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51.

following table shows the weights of 12 studentsweights(in kgs) : 67,70,72,73,75number of students:4,3,2,2,1find the mean in weight?

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thanks

52.

kg. If the empty drum weights13kg, find the weight of rice in12. A drum full of rice weights 40the drum.

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53.

The weights of 15 students are given. Find the average weightof each student.Weights40 42 45 50Number of students 3 | 5 | 6 | 1

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wt no of students total wt40. 3. 120 kg42. 5. 21045. 6. 27050. 1. 50total students=15total wt = 650 kgavg wt = 650/15;kg

54.

0 first10compositenumbers2. The weight of 12 students was measured (in kg) and the results are as follows :41, 36, 38, 45, 42, 37, 49, 50, 52, 44, 48, 46(i) Arrange the weights in ascending order(i) What is the range of their weights?(ii) Find the mean weight.(iv) How many boys have weights more than the mean weight?

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(i) In ascending order36, 37, 38, 41, 42, 44, 45, 46, 48, 49, 50, 52

(ii) Range of weights = (52 - 36) = 16

(iii) Mean weight = (36+37+38+41+42+44+45+46+48+49+50+52) /12= 528/12 = 44

(iv) 6 boys have weight more than mean weight

55.

9. The following table shows the weights of 12 persons.58Weight (in kg):Number of persons:Find the median and mean weights. Using empirical relation, calculate its mode.484503522542

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56.

5The following table gives the distribution of students of two sections according tothe marks obtained by them:Section ASection BMarksFrequencyMarksFrequency0- 1010-2020-3030-4040-500-1010-2020-3030-4040-5017121915Represent the marks of the students of both the sections on the same graph by twofrequency polygons. From the two polygons compare the performance of the twosections.

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57.

The following table gives the distribution of students of two sectionsthe marks obtained by them:according toSection BSection AMarksFrequencyMarksFrequeney0-1010-2020-3030-4040-500-1010-2020-3030-4040-501915101712Dr9. 10disasRepresent the marks of the students of both the sections on the same graph by twofrequency polygons. From the two polygons compare the performance of the twosections

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58.

9. In a class test, the sum of the marks obtained by P in Mathematics and scdenceHad he got 3 marks more in Mathematics and 4 marks less in Science Thehis marks, would have been 180. Find his marks in the two subjects.ICaSE2

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59.

Dul, Tlnd the numbers give an examination. One of them secured 9 marks more than the othermarks were 56% of the sum of their marks. Find the marks obtained by them5. Twoand his

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Let the marks of the 1st guy = xLet the marks of 2nd guy = x + 9Sum of their marks = x + (x+9) = 2x+9According To the Question -(x+9) = 56% of 2x+9=> (x+9) = 56/100 * 2x+9=> 25(x+9) = 14(2x+9)=> 25x + 225 = 28x + 126=> 3x = 99=> x = 33

60.

In a class test the sum of the marks obtained by Puneet in Maths and Science is 28. Had hegot 3 marks more in Maths and 4 marks less in Science, the product of their marks wouldhave been 180. Find his marks in two subject:s

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61.

3. Subtract the sum of (- 1250 ) and 1138 from the sum of 1136 and (-1272)

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62.

2016las est, the sum of the marks ohtained by P in Mathematics and scieeofmarks more in Mathematics and 4 marks less in Science. The productHad he got 3his marks, would have been 180. Find his marks in the two subjects. ICBSE 2008

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63.

If 7 kg onion cost 140 ,then how many cost of 12 kg onion .

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Cost of 7 Kg Onion is ₹140Cost of 1 Kg Onion is ₹140/7= ₹20Cost of 12 Kg Onion is ₹ 20 × 12 kg = ₹240

64.

if 7 kg onion cost 140 rupees ,then how many cost of 12 kg onion.

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Cost of 7 Kg Onion is ₹140Cost of 1 Kg Onion is ₹140/7= ₹20Cost of 12 Kg Onion is ₹ 20 × 12 kg = ₹240

if 7 kg onion costs 140 rupees then 1 kg of onion will cost 140÷7=20 rupees ...

so 12 kg of onion will costs =12×20=240 rupees.

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65.

anurag buys coffee and tea at 37.5 and 35.0 rupees per kg, respectively; he spends 870 rupees and gets 24 kg in all , how much of each kind did he buy?

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let Amount of coffee he buy = x kgand that of tea = y kgx + y = 24 (given) x = 24 - y ... (i) cost of x kg coffee =Rs37.5xcost of y kg tea = Rs 35y37.5x + 35y = 870 (given) using (i) equation 37.5(24-y) + 35y = 870900 - 37.5y + 35y = 870-2.5y = -30y = 30/2.5y = 12 kgand x = 24 - 12 =12kgcoffee= 12kg and tea = 12 kg answer

66.

m-(a2+82+ 16)รท(2+4)

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67.

CHALLENGERSwhendi d laves are in cach cases 135,1272 ving themethewwww1.12 kesthereLCMERCEThes

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send picture clarity

68.

(a) 3510. If sec 4A cosec (A-10") and 4A is acute then(c) 40°(b) 30°(d) 50(a) 20°

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We know sec ( 90 - A ) = cosec (A)

sec(4A) = cosec ( A - 10)

sec(4A) = sec ( 90 - ( A - 10))

Comparing both sides

4A = 90 - A + 10

5A = 100

A = 100/5

A = 20

Option (a) is correct

69.

eces.of china tea-sets for 18,000. When the goods ariest,ets were damaged. At what price per set should he sell the remainingl the whole?3.crockery-dealer ordered for 50teath2 tea-setstea-sets to earn a total profit of200?

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70.

A stone thrown vertically upwards withinitial velocity u reaches a height 'h'before coming down. Show that thetime taken to go up is same as the timetaken to come down.

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Let us consider an object which is projected vertically upwards with initial velocity u which reaches a maximum height h.Acceleration due to gravity=-gEquation of Motion for a body projected thrown upwards :V=u-gt----------(1)h=ut-1/2gt² -----------(2)v²-u²=-2gh --------(3)

Equations of motion for freely falling body :for free fall :Initial velocity=u=0g=gV=gt----------(4)h=1/2gt²-------(5)v²=2gh------(6)Time of Ascent is the time taken by body thrown up to reach maximum height hAt maximum height , V=0

Equation (1) turns to u=gt1t1=u/g ----------(7)Maximm height h=u²/2g ----------(8)

Time of descent : After reaching maximum height , the body begins to travel downward like free fallso equation (5) h=1/2gt₂²t₂²=2h/gt₂=√2h/gbut from equation (9)t₂=√2xu²/2g²t₂=u/g -----------------equation (10)∴t₁=t₂The time ascent is equal to time of descent in case of bodies moving under gravity

71.

To is monthly saving.che sum of three numbers is 2 If two of them are6. The sum ofandfind the thirdof te

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Sum of two nmbr=7/12+11/18L.c.m of 12,18=36So,(7*3/36)+(11*3/36)21/36+33/36=54/36Sum of 3 nmbr=73/36So,diffrence=3rd nmbr=73/36-54/36=19/36

5/6 is the right answer

72.

3(a+b) -2(a-b) +4a-7

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Thnku ji

73.

(J.A.C(J.A.C12. How many numbers between 100 and 1,000 can be made with the digits 1,2,3,4,5,6,7,8,9?1.Te

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991 is the answer because

The numbers range from 100 to 999 are 3 digit numbers. There is no restriction. Any digit can appear any number of times.

So ways of selecting 3 digits out of 7 digits : 9*9*9= 729

74.

\left[ 2 ^ { 2 } + 3 ^ { 2 } + 4 ^ { 2 } + 5 ^ { 2 } + 6 ^ { 2 } + 7 ^ { 2 } + 8 ^ { 2 } + 9 ^ { 2 } + 10 ^ { 2 } \right]-

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75.

Calculate the mass of the earth using thefollowing data. Radius of the earth is6400 km. Acceleration due to gravity is9.8 m s2 and the gravitational constant is6.673x10 N m2 kg2

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5.96×1024kg.

F=Gm1m2/r2=maGm/r2=gm=gr2/Gm=(9.8m/s2)(6.37×106m)2/(6.673×10−11Nm2/kg2)m=5.96×1024kg

76.

\left[2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}+9^{2}+10^{2}\right]

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77.

\left. \begin array l ( a , b ) \text and cuts orthogonally to circle x ^ 2 %2B y ^ 2 = \\ \text is \\ \text (a) 2 a x %2B 2 b y - ( a ^ 2 %2B b ^ 2 %2B p ^ 2 ) = 0 \\ \text (b) 2 a x %2B 2 b y - ( a ^ 2 - b ^ 2 %2B p ^ 2 ) = 0 \\ \text (c) x ^ 2 %2B y ^ 2 - 3 a x - 4 b y %2B ( a ^ 2 %2B b ^ 2 - p ^ 2 ) = 0 \\ \text (d) x ^ 2 %2B y ^ 2 - 2 a x - 3 b y %2B ( a ^ 2 - b ^ 2 - p ^ 2 ) = 0 \end array \right.

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Let the centre be (α, β)∵It cut the circle x^2+ y^2= p^2orthogonally2(-α) × 0 + 2(-β) × 0 = c1 – p2c1 = p2Let equation of circle is x^2+ y^2- 2αx - 2βy + p2= 0It pass through (a, b) ⇒ a2+ b2- 2αa - 2βb + p2= 0Locus ∴ 2ax + 2by – (a^2+ b^2+ p^2) = 0.

Option=A

78.

What are some of the physical benefits of playing basketball?

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Following are the health benefits of basketball. Promotes Cardiovascular Health. Burns Calories. Builds Bone Strength. Boosts The Immune System. Provides Strength Training. Boosts Mental Development. Develops Better Coordination And Motor Skills. Develops Self-Discipline And Concentration

79.

Q.12A train is moving with uniform velocity on a straighttrack. A boy moving at 4 km/h due east observes thatthe train is moving due north. Another boy moving at 1km/h due south claims that the train is moving towardsnorth-east. What is the speed of the train in groundframe?(1) 3 km/hr(2)4 km/hr(3) 5 km/hr(4) 6 km/hr

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5km/hr train moving on ground , boy moving due north and another boy moving south ,boy move your opposite direction to increased train moving speed

80.

Expmess te Follawing numbers m5.08*10^-67.9*10^4

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i)0.00000508ii) 79000

for first question...00000508

for second question..79000

81.

18. A cylindrical bucket, 28 cm in diameter and 72 cm high, is full of waeThe water is emptied into a rectangular tank, 66 cm long and 28wide. Find the height of the water level in the tank.m iong

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82.

FindalsothecostOICUmUinwatecubic metre.A cylindrical bucket, 28 cm in diameter and 72 cm high, is full of water. Theinto a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water l.e water leyank.

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thank you

83.

A cylindrical bucket, 28 cm in diameter and 72 cm high, is full of water. The water is encubic meter. Find also the cost of cementing its inner curved surface at 2.503.60into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level bestrank

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84.

Write four field applications of worm and worm wheel.

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Worm gear (or worm drive) is a specific gear composition in which a screw (worm) meshes with a gear/wheel similar to a spur gear. The set-up allows the user to determine rotational speed and also allows for higher torque to be transmitted. This mechanism can be found in devices both at home and in heavy machinery; the simplest form evident in the tuning mechanism of an acoustic guitar.

85.

1. Pankaj got 15 marks less than Charu and Charu got 5 marks more than Kanta. If the(2013)sum of their marks (all three) is 112, how many marks did Kanta get?(1) 29 (2) 39 (3) 44 (4) 45

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Answer :2)39Explanation Let kanta mark be xcharu mark =x+5pankaj mark=x-10sum =112Thus x+x+5+x-10=112x=117/3x=39thus kanta mark is 39

86.

Q13. There are 2000 students in a school, out of these 1000 play cricket, 600 play basketball and 5s0play football, 120 play cricket and basketball 80 play basketball and football, 150 plfootball and 45 play all the three games. How many students play(i) none of the three games? t(ii) exactly one of the three games ?iii) at least one game

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U=2000

Cricket(c)=1000

Basketball(b)=600

Football(f)=550

C nB nF'=120

BnFnC'=80

CnFnB'=150

CnBnF=45

Secondly,put variables in the set of cricket,football and basketball only respectively

A for cricket only;B for basketball only and C for cricket only

a

A+105+75+45=1000

A+225=1000 (use the principle of change of subject then arrive at)

A=775

B+75+45+35=600

B=445

C+105+45+35=550

C=365

Now to find the students who play none of the games, add everything in the set and add another variable(x) to replace the number of students who play none of the games

It becomes A+B+C+x+45+35+75=2000

775+445+365+260+x=2000

1845+x=2000(make x the subject)

x=1845

therefore 1846 students played none of the games

(ii)exactly one=A+B+C

775+445+365=1585

therefore 1585 played exactly one games

(iii)exactly two=105+75+35

=215

therefore 215 played exactly two games

87.

13. In a class of 50 students, every student plays lfootball or cricket. If 32 of them play footballand 35 play cricket, then find the numbersof students who play both the games and onlyfootball. (50 4- DR

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Total n = 50and n (football ) = 32and n (cricket) = 35

so, n( football and cricket) = 35+32-50 = 67-50 = 17

so, n ( football only) = n(football) - n( football and cricket) = 32-17 = 15.

88.

13, In a class of 50 students, every student plays |football or cricket. If 32 of them play football Iand 35 play cricket, then find the numbersof students who play both the games and onlyfootball. (50a-za ca ab

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By properties of Venn diagram,n(AUB) = n(A)+n(B)-n(A intersection B)50=32+35+xx=17So, 17 students play both the games.

89.

In a school 40% of the studentsplay football and 50% playcricket. If 18% of the studentsneither play football nor cricketthe percentage of the students playing both is?

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No. of student who play football=40%no. of student who play cricket=50%18%student neither play cricket nor play footballthen , 40+50+18=108%108-100=8%

90.

50. Average of 10 numbers 4, 6, 8, 10,12, 14, 16, 18, 20 and 22 is 13. If 5 is added to eachnumber then what is the new average of all numbers?

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If 5 is added to each numberthen avg also will be increased by 5and now new avg will be 13+5= 18thanks

91.

Expressions & Equations 6.EE.Sko abesse he wraph represents theA-24 6 8 10 12 14 16 18 2024 6 8 10 12 14 16 18 2024 6 8 10 12 14 16 18 20024 6 8 10 12 14 16 18 20the four books on the

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casas tipicas 3 * 4 equals 12 and 12 is the answer to the problem in which one equals three three times for would be 12

92.

2.Raghav got 650 marks out of 1000. Rahul got 275 marks out of 500. Express their marks in ratio.

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Their marks ratio 13:11

93.

Entrance1. Ram got 8 marks more than Shyam in anexamination. Anil got 4 marks more thanRam in the same examination. If all threeof them got 128 marks together as a total,Ram's marks would beINV 2016](1) 36(2) 44(3) 48(4) 54

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54 is right answer bro

44 is your answer brother

44 is the answer of the following

44is our answers to questions

94.

Q.9.In an examination A got 10% marks less than B, who got 25% marks more than C, who got 20%marks less then D. If A got 360 marks out of 500, then percentage of marks obtained by D was?

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A got 369/500 * 100 = 72%This was 90% of B. So B = 72 * 100 / 90 = 80%.

This was 125% of C. So C = 80 * 100 / 125 = 64%.

This was 80% of D. So D = 64 * 100 / 80 = 80%.So D got 80% marks.

95.

In a Science Olympiad, 5 marks are given for each correct answer and 2 for each incorrect answer. Pankhurianswered all the questions and scored 26 marks, though she got 2 in correct answers. How many questionsdid Pankhuri answer correctly?

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she answered 6 questions correctly as 5*6 = 30 marks and 2 questions incorrect so negative marks = -4 total marks got = 30-4= 26

96.

(b)32. Which of the following is prime factorisation of 60.(a) 2 x3 x 5(b) 2 x6 x5(c) 2 x3 x 15

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Prime factors means factors which are prime numbers

60 = 2*2*3*5

Prime factors of 60 are 2, 3, 5

97.

Attempt any five questions from this sectionFind the rank of the matrix-2 6 5

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Hit like button if you like my solution !

98.

Answer any five from questions 5 to 11. Each question series 3 scores (a) Find the area and perimeter of the rectangle in the figure(b) What is the length of AC cm

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Thnks

99.

Attempt any five questions from this sectionFind the rank of the matrix-

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100.

Mukesh t34100 to Rshen for certaintime period at 8% Afler the end of thatperind, Mukesh got back 58652 fromRoshan Calculae the time period for whicthe time period for which1. Kanta investedanom for 215% per annumend of S years, he returned 42630 toThe interest othe money was lent7 Mohir borrowedt 24360 from Rohit for5 years at a vertain rate of interest. At theRohit Find the rate of interestnterest per cent10% for 2 years and lenafter paying back to MohMaya wants to earn 600 on 6000in2 years What rate of interest will fetch the 12. Kamal borrowedrequired interest?20009 A sum of money invested at 14%-per annumamounted to 10400 in S years. CalculateInterestThe additional money paid for using the money taken fromPEEDsome gue is called the interestPrincipalThe money borrowed is called the principalAmountThe money paid back is called the amountAmount Principal +InterestRate of InterestThe interest paid per &100 per year is called the simpleiOO TimeThe period for which money is borrowed is called the timeO Simple interestPrincipal x Rate of interest Time100speed of thIt is same for each year126

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