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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.
| 11201. |
Find:i) 1.7 x 3 ii) 2.0 x 1.5 iii) 2.3 x 4.35 |
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Answer» i) 1.7 x 3 = 5.1 ii) 2.0 x 1.5 = 3.00 iii) 2.3 x 4.35 = 10.005 1) 1.7*3 = 5.1 2) 2.0*1.5 = 3 3) 2.3*4.35 = 10.005
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| 11202. |
Change the following into like decimal fractions. i) 0.802, 54.32, 873.274 ii) 4.78, 9.193, 11.3 iii) 16.003, 16.9, 16.19 |
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Answer» i) 0.802, 54.32, 873.274 Given decimal fractions are 0.802, 54.32, 873.274 The greatest number of decimal places is 3. So, we convert all of them to equivalent decimals with 3 decimal places by placing sufficient zeroes. 0.802 = 0.802 54.32 = 54.320 873.274 = 873.274 Thus, 0.802, 54.32, 873.274 when converted to like decimals becomes 0.802, 54.320, 873.274. ii) 4.78, 9.193, 11.3 Given decimal fractions are 4.78, 9.193, 11.3. So, we convert all of them to equivalent decimals with 3 decimal places. 4.78 = 4.780 9.193 = 9.193 11.3 = 11.300 Thus, 4.78, 9.193, 11.3 when converted to like decimals becomes 4.780, 9.193, 11.300. iii) 16.003, 16.9, 16.19 Given decimal fractions are 16.003, 16.9, 16.19 So, we convert all of them to equivalent decimals with three decimal places by providing sufficient zeroes. 16.3 = 16.003 16.9 = 16.900 16.19 = 16.190 Thus, 16.003, 16.9, 16.19 when converted to like decimals becomes 16.003, 16.900, 16.190. |
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| 11203. |
20, 10, 10, 20, 80, ………. (a) 320 (b) 640 (c) 400 (d) 80 |
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Answer» Correct option is : (b) 640 20, 10, 10, 20, 80, …………… 20, 20 × 2-1 , 10 × 20, 10 × 21 , 20 × 22 , So, next number is 80 × 23 = 80 × 8 = 640 |
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| 11204. |
Which of the following are unlike decimal fractions ? i) 5.03, 6.185ii) 42.7, 7.42 iii) 16.003, 5.301 iv)15.81, 1.36 |
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Answer» i) 5.03, 6.185 5.3 has 2 decimal places 6.185 has 3 decimal places Their decimal places are not equal. So, these decimal fraction are unlike decimal fractions. ii) 42.7, 7.42 42.7 has one decimal place 7.42 has two decimal places Their decimal places are not equal. So, these are unlike decimal fractions. iii) 16.003, 5.301 16.3 has three decimal places 5.301 has three decimal places Their decimal places are equal. So, these are like decimal fractions. iv) 15.81, 1.36 15.81 has two decimal places 1.36 has two decimal places Their decimal places are equal. So, these are like decimal fractions. |
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| 11205. |
If a picture chart costs 74.25. Amrutha wants to buy 16 charts to make an album. How much money does she has to pay ? |
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Answer» Given, cost of one picture chart = ₹ 4.25 Cost of 16 charts = 4.25 × 16 = \(\frac {425}{100}\times16\) = \(\frac {425\times16}{100}\) ∴ Cost of 16 charts to make album = 6800/100 = ₹ 68 |
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| 11206. |
Calculate : \(\frac 35 + \frac 74\) |
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Answer» Given \(\frac 35 + \frac 74\)= \(\frac {12+35}{20} = \frac {47}{20}\) |
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| 11207. |
Add the pattern: Eg: 10, 20, 40, 70, 110, … (a) 160 (b) 180 (c) 150 (d) 210 |
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Answer» Correct option is : (a) 160 (10 + 10,), (20 + 20), (40 + 30), (70 + 40) so, next number is (110 + 50) = 160 |
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| 11208. |
240, 240, 120, 40, ……………., 2. (a) 10 (b) 20 (c) 18 (d) 35 |
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Answer» Correct option is : (a) 10 240, 240, 120, 40, ……………….,2. 240, 240 ÷ 1, 240 ÷ 2, 120 ÷ 3, 40 ÷ 4, 10 ÷ 5 So, next number is 40 ÷ 4 = 40/4 = 10 |
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| 11209. |
Adding or subtract of natural numbers: Eg: 6, 7, 9, 12, 16, 21,… (a) 21 (b) 25 (c) 27 (d) 28 |
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Answer» Correct option is : (c) 27 (6 + 1), (7 + 2,), (9 + 3), (12 + 4), (16 + 5) so, next number is (21 + 6) = 27 |
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| 11210. |
5, 15, 35, 75, 155, _____(a) 215 (b) 305 (c) 315 (d) 265 |
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Answer» Correct option is : (c) 315 5, 15, 35, 75, 155, ……….. 5 × 1, 5 × 3, 5 × 7, 5 × 15, 5 × 31, ……………. 5(21 – 1), 5(22 – 1), 5(23 – 1), 5(24 – 1), 5(25 – 1), ……….. So, next number is 5 x (26 – 1) = 5 × 63 = 315 |
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| 11211. |
Find the smallest number by which 28812 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number. |
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Answer» Factors of 28812 are: 28812 = 2 × 2 × 3 × 3 × 3 × 17 × 17 Pairs = 22 × 33 × 172 Hence, 28812 should be divided by 3 in order to get a perfect square when divided by the least number The square root will be: 2 × 3 × 17 = 102 |
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| 11212. |
Write the following numbers In ascending order.i) -5,-75,3,-2,4, 3/2ii) 2/3, 3/2, 0, -1, -2, 5 |
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Answer» i) -5,-75,3,-2,4, 3/2 Ascending order = -75 <-5 <-2 < 3/2 <3 < 4 or -75, -5, -2, 3/2 , 3, 4 ii) 2/3, 3/2, 0, -1, -2, 5 Ascending order = -2, -1, 0, 2/3, 3/2, 5 |
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| 11213. |
Find i) 5 x 3/2 =ii) 4 x 7/5 =iii) 7 x 8/3 |
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Answer» i) 5 x 3/2 = \(\frac {5\times3}{2} \frac {15}{2} = 7\frac 12\) ii) 4 x 7/5 = \(\frac {4\times7}{5} \frac {28}{5} = 5\frac 35\) iii) 7 x 8/3 = \(\frac {7\times8}{3} \frac {56}{3} = 18\frac 23\) |
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| 11214. |
fill in the blanks using ‘>’, ‘<’ or ‘=’ : 12/75...32/200 |
| Answer» The correct answer is 12/75 = 32/200 | |
| 11215. |
Find the squares of the following numbers by visual method:(i) 52 (ii) 95 (iii) 505 (iv) 702 (v) 99 |
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Answer» (i) 52, (52)2 = (50 + 2)2 = 502 + 22 + (2 × 50 × 2) = 2500 + 4 + 200 = 2704 (ii) 95, (95)2 = (100 - 5)2 = 1002 + 52 - (2 × 5 × 100) = 10000 + 25 - 1000 = 9025 (iii) 505, (505)2 = (505 + 5)2 = 5002 + 52 + (2 × 500 × 5) = 250000 + 25 + 5000 = 255025 (iv) 702, (702)2 = (700 + 2)2 = 7002 + 22 + (2 × 700 × 2) = 140000 + 4 + 2800 = 142804 (v) 99, (99)2 = (100 - 1)2 = 1002 + 12 - (2 × 100 × 1) = 10000 + 1 - 200 = 9301 |
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| 11216. |
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number. |
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Answer» The prime factors for 1152 is 28812 = 2×2×2×2×2×2×2×3×3 (grouping the prime factors in equal pairs we get) = (2×2) × (2×2) × (2×2) × (3×3) × 3 Hence,the smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2. The number after division, 1152/2 = 576 576 = 2×2×2×2×2×2×3×3 = (2×2) × (2×2) × (2×2) × (3×3) (grouping the prime factors in equal pairs we get) = 26 × 32 = 242 Hence,the resulting number is the square of 24. |
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| 11217. |
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square. |
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Answer» The prime factors for 28812 28812 = 2×2×3×3×3×17×17 = (2×2) × (3×3) × (17×17) × 3 (grouping the prime factors in equal pairs we get) Hence, the smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3 |
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| 11218. |
Find the smallest number by which 1152 must be divided so that it becomes a square. Also, find the square root of the number so obtained. |
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Answer» As understood we calculate prime factor of, 1152 = (2 × 2) × (2 × 2) × (2 × 2) × 2 × (3 × 3) =22 × 22 × 22 × 32 × 2 To make the unpaired 2 into paired, the number 1152 has to be divided by 2 1152 ÷ 2 = 22 × 22 × 22 × 32 ∴ Square root of √(1152 ÷ 2) = 2 × 2 × 2 × 3 = 24 |
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| 11219. |
Find:(i) 1.07 × 0.02(ii) 10.05 × 1.05(iii) 101.01 × 0.01(iv) 100.01 × 1.1 |
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Answer» (i) 1.07 × 0.02 1.07 × 0.02 = 0.0214 (ii) 10.05 × 1.05 10.05 × 1.05 = 10.5525 (iii) 101.01 × 0.01 101.01 × 0.01 = 1.01011 (iv) 100.01 × 1.1 100.01 × 1.1 = 110.011 |
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| 11220. |
The value of \(\frac{(67.542)^2 - (32.458)^2}{45.458 - 40.374}\) is (a) 1 (b) 10 (c) 100 (d) 0.1 |
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Answer» (c) 100 Given exp. = \(\frac{(67.542+320458)(67.542 - 32.458)}{35.084}\) = \(\frac{100\times35.084}{35.084}\) = 100. |
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| 11221. |
Add the following: 20.02 and 2.002 |
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Answer» The correct answer is 22.022 |
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| 11222. |
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number. |
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Answer» Given that, 3645 = (3 × 3) × (3 × 3) × 5 × 3 =32 × 32 × 5 × 3 To make the unpaired 5 and 3 into paired, the number 3645 has to be divided by 5 × 3=15 3645 ÷ 15 = 32 × 32 ∴ Square root , √(3645 ÷ 15) = 3 × 3 = 9 |
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| 11223. |
\(\frac{0.1\times0.1\times0.1 + 0.02\times0.02\times0.02}{0.2\times0.2\times0.2+0.04\times0.04\times0.04} \) is equal to(a) 0.0125 (b) 0.125 (c) 0.25 (d) 0.5 |
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Answer» (b) 0.0125 Given exp. = \(\frac{0.1\times0.1\times0.1 + 0.02\times0.02\times0.02}{8(0.1\times0.1\times0.1)+8(0.02\times0.02\times0.02)} \) = \(\frac18\bigg(\frac{0.1\times0.1\times0.1 + 0.02\times0.02\times0.02}{0.1\times0.1\times0.1+0.02\times0.02\times0.02}\bigg)\) = \(\frac18\) = 0.0125. |
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| 11224. |
Without adding, find the sum (1+3+5+7+9+11+13). |
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Answer» Given (1+3+5+7+9+11+13) According to the property of perfect square, for any natural number n, we have sum of first n odd natural numbers = n2 But here n = 7 Applying the above law we get Therefore (1+3+5+7+9+11+13) = 72 = 49 |
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| 11225. |
Without adding, find the sum (1+3+5+7+9+11+13+15+17+19). |
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Answer» Given (1+3+5+7+9+11+13+15+17+19) According to the property of perfect square, for any natural number n, we have sum of first n odd natural numbers = n2 But here n = 10 Applying the above law we get Therefore (1+3+5+7+9+11+13+15+17+19) = 102 = 100 |
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| 11226. |
Write true (T) or false (F) for the following statements. (i) The number of digits in a square number is even. (ii) The square of a prime number is prime (iii) The sum of two square numbers is a square number.(iv) The difference of two square numbers is a square number. (v) The product of two square numbers is a square number. (vi) No square number is negative. (vii) There is no square number between 50 and 56. (viii) There are fourteen square number upto 200. |
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Answer» (i) False: Because 169 is square number with odd digit (ii) False: Square of 3 (Prime) is 9 (not prime) (iii) False: Sum of 22 and 32 is 13 which is not square number (iv) False: Difference of 32 and 22 is 5, which is not square number (v) True: Because the square of 22 and 32 is 36 which is square of 6 (vi) True: As (-2)2 is 4, i.e. not negative (vii) True: As there is no square number between them (viii) True: The fourteen numbers upto 200 are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 |
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| 11227. |
Without adding find the sum(i) 1 + 3 + 5 + 7 + 9(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 |
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Answer» We know that the sum of first n odd natural numbers is n2. (i) Here, we have to find the sum of first five odd natural numbers. Therefore, 1 + 3 + 5 + 7 + 9 = 52 = 25 (ii) Here, we have to find the sum of first ten odd natural numbers. Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 102 = 100 (iii) Here, we have to find the sum of first twelve odd natural numbers. Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 +17 + 19 + 21 + 23 = 122 = 144 |
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| 11228. |
\(\sqrt{(0.798)^2 + 0.404\times0.798 + (0.202)^2} + 1\) is equal to(a) 0 (b) 2(c) 1.596 (d) 0.404 |
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Answer» (b) 2 \(\sqrt{(0.798)^2 + 0.404\times0.798 + (0.202)^2} + 1\) = \(\sqrt{(0.798)^2 + 2\times0.202\times0.798 + (0.202)^2} + 1\) = \(\sqrt{(0.798 +0.202)^2} + 1\) = √1 + 1 = 1 + 1 = 2. |
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| 11229. |
Find the sum 1 + 3 + 5 + _____ +51 (the sum fo all odd numbers from 1 to 51) without actually adding them. |
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Answer» The sum of first n odd numbers is n2 ∴ 1 + 3 + 5 + _____ + 51 = 262 = 676. |
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| 11230. |
(x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x. |
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Answer» (x + 3) : (x + 11) = (x- 2) : (x+ 1) \(\frac{(x\, +\,3)}{(x\,+\,11)}\) = \(\frac{(x\, -\,2)}{(x\,+\,1)}\) ∴ (x + 3)(x +1) = (x – 2)(x + 11) ∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11) ∴ x2 + x + 3x + 3 = x2 + 11x – 2x – 22 ∴ x2 + 4x + 3 = x2 + 9x – 22 ∴ 4x + 3 = 9x – 22 ∴ 3 + 22 = 9x – 4x ∴ 25 = 5x ∴ x = 5 |
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| 11231. |
Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit. |
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Answer» Any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number. Hence, The five examples are: (i) 2061 The ending digit is 1. Hence, it may or may not be a square number (ii) 1069 The ending digit is 9. Hence, it may or may not be a square number (iii) 1234 The ending digit is 4. Hence, it may or may not be a square number (iv) 56790 The ending digit is 0. Hence, it may or may not be a square number (v) 76555 The ending digit is 5. Hence, it may or may not be a square number |
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| 11232. |
\(\sqrt{(0.798)^2+0.404\times 0.798+(0.202)^2}\) is equal to(a) 0 (b) 2 (c) 1.596 (d) 0.404 |
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Answer» (b) 2 Given exp. = \(\sqrt{(0.798)^2+2\times0.202\times0.798+(0.202)^2}\) +1 = \(\sqrt{(0.798+0.202)^2}+1\) = \(\sqrt1\) + 1 = 1 + 1 = 2 |
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| 11233. |
Without adding, find the sum (1+3+5+7+9+11+13+15+17+19+21+23). |
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Answer» Given (1+3+5+7+9+11+13+15+17+19+21+23) According to the property of perfect square, for any natural number n, we have the sum of first n odd natural numbers =n2 But here n = 12 Applying the above law we get Therefore (1+3+5+7+9+11+13+15+17+19+21+23) = 122 = 144 |
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| 11234. |
Without adding, find the sum: (i) (1 + 3 + 5 + 7 + 9 + 11 + 13) (ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) |
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Answer» (i) We know that, Sum of first n odd numbers = n2 Applying this formula in the question, we get (1 + 3 + 5 + 7 + 9 + 11 + 13) = (7)2 = 49 (ii) We know that, Sum of first n odd numbers = n2 Applying this formula in the question, we get (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) = (10)2 = 100 (iii) We know that, Sum of first n odd numbers = n2 Applying this formula in the question, we get (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23) = (12)2 = 144 |
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| 11235. |
Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit. |
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Answer» As we know that any natural number ending with 0, 1, 4, 5, 6 or 9 can be or cannot be a square number. Here are the five examples which you cannot decide whether they are square or not just by looking at the units place: (i) 2061 \(\therefore\)unit digit is 1. So, it may or may not be a square number (ii) 1069 \(\therefore\) unit digit is 9. So, it may or may not be a square number (iii) 1234 \(\therefore\)unit digit is 4. So, it may or may not be a square number (iv) 56790 \(\therefore\) unit digit is 0. So, it may or may not be a square number (v) 76555 \(\therefore\) unit digit is 5. So, it may or may not be a square number |
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| 11236. |
Assume that \(\sqrt {13}\) = 3.605 (approx.) and \(\sqrt{130}\) = 11.40 (approx.). Find the value of \(\sqrt{1.3}\) + \(\sqrt {1300}\) + \(\sqrt {0.013}\) (a) 36.164 (b) 37.304 (c) 36.304 (d) 37.164 |
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Answer» (b) 37.304 \(\sqrt{1.3} +\sqrt{1300}+\sqrt{0.013}\) = \(\sqrt{\frac{130}{100}}\) + \(\sqrt{13\times100}\) + \(\sqrt{\frac{130}{10000}}\) = \({\frac{\sqrt{130}}{\sqrt{100}}}\) + \(\sqrt{13}\) x \(\sqrt{100}\) + \(\frac{\sqrt{130}}{\sqrt{10000}}\) = \(\frac{11.40}{10}\) + 3.605 x 10 + \(\frac{11.40}{100}\) = 1.14 + 36.05 + 0.114 = 37.304 |
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| 11237. |
Which of the numbers for which you cannot decide whether they are squares. |
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Answer» As we know that the natural numbers such as 0, 1, 4, 5, 6 or 9 cannot be decided surely whether they are squares or not. |
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| 11238. |
Evaluate:\(\frac{\sqrt{1183}}{\sqrt{2023}}\)√1183/√2023 |
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Answer» We have, \(\frac{\sqrt{1183}}{\sqrt{2023}}=\sqrt{\frac{1183}{2023}}\) = \(\sqrt{\frac{169}{289}}\) = \(\frac{\sqrt{169}}{\sqrt{289}}\) = \(\sqrt{\frac{13\times13}{17\times17}}\) = \(\frac{13}{17}\) |
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| 11239. |
Find the squares of the following numbers:(i) 425(ii) 575(iii)405(iv) 205(v) 95 |
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Answer» (i) 425 4252 = 425 × 425 = 180625 (ii) 575 5752 = 575 × 575 = 330625 (iii) 405 4052 = 405 × 405 = 164025 (iv) 205 2052 = 205 × 205 = 42025 (v) 95 952 = 95 × 95 = 9025 |
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| 11240. |
The square root of the expression \(\frac{(12.1)^2-(8.10^2}{(0.25)^2+(0.25)(19.95)}\) is(a) 1(b) 2(c) 3(d) 4 |
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Answer» The answer is: (d) 4 |
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| 11241. |
If \(\sqrt{1369}\) + \(\sqrt{0.0615+X}\) , =37.5, then x is equal to (a) 10–1 (b) 10–2 (c) 10–3 (d) 10 |
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Answer» (c) 10-3 \(\sqrt{1369} +\sqrt{0.0615+X}\) = 37.25 \(\Rightarrow\) \(\sqrt{0.0615+X}\) = 37.25 - \(\sqrt{1369}\) = 37.25 - 37 = 0.25 \(\Rightarrow\) 0.0615 + x = (0.25)2 = 0.0625 \(\Rightarrow\) x = 0.0625 - 0.0615 = 0.001 = \(\frac{1}{10^3}\) = 10-3. |
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| 11242. |
By just examining the units digits, can you tell which of the following cannot be whole squares?(i) 1026(ii) 1028(iii)1024(iv) 1022(v) 1023(vi) 1027 |
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Answer» As we know that numbers ending with 2, 3, 7, 8 cannot be a perfect square. ∴ 1028, 1022, 1023, and 1027 cannot be whole squares. |
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| 11243. |
Evaluate:\(\sqrt{\frac{80}{405}}\)√80/405 |
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Answer» We have, \(\sqrt{\frac{80}{405}}\) = \(\sqrt{\frac{16}{81}}\) = \(\frac{\sqrt{16}}{\sqrt{81}}\) = \(\frac{4}{9}\) |
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| 11244. |
If \(\sqrt {0.04 \times 0.4\times a}\) = 0.004 x 0.4 x \(\sqrt b\) ,then \(\frac{a}{b}\) is(a) 16 × 10–3(b) 16 × 10–4(c) 16 × 10–5(d) 16 × 10–2 |
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Answer» (c) 16 x 10-5 Given exp.\(\Rightarrow\) \(\sqrt {0.016\times a}\) = 0.0016 x \(\sqrt b\) \(\Rightarrow\) \(\sqrt {0.016}\times \sqrt a\) = 0.0016 x \(\sqrt b\) \(\Rightarrow\) \(\frac{\sqrt a}{\sqrt{b}}\) = \(\frac{0.0016}{\sqrt{0.016}} \) \(\Rightarrow\) \(\frac{a}{b}\) = \(\frac{0.0016\times0.0016}{0.016}\) = 0.00016 = \(\frac{16}{100000}\) = 16 x 10-5. |
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| 11245. |
Consider the following values of the three given numbers:\(\sqrt{103}\) , \(\sqrt{99.35}\) , \(\sqrt{102.20}\)1. 10.1489 (approx.) 2. 10.109 (approx.) 3. 9.967 (approx.)The correct sequence of these values matching with the above numbers is: (a) 1, 2, 3 (b) 1, 3, 2 (c) 2, 3, 1 (d) 3, 1, 2 |
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Answer» The answer is: (b) 1, 3, 2 |
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| 11246. |
If \(\sqrt {(X-1)(y+2)}\) = 7, x and y being positive whole numbers, then the values of x and y are respectively(a) 8,5(b) 15,12(c) 22,19(d) 6,8 |
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Answer» (a) 8, 5 \(\sqrt{(X-1)(y+2)}\) = 7 \(\Rightarrow\) (x - 1) (y + 2) = 72 \(\Rightarrow\) (x - 1) = 7 and (y + 2) = 7 \(\Rightarrow\) x = 8 and y = 5. |
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| 11247. |
If \(\sqrt {(X-1)(y+2)}\) = 7 and x and y are positive whole numbers, their values respectively are(a) 8, 5(b) 15, 12(c) 22, 19(d) None of these |
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Answer» The answer is: (a) 8, 5 |
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| 11248. |
If a = 0.1039, then the value of \(\sqrt{4a^2- 4a +1}\) + 3a is(a) 0.1039(b) 0.2078(c) 1.1039(d) 2.1039 |
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Answer» (c) 1.1039 \(\sqrt{4a^2-4a+1}+3a\) = \(\sqrt{(1)^2-2\times 2a\times 1 +(2a)^2}+3a\) = \(\sqrt{(1-2a)^2}+3a\) = (1 - 2a) + 3a = 1 + a = 1 + 0.1039 = 1.1039. |
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| 11249. |
Find the square root of 9216 by using the method of prime factorization. |
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Answer» Given 9216 A square can always be expressed as a product of pairs of equal factors Resolve the given number into prime factors We get 9216 =2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 √9216 = 2 × 2 × 2 × 2 × 2 × 3 = 96 |
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| 11250. |
Evaluate: \(\sqrt{\frac{16}{81}}\)√(16/81) |
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Answer» We have, \(\sqrt{\frac{16}{81}}={\frac{\sqrt{16}}{\sqrt{81}}}\) We know that, \(\sqrt{16}\) = 4 And, \(\sqrt{81}\) = 9 Therefore, \(\sqrt{\frac{16}{81}}={\frac{\sqrt{16}}{\sqrt{81}}}\) \(=\frac{4}{9}\) |
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