9 = 16 and 16 is divisible by 4. 9 + 11 = 20 and 20 is divisible by 4. |
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| 185. |
Ravi purchases 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases. |
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Answer» Weight of Rice = 5 kg 400 g = 5.400 kg Weight of Sugar = 2 kg 20 g = 2.020 kg Weight of Flour = 10 kg 850 g = 10.850 kg Total weight = 5.400 + 2.020 + 10.850 = 18.270 kg Therefore total weight of Ravi’s purchase = 18.270 kg |
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| 186. |
Determine if 25110 is divisible by 45.[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9.] |
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Answer» The prime factorization of 45 = 5 x 9 25110 is divisible by 5 as ‘0’ is at its unit place. 25110 is divisible by 9 as sum of digits is divisible by 9. Therefore, the number must be divisible by 5 x 9 = 45 |
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| 187. |
Which of the following numbers are co-prime:(a) 18 and 35 (b) 15 and 37 (c) 30 and 415 (d) 17 and 68(e) 216 and 215 (f) 81 and 16 |
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Answer» (a) Factors of 18 = 1, 2, 3, 6, 9,18 Factors of 35 = 1, 5, 7, 35 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. (b) Factors of 15 = 1, 3, 5,15 Factors of 37 = 1, 37 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. (c) Factors of 30 = 1, 2, 3, 5, 6, 15, 30 Factors of 415 = 1, 5, …….., 83,415 Common factor = 1, 5 Since, both have more than one common factor, therefore, they are not co-prime numbers. (d) Factors of 17 = 1, 17 Factors of 68 = 1, 2, 4, 17, 34, 86 Common factor = 1, 17 Since, both have more than one common factor, therefore, they are not co-prime numbers. (e) Factors of 216 = 1, 2, 3, 4, 6, 8, 36, 72, 108, 216 Factors of 215 = 1, 5, 43, 215 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. (f) Factors of 81 = 1, 3, 9, 27, 81 Factors of 16 = 1, 2, 4, 8, 16 Common factor = 1 Since, both have only one common factor, i.e., 1, therefore, they are co-prime numbers. |
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| 188. |
Find the value of the following:81265 x 169 – 81265 x 69 |
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Answer» 81265 x 169 – 81265 x 69 = 81265 x (169 – 69) = 81265 x 100 = 8126500 |
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| 189. |
Write the natural numbers from 102 to 113. What fraction of them are prime numbers? |
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Answer» Natural numbers from 102 to 113: 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113 Prime numbers from 102 to 113: 103, 107, 109, 113 Hence fraction of prime numbers = 4/12 = 1/3 |
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| 190. |
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted. |
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Answer» Length of floor = 5 m and breadth of floor = 4 m = 5 m x 4 m = 20 m2 Now, Side of square carpet = 3 m Area of square carpet = side x side = 3 x 3 = 9 m2 Area of floor that is not carpeted = 20 m2 – 9 m2 = 11 m2 |
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| 191. |
What is the H.C.F. of two consecutive:(a) numbers?(b) even numbers?(c) odd numbers? |
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Answer» (a) H.C.F. of two consecutive numbers be 1. (b) H.C.F. of two consecutive even numbers be 2. (c) H.C.F. of two consecutive odd numbers be 1. |
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| 192. |
Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times. |
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Answer» For finding maximum weight, we have to find H.C.F. of 75 and 69. Factors of 75 = 3 x 5 x 5 Factors of 69 = 3 x 69 H.C.F. = 3 Therefore the required weight is 3 kg. |
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| 193. |
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land? |
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Answer» Side of square bed = 1 m Area of square bed = side x side = 1 m x 1 m = 1 m2 Area of 5 square beds = 1 x 5 = 5 m2 Now, Length of land = 5 m and breadth of land = 4 m Area of land = length x breadth = 5 m x 4 m = 20 m2 Area of remaining part = Area of land – Area of 5 flower beds = 20 m2 – 5 m2 = 15 m2 |
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| 194. |
H.C.F. of co-prime numbers 4 and 15 was found as follows by factorization:4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F.? |
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Answer» No. The correct H.C.F. is 1. |
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| 195. |
When the integers 10, 0, 5, – 5, – 7 are arranged in descending or ascending order, them find out which of the following integers always remains in the middle of the arrangement.(a) 0 (b) 5 (c) – 7 (d) – 5 |
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Answer» After arranging ascending or decending 0 lies in the middle Therefore correct answer is (a) 0 |
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| 196. |
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows) |
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Answer» Number of rows = n Cadets in each row = 5 Therefore, total number of cadets = 5n |
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| 197. |
Find the product by suitable arrangement:8 x 291 x 125 |
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Answer» 8 x 291 x 125 = (8 x 125) x 291 = 1000 x 291 = 291000 |
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| 198. |
Find the value of the following:297 x 17 + 297 x 3 |
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Answer» 297 x 17 + 297 x 3 = 297 x (17 + 3) = 297 x 20 = 5940 |
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| 199. |
Find the sum by suitable rearrangement:837 + 208 + 363 |
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Answer» 837 + 208 + 363 = (837 + 363) + 208 = 1200 + 208 = 1408 |
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| 200. |
Find the sum by suitable rearrangement:1962 + 453 + 1538 + 647 |
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Answer» 1962 + 453 + 1538 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600 |
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