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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.
| 51. |
If `sum_(k=4)^(143) (1)/(sqrt(k)+sqrt(k+1))=a-sqrt(b)` then a and b respectively areA. 10 and 0B. `-10 and 4`C. 10 and 4D. `-10 and 0` |
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Answer» Correct Answer - A (i) Substitute the values of k and rationalize every term of LHS. (ii) `underset(k=4)overset(143)sum(1)/(sqrt(k)+sqrt(k+1))=(1)/(sqrt(5)+sqrt(4))+(1)/(sqrt(6)+sqrt(5))+...+(1)/(sqrt(144)+sqrt(143))` (iii) Rationalize the denominator two times. |
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| 52. |
Express the surd `(root(3)(5))/(root(3)(6))` with rational denominator. |
| Answer» Correct Answer - `(root(3)(180))/(6)` | |
| 53. |
`(4)/(sqrt(10-2sqrt(21)))`=______.A. `(1)/(4)(sqrt(7)+sqrt(3))`B. `(1)/(4)(sqrt(7)-sqrt(3))`C. `sqrt()7+sqrt()3`D. `sqrt(7)+sqrt(3)` |
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Answer» Correct Answer - C `(4)/(sqrt(10-2sqrt(21)))=(4)/(sqrt((sqrt(7))^(2)+(sqrt(3))^(2)-2sqrt(7xx3)))` `=(4)/(sqrt((sqrt(7)-sqrt(3))^(2)))=(4)/(sqrt(7)-sqrt(3))` `(4(sqrt(7)+sqrt(3)))/((sqrt(7)-sqrt(3))(sqrt(7)+sqrt(3)))` `(4(sqrt(7)+sqrt(3)))/(4)=sqrt(7)+sqrt(3)` |
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| 54. |
Express the surd `(3)/(sqrt(11)` with rational denominator . |
| Answer» Correct Answer - `(3sqrt(11))/(11)` | |
| 55. |
The surd `(12)/(3+sqrt(5)+2sqrt(2))`, after rationalizing th denominator becomesA. `sqrt(5)-sqrt(2)+sqrt(10)+1`B. `sqrt(5)+sqrt(10)+sqrt(2)+1`C. `sqrt(10)+sqrt(2)+sqrt(5)+1`D. `sqrt(5)-sqrt(10)-sqrt(2)-1` |
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Answer» Correct Answer - B (i) Rationalize the denominator two times (ii) RF of denominator is `(3+sqrt(5)-2sqrt(2))`. (iii) Rationalize the denominator twice and simplify. |
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| 56. |
`x+sqrt(y)` is a pure surd, if x=______ (zero/one) |
| Answer» Correct Answer - zero | |
| 57. |
The surd obtained after rationalizing the numberator of `(4-sqrt(25-a))/(a-9)` is equal toA. `(a-9)/(4-sqrt(25-a))`B. `(1)/(4-sqrt(25-a))`C. `(1)/((a-9)(4+sqrt(25-a)))`D. `(1)/(4+sqrt(25-a))` |
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Answer» Correct Answer - B R F of `a-sqrt(b) ` is `a+sqrt(b)`. |
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| 58. |
Given `sqrt(3)=1.7321`, find the value of the following surd, correct to three decimal places. `(sqrt(3)+1)/(sqrt(3)+1)+(sqrt(3)-1)/(sqrt(3)+1)+(4+sqrt(3))/(4-sqrt(3))` |
| Answer» Correct Answer - 6.527 | |
| 59. |
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of `1/(17)`? Perform the division to check your answer. |
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Answer» When we calculate `1/17`, it comes `0.0588235294117647`. After that, if we still continue the division, it starts with the same numbers after decimal. So, we can count the number of digits after decimal point in `0.0588235294117647` to find the maximum number of digits in the repeating block of digits. It comes 16 that is required number. |
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| 60. |
PQRS is a parallelogram. PS is produced to M, so that SM = SR and MR produced meets PQ produced at N. Prove that QN = QR. |
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Answer» With the given details, we can create a diagram. Please refer to video for the diagram. From the figure,`SM||QR||PS` `:. /_QRN =/_SMR` Also, `PQ||SR||QN` `:./_RSM = /_RQN` In `Delta RSM and Delta NQR`, `/_QRN =/_SMR` `/_RSM = /_RQN` As two of their angles are equal, third angle will also be equal. So, `Delta RSM ~=Delta NQR` `:. (SM)/(QR) = (SR)/(QN)=> (SM)/(SR)= (QR)/(QN)` As, `SM =SR` , it means ,`QR = QN` |
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| 61. |
A right circular cylinder circumscribes a sphere of radius r. Find the ratio of the surface area of the sphere to the curved surface area of the cylinder. |
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Answer» As cylinder circumscribes a sphere of radius `r`. `:.` Height of cyllinder, `h = 2r` Surface area of sphere `S_s = 4pir^2` Surface area of cyllinder `S_c = 2pirh = 2pir(2r) = 4pir^2` `:.` Ratio of the surface area of the sphere to the curved surface area of the cylinder `= S_s/S_c = (4pir^2)/(4pir^2) = 1/1` |
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| 62. |
Calculate the amount and the compound interest on Rs. 20000 in 3 years when the rate of interest for the successive years is 10%, 10% and 12% per annum respectively. |
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Answer» Here, Principal amount `(P) = 20000` Rs So, amount after first year ` = 20000(1+10/100)^1` Amount after second year ` = 20000(1+10/100)(1+10/100)^1` Amount after third year ` = 20000(1+10/100)(1+10/100)(1+12/100)^1` ` = 20000**11/10**11/10**112/100 = 224**121 = 27104` Rs `:.` Compound interest ` = 27104-20000 = 7104` Rs |
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| 63. |
Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half- yearly. If Geeta gets Rs. 33 more than Ashok in 18 months, calculate the money invested. |
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Answer» Let money invested=P Amount`(A_1)=P(1+R/100)^t` `A_1=P(1+20/100)^(3/2)` `A_1=P(1+1/5)^(3/2)` `A_1=P(6/5)(11/10)` `A_2=P(1+20/(200))^(3/2*2)` `A_2=P(1+1/10)^3` subtracting equation 1 from equation 2 `P(1331/1000)-P(6/5)(11/10)=33` `P(1.331-66/50)=33` `P(1.331-1.32)=33` `P=3000Rs`. |
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| 64. |
Find LCM of `12^16 , 18^18 , 30^18` ? |
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Answer» `LCM(12^16)=2^16*2^16*3^16` `LCM(18^18)=2^18*3^18*3^18` `LCM(30^18)=2^18*3^18*5^18` `LCM(12^16,18^18,30^18)=2^32*3^36*5^18`. |
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| 65. |
Simplify the following (i) `sqrt45-3sqrt20+4sqrt5` (ii) `sqrt(24)/8 + sqrt54/9` (iii) `root4(12) xx root7(6)` (iv) `4sqrt28 div 3sqrt7 div root3(7)` (v)` 3sqrt3+2sqrt27 + 7/(sqrt3)` (vi) `(sqrt3-sqrt2)^(2)` (vii) `root4(81)-8root3(216)+15root5(32)+ sqrt225` (viii) `3/sqrt8+ 1 / sqrt2` (ix) `(2sqrt3)/3- (sqrt3)/6` |
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Answer» `sqrt45-3sqrt20+4sqrt5= sqrt(3xx3xx5)-3sqrt(2xx2xx2)+4sqrt5` ` 3sqrt5-3xx2sqrt5+4sqrt5` `3sqrt5-6sqrt5+4sqrt5=sqrt5` `sqrt24/8+sqrt54/9=sqrt(2xx2xx2xx2xx3)/8+sqrt(3xx3xx3xx2)/9` `(2sqrt6)/8 + (3sqrt6)/9=sqrt6/4+sqrt6/3` `(3sqrt6+4sqrt6)/12=(7sqrt6)/12` `root4(12)xxroot7(6)=(12)^(1//4)xx(6)^(1//7)` `(2xx2xx3)^(1//4)xx(2xx3)^(1//7)=2^(1//4).2^(1//4).3^(1//4).2^(1//7).3^(1//7)` `2^(1/4+1/4+1/7)xx3^(1/4+1/7)` `2^(9//14)xx3^(11//28)=root14(2^(9))root28(3^(11)) " " [ x^(m).x^(n)=x^(m+n)]` `root28(2^(18))root28(33^(11))=root28(2^(18)xx3^(11)) " " [ root na= root(mn)(a^(m))` (iv) `4sqrt28+3sqrt7 div root3(7) =(4sqrt(4xx7)div3sqrt7)divroot3(7)` ` ((8sqrt7)/(3sqrt7))div(7)^(1/3)` `8/3div7^(1/3)= 8/(3root3(7))` (v) `3sqrt3+2sqrt27+7/sqrt3=3sqrt3+2sqrt(3xx3xx3)+7/sqrt3xxsqrt3/sqrt3` `3sqrt3+ 6sqrt3+(7sqrt3)/3=9sqrt3+(7sqrt3)/3` `3sqrt3+6sqrt3+(7sqrt3)/3=9sqrt3+(7sqrt3)/3` `(27sqrt3+7sqrt3)/3=(34sqrt3)/3` (vi) `(sqrt3-sqrt2)^(2)+(sqrt2)^(2)-2sqrt3xxsqrt2 ` [ using identity , `(a-b)^(2)= a^(2+b^(2)-2ab]` `3+2-2sqrt(3xx2)=5-2sqrt6` (vii) `root4(81)-8root3(216)+15root5(32)+sqrt225` `(81)^(1/4)-8xx(6^(3))^(1/3)15xx(32)^(1/5)+sqrt((15)^(2))` ` (3^(4))^(1/4)-8xx(6^(3))^(1/3)+15xx2^(5xx(1/5))+15` `3^(1)-8xx6^(1)+15xx2^(1)+15` 3-48+30+15 48-48=0 `3/sqrt8+1/sqrt2=3/(sqrt(2xx2xx2))+1/sqrt2=3/(2sqrt2)+1/sqrt2` ` (3+2)/(2sqrt2)+1/sqrt2` `(5sqrt2)/(2xx2)=5/(2sqrt2)xxsqrt2/sqrt2` `(5sqrt2)/(2xx2)=(5sqrt2)/4` (ix) `(2sqrt3)/3-sqrt3/6=(4sqrt3-sqrt3)/6=(3sqrt3)/6=sqrt3/2` |
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| 66. |
if `log 15` base 16 =a and log 18 base 12 = b then show that log 24 base 25 = (5-b)/(16a - 8ab -4b +2 ) |
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Answer» `(log_2^15/log_2^16) =a` `(log_2^3+log_2^5)/4=a` `(log_2^18/log_2^12)=b` `(3log_2^3+1)/(2+log_2^3)=b` `(log_2^24/log_2^25)=(log_2^8+log_2^3)/(2log_2^5)=(3+log_2^3)/(2(4a-log_2^3)` `log_2^3=t` `(3+t)/(2(4a-t))` `(3+(2b-1)/(2-b))/(2(4a-(2b-1)/(2-b))` `((6-3b+2b-1)/(2-b))/((8a-4ab-2b+1)/(2-b))` `(5-b)/(16a-8ab-4b+2)`. |
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| 67. |
In the given figure, O is the centre of a circle such that OA = 5 cm, AB = 8 cm and `OD_|_AB` meeting AB at C. Find the length of CD. |
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Answer» As `OD _|_ AB`, `:. AC = 1/2AB = 1/2*8 = 4cm` Here, `OA = OD = 5cm` `:. OC = sqrt(OA^2-AC^2) = sqrt(5^2-4^2) = 3cm` Now, `CD = OD - OC = 5-3 = 2cm` |
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| 68. |
AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that AB=10 cm, CD=24 cm and the distance between AB and CD is 17 cm.Find the radius of the circle. |
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Answer» Let `r` is the radius of the circle and `O` is the center of fthe circle. We can draw a diagram with the given details. Please refer to video for the diagram. From the diagram, In `Delta OAN`, `r^2 = 5^2+(17-x)^2->(1)` In `Delta OMC`, `r^2 = 12^2+x^2->(2)` From (1) and (2),` 5^2+(17-x)^2 = 12^2+x^2` `=> 25+289+x^2-34x = 144+x^2` `=>34x = 170=> x = 5` Putting value of `x` in (2), ` r^2 = 12^2+5^2 = 169` `r = 13` cm So, radius of the circle is `13` cm. |
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| 69. |
In the adjoining figure chords AC and BD of a circle O, intersect at right angles at E. If `/_ AOB = 25^@` calculate `/_EBC` |
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Answer» OA=OB `/_OAB=/_OBA=25^0` `/_OAB` `/_A+/_O+/_B=180^0` `50^0+/_AOB=180^0` `/_AOB=130^0` `/_ACB=1/2*/_AOB` `/_ACB=1/2*130=65^0` In`/_BEC` `/_ECB+/_EBC+/_CEB=180^0` `65^0+/_EBC+90^0=180^0` `/_EBC=25^0`. |
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| 70. |
Chords equidistant from the center of a circle are equal in length. |
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Answer» OX=OY `In/_OAX and /_ODY` OA=OD(radii) OX=OY(given) `/_OXA-/_OYD(90^0)` `/_OAX and /_ODY` AX=DY AB=CD=PQ so, chords are equal. |
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| 71. |
If `x=(1)/(7+4sqrt(3)), gamma=(1)/(7-4sqrt(3))`, find the value of `5x^(2)-7xy-5gamma^(2)` |
| Answer» Correct Answer - `-7(1+80sqrt(3))` | |
| 72. |
Visualise 3.765 on the number line, using successive magnification. |
| Answer» Please refer to video for the first question to visualise 3.765 on the number line. | |
| 73. |
The horizontal cross-section of a water tank is in the shape of a rectangle with a semicircle at one end, as shown in fig. 18.20. The water is 2.4m deep in tank. Calculate the volume of the water in gallons. |
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Answer» Here, Area of water tank = Area of rectangle+ Area of semicircle `:.` Area of tank,`A = (21**7)+(1/2**22/7**7/2**7/2)` `A = 147+77/4 = 166.25 m^2` As, water tank is `2.4` m deep in tank. `:.` Volume of tank, `V = 166.25**2.4 = 399m^3` Volume of tank in litres ` = 399**1000 = 399000` Volume of tank in gallons`= 399000/3.78 = 105555.55` gallons |
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| 74. |
There are 3 books of mathematics, 4 books of Physics and 5 on English. How many different collection can be made such that each collection consist of : |
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Answer» M=3,P=4,E=5 a)`3C_1*4C_1*5C_1=60` 3)`3C_0+3C_1+..3C_3=1+3+3+1=8` `8(4C_0+4C_1+4C_2+4C_3+4C_4)` `8*16=144` Total=`128(5C_1+5C_2+5C_3+5C_4+5C_5)` `=128(5+10+10+5+1)`=3468 2)`M=3C_1+3C_2+3C_3=3+3+1=7` `P=4C_1+4C_2+4C_3+4CC_4=4+6+4+1=15` `E=31` Ways=31*15*7=3255 |
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| 75. |
If `[{((1)/(7^(2)))^(-2)}^(-1//3)]^(1//4)=7^(m)` , then m= _____.A. `(-1)/(3)`B. `(1)/(4)`C. `-3`D. 2 |
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Answer» Correct Answer - A Apply laws of indices. |
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| 76. |
Which of the following is irrational?`0. 14`(b) `0. 14 16 ``0. 1416 `(d) `0. 1014001400014. .`A. `0.14`B. `0.14bar(16)`C. `0.bar(1416)`D. `0.4014001400014` |
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Answer» Correct Answer - D An irrational number is non-terminating non-recurring which is `0.4014001400014"….."`. Hence, `0.14` is terminating and `0.14bar(16)`, . `bar(1416)` are non-terminating recurring. |
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| 77. |
`[{(1/(x^(a^(2)-b^2)))^(1/(a-b))} ^(a+b)]^(1/((a+b)^2))=_________`A. `x^(2)`B. `(1)/(x)`C. `7^(3)`D. `(1)/(x^(2))` |
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Answer» Correct Answer - B (i) `(a^(m))^(n)=a^(mn)` (ii) Use `((a^(m))^(n))^(s)=a^(mns)` |
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| 78. |
If `(2^(m+n))/(2^(m-n))=16 and a=2^(1/10)` then `((a^(2m+n-p))^2)/((a^(m-2n+2p)- 1)^-1)=`A. 2B. `(1)/(4)`C. 9D. `(1)/(8)` |
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Answer» Correct Answer - A (i) Apply laws of indices (ii) `(2^(m+n))/(2^(m-n))=16 implies 2^(2m)=2^(4)` . Find the value of m and apply in given expression. |
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| 79. |
If `(9sqrt3+11sqrt2)^[1/3]=sqrta+sqrtb`,where a and b are rational with a > b.Then |
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Answer» `(9sqrt3+11sqrt2)^(1/3) =sqrta+sqrtb` Taking cube both sides, `(9sqrt3+11sqrt2) = (sqrta+sqrtb)^3``(9sqrt3+11sqrt2) = (sqrta)^3+(sqrtb)^3 + 3(sqrta)^2b+3a(sqrtb)^2` `(9sqrt3+11sqrt2) = asqrta+bsqrtb+3asqrtb+3sqrtab` `(9sqrt3+11sqrt2) = (a+3b)sqrta+(b+3a)sqrtb` If we put, `a = 3 and b = 2`, it satisfies the above equation. So, option `B` is the correct option. |
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| 80. |
The sum of n terms of two arithmetic progression is in the ratio 2n+1 : 2n-1. Find the ratio of `10^th` term. |
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Answer» Let first term in first AP is `a_1` and common difference is `d_1`. Let first term in second AP is `a_2` and common difference is `d_2`. Then, we are given, `S_1/S_2 = (2n+1)/(2n-1)` `(n/2(2a_1+(n-1)d_1))/(n/2(2a_2+(n-1)d_2)) = (2n+1)/(2n-1)` `=>(2a_1+(n-1)d_1)/(2a_2+(n-1)d_2) = (2n+1)/(2n-1)->(1)` Now, we have to find ratio of `10th ` terms of both AP. If we put `n = 19` in (1), `(2a_1+18d_1)/(2a_2+18d_2) = 39/37` `=> (a_1+9d_1)/(a_2+9d_2) = 39/37->(2)` `10th` term of an AP is given by, `T_10 = a+9d` So, equation `(2)` gives us the required ratio that is `39:37`. |
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| 81. |
Find six rational numbers between 3 and 4. |
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Answer» Difference between two given rational numbers `(b-a) = 4-3= 1` So, we will take `6` numbers with difference ,`(b-a)/(n+1) = 1/(6+1) = 1/7`. `:.` First number will be ` = 3+1/7 = 22/7` So, `6` rational numbers between `3` and `4` using mean method are, `22/7,23/17,24/7,25/7,26/7,27/7.` |
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| 82. |
Solve by componendo dividendo method |
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Answer» `((3x-4)^3-(x+1)^3)/((3x-4)^3+(x+1)^3) = 61/89` Using componendo and dividendo both sides, `((3x-4)^3-(x+1)^3+(3x-4)^3+(x+1)^3)/((3x-4)^3-(x+1)^3-(3x-4)^3-(x+1)^3) = (61+189)/(61-189)` `=>(2(3x-4)^3)/(-2(x+1)^3) = -250/128` `=>(3x-4)^3/(x+1)^3 = 125/64` `=>((3x-4)/(x+1))^3 = (5/4)^3` `=>(3x-4)/(x+1) = 5/4` `=>12x-16 = 5x+5` `=>7x = 21=> x = 3` |
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| 83. |
In `Delta ABC` the lines are drawn parallel to BC,CA and AB respectively through A,B and C intersecting at P, Q and R. Find the ration of perimeter of `Delta PQR` and `DeltaABC` |
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Answer» With the given details, we can create a diagram. Please refer to video to see the diagram. It is given that `PQ || BC and PR||AC` `:. PA||BC and PB||AC` It mean `APBC` is a parallelogram. `:. AC = PB, AP = BC` It means, all sides of `Delta APB and Delta ABC` are equal So, their parimeters will be equal. Similarly, `AQCB` and `ABRC` are also parallelogram. `:.` Perimeter of `Delta PQR = 2(AB+BC+AC)` Perimeter of `Delta ABC = AB+BC+AC` So, ratio of their perimeters ` = 2:1.` |
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| 84. |
Among the 100 employees, the average salary of 99 employees is Rs. 50. If the `100^th` has the salary exceeding the average of 100 employees by Rs. 49.50, then the average salary of all the employees is: |
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Answer» Let average salary of all employees in rupees `= x` Then, Sum of salaries of all employees ` = (99**50)+(x+49.5) = 4999.5+x` `:. (4999.5+x)/100 = x` `=> 4999.5+x = 100x` `=> 99x = 4999.5 => x = 4999.5/99 = 454.5/9 = 50.50` `:.` Average salary of all employees is `50.50` Rs. |
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| 85. |
`(21x)/20+(11y)/10-x-y` |
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Answer» `21/20x+11/10y-x-y` `21/x-x+11/10y-y` `(21x-20x)/20+(11y-10y)/10` `x/20+y/10` `(x+2y)/20`. |
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| 86. |
Identify the following types of surds : `(a) sqrt(6)+5sqrt(3)" "(b)sqrt(15)+sqrt(8)-sqrt(11) " "(c)sqrt(5)" "(d) 5+sqrt(7)` |
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Answer» `(a) sqrt(6)+5sqrt(3)` . It is the sum of two surds. `therefore ` It is the compound surd of two surds, i.e., a binomial surd. (b) `sqrt(15)+sqrt(8)-sqrt(11)` It is the combination of three surds. `therefore ` It is a compound surd. (c) `sqrt(5)`. It is a monomial surd or a simple surd. `(d) 5+sqrt(7)`. It is the sum of a rational number and a surd. ` therefore ` It si a mixed surd. |
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| 87. |
Find two irrational numbers lying between 0.1 and 0.12 |
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Answer» two irrational number between 0.1 and 0.2 are 1)0.10100100010000 2)0.110100100010000. |
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| 88. |
Find the mode of the following data 7, 9 12, 13, 7, 12, 15, 7 ,12, 7, 25, 18, 7 |
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Answer» In the given data, `7` has maximum frequency as it has appeared `5` times in the given data. So, mode of the given data is `7`. |
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| 89. |
A shopkeeper cheats to the extent of 10% while buying as well as selling by using false weights.His total gain is: |
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Answer» Let shopkeeper buy 100 units with normal weights. Let price of each unit is `1` rupee. As he is cheating to the extent of `10%` while buying, he will buy `= 100**(100+10)/100 = 110` units So, in `100` rupees, he will buy `110` units. Now, when he is selling these `110` units, again he is getting `10%` gain. So, selling price in rupees ` = 110**(100+10)/100 = 121` So, his total gain in rupees` = 121-100 = 21` Total gain in `% = 21/100**100 = 21%` |
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| 90. |
The radius and height of a cone are in the ratio 3:4. If its volume is 301.44` cm^3` Find the radius of the cone. |
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Answer» Let radius of the cone is `r` and height is `h` in cm. Then, we are given, `r/h = 3/4=> h = (4r)/3->(1)` Volume of a cone ` = 1/3pir^2h = 301.44cm^3` `=>1/3pir^2((4r)/3) = 301.44` `=>(4pir^3)/9 = 301.44` `=>r^3 = (301.44**9)/(4**3.14) = 216` `r = 6cm` So, radius of the cone is `6` cm. |
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| 91. |
If the radius and height of the cone both are increased by 10%, then volume and surface area of cone is increased by: |
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Answer» Let radius of cone is `r` and height is `h`. Then, radius after increase ` = r+10/100 = 1.1r` Similarly, height after increase `= 1.1h` Initial surface area of cone`(S_1) = pir(r+l)= pir^2+pirsqrt(h^2+r^2)` Initial volume of cone `(V_1) = 1/3pir^2h` Surface area after increase `(S_2) = pi(1.1r)^2+pi(1.1r)sqrt((1.1h)^2+(1.1)r^2)` `S_2 = (1.1)^2(pir^2+pisqrt(h^2+r^2)) = 1.21 S_1` `:. %` increase in Surface area `= (1.21S_1-S_1)/S_1**100 = 21%` Volume after increase`(V_2) = 1/3pi(1.1r)^2(1.1)h` `V_2 = (1.1)^3(1/3pir^h) = 1.331V_1` `:. %` increase in Volume `= (1.331V_1-V_1)/V_1**100 = 33.1%` |
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| 92. |
If the breadth of a rectangle is increased by 5 cm, its area increases by `25cm^2`.If increased by 5 cm, its area increases by `20cm^2`,Fins the area of the rectangle (In `cm^2` |
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Answer» Area=`l*b=Acm^2-(1)` `l(b+5)=(A+25)-(2)` `(l+5)b=(A+20)-(3)` subtracting equation 1 from equation 2 5l=25 l=5cm subtracting equation 1 from equation 3 `5b=20cm^2` b=4cm Area=`l*b=4*5=20cm^2` Option 1 is correct |
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| 93. |
Prove ` ( 1 - sin 60^@)/ ( cos 60^@) = ( 1 - tan 30^@)/ ( 1+ tan 30^@)` |
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Answer» `L.H.S. = (1-sin60^@)/cos60^@` `=(1-sqrt3/2)/(1/2) = 2-sqrt3` `R.H.S. = (1-tan30^@)/(1+tan30^@)` `=(sqrt3-1)/(sqrt3+1)**(sqrt3-1)/(sqrt(3-1)` `=(4-2sqrt3)/2 = 2-sqrt3` `:. L.H.S. = R.H.S.` |
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| 94. |
The surface area of a sphere is 346.5 `cm^2`, calculate its radius and the volume. |
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Answer» Surface area of sphere=`4pir^2` `4pir^2=346.5` `4*3.14*r^2=346.5` `r^2=346.5/(4*3.14)` `r=5.2 cm` Volume of sphere=`4/3pir^3` `=4/3*22/7*(5.2)^2` `=589.21 cm^3`. |
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| 95. |
Find the mean of the daily wages of 60 workers in a factory as per given below |
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Answer» Mean=`(sumx_i f_i)/(sumf_i)` Mean=7110/60=118.5 Rs. |
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| 96. |
Evaluate: ` 3 sin^2 30^@ + 2 tan^2 60^@ - 5 cos^2 45^@` |
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Answer» `3sin^2 30^o +2tan^2 60^o-5cos^2 45^o` `3(1/2)^2+2(sqrt3)^2-5(1/sqrt2)^2` `3*1/4+2*3-5*1/2` `17/4`. |
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| 97. |
If the external diameter of a pipe is 25 cm and the thickness is 1 cm , find the total surface area. |
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Answer» Total surface area=CSA of outside cylinder+ CSA of inside cylinder+2*area of ring. `=2piRh+2pirh+2pi(r^2-r^2)` `=2pi(25/2h+23/2h+(25/2+23/2)(25/2-23/2))` `TSA=48pi(h+1)`. |
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| 98. |
State whether the following statements are ture or false ? Justify your answer. `(i) (sqrt2)/3` is a rational number. (ii) there are infinitely many integers between any two intergers. (iii) Number of rational numbers between 15 and 18 is finite. (iv) there are numbers which cannot be written in the form `p/q q ne 0,` p and q both are intergers. (v) the square of an irrational number is always rational. (vi) `(sqrt12)/sqrt(3)` is not a rational as `sqrt12 and sqrt3` are not integers. (vii) `sqrt (15)/sqrt3` is written in the form `p/q , q ne 0 ` and so it is a rational number. |
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Answer» (i) False , here `sqrt (2)` is an irrational number and 3 is a rational number, we know that when we divide irrational number by non - zero rational number it will always give an irrational number . (ii) False, because between two consecutive integers (like 1 and 2) there does not exist any other interger. (iii) False, becouse there any two rational numbers there exist infinitely many numbers. (iv) true, because there are infinitely many numbers which cannot be written in the form `p/q q ne 0.` p,q both are integers and these numbers are called irrational numbers. (v) False, e.g., Let an irrational number be `sqrt2 and root4(2)` (a) ` (sqrt(2))^(2)=2` which is a rational number. ` root4(2)^(2)=sqrt2` , which is not a rational number. Hence, square of an irrational number is not always a rational number. (vi) False, `sqrt12/sqrt3=sqrt(4xx3)/sqrt3=(sqrt4xxsqrt3)/sqrt3=2xx1 = 2` which is a rational number. `sqrt15/sqrt3=sqrt(5xx3)/sqrt3=(sqrt5xxsqrt3)/sqrt5= sqrt5` which is an irrational number. |
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| 99. |
How many prime numbers are there between 110 and 120? |
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Answer» 111,113,115,117,119 there are total of 5 numbers. |
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| 100. |
If a, b, c are non-zero real numbers and `a +b+ c= 0`, then the value of `a^2/(bc)+b^2/(ca)+c^2/(ab)` is (A) 0 (B) 2 (D) 4 (C) 3 |
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Answer» `a+b+c=0` `a^2/(bc)+b^2/(ca)+c^2/(ab)` `(a^3+b^3+c^3)/(abc)` `((a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc)/(abc)` `(3abc)/(abc)` `3`. |
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