26:I[81166,["/_next/static/immutable/chunks/2corl6wyj4669.js","/_next/static/immutable/chunks/38n271cv4vmyv.js","/_next/static/immutable/chunks/2pf73jja2520w.js"],"default"] 27:T66d,Fractions and percentage Let the original fraction be

\frac{x}{y}

. New numerator

= x + \frac{40 x}{100} = \frac{7 x}{5}

New denominator

= y - \frac{20 y}{100} = \frac{4 y}{5}

New fraction

= \frac{7 x / 5}{4 y / 5} = \frac{7 x}{4 y}

\frac{7 x}{4 y} = \frac{7}{4} \times \frac{x}{y}

, so the new fraction is

\frac{7}{4}

times the original fraction. Hence the correct option is (d).28:T6eb,भिन्न और प्रतिशत माना कि मूल भिन्न

\frac{x}{y}

है। नया अंश

= x + \frac{40 x}{100} = \frac{7 x}{5}

नया हर

= y - \frac{20 y}{100} = \frac{4 y}{5}

नई भिन्न

= \frac{7 x / 5}{4 y / 5} = \frac{7 x}{4 y}

\frac{7 x}{4 y} = \frac{7}{4} \times \frac{x}{y}

, अतः नई भिन्न मूल भिन्न की

\frac{7}{4}

गुनी है। इसलिए सही विकल्प (d) है।29:T4d2,Average speed Going and returning cover the same distance, so the average speed is the harmonic mean of the two speeds. Formula: average speed

= \frac{2 x y}{x + y}

, where x=23 and y=33.

= \frac{2 \times 23 \times 33}{23 + 33} = \frac{1518}{56}

= \frac{759}{28} = 27 \frac{3}{28}

km/h The plain average

\frac{23 + 33}{2} = 28

is not valid here, because the car travels for different times at the two speeds. Hence the correct option is (a).2a:T5fb,औसत चाल जाते और लौटते समय दूरी समान रहती है, इसलिए औसत चाल दोनों चालों का हरात्मक माध्य होती है। सूत्र: औसत चाल

= \frac{2 x y}{x + y}

, जहाँ x=23 और y=33 है।

= \frac{2 \times 23 \times 33}{23 + 33} = \frac{1518}{56}

= \frac{759}{28} = 27 \frac{3}{28}

km/h सामान्य औसत

\frac{23 + 33}{2} = 28

यहाँ मान्य नहीं है, क्योंकि दोनों चालों पर लगने वाला समय अलग-अलग है। इसलिए सही विकल्प (a) है।2b:T4e6,LCM of numbers A number divisible by 48, 72, 16 and 15 must be a multiple of their LCM, so the smallest such natural number is the LCM itself.

LCM

= 2^{4} \times 3^{2} \times 5 = 16 \times 9 \times 5 = 720

Check:

720 \div 48 = 15

720 \div 72 = 10

720 \div 16 = 45

720 \div 15 = 48

. Hence the correct option is (b).2c:T59c,संख्याओं का ल.स. 48, 72, 16 और 15 से विभाज्य संख्या इनके ल.स. का गुणज होगी, अतः ऐसी सबसे छोटी प्राकृत संख्या स्वयं ल.स. है।

ल.स.

= 2^{4} \times 3^{2} \times 5 = 16 \times 9 \times 5 = 720

जाँच:

720 \div 48 = 15

720 \div 72 = 10

720 \div 16 = 45

720 \div 15 = 48

इसलिए सही विकल्प (b) है।2d:T6a5,CI and SI difference For three years, difference

= P ({(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100})

The same formula in short form: difference

= \frac{P R^{2} (300 + R)}{10^{6}}

Putting R=10: difference

= \frac{P \times 100 \times 310}{10^{6}} = \frac{31 P}{1000}

\frac{31 P}{1000} = 217

P = \frac{217 \times 1000}{31} = 7000

So the principal is ₹7,000, and the correct option is (d).2e:T761,चक्रवृद्धि और साधारण ब्याज का अंतर तीन वर्षों के लिए, अंतर

= P ({(1 + \frac{R}{100})}^{3} - 1 - \frac{3 R}{100})

यही सूत्र संक्षेप में: अंतर

= \frac{P R^{2} (300 + R)}{10^{6}}

R=10 रखने पर: अंतर

= \frac{P \times 100 \times 310}{10^{6}} = \frac{31 P}{1000}

\frac{31 P}{1000} = 217

P = \frac{217 \times 1000}{31} = 7000

अतः मूलधन ₹7,000 है, इसलिए सही विकल्प (d) है।2f:T6f2,Compound interest and ratio Under compound interest an amount X becomes

X {(1 + \frac{4}{100})}^{n}

in n years. Nitin's share grows for 19 years and Pravin's for 20 years, and the two amounts turn out equal.

N {(\frac{26}{25})}^{19} = P {(\frac{26}{25})}^{20}

\frac{N}{P} = \frac{26}{25}

, so Nitin : Pravin =26:25.

26+25=51 units = ₹1,224, so 1 unit

= \frac{1224}{51} = 24

, that is ₹24. Pravin's share

= 25 \times 24 = 600

So Veer gave Pravin ₹600, and the correct option is (c).30:T850,चक्रवृद्धि ब्याज और अनुपात चक्रवृद्धि ब्याज में राशि X, n वर्षों में बढ़कर

X {(1 + \frac{4}{100})}^{n}

हो जाती है। Nitin का भाग 19 वर्ष तक और Pravin का भाग 20 वर्ष तक बढ़ता है, और दोनों राशियाँ बराबर हो जाती हैं।

N {(\frac{26}{25})}^{19} = P {(\frac{26}{25})}^{20}

\frac{N}{P} = \frac{26}{25}

, अतः Nitin : Pravin =26:25

26+25=51 इकाई = ₹1,224, अतः 1 इकाई

= \frac{1224}{51} = 24

, अर्थात् ₹24 Pravin का भाग

= 25 \times 24 = 600

अतः Veer ने Pravin को ₹600 दिए, इसलिए सही विकल्प (c) है।31:T694,Average speed Take the whole journey as 8 units of distance. Time for the first

\frac{3}{8}

part

= \frac{3}{60}

h; time for the remaining

\frac{5}{8}

part

= \frac{5}{40}

h. Total time

= \frac{1}{20} + \frac{1}{8} = \frac{2 + 5}{40} = \frac{7}{40}

h Average speed = total distance ÷ total time

= \frac{8}{7 / 40} = \frac{320}{7}

\frac{320}{7} = 45.71 \dots \approx 46

km/hr Hence the correct option is (b).32:T73f,औसत चाल माना कि पूरी यात्रा की दूरी 8 इकाई है। पहले

\frac{3}{8}

भाग का समय

= \frac{3}{60}

घंटे; शेष

\frac{5}{8}

भाग का समय

= \frac{5}{40}

घंटे कुल समय

= \frac{1}{20} + \frac{1}{8} = \frac{2 + 5}{40} = \frac{7}{40}

घंटे औसत चाल = कुल दूरी ÷ कुल समय

= \frac{8}{7 / 40} = \frac{320}{7}

\frac{320}{7} = 45.71 \dots \approx 46

km/hr इसलिए सही विकल्प (b) है।33:T566,Surface area of a sphere Curved surface area of a sphere

= 4 π r^{2}

4 \times \frac{22}{7} \times r^{2} = 9856

r^{2} = \frac{9856 \times 7}{88} = 784

r = \sqrt{784} = 28

cm Diameter

= 2 r = 2 \times 28 = 56

cm The answer is asked in metres, so 56 cm = 0.56 m (not 5.6 m). Hence the correct option is (d).34:T615,गोले का पृष्ठीय क्षेत्रफल गोले का वक्र पृष्ठीय क्षेत्रफल

= 4 π r^{2}

4 \times \frac{22}{7} \times r^{2} = 9856

r^{2} = \frac{9856 \times 7}{88} = 784

r = \sqrt{784} = 28

cm व्यास

= 2 r = 2 \times 28 = 56

cm उत्तर m में पूछा गया है, अतः 56 cm = 0.56 m (5.6 m नहीं)। इसलिए सही विकल्प (d) है।35:T528,Work done in rotation Take the total work as the LCM of 25, 30, 36 and 40, that is 1800 units. One day's work: P = 72, Q = 60, R = 50 and S = 45 units. The rotation is Q, S, P, R, so one cycle of 4 days does 60+45+72+50=227 units.

227 \times 7 = 1589

units, so 7 complete cycles (28 days) leave the work unfinished. Remaining work =1800-1589=211 units

Day 29 (Q) leaves 211-60=151; day 30 (S) leaves 151-45=106; day 31 (P) leaves 106-72=34. R does 50 units a day, so R needs

\frac{34}{50} = \frac{17}{25}

of the 32nd day. Total time

= 31 + \frac{17}{25} = 31 \frac{17}{25}

days Hence the correct option is (b).36:T74b,बारी-बारी से कार्य कुल कार्य को 25, 30, 36 और 40 का ल.स., अर्थात् 1800 इकाई मानिए। एक दिन का कार्य: P = 72, Q = 60, R = 50 तथा S = 45 इकाई क्रम Q, S, P, R है, अतः 4 दिन के एक चक्र में 60+45+72+50=227 इकाई कार्य होता है।

227 \times 7 = 1589

इकाई, अर्थात् 7 पूरे चक्रों (28 दिन) के बाद भी कार्य शेष रहता है। शेष कार्य =1800-1589=211 इकाई

29वें दिन (Q) के बाद शेष 211-60=151; 30वें दिन (S) के बाद 151-45=106; 31वें दिन (P) के बाद 106-72=34 R एक दिन में 50 इकाई करता है, अतः 34 इकाई के लिए 32वें दिन का

\frac{34}{50} = \frac{17}{25}

भाग चाहिए। कुल समय

= 31 + \frac{17}{25} = 31 \frac{17}{25}

दिन इसलिए सही विकल्प (b) है।1c:["$","$L26",null,{"mode":"practice","dropped":0,"bilingual":true,"choiceSubjects":[],"questions":[{"id":8941,"n":41,"subject":"Elementary Mathematics","stem":"In a fraction, the numerator is increased by 40% and the denominator is diminished by 20%. What fraction of the original is the new fraction?","stemHi":"एक भिन्न में अंश को 40% बढ़ा दिया जाता है और हर को 20% घटा दिया जाता है। नई भिन्न, मूल भिन्न की कितनी गुनी है?","context":null,"contextHi":null,"options":[{"key":"a","text":"

\frac{4}{7}

","textHi":null,"imageUrl":null},{"key":"b","text":"

\frac{2}{5}

","textHi":null,"imageUrl":null},{"key":"c","text":"

\frac{3}{4}

","textHi":null,"imageUrl":null},{"key":"d","text":"

\frac{7}{4}

","textHi":null,"imageUrl":null}],"correct":"d","sol":"$27","solHi":"$28","pos":2,"neg":0.25},{"id":8942,"n":42,"subject":"Elementary Mathematics","stem":"A motor car goes to a place at a speed of 23 km/h and returns at a speed of

33 km / h

. The average speed (in km/h) of the motor car in the entire journey is:","stemHi":"एक मोटरकार किसी स्थान तक 23 km/h की चाल से जाती है और

33 km / h

की चाल से वापस लौटती है। संपूर्ण यात्रा में मोटरकार की औसत चाल (km/h में) कितनी है?","context":null,"contextHi":null,"options":[{"key":"a","text":"

27 \frac{3}{28}

","textHi":null,"imageUrl":null},{"key":"b","text":"

29 \frac{3}{31}

","textHi":null,"imageUrl":null},{"key":"c","text":"

28 \frac{2}{13}

","textHi":null,"imageUrl":null},{"key":"d","text":"

30 \frac{1}{12}

","textHi":null,"imageUrl":null}],"correct":"a","sol":"$29","solHi":"$2a","pos":2,"neg":0.25},{"id":8943,"n":43,"subject":"Elementary Mathematics","stem":"The smallest natural number which is divisible by 48, 72, 16 and 15 is:","stemHi":"वह सबसे छोटी प्राकृत संख्या कौन-सी है, जो 48, 72, 16 और 15 से विभाज्य है?","context":null,"contextHi":null,"options":[{"key":"a","text":"751","textHi":null,"imageUrl":null},{"key":"b","text":"720","textHi":null,"imageUrl":null},{"key":"c","text":"710","textHi":null,"imageUrl":null},{"key":"d","text":"784","textHi":null,"imageUrl":null}],"correct":"b","sol":"$2b","solHi":"$2c","pos":2,"neg":0.25},{"id":8944,"n":44,"subject":"Elementary Mathematics","stem":"The LCM of 21,24,104 and 312 is:","stemHi":"21, 24, 104 और 312 का ल.स. (LCM) क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"2184","textHi":null,"imageUrl":null},{"key":"b","text":"2253","textHi":null,"imageUrl":null},{"key":"c","text":"2229","textHi":null,"imageUrl":null},{"key":"d","text":"2149","textHi":null,"imageUrl":null}],"correct":"a","sol":"LCM of numbers\nWrite each number as a product of primes.\n $\"figure\"/$ \nLCM = highest power of each prime

= 2^{3} \times 3 \times 7 \times 13

= 8 \times 3 \times 7 \times 13 = 2184

\nHence the correct option is (a).","solHi":"संख्याओं का ल.स.\nप्रत्येक संख्या को अभाज्य गुणनखंडों के रूप में लिखिए।\n $\"figure\"/$ \nल.स. = प्रत्येक अभाज्य की उच्चतम घात

= 2^{3} \times 3 \times 7 \times 13

= 8 \times 3 \times 7 \times 13 = 2184

\nइसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":8945,"n":45,"subject":"Elementary Mathematics","stem":"100 guavas were bought at ₹ 1210 per hundred and were sold at a profit of ₹965. Find the selling price (in ₹) per dozen of guavas.","stemHi":"100 अमरूद ₹1210 प्रति सैकड़ा की दर से खरीदे गए और ₹965 के लाभ पर बेच दिए गए। अमरूदों का प्रति दर्जन विक्रय मूल्य (₹ में) ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"256","textHi":null,"imageUrl":null},{"key":"b","text":"266","textHi":null,"imageUrl":null},{"key":"c","text":"261","textHi":null,"imageUrl":null},{"key":"d","text":"271","textHi":null,"imageUrl":null}],"correct":"c","sol":"Profit and selling price\nCost price of 100 guavas = ₹1,210\nSelling price of 100 guavas = cost price + profit = ₹1,210 + ₹965 = ₹2,175\nSelling price of 1 guava

= \frac{2175}{100} = 21.75

\nSelling price of one dozen

= 21.75 \times 12 = 261

\nSo one dozen guavas sell for ₹261, and the correct option is (c).","solHi":"लाभ और विक्रय मूल्य\n100 अमरूदों का क्रय मूल्य = ₹1,210\n100 अमरूदों का विक्रय मूल्य = क्रय मूल्य + लाभ = ₹1,210 + ₹965 = ₹2,175\n1 अमरूद का विक्रय मूल्य

= \frac{2175}{100} = 21.75

\nएक दर्जन का विक्रय मूल्य

= 21.75 \times 12 = 261

\nअतः एक दर्जन अमरूदों का विक्रय मूल्य ₹261 है, इसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":8946,"n":46,"subject":"Elementary Mathematics","stem":"The population of a district is 335000, out of which 214000 are males. 31% of the population is literate. If 31% males are literate, then what percentage of females are literate?","stemHi":"एक जिले की जनसंख्या 335000 है, जिसमें से 214000 पुरुष हैं। जनसंख्या का 31% साक्षर है। यदि 31% पुरुष साक्षर हैं, तो कितने प्रतिशत महिलाएँ साक्षर हैं?","context":null,"contextHi":null,"options":[{"key":"a","text":"31%","textHi":null,"imageUrl":null},{"key":"b","text":"30%","textHi":null,"imageUrl":null},{"key":"c","text":"33%","textHi":null,"imageUrl":null},{"key":"d","text":"28%","textHi":null,"imageUrl":null}],"correct":"a","sol":"Percentage of literates\nNumber of females = 335000 − 214000 = 121000\nTotal literates = 31% of 335000 = 103850\nLiterate males = 31% of 214000 = 66340\nLiterate females = 103850 − 66340 = 37510\n $\"figure\"/$ \nRequired percentage

= \frac{37510}{121000} \times 100 = 31 %

\nShortcut: if the whole population and the male group are both 31% literate, the remaining group must also be exactly 31% literate.\nHence the correct option is (a).","solHi":"साक्षरता प्रतिशत\nमहिलाओं की संख्या = 335000 − 214000 = 121000\nकुल साक्षर = 335000 का 31% = 103850\nसाक्षर पुरुष = 214000 का 31% = 66340\nसाक्षर महिलाएँ = 103850 − 66340 = 37510\n $\"figure\"/$ \nअभीष्ट प्रतिशत

= \frac{37510}{121000} \times 100 = 31 %

\nसंक्षिप्त विधि: जब पूरी जनसंख्या तथा पुरुष दोनों 31% साक्षर हों, तो शेष वर्ग अर्थात् महिलाएँ भी ठीक 31% साक्षर होंगी।\nइसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":8947,"n":47,"subject":"Elementary Mathematics","stem":"The fourth proportional to 2.5, 9.6 and 75 is:","stemHi":"2.5, 9.6 और 75 का चतुर्थानुपाती क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"212","textHi":null,"imageUrl":null},{"key":"b","text":"288","textHi":null,"imageUrl":null},{"key":"c","text":"198","textHi":null,"imageUrl":null},{"key":"d","text":"232","textHi":null,"imageUrl":null}],"correct":"b","sol":"Fourth proportional\nIf a:b::c:d, then d is the fourth proportional and

a \times d = b \times c

.\nHere a=2.5, b=9.6 and c=75.\n

d = \frac{b \times c}{a} = \frac{9.6 \times 75}{2.5}

= \frac{720}{2.5} = 288

\nHence the correct option is (b).","solHi":"चतुर्थानुपाती\nयदि a:b::c:d हो, तो d चतुर्थानुपाती कहलाता है और

a \times d = b \times c

होता है।\nयहाँ a=2.5, b=9.6 तथा c=75 है।\n

d = \frac{b \times c}{a} = \frac{9.6 \times 75}{2.5}

= \frac{720}{2.5} = 288

\nइसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8948,"n":48,"subject":"Elementary Mathematics","stem":"The fourth proportional to 16, a and

9 a

is 324 . What is the value of

a

?","stemHi":"16, a और

9 a

का चतुर्थानुपाती 324 है।

a

का मान क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"28","textHi":null,"imageUrl":null},{"key":"b","text":"20","textHi":null,"imageUrl":null},{"key":"c","text":"24","textHi":null,"imageUrl":null},{"key":"d","text":"32","textHi":null,"imageUrl":null}],"correct":"c","sol":"Fourth proportional\nThe fourth proportional to 16, a and 9a is the number d for which 16:a::9a:d.\nProduct of extremes = product of means, so

16 \times d = a \times 9 a

.\n

16 \times 324 = 9 a^{2}

a^{2} = \frac{5184}{9} = 576

a = \sqrt{576} = 24

\nHence the correct option is (c).","solHi":"चतुर्थानुपाती\n16, a और 9a का चतुर्थानुपाती वह संख्या d है, जिसके लिए 16:a::9a:d हो।\nबाह्य पदों का गुणनफल = मध्य पदों का गुणनफल, अतः

16 \times d = a \times 9 a

16 \times 324 = 9 a^{2}

a^{2} = \frac{5184}{9} = 576

a = \sqrt{576} = 24

\nइसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":8949,"n":49,"subject":"Elementary Mathematics","stem":"When difference between compound and simple interest for three years is ₹217 at 10% interest per annum, the principal is ₹

____

","stemHi":"जब तीन वर्षों के लिए 10% वार्षिक ब्याज दर पर चक्रवृद्धि ब्याज और साधारण ब्याज का अंतर ₹217 है, तो मूलधन ₹

____

है।","context":null,"contextHi":null,"options":[{"key":"a","text":"7420","textHi":null,"imageUrl":null},{"key":"b","text":"6425","textHi":null,"imageUrl":null},{"key":"c","text":"8200","textHi":null,"imageUrl":null},{"key":"d","text":"7000","textHi":null,"imageUrl":null}],"correct":"d","sol":"$2d","solHi":"$2e","pos":2,"neg":0.25},{"id":8950,"n":50,"subject":"Elementary Mathematics","stem":"The average weight of Mohan, Mohammad and Ashish is 46 kg. If the average weight of Mohan and\nMohammad be 39 kg and that of Mohammad and Ashish be 47 kg, then the weight of Mohammad (in kg ) is:","stemHi":"Mohan, Mohammad और Ashish का औसत भार 46 kg है। यदि Mohan और\nMohammad का औसत भार 39 kg तथा Mohammad और Ashish का औसत भार 47 kg हो, तो Mohammad का भार (kg में) कितना है?","context":null,"contextHi":null,"options":[{"key":"a","text":"44","textHi":null,"imageUrl":null},{"key":"b","text":"49","textHi":null,"imageUrl":null},{"key":"c","text":"54","textHi":null,"imageUrl":null},{"key":"d","text":"34","textHi":null,"imageUrl":null}],"correct":"d","sol":"Averages of weights\nTotal weight of the three

= 46 \times 3 = 138

kg\nMohan + Mohammad

= 39 \times 2 = 78

kg\nMohammad + Ashish

= 47 \times 2 = 94

kg\n $\"figure\"/$ \nAshish =138-78=60 kg\nMohammad =94-60=34 kg\nHence the correct option is (d).","solHi":"भार का औसत\nतीनों का कुल भार

= 46 \times 3 = 138

kg\nMohan + Mohammad का कुल भार

= 39 \times 2 = 78

kg\nMohammad + Ashish का कुल भार

= 47 \times 2 = 94

kg\n $\"figure\"/$ \nAshish का भार =138-78=60 kg\nMohammad का भार =94-60=34 kg\nइसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":8951,"n":51,"subject":"Elementary Mathematics","stem":"Veer has ₹1224 with him. He divided it amongst his sons Nitin and Pravin and asked them to invest it at 4% rate of interest compounded annually. It was seen that Nitin and Pravin got same amount after 19 and 20 years respectively. How much (in ₹) did Veer give to Pravin?","stemHi":"Veer के पास ₹1224 हैं। उसने यह राशि अपने पुत्रों Nitin और Pravin में बाँट दी और उनसे कहा कि वे इसे 4% वार्षिक चक्रवृद्धि ब्याज दर पर निवेश करें। देखा गया कि Nitin और Pravin को क्रमशः 19 और 20 वर्ष बाद समान राशि प्राप्त हुई। Veer ने Pravin को कितनी राशि (₹ में) दी?","context":null,"contextHi":null,"options":[{"key":"a","text":"450","textHi":null,"imageUrl":null},{"key":"b","text":"624","textHi":null,"imageUrl":null},{"key":"c","text":"600","textHi":null,"imageUrl":null},{"key":"d","text":"724","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2f","solHi":"$30","pos":2,"neg":0.25},{"id":8952,"n":52,"subject":"Elementary Mathematics","stem":"Three-eighth of a journey is covered at a speed of

60 km / hr

, and the remaining distance is covered at the rate of

40 km / hr

. The average speed for the whole journey (in km/hr, rounded off to the nearest integer) is :","stemHi":"एक यात्रा का तीन-आठवाँ भाग

60 km / hr

की चाल से तय किया जाता है, और शेष दूरी

40 km / hr

की दर से तय की जाती है। संपूर्ण यात्रा की औसत चाल (km/hr में, निकटतम पूर्णांक तक पूर्णांकित) कितनी है?","context":null,"contextHi":null,"options":[{"key":"a","text":"38","textHi":null,"imageUrl":null},{"key":"b","text":"46","textHi":null,"imageUrl":null},{"key":"c","text":"33","textHi":null,"imageUrl":null},{"key":"d","text":"50","textHi":null,"imageUrl":null}],"correct":"b","sol":"$31","solHi":"$32","pos":2,"neg":0.25},{"id":8953,"n":53,"subject":"Elementary Mathematics","stem":"The curved surface area of a sphere is

9856 {cm}^{2}

, find its diameter in m.","stemHi":"एक गोले का वक्र पृष्ठीय क्षेत्रफल

9856 {cm}^{2}

है, तो इसका व्यास m में ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"2.8","textHi":null,"imageUrl":null},{"key":"b","text":"0.28","textHi":null,"imageUrl":null},{"key":"c","text":"5.6","textHi":null,"imageUrl":null},{"key":"d","text":"0.56","textHi":null,"imageUrl":null}],"correct":"d","sol":"$33","solHi":"$34","pos":2,"neg":0.25},{"id":8954,"n":54,"subject":"Elementary Mathematics","stem":"The cost price of 12 pens is equal to the selling price of 9 pens. Find the profit percentage.","stemHi":"12 पेनों का क्रय मूल्य 9 पेनों के विक्रय मूल्य के बराबर है। लाभ प्रतिशत ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"21%","textHi":null,"imageUrl":null},{"key":"b","text":"

31 %

","textHi":null,"imageUrl":null},{"key":"c","text":"

33 \frac{1}{3} %

","textHi":null,"imageUrl":null},{"key":"d","text":"

27 \frac{1}{4} %

","textHi":null,"imageUrl":null}],"correct":"c","sol":"Profit percentage\nLet the cost price of one pen be x and its selling price be y.\n12x=9y, so

\frac{y}{x} = \frac{12}{9} = \frac{4}{3}

.\nTake CP = ₹3 and SP = ₹4 per pen, so the profit is ₹1 per pen.\nProfit percentage

= \frac{1}{3} \times 100 = 33 \frac{1}{3} %

\nHence the correct option is (c).","solHi":"लाभ प्रतिशत\nमाना कि एक पेन का क्रय मूल्य x और विक्रय मूल्य y है।\n12x=9y, अतः

\frac{y}{x} = \frac{12}{9} = \frac{4}{3}

\nमाना प्रति पेन क्रय मूल्य ₹3 और विक्रय मूल्य ₹4, अतः प्रति पेन लाभ ₹1\nलाभ प्रतिशत

= \frac{1}{3} \times 100 = 33 \frac{1}{3} %

\nइसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":8955,"n":55,"subject":"Elementary Mathematics","stem":"The simple interest on a principal amount (in ₹) is ₹487 for a period of 2 years at the rate of

5 %

per annum. The principal amount (in ₹) is:","stemHi":"किसी मूलधन (₹ में) पर 2 वर्ष की अवधि के लिए

5 %

वार्षिक ब्याज दर से साधारण ब्याज ₹487 है। मूलधन (₹ में) कितना है?","context":null,"contextHi":null,"options":[{"key":"a","text":"4870","textHi":null,"imageUrl":null},{"key":"b","text":"4877","textHi":null,"imageUrl":null},{"key":"c","text":"4861","textHi":null,"imageUrl":null},{"key":"d","text":"4873","textHi":null,"imageUrl":null}],"correct":"a","sol":"Simple interest\nFormula: simple interest

= \frac{P \times R \times T}{100}

487 = \frac{P \times 5 \times 2}{100}

P = \frac{487 \times 100}{10} = 4870

\nSo the principal is ₹4,870, and the correct option is (a).","solHi":"साधारण ब्याज\nसूत्र: साधारण ब्याज

= \frac{P \times R \times T}{100}

487 = \frac{P \times 5 \times 2}{100}

P = \frac{487 \times 100}{10} = 4870

\nअतः मूलधन ₹4,870 है, इसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":8956,"n":56,"subject":"Elementary Mathematics","stem":"P, Q, R and S can complete a piece of work in 25 days, 30 days, 36 days and 40 days, respectively. They work on a rotation basis with

Q

working on the first day, S on the second day, P on the third day and

R

on the fourth day, then again

Q

on the fifth day and so on. In how much time will the work be completed?","stemHi":"P, Q, R और S किसी कार्य को क्रमशः 25 दिन, 30 दिन, 36 दिन और 40 दिन में पूरा कर सकते हैं। वे बारी-बारी से इस प्रकार कार्य करते हैं कि पहले दिन

Q

, दूसरे दिन S, तीसरे दिन P और चौथे दिन

R

कार्य करता है, फिर पाँचवें दिन पुनः

Q

कार्य करता है और इसी क्रम में आगे। यह कार्य कितने समय में पूरा होगा?","context":null,"contextHi":null,"options":[{"key":"a","text":"

29 \frac{4}{27}

days","textHi":"

29 \frac{4}{27}

दिन","imageUrl":null},{"key":"b","text":"

31 \frac{17}{25}

days","textHi":"

31 \frac{17}{25}

दिन","imageUrl":null},{"key":"c","text":"

23 \frac{11}{15}

days","textHi":"

23 \frac{11}{15}

दिन","imageUrl":null},{"key":"d","text":"

26 \frac{7}{30}

days","textHi":"

26 \frac{7}{30}

दिन","imageUrl":null}],"correct":"b","sol":"$35","solHi":"$36","pos":2,"neg":0.25},{"id":8957,"n":57,"subject":"Elementary Mathematics","stem":"Evaluate:\n

(- 9) - (- 60) \div (- 15) + (- 2) \times 7

","stemHi":"मान ज्ञात कीजिए:\n

(- 9) - (- 60) \div (- 15) + (- 2) \times 7

","context":null,"contextHi":null,"options":[{"key":"a","text":"-27","textHi":null,"imageUrl":null},{"key":"b","text":"-29","textHi":null,"imageUrl":null},{"key":"c","text":"-30","textHi":null,"imageUrl":null},{"key":"d","text":"-26","textHi":null,"imageUrl":null}],"correct":"a","sol":"BODMAS simplification\nApply BODMAS: division and multiplication first, then addition and subtraction.\n

(- 60) \div (- 15) = 4

and

(- 2) \times 7 = - 14

\n(-9)-4+(-14)\n=-13-14=-27\nHence the correct option is (a).","solHi":"BODMAS सरलीकरण\nBODMAS नियम लगाइए: पहले भाग और गुणा, फिर जोड़ और घटाव।\n

(- 60) \div (- 15) = 4

तथा

(- 2) \times 7 = - 14

\n(-9)-4+(-14)\n=-13-14=-27\nइसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":8958,"n":58,"subject":"Elementary Mathematics","stem":"An article is marked at ₹950. If it is sold at a discount of

20 %

, then the selling price becomes 2.5 times of its cost price. What is the cost price ?","stemHi":"एक वस्तु का अंकित मूल्य ₹950 है। यदि इसे

20 %

की छूट पर बेचा जाता है, तो विक्रय मूल्य इसके क्रय मूल्य का 2.5 गुना हो जाता है। क्रय मूल्य क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"₹318","textHi":null,"imageUrl":null},{"key":"b","text":"₹304","textHi":null,"imageUrl":null},{"key":"c","text":"₹312","textHi":null,"imageUrl":null},{"key":"d","text":"₹299","textHi":null,"imageUrl":null}],"correct":"b","sol":"Discount and cost price\nSelling price

= 950 \times \frac{100 - 20}{100} = 950 \times \frac{80}{100} = 760

\nGiven: selling price = 2.5 × cost price\nCost price

= \frac{760}{2.5} = 304

\nSo the cost price is ₹304, and the correct option is (b).","solHi":"छूट और क्रय मूल्य\nविक्रय मूल्य

= 950 \times \frac{100 - 20}{100} = 950 \times \frac{80}{100} = 760

\nदिया है: विक्रय मूल्य = 2.5 × क्रय मूल्य\nक्रय मूल्य

= \frac{760}{2.5} = 304

\nअतः क्रय मूल्य ₹304 है, इसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8959,"n":59,"subject":"Elementary Mathematics","stem":"Evaluate:

16 + 9 \div 3 - 3 \times 3

","stemHi":"मान ज्ञात कीजिए:

16 + 9 \div 3 - 3 \times 3

","context":null,"contextHi":null,"options":[{"key":"a","text":"13","textHi":null,"imageUrl":null},{"key":"b","text":"10","textHi":null,"imageUrl":null},{"key":"c","text":"9","textHi":null,"imageUrl":null},{"key":"d","text":"12","textHi":null,"imageUrl":null}],"correct":"b","sol":"BODMAS simplification\nBODMAS: solve

\div

and

\times

before + and -.\n

9 \div 3 = 3

and

3 \times 3 = 9

\n16+3-9=10\nHence the correct option is (b).","solHi":"BODMAS सरलीकरण\nBODMAS नियम: + और - से पहले

\div

तथा

\times

हल कीजिए।\n

9 \div 3 = 3

तथा

3 \times 3 = 9

\n16+3-9=10\nइसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8960,"n":60,"subject":"Elementary Mathematics","stem":"Kamal travels from City A to City B.\nIf Kamal drives his car at

\frac{1}{4}

of his normal speed, then he reaches City B 54 minutes late. Find the time (in minutes) that Kamal would have taken to travel from City A to City B if he drove at his normal speed.","stemHi":"Kamal शहर A से शहर B तक यात्रा करता है।\nयदि Kamal अपनी कार को अपनी सामान्य चाल के

\frac{1}{4}

भाग की चाल से चलाता है, तो वह शहर B पर 54 मिनट देरी से पहुँचता है। यदि वह अपनी सामान्य चाल से चलता, तो शहर A से शहर B तक की यात्रा में Kamal को लगने वाला समय (मिनट में) ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"13","textHi":null,"imageUrl":null},{"key":"b","text":"23","textHi":null,"imageUrl":null},{"key":"c","text":"10","textHi":null,"imageUrl":null},{"key":"d","text":"18","textHi":null,"imageUrl":null}],"correct":"d","sol":"Speed, time and distance\nFor a fixed distance, time is inversely proportional to speed.\nThe new speed is

\frac{1}{4}

of the normal speed, so the new time is 4 times the normal time.\n $\"figure\"/$ \nExtra time =4t-t=3t, and this extra time is 54 minutes.\n3t=54, so t=18\nSo at his normal speed Kamal takes 18 minutes, and the correct option is (d).","solHi":"चाल, समय और दूरी\nदूरी नियत होने पर समय, चाल के व्युत्क्रमानुपाती होता है।\nनई चाल सामान्य चाल की

\frac{1}{4}

है, अतः नया समय सामान्य समय का 4 गुना हो जाएगा।\n $\"figure\"/$ \nअतिरिक्त समय =4t-t=3t, और यह अतिरिक्त समय 54 मिनट है।\n3t=54, अतः t=18\nअतः सामान्य चाल से Kamal को 18 मिनट लगते हैं, इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25}],"test":{"slug":"ssc-gd-constable--19-feb-2025-shift-2-maths","title":"SSC GD Constable 19 February 2025 Shift-2 — Elementary Mathematics Sectional Test","examName":"SSC GD","paperLabel":"19 February 2025 · Shift 2","sectionSubject":"Elementary Mathematics","durationMinutes":15,"format":"sectional_test"}}]