InterviewSolution
Saved Bookmarks
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.
| 1. |
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years. |
|
Answer» Radha’s age = x years Therefore, Leela’s age = (x - 4) years |
|
| 2. |
Determine the following are in proportion: 33,121,9,96 |
|
Answer» 33 : 121 = 33/121 = 3/11 = 3:11 9 : 96 = 9/96 = 3/32 = 3:32 Since 33:121 ≠ 9:96 Therefore, 33, 121, 9, 96 are not in proportion |
|
| 3. |
Express the following as cm using decimals:4 cm 2 mm |
|
Answer» 4 cm + 2/10 cm 4+0.2 = 4.2 cm |
|
| 4. |
Express the following as cm using decimals:116 mm |
|
Answer» 10 mm = 1 cm 1 mm = 1/10 cm 116 mm = 1/10 x 116 = 11.6 cm |
|
| 5. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:2.5 |
|
Answer» 2.5 = 25/10 = 5/2 |
|
| 6. |
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need? |
|
Answer» Runs to achieve = 10,000 Runs scored = 6,980 Runs required =10000-6980= 3,020 Therefore, he needs 3,020 more runs. |
|
| 7. |
Express the following as cm using decimals:30 mm |
|
Answer» 10 mm = 1 cm 1 mm = 1/10 cm 30 mm = 1/10 x 30 = 3.0 cm |
|
| 8. |
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished? |
|
Answer» Total number of dresses = 30 Work finished = 20 Fraction of finished work = 20/30 = 2/3
|
|
| 9. |
Express the following as cm using decimals:2 mm |
|
Answer» 10 mm = 1 cm 1 mm = 1/10 cm 2 mm = 1/10 x 2 = 0.2 cm |
|
| 10. |
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts? |
|
Answer» Total number of CDs = 3 + 5 = 8 Number of CDs purchased = 3 Fraction of CDs purchased = 3/8 Fraction of CDs received as gifts = 5/8 |
|
| 11. |
If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples. |
|
Answer» Yes, if we multiply any number with zero the resultant product will be zero. Example: 2 x 0 = 0, 5 x 0 = 0, 9 x 0 = 0 If both numbers are zero, then the result also be zero. 0 x 0 = 0 |
|
| 12. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:6.4 |
|
Answer» 6.4 = 64/10 = 32/5 |
|
| 13. |
Write the fractions. Are all these fractions equivalent: |
|
Answer» (a) 1/2, 2/4, 3/6, 4/8 Yes, all of these fractions are equivalent. (b) 4/12, 3/9, 2/6, 1/3, 6/15 No, these fractions are not equivalent. |
|
| 14. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:21.2 |
|
Answer» 21.2 = 212/10 = 106/5 |
|
| 15. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:13.7 |
| Answer» 13.7 = 137/10 | |
| 16. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:3.8 |
|
Answer» 3.8 = 38/10 = 19/5 |
|
| 17. |
Write the following decimals as fraction. Reduce the fractions to lowest terms:1.0 |
|
Answer» 1.0 = 10/10 = 1 |
|
| 18. |
If the product of two whole number is 1, can we say that one or both of them will be 1? Justify through examples. |
|
Answer» If only one number be 1 then the product cannot be 1. Examples: 5 x 1 = 5, 4 x 1 = 4, 8 x 1 = 8 If both number are 1, then the product is 1 1 x 1 = 1 |
|
| 19. |
Find using distributive property:5437 x 1001 |
|
Answer» 5437 x 1001 = 5437 x (1000 + 1) = 5437 x 1000 + 5437 x 1 = 5437000 + 5437 = 5442437 |
|
| 20. |
Write the fraction and pair up the equivalent fractions to each row: |
|
Answer» (a) 1/2 (ii) 4/8 = 1/2 (b) 4/6 = 2/3 (iv) 8/12 = 2/3 (c) 3/9 = 1/3 (i) 6/18 = 1/3 (d) 2/8 = 1/4 (v) 4/16 = 1/4 (e) 3/4 (iii) 12/16 = 3/4 |
|
| 21. |
Find using distributive property: 728 x 101 |
|
Answer» 728 x 101 = 728 x (100 + 1) = 728 x 100 + 728 x 1 = 72800 + 728 = 73528 |
|
| 22. |
Find using distributive property: 504 x 35 |
|
Answer» 504 x 35 = (500 + 4) x 35 = 500 x 35 + 4 x 35 = 17500 + 140 = 17640 |
|
| 23. |
Write all the factors of the following numbers:24 |
|
Answer» 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 = 6 x 4 Factors of 24 = 1, 2, 3, 4, 6, 12, 24 |
|
| 24. |
We already know the rule for the pattern of letter L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen? |
| Answer» The letter ‘T’ and ‘V’ that has pattern 2 n, since 2 matchsticks are used in all these letters. | |
| 25. |
Study the pattern: 1 x 8 + 1 = 9; 12 x 8 + 2 = 98; 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876; 12345 x 8 + 5 = 98765 Write the next two steps. Can you say how the pattern works? |
|
Answer» 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9876543 Pattern works like this: 1 x 8 + 1 = 9 12 x 8 + 2 = 98 123 x 8 + 3 = 987 1234 x 8 + 4 = 9876 12345 x 8 + 5 = 98765 123456 x 8 + 6 = 987654 1234567 x 8 + 7 = 9875643 |
|
| 26. |
In the given diagram, name the point(s): (a) In the interior of ∠DOE. (b)In the exterior of ∠EOF. (c) On ∠EOF. |
|
Answer» (a) Point interior of ∠DOE : A (b) Points exterior of ∠EOF : C, A, D (c) Points on ∠EOF : E, O, B, F |
|
| 27. |
Using divisibility test, determine which of the following numbers are divisibly by 4; by 8:(a) 572 (b) 726352 (c) 5500 (d) 6000(e) 12159 (f) 14560 (g) 21084 (h) 31795072(i) 1700 (j) 2150 |
|
Answer» (a) 572 --> Divisible by 4 as its last two digits are divisible by 4. Not divisible by 8 as its last three digits are not divisible by 8. (b) 726352 --> Divisible by 4 as its last two digits are divisible by 4. Divisible by 8 as its last three digits are divisible by 8. (c) 5500 --> Divisible by 4 as its last two digits are divisible by 4. Not divisible by 8 as its last three digits are not divisible by 8. (d) 6000 --> Divisible by 4 as its last two digits are 0. Divisible by 8 as its last three digits are 0. (e) 12159 --> Not divisible by 4 and 8 as it is an odd number. (f) 14560 --> Divisible by 4 as its last two digits are divisible by 4. Divisible by 8 as its last three digits are divisible by 8. (g) 21084 --> Divisible by 4 as its last two digits are divisible by 4. Not divisible by 8 as its last three digits are not divisible by 8. (h) 31795072 --> Divisible by 4 as its last two digits are divisible by 4. Divisible by 8 as its last three digits are divisible by 8. (i) 1700 --> Divisible by 4 as its last two digits are 0. Not divisible by 8 as its last three digits are not divisible by 8. (j) 5500 --> Not divisible by 4 as its last two digits are not divisible by 4. Not divisible by 8 as its last three digits are not divisible by 8. |
|
| 28. |
Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 3:(a) __________ 6724 (b) 4765 __________ 2 |
|
Answer» (a) We know that a number is divisible by 3 if the sum of all digits is divisible by 3. Therefore, Smallest digit : 2 --> 26724 = 2 + 6 + 7 + 2 + 4 = 21 Largest digit : 8 --> 86724 = 8 + 6 + 7 + 2 + 4 = 27 (b) We know that a number is divisible by 3 if the sum of all digits is divisible by 3. Therefore, Smallest digit : 0 --> 476502 = 4 + 7 + 6 + 5 + 0 + 2 = 24 Largest digit : 9 --> 476592 = 4 + 7 + 6 + 5 + 0 + 2 = 33 |
|
| 29. |
(a) The length of Ramesh’s notebook is 9 cm and 5 mm. What will be its length in cm? (b) The length of a young gram plant is 65 mm. Express its length in cm. |
|
Answer» (a) 9 cm 5 mm = 9 cm + 5 mm = 9+5/10 = 9.5 cm (b) 65 mm = 65/10 cm = 6.5 cm |
|
| 30. |
Find the ratio of the following:98 to 63 |
|
Answer» Ratio of 98 to 63 = 98/63 = 14/9 = 14:9 |
|
| 31. |
From the figure, identify: (a) The centre of circle. (b) Three radii. (c) A diameter. (d) A chord. (e) Two points in the interior. (f) A point in the exterior. (g) A sector. (h) A segment. |
|
Answer» (a) O is the centre. (b) Three radii: OA, OB and OC (c) A diameter: AC (d) A chord: ED (e) Interior points: O, P (f) Exterior point: Q (g) A sector: OAB (h) A segment: ED |
|
| 32. |
Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisibly by 11:(a) 92 __________ 389 (b) 8 __________ 9484 |
|
Answer» (a) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 928389 --> Odd places = 9 + 8 + 8 = 25 Even places = 2 + 3 + 9 = 14 Difference = 25 – 14 = 11 (b) We know that a number is divisible by 11 if the difference of the sum of the digits at odd places and that of even places should be either 0 or 11. Therefore, 869484 --> Odd places = 8 + 9 + 8 = 25 Even places = 6 + 4 + 4 = 14 Difference = 25 – 14 = 11 |
|
| 33. |
(a) Is every diameter of a circle also a chord? (b) Is every chord of a circle also a diameter? |
|
Answer» (a) Yes, every diameter of a circle is also a chord. (b) No, every chord of a circle is not a diameter. |
|
| 34. |
Write all the factors of the following numbers:27 |
|
Answer» 27 = 1 x 27 = 3 x 9 = 9 x 3 Factors of 27 = 1, 3, 9, 27 |
|
| 35. |
Find the common factors of:20 and 28 |
|
Answer» Factors of 20 = 1, 2, 4, 5, 10, 20 Factors of 28 = 1, 2, 4, 7, 14, 28 Common factors = 1, 2, 4 |
|
| 36. |
A cube is a three-dimensional figure. It has six faces and all of them are identical squares.The length of an edge of the cube is given by l. find the formula for the total length of the edges of a cube. . |
|
Answer» Length of one edge of cube = l Number of edges in a cube = 12 Therefore, total length = 12 x l = 12 l |
|
| 37. |
Find the H.C.F. of the following numbers:27, 63 |
|
Answer» Factors of 27 = 3 x 3 x 3 Factors of 63 = 3 x 3 x 7 H.C.F. (27, 63) = 3 x 3 = 9 |
|
| 38. |
The diameter of a circle is a line, which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figureAB is a diameter of the circle; C is its centre). Express the diameter of the circle (d) in terms of its radius (r). |
|
Answer» Since, length of diameter is double the length of radius. Therefore, d = 2r |
|
| 39. |
Find the common factors of:56 and 120 |
|
Answer» Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56 Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 60, 120 Common factors = 1, 2, 4, 8 |
|
| 40. |
Mother wants to divide Rs.36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get. |
|
Answer» Ratio of the age of Shreya to that of Bhoomika = 15/12 =5/4 = 5 : 4 Thus, Rs.36 divide between Shreya and Bhoomika in the ratio of 5 : 4. Shreya gets = 5/9 of Rs.36 = 5 x 36/9 = Rs.20 Bhoomika gets = 4/9 of Rs.36 = 4 x 36/9 = Rs.16 |
|
| 41. |
Find the common factors of:35 and 50 |
|
Answer» Factors of 35 = 1, 5, 7, 35 Factors of 50 = 1, 2, 5, 10, 25, 50 Common factors = 1, 5 |
|
| 42. |
Find the common factors of:15 and 25 |
|
Answer» Factors of 15 = 1, 3, 5, 15 Factors of 25 = 1, 5, 25 Common factors = 1, 5 |
|
| 43. |
Find the common factors of:4, 8 and 12 |
|
Answer» Factors of 4 = 1, 2, 4 Factor of 8 = 1, 2, 4, 8 Factor of 12 = 1, 2, 3, 4, 6, 12 Common factors of 4, 8 and 12 = 1, 2, 4 |
|
| 44. |
Find the areas of the squares whose sides are:14 cm |
|
Answer» Area of square = side x side = 14 cm x 14 cm = 196 cm2 |
|
| 45. |
Present age of father is 42 years and that of his son is 14 years. Find the ratio of: (a) Present age of father to the present age of son. (b) Age of the father to the age of the son, when son was 12 years old. (c) Age of father after 10 years to the age of son after 10 years. (d) Age of father to the age of son when father was 30 years old. |
|
Answer» (a) Ratio of father’s present age to that of son = 42/14 = 3/1 = 3:1 (b) When son was 12 years, i.e., 2 years ago, then father was (42 – 2) = 40 years Therefore, the ratio of their ages = 40/12 = 10/3 = 10 : 3 (c) Age of father after 10 years = 42 + 10 = 52 years Age of son after 10 years = 14 + 10 = 24 years Therefore, ratio of their ages = 52/24 = 13/6 = 13 : 6 (d) When father was 30 years old, i.e., 12 years ago, then son was (14 – 12) = 2 years old Therefore, the ratio of their ages = 30/2 =15/1 = 15 : 1 |
|
| 46. |
To find sum of three numbers 14, 27 and 13. We can have two ways.(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54, or(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus(14 + 27) + 13 = 14 + (27 + 13)This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numebrs, in a general way, by using variables a , b and c. |
| Answer» (a + b ) + c = a + (b + c) | |
| 47. |
The length and the breadth of three rectangles are as given below: (a) 9 m and 6 m (b) 17 m and 3 m (c) 4 m and 14 m Which one has the largest area and which one has the smallest? |
|
Answer» (a) Area of rectangle = length x breadth = 9 m x 6 m = 54 m2 (b) Area of rectangle = length x breadth= 3 m x 17 m = 51 m2 (c) Area of rectangle = length x breadth= 4 m x 14 m = 56 m2 Thus, the rectangle (c) has largest area, i.e. 56 m2 and rectangle (b) has smallest area, i.e., 56 m2 |
|
| 48. |
Write all the factors of the following numbers:36 |
|
Answer» 36 = 1 x 36 = 2 x 18 = 3 x 12 = 4 x 9 = 6 x 6 Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 |
|
| 49. |
Find the common factors of:5, 15 and 25 |
|
Answer» Factors of 5 = 1, 5 Factor of 15 = 1, 3, 5, 15 factor of 25 = 1, 5, 25 Common factors of 5, 15 and 25 = 1, 5 |
|
| 50. |
See Fig.3.14, and write the following:(i) The coordinates of B.(ii) The coordinates of C.(iii) The point identified by the coordinates (–3, –5).(iv) The point identified by the coordinates (2, – 4).(v) The abscissa of the point D.(vi) The ordinate of the point H.(vii) The coordinates of the point L.(viii) The coordinates of the point M. |
|
Answer» We need to consider the given below figure to answer the following questions. (i) The coordinates of point B in the above figure is the distance of point B from xaxis and y-axis. Therefore, we can conclude that the coordinates of point B are (―5, 2). (ii) The coordinates of point C in the above figure is the distance of point C from xaxis and y-axis. Therefore, we can conclude that the coordinates of point C are (5, ―5). (iii) The point that represents the coordinates (―3, ―5) is E. (iv) The point that represents the coordinates (2, ―4) is G. (v) The abscissa of point D in the above figure is the distance of point D from the yaxis. Therefore, we can conclude that the abscissa of point D is 6. (vi) The ordinate of point H in the above figure is the distance of point H from the xaxis. Therefore, we can conclude that the abscissa of point H is ―3. (vii) The coordinates of point L in the above figure is the distance of point L from xaxis and y-axis. Therefore, we can conclude that the coordinates of point L are (0, 5). (viii) The coordinates of point M in the above figure is the distance of point M from xaxis and y-axis. Therefore, we can conclude that the coordinates of point M are (―3, 0). |
|