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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. |
The cost of a pen is Rs.`16(3/5)` and that of a pencil is Rs. `4(3/4)`. Which costs more and by how much? |
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Answer» Cost of a pen in rupees ` = 16 3/5 = 16.60` Cost of a pencil in rupees ` = 4 3/4 = 4.75` So, cost of a pen is more. Difference in cost of pen and pencil ` = 16.60 -4.75 = 11.85 = 11 85/100 = 11 17/20` `:.` Cost of a pen is `11 17/20` is more than that of cost of a pencil. |
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| 152. |
Prove that `2sqrt3` is an irrational number |
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Answer» Rational numbers can be written in the form `p/q` where `p` and `q` are intergers and `q !=0`. Now, given number is , `2sqrt3 = 2sqrt3**sqrt3/sqrt3 = 6/sqrt3` So, it is in `p/q` form and `q !=0`. But, as `q` is not an integer, given number is not a rational number. Hence, `2sqrt3` is an irrational number. |
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| 153. |
A bag contains some square cards. A prime number between 1 and 100 has been written on each card. Find the probability of getting a card that the sum of the digits of prime number written on it is 8. |
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Answer» Total terms=25 Favorable terms=3 Prob.=3/25. |
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| 154. |
Show that ` 2sqrt3` is irrational. |
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Answer» Let us assume to the contrary, that ` 2sqrt3` is rational. Then, there exist co-prime a and b (` b ne 0)` such that ` 2 sqrt3= a/b Rightarrow sqrt3 = a/(2b)` Thus, ` sqrt3` is also rational. But, this contradicts that fact that ` sqrt3` is irrational,. So, our assumption is incorrect. Hence ` 2sqrt3` is irrational. |
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| 155. |
`sqrt((324)/(49))xxsqrt((676)/(169))xx(126)/sqrt(81) = x` |
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Answer» `x=sqrt(324/49)*sqrt(676/169)*(126/sqrt81)` `x=18/7*26/13*126/9` `x=72`. |
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| 156. |
Which of the following is a pair of co-primes ?A. (14,35)B. (18,25)C. (31,93)D. (32,62) |
| Answer» Correct Answer - B | |
| 157. |
The value of `sqrt(5+2sqrt(6))-1/(sqrt(5+2sqrt(6)))` is |
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Answer» `5 + 2sqrt6 = (sqrt2)^2 + (sqrt3)^2 + 2sqrt2sqrt3` so,`5 + 2sqrt6 = (sqrt2 + sqrt3)^2` now, `sqrt((sqrt2 + sqrt3)^2) - 1/sqrt((sqrt2 + sqrt3)^2)` `sqrt2 + sqrt3 - (1/(sqrt2 + sqrt3)) xx (sqrt2 - sqrt3)/(sqrt2 - sqrt3)` `sqrt2 + sqrt3 - (sqrt2 - sqrt3)/(2-3)` `sqrt2 + sqrt3 + sqrt2 - sqrt3= 2sqrt2` Answer |
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| 158. |
A bag contains 18 balls.out of which x are red. (i) If one ball is drawn at random from the bag, the what is the probability that it is a red ball ? (ii) If 2 more red balls are put in the bag, then probability of drawing a red ball will be 9/8 times that of red ball coming in part (i), find value of x. |
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Answer» Probability ball is red=>`x/18` new probability of red ball=`(x+2)/20=9/8xxx/18=x/16` `x=8balls` |
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| 159. |
If `log(a^3-b^3)-log3=3/2(log a+log b),` find the value of `(a/b)^3+(b/a)^3.` |
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Answer» `log(a^3 - b^3) - log3 = 3/2( log a + log b)` `log((a^3-b^3)/3) = 3/2(log ab)` `log ((a^3-b^3)/3) = log (ab)^(3/2)` `(a^3- b^3)/3 = (ab)^(3/2)` `a^3 - b^3 = 3(ab)^(3/2)` eqn1 `(a/b)^3- 1 = 3(ab)^(3/2)` `(a/b)^3 = 1 + 3(a/b)^(3/2)` `1 - (b/a)^3 = (3(ab^(3/2)))/a^3` `1 - 3(b/a)^(3/2) = (b/a)^3` `(a/b)^3 + (b/a)^3 = 1 + 3(a/b)^(3/2) + 1 - 3(a/b)^(3/2)` `= 2 + 3 ((a^3 - b^3)/(ab)^(3/2))` `= 2 + 3 [(3(ab)^(3/2))/(ab)^(3/2)]` `= 2 + 3xx 3` `= 11` Answer |
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| 160. |
Prove that ` (3 + 5sqrt2)` is irrational. |
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Answer» Let us assume, to the contrary , that ` (3 + 5sqrt2)` is rational. Then , there exist co-primes a and b ` ( b ne0)` such that ` 3 +5sqrt2= a/b` ` Rightarrow 5sqrt2 = a/b -3 = (a -3b)/b` ` Rightarrow sqrt2 = (a-3b)/(5b)` is rational. Thus, `sqrt2` is also rational. But, this contradicts the fact that ` sqrt2` is irrational. So, our asumption is incorrect. Hence, ` ( 3+5sqrt2)` is irrational. |
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| 161. |
`sqrt2` isA. a rational numberB. an irrational numberC. a terminating decimalD. a nonterminating repeating decimal |
| Answer» Correct Answer - B | |
| 162. |
Prove that ` (sqrt2+ 5sqrt2)` is irrational. |
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Answer» Let us assume that ` ( sqrt2 +sqrt3)` is irrational. Then, there exist co-primes a and b such that ` sqrt2 + sqrt3= a/b` ` Rightarrow sqrt3 = a/b -sqrt2` ` Rightarrow (sqrt3)^(2) = ( a/b -sqrt2)^(2)` ` Rightarrow 3 = a^(2)/b^(2) - (2a)/b sqrt2 +2` ` Rightarrow (2a)/b sqrt2 = a^(2)/b^(2) -1` ` Rightarrow sqrt2 = (a^(2) -b^(2))/(2ab)` Since a and b are intergers, so ` (a^(2) -b^(2))/(2ab)` is rational. Thus, ` sqrt2` is also rational. But, this contradicts the fact that ` sqrt2` is irrational , so, our assumption is incorrect. Hence ` (sqrt2+ sqrt3)` is irrational. |
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| 163. |
Show that ` 1/sqrt2` is irrational. |
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Answer» Let us assume, to the contrary that` 1/sqrt2` is rational. Then , there exist co-prime a and b ` (b ne 0)` such that ` 1/sqrt2 = a/b Rightarrow sqrt2 = b/a` Since a and b are integers , so ` a/b` is rational. Thus, `sqrt2` is also rational. But, this contradicts the fact that ` sqrt2` is irrational. So, our assumption is incorrect. Hence, `1/sqrt2` is irrational. |
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| 164. |
In a G.P. the 3rd term is 8 times the 6th term and the 4th term is 4 times the 6th term.Find the common ratio of the G.P. |
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Answer» Let `a` is the first term and `r` is the common ratio of the given GP. Then, we are given, `ar^2 = 8ar^5` `=>r^3 = 1/8` `r = 1/2->(1)` Also, we are given, `ar^3 = 4ar^5` `=>r^2 = 1/4` `=>r = +-1/2->(2)` From (1) , we can see that r can not be negative. So, common ratio of given GP is `1/2`. |
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| 165. |
`1/sqrt2` isA. a fractionB. a rational numberC. an irrational numberD. none of these |
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Answer» Correct Answer - C ` 1/sqrt2 = 1/sqrt2 xx sqrt2/sqrt2 = 1/2 = sqrt2` Here, `1/2` is rational and ` sqrt 2` is irrational. And, the product of a rational and on irrational irrational. `1/2. sqrt2 and " hence" sqrt2` is irrational. |
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| 166. |
Two positive integers p and q can be expressed as ` p = a b^2 and q = a^2 b`, and b are prime numbers. what is L.C.M of p and q. |
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Answer» Here, `p = ab^2 = a**b**b` `q = a^2b = a**a**b` `:. LCM` of `p and q = b**b**a**a = a^2b^2`. |
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| 167. |
p and q are two positive integers such that the least prime factor of p is 3 and the least prime factor of q is 5 . Find the least prime factor of (p + q).A. 2B. 3C. 5D. 8 |
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Answer» Correct Answer - A Clearly, 2 is neither a factor of a nor that of b. a and b are both odd. Hence , (a +b) is even. least prime factor of ( a+b) is 2. |
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| 168. |
If `5 leq 4K +1 leq 29`, then find the maximum num- ber of co-primes formed from the possible valuesof K?A. 15B. 11C. 13D. 17 |
| Answer» Correct Answer - D | |
| 169. |
Find the number of factors of `1498176` and also their sum. |
| Answer» Correct Answer - `105,4717669` | |
| 170. |
A fruit vendor has 260 mangoes , 292 oranges and 220 apples. He cells each of these fruits in a package containing the same number of fruits in a package containing the same number of fruits of the same kind and has found that 4 apples, 4 orangles and 4 mangoes are left out. Find the greatest possible size of the package. where size of the package is defined as the number of fruits in each package. |
| Answer» Correct Answer - 8 | |
| 171. |
The GCD of two numbers is 128 and their LCM is 256. Then their product isA. 38028B. 36868C. 32768D. Data inconsistent |
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Answer» Correct Answer - C LCM x GCD = twin primes |
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| 172. |
The LCM of `(5)/(12),(6)/(5),(3)/(2)and (4)/(17)` isA. 60B. `(1)/(60)`C. 180D. None of these |
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Answer» Correct Answer - A LCM of fractions `("LCM of numerator")/("HCF of denominator")` |
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| 173. |
Find the LCM of 12, 48 and 36. The following are the sequential order from the first to last. (A) `12=2^2xx3^1,48=2^4xx3^1,36=2^2xx3^2` (B) `LCM =2^4xx3^2=16xx9=144` (C) Resolving the given into prodct of prime factors.A. ACBB. ABCC. CBAD. CAB |
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Answer» Correct Answer - D cab is the sequential order from first to last. |
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| 174. |
The LCM of 100 and 101 is ____A. 101000B. 10001C. 10101D. None of these |
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Answer» Correct Answer - A The LCM of any two consecutive integers is their proudct. `therefore LCM =100 xx 101 =10100` |
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| 175. |
The HCF of 100 and 101 is ____A. 1B. 7C. 37D. None of these |
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Answer» Correct Answer - A The HCF of any two consecutive integers is 1. |
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| 176. |
In the given figure, D, E and F are the points where the incircle of the `Delta`ABC touches F the sides BC, CA and AB respectively. Show that AF+ BD + CE = AE + BF + CD = (perimeter of `Delta ABC`) |
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Answer» We know, length of tangents from any common point outside a circle are always equal. `:. AF = AE ->(1)` `BD = BF->(2)` `CE = CD ->(3)` Adding (1),(2) and (3), `AF+BD+CE = AE+BF+CD` Now, perimeter of `ABC = AB+BC+AC` `=AF+BF+BD+DC+CE+AE` `=2(AF+BD+CE)` `:. 1/2 (` Perimeter of `ABC) = AF+BD+CE = AE+BF+CD.` |
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| 177. |
In a right angle triangle `Delta ABC` is which `/_ B = 90^@` a circle is drawn with AB diameter intersecting the hypotenuse AC at P.Prove that the tangent to the circle at PQ bisects BC. |
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Answer» BQ=PQ-(1) `/_APB=90^o` `/_BPC=180-90=90^o` `/_BPC=90^0` `/_BPQ+/_QPC=90^o` from equation 1 `/_BPQ=/_PBQ` `/_PBQ+/_PCQ=90^o-(3)` from equation 2 and 3 `/_BPQ+/_QPC=/_PBQ+/_PCQ` `/_QPC=/_PCQ` PQ=QC From 1 BQ=QC Q is mid point of BC. |
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| 178. |
Jacks, Queens and Red Kings are removed from a pack of 52 playing cards and then well shuffled .A card is drawn from the remaining cards. Find the probability of gettinga) a kingb) a red cardc) a spade |
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Answer» 52 cards Total=52-10-42 1)P=favorable outcomes/Total outcome P=`(2C_1)/(42C_1)=2/42=1/21` 2)P=`(20C_1)/(42C_1)=20/42=10/21` 3)P=`(11C_1)/(42C_1)=11/42`. |
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| 179. |
Can two number have 18 as their HCF and 380 as their LCM? Give reason |
| Answer» No, because HCF is always a factor of LCM but here 18 is not a factor of 380 | |
| 180. |
Can two numbers have `18` as their HCF and `380` as their LCM? Give reason. |
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Answer» We know that the L.C.M of two numbers is always divisible by their H.C.F So. `(380)/(18)` should be a natural number. But it is not a natural number. |
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| 181. |
Write a rational number between ` sqrt3 and 2` |
| Answer» Clearly, ` sqrt3 = 1.732` ….., so, we may take 1.8 as the required rational number between ` sqrt3 and 2`. | |
| 182. |
The number 525 and 3000 are both divisible only 3,5,15,25,75. What is HCF (525, 3000)? Justify your answer. |
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Answer» The numbers 525 and 3000 both are divisible by 3,5,15,25 and 75`:.` Highest common factor is 75. Hence H.C.F. (3000, 525 )=75 Verification : `3000=525 xx 5 +375` `525 =375 xx 1+150` `375 =150 xx 2 +75` `150 =75 xx 2+0` `:. H.C.F. (3000 ,525 )=75` |
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| 183. |
Show that `12^n` cannot end with the digits `0` or `5` for any natural number `n` |
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Answer» If any number ends with the digit 0 or 5, it is always divisible by 5. This is possible only if prime factorisation of `12^(n)` contains the prime number 5. Now, `12=2 xx 2 xx 3=2^(2)xx3` `Rightarrow 12^(m)=(2^(2)xx3)^(n) =2^(2n)xx3^(m)` [Since, there is no term contains 5] Hence, there is no value of n `in` N for which `12^(n)` ends with digit zero or live. |
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| 184. |
Prove that `sqrt3+sqrt5` is irrational |
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Answer» Let us suppose that `sqrt3+sqrt5` is rational Let `sqrt3+sqrt5=a`, where a is rational Therefore, `sqrt3=a-sqrt5` On squaring both sides, we get `(sqrt3)^(2)=(a-sqrt5)^(2)` `Rightarrow 3=a^(2)+5-2asqrt5` `"["therefore (a-b)^(2)=a^(2)+b^(2)-2ab]` `Rightarrow 2a sqrt5=a^(2)+2` `Therefore, sqrt5=(a^(2)+2)/(2a)` which is contracdiction. As the right hand side is rational number while `sqrt5` is irrational. Since 3 and 5 are prime number. Hence, `sqrt3 +sqrt5` is irrational. |
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| 185. |
Prove that if `xa n dy`are odd positive integers, then `x^2+y^2`is even but not divisible by 4. |
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Answer» Let x=2m+1 and y=2m+3 are odd integers, for every positive integers, for every positive integer m, Then, `x^(2)+y^(2)(2m+1)^(2)+(2m+3)^(2)` `=4m^(2) =(2m+1)^(2)+(2m+3)^(2)` `=4m^(2)+1+4m+4m^(2)+9+12m [therefore (a+b)^(2)=a^(2)+2ab+b^(2)]` `=8m^(2)+16m+10="even"` `=2(4m^(2)+8m+5) or 4(2m^(2)+4m+2)+1` Hence, `x^(2)+y^(2)` is even for every positive integer m but not divisible by 4. |
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| 186. |
The decimal expansion of the number ` 4753/1250` will terminate afterA. one decimal placeB. two decimal placesC. three decimal placesD. four decimal places |
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Answer» Correct Answer - D `147531250 xx 8/8 = ( 14753xx 8) / 1000 = (14753 xx 8)/((10)^(4))` So, it will terminate after 4 decimal places. |
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| 187. |
The decimal expansion of the rational number ` 43/((2^(4). 5^(3))` will termiate after how many places of decimals. |
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Answer» We have `43/(2^(4).5^(3))= (43xx5)/(2^(4)xx5^(4))= 215/((2xx5)^(4))= 215/(10^(4))= 215/(10000)= 0.0215` So, it will terminate after 4 places of decimals. |
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| 188. |
Express `0.overline(32)` as a fraction in simplest form. |
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Answer» Let x= `0.overline(32)`,then x = 0.3232… 100x= 32.3232 On subtracting (i) from(ii) ,we get `99x= 32 Rightarrow x = 32/99` Hence , `0.overline(32) = 32/99` |
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| 189. |
Express each of the following as a fraction in simplest form : ` 0.overline(8)` (ii)` 2.overline(4)` (iii) ` 0.overline(24)` (iv) ` 0.1overline2` (vi) 0.00608 |
| Answer» Correct Answer - (i) ` 8/9 (ii) 22/9 (iii) 8/33 (iv) 11/90 (v) 101/45 (vi) 181/495 ` | |
| 190. |
`0.overline(68) + 0. overline(73) ` = ?A. `1.overline(41)`B. `1.overline(42)`C. `0.overline(141)`D. none of these |
| Answer» Correct Answer - B | |
| 191. |
In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT=24 cm.If O is the center of circle then length OP=? |
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Answer» `In/_PTO` by Pythagoras theorem `OP^2=24^2+7^2` `OP^2=625` `OP=25 cm`. |
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| 192. |
(i) Given an example of two irrational whose sum is rational. (ii) Give an example of two irrationals whose product is rational. |
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Answer» Take `(2+sqrt(3))` and `(2- sqrt(3))`. (ii) Take `(3+sqrt(2))` and `(3- sqrt(2))` |
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| 193. |
Which of the following numbers are rational? Also, identify the irrational numbers. (a) `0 -1 +1` (d) `(3)/(2)` (e) `(-3)/(2)` (f) `sqrt(2)` (g) `-sqrt(2)` (h) `sqrt(2)xxsqrt(8)` (i) `-sqrt(2)xxsqrt(8)` (j) `(4-sqrt(2))(2+(1)/sqer(2))` (k) `9990` (l) `14^(-28)` |
| Answer» Correct Answer - `{:("Rational Numbers",,"Irrational Numbers"),("(a),(b),(c)",,"(f),(g)"),("(d),(e),(h)",,),("(i),(j),(k),(l)",,):}` | |
| 194. |
A rectangular parking lot is 30 units long and 20 units wide. An equal no. of units are to be added to the length and to the width so that the area of a lot is doubled. Find the dimension of new parking lot. |
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Answer» Area of rectangular park ` = 30**30 = 600` square units Let `x` unit of length and width is added, then, Area of new rectangular park ` = (30+x)(20+x)` square units New area is double than previous area, `:. (30+x)(20+x) = 2**600` `=>x^2+50x+600-1200 = 0` `=>x^2+50x-600 = 0` `=>x^2+60x-10x-600 = 0` `=>(x+60)(x-10) = 0` As, `x` can not be negative, so, `x = 10` So, New length of the park `= 30+10 = 40` units New width of the park `= 20+10 = 30` units |
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| 195. |
Give an example of two irrational numbers whose sum is rational |
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Answer» Let `x` and `y` are two irrational number such that, `x = 10+2sqrt5` and `y = 10-2sqrt5` Their sum will be, `x+y = 10+2sqrt5+10-2sqrt5 = 20`, which is rational. |
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| 196. |
A drinking glass is in the shape of a frustum of a cone of height 14 cm.The diameter of its two circular ends are 4 cm and 2 cm respectively.Find the capacity of the glass. |
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Answer» The volume of a frustum can be given as `V = (pi h)/3(R^2+Rr+r^2)` Here, `h = 14` cm, `R =4/2 = 2` cm, `r = 2/2 = 1` cm `: V = 22/7**14/3(4+2+1)` `V = (44**7)/3 = 308/3cm^3`. So, capacity of the given glass is `308/3 cm^3`. |
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| 197. |
From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a circular cone of the same height and same base is removed. Find the volume of remaining solid and also find the whole surface area. |
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Answer» Given, r=6cm, h=10cm. Volume of remaining solid= `pir^2h - 1/3pir^2h`=`2/3pir^2h` =`2/3*22/7*6*6*10=5280/7cm^3` Total Surface area=`2pirh+ pir^2+ pirsqrt(r^2+h^2)`=`22/7(2*6*10+6*6+6*sqrt(6*6+6*10))`=`22/7(156+ 12sqrt34)cm^2` |
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| 198. |
Is it possible to have two numbers whose HCF is 25 and LCM is 520 ? |
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Answer» Correct Answer - No HCF always divides the LCM completely. |
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| 199. |
If the number `2345p60q` is exactly divisible by 3 and 5, then the maximum value of `p+q` isA. 12B. 13C. 14D. 15 |
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Answer» Correct Answer - B (i) Recall the divisibility rules of 3 and 5. (ii) `q=0 or 5` , But q has to 5, because `p+q` has to be maximum. (iii) Sum of the digits of `2345p60q` is divisible by 3 . Find p, when `q=5`. |
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| 200. |
`73412130` is exactly divisible byA. 3B. 11C. 7D. Both (a) and (b) |
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Answer» Correct Answer - D Recall the divisibility rules |
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