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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.
| 1851. |
Convert into percentage.502 Convert into fractions and2(b) 24 |
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| 1852. |
1. Convert the following fractions into percentage(a)VE) 13IN(d)102Convert the fort |
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Answer» 1a 80%b 60%c 75%d 18%e 140%f 110% 80%60%75%18%140%110% |
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| 1853. |
indl9gel tol |
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| 1854. |
95. What would be the simple interest obtainedon an amount of Rs. 8,930 at the rate of 8% perannum after 5 years? |
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Answer» answer |
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| 1855. |
15Which answer is equal tol? |
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Answer» We know i² = - 1 So, raise 8 both sides i^(16) = 1 Second option is correct |
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| 1856. |
22. If LLTS = 38", find the external < TOL. |
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Answer» As RS is tangent to OT then OTS is 90°.so OTL+ LTS = 90°thereforeOTL = 90° - 38° = 52°angle OTL = angle OLT as OT = OL (radius)angle OTL = 52° please send the correct answer extrerior angle TOL |
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| 1857. |
A sum of at compound interest amounts to Rs.6600 in 1 year and to Rs.7986 in 3 years. Find the sum and rate of interest. |
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| 1858. |
2.Find the sum which amounts to? 1352 m 2 years at 4% compound interest. |
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Answer» thanks bro |
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| 1859. |
7. A principal amounts to 213924 in 2 years by compound interest at 18 p.c.p.a.Find the principal.of increase in the |
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Answer» Given, amount, A = ₹ 13924 time period , n = 2 interest rate , r = 18 % per annum Let principal is P use formula, so, 13924 = P(1 + 18/100)² => 13924 = P(118/100)² => P = 13924 × 10000/118 × 118 P = 10,000 hence, principal is ₹ 10, 000 |
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| 1860. |
DatePage4 km. Rajahat 6:50pm.0 minutes.1. Convert into m/s :(a) 54 km/hr(c) 126 km/hr2. Convert into km/hr. |
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Answer» first answer is a 5m/s 126 km/hr is correct answer second answer is 7m/s 54*5/18=15m/sec126*5/18=35m/sec (a)54000m/s (b)126000m/s a. 54/36 m/sb. 126/36m/s 54*5/18 = 15 m/ s126*5/18 = 35m/ s 15 54 × (5/18 ) = 15 m/s126 × (5/18 ) = 35 m/s are the right answer 126 km this is correct answer please like 15 m/s 35 m/s is the best answer. |
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| 1861. |
A invested 125% as much money as B. C invested 80% as much money as B. The total of all the three is 61000. How much did C invest ?(A) ₹25000 (B)₹16000 (C) ₹20000 (D)₹45000 |
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Answer» Let B invests Rs. xSo, investment by A= 125% of x = 125x/100Investment by C= 80% of x= 80x/100total investment = 125x/100 + x + 80x/100= 305x/100Now , 305x/100= 61000which gives x=20000so, investment by C = 80* 20000/100= 16000 |
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| 1862. |
(a) 120(b) 110(c) 135 |
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Answer» माना बाद संख्या x thenx-(27.5/100*x)= 87x= 120 |
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| 1863. |
a man invested 2500012%per annum compounded half yearly.what amount would he get after a year???? |
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Answer» 25000(1+12/100)² = 25000(112*112/100*100) = 31360 rupees |
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| 1864. |
find the compound interest in rupees 25000 for one and half years at 8per pa when the interest is compound ef half yearaly |
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| 1865. |
activity 1: Amita inte pear. Altogetherwo different ratesita invested some part of 35000 rupees at 4% and the rest at 5% interest forother her gain was Rs. 1530. Find out the amounts she had invested at thent rates. Write your answer in words.Invested Rs. x at 4% rateInvested Rs. y at 5% rate>42 +52 = 35000Interest100* + jรณo = 1530 ... (1)- U21v=207 |
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Answer» answer s 60 is the answer |
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| 1866. |
Sandeep deposited 25000 rupees in a bank which pays 8% interestwould he get back after two yeans?compounded annually. How much would he get back after two years?(I) |
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| 1867. |
3 Shweta, Piyush & Nachiket together invested 8000rupees & Started a busiess of selling Sheets & towefrom solapur. Shweta's sone of th capital was.30000 rupees & Piyush's 12000. At the end of theyear they had made a profit of 24% whatwas Nochiket's investment & what was hisShare of the profet? |
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Answer» Nachiket's investment was 38,000 rupees and share of his profit was 9,120 rupees. Step-by-step explanation: OK.So lets begin. . . So Shweta, Piyush and Nachiket invested 80,000 rupees and started a business. ∴ Shweta & Piyush's total share is = (30,000+12,000) = 42,000 rupees. ∴ Nachiket's share should be = (80,000-42,000) = 38,000 rupees. Now, ∴ Ratio of Shweta,Piyush and Nachiket's share = 30,000 : 12,000 : 38,000 = 15 : 6: 19 15+6+19 = 40. ∴ Profit of 24% on 80,000 rupees = 19,200 rupees. ∴ Nachiket's profit on 19,200 rupees was = 19200×19÷40 = 9,120 rupees. (19 is ratio of Nachiket's share and 40 is the sum of the share ratios of the three person). your answer is 192000.... 9,120 is the profit.... |
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| 1868. |
amount to rea10220!s. An amount of money invested trebled in 6 years. Find the rate of interestearned. |
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| 1869. |
35.The simple interest obtained when a certain sum ofmoney is invested in scheme A for 3 years is ?38.8less than the simple interest obtained when thesame sum of money is invested in scheme B for 4years. If the rates of interest offered by scheme Aand scheme B are 14% pa and 12.5% parespectively, how much money was invested inthese schemes individually? |
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| 1870. |
7. A sum of money invested at 8%12122 in 2 years. What willper annum amounts toamount to in 2 years 8 months at 9% per annum? |
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| 1871. |
17. A sum of money invested at 8% per annum amounts toamount to in 2 years 8 months at 9% per annum?12122 in 2 years . What will it |
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Answer» Thanks |
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| 1872. |
2A sums of money invested at 8% per annum amounts toamount to in 2 years 8 months at 9% per annum?12 122 in 2 years. What wilt |
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| 1873. |
3. (a) A sum of money invested at compound interestamounts to Rs. 2,200 in one year and to Rs. 2,662in three years. Find the principal and the rate percent |
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| 1874. |
17. A sum of money invested at 8% per annum amounts to? 12122 in 2 years . What will itamount to in 2 years 8 months at 9% per annum? |
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| 1875. |
In a retail market, fnsit veados were selling oranges kept in packing baskets. Thesebaskets contained varying number of oranges. The following was the distribution oforangeNumber of oranges 10-14Number of baskets 1525-29 30-342515-19 20-241101351iSFind the mean sumber of onanges kept in cach basket. Which method of finding themsan did you choose |
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| 1876. |
reduction of 50 paise per dozen in the price of orange means that adozen more orange can be bought for R 78. Find the original price oforange per dozen is. |
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| 1877. |
of part of profit as an active partner. Remaining profit is distributed1023. In a business, 'x' receivedbetween âX' and 'Y, in the ratio 4:5. If part of the profit of 'X' is Rs. 325, then the part of the profit "Yis? |
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Answer» Thankzz |
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| 1878. |
Write the answer in simplest form(i) Mira ate four-ninths of her orangebefore lunch and two-ninths of herorange after lunch. How much of horange did she eat in all? |
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Answer» 4/9+2/9=6/9 is the correct answer sir answer me diya hai iska Mere book Kaise Sapne Diya Hai 2 by 3 correct answer properly nahi aa raha hai |
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| 1879. |
At the beginning of the season, MacDonald had to remove 5 orange trees from his farm. Each of theremaining trees produced 210 oranges for a total harvest of 41790 oranges.Write an equation to determine the initial number of orange trees (t) on MacDonald's farm.Find the initial number of orange trees on MacDonald's farm.orange trees |
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Answer» Let there be x orange trees in the beginning(x - 5) × 210 = 41790 x - 5 = 41790/210 = 199 x = 199 + 5 = 204 |
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| 1880. |
-garden has 2000 trees of these areorange trees. Find the number of orangethree |
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| 1881. |
A.Complete the table.S.NoCP (R)SP (R)Profit or Loss (R)Profit % or Loss %1.2.3.4.800010,000Os6890390 (profit)1000 (loss)25,00048,560 19719-28,3202000 (profit %)26,1963400416.50 (profit)7820800 (loss %)6245749.40 (loss) |
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| 1882. |
Complete the table.A.CP (R)8000S.No.SP R)10,0006890Profit or Loss ()Profit % or Loss %1.2.3.4.5.6.390 (profit)1000 (loss)25,00048,56028,320340020% (profit %)26,196416.50 (profit)800 (loss %)7820749.40 (loss)8.6245 |
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| 1883. |
15. Two complementary angles differ by 20°. Find the measure of each angle. |
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Answer» Sum of two complementary angles= 90°let one of the complementary angles be x and second angle be x-20then a/q x+ x-20=90 =2x = 90+20 = x= 110/2 ∴x= 55so measure of one angle is 55°andthe second angle is 55-20= 35°and 55+35= 90° |
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| 1884. |
A shopkeeper sells a pen set at 20% profitto another shopkeeper who sells it at a lossof 20%. If price of pen set was Rs.112,what is net profit, or loss on totaltransaction?2% profit4% loss(A)2% 10(B).(C)4% profit |
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Answer» Cost price of pen set CP = 112First shopkeeper sell it for SP1= 112(1 + 20/100) = 112(6/5) = 134.4 Other shopkeeper sell it forSP2 = 134.4(1 - 20/100) = 134.4(4/5) = 107.5 Total loss % in transaction=[ (CP - SP2)/CP]*100= 4.5*100/112= 4.01 THERE IS LOSS OF 4% |
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| 1885. |
Two complementary angles differ by 10 degree. What is the larger angle? |
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| 1886. |
[3] Cost price of cell phones 72500Selling price of cell phonesNet Profit-S.P.-C.P.2750= 2750- 2500= 250Profit x 100C.P250x1002500% Profit = |
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Answer» Cost Price of cell phone = 2500 Selling price of cell phone = 2750 Net Profit = S.P. - C.P. = 2750 - 2500 = 250 % Profit = Profit×100/C.P. = 250×100/2500 = 10% |
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| 1887. |
Assess Yourself 7.2percentage profit or loss :1. CP = 28, SP = * 352. CP = 60, SP = 503. CP = 40, Profit = 54. SP = 30, Profit = 25. CP = 16.25, SP = * 18.50 6. CP = 790. Profit = *2Simple Interestwe borrow or lend money on interest the following terms are involved : |
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| 1888. |
42. A man makes a profit of 20%cent if he sells them for 2840?I be his profita profit of 20% by selling sweets for 960. What will be hi |
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Answer» So 120 percentage = 9001 percentage = 900/12In 900 = 120 percentageIn 1 rupee = 120/900 percentageIn 840= 120/900*(840)% cp. sp100. 120x. 960100/x= 120/960100/x= 1/8x= 800cp = 800₹ sp = 840profit= sp-cp = 840-800= ₹ 40% profit = 100×profit /cp = 100×40/800 . = 5% increases by profit 5% this is answer |
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| 1889. |
Asels some articles to B at a profit of 5%Bsells the same to C at a profit of 20% and Csells them to D at a profit of 25%. What wouldA's percenitage of profit be, if he sold thearticles directly to D, at the price D paid forthem? |
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| 1890. |
4. Find the area of the following polygon, ifAL=10cm, AM = 20 cm, AN = 50 cm,AC): 60 cm and AD 90 cm.20cm60 cmE T..W" 가)30 cm40 cm |
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| 1891. |
In 1980, there was a net profit of Rs. 46.67crores to Oil and Natural Gas CommissionAfter seven years in 1987 profit wasamounted to Rs. 1508 crores. If the profit isreckoned yearly, find the rate of profit peryear.1508 746-571-64345uiven(A) 64.8%(C) 66.275%(B) 64.345%(D) 76.28%0, Mo |
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| 1892. |
36 cm0.4. Find the area of the following polygon, if50 cm,AL10 cm, AM20 cm, ANAO-60 cm and AD = 90 cm.20 cm60 cm30 cm40 cm |
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| 1893. |
44. Find the area of the following polygon, ifAL-10 cm. AM = 20 cm, AN-50 cm,AO-60 cm and AD = 90 cm.20 cm60 cmN i30 cm40 cm |
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| 1894. |
d the area of the following triangles whose sides are60 cm, 153 cm and 111 cmiii) 50 cm, 78 cm and 112 cm(ii) 5 m, 7 m and m(iu) 60 cm, 61 em and 1l on |
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Answer» formula Area of Triangle 1/2b*h |
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| 1895. |
.4. Find the area of the following polygon, jrAL-10 cm, AM-20 cm, AN-50 cm,AO 60 cmand AD 90cm20 cm60 cm30 cm40 em |
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| 1896. |
8. Triangle ABC is right angled at A. AD is drawn perpendicular to BC. If AB = 5 cm andAC12 cm, find the area of AABC. Also find the length of AD. |
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| 1897. |
pole, given that Iemppl7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standinga hemisphere of radius 60 cm is placed upright in a right circular cylinder full of wesuch that it touches the bottom. Find the volume of water left in the cylinder, if the radiaof the cylinder is 60 cm and its height is 180 cm.? cm in diameter, the diame |
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| 1898. |
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing ona hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water, suchthat it touches the bottom. Find the volume of water left in the cylinder, if the radius of thecylinder is 60 cm and its height is 180 cm. |
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Answer» We have,Radius of cylinder = Radius of cone = Radius of hemisphere = 60 cmHeight of cone = 120 cmHeight of cylindrical vessel = 120 + 60 = 180 cm Volume of cylinder =πr^2h =22/7 x(60)^2x180 Now, volume of the solid= Volume of cone + Volume of hemisphere =1/3*22/7*3600*120+2/3*22/7*60^3Volume of water left in the cylinder= Volume of the cylinder – Volume of the solid=22/7*60^2*180-1/3*22/7*60^2*120-2/3*22/7*60^3=22/7*3600(180-40-40)=7920000/7=792/700m^3 |
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| 1899. |
Fig. 10.1112. The sides of a right-angled triangle containing theright angle are Sx and (3x -1) cm. If the area of thetriangle be 60 cm2, calculate the lengths of the sides(ICSE)of the triangle. |
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| 1900. |
The lengths of two sides of a right-angled triangle containing the right angle differ by2 em. If the area of the triangle is 24 cm2, find the perimeter of the triangle.13Tf the diagonal of the square |
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Answer» Thanks |
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