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68951.

4. Underwhat condition will ax2 +5x + 7-0 be a quadraticequation?

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68952.

18.Find the relation between the co-efficients of the quadratic equationax2 + bx + c # 0 if one root be n times the other.TuRIT 59:R. U. 691

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wroung solution

68953.

Fig. 8.7, AB = AC and AP丄BC and LB-60°. Then find (i) <BAP (ii) <ACB.Fig. 8.7

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68954.

4. One root of the quadratic equation ax2 +bx+C 0 isthree times of the other. From the followings whichcondition is correct ?(1) b 16ac(3) 3b2 16ac (4) 16b2 3ac(2) b2 -3ac

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68955.

8CDh and oceived 5 others as gifts. What fraction of her total CDs did she bKrisinreceived a CD player for her birthday. She boughtand whut fraction did she receive as gifts?

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68956.

, A plott is in the form of a rectangle ABCD having semi-citr umFig 20.23 If AB-60 m and BC = 28 m, find the area of the plot.28 m60 mFig. 20 23circles on its smaller sides

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68957.

2. A plot is in the form of a rectangle ABCD having semi-circle on BC as shown inFig. 20.23. If AB 60 m and BC-28 m, find the area of the plot.28 m60 mFig. 20.23los on its smaller sides

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full page

full page answer

this is the answer

68958.

8. Simplify 2x-[3x-4x-(3y-2x-2ym32 A hog sontains zs our timos that of 50 pauise coinsbag contains 25 paise and 50 paise coins whose total valuc is 30. If the number ofpaise coins is four times that of 50 paise coins, find the number of each type of25coins

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68959.

A bag contains one rupee, 50 paise and 25 paise coinsin the ratio 5 6: 8. If the total amount is 420,then find the total number of coins.

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900

68960.

54. A box contains 280 coins of one-rupee, 50-paiseand 25-paise. The values of each kind of the coinsare in the ratio of 8 :4:3. Then the number of50-paise coins is(a) 80(c) 70(b) 90(d) 60

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Rs. 1 : 50P : 25P

Value of coins 8x : 4x : 3x

No. of coins8x×1:4x×2:3x×4

8x:8x:12x

∴Totalcoins∴Totalcoins

=> 8x + 8x + 12x = 28x

28x = 280 (given)

x=10

∴Number of 50 P coins=8x=80

Option a is correct.

68961.

. 70 coins of 10-paise and 50-paise are mixed ina purse. If the total value of the money in thepurse is 19, find the number of each type ofcoins.

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68962.

4. What is the least number that should be added to 47580 so that the result-ing number is a perfect square?

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number near to 47580 perfect square is 47961so minimum number should be added=47961-47580=381

68963.

wo gases A and B have critical temperatures as 250 K and 125 K respectively. Whichne of these can be liquefied easily and why?

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as A has higher critical temperaturehenceit has greater intermolecular forceshence A can be liquified easily

68964.

(1) If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to thesum of the squares of their reciprocals, then prove that 2a c = cb + b-a.

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68965.

16. In the given figure a rectangular plot of land measures 8 m by p2m6 m. In each of the corners, there is a flower bed in the form of aquadrant of a circle of radius 2 m. Also, there is a flower bed inthe form of a circle of radius 2 m in the middle of the plot. Findthe area of the remaining plot.EHint. Required area-I(8x6)-(4x4×22×2x2)-(큭x2x2h1meN(8x6-t4 x 4x7x2x2)+げx22/Hint. Required areaA 2 mm-

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68966.

is the centre of the circle in Fig. 11.28. Findthe measures of LACB, LABC and 4BAC.60Fig. 1128

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angle ACB=90°as it is angle subtended by diameter it is 90°hence angle OCB=90-60=30°nowangle ABC=angle CAB..as equal angle is subtended on both endshence as its a right angled triangle sum of these angles will be 90 and as they are equal it will be 45° each.

68967.

/16.Inthegiven figure a rectangular plot of land measures 8 m byozn6 m. In cach of the corners, there is a flower bed in the form of a Equadrant of a circle of radius 2 m. Also, there is a flower bed inthe form of a circle of radius 2 Im in the middle of the plot. Findthe area of the remaining plot.e mHint. Requtred area (8x6)-422 2222m2A 2 m

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68968.

(Question numbers 1 to 6 carry 1 mark each)1. If P(E) 0.05, what is the probability of 'not-E?

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Probability = 1 (always)

If prob(E) = 0.05

then,

prob(not E) = 1-0.05

= 0.95

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thanks

68969.

Section 'AQuestion numbers 1 to 6 carry 1 mark eacl1. What is the HCFof the smallest composite number and the smallest prime number ?

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Smallest Composite no. = 4

Smallest Prime no.= 2

4 = 2*2

2=2*1

therefore they have only one common factor i.e. 2.

H.C.F. =2

68970.

198198One compartment of a purse contains three 25 paise coins and 2 one rupee coins and the other ccontains two 25 ps. coins and 3 one rupee coins. The probability of drawing a rupce coin fra) 1/5ompartub) 2/5c) 3/5d) 1/2ld 1

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Probability = possible outcomes /total outcomes

For drawing one rupee coin

Possible outcomes = 5Total outcomes = 10

Therefore, Probability of drawing a one rupee coin from purse= 5/10 = 1/2

68971.

estion numbers 1 to 6 carry 1 mark each.Find the nature of the roots of the quadratic equation 4x2 + 4-3 x + 3 = 0.

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D= b^2-4ac= 48- 48= 0equal roots

68972.

Fig 6.1r> 10Example s Solve.1. Show the graph of the solutions on number line. ing 124

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68973.

.1Write four solutions of 2x +3y -8.

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Four solutions in the form of x,y are:-

1.1,22.4,03.-2,44.6,-2

68974.

.Find four solutions of 2x-y = 4.

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68975.

Question numbers 1 to 6 carry 1 mark each.I fx = 3 is one root of the quadratic equation x2-2ke-6 = 0, then find the valueof k.1.

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68976.

Question mambers 1 to 6 carry 1 mark each.Nt it : 3 is one root of the quadratic equation x2-2kr-6, then find the value ofk

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x²-2kx-6If x = 33²-2(3) k-6 = 09-6k-6=03=6k1/2 = k

x square- 2kx- 6of x=33× 3- 2×3 k- 6= 0 9-6k-6=03=6k 1/2=Ik= 1/2

68977.

If S, denote the sum of n terms of an A.P. with first term a and common difference dS,such that s is independent of x, then(a) d=ate(b) d=2a-(c) a=2d(d)d=-a

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68978.

1. Expand

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Thanks

68979.

An arc of a circle measures 2.4 radians. To thenearest degree, what is the measure, in degrees, ofthis arc? (Disregard the degree sign when griddingyour answer.)

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tysm

68980.

bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball isselected at random. Find the probability that it is(i) white(ii) not red.

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thanks boss

68981.

24 = mसिद्ध कीजिए कि :cos? 0.cosec 0 + sin 6 = cosec 0,

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cos² cosec + sin

(cos² + sin²)/sin

1/sin

cosec

68982.

30. Mr. Smith has a bag A. Bag A has n numbers ofbag B. Each bag B has n numbers of bag C. Eachbag C has n numbers of one rupee coins in it. If onebag B is removed from bag A, then the total numberof coins left in bag A isnin-1+ 1)(2) n(n- 1)(3) n2(n + 1)(4) n(n-1)134

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68983.

IS I two variablesand finding their solutions frimthe graph 2x t y 5 and 3x - 2y4une 2015If the present ages of A and 'B are in the ratio 9:4 and seven years hence, the ratio of theirages will be 5:3 then find their present ages.arch 20164. For what values of m,the pair ot eguntinms Rr

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Given,9x+7/4x+7 = 5/3

3(9x+7) = 5(4x+7)27x+21 = 20x+3527x-20x = 35-217x = 14x = 14/7x = 2

A's age = 9x = 9*2 = 18B's age = 4x = 4*2 = 8

therefore A is 18 years old and B is 8 years old.

68984.

sece--7-2 secIf cos 0find, cosec θ

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We know that Cos (60°) = 1/2theta= 60°so we will put the value of theta sec(60)=2tan(60)= √3cosec(60)= 2/√3= 4+3/7-4.(2/√3)7/7-(8/√3)7/2.382.39 Approximately

68985.

Solve the following simultaneous equations:ar + by a4bwhere a and b' are non - zero constants.

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thanks

68986.

Section 'AQuestion numbers 1 to 6 carry 1 mark1. Simplify 143)Simplif13222. Factorize: 6-x-x2

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2)

68987.

0 Find the term independent of x in the expansion of-3x01

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68988.

Find the coordinates of the point on y-axis which is nearest to the point (-2, 5).

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the answer is 4 and 6

68989.

9.,Thevolumeof a cylinder of height 8 cm is 1232 cm3. Find its curved surface area and thetotal surface area.

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68990.

9., The volume of a cylinder of height 8 cm is 1232 cm3. Find its curved surface area and thetotal surface area.

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68991.

by four. How many balls does the bag contaln24. A number consists of three digits, the right-hand one beingzero. If the left hand and middle digits be interchanged the numberis diminished by 180; if the left-hand digit be halved and the middieand right-hand digits be interchanged the number is diminished by454. Find the number.umount to Rs 37; if 4 men

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Let x be in hundred place

y be in tens place

third digit is 0 so dont mind it

original number =100x+10y

as per first condition new number

2nd digit becomes first digit

so new number =100y + 10x + 0

EQUATION 1

100x + 10y-180=100y+10x

simplifying we get 90x - 90y=180

90(x - y)=180

x-y=180/90=2

as per second condition new number

fiest digit gets halfed

100x/2=50x

next 2nd digit becomes 1st and viceversa

so 0+1*y

EQUATION 2

100x + 10y -454=50x + y

simplifying we get 50x + 9y=454

third step = multiply 50 in equation 1

so50(x - y)=50*2

50x-50y=100

next subtract thirs step from equation 2

50x + 9y=454

--------

50x-50y=100

___________

0+ 59y =354

we get 59y=354

so y = 6

putting this value to equation 1 we get x as 8

so original number

100x+10y = 800 + 60=860

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68992.

Find the measure of the angle MNP to the nearest degree110 10090 8070

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125° is the measurement of < MNP

68993.

17. Find the values ot a and b if 3a + b V7

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tqu

68994.

Find the eleveHul tellIuFind the coordinates of the point on y-axis which is nearest to the point (-2, 5).

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68995.

In APQR. 2Q = 90. If PQ = 3 V5, LPRQ = 0 and PR - QR = 3. Find cosec 0 + cot 0 = ?ta) 205 (b) V5 (c) V7 (d)

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68996.

arte Constants

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68997.

ay an afb are rational no find the valuea&amp;b in each of the following equalitiesa) 13-1 utb V3143 t)1 62 |=a +b 173+153- V7

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68998.

Independent PracticeFind the value of x. Round to the nearest degree.2.1.14

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Use trignomerty ratios

1. sin x = 5/14

So, x = sin inverse 5/14

x = 20.9° ( approx)

2. tan x = 8/5

So, x = tan inverse 8/5

x = 59.7° ( approx)

Hit like and BeScholr

68999.

vuvuruJILI YUCUFind the area of each figure. Round to the nearest tenth1.7 in

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area of square will 7squar

69000.

find the coordinates of the point on y-axis which is nearest to the point (-2,5).

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