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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.
| 201. |
Find the largest number which divides 438 and 606, leaving remainder 6 in each case. |
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Answer» Correct Answer - 24 Required number = HCF ( 432, 600) = 24 |
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| 202. |
Prove that: `(a-b)^2=a^2-2ab+b^2` |
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Answer» `L.H.S. = (a-b)^2 = (a-b)(a-b)` `=a*a-a*b-b*a-b*b` `=a^2-2ab+b^2` `=R.H.S` |
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| 203. |
Find the largest numberwhich divides 245 and 1029 leaving remainder 5 in each case.A. 15B. 16C. 9D. 5 |
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Answer» Correct Answer - B Required number = HCF {(245 -5) ,( 1029 -5)}= HCF ( 240, 1024) = 16 |
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| 204. |
Find the digit in the units place of `(676)^99`.A. 9B. 2C. 4D. 6 |
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Answer» Correct Answer - D Units digit of `(ab6)^n is (n in z^+)6` |
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| 205. |
The units digit of the number `9^26` is _______ |
| Answer» Correct Answer - 1 | |
| 206. |
The product of two numbers is 1600 and their HCF is 5. Find the lcm of the numbers |
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Answer» We know, Product of two numbers = Product of their `LCM and HCF` `:. 1600 = 5**LCM` `=>LCM = 160/5 = 320` So, `LCM` of the given numbers will be `320`. |
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| 207. |
What is the digit in the tens in the product of the first 35 even natural numbers ?A. 6B. 2C. 0D. 5 |
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Answer» Correct Answer - C Check how many zeroes are involved in the product. |
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| 208. |
The simplest from of ` 1095/1168` usA. `17/26`B. `25/26`C. `13/16`D. `15/16` |
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Answer» Correct Answer - D HCF (1095,1168) = 73 `1095/1168 = (1095 div 73)/(1168 div73) = 15/16` |
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| 209. |
In the given figure, PQ is tangent at a point R of the circle with centre O. If , find `m anglePRS`. |
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Answer» As `ST` is passing through center of the circle, `:. ST` is the diameter of the circle. `PQ` is the tangent at point `R` of the circle. So, `/_TRS = 90^@` Also, as `PRQ` is a straight line, `/_PRS+/_SRT+/_QRT = 180^@` `/_PRS+90^@+30^@ = 180^@` `:. /_PRS = 60^@` |
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| 210. |
If `2^(2x+y)=4^(x-y-3)=1`, then `(x, y)` is |
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Answer» `2^(2x+y)=2^(2x-2y-6)=1` `2x+y=2x-2y-6` `3y=-6` `y=-2` `2^(2x-2y-6)=1=2^0` `2x-2y-6=0` `2x+4-6=0` `x=1` `(x,y)=(1,-2)`. |
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| 211. |
Property 7 (DIvision Algorithm) If a whole number a is divided by a non-zero whole number b then there exists whole numbers q and r such that `a = bq + r` whole either `r = 0 or r < b`.A. 1lt rlt bB. `0 lt r lt b`C. `0 ler le b`D. `0 lt r lt b` |
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Answer» Correct Answer - C On dividing a by b, let q be the quotient and r be the remiander. Then ,we have a= bq +r where ` 0 le r lt b` |
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| 212. |
In the given fig, AB is the diameter of the circle with centre O. The point C, D and E are on the circle such that AC = CD = DE = EB. Find `angleEBA`. |
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Answer» In the given figure, we can join `OC,OD and OE.` Please refer to video to see the diagram. As, `AC = CD = DE = EB` `:. /_COA=/_DOC=/_EOD=/_BOE` Let `/_COA=/_DOC=/_EOD=/_BOE = theta` Then, `theta+theta+theta+theta = 180^@` `=>theta = 45^@` Now, in `Delta EOB`, `OB = OE = ` Radius of the circle `:. /_OEB = /_OBE` Also,`/_OEB+/_OBE + /_BOE = 180^@` `=>/_OBE+/_OBE +45^@ = 180^@` `=>2/_OBE = 135^@` `=>/_OBE = 67.5^@` `:. /_EBA = /_OBE = 67.5^@` |
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| 213. |
A passenger train takes 3 hours less for a journey of 360 km if its speed is increased by 20km/hr from its usual speed. We need to find its usual speed. Represent the problem situation in the form of a quadratic equation. |
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Answer» Let usual speed of train is `x` km/h. Let the time taken by the train to cover `360` km distance with normal speed is `t` hour. Then, `360/x = t->(1)` When speed is increased by `20` km/h, then, `360/(x+20) = t-3->(2)` Subtracting (1) from (2), `360/(x+20) - 360/x = -3` `=>360x -360x-7200 = -3x(x+20)` `=>3x^2+60x-7200 = 0` `=>x^2+20x-2400 = 0`, which is the required quadratic equation. Now, we will solve it to find the speed of the train. `=> x^2+20x-2400 = 0 ` `=>x^2+60x-40x-2400 = 0` `=>(x+60)(x-40) = 0` `=> x = -60 and x = 40` But, `x` can not be negative, so, `x = 40` km/h. `:.` Usual speed of the train is `40` km/h. |
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| 214. |
The inner diameter ofa cylindrical container is 7 cm and its top is of the shape of a hemisphere, as shown in the figure. If the height of the container is 16 cm, find the actual capacity of the container. |
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Answer» Height of cylinder= `16-7/2=25/2cm`Radius=`7/2cm`Capacity of container = `pir^2h+ 2/3pir^3` = `pir^2(h+2r/3)=22/7*(7/2)^2*(25/2+2/3*(7/2))cm^3`= `6853/12= 571.08cm^3` |
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| 215. |
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they all change simultaneously at 8 a.m . Then at what time will they again change simultanously ? |
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Answer» Correct Answer - `8:7:21` hrs Interval of change= LCM ( 48 seconds, 72 seconds, 108 seconds) = 432 seconds = 7 min 12 seconds. Required time of simultaneous change = 8:7:12 hours. |
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| 216. |
Find the area of the shaded region, if the diameter of the circle with centre O is 28 cm and `AQ=1/4AB` [Use `pi=22/7`] |
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Answer» `AQ= 7cm``BQ= AB-AQ= 21cm`Area of shaded region=`pi/2(7/2)^2+ pi/2(21/2)^2= pi/2(49/4+441/4)` =`22/(7*2)*490/4= 192.5cm^2` |
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| 217. |
Find the lengths of the arcs cut off from a circle of radius `12cm` by a chord `12cm` long. Also find the area of the minor segment. [use `pi=3.14`, `sqrt(3)=1.732`] |
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Answer» Area of arc=`1/2r^2a` `=1/2r^2pi/3` Area of arc=`sqrt3/4a^2=sqrt3/4r^2` Area of minar segment=`1/2*12^2pi/3=sqrt3/4*12^2` `=13.008cm^2`. |
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| 218. |
Prove that `tan^(-1)(x/y)- tan^(-1)((x-y)/(x+y)) ` is ` (pi)/4` and Not `(-3pi)/4` . |
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Answer» `tan^(-1)(x/y)-tan^(-1)((x-y)/(x+y))` `tan^(-1)((x/y)-((x-y)/(x+y)))/((x/y)*((x-y)/(x+y))` `tan^(-1)((x^2+xy-xy+y^2)/(xy+y^2+x^2-xy))` `tan^(-1)((x^2+y^2)/(x^2+y^2))` `tan^(-1)(1)` `pi/4`. |
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| 219. |
If a=` ( 2^(2) xx 3^(3) xx 5^(4)) and b = (2^(3) xx 3^(2) xx 5) ` then HCF (a ,b) ?A. 90B. 180C. 360D. 540 |
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Answer» Correct Answer - B HCF (a,b) = product of common terms with lowest power , `(2^(2) xx 3^(2) xx5) = (4xx 9xx 5) =108` |
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| 220. |
HCF of ` (2^(3)xx 3xx 5) , (2^(2) xx 3^(3) xx5^(2)) and (2^(4) xx3xx 5^(3) xx7)` isA. 30B. 48C. 60D. 105 |
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Answer» Correct Answer - C HCF = product of common terms with lowest power ` = (2^(2) xx 3xx 5) = ( 4xx 3 xx 5) = 60 ` |
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| 221. |
If `8 cos^2 theta+8 sec^2 theta=65` and `0^@` < `theta` |
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Answer» `8cos^theta+8sec^2theta=65` `Let cos^2theta=t` `8t+8/t=65` `8t^2-65t+8=0` `8t(t-8)-1(t-8)=0` `(8t-1)(t-8)` `t=1/8,8` `cos^2theta=1/8` `costheta=1/(2sqrt2)` `4cos2theta=4(2cos^2theta-1)` `=4(2*1/8-1)` `=-3`. |
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| 222. |
Find the number of bricks, each measuring 25 cm, 12.5 cm and 7.5 cm required to construct a wall 6 m long , 5 m high and 50 cm thick, while the cement and sand mixture occupies `1/20 th` of the volume of the wall. |
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Answer» Volume of the wall `(V_w) = l**b**h = 6*5*50/100 = 15m^3` Volume of the cement and the sand mixture `(V_m) = 1/20*15 = 3/4m^3` So, Volume of the bricks `(V_b) = V_w - V_m = 15-3/4 = 57/4m^3` Now, volume of a single brick `(V_s) = 25/100**12.5/100**7.5/100 = 3/1280m^3` So, total number of bricks required `= V_b/V_s = (57/4)/(3/1280) = (57**1280)/(4**3) = 19**320 = 6080` So, `6080` bricks will be required to make a wall. |
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| 223. |
If the number `8764xx5` is divisble by 9, then find the least possible value of x, where x is a two -digit numbers. |
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Answer» The least possible two digit number of x is 15. `therefore` The number is `8764155` which is divisible by 9 . `therefore 8+7+6+4+1+5+5=36` `rArr` divisible by 9) . |
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| 224. |
Show that and only one out of n, n+2 , n+4, is divisble by, 3 , where n is any postivie integer. |
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Answer» On dividing n by 3, let q be the quotient and r be the remainder. Then , n+ 3q +r, where ` 0 le r gt 3` ` Rightarrow n = 3q +r ," where" r= 0,1,2` ` Rightarrow n = 3q or n = 3q +1 or n = 3q +2` Case I if n = 3q then n is divisible by3. Case II If n = 3q +1 then ( n +2) =3q +3 = 3( q +1) , which is divisible by 3. So , in the case, (n +2) is divisible by 3. Case III when n = 3q +2 then (n+4) = 3q + 6 =3 ( q+2), which is divisible by 3. Hence, one and only one out of n , n+2, n+4 is divisible by 3. |
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| 225. |
The angle of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complementary.Prove that height of the tower is` sqrt(ab)`. |
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Answer» Let height of tower is h. Let angle at distance a is `alpha`. Then, from distance b, it will be `90-alpha`. Then, `tanalpha = h/a` (eq 1) `tan(90-alpha) = h/b ``cotalpha = h/b` `tanalpha = b/h` (eq 2) From equations1 and 2, `h/a = b/h` `h^2 = ab` `h=sqrt(ab)` |
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| 226. |
Check whether `6^n`can end with the digit `0` for any natural number `n`. |
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Answer» We can write `6^n` as `(2**3)^n` or `2^n3^n`. Now, for the end digit to be `0`, number should be divided by 10 or it should be divided by 2 and 5. In our case, it can be divided by `2` and `3`, but, not by `5`. So, given number can not end with 0. |
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| 227. |
Given that HCF (`306, 657`) `= 9`, find LCM (`306, 657`). |
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Answer» We know, `LCM(a,b)**HCF(a,b) = a**b` Here, `LCM**9 = 306**657` `LCM = (306**657)/9 = 306**73 = 22338 ` |
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| 228. |
What is the smallest number that, when divided by 35,56 and 91 leaves remainder of 7. |
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Answer» LCM of 35,56 and 91 is 3640 we have to find multiples of 3640 but since smallest number is required we will take 3640 itself. required number=3640+7=3647. |
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| 229. |
( 1) `4z + 3 = 6 + 2z`(2) `(2x)/3 +1 = (7x)/15 + 3` |
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Answer» `1) 4z+3=6+2z` `4z-2z=6-3` `2z=3` `z=3/2` `2) 2/3x+1=7/15x+3` `2/3x-7/15x=3-1` `(10x-7x)/15=2` `3x/15=2` `x=10`. |
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| 230. |
Find the least number that will divide 43, 91 and 183 so as to leave the same remainder 4erin each case. |
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Answer» Correct Answer - 4 Required number = HCF of ( 91 -43) , (183-91) and ( 183-43) i.e, HCF of 48,92 and 140. |
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| 231. |
Three athletes A, B and C run a race, B finished `24` meters ahead of C and `36` meters ahead of A, while C finished `16` meters ahead of A. If each athlete runs the entire distance at their respective constant speeds, what is the length of the race ? |
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Answer» Length of the race be `D_m`,`t_b` units `(V_b-V_c)t_b=24m-(1)` `(V_c-V_a)t_b=12m-(2)` `(V_c-V_a)(t_b+24/V_c)=16m-(3)` `D=V_bt_b` `V_bt_b-V_ct_b=24` `D-V_ct_b=24-(4)` `V_a/V_c=5/6` `(V_b-V_a)t_b=36` `V_bt_b-V_at_b=36-(5)` `D-24=V_ct_b-(6)` `D-36=V_at_b`-(7) `(D-24)/(D-26)=V_c/C_a=6/5` `(D-24)5=(D-36)6` D=96m. |
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| 232. |
Find the HCF and LCM of `(4)/(5),(2)/(5)and (3)/(4)` |
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Answer» `LCM ((4)/(5),(2)/(5),(3)/(4))=(LCM(4,2,3))/(HCF(5,5,4))=(4xx3)/(1)=12` `HCF ((4)/(5),(2)/(5),(3)/(4))=(HCF(4,2,3))/(LCM(5,5,4))=(1)/(5xx4)=(1)/(20)` |
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| 233. |
Find the LCM of 144 and 156. |
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Answer» Using division , we have, `therefore LCM =2xx2xx3xx12xx13=1872`. |
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| 234. |
Three bells toll at intervals of 12 min, 24 min and 9 min respectively. If they toll togather at `11.00` a.m. then find the time at whcih they toll togather again for the first time. The Following are the steps involved in solving the above problem. Arrange them in the sequential order.(A) We know that the three bells toll togather at the multiples of LCM of 12 min, 24 min and 9 min.(B) After 11 a.m. they toll togather at (11 a.m.+ 72 min ) i.e., 12.12 p.m.(C) THe LCM of 12, 24 and 9 is 72.(D) Therefore, all the three bells toll togather for every 72 min.A. ABCDB. ACDBC. DBACD. DABC |
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Answer» Correct Answer - B acdb is the requried sequential order. |
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| 235. |
Five bells begins to toll together and toll respectively at intervals 6,7,8,9 and 12 seconds. How many times will they toll together in one hour, excluding the one at the start? |
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Answer» LCM of 6,7,8,9 and 12=504 So, 504 is the smallest number which is divided by 6,7,8,9,12 completely. Hence, all the bells will ring together after each 504 seconds. So, in 1hr (`60xx60` seconds) they will ring together` =(3600)/(504)=7.14` i.e. 7 times. |
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| 236. |
Three bells ring at intervals of `4, 7 and 14` minutes. All three range at `6` am. When will they ring together again ? |
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Answer» We have to find a number (in time ) which is divided by all the 3 numbers completely. It means `to LCM ` of 3 number. `4=2xx2` `7=7` `14=2xx7` `:. LCM=2xx2xx7=28` So, 28 is the smallest number which is divided by 4,7 and 14 completely. Hence, all the three bells will ring together at 6:28 am. |
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| 237. |
Find the greatest number that divides 59 and 54 leaving remainders 3 and 5 respecitvely.A. 3B. 7C. 8D. 5 |
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Answer» Correct Answer - B Use GCD concept. |
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| 238. |
Data: 10, 15, 20, 15, 25, x; mode=median, find x, mean |
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Answer» mean=`(10+15+20+15+25+15)/6` mean=`100/6=16.66. |
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| 239. |
Tangents PA and PB are drawn to a circle with point P outside the circle, line joining P and centre intersect chord AB at M then, Prove that `/_OPA = /_OAM` |
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Answer» `/_PAM` and `/_PBM` PA=PB PM=PM `/_APM=/_BPM` `/_PAM cong /_PBM` `/_OAM=x` `/_PAO=90` `/_PAM=90-/_OAM` `/_APM=/_OAM` `/_APM=/_APO=/_OPA` `/_OPA=/_OAM`. |
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| 240. |
A path separates two walls. A ladder leaning against one wall rests at a point on the path. It reaches 90m on the wall and makes an angle of `60^@` degree with the ground. If while resting at the same point on the path, it were made to lean against the other wall it would have made an angle of `45^@` with the ground. Find the height it would have reached on the second wall. |
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Answer» We can draw a diagram with the given details. Please refer to video to see the diagram. From the diagram, In `Delta ABC`, `(AB)/(BC) = sin60^@ =>AB = BC sin60^@` `=>90 = x(sqrt3/2) =>x = 180/sqrt3 = 60sqrt3m` Now, in `Delta CDE`, `=>(PE)/(CE) = sin45^@ =>PE = CEsin45^@` `=>PE = x/sqrt2 = (60sqrt3)/(sqrt2) = 30sqrt6m = 73.47 m` So, the required height is `73.47m.` |
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| 241. |
Find the LCM of 1296 and 1728 using synthetic division method. |
| Answer» Correct Answer - 5184 | |
| 242. |
(a) Find the HCF of `(2)/(5),(12)/(5) and (3)/(4)` |
| Answer» Correct Answer - `(1)/(60)` | |
| 243. |
Reduce `(6912)/(6561)` to its lowest terms. |
| Answer» Correct Answer - `(256)/(243)` | |
| 244. |
An electronic device makes a beep after every `60` sec. Another device makes abeep after every `62` sec. They beeped together at `10` a.m. The next time, whenthey would beep together at the earliest is |
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Answer» Minimum time interval of beeping together = L.C.M. of 60 sec and 62 sec ltb rgt Now , `" " 60 = 2 xx 2 xx 3 xx 5 , 62 = 2 xx 31` L.C.M . = `2 xx 2 xx 3 xx 5 xx 31 = 1860` seconds `= (1860)/(60)` min = 31 min So , they will beep together again at `5 : 31` am . |
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| 245. |
A sweet seller has `420`Kaju burfis and `130` Badam burfis she wants to stack them in such a way thateach stack has the same number, and they take up the least area of the tray.What is the number of burfis that can be placed in each stack for thispurpose? |
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Answer» Maximum number of burfis in each stack = H.C.F. of 420 and 150 `420 = 2 xx 2 xx 3 xx 5 xx 7 , 150 = 2xx 3 xx 5 xx 5` `therefore " "` H.C.F. = ` 2 xx 3 xx 5 = 30` `therefore` Maximum number of burfis in each stack = 30 `therefore` Also , number of stacks = `(420)/(30) + (150)/(30) = 14 + 5 = 19` |
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| 246. |
`sqrt((1+cos30)/(1-cos30))=sec60+tan60` |
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Answer» `(1+cos30)/(1-cos30)=(1+cos30)^2/(1-cos^30)` `=(1+cos30)^2/sin^30` LHS=`sqrt((1+cos30)^2/sin^230)=(1+cos30)/sin30` `=cosec30+cot30` `=sec(90-30)+tan(90-30)` `=sec60+tan60=RHS`. |
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| 247. |
Show that any number of the form ` 6^(n)` , where ` n ne N` can never and with the digit 0. |
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Answer» If `6^(n)` end with 0 then it must have 5 as a factor. Butm, ` 6^(n)= ( 2 xx 3)^(n) = ( 2^(n) xx 3^(n) )` shows that 2 and 4 are the only prime factors of ` 6^(n)`. Also, we know form the fundamental theorem of arithmetic that the prime factorisation of each number is unique. SO, is not a factor of ` 6 ^(n)` Hence,` 6^(n)` can not a factor of ` 6^(n)` . Hence,` 6^(n)` can never end with the digit 0. |
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| 248. |
If the rational number ` a/b` has a terminating decimal expansion, what is the condition to be satisfied by b ? |
| Answer» Correct Answer - b = `( 2^(m) xx 5^(m))` , where m and n are some non-negative integers | |
| 249. |
which of the following has a terminating decimal expansion ?A. `32/91`B. `19/80`C. `23/45`D. `25/42` |
| Answer» Correct Answer - B | |
| 250. |
Without actual division, show that each of the following rational numbers is a terminating decimal. Express each in decimal form. ` 23/((2^(3) xx 5^(2))` (ii) ` 24/125` (iii) ` 171/800` (iv) `15/1600` (v) ` 17/320` (vi)` 19/3125` |
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Answer» Correct Answer - (i) 0.115 (ii) 0.192 (iii) 0.21375 (iv) 0.009375 (v) 0.053125 (vi) 0.00608 `(i)23/((2^(3)xx5^(2)))=(23xx5)/((2^(3)xx5^(3)))=115/((10)^(3))= 115/1000= 0.115` `(ii) 24/124 = 24/(5^(3))xx 2^(3)/2^(3) = (24xx8)/((5xx2)^(3))= 192/((10)^(3))= 192/1000= 0.192` (iii) ` 171/800 = 171/16 xx1/100 = 21.375/100 = 0.21375` `(iv) 15/1600 = 15/16xx1/100 = (0.9375)/100 = 0.009375` `(v) 17/320 = ( 17xx5)/(320xx5) = 85/16 xx1/100 = (5.3125)/100 = 0.053125` `(vi) 19/3125 = (19xx8)/(3125xx8) = 152/25000 = 152/25 xx 1/1000 = 6.08/1000 = 0.00608` |
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