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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.
| 80551. |
c13Ft7 R01(Short Answer Type Questions)3UP 2008] |
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| 80552. |
.A, B and C can finish a job in 2 hours. If A alone can finish the job in 5 hours, B alone finishesin 6 hours, how long will it take for C to finish the job alone? |
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Answer» 1/2-11/3030/47.5 hours |
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| 80553. |
below and tell what will come next: 2. 1 x 9 + 2 |
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Answer» 1*9+2 9 + 2 = 11 |
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| 80554. |
dimensions of a cuboidal iron slab are 1.5 mx25 cm X 150 cm. Find its volume in cubic metres.. The |
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Answer» l=1.5 m = 150 cm So volume in cubic cm is 150×25×150 = 5,62,500 cm^3= 0.5625 m^3 |
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| 80555. |
How many millilitres of milk does each bottle have?Shan poured 1 litre of milk into twobottles so that the first bottle holds1 litre and the other holds litre.+ Shade the level of milk in each bottle.+ How many millilitres of milk doeseach bottle hold? |
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Answer» 1 ltr = 1000 mililitre 1st bottle = (3/4) x 1 = 0.75 litre = 750 mililitre 2nd bottle = (1/4) x 1 = 0.25 litre = 250 mililtre |
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| 80556. |
showlues 23 - without__(d) bcI cza?I arbaexpanding_bc _btc ।ca ctaab atb ,o |
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| 80557. |
Time and Workans1) A does 20% less work than B. If A can finish a piece of work inhours, then Beanfinish it in(a) 5 hours(b) 5 hours2td16-hours26 hours |
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| 80558. |
The radius of a spherical balloon increase from 6cm to 12 cm as air isbeing pumped into it. Find the ratio of the surface areas of the balloonin two cases. |
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Answer» wrong |
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| 80559. |
- A man walks at a speed of 2 km/hr for 2 hours and then at a speed of 1- km/hr for next 2 hours. Findthe distance travelled by the man in 4 hours. |
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Answer» Distance = speed x time= 7/3*2+ 3/2*2= 14/3+3= 14+9/3= 23/3km Distance covered in 1 hour=66.6÷12km Distance covered in 5 hours= |
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| 80560. |
9. A swimming pool is 260 m long and 140 m wide. If 54600 cubic metres of water is pumpedinto it, find the height of the water level in it. |
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| 80561. |
A swimming pool is 30m long and 15m wide. How many kilolitres ofwater must be pumped into it so as to raise the level of water by 4m?(1000cm3 1 litre) |
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Answer» volume of cuboid=lbh=30*15*4=1800m3since 1m3=1000 Litres1800m3=1800*1000 litres=1800000 litressince 1000 litres=1KiloLitre1800000=1800 Kilo Litres. |
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| 80562. |
(villf α, β are the roots of the quadratic equation x2 + bx-c = 0, the equation whose roots are b and c, is(a) x2 + α-β=09·blx2 |
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| 80563. |
25. If /2 sin θ = 1, find the value of sec2 θ[CBSE 2012cosec2 θ. |
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| 80564. |
5. In a pond there were 1,37,896 fishes. Out of them 49579 died due to heat, find thefishes left in the pond. |
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Answer» an objects creat an image on a plane mirror. if the mirror now movw backwards with an acceleration f then what will be the acceleration of the image? |
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| 80565. |
Merive ist, and ond 8nd equaton |
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| 80566. |
(iv)Without expanding at any stage, find the value of: \left| \begin{array}{ccc}{a} & {b} & {c} \\ {a+2 x} & {b+2 y} & {c+2 z} \\ {x} & {y} & {z}\end{array}\right| |
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| 80567. |
1S. tn the siven figure, ABCD is a trapezium in whichAB II DC. P and Q are the mid-points of AD andBC respectively. DQ and AB when produced meetat E. Prove that08(ii) AR= CR,PR II AB |
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| 80568. |
The number of permutations formed withoutchanging the position of vowel and consonants ofthe letter of word 'ALGEBRA: |
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| 80569. |
You spent 1 1/2 hours on homework last night and 1 1/4 of that time was spent on mathematics. What fractionof an hour did you spend on mathematics ? |
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Answer» Hours spent on Homework = 11/21/4 of 11/2 on mathematics1/4×11/2=1.375 hours spent on mathematics |
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| 80570. |
(Parunw) All squares are...In figure, BESTjares are .......... (Congruent/similar)ure, BEST is a parallelogram. Find thevalues of x, y, z.coyo110°toVin the following parallelograms, find the |
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Answer» as BEST is a parallelogram so angle S is equal to angle B. so the value of angle S is 110 degree. as BEST is a parallelogram so the sum of all it's angle is 360 degree because of angle sum property. so angle B +E+S+T is equal to 360 degree.110 + x + 110 + y = 360220+x+y=360x+y=360-220x+y=140x+x=140 (as best is a parallelogram so angle E= angle T or x = y)2x=140x=70and y also equal to 70. |
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| 80571. |
raulus item (UseçŚ= 3.14)4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pampedinto it. Find the ratio of surface areas of the balloon in the two cases. |
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| 80572. |
1. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumpedesof thalon in the two caves.into it. Find the ratio of surface areas of the balloon in the two cases. |
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| 80573. |
14cma),nd the totalFind the total)21 cm2-al surface area of a hemisphere of radius 10 cm. (Use T -3.14)f a spherical balloon increases S(iii) 35m |
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| 80574. |
29. Solve the quadratic equationb 0.[CBSE March 2012] |
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Answer» 9a^2b^2x^2 -16abcdx -25c^2 d^2 = 09a^2b^2x^2 -25abcdx + 9abcdx - 25c^2d^2 = 0abx (9abx -25cd)+cd(9abx - 25cd)=0(abx+cd)(9abx-25cd)=0x= -cd/ab, 25cd/9ab |
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| 80575. |
In the AP 2, x, 26, find the value of x.[CBSE 2012] |
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Answer» like if you find it useful |
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| 80576. |
37. For an AP show that a+ ap 2-2ap[CBSE 2012] |
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| 80577. |
a +1 ab acabса сЬсг +1 |
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| 80578. |
Example 15 Find AB, if A=Example 1s Find AB i AC ) and B-C: -and B=1 |
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| 80579. |
Q.3 Using the property of determinant and without expanding prove1 be a(b + c)1 ca b(c+a) 01 ab c(a + b)-a2 ab acQ.4 Prove that ba b be 4abca cb cClc |
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Answer» Q.3 Q.3 continuing... Q.4 Q.4 continuing.... Q.4 continuing... |
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| 80580. |
rs(6) Determine how many words can be formedusing the letters of the word LOGARITHM.if(a) vowels are always together.(b) no two vowels are together.(c) consonants occupy even positions.(d) begin with O and end with T. |
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| 80581. |
21. In the adjoining figure, find the area of the shaded regionenclosed between two concentric circles of radii 7 cm and14 cm, where <AOC 40°. | Use-22 |
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Answer» The Area of Major Sector of bigger circle =360 -theta / 360 × 22/7 × r × r360- 40 /360 × 22/7 × 14× 14By solving it you will get answer as,547.55 cm^2 The Area of major sector of smaller circle =360- theta / 360 × 22/7 × r × r360 - 40 /360 ×22/7×7×7By solving it you will get answer as,136.88cm^2The area of shaded region = area of bigger cicles major sector - area of smaller cicles major sectorArea of shaded region = 547.55 - 136.88= 410.67 cm ^2 I am not satisfied |
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| 80582. |
The radius of a spherical balloon increases from 7 em to 14 cm as air is being pumpedinto it. Find the ratio of surface areas of the balloon in the two cases. |
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| 80583. |
In fig. 6, find the area of the shaded region, enclosed between two concentriccircles of radii 7 cm and 14 cm where 2A0C-40°. (Use π22)Figure 6 |
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Answer» The Area of Major Sector of bigger circle =360 -theta / 360 × 22/7 × r × r360- 40 /360 × 22/7 × 14× 14By solving it you will get answer as, 547.55 cm^2 The Area of major sector of smaller circle =360- theta / 360 × 22/7 × r × r360 - 40 /360 ×22/7×7×7By solving it you will get answer as,136.88cm^2The area of shaded region = area of bigger cicles major sector - area of smaller cicles major sectorArea of shaded region = 547.55 - 136.88= 410.67 cm ^2 Like my answer i you find it useful! |
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| 80584. |
ŠDr.5% 5% than find Su-151. |
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| 80585. |
integer m.Show that the square of any positive odd integer is of the form 8q+1 for someinteger q.[CBSE 2009, CBSE SP 2012] |
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| 80586. |
show that any positive odd integer is of the form 6q+1or6q+3or6q+5where q is some integer |
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Answer» Let a be any positive integer a=6q+r where 0< or equal to r <6 put r=1: a=6q+1 =odd integer r=3: a=6q+3=odd integer r=5: a=6q+5=odd integer Therefore , any positive odd integer is of the form 6q+1, 6q+3 or 6q+5,where q is some integer . |
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| 80587. |
In the adjoining figure, O is the centreof a circle. If AB and AC are chords ofthe circle such that AB =AC, OP丄ABand OQ丄 .AC, prove that PB QC |
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| 80588. |
16. Draw a triangle ABC, with AC 4 cm, BC-3 cm and L C-80°. Measure AB. Is (ABAC2+(B2? If not, which one of the following is true:(AB)2> (AC)+ (BOP or (AB) (AC?+ (BO? |
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| 80589. |
Using properties of determinants, prove thata2+1 ab acab b+1 be1 a b2+c2ac be c2 +1 |
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| 80590. |
uiturucuolal sulauulindrical vessel open at the top has a base diameter of 21 cm and height 14 cm. Find the costof tin plating its inner part at the rate of 5 per 100 cm |
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Answer» diameter= 21 cm radius=21/2=10.5 height=14 cm curved surface area of cylinder= 2πrh 2×22/7×10.5×14 44/7×10.5×14 (14 will be eliminate with 7 ) 44×10.5×2=88×10.5 924cm squarearea of base circle= πr square 22/7×14 (14 will be eliminate with 7) 22×2=44924+44=968cm square 968×5/100=4840/100 answer=48.4 rupees |
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47) In a city, 70% people eat fish and 40% people eatchicken. If 25% people eat both chicken and fish, then thepercentage of people who neither eat chicken nor eat fishisA) 45%B) 30%C) 5%D) 15% |
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Answer» answer is d ok sir this question are simply |
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| 80592. |
2. Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameterofnd the height is 14 cm. Diameter ofcylinder B is 14 cm and height is 7 cm. Without doing any calculationscan you suggest whose volume is greater? Verify it by finding thevolume of both the cylinders. Check whetherthe cylinder with greater14 cm7 em14 cm-volume also has greater surface area?7 cm |
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Answer» Radiusof cylinder A is half of radius of cylinder B it means the radius of cylinder Bis Greater . While finding the volume we need to square the radius & squareof radius gives more value . Thevolume of cylinder B is Greater then Volume of Cylinder A. Diameterof cylinder A= 7 cm Radiusof cylinder A= 7/2 cm Heightof cylinder A= 14 cm Volume of cylinder = πr2H Volume of Cylinder A= (22/7)(7/2)²14 = (22/7) (49/4)× 14 = 539cm³ Diameterof cylinder B= 14 cm Radiusof cylinder B= 14/2=7 cm Heightof cylinder B= 7 cm Volume of Cylinder B= (22/7)(7)27 =22×49 =1078m³ So Volume Of Cylinder B is greater than Volumeof cylinder A Total Surface Area of Cylinder = 2πr(r+H) Total Surface Area of Cylinder A=2×(22/7)×(7/2)(7/2 +14) = 22((7+28)/2) =22(35)/2 =11×35 = 385cm² Total Surface Area of Cylinder B= 2×(22/7)×(7)(7+7) = 44× 14 = 616cm² Hence, the surface area of Cylinder B is also greater |
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| 80593. |
In Fig. 10.39, A, B, C and D are four points on acircle. AC and BD intersect at a point E suchthat 4 BEC 130° and Z ECD 20°, FindZ BAC5.D)360 |
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| 80594. |
gate in 40 minutes, if 10 cm of standing water is requORnd the area of the shaded region(Use14 cm#3.14] |
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| 80595. |
Show that any positive odd integer is of the form 4q + 1 or 4q +3, where q is a positiveinteger. |
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| 80596. |
-y-, ifx = at, ydrFind2a1. |
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| 80597. |
eachofIf a = 5, b = 3the following:and c -3, evaluate\begin{array}{l}{\frac{4 a b+b c^{2}}{2 a+2 b-3 c}} \\ {\frac{3 a^{2} b c}{a b+b c}} \\ {\frac{5 a^{3} b^{2} c}{a b+b c+a c}} \\ {\frac{a b}{c}+\frac{a c}{b}}\end{array} |
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Answer» you Don't know how much you helped me thank you verymuch |
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| 80598. |
10.5 cm21 cmIn the above figure, XLMT is a rectangle. LM 21 cm, XL 10.5 cmDiameter of the smaller samicirele is half the diametar of the laregersemicircle. Find the area of non-shaded region. |
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Answer» Area of semicircle with diameter( AB)= 21cm.Radius of semicircle AB = 21/2 cmDiameter AM= MN= NB (GIVEN)AB = AM+MN+NBAB = AM+AM+AMAB = 3AM21 = 3AMAM= 21/3= 7 cmAM= MN= NB= 7cmArea of Semicircle of diameter AB=πr²/2=[ (22/7)× 21/2×21/2]/2= 693/4 cm²Area of semicircle with diameter AM= area of semicircle diameter MN= Area of semi circle with diameter NBRadius of semicircle with diameter AM,MN,NB= 7/2Area of semicircle with diameter AM,MN,NB=πr²/2= [(22/7)×7/2×7/2]/2= 77/4 cm²Area of shaded region = Area of semicircle with diameter AB - (Area of semicircle with diameter MN) +(Area of semicircle with diameter AM+Area of semicircle with diameter NB)= 693/4 -(77/4) +(77/4 +77/4)= 693/4+77/4= 770/4= 192.5 cm²Hence, the area of the shaded region is 192.5 cm² |
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2.Find the surface area of a sphere of diameter:14 cm(ii) 21 cm(irj 3.5m |
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| 80600. |
Exapr le 3 : In right angled ΔΡΟΕ, <p is a right angle and PM is the altitude on the hypotenuse.If PQ 8, PR 6. find PM. |
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