InterviewSolution
Saved Bookmarks
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.
| 1301. |
The following data show the expenditure of an individual over various itemsems Education food Rent Clothingxpenditure | 1600 | 3200 -400024003200Draw a pie chart |
| Answer» | |
| 1302. |
A rectangular field is 72m by 58m . Keya walks around it at the rate of 3km per hour . How long will she take 3 rounds? |
|
Answer» Perimeter=2(72+58)=2(130)=260m 3 rounds=3*260=780m=.780Km Speed=Distance/time3=.780/timetime=.780/3=.26hours=15.6minutes |
|
| 1303. |
articular wear a company produced 8765435 bicycles. Next year, the number ofbicycles produced was 1378689 more than those produced in the preceding vear.How many bicycles were produced during the second year?How many bicycles were produced during these two years?non durinder was 20056290 Naut |
| Answer» | |
| 1304. |
rectangular field is 72 m by 58 m. Keya walks around it at the rate of 3 km per hourHow longwill she take to make 3 rounds? |
|
Answer» Length of one round (perimeter of rectangle) = 2 (72 + 58) = 2(130) = 260mTotal length of three rounds = 3*260 = 780m = 780/1000 = 0.78kmTime to make three rounds = Distance/Speed = 0.78/3 = 0.26 hours = 15.6mins |
|
| 1305. |
A rectangular field is 72 m by 58 m. Keya walks around it at the rate of 3 km per hour.How long will she take to make 3 rounds? |
|
Answer» Dimensions of a rectangular field= 72cm y 58cmperimeter of the rectangular field= 2( l+b ) =2×( 72+58) = 2×130 = 260cm∴ Distance taken by Amar in taking two rounds= 3×260m = 780mSpeed of Amar= 3km/hr =3×1000/3000 m/sec = 5/6 m/secTotal time taken by Amar to cover the distance= Distance/speed = 780/5/6 = 780×6/5 = 156×6 = 156×6/60 min. = 15.6 min. |
|
| 1306. |
A rectangular field is 72 m by 58 m. Amar walksround it at the rate of 3 km per hour. What timewill he take in taking 2 rounds. |
|
Answer» sorry but ,answer is coming 10 minutes 24 second |
|
| 1307. |
Evaluate the integral f: foz fol (x+y +2z)dz dx.dyि [N+ y + 2)dz. dx. dy का मान ज्ञात कीजिए। |
| Answer» | |
| 1308. |
5. A rectangular field is 72 m by 58 m. Amar walksround it at the rate of 3 km per hour. What timewill he take in taking 2 rounds. |
|
Answer» The perimeter is 2(72+58) = 260m 260(3) = 780 m total distance = 0.78 km km/(km/hr) = hours 0.78/3 = 0.26 hours (0.26 hours)(60 min/hr) = 15.6 minutes (0.6 min)(60 sec/min) = 36 seconds 15 minutes 36 seconds |
|
| 1309. |
A trainuhichSpeed of ta km Per hour takesandl o minides from |
|
Answer» Speed=70 km/hrTime=2.5 hr:. Distance between the two stations=Speed×time=70×2.5=175 km |
|
| 1310. |
- नकाराo) R(-=dz-v |
|
Answer» x-2y=-32x+y=4 Multiplying second equation by 2 and adding to first x-2y=-34x+2y=8 5x=5x=1y=2 |
|
| 1311. |
8. | Ď sin x cos x dz |
|
Answer» mai apna answer match kr rha tha sahi h |
|
| 1312. |
(S + b+ dz) (17 फणा |
|
Answer» (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca(2p+q+5)^2=4p^2+q^2+25+4pq+10q+20p |
|
| 1313. |
-4*z^2 %2B 4*(y*z) %2B 6*x^2 - y^2 |
| Answer» | |
| 1314. |
( x %2B 2 y %2B 4 z ) ^ 2 |
|
Answer» x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz is question me(a+b+c) square formula apply karo |
|
| 1315. |
5 x - y - 5 = 0 , \quad 3 x - y - 3 = 0 |
| Answer» | |
| 1316. |
2 x + \frac { 3 } { y } = 4 , \quad 3 x + \frac { 2 } { y } = 5 |
| Answer» | |
| 1317. |
5 y = 20 , \quad 3 x + 5 y = 15 |
|
Answer» 4x - 5y = 20......(1)3x + 5y = 15.......(2)(1)+(2);7x = 35x=5then y=0 |
|
| 1318. |
3 x + 4 y = 10 \quad 3 \pi / 2 x - 2 y = 2 |
|
Answer» 3x+4y=10and 2x-2y=2multiply equation 2 by 23x+4y=104x-4y=4add them7x=14X=2put in equation 24-2y=22=2yy=1 |
|
| 1319. |
Manav takes 20 minutes to reach hisschool from home, walking at the rateof 4 km per hour. How long will it take |
|
Answer» thank you |
|
| 1320. |
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions doeach wheel make in 10 minutes when the car is travelling at a speed of 66 km per hou |
| Answer» | |
| 1321. |
A train 80 m long is moving at 80 km/h. If it takes 18 seconds to cross a bridge, then find the length of the bridge. |
|
Answer» length will be 320 m |
|
| 1322. |
Z+4Z+xZ+ÂŁ+x |
|
Answer» x + y + z = 9.........(1)x + z = 5...............(2)y + z = 7................(3)Put (3) in (1)x + 7 = 9x = 2put (2) in (1)y = 4then z = 3 |
|
| 1323. |
\frac{4 z-3}{4}-3=\frac{5 z-7}{3}-4 z-1 |
| Answer» | |
| 1324. |
(text*((b*y)*(5*z(z %2B 4))))*z(5*z^2 - 80) |
|
Answer» z(5z^2 - 80) ÷ 5z(z + 4) = 5z(z^2 - 16) / 5z(z + 4) Usinga^2 - b^2 = (a + b)(a - b) = 5z(z + 4)(z - 4) / 5z(z + 4) = (z - 4) |
|
| 1325. |
\left. \begin{array} { l } { 25 x ^ { 2 } + 16 y ^ { 2 } + 4 z ^ { 2 } - 40 x y + 16 y z - 20 x z } \\ { 16 x ^ { 2 } + 4 y ^ { 2 } + 9 z ^ { 2 } - 16 x y - 12 y z + 24 x z } \end{array} \right. |
|
Answer» or sand kar Bhai maths ka |
|
| 1326. |
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of itare observed to be 30° and 600. If the height of the lighthouse is h metres and the line joining19.the ships passes through the foot of the lighthouse, show that the distance between the ships4is--h metres.3 |
| Answer» | |
| 1327. |
Determine the locus of z, z # 21, such that Re\left(\frac{z-4}{z-2 i}\right)=0 |
| Answer» | |
| 1328. |
x + 2 y + 5 z = 4 , \quad 3 x + y + 4 z = 6 \text { and } - x + y + z = - 1 |
| Answer» | |
| 1329. |
(-x %2B z)^3 %2B (x - y)^3 %2B (y - z)^3 |
|
Answer» a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)when a + b + c = 0a³ + b³ + c³ - 3abc = 0=> a³ + b³ + c³ = 3abc Let a = x - yb = y - zc = z - xa + b + c = x - y + y - z + z - x => the above equation turns out to be 3(x - y)(y - z)(z - x)Expand it to find the answer (-3x²y + 3xy² + 3x²z - 3y²z - 3xz² + 3yz²) |
|
| 1330. |
-9*x*y*z %2B z^3 %2B 27*x^3 %2B y^3 |
|
Answer» Using identity a^3 + b^3 + c^3 - 3abc = (a + b + c )(a^2 + b^2 + c^2 - ab - ac - bc) Then,27x^3 + y^3 + z^3 - 9xyz= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - 3xz - yz) |
|
| 1331. |
-15*x*y*z %2B z**3 %2B 125*x**3 %2B y**3 |
|
Answer» yaar question glt h kuuki (a+b+c)³ ke form me nii aa pa rha h |
|
| 1332. |
(-x %2B 3*z)^3 %2B (x - 2*y)^3 %2B (2*y - 3*z)^3 |
| Answer» | |
| 1333. |
paianelogramn6. The following data show the expenditure of an individual over various itemsitemsEducation foodRent |Clothing Others2400Expenditure 1600 320040093200Draw a pie chart |
| Answer» | |
| 1334. |
u. ihe talowing data show the expenditure of an individual over various itemsEducation foodure03200itemsExpenditure400024003200Draw a pie chart7. Constructa guacrieteral ABD inwhichrilateralABCD in which AR |
| Answer» | |
| 1335. |
1. The monthly income of a family is 28800. The monthly expenditure of the family onvarious items is given belowClothing EducationSavingsFood10800Represent the above data by a pie chart.itemRentExpenditurein v800056003600800 |
| Answer» | |
| 1336. |
Two ships leme aportat'the same time. One goes 2A km per hour in thedirection N 45° E and the other travels 32 km per hour in the directionS 75° E. Find the distance between the ships at the end of 3 hoursAMPLE 11つ |
| Answer» | |
| 1337. |
line, then the speed106. A train is 250 m long. If the train takesof the train in km/hr is:rain is 250 m long. If the train takes 50 seconds to cross a tree by the railway line, then th(A) 10(B) 9(C) 5(D) 18एक रेलगाड़ी 250 मीटर लंबी है। यदि वह रेलवे लाई के बगल में स्थित एक वृक्ष को पार करने में 50 मेंलेती है तो उसकी गति कितने किमी/घंटा है?(A) 10(B) 9(C) 5(D) 18=== |
|
Answer» 250/50= 5m/s....convert it to km/h18 km/h c) 5 is the correct answer c) 5 is the correct answer |
|
| 1338. |
The following pie-chart shows the monthly expenditureof Sufia on various items. If she spends Rs.16000per month, answer the following questions: 20-2236EducationFood13581Rent108Miscellaneous20. How much does the spend on rent? |
| Answer» | |
| 1339. |
a river 2 metre Deep and 45m wide is running at the rate of 3 km per hour the amount of water that Run into the sea per minute |
| Answer» | |
| 1340. |
67. Sabarmati express takes 18 seconds to pass completelythrough a station 162 m long and 15 seconds throughanother station 120 m long. The length of the Sabarmatiexpress is :(a) 132 m(b) 100 m(c) 80 m(d) 90 m |
|
Answer» 80m is the correct answer C is the right answer. option C (80m) is the correct answer |
|
| 1341. |
7. A train. 150 m long, takes 30seconds to cross a bridge 500 mlong. How much time will the traintake to cross a platform 370 mlong?(1) 36 secs (2) 30 secs(3) 24 secs 14) 18 secsone |
|
Answer» Time taken by train to cross the plateform = (30/650)*520 = 24 sec 24 sec is the correct answer.... (2): 30 secs .is the right answer of your given question |
|
| 1342. |
15.Ashok and Sumitrana 100mrace. Ashoktook18 seconds and Sumit took 21 seconds.How much longer did Sumit take ?4 |
| Answer» | |
| 1343. |
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour Howill fall into the sea in a minute? |
| Answer» | |
| 1344. |
The angles of a heptagon are (x + 3)°, (2x + 5)°. ( +810, (3x(x - 5)°. Calculate x.1)°, (5x 6), (2x+9e andTill the remaining angles have |
|
Answer» As we all know a heptagon has 7 sides. Hence, it also has 7 angles. We also know that sum of interior angles of a polygon is equal to 360°. Then, => (x+3)° + (2x+5)° + (x+8)° + (3x+1)° + (5x-6)° + (2x+9)° + (x-5)° = 360° On breaking the brackets, we get:- => x + 3 + 2x + 5 + x + 8 + 3x + 1 + 5x - 6 + 2x + 9 + x - 5 = 360° On transposing, all constants from LHS to RHS, we get:- => x + 2x + x + 3x + 5x + 2x + x = 360 - 3 - 5 - 8 - 1 + 6 - 9 + 5 => 15x = 345 On transposing 15 from LHS to RHS, we get:- => x = 345/15 => x = 23 Therefore, x = 23 |
|
| 1345. |
9.Considerf: R+ →1-s,co) given by f(x)#9e+6x.-5.Show that/is invertible10, |
| Answer» | |
| 1346. |
9E=Ag-D+A5+% 9 |
|
Answer» Like if you find it useful |
|
| 1347. |
3-74&”774%4&&7,%[}0! wch thol 9e=RE.ik s7_ and E. g L ssi] = i दा o dus g POES.iy, oh :e e o |देन - — — SR| % -औके का.के... - > gi |
|
Answer» By mid-point theorem in (PEQ) F is the mid point of PE.[Since R is mid point of QE (QR=RE) and SR||PQ (||gm PQRS).]So =>2RF=PQ.PQ=SR (||gm)=> 2RF=SRHence F is the midpoint of SR. In (QSR) QF is the median. Hence SQF=QFR= 3cm square In || PQRS, diagonal QS divides it into two equal triangles i.e. QPS=QSR QSR= QSF+QFR=> 3+3= 6 cm square.2 ar(QSR) = ||gm PQRS (QPS=QSR) => PQRS= 12 cm square. Hence the area of parallelogram PQRS is 12 cm square. |
|
| 1348. |
x(-z^2 %2B x^2 %2B y^2) %2B y(z^2 - x^2 - y^2) - z(-z^2 %2B x^2 %2B y^2) |
| Answer» | |
| 1349. |
21.A bookseller sells a book at a gain of 10%. If he had bought it at 4% less sold it for RS 6 more,he would have gained 184% . The cost price of the book isÎą) Rs 130b) Rs 140c) Rs 150d) Rs 160 |
|
Answer» Let the C.P of the book= ₹ x gain = 10% S.P =110x/ 100 If he had bought it at 4% less and sold it for Rs 6 more, C.P=96x/ 100 S.P =(110x/100)+6 Gain=((110x/100)+6)− 96x/100 = (14x/100)+6 Now, the gain=18 3/4%=75/ 4% Therefore, (14x/100)+6=(96x/100)×75/400 (14x/100)+6=(96x/4)×3/ 400 14x+600=(96x/ 4)×3/ 4 14x+600=6 x(×3) 14x+600=18x 18x-14x=600 4x=600 x=150 The C.P of the book is ₹ 150 |
|
| 1350. |
NUW many Uley Will they meet again at the starts8. Four boys started walking together after the school break.boys is 40, 45, 50 and 60 cm respectively. If they had simuch least distance will they start together.Pgether after the school break. The length of steps of therespectively. If they had started together, after how |
|
Answer» prime factors of 8,12 |
|