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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.
| 101. |
Find:(–13) + 32 – 8 – 1 |
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Answer» (–13) + 32 – 8 – 1 = –13 + 32 – 8 – 1 = 32 – 22 = 10 |
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| 102. |
Solve the question:321/5 |
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Answer» Solution is 321/5 We know that a1/n = n√a,where a > 0. we conclude that 321/5 can also be written as 5√32 = 2√2 x 2 x 2 x 2 x 2 = 2 Therefore the value of 321/5 will be 2. |
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| 103. |
On the following number line value ‘Zero’ is shown by the point(a) X (b) Y (c) Z (d) W |
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Answer» On the number line value of 'Zero' shown by Z |
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| 104. |
Which of the following statements is not true?(a) When two positive integers are added, we always get a positive integer.(b) When two negative integers are added we always get a negative integer.(c) When a positive integer and a negative integer is added we always get a negative integer.(d) Additive inverse of an integer 2 is (– 2) and additive inverse of ( – 2) is 2. |
| Answer» (c) When a positive integer and a negative integer is added we always get a negative integer. | |
| 105. |
The next number in the pattern – 62, – 37, – 12 _________ is(a) 25 (b) 13 (c) 0 (d) –13 |
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Answer» the pattern – 62, – 37, – 12 ____13_____ |
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| 106. |
Next three consecutive numbers in the pattern 11, 8, 5, 2, --, --, --are(a) 0, – 3, – 6 (b) – 1, – 5, – 8(c) – 2, – 5, – 8 (d) – 1, – 4, – 7 |
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Answer» consecutive numbers in the pattern 11, 8, 5, 2, -(-1)-, -(4)-, -(-7)- |
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| 107. |
By observing the below number line (Fig. 1.2), state which of the following statements is true. |
| Answer» correct answer is (d) B is – 4 | |
| 108. |
By observing the number line (Fig. 1.2), state which of the following statements is not true.(a) B is greater than –10 (b) A is greater than 0(c) B is greater than A (d) B is smaller than 0 |
| Answer» (c) B is greater than A | |
| 109. |
Find six rational numbers between 3 and 4. |
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Answer» We know that there are infinite rational numbers between any two numbers. A rational number is the one that can be written in the form of p/q, where p and q are integers and q ≠ o. We know that the numbers 3.1,3.2, 3.3, 3.4,3.5 and 3.6 all lie between 3 and 4. We need to rewrite the numbers 3.1,3.2, 3.3, 3.4,3.5 and 3.6 in p/q form to get the rational numbers between 3 and 4. 31/10, 32/10, 33/10, 34/10, 35/10, 36/10 We can further convert the rational numbers 32/10, 34/10, 35/10 and 36/10 into lowest fractions. On converting the fractions into lowest fractions, we get 16/5, 17/5, 7/2 and 18/5. Therefore, six rational numbers between 3 and 4 are 31/10, 16/5, 33/10, 17/5, 7/2 and 18/5. |
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| 110. |
Subtract:(-15) - (-18) |
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Answer» (–15) – (–18) = –15 + 18 = 3 |
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| 111. |
Subtract:72-(90) |
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Answer» 72 – 90 = –18 |
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| 112. |
Subtract:35-(20) |
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Answer» 35– 20 = 15 |
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| 113. |
Encircle the odd one of the following(a) (–3, 3) (b) (–5, 5) (c) (–6, 1) (d) (–8, 8) |
| Answer» (d) (–8, 8) is different from others | |
| 114. |
For the following statements write True (T) or False (F). If the statement is false, correct the statement: (a) –8 is to the right of –10 on a number line. (b) –100 is the right of –50 on a number line. (c) Smallest negative integer is –1. (d) –26 is larger than –25. |
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Answer» (a) True (b) False (c) False (d) False |
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| 115. |
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is,π = c/d.This seems to contradict the fact that π is irrational. How will you resolve this contradiction? |
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Answer» We know that when we measure the length of a line or a figure by using a scale or any device, we do not get an exact measurement. In fact, we get an approximate rational value. So, we are not able to realize that either circumference (c) or diameter (d) of a circle is irrational. Therefore, we can conclude that as such there is not any contradiction regarding the value of π and we realize that the value of π is irrational. |
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| 116. |
Find the ratio of the following:500 ml to 2 litres |
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Answer» 500 ml to 2 liters 2 liters = 2 x 1000 ml = 2000 ml [∵ 1 litre = 1000 ml] Now, ratio of 500 ml to 2 liters = 500 ml : 2 liters ⇒ 500 ml : 2000 ml = 500/2000 = 1/4 = 1 : 4 |
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| 117. |
There are 20 girls and 15 boys in a class. (a) What is the ratio of number of girls to the number of boys? (b) What is the ratio of girls to the total number of students in the class? |
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Answer» (a) The ratio of girls to that of boys = 20/15 = 4/3 = 4:3 (b) The ratio of girls to total students = 20/20+15 = 20/35 = 4/7 = 4:7 |
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| 118. |
In a college out of 4320 students, 2300 are girls. Find the ratio of: (a) Number of girls to the total number of students. (b) Number of boys to the number of girls. (c) Number of boys to the total number of students. |
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Answer» Total number of students in school = 4320 Number of girls = 2300 Therefore, number of boys = 4320 – 2300 = 2020 (a) Ratio of girls to total number of students = 2300/4320 =115/216 = 115 : 216 (b) Ratio of boys to that of girls = 2020/2300 = 101/115 = 101 : 115 (c) Ratio of boys to total number of students = 2020/4320 = 101/216 = 101 : 216 |
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| 119. |
In a year, Seema earns Rs.1,50,000 and saves Rs.50,000. Find the ratio of: (a) Money that Seema earns to the money she saves. (b) Money that she saves to the money she spends. |
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Answer» Total earning = Rs.1,50,000 and Saving = Rs.50,000 Money spent = Rs.1,50,000 - Rs.50,000 = Rs.1,00,000 (a) Ratio of money earned to money saved = 150000/50000 = 3/1 = 3 : 1 (b) Ratio of money saved to money spend = 50000/100000 = 1/2 = 1 : 2 |
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| 120. |
Find the H.C.F. of the following numbers:18, 54, 81 |
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Answer» Factors of 18 = 2 x 3 x 3 Factors of 54 = 2 x 3 x 3 x 3 Factors of 81 = 3 x 3 x 3 x 3 H.C.F. = 3 x 3 = 9 |
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| 121. |
Find the H.C.F. of the following numbers:91, 112, 49 |
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Answer» Factors of 91 = 7 x 13 Factors of 112 = 2 x 2 x 2 x 2 x 7 Factors of 49 = 7 x 7 H.C.F. = 1 x 7 = 7 |
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| 122. |
Find the odd one out of the four options given below: (a) (–3, – 6) (b) (+1, –10) (c) (–2, –7) (d) (–4, –9) |
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Answer» Here –3 + (–6) = –9, +1 + (–10) = –9 and – 2 + (–7) = –9 All the above pairs i.e. (–3, – 6); (+1, –10); (–2, –7) give same answer on adding, whereas – 4 + (– 9) = –13, gives a different answer. So, odd one out is (d). |
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| 123. |
Find the odd one out of the four options in the following:(a) (–2, 24) (b) (–3, 10) (c) (–4, 12) (d) (–6, 8) |
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Answer» Here – 2 × 24 = – 48, – 4 × 12 = – 48 and – 6 × 8 = – 48 |
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| 124. |
–16 ÷ [8 ÷ (–2)] is equal to(a) –1 (b) 1 (c) 4 (d) –4 |
| Answer» Correct answer is (c), | |
| 125. |
[(– 10) × (+ 9)] + ( – 10) is equal to(a) 100 (b) –100 (c) – 80 (d) 80 |
| Answer» Correct answer is (b) | |
| 126. |
Madhre is standing in the middle of a bridge which is 20 m above the water level of a river. If a 35 m deep river is flowing under the bridge (see Fig. 1.1), then the vertical distance between the foot of Madhre and bottom level of the river is:(a) 55 m (b) 35 m (c) 20 m (d) 15 m |
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Answer» The correct answer is (a) [Vertical distance = 20 m + 35 m = 55m] |
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| 127. |
Rationalize the denominators of the following:1/√7 |
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Answer» 1/√7 We need to multiply the numerator and denominator of 1/√7 by √7, to get 1/√7 x √7/√7 = √7/7 Therefore, we conclude that on rationalizing the denominator of 1/√7, we get √7/7. |
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| 128. |
Rationalize the denominators of the following:1 / (√7 - 2) |
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Answer» 1/(√7 - 2) We need to multiply the numerator and denominator of 1/(√7 - 2) by √7 + 2, to get 1/(√7 - 2) x (√7 + 2)/(√7 + 2) = (√7 + 2)/(√7 - 2)(√7 + 2) We need to apply the formula (a - b)(a + b) = a2 - b2 in the denominator to get 1/(√7 - 2) = (√7 + 2)/(√7)2 - (2)2 = (√7 + 2)/(7 -4) = (√7 + 2)/3 Therefore, we conclude that on rationalizing the denominator of 1/(√7 - 2), we get (√7 + 2)/3. |
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| 129. |
The value of 5 ÷ (–1) does not lie between(a) 0 and – 10 (b) 0 and 10 (c) – 4 and – 15 (d) – 6 and 6 |
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Answer» The value of 5 ÷ (–1) =-5 does not lie between 0 and 10 ans (b) |
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| 130. |
The temperature dropped 15 degree Celsius in the last 30 days. If the rate of temperature drop remains the same, how many degrees will the temperature drop in the next ten days? |
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Answer» Degree of temperature dropped in last 30 days = 15 degrees Degree of temperature dropped in last 1 days = 15/30 = 1/2 degree Degree of temperature dropped in last 10 days = 1/2 x 10 = 5 degree Thus, 5 degree Celsius temperature dropped in 10 days. |
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| 131. |
The sale of electric bulbs on different days of a week is shown below:What can be conclude from the said pictograph? |
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Answer» (a) Number of bulbs sold on Monday are 12. Similarly, number of bulbs sold on other days can be found. (b) Maximum number of bulbs were sold on Sunday. (c) Same number of bulbs were sold on Wednesday and Saturday. (d) Then minimum number of bulbs were sold on Wednesday and Saturday. (e) The total number of bulbs sold in the given week were 86. |
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| 132. |
Observe this paragraph which shows the marks obtained by Aziz in half yearly examination in different subjects: Answer the given questions:(a) What information is does the bar graph give?(b) Name the subject in which Aziz scored maximum marks.(c) Name the subject in which he has scored minimum marks.(d) State the name of the subjects and marks obtained in each of them. |
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Answer» (a) The paragraph shows the marks obtained by Aziz in half yearly examination in different subjects. (b) Hindi. (c) Social Studies. (d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40. |
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| 133. |
Match the integer in Column I to an integer inColumn II so that the sum is between –11 and –4Column IColumn II(a) –6(i) –11(b) +1(ii) –5(c) +7(iii) +1(d) –2(iv) –13 |
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Answer» (a) ↔ (iii) because –6 + (+1) = –5, which lies between –11 and –4. (b ) ↔(i) because +1 + (–11) = –10 which lies between –11 and –4. (c) ↔ (iv) because +7 + (–13) = –6 which lies between –11 and –4 (d) ↔(ii) because –2 + (–5) = –7 which lies between –11 and –4. |
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| 134. |
Write a pair of integers whose sum is zero (0) but difference is 10. |
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Answer» Since sum of two integers is zero, one integer is the additive inverse of other integer, like – 3, 3; – 4, 4 etc. But the difference has to be 10. So, the integers are 5 and – 5 as 5–(–5) is 10. |
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| 135. |
In an objective type test containing 25 questions. A student is to be awarded +5 marks for every correct answer, –5 for every incorrect answer and zero for not writing any answer. Mention the ways of scoring 110 marks by a student. |
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Answer» Marks scored = +110 So, minimum correct responses = 110 ÷ (+5) = 22 Case 1 Correct responses = 22 Marks for 1 correct response = + 5 Marks for 22 correct respones = +110 (22 × 5) Marks scored = +110 Marks obtained for incorrect answer = 0 So, no incorrect response And, therefore, 3 were unattempted Case 2 Correct responses = 23 Marks from 23 correct responses = + 115 (23 × 5) Marks scored = + 110 Marks obtained for incorrect answers = 110 – (+115) = –5 Marks for 1 incorrect answer = –5 Number of incorrect responses = (–5) ÷ (–5) = 1 So, 23 correct, 1 incorrect and 1 unattempted. Case 3 Correct responses = 24 Marks from 24 correct responses = + 120 (24 × 5) Marks scored = + 110 Marks obtained for incorrect answers = +110 – (+120) = –10 Number of incorrect responses = (–10) ÷ (–5) = 2 Thus the number of questions = 24 + 2 = 26. Whereas, total number of questions is 25. So, this case is not possible. So, the possible ways are: • 22 correct, 0 incorrect, 3 unattempted • 23 correct, 1 incorrect, 1unattempted. |
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| 136. |
Classify the following numbers as rational or irrational:(i) 1/√2 (ii) 2π |
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Answer» (i) 1/√2 we know that √2 = 1.414..., which is an irrational number . We can conclude that, when 1 is divided by √2 we will get an irrational number. Therefore, we conclude that 1/√2 is an irrational number. (ii) 2π We know that π = 3.1415...., which is an irrational number. We can conclude that 2π will also be an irrational number. Therefore, we conclude that 2π is an irrational number. |
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| 137. |
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again? |
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Answer» L.C.M. of 48, 72, 108 = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 sec. After 432 seconds, the lights change simultaneously. 432 second = 7 minutes 12 seconds Therefore the time = 7 a.m. + 7 minutes 12 seconds = 7 : 07 : 12 a.m. L.C.M of 48,72,108 can be taken to find the intervals of change =432 secAfter 432 sec = 432÷(60×60) =432÷3600 =0.12 hrs Hence, next simultaneous change will take place at = 8.00 am +0.12 =8.12 am |
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| 138. |
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed? |
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Answer» Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle. Perimeter of field = 2 x (length + breadth) = 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km = 2.4 x 1000 m = 2400 m Thus, the length of wire = 4 x 2400 = 9600 m = 9.6 m |
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| 139. |
If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate. |
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Answer» Rain in 3 days = 276 mm Rain in 1 day = 276/3 = 92 mm Rain in 7 days = 92 x 7 = 644 mm Thus, the rain in 7 days is 644 mm |
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| 140. |
Water level in a well was 20m below ground level. During rainy season, rain water collected in different water tanks was drained into the well and the water level rises 5 m above the previous level. The wall of the well is 1m 20 cm high and a pulley is fixed at aheight of 80 cm. Raghu wants to draw water from the well. The minimum length of the rope that he can use is(a) 17 m (b) 18 m (c) 96 m (d) 97 m |
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Answer» Correct answer is (a) 17 m |
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| 141. |
Write the predecessor of: (a) 94 (b) 10000 (c) 208090 (d) 7654321 |
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Answer» (a) The predecessor of 94 is 94 – 1 = 93 (b) The predecessor of 10000 is 10000 – 1 = 9999 (c) The predecessor of 208090 is 208090 – 1 = 208089 (d) The predecessor of 7654321 is 7654321 – 1 = 7654320 |
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| 142. |
Match the following: (i) 425 x 136 = 425 x (6 + 30 + 100)(a) Commutativity under multiplication(ii) 2 x 49 x 50 = 2 x 50 x 49(b) Commutativity under addition(iii) 80 + 2005 + 20 = 80 + 20 + 2005(c) Distributivity multiplication under addition |
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| 143. |
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l. |
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Answer» Side of equilateral triangle = l Therefore, Perimeter of equilateral triangle = 3 x side = 3l |
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| 144. |
What is the perimeter of each of the following figures? What do you infer from the answer? |
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Answer» (a) Perimeter of square = 4 x side = 4 x 25 = 100 cm (b) Perimeter of rectangle = 2 x (length + breadth) = 2 x (40 + 10) = 2 x 50 = 100 cm (c) Perimeter of rectangle = 2 x (length + breadth) = 2 x (30 + 20) = 2 x 50 = 100 cm (d) Perimeter of triangle = Sum of all sides = 30 cm + 30 cm + 40 cm = 100 cm Thus, all the figures have same perimeter. |
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| 145. |
The perimeter of a regular pentagon is 100 cm. How long is its each side? |
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Answer» Perimeter of regular pentagon = 100 cm ⇒ 5 x side = 100 cm ⇒ side = 100/5 = 20 cm Thus, the side of regular pentagon is 20 cm. |
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| 146. |
Find the side of the square whose perimeter is 20 m |
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Answer» Perimeter of square = 4 x side ⇒ 20 = 4 x side ⇒ side = 20/4 = 5 cm Thus, the side of square is 5 cm. |
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| 147. |
A piece of string is 30 cm long. What will be the length of each side if the string is used to form: (a) a square (b) an equilateral triangle (c) a regular hexagon? |
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Answer» Length of string = Perimeter of each figure (a) Perimeter of square = 30 cm ⇒ 4 x side = 30 cm ⇒ side = 30/7 = 7.5 cm Thus, the length of each side of square is 7.5 cm. (b) Perimeter of equilateral triangle = 30 cm ⇒ 3 x side = 30 cm ⇒ side = 30/3 = 10 cm Thus, the length of each side of equilateral triangle is 10 cm. (c) Perimeter of hexagon = 30 cm ⇒ 6 x side = 30 cm ⇒ side = 30/6 = 5 cm Thus, the side of each side of hexagon is 5 cm. |
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| 148. |
Find the cost of fencing a square park of side 250 m at the rate of Rs.20 per meter. |
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Answer» Side of square = 250 m Perimeter of square = 4 x side = 4 x 250 = 1000 m Since, cost of fencing of per meter = Rs.20 Therefore, cost of fencing of 1000 meters = 20 x 1000 = Rs.20,000 |
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| 149. |
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is the third side? |
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Answer» Let the length of third side be x cm. Length of other two side are 12 cm and 14 cm. Now, Perimeter of triangle = 36 cm ⇒ 12 + 14 + x = 36 ⇒ 26 + x = 36 ⇒ x = 36 - 26 ⇒ x = 10 cm Thus, the length of third side is 10 cm. |
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| 150. |
Find the number of lines of symmetry for each of the following shapes: |
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Answer» (a) 4 (b) 4 (c) 4 (d) 1 (e) 6 (f) 4 (g) 0 (h) 0 (i) 3 |
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