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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.
| 78001. |
Divide a rope of length 560 cm into 2 parts such that twice the length of the smallers equal to 3 of the larger part. Then find the length of the larger part. |
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Answer» So the larger part is 480 cm. So larger part is 420cm |
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| 78002. |
66) A plane left 40 minutes late due to bad weatherand in order to reach its destination, 1600km awayin time, it had to increase its speed by 400km/hrfrom its usual speed. Find the usual speed of theplane. |
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Answer» Distance = 1600 km Let the usual speed be = x km/h now, we know that speed = distance /timetime = distance / speed Usual time = 1600/x Now, due to bad weather the speed is increased b 400 km/ hr which means new speed = 1600/x+400 Now it's given that the plain left 40 minutes late which means new time = 1600/(x+400) + 2/3 (40/60 = 2/3 hours) 1600/x = 1600/(x+400) + 2/3 1600/x-1600/(x+400) = 2/3 1600(x+400)-1600x/x(x+400) = 2/3 1600x + 640000-1600x/x²+400x = 2/3 640000 = 2/3(x²+400x) 640000 * 3/2 = x²+400x 960000 = x² +400x 0 = x²+400x-960000 0 = x² +1200x-800x-960000 0 = x( x+ 1200) - 800(x+1200) 0= (x+1200)(x-800) x= -1200 or x= 800 Speed cannot be negative∴ usual speed = 800 km/h |
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| 78003. |
Find the'selling price if the cost price is 1200 and loss percent is 25 |
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Answer» CP = ₹ 1200 Loss % = 25 % So, Loss price = 25% of 1200 = 25/100 × 100 = 25 So, SP = CP - Loss = ₹1200 - ₹25 = ₹1175 |
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| 78004. |
find gain or loss percent when:CP=250, overhead expenses= 80 and SP= 600 |
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| 78005. |
Hari bought rice worth750 and spent 50on transport and packing. He sold three-fourths of it at a loss of 10% and theremainderat a gain of 10%. Find his gain orosson the whole transaction.Afan is sold for 644. The gain is one-sixthf the cost price of the fan. Find the gainer cent. |
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Answer» Total 8003/4 of 800=600600 loss of 10%10/100*600=60he sold for 540 then 200 for 10%gainhe sold that for 220 rs total rice is sold for 540+220=760 |
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| 78006. |
10, Anil gets 85% marks in an examination. Ifhis actual marks are 425, what are the totalmarks of the examination? In the sameexamination, Sunita gets 360 marks. Findher percentage. |
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Answer» = Let the total number of marks in the examination = x. = Anil gets 85 % of marks, which is equal to 425. Hence,= 85 / 100 * x = 425 = x = 425 * 100 / 85 = 500 marks. = Sunita scores 360. Her percentage: = 360 / 500 * 100 = 72 %. |
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| 78007. |
An electric bulb is connected to a 220 V power supply line. If the bulb draws a current of 0.5 A, calculate the power of the bulb. |
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Answer» V=220I=0.50A P=V×IP=220×0.50P=110W Like my answer if you find it useful! |
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| 78008. |
If= , then the value4) 27of x is:(2) 2(3) 4(4) 3 |
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| 78009. |
what is the comman difference of an a.p in which a21-a7=84 |
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Answer» Let a is the first term and d is the common difference of an APa21-a7= 84a21= a +20da7= a + 6da + 20d - a - 6d= 8414d = 84d = 84/14d = 6 |
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| 78010. |
9. In the adjoining figure, ABCD is a parallelogramwhose diagonals intersect each other at O. A linesegment EOF is drawn to meet AB at E and DC atF. Prove that OE OF. |
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| 78011. |
4. A rectangular hall is 12 m long and 6 m broad. Its flooring is to be made of squaretiles of side 30 cm. How many tiles will fit in the entire hall? How many would berequired if tiles of side 15 cm were used? |
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Answer» L=12m=12×100cm=1200cm. (1m=100cm) b=6m=6×100cm=600cm area of room=l×b 1200×600=720000cm² a=30cm area of square tile=a×a 30×30=900cm² tiles required=area of room/area of tiles 720000/900=800tiles required in case of 15 cm tiles a=15cm area of tile=15×15=225cm² required tiles=area of room/area of tile 720000/225=3200tiles required |
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| 78012. |
4. A rectangular hall is 12 m long and 6 m broad. Its looring is to be made of squarstiles of side 30 cm. How many tiles will it in the entire hall? How many would brequired if tiles of side 15 cm were used? |
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Answer» l=12m=12×100cm=1200cm. (1m=100cm) b=6m=6×100cm=600cm Area of room=l×b 1200×600=720000cm²a=30cm Area of square tile=a×a30×30 = 900cm² Tiles required=area of room/area of tiles720000/900=800 tiles In case of 15 cm tilesa=15 cmarea of tile=15×15=225 cm² Required tiles= area of room/area of tile720000/225= 3200 tiles required thanks |
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| 78013. |
Divide Rs 1500 into two parts so that 10% of the larger part exceeds 8% orsmaller part by Rs 60. The value of larger and smaller parts are(a) Rs 1200 and Rs 300 (b) Rs 850 and Rs 650(c) Rs 900 and Rs 600 (d) Rs 1000 and Rs 50010. |
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Answer» Let large part be xThen smaller part = 1500-xAs per given conditionx(10/100) = (1500-x)8/100 + 60x(18/100) = 15*8 + 60x (6/100) = 5*8 + 20x (3/50) = 40 + 20x = (60/3)*50x = 1000 Large part x = 1000Smaller part (1500-x) = 500 |
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| 78014. |
In the given figure, two circles toucheach other at the point C. Prove thatthe common tangent to the circles at C,bisects the common tangent at P and QP RА-С -В |
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| 78015. |
RS. I JUU, 1 .D (0)able together for Rs. 760 thereby making a profit of 25% on18. A man sold a chair and a table together for Rs. 760 thereby makithem together for Rs. 767.50, he would have made achair and 10% on table. By selling them together for Rs. 767.50. he wprofit of 10% on chair and 25% on table. Find the cost price of each(A) Rs 300, Rs.350 (B) Rs.250, Rs.200 (C) Rs.250, Rs.350 (D) Rs 200 25o nart of the monthly expenses of a family is constant and the remaining varies with theprice of wheat. When the rate of wheat is Rs. 250 a quintal, the monthly expenses of thily is Rs 1000 and when it is Rs. 240 a quintal, the monthly expenses is Rs. 980 Finaly expenses of the family on wheat when the cost of wheat is Rs. 350 a quinta(A) Rs.2 Q, Rs. 70 0(B) Rs. 3 Q, Rs.60 Q(C) Rs. 1 Q, Rs. 50 Q(D) Rs.4 Q, Rs 70 QA plane left 30 minutes later than the scheduled time and in order to reach the destina500 km away in time, it has to increase the speed by 250 km/hr from the usual seind its usual speed.3) 600 km/hr (B) 650 km/hr (C) 700 km/hr (D) 750 km/hr |
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Answer» 1 st a 2 st b 3 st c vvddfggggddfffhgffffffff |
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| 78016. |
Find gain or loss percent if,C.P.=Rs.600 , S.P. = Rs. 750 |
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Answer» GivenCP = ₹600SP =₹ 750SP > CPso there is a profit.Profit = SP - CP = 750 - 600 = ₹150So profit % is150/600 × 100 = 25%So there is a profit of 25% |
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| 78017. |
an article is bought for Rs. 600 and sold for Rs. 750.find the gain%. |
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Answer» Total Profit = 750 - 600 = 150 rs Percentage of Profit = 150/600 = 25% |
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| 78018. |
51.A gottwice as many marks inEnglish as in Science. His total marks in English, Science and Mathemali I180. If the ratio of his marks in English and Mathematics is 2:3, what is his marks in Science?(1) 20(2) 60(3) 30(4) 40- |
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Answer» Answer: 3) 30 Like if you find it useful |
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| 78019. |
The sum of thind and serenth termm of an AP is s adtheir produch is & find the Airst ten and treLurDterm and the commanF ference OE an A.P |
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| 78020. |
7.) The difference of squares of two numbers is 180. The squareoe an the shorter side, find the sides of the neid.times the larger number. Find the two numbers.of the smaller number is 8So, a qu(i) tw |
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| 78021. |
f m long and 8 m wide?17. How many rectangular tiles 10 cm by 7 cm will be required to cover a floor 14 m by 12% m andi what will be the cost at 5 per hundred tiles?18. A lane 180 m long and 5 m wide is to be paved with bricks, of length 20 cm and breadth 15 cm.Find the cost of the bricks that are required, at the rate of750 per thousand.A. |
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| 78022. |
4. In how many years will Rs 750 amount to Rs 930 at 6% per annum? |
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Answer» S.I will be = 930-750 = 180 so, S.I = prt/100=> 750*6*T/100 = 180=> T = 1800/450 = 4 years |
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| 78023. |
10. Find the area of the following compositefigure17 cm5 cmăăź3 cm-t10 cm |
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| 78024. |
Question 10Question Id: 43317 (2 marks)Find the simple interest, when Principal = Rs 12000, Rate of Interest = 18% per amm and Time = 4 months(a) Rs 720(b) Rs 800(c) Rs 750() Rs 650 |
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Answer» Please hit the like button if this helped you |
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| 78025. |
The given figure shows two circles withcentres A and B; and radii 5 cm and 3 cmrespectively, touching each other internally. Ifthe perpendicular bisector of AB meets thebigger circle in P and Q, find the length ofPQ. |
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| 78026. |
If sec 4A =cosec(A-20°), where 4A is an acute angle, find the value of Asarty Education Management Pyt. Ltd. |
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| 78027. |
The sum of 5 and third part of a number is 25 what is the number? |
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Answer» let the no be xso 5+×/3=25 X/3=25-5 X/3 =20 X=20×3 X=60 Let the number is X,A/Q5 + X/3 = 25X/3 = 25 - 5X/3 = 20X = 20 ✖ 3X = 60 Hence, required number is 60 ( ANS ) x=60 is the correct answer |
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| 78028. |
RICE is a Rhombus. Find OE and OR. Justify your findings. |
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| 78029. |
why did helen divide the period in three parts? |
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Answer» because she knows that it is not easy to see all thing in 3 days so she wants that she can touch the thing which is most important to her |
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| 78030. |
36) The angle of elevation θ of the top ofa lighthouse as seen by a person on the ground is suchthat tan 8When the person movdistance of 240 m towards the light house, theangle of elevation becomes o, such that tan o-3Find the height of the light house.12 |
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| 78031. |
6. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cnm.Find the cost of bricks that are required, at the rate of Rs 750 per thousand |
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| 78032. |
. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre Oa point Q so that OQ-12 cm. Length PQ is:(A) 12cm(C) 8.5 cm(D) 119 cm.(B) 13 cnm |
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| 78033. |
then its area is(a) 24 35 cm (6) 12 35 cm2 (c) 6/35 cm2 (d) 150 cm13. The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest altitude is( d) 165 cm (b) 10 5 cm245 cm (d) 28 cm(NCERT Exemplar |
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Answer» In a triangle the shortest side acts as the base for the longest altitude. Here, in the given question the side of 35 cm length is the shortest. Hence it will act as the base for the longest altitude. Thee altitude divides its base into two parts.Let's assume the length of one part be x. Then the length of the second part will become (35 - x). Now by using Pythagoras theorem the length of the altitude is given by.L²= [54²-x²]........(i) and also byL²= [61²-(35-x)²]........(ii) By substracting eq(i) from eq (ii) we get61²-54²-(35-x)²+x²=070x =420x=6 Now substituting the value of x in eq (i) we getL² =54²-6²L=√2880L =24√5 (c) is correct option correct option |
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| 78034. |
5.What should be the least number of years in which2the simple interest on Rs.2600 at 61% will be an exactnumber of rupees ?(a) 2(b) 3(c) 4(d) 5 |
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Answer» Option-B SI = Pnr/100, P=Interest, n= Number of years, r = rate of interestHere P = 2600, r = 20/3For n=1 and n=2, the interest amount will be 173.33 and 346.66 respectively.For n = 3, Interest = 520The least number of years at which the interest amount is whole number is Rs.520. please like the solution 👍 ✔️ |
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| 78035. |
he second part and the ratio between thesecond and third part is 4: 5. Find each part. |
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Answer» suppose second part is xso first part=2/3xnow ratio of 2nd and 3rd part is 4:5so third part is 5/4xso 1400=x+(2/3)x+(5/4)x12*1400=12x+8x+15x=35xso x=40*12=480so first number is 2/3*480=320third number is 5/4*480=600so numbers are 480,320,600 |
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| 78036. |
TICILE LUI HUIUI 10 J.Example 3. Divide 16 into two parts such that twice the square of the larger partexceeds the square of the smaller part by 164. |
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Answer» Let two parts of 16 are x and (16 - x) As per given condition2x^2 = (16 - x)^2 + 1642x^2 = x^2 + 256 - 32x + 164x^2 + 32x - 420 = 0x^2 + 42x - 10x - 420 = 0x(x + 42) - 10(x + 42) = 0(x - 10)(x + 42) = 0x = 10, - 42 Negative value not possible Thus, x = 10 Two parts of 16 are 10, 6 |
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| 78037. |
one-thirdltd1. Divide 243 into three parts such that half of the first part, one.second part and one-fourth of the third part are all equal.s The sum of the coins is e |
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Answer» Let first part be x, second part be y, third part 243 - (x + y) As per given condition x/2 = y/3 x = 2y/3....(1) Y/3 = [243 - ( x + y)] /4.....(2) Put Value of x from eq(1) in eq(2)4y/3 = 243 - (2y/3 +y)4y/3 + 5y/3 = 2439y/3 = 2433y = 243y = 81 x = 2*81/3 = 2*27 = 54 Three parts arex/2 = 54/2 = 27y/3 = 81/3 = 27[243 - (x + y)]/4 = (243 - 135) /4 = 108/4 = 27 |
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| 78038. |
11. Divide 243 into three parts such that half of the first part, one thrd ofsecond part and one-fourth of the third part are allequal. |
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| 78039. |
Divide Rs 645 into three parts such that the firstpart is 2/5 of the second part and the ratiobetween second and third part is 4:3. |
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| 78040. |
10 a ^ { 2 } - 15 b ^ { 2 } + 20 c ^ { 2 } |
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Answer» 10a^2-15b^2+20c^2=5(2a^2-3b^2+4c^2) |
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| 78041. |
a)20cm(b)TCth, what is the breadth ofThe area of a rectangle is 460 cmthe rectangle?if the length IS 15% more than the bread(c) 34.5 cnm(d) 20 cm(a) 15 cm(b) 36 cm |
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| 78042. |
A olles Ism lona har a diarovnusuoluten will it amake to level a play goundmeastusina 30m X 33m? |
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| 78043. |
The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, findthe minimum length of the shortest side25. |
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| 78044. |
A,d,в,u,Long Answer Questions:Explain interference oflight and hence, give expression for path difference in constructive anddestructive interference pattern.Refer Short Answer Question no. 2 and 3(4 Marks)Chapter - 11 Interference and Diffraction |
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Answer» Interference of Light Waves is defined as the modification in the distribution of light energy when two or more waves superimpose each other. |
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| 78045. |
The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, findthe minimum length of the shortest side.15. |
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| 78046. |
11. The king, queen and jack of clubs are removed from a deck of 52 playing cardsand the remaining cards are shuffled. A card is drawn from the remaining cards.Find the probability of getting a card of (i) heart; (l) queen; (i) Clubs.12 A ho |
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| 78047. |
2. What number is that of which the third part exceeds the fifth part by 4? |
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| 78048. |
10. Two-thirds of a nuli1i. The fifth part of a number when increased by 5 equals its fourth part decreased by 5. Find |
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| 78049. |
a pungiven figure, find the length of sides AB and BD.b) an odd numberFrom the20ODC15B |
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Answer» As triangle ABC is right angledhenceAB^2=AC^2+CB^2hence20^2+15^2=AB^2AB=√425cm |
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| 78050. |
(C) 34Which of the following is true about17x41x43x61+ 43?(D) 4832(A) It is a prime number(B) It is a composite number.(C) It is an odd number(D) Both (A)and (C)ofGeld har acircumference of 360 |
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