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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.
| 1201. |
Example 17 If "C, = "C,, find"Solution We have "C, = "C,n! n!i.e.9!(n -9)! (n-8)! 8orn-8= 9orn= 17Therefore "C, = "C,7 = 1 |
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| 1202. |
\left(\frac{4}{3} m-\frac{3}{4} n\right)^{2}+2 m n=\frac{16}{9} m^{2}+\frac{9}{16} n^{2} |
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Answer» (4m/3 - 3n/4) ² + 2mn16m²/9 + 9n²/16 - 2mn + 2mn16m²/9 + 9n²/16 |
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| 1203. |
18000 (-1D) of these9. The mean of the following data 1222, 32........ n?is -(D)39. FF(n +1) (2n+1)n(n-1)(2n +1)(AM6(B)(A)6on(n + 1/21 - n(n â 1/20 - 1)(C)40.9. A fair die is thrown once. The probability of getting aless than 5 is10 |
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Answer» sum of squares of first n natural numbers = n(n + 1)(2n + 1)/6mean = (n + 1)(2n + 1)/6 |
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| 1204. |
9.If, for a positive integer n, the quadratic equation,X(X ++x+ n-1) (+n)- 10nhas two consecutive integral solutions, then n isequal to:(1) 115 (2) 12 (3) 9x + n) - 10nJEE (MAIN)-2017](4) 10 |
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| 1205. |
Given n(U) 20, /n(A)- 12, n(B)-9,subsets of U, then n[(A U B)]-nAnB)4, where U is the universal set, A and B area.b. 9d. 3e. 16(Kerala EEE 2004) |
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Answer» but the answer given is 3 |
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| 1206. |
\frac { ( 243 ) ^ { \frac { n } { 5 } } \cdot 3 ^ { 2 n + 1 } } { 9 ^ { n } \cdot 3 ^ { n - 1 } } |
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| 1207. |
\frac{5^{n+3}-6 \times 5^{n+1}}{9 \times 5^{n}-2^{2} \times 5^{n}} |
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Answer» Thank you |
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| 1208. |
Q4.The product of two consecutive odd numbers is 483. Find the numbers |
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Answer» answer is correct or not |
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| 1209. |
The sum of the square of two consecutive odd numbers is 394. Pind the numbers |
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| 1210. |
WCIUCU IS 20!The sum of two consecutive odd numbers is 56. Find the numbers.Ti01 |
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| 1211. |
\sqrt { 3 } \cosec 20 ^ { \circ } - \sec 20 ^ { \circ } = 4 |
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| 1212. |
The population of town is 8000. It increases by 10% during first year and by 20% during secyear. Find the population after 2 years.6 |
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Answer» population is 8000increases 10% during first year so it will be 8000*1.1=8800increases 20% during second year it will be 8800*1.2=10560 |
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| 1213. |
A car covers a certain distance in 40 minutes, if it runs at 54 km/hr . how much time will it take to cover the same distance at a speed 45 km/hr |
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| 1214. |
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum( find the population in 2001. |
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Answer» less than 2003 in two years. A(2003)= P(2001)(1+R/100)^n 54000 = P (2001)(1+5/100)² 54000 = P(2001)(1+1/20)² 54000= P(2001)(21/20)² 54000 = P(2001)((21×21)/(20×20)) P(2001)=( 54000×20×20)/21×21 Population in 2001 = 48979.59= 48980(approx) ER. RAVI KUMAR ROY (ii)ATQ, population is increasing. Therefore population in 2005, A(2005)= P(1+R/100)^n = 54000(1+5/100)² = 54000(1+1/20)² = 54000( 21/20)² = 54000(21/20)× (21/20) =( 54000× 441)/ 400 =( 540×441)/4 = 238140/4 = 59535 Population in 2005=59535 let population in 2001 =x54000=x[1+(5/100)]^254000=x[1+(1/20)]^254000=x(21/20)^254000×(20/21)^2=xx=54000×20/21×20/21x=48980 (approx)so, population in 2001 is 48980. answer 48980 the correct answer of the given question |
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| 1215. |
ID:66 A train is moving at the speed of 54 km/hr. How many seconds it will take to cover a distance of 450 metres?ชาptions: |
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Answer» Speed of train = 54km/hr= 54*1000/(3600)= 54*5/18= 15 m/s Time taken to cover 450 s= 450/15= 30s Ans: The train covers the distance of 450 m in 30 seconds. |
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| 1216. |
Two buses cover a distance between two stations A and B. First bus moves at an average speedof 54 km/h and the second bus at an average speed of 60 km/h. If the first bus takes 10 hours tocomplete the journey, how long will the second bus take? |
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| 1217. |
MATII NOTESHICS Coaching CentreState Bank Wali Gali, Railway Road, Sonepat Ph.No.9812009696Two trains of 300 M long and 200 M long is moving on parllel lines at a speedkm/h.and 40 km/h. if both the trains is going in same direction then find theto cross each other.दो रेलगाडियो 300 मी. लम्बी तथा 200 मी. लम्बी समांतर पटरियों पर क्रमशः 60 km/40 km/h की चाल से चल रही है। यदि ये दोनो एक ही दिशा में जा रही हो तो एटको पार करने का समय ज्ञात कीजिए।(a)80 sec (b) 60 sec(c) 85 sec (d) 90 sec |
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Answer» the awnser is 85 sec The answer is 85sec. the correct answer is 85 sec the answer is 85............okay |
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| 1218. |
10. A train 150 m long is running at 40 km/h. In what time will it pass a man running at 4 km/h inthe same direction in which the train is going?(1) 16 sec(2) 12 sec(3) 15 sec(4) 18 sec |
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| 1219. |
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum(1) find the population in 2001.(ii) what would be its population in 2005? |
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Answer» option I is the correct answer 1) 2003 =5%2002= 5% =10%so population in 2001 54000× 10% = 5400=54000 - 5400 = 48,6002)2004 = 5%2005 = 5% = 10%54,000 × 10% = 5,400so population in 2005 =54,000 + 5,400 = 59,400 the correct answer is 59400 people in 2005 (1) let the population in 2001 = X A/Q,X (1+5/100)^2 = 54000X(21/20)^2 = 54000X = (54000 x 400)/441X = 48979.59 = 48980 (approx) ANS.. (2)population in 2005 = 54000 ( 1 + 5 /100)^2 = (54000 X 441)/400 = 59535. (ANS) the population in 2001is 52001 and the population in 2005 is 56005 59; 535 is right answer 5400 and 540000 it is answer The answer is 59400 people 1-2003=5% 2002=5%=10% by this population in 2001 =54,000 - 5,400 = 48,6002-2004 = 5% 2005=10% 54,000×10%=5,400 by this population of 2005=54,000+5,400=59,400this is right answer let the no be x the population in 2001 2003=5/:2002 =5 ==10 49600 in year 200159400 in year 2005 59400 people in 2005 3 ANSWERhusuevvsgzhzhdbdbdjdjdjjd The population in 2001 is 52001and the population in 2005 is 56005 the population in 2001 is 52001 and the population in 2005 is 56005 1)2003=5%2002=5%=10%so population in 200154000×10%=5400=54000-5400=48,6002)2004=5% 2005=5%=10%54,000×10%=5400so population in 2005=54,000+5,400=59,400 1 answer is 48600 and 2 answer is 59400 1)486002)59400 are the best answers The 15th Indian Census was conducted in two phases, house listing and .... The population of India as per 2011 census was 1,210,193,422. India added 181.5 ..... See also[edit]. flag... Total population:1,210,193,422 Most populous state:Uttar Pradesh(199,812,341) Least populous state:Sikkim(610,577) Date taken:2010–2011 2003=5%2002=5%=10%so population in 200154000×10%=5400=54000-5400=48,6002)2004=5%2005=5%=10%54000×10%=5,400so population in 2005=54,000+5,400=59,400.plz mark me best answer and like my answer. 1)2003=5%2002=5%=10%so population in 200154000×10%=5400there for 54000-5400 =48,600 2)2004=5%2002=5%=10%54,000×10%=5,400so the population in 2005=54,000+5,400=59,400 is the answer 2003=5%2002=5%=10%so population in 200154000 × 10%=5400=48,600) I I) a = p - I I = 54000*2*5/100 = 5400a = 54000-5400 = 48600ii) a = p + I I = 54000*2*5/100= 5400a = 54000 + 5400= 59400 1 Solution:- 2003=5%2002=5%=10%So Population In 200154000×10%=5400=54000-5400=48600 2 Solution:- 2004=5%2005=5%=10%54000×10%=5400So Population In 2005=54000+5400=59400 is the best answer 1) 2003 = 5℅. 2002= 5℅. so population in 2001. 54000×10℅=5400. =54000-5400=48,600. 2).2004 =5℅. 2005=5℅=10℅. so population in 2005. =54000+5,400= 59,400 first answer is48600 and second answer is 59400 1)2003=5 percentage 2002=5 percentage=10 percentage54000 into 10 percentage is equals to 5400is equals to 54000 - 5400 is equals to 4860second problem2004 =25 percentage2005 is equals to 5% is equals to 10 percentage54000 + 5400 is equals to 59400 I also don't Mysore sandal answers 1. 48,600. 2. 59,400 48600 in 2001 population and 5940 in 2005 59400 is the right answer of the following 59400. people in 2005is the best answer ¡ ) 2003 = 5% 2002 = 5% = 10% So" population in 2001 54000 × 10% = 5400 =54000 - 5400 = 48,600 ;¡¡ ) 2004 = 5% 2005 = 5% = 10% 54,000 × 10% = 5,400So" population in 2005 = 54,000 + 5,400 = 59,400 ; 《~Now everything is clear~》 59400 is the best answer 59400 is the best answer 59400. is your answer 59400 people in 2005 59400 is the best a swer a) 2003=5%2002=5%=10%So, population in 200154000×10%=5400=54000-5400=48600b) 2004=5% 2005=5%=10% 54000×10%=5400So, population in 2005=54000+5400=59400 48600 and 59400 is the best answer option 2 is correct answer population in 2003 =54,000increasing rate 5% per annum =(54000×5/100)=540×5=2700 2700 for 1 year (per annum)so for (2003-2001) or (2005-2003)=2 years =2700×2=5400 (i) population in 2001=(54000-5400)=48600 ✔(ii) population in 2005=(54000+5400)=59,400✔ 59400 is right answer 1)population in 2003 = 5%population in 2002= (d) is right Best andwer Given: we use formula of amount of compound interest to find population A( 2003)= 54,000, R = 5%, n= 2 years In 2001 Population would be less than 2003 in two years. A(2003)= P(2001)(1+R/100)^n 54000 = P (2001)(1+5/100)² 54000 = P(2001)(1+1/20)² 54000= P(2001)(21/20)² 54000 = P(2001)((21×21)/(20×20)) P(2001)=( 54000×20×20)/21×21 Population in 2001 = 48979.59= 48980(approx) (ii)ATQ, population is increasing. Therefore population in 2005, A(2005)= P(1+R/100)^n = 54000(1+5/100)² = 54000(1+1/20)² = 54000( 21/20)² = 54000(21/20)× (21/20) =( 54000× 441)/ 400 =( 540×441)/4 = 238140/4 = 59535 Population in 2005=59535 ==========================================================Hope this will help you.... plz like my answer and accept as best 1)48,600="population in 2001"2)59,400="population in 2005" Population in 2001=48,600Population in 2005=59,400 population in 2001 is 48600 in 2005 answer is 59400 2003=5% population increase in 2002 and. 2003 is =10%so in 2001 = 54000 x 10%= 5400and in 2005 =10%=5400=54000 +5400=59400 |
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| 1220. |
6. if 1 kg of ghee contains 0.582 kg fat, how much fat is there in 9.603 ko ghee? |
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Answer» 1kg ghee contains 0.582kg fatso9.603 kg of ghee conatins 9.603× 0.582kg fat = 5.588946 kg |
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| 1221. |
6. If 1 kg of ghee contains 0.582 kg fat, how much fat is there in 9.603 kg of ghee? |
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Answer» 1kg ghee contains 0.582kg fatso9.603 kg of ghee conatins 9.603× 0.582kg fat = 5.588946 kg |
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| 1222. |
(a-JJUJU10. Population in a city wasity was 8,60,000. In the next year, if it is increased by 14.000evious population, the present population in the city isthan one-tenth of the previous population, the present. De77 |
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Answer» 9,60,000 is the correct answer 960000 is right answer of this question. please like my answer |
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| 1223. |
ft械f(Prove, by vector method, that)sin(A -B)sinA cos B-cosA sin B. |
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| 1224. |
10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum(i) find the population in 2001.(ii) what would be its population in 2005? |
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| 1225. |
10. The population of a place increased to 54,000 in 2003at a rate of 5% per annum() find the population in 2001(i) what would be its population in 2005? |
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| 1226. |
\left| \begin{array}{lll}{y+z} & {x} & {y} \\ {z+x} & {z} & {x} \\ {x+y} & {y} & {z}\end{array}\right|=?\begin{array}{l}{\text { (a) }(x+y+z)(x-z)} \\ {\text { (b) }(x+y+z)(y-z)} \\ {\text { (c) }(x+y+z)(x-z)^{2}} \\ {\text { (d) }(x+y+z)(y-z)^{2}}\end{array} |
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| 1227. |
sinA +sin B A+BcosA+cosB ~ 2 |
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Answer» (sinA+sinB)/(cosA+cosB) = (2sin(A+B)/2*cos(A-B)/2)/(2cos(A+B)/2*cos(A-B)/2) = tan(A+B)/2 |
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| 1228. |
Prove that\left| \begin{array}{lll}{x} & {x^{2}} & {x^{3}} \\ {y} & {y^{2}} & {y^{3}} \\ {z} & {z^{2}} & {z^{3}}\end{array}\right|=x y z(x-y)(y-z)(z-x) |
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Answer» hlo ji app ne je answers kaha c lea hai |
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| 1229. |
(z^2 %2B y^2 %2B y*z)/(((x - y)*(x - z))) %2B (x^2 %2B x*z %2B z^2)/(((-x %2B y)*(y - z))) %2B (y^2 %2B x^2 %2B x*y)/(((-x %2B z)*(-y %2B z))) |
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Answer» 0 is the correct answer of the given question. |
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| 1230. |
( x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z ) = ( x + y + z ) ( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - x y - y z - z x ) |
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| 1231. |
- z ^ { 2 } + 13 z ^ { 2 } - 5 z + 7 z ^ { 3 } - 15 z |
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| 1232. |
z _ { 1 } = 2 - i , z _ { 2 } = 1 + i , \text { find } | \frac { z _ { 1 } + z _ { 2 } + 1 } { z _ { 1 } - z _ { 2 } + i } | |
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| 1233. |
z _ { 1 } = 2 + i , z _ { 2 } = 2 - 3 i , z _ { 3 } = 4 + 5 i , \text { evaluate Re } \frac { z _ { 1 } \overline { z } _ { 2 } } { z _ { 3 } } |
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Answer» z1 z2*=(2+i)(2+3i)=4+8i-3=1+8iz1 z2* /z3 =(1+8i)/(4+5i)=(1+8i)(4-5i)/(4+5i)(4-5i)=(44-27i)/41real part=44/41 |
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| 1234. |
\begin{array} { l } { \text { The solution of } } \\ { \left( x ^ { 3 } - 3 x y ^ { 2 } \right) d x = \left( y ^ { 3 } - 3 x ^ { 2 } y \right) d y } \\ { y ^ { 2 } - x ^ { 2 } = c \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } \text { ) } y ^ { 3 } - x ^ { 3 } = c \left( x ^ { 2 } + y ^ { 2 } \right) } \\ { y ^ { 2 } + x ^ { 2 } = c \left( x ^ { 2 } - y ^ { 2 } \right) 4 ) y ^ { 3 } + x ^ { 3 } = c \left( x ^ { 2 } - y ^ { 2 } \right) } \end{array} |
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Answer» Taking - common we gety² - x² = c ( x² + y²)²so option (1) is correct |
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| 1235. |
Find arational e2andernumbers.Alviteenirrational |
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Answer» Irrational number between √2&√3 √(√2√3)=√√6=6¼Irrational number between √2&6¼=√(√2× 6¼)=2¼×6⅛ Hence Irrational numbers between √2&√3 are 6¼ and 2¼×6⅛ |
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| 1236. |
y3. Swathi said that (6x)4 and 6x4 are same. Do you agree with Swathi. Justify youranswer.[Contin..2 |
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Answer» No it's not Same if you knows maths then you will understand by yourself |
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| 1237. |
117 sin2θ + 3 cosa4, show that tan θ3 |
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| 1238. |
Madhu bought a house forăŚ1,31,25,000. If its value depreciates at the rate of10% per annum, what will be its sale price after three years?3. |
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| 1239. |
lible byPractice Time 2.5Find the common factors of the following:(a) 12 and 16(b) 15 and 21(c) 63 and 210(d) 4, 6 and 8(e) 15, 25 and 30(f) 24, 32, 442. Find the first three common multiples of the following:(a) 2 and 3(b) 15 and 203 Write all the numbers less than 100 which are common multiples of 4 and 5.4. Which of the following numbers are co-prime?(a) 4,9 (b) 8,15 (c) 18,81 (d) 40,515 (e) 73,213(6) 399,437-----PRIME FACTORISATIONThe factors of a number which are prime numbers are called prime factors.or erample, the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. You can see that only 2 and 3 are priumbers in all the factors. So, 2 and 3 are called prime factors of 48.milarly, the factors of 50 are 1, 2, 5, 10, 25 and 50.e, 2 and 5 are the only prime numbers. So, 2 and 5 are prime factors of 50.e factorisation is expressing a given number as the product of its prime factors. |
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Answer» 1st find the components of the numbers and the common one will be the common factorsfor example..10 and 3010=5×230=3×5×2so 5×2 are common hence the commo factor |
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| 1240. |
In a showroom, the sale of refrigerators increased from 10.000 to 13.310 in 3 years. Find the rate of increase in sale per annum. |
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| 1241. |
The simple interest on a certain sum of money for two years at 5% is 6,600.The simple interest on a certain sum of money for two years at 5 26 isWhat will be the compound interest on that sum at the same rate for the sametime period? |
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Answer» p=6,600r=11/2=5 .5t=2CI=6,600×5.5×1÷100=363A=6,600+363=6,963CI=6,963×5.5÷100=382.9A=6,963+382.9=7,345.9 |
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| 1242. |
If ZA and ZX are acute angles such that cosA- cosX, then show that |
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| 1243. |
3X2=bCOSA = 4x then show that cosecA + cotA=of a1 + 2x1-2xtreets |
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| 1244. |
In figure, ABCD is a parallelogram and CEfind DFAD, DF1 AB. If AD-20 m, CE-8 m, AB-16 m, |
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Answer» Area of parallelogram = base × height= AD × CE= 20×8=160Therefore, 160 = 16× DFDF = 10 |
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| 1245. |
godown is 50 m long, 40 m broad and 10 m high. Find the cost of whitewashingllsand ceiling at20 per square metre. |
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| 1246. |
TespoctiveIf AD 6cm, DB-9am and AE 8 cm, then find ACIFAD 8 cm. AB 12cm and AE 12 cm, then find CE |
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| 1247. |
10 cm and lls nergiit IU3. Length of the fence of a trapezium shaped field ABCD is 120 m.IfBC 48 m, CD 17 m andAD 40 m, find the area of this field. Sidec AB is perpendicular to the parallel sides AD and BC |
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| 1248. |
the sale receipt of a company during a year was rupees 30587850 .next year increase by rupees 6375490 what was totally sale two years |
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Answer» total of sale of two years is 36,963,340 |
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| 1249. |
How many bicycles were produced during these two years5. The sale receipt of a company during a year was 20956480. Next year, it increased by6709570. What was the total sale receipt of the company during these two years?of a city is 28756304. If the number of males is 16987059, find the |
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| 1250. |
2. sin(n+1)A•sin(n+2)A+cos(n+1)A•cos(n+2)A=cosA. |
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