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.
| 65701. |
Solveforn,if(15-n):-5!(12-n)? |
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| 65702. |
he ratio of zine and copper in an alloy is 7:9. If th1.7 kg, find the weight of zinc in the alloyc weight of copper in the alloRATIO AND PROPSRTIO |
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Answer» thanku sir sorry mam |
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| 65703. |
20Define the following(a Metal refining(b) Alloy(c) Amalgum |
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Answer» Metal refining:-In metallurgy, refining consists of purifying an impure metal. It is to be distinguished from other processes such as smelting and calcining in that those two involve a chemical change to the raw material, whereas in refining, the final material is usually identical chemically to the original one, only it is purer. Alloy:-An alloy is a combination of metals or of a metal and another element. Alloys are defined by a metallic bonding character. An alloy may be a solid solution of metal elements (a single phase) or a mixture of metallic phases (two or more solutions). Amalgam:-An amalgam is an alloy of mercury with another metal, which may be a liquid, a soft paste or a solid, depending upon the proportion of mercury. |
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| 65704. |
howThe ratio of copper and zinzinc does it contain?c in an alloy is 5: 7. If the alloy contains 35.5 g of copper, |
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Answer» Let amount of copper be 5x and zinc be 7x Given5x = 35.5x = 7.1 Weight of zinc = 7*7.1 = 49.7 gm thanks |
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| 65705. |
(3) If n (A)= 15, n (AUB)-29, n(An B)-7, then n(B)? |
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Answer» n(A U B) = n(A) + n(B) - n(A ∩ B)n(B) = n(A U B) + n(A ∩ B) - n(A) = 29 + 7 - 15 = 21 |
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| 65706. |
(3) If n(A) 15, n(AU B) 29, n(AnB) 7, then n(B)? |
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Answer» We know, n(A U B) = n( A) + n(B) - n(A ∩ B)So,29 = 15 + n(B) - 29n(B) = 29 + 14n(B) = 43 |
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| 65707. |
(जीवन से26. यदि . = 0:2689 हो, तो oo(a) 02689 (b) 2:689, का मान क्या होगा?(c) 2689 |
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Answer» b is the right answer |
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| 65708. |
PART-Aof 5 students in a subject out of 50 are 32,48, 50, 27,37.Ff marks. |
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Answer» marks in increasing order: 27,32,37,48,50range: 27-50 |
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| 65709. |
AB =AC in triangle ABC.If angle A =50,find angle B and angle C |
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| 65710. |
In triangle PQR, if PQ=QR and angle Q=50°. Find angle P and angle R. |
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Answer» angles opposite to equal sides are equal.so P and R are equal.so if Q=50P+R=130so each angle is 65° |
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| 65711. |
area covered by a circular plate whose radius is 4.9 cm. |
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| 65712. |
Find the area of a circular park whose radius is 4.5 m. |
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| 65713. |
-In a box the ratio of nails to screws is 7: 6. If there are 12 screwhow many nails are there? |
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| 65714. |
अभ्यास 3(b)'म केन्द्रीय खाने की संख्या उसके चारों ओर के पटलों पर अंकित संख्याओं से किसी(?) वाली संख्या ज्ञात कीजिए :X-15)(ख) 33.\21X11/(ग)।(ii) 2(iii) 3(iv) -3 |
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Answer» I don't know of this answer |
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| 65715. |
6. Find the area of a circular park whose radius is 4.5 m |
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Answer» Like if you find it useful |
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| 65716. |
1. यदि १७* * * * 6) तथा--20 का म. सा.(४) 2 है, तो| 1. यदि (x + 5| मान होगा।+ 6) तथा (12-x-) का म. स. { ।का+ 2) है, तो-k) का म. स. ((पॉलिटेक्निक 2011] |
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| 65717. |
The measure of the angles of a triangleare in the ratio 2:7:11. Measures ofangles are:एक त्रिभुज के कोणों का माप का 2:7:11 अनुपातहै, तो कोणों का माप ज्ञात करें। |
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Answer» 18 63 and99 right ans |
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| 65718. |
1. Observe the measures of pots in figure A and B. How many jugs ofwater can the cylindrical pot hold?35m10cm-7 emA. Conical water jugB. Cylindrical water pot |
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| 65719. |
two complementary angles are such that their ratio 2:7 find the angle |
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| 65720. |
2.A and B are two alloys of gold and copperprepared by mixing these metals in the ratio7:2 and 7:11 respectively. If equal quantitiesof alloys are melted to form a third alloy C theratio of gold and copper in the alloy C?2-a) 5:7b) 7:5c) 2:5d) 5:2e) of these |
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Answer» Solution Let 1 kg of each one of A and B be taken to form 2 kg of C.Gold in 1 kg of A = 7/9, copper in 1 kg of A = 2/9Gold in 1 kg of B = 7/18, copper in 1 kg of B = 11/18Gold in 2 kg of C = (7/9 + 7/18) = 21/18 = 7/6Copper in 2 kg of C = (2/9 + 11/18) = 15/18 = 5/6Ratio of gold and copper in C = 7/6 : 5/6 = 7:5 please like the solution 👍 ✔️ |
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| 65721. |
tary angios are in the ratio 2:7. Find the tmeazures ofangles. |
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Answer» Let one of the angles be x° and the other supplementary angle be (180 - x)°. It has been given that these two angles are supplementary and are in the ratio 2:7. ∴x/(180-x)=2/7=>7x=2(180-x)=>9x=360=>x=40 ∴the two angles are X equals to 40° and the other angle equals to (180°- x) equals to (180°-40°) equals to 140°. |
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| 65722. |
In each of the following , which ratio is smaa) 7: 13 or 2:7) 9:7 or 22 : 15 |
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Answer» a) 2:7 is smallerb) 22:15 is smaller |
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| 65723. |
(d) Simran's latner is I yeals ao) In an isoceles trängle, the base angles are equal. Each of the base angle is doublevertex angle(f) The sum of two consecutive odd numbers is 44 |
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| 65724. |
. 4F (рек) sin = 3 for some angle 6. |
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| 65725. |
f O is centre of a circle, PQ is a chord and the tangent PR at P makes an anglef 50 with PQ, then find LPoQ |
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Answer» 90-50=40 angle poq=180-80=100 |
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| 65726. |
sExercise 13.1If an angle of a parallelogram is two-thirds of its adjacent angle, farallelogram.driloteral are 1100 689 829 resnectively. Find t |
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Answer» Let x and y be the two angles of a parallelogram.It is given that one angle istwo-third of its adjacent angle. So, we assume that angle"x" is two-third of angle "y" .which can be written as,x = (2/3). y ........ (1)We also know that the adjacent sides of a parallelogram are supplementary. It means that the sum of adjacent angles is equal to 180.Hence, x + y = 180 ........ (2)Put the value of x from equation (1) in equation (2):(2/3) y + y = 180(2/3 +1) y = 180(5/3) y = 180y = 180. (3/5)y =108Now put this value in equation (2) to get the value of x:x + 108 = 180x = 180 - 108x = 72Hence, the adjacent angles of a parallelogram are 72° and 108° |
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| 65727. |
9. A machine manufactures 5,782 screws every day. How many screws will it manufacture inthe month of April ? |
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Answer» the day in the month of April=30a machine manufacturers 5782 screws in a day,so total screws =30*5782=173460 In one day number of screws manufactured by machine = 5782 In month of April number of days are 30 Therefore, Number of screws manufactured in month of April= 30*5782= 173460 |
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| 65728. |
machine, on an average, nit produce in the month of January 2006?%, Aanufactures 2,825 screws a day. How many screws did |
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| 65729. |
11. Infront of a house, flower plants are grown in a circular quardantshaped pot whose radius is 2 feet. Find the area of the pot in whichthe plants grow. (1 = 3.14) |
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Answer» area=πr^2=3.14×2×2=12.56 area=nr^2=3.14×2×2=3.14×4=12.56 area =π^2=3.4x2x2=12.56 Area of the pot = πr squareArea of the pot = 3.14 × 2×2Area of the pot = 3.14×4Area of the pot = 12.56 sq. feet area= |
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| 65730. |
Observe the measures ofpots in figure 7.8 and 79.i How many jugs of watercan the cylindrical pot10 cm10 cm-1.2.cFig 7.9Fig 7.8conical water jugcylindrical water pot hold? |
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Answer» thanks |
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| 65731. |
(4) x2 + (x + 5)2-625 |
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| 65732. |
- - , und ZDUC and ZAUD.Two lines AB and CD intersect at O.(1) If ZAOC + ZBOD = 250°, find ZAOC, ZBOD, ZAOD and ZBOC(ii) If ZAOC + ZBOC + ZBOD = 320°, findZAOC, ZBOC, ZBOD and ZAOD(iii) If ZAOC = (2x - 3)º and ZBOC = (3x - 2)°, findZAOC, ZBOC, ZBOD and ZAOD. |
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| 65733. |
6.संलग्न चित्र में ZAOB तथा ZBOC एकरैखिक युग्म बना रहे हैं। यदि /AOB=60° होतब ZBOC की माप क्या क्या होगी? |
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Answer» AOB+BOC=180°60°+BOC=180°BOC=180°-60°BOC=120°60°+BOC =180° BOC=180°-60° |
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| 65734. |
f'manravels65kmin3days by walking 7 hours in a day, in how manydays will he tuavel2km by walking 8 hours a day? |
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Answer» Every day he walks 7.5 hrsSo in three days he walks 7.5×3 =22.5hrsAnd he travelled 65km in these 22.5hrsSo the speed = distance/time = 65/22.5 = 2.88km/hr Speed of the man will remain same,So time = distance/speed = 156/2.88 = 54 hrs Everyday he walks 8hrs, so for 54 hrs, he has to walk 54/8 = 6.75 days |
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| 65735. |
6. If a man travels 65 km in 3 days by walking 7hors a day, in how many days will he travel 156 kmby walking 8 hours a day? |
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Answer» Every day he walks 7.5 hrsSo in three days he walks 7.5×3 =22.5hrsAnd he travelled 65km in these 22.5hrsSo the speed = distance/time = 65/22.5 = 2.88km/hr Speed of the man will remain same,So time = distance/speed = 156/2.88 = 54 hrs Everyday he walks 8hrs, so for 54 hrs, he has to walk 54/8 = 6.75 days |
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| 65736. |
2.What is the name of angle, if the angle is more than 180°?MORE THAN 180BUTLESS THAN 360a) Acute anglec) Obtuse angleb) Reflex angled) Straight angle |
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Answer» option b ) reflex angle hit like if you find it useful this is a obtuse angle |
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| 65737. |
\angle A + \angle B + \angle C + \angle D + \angle E + \angle F |
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Answer» In triangle ABE and triangle DCF,ANGLE A+ANGLE B+ANGLE E=180(ANGLE SUM PROPERTY)-(1)ANGLE D+ANGLE C+ANGLE F=180(ANGLE SUM PROPERTY)-(2) (1)+(2)ANGLE A+ANGLE B+ANGLE E+ANGLE D+ANGLE C+ANGLE F=180+180ANGLE A+ANGLE B+ANGLE C+ANGLE D+ANGLE E+ANGLE F=360 |
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| 65738. |
Find the angle which is equal to its complement.i) Pind the angle which is equal to twice of its complement,Find the angle which is equal to four times its supplement.loe re in the ratio 2 7. Find the angles. |
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Answer» i) let the angle be x so x+x =90 which x= 45 I want the answer of 3rd part the angle which is equal to its complement is 45 |
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| 65739. |
In Fig 10. 16, 20AB = 3or and ZOCÂť-57. Find ZBOC and ZAOCFe. 10.16 |
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| 65740. |
x contains 600 screws, one-tenth are rusted. One screw is takenfrom this box. Find the probability that it is a good screw.(0 yellow ball?()red ball?out at rand2 A bod 995 blank tickets. A person buys a loe |
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Answer» If 1/10 are rusted in a box of 600 screws it means that 60 are rusted.the probability of finding a food screw is:600-60/600=540/600=9/10so the probability of finding a good screw is 9/10 thanks sir |
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| 65741. |
From the given figure, find65°O(a) ZAOD(b) ZBOC(c) ZDOC |
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Answer» 《AOD=115《BOC=115《DOC=65 |
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| 65742. |
(6 9/ e ;रहlfgdarce g it न fhow |
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Answer» P(x,y) O (0,0) PO = √(x -0)² + (y-0)² PO = √(x² + y²) |
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| 65743. |
8. Find the probability of gettinga'Y' on spinning the wheel.оо |
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Answer» Total possibility = 4 A = Event of getting Y = 1 P(A) = 1/4 |
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| 65744. |
(4) x2 + (x + 5)2 = 625 |
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| 65745. |
EXERCISE 5.3t the outcomes you can see in these experimentsSpinning a wheel(b) Tossing two coins togetheAİB |
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| 65746. |
A man travel from A to B at 20 km/hr. andcome back from B to A at 30 km/hr. Find theaverage speed of the whole journey? |
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Answer» let the distance between A to B be x kmtime(AB)=x/20 hrtime(BA)=x/30 hraverage speed= total distance/ total time =(2x)/(x/20 + x/30) =(2×20×30)/(20+30) =120/5=24 km/hr |
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| 65747. |
Twould angle of a quadrilateral are measure 65 each and the other two angle are equal .Wath is the measure of either of these two angle |
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| 65748. |
A motorbike rider covered 200 km in 3 hours.He then continued with the journey at a speedof 60 km/h for 2 hours. What was hisaverage speed for the whole journey? |
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Answer» At first he goes 200km. Now he goes 2 hours in 60km/h in second time he goes(60*2)=(120)km so his average velocity is =(200+120)/(3+2)=320/5=64km/h 64 km/hr is the best answer of the que |
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| 65749. |
two angles are adjacent and form an angle of 150.The larger angle is 30 less than two times the smaller.Find the angles |
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| 65750. |
Example 6: Prove that the quadrilateral formed (if possible) by the internal anglebisectors of any quadrilateral is cyclic. |
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Answer» Given: ABCD is a quadrilateral AH, BF, CF, DH are bisectors of∠A,∠B∠C∠D respectively To prove: EFGH is cyclic quadrilateral Proof: To prove EFGH is a cyclic quadrilateral, we prove that sum of one pair of opposite angles is 180 degrees. in triangle AEB ∠ABE +∠ BAE +∠AEB = 180 degrees ∠AEB = 180 degress -∠ABE -∠BAE ∠AEB = 180 degress (1/2∠B +1/2∠A) ∠AEB = 180 degrees -1/2 (∠B +∠A) ....(1) Now Lines AH and BF intersect So∠ FEH =∠AEB ∠ FEH = 180 degrees - 1/2 (∠B +∠ A ) ....(2) Similarly we can prove that∠FGH = 180 degrees = 1/2 (∠C +∠ D).... (3) addding(2) and (3) ∠FEH +∠ FGH = 180 degrees -1/2 (∠A +∠D) +180 degress - 1/2 (∠C +∠ B) ∠FEH +∠FGH = 180 degrees + 180 Degrees - 1/2 (∠A +∠D +∠C +∠B) ∠ FEH +∠FGH = 360 degrees - 1/2 (∠A +∠B +∠C +∠D) ∠ FEH +∠ FGH = 360 degrees -1/2 (∠A +∠B +∠C +∠D) Since ABCD is a quadrilateralSum of angles of Quadrilateral = 360degress∠A +∠ B +∠c +∠ D = 360 Degrees ∠FEH +∠FGH = 360 degrees -1/2 x 360 degrees ∠ FEH + FGH = 360 degrees - 180 degrees ∠ FEH +∠ FGH = 180 degrees Thus in EFGHSince sum of one pair of opposite angles is 180 degrees |
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