26:I[81166,["/_next/static/immutable/chunks/2corl6wyj4669.js","/_next/static/immutable/chunks/38n271cv4vmyv.js","/_next/static/immutable/chunks/2pf73jja2520w.js"],"default"] 27:T62a,Area of a rhombus The diagonals of a rhombus bisect each other at right angles, so the rhombus splits into 4 equal right triangles.

Formula: area of a rhombus

= \frac{1}{2} \times d_{1} \times d_{2}

. Substituting

d_{1} = 15

and

d_{2} = 24

: area

= \frac{1}{2} \times 15 \times 24

= \frac{360}{2} = 180

So the area is

180 {c m}^{2}

and the correct option is (b).28:T7cc,समचतुर्भुज का क्षेत्रफल समचतुर्भुज के विकर्ण एक-दूसरे को समकोण पर समद्विभाजित करते हैं, अतः समचतुर्भुज 4 बराबर समकोण त्रिभुजों में बँट जाता है।

सूत्र: समचतुर्भुज का क्षेत्रफल

= \frac{1}{2} \times d_{1} \times d_{2}

।

d_{1} = 15

और

d_{2} = 24

रखने पर: क्षेत्रफल

= \frac{1}{2} \times 15 \times 24

= \frac{360}{2} = 180

अतः अभीष्ट क्षेत्रफल

180 {c m}^{2}

है। इसलिए सही विकल्प (b) है।29:T831,Time and work in cycles Take the total work as 4 times the LCM of 6, 8, 12 and 16, i.e.

4 \times 48 = 192

units, so that a quarter-day of work is a whole number. One day's work: Avi

= \frac{192}{6} = 32

, Avni

= \frac{192}{8} = 24

, Aarti

= \frac{192}{12} = 16

, Reeta

= \frac{192}{16} = 12

units. Each of them works only

\frac{1}{4}

day per turn, so in one turn they do 8, 6, 4 and 3 units respectively.

One complete cycle (Avi, Avni, Aarti, Reeta) =8+6+4+3=21 units.

192 \div 21

gives 9 complete cycles, work done

= 9 \times 21 = 189

units. Work still left =192-189=3 units. The next turn is Avi's, and he does 8 units in his quarter-day, which is more than the 3 units left, so the work ends inside his turn. So Avi was working when the work finished and the correct option is (c).2a:Tb60,चक्रों में समय और कार्य कुल कार्य को 6, 8, 12 और 16 के ल.स. का 4 गुना, अर्थात्

4 \times 48 = 192

इकाई मान लीजिए, ताकि एक-चौथाई दिन का कार्य भी पूर्णांक रहे। एक दिन का कार्य: Avi

= \frac{192}{6} = 32

, Avni

= \frac{192}{8} = 24

, Aarti

= \frac{192}{12} = 16

, Reeta

= \frac{192}{16} = 12

इकाई। इनमें से प्रत्येक एक बारी में केवल

\frac{1}{4}

दिन कार्य करता है, अतः एक बारी में वे क्रमशः 8, 6, 4 और 3 इकाई कार्य करते हैं।

एक पूरा चक्र (Avi, Avni, Aarti, Reeta) =8+6+4+3=21 इकाई।

192 \div 21

से 9 पूरे चक्र मिलते हैं, किया गया कार्य

= 9 \times 21 = 189

इकाई। शेष कार्य =192-189=3 इकाई। अगली बारी Avi की है, और वह अपने एक-चौथाई दिन में 8 इकाई कार्य करता है, जो शेष 3 इकाई से अधिक है, अतः कार्य उसी की बारी में समाप्त हो जाता है। अतः कार्य समाप्त होते समय Avi कार्य कर रहा था। इसलिए सही विकल्प (c) है।2b:T584,Percentage of literates Number of females =385000-168000=217000 . Total literates

= 52 %

385000 = \frac{52}{100} \times 385000 = 200200

. Literate males

= 21 %

168000 = \frac{21}{100} \times 168000 = 35280

Literate females =200200-35280=164920 . The percentage asked for is out of the female population, not the whole district:

\frac{164920}{217000} \times 100

= 76 %

, so the correct option is (c).2c:T5b9,साक्षरता का प्रतिशत महिलाओं की संख्या =385000-168000=217000 । कुल साक्षर =385000 का

52 % = \frac{52}{100} \times 385000 = 200200

। साक्षर पुरुष =168000 का

21 % = \frac{21}{100} \times 168000 = 35280

।

साक्षर महिलाएँ =200200-35280=164920 । अभीष्ट प्रतिशत पूरे जिले पर नहीं, महिलाओं की जनसंख्या पर निकालना है:

\frac{164920}{217000} \times 100

= 76 %

। इसलिए सही विकल्प (c) है।2d:T51c,Cost price from gain percent Formula: selling price = cost price

\times (1 + \frac{20}{100})

, because the gain is

20 %

of the cost price.

222 = CP \times \frac{6}{5}

CP = 222 \times \frac{5}{6}

CP = 185

So the cost price is ₹185 and the correct option is (c); dividing by

\frac{6}{5}

is the step, not taking

20 %

off ₹222.2e:T5d4,लाभ प्रतिशत से क्रय मूल्य सूत्र: विक्रय मूल्य = क्रय मूल्य

\times (1 + \frac{20}{100})

, क्योंकि लाभ क्रय मूल्य का

20 %

होता है। 222= क्रय मूल्य

\times \frac{6}{5}

क्रय मूल्य

= 222 \times \frac{5}{6}

क्रय मूल्य =185 अतः क्रय मूल्य ₹185 है। इसलिए सही विकल्प (c) है; यहाँ

\frac{6}{5}

से भाग देना है, ₹222 में से

20 %

घटाना नहीं।2f:T5c8,Average speed Average speed is total distance divided by total time; it is never the plain average of the four speeds.

Total distance =2+2+2+2=8 km. Total time

= \frac{2}{10} + \frac{2}{20} + \frac{2}{30} + \frac{2}{40} = 2 \times \frac{12 + 6 + 4 + 3}{120} = \frac{50}{120} = \frac{5}{12}

hours. Average speed

= 8 \div \frac{5}{12} = 8 \times \frac{12}{5} = \frac{96}{5}

=19.2 km/hr, so the correct option is (c); the plain average of 10, 20, 30 and 40 would be 25, which is not asked.30:T6d0,औसत चाल औसत चाल, कुल दूरी में कुल समय का भाग देकर निकलती है; यह चारों चालों का साधारण औसत कभी नहीं होती।

कुल दूरी =2+2+2+2=8 km। कुल समय

= \frac{2}{10} + \frac{2}{20} + \frac{2}{30} + \frac{2}{40} = 2 \times \frac{12 + 6 + 4 + 3}{120} = \frac{50}{120} = \frac{5}{12}

घंटे। औसत चाल

= 8 \div \frac{5}{12} = 8 \times \frac{12}{5} = \frac{96}{5}

=19.2 km/hr। इसलिए सही विकल्प (c) है; 10, 20, 30 और 40 का साधारण औसत 25 होता, जो पूछा नहीं गया है।31:T878,Compound interest sharing Mandar is 13, so his money grows for 23-13=10 years; Ketan is 14, so his grows for 23-14=9 years. Formula: amount

= P {(1 + \frac{5}{100})}^{n} = P {(\frac{21}{20})}^{n}

Equal amounts at age 23 give

M {(\frac{21}{20})}^{10} = K {(\frac{21}{20})}^{9}

. Dividing both sides by

{(\frac{21}{20})}^{9}

M \times \frac{21}{20} = K

, so M:K=20:21 . 20+21=41 units correspond to ₹24600, so 1 unit

= \frac{24600}{41} = 600

. Ketan's share

= 21 \times 600 = 12600

, i.e. ₹12600, so the correct option is (c).32:Ta27,चक्रवृद्धि ब्याज पर धन-विभाजन Mandar 13 वर्ष का है, अतः उसका धन 23-13=10 वर्ष तक बढ़ता है; Ketan 14 वर्ष का है, अतः उसका धन 23-14=9 वर्ष तक बढ़ता है। सूत्र: मिश्रधन

= P {(1 + \frac{5}{100})}^{n} = P {(\frac{21}{20})}^{n}

।

23 वर्ष की आयु पर समान राशि से

M {(\frac{21}{20})}^{10} = K {(\frac{21}{20})}^{9}

। दोनों पक्षों में

{(\frac{21}{20})}^{9}

का भाग देने पर:

M \times \frac{21}{20} = K

, अतः M:K=20:21 । 20+21=41 इकाई ₹24600 के तुल्य हैं, अतः 1 इकाई

= \frac{24600}{41} = 600

। Ketan का भाग

= 21 \times 600 = 12600

, अर्थात् ₹12600। इसलिए सही विकल्प (c) है।33:T57a,LCM of several numbers The smallest number divisible by all of them is their LCM. Prime factors:

12 = 2^{2} \times 3

28 = 2^{2} \times 7

36 = 2^{2} \times 3^{2}

27 = 3^{3}

. The LCM takes the highest power of each prime:

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

= 4 \times 27 \times 7

=756 , so the correct option is (c); 756 is also the only option divisible by 27.34:T6bc,कई संख्याओं का ल.स. इन सभी से विभाज्य छोटी से छोटी संख्या इनका ल.स. होती है। अभाज्य गुणनखंड:

12 = 2^{2} \times 3

28 = 2^{2} \times 7

36 = 2^{2} \times 3^{2}

27 = 3^{3}

। ल.स. में प्रत्येक अभाज्य की उच्चतम घात ली जाती है:

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

।

= 4 \times 27 \times 7

=756 । इसलिए सही विकल्प (c) है; 756 ही एकमात्र ऐसा विकल्प है जो 27 से विभाज्य है।35:T67c,Successive discounts Successive discounts act one after another, so their multipliers are multiplied; they are never simply added. Selling price

= 60000 \times (1 - \frac{10}{100}) \times (1 - \frac{15}{100}) \times (1 - \frac{20}{100})

= 60000 \times \frac{9}{10} \times \frac{17}{20} \times \frac{4}{5}

= 54000 \times \frac{17}{20} = 45900

, and then

45900 \times \frac{4}{5} = 36720

So the selling price is ₹36,720 and the correct option is (a).36:T765,क्रमिक छूट क्रमिक छूटें एक के बाद एक लगती हैं, अतः उनके गुणक गुणा किए जाते हैं; इन्हें कभी जोड़ा नहीं जाता। विक्रय मूल्य

= 60000 \times (1 - \frac{10}{100}) \times (1 - \frac{15}{100}) \times (1 - \frac{20}{100})

= 60000 \times \frac{9}{10} \times \frac{17}{20} \times \frac{4}{5}

= 54000 \times \frac{17}{20} = 45900

, और फिर

45900 \times \frac{4}{5} = 36720

अतः विक्रय मूल्य ₹36,720 है। इसलिए सही विकल्प (a) है।37:T653,Difference of CI and SI For 2 years the simple interest is the same in each year, so one year's interest

= \frac{40}{2} = 20

. For 2 years, compound interest minus simple interest is exactly the interest earned on the first year's interest.

45 - 40 = 20 \times \frac{R}{100}

5 = \frac{20 R}{100}

, so

R = 25 %

. Now use simple interest

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

40 = \frac{P \times 25 \times 2}{100}

40 = \frac{P}{2}

P=80 , so the principal is ₹80 and the correct option is (a).38:T83a,चक्रवृद्धि और साधारण ब्याज का अंतर 2 वर्षों में साधारण ब्याज प्रत्येक वर्ष समान होता है, अतः एक वर्ष का ब्याज

= \frac{40}{2} = 20

। 2 वर्षों के लिए, चक्रवृद्धि ब्याज में से साधारण ब्याज घटाने पर ठीक वही राशि बचती है जो पहले वर्ष के ब्याज पर मिला ब्याज है।

45 - 40 = 20 \times \frac{R}{100}

5 = \frac{20 R}{100}

, अतः

R = 25 %

। अब साधारण ब्याज

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

का प्रयोग कीजिए:

40 = \frac{P \times 25 \times 2}{100}

।

40 = \frac{P}{2}

P=80 , अतः मूलधन ₹80 है। इसलिए सही विकल्प (a) है।39:T4b1,Profit by false weight Take the marked weight as 1000 g. It is

32 %

light, so the customer actually receives 1000-320=680 g. The shopkeeper pays for 680 g but charges the price of 1000 g, so his cost is 680 and his gain is 1000-680=320 . Profit percentage

= \frac{320}{680} \times 100

= \frac{3200}{68} = \frac{800}{17}

= 47 \frac{1}{17} %

, so the correct option is (c). The gain must be divided by 680, his own cost, and not by 1000; dividing by 1000 would wrongly give

32 %

.3a:T663,कम बाट से लाभ अंकित भार 1000 g लीजिए। यह

32 %

कम है, अतः ग्राहक को वास्तव में 1000-320=680 g मिलता है। दुकानदार 680 g का मूल्य चुकाता है परंतु 1000 g का मूल्य वसूलता है, अतः उसका क्रय मूल्य 680 और लाभ 1000-680=320 है। लाभ प्रतिशत

= \frac{320}{680} \times 100

= \frac{3200}{68} = \frac{800}{17}

= 47 \frac{1}{17} %

। इसलिए सही विकल्प (c) है। लाभ में 680, अर्थात् उसके अपने क्रय मूल्य, का भाग देना चाहिए, 1000 का नहीं; 1000 से भाग देने पर ग़लत रूप से

32 %

आता।1c:["$","$L26",null,{"mode":"practice","dropped":0,"bilingual":true,"choiceSubjects":[],"questions":[{"id":13541,"n":41,"subject":"Elementary Mathematics","stem":"If the diagonal of a rhombus are 15 cm and 24 cm , then its area will be:","stemHi":"यदि एक समचतुर्भुज के विकर्ण 15 cm और 24 cm हैं, तो उसका क्षेत्रफल होगा:","context":null,"contextHi":null,"options":[{"key":"a","text":"

160 {cm}^{2}

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

180 {cm}^{2}

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

360 {cm}^{2}

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

90 {cm}^{2}

","textHi":null,"imageUrl":null}],"correct":"b","sol":"$27","solHi":"$28","pos":2,"neg":0.25},{"id":13542,"n":42,"subject":"Elementary Mathematics","stem":"Avi can finish a piece of work in 6 days, Avni can finish it in 8 days, Aarti can do it in 12 days and Reeta can do it in 16 days. They all decided to work one by one, Avi starts the work and works for one-fourth of a day, then Avni works for one-fourth of a day, then Aarti works for one-fourth of a day and then Reeta works for the remaining one-fourth of a day. The cycle is repeated till the work finished. Who was working when the work finished ?","stemHi":"Avi एक कार्य को 6 दिनों में, Avni उसे 8 दिनों में, Aarti उसे 12 दिनों में और Reeta उसे 16 दिनों में पूरा कर सकती है। उन्होंने बारी-बारी से कार्य करने का निश्चय किया; Avi कार्य आरंभ करता है और एक-चौथाई दिन कार्य करता है, फिर Avni एक-चौथाई दिन कार्य करती है, फिर Aarti एक-चौथाई दिन कार्य करती है और फिर Reeta शेष एक-चौथाई दिन कार्य करती है। यह चक्र कार्य पूरा होने तक दोहराया जाता है। कार्य समाप्त होते समय कौन कार्य कर रहा था?","context":null,"contextHi":null,"options":[{"key":"a","text":"Aarti","textHi":null,"imageUrl":null},{"key":"b","text":"Reeta","textHi":null,"imageUrl":null},{"key":"c","text":"Avi","textHi":null,"imageUrl":null},{"key":"d","text":"Avni","textHi":null,"imageUrl":null}],"correct":"c","sol":"$29","solHi":"$2a","pos":2,"neg":0.25},{"id":13543,"n":43,"subject":"Elementary Mathematics","stem":"What will come in place of '

A

' in the given equation ?\n

\frac{A}{192} = \frac{48}{A}

","stemHi":"दी गई समीकरण में '

A

' के स्थान पर क्या आएगा?\n

\frac{A}{192} = \frac{48}{A}

","context":null,"contextHi":null,"options":[{"key":"a","text":"86","textHi":null,"imageUrl":null},{"key":"b","text":"84","textHi":null,"imageUrl":null},{"key":"c","text":"74","textHi":null,"imageUrl":null},{"key":"d","text":"96","textHi":null,"imageUrl":null}],"correct":"d","sol":"Mean proportional\nA is the mean proportional of 192 and 48, so cross-multiply

\frac{A}{192} = \frac{48}{A}

.\n

A \times A = 192 \times 48

A^{2} = 9216

A = \sqrt{9216} = 96

\nSo A=96 and the correct option is (d).","solHi":"मध्यानुपाती\nA, 192 और 48 का मध्यानुपाती है, अतः

\frac{A}{192} = \frac{48}{A}

का तिर्यक गुणन कीजिए।\n

A \times A = 192 \times 48

A^{2} = 9216

A = \sqrt{9216} = 96

\nअतः A=96 है। इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":13544,"n":44,"subject":"Elementary Mathematics","stem":"The population of a district is 385000, out of which 168000 are males. 52% of the population is literate. If 21% males are literate, then what percentage of females are literate?","stemHi":"एक जिले की जनसंख्या 385000 है, जिनमें से 168000 पुरुष हैं। जनसंख्या का 52% साक्षर है। यदि 21% पुरुष साक्षर हैं, तो कितने प्रतिशत महिलाएँ साक्षर हैं?","context":null,"contextHi":null,"options":[{"key":"a","text":"78%","textHi":null,"imageUrl":null},{"key":"b","text":"

73 %

","textHi":null,"imageUrl":null},{"key":"c","text":"76%","textHi":null,"imageUrl":null},{"key":"d","text":"74%","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2b","solHi":"$2c","pos":2,"neg":0.25},{"id":13545,"n":45,"subject":"Elementary Mathematics","stem":"Hari travels from City A to City B. If\nHari drives his car at

\frac{1}{8}

of his normal speed, then he reaches City B 21 minutes late. Find the time (in minutes) that Hari would have taken to travel from City A to City

B

if he drove at his normal speed.","stemHi":"Hari, City A से City B तक यात्रा करता है। यदि\nHari अपनी कार को अपनी सामान्य चाल की

\frac{1}{8}

चाल से चलाता है, तो वह City B पर 21 मिनट देर से पहुँचता है। वह समय (मिनटों में) ज्ञात कीजिए जो Hari को City A से City

B

तक की यात्रा में लगता, यदि वह अपनी सामान्य चाल से चलता।","context":null,"contextHi":null,"options":[{"key":"a","text":"1","textHi":null,"imageUrl":null},{"key":"b","text":"3","textHi":null,"imageUrl":null},{"key":"c","text":"7","textHi":null,"imageUrl":null},{"key":"d","text":"11","textHi":null,"imageUrl":null}],"correct":"b","sol":"Speed, time and distance\nThe distance is fixed, so speed and time are inversely proportional: speed

\times

time is constant.\nThe new speed is

\frac{1}{8}

of the normal speed, so speed ratio (normal : new) =8:1 and time ratio =1:8 .\n $\"figure\"/$ \nLet 1 unit of time be t minutes; the delay is 8t-t=7t .\n7t=21\nt=3\nSo at his normal speed Hari would take 3 minutes and the correct option is (b).","solHi":"चाल, समय और दूरी\nदूरी नियत है, अतः चाल और समय व्युत्क्रमानुपाती हैं: चाल

\times

समय अचर रहता है।\nनई चाल सामान्य चाल की

\frac{1}{8}

है, अतः चाल का अनुपात (सामान्य : नई) =8:1 और समय का अनुपात =1:8 ।\n $\"figure\"/$ \nमाना कि समय की 1 इकाई t मिनट है; विलंब =8t-t=7t ।\n7t=21\nt=3\nअतः सामान्य चाल से Hari को 3 मिनट लगते। इसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":13546,"n":46,"subject":"Elementary Mathematics","stem":"LCM and HCF of two numbers are 198 and 9 respectively. If one of the numbers is 99 , then find the other number.","stemHi":"दो संख्याओं के LCM और HCF क्रमशः 198 और 9 हैं। यदि उनमें से एक संख्या 99 है, तो दूसरी संख्या ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"16","textHi":null,"imageUrl":null},{"key":"b","text":"20","textHi":null,"imageUrl":null},{"key":"c","text":"18","textHi":null,"imageUrl":null},{"key":"d","text":"17","textHi":null,"imageUrl":null}],"correct":"c","sol":"LCM and HCF relation\nRule: for two numbers, LCM

\times

HCF = first number

\times

second number.\n

198 \times 9 = 99 \times x

\n1782=99x\n

x = \frac{1782}{99} = 18

\nSo the other number is 18 and the correct option is (c).","solHi":"ल.स. और म.स. का संबंध\nनियम: दो संख्याओं के लिए LCM

\times

HCF = पहली संख्या

\times

दूसरी संख्या।\n

198 \times 9 = 99 \times x

\n1782=99x\n

x = \frac{1782}{99} = 18

\nअतः दूसरी संख्या 18 है। इसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":13547,"n":47,"subject":"Elementary Mathematics","stem":"A man sold an article for ₹222, there by gaining

20 %

. The cost of the article (in ₹) was:","stemHi":"एक व्यक्ति ने एक वस्तु ₹222 में बेची और इस पर उसे

20 %

का लाभ हुआ। वस्तु का क्रय मूल्य (₹ में) था:","context":null,"contextHi":null,"options":[{"key":"a","text":"266","textHi":null,"imageUrl":null},{"key":"b","text":"178","textHi":null,"imageUrl":null},{"key":"c","text":"185","textHi":null,"imageUrl":null},{"key":"d","text":"278","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2d","solHi":"$2e","pos":2,"neg":0.25},{"id":13548,"n":48,"subject":"Elementary Mathematics","stem":"A car travels four consecutive distances of 2 km each at speeds of 10 km/hr, 20 km/hr, 30 km/hr, and 40 km/hr, respectively. What is the average speed of the car (in

km / hr

) over this total distance?","stemHi":"एक कार 2 km की चार क्रमागत दूरियाँ क्रमशः 10 km/hr, 20 km/hr, 30 km/hr और 40 km/hr की चाल से तय करती है। इस कुल दूरी पर कार की औसत चाल (

km / hr

में) क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"21.2","textHi":null,"imageUrl":null},{"key":"b","text":"18.3","textHi":null,"imageUrl":null},{"key":"c","text":"19.2","textHi":null,"imageUrl":null},{"key":"d","text":"17.36","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2f","solHi":"$30","pos":2,"neg":0.25},{"id":13549,"n":49,"subject":"Elementary Mathematics","stem":"The average age of 48 students of a class is 10 years. If the age of the teacher is also included, the average age of the whole group becomes 11 years. The age (in years) of the teacher is:","stemHi":"एक कक्षा के 48 विद्यार्थियों की औसत आयु 10 वर्ष है। यदि शिक्षक की आयु भी सम्मिलित कर ली जाए, तो पूरे समूह की औसत आयु 11 वर्ष हो जाती है। शिक्षक की आयु (वर्षों में) है:","context":null,"contextHi":null,"options":[{"key":"a","text":"65","textHi":null,"imageUrl":null},{"key":"b","text":"62","textHi":null,"imageUrl":null},{"key":"c","text":"59","textHi":null,"imageUrl":null},{"key":"d","text":"60","textHi":null,"imageUrl":null}],"correct":"c","sol":"Average age\nTotal age of the 48 students

= 48 \times 10 = 480

years.\nWith the teacher the group has 49 members, so the new total

= 49 \times 11 = 539

years.\nTeacher's age =539-480\n=59 years, so the correct option is (c).","solHi":"औसत आयु\n48 विद्यार्थियों की कुल आयु

= 48 \times 10 = 480

वर्ष।\nशिक्षक सहित समूह में 49 सदस्य हैं, अतः नई कुल आयु

= 49 \times 11 = 539

वर्ष।\nशिक्षक की आयु =539-480\n=59 वर्ष। इसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":13550,"n":50,"subject":"Elementary Mathematics","stem":"A refinery produces 3 grades of gasoline: Regular (85% octane), Mid-grade (90% octane), and Premium (95% octane). If 200 barrels of Regular, 100 barrels of Mid-grade, and 100 barrels of Premium are mixed, what is the average octane rating ?","stemHi":"एक रिफाइनरी 3 ग्रेड का पेट्रोल बनाती है: Regular (85% ऑक्टेन), Mid-grade (90% ऑक्टेन) और Premium (95% ऑक्टेन)। यदि Regular के 200 बैरल, Mid-grade के 100 बैरल और Premium के 100 बैरल मिलाए जाएँ, तो औसत ऑक्टेन रेटिंग क्या होगी?","context":null,"contextHi":null,"options":[{"key":"a","text":"88.75%","textHi":null,"imageUrl":null},{"key":"b","text":"

88.25 %

","textHi":null,"imageUrl":null},{"key":"c","text":"89%","textHi":null,"imageUrl":null},{"key":"d","text":"88.5%","textHi":null,"imageUrl":null}],"correct":"a","sol":"Weighted average\nThe three quantities are unequal, so a weighted average is needed, with the number of barrels as the weights.\nWeighted total

= 200 \times 85 + 100 \times 90 + 100 \times 95

\n=17000+9000+9500=35500\nTotal barrels =200+100+100=400\nAverage octane

= \frac{35500}{400} = 88.75 %

\nSo the correct option is (a); the plain average of 85, 90 and 95 would be 90, which ignores the extra Regular barrels.","solHi":"भारित औसत\nतीनों मात्राएँ असमान हैं, अतः बैरलों की संख्या को भार मानकर भारित औसत लेना होगा।\nभारित योग

= 200 \times 85 + 100 \times 90 + 100 \times 95

\n=17000+9000+9500=35500\nकुल बैरल =200+100+100=400\nऔसत ऑक्टेन

= \frac{35500}{400} = 88.75 %

\nइसलिए सही विकल्प (a) है; 85, 90 और 95 का साधारण औसत 90 होता, जिसमें Regular के अतिरिक्त बैरल छूट जाते हैं।","pos":2,"neg":0.25},{"id":13551,"n":51,"subject":"Elementary Mathematics","stem":"Find the third proportional of 18 and 30 .","stemHi":"18 और 30 का तृतीयानुपाती ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"48","textHi":null,"imageUrl":null},{"key":"b","text":"53","textHi":null,"imageUrl":null},{"key":"c","text":"60","textHi":null,"imageUrl":null},{"key":"d","text":"50","textHi":null,"imageUrl":null}],"correct":"d","sol":"Third proportional\nIf c is the third proportional to a and b , then a:b=b:c , which gives

c = \frac{b^{2}}{a}

.\n

c = \frac{30^{2}}{18} = \frac{900}{18}

\nc=50\nSo the third proportional is 50 and the correct option is (d).","solHi":"तृतीयानुपाती\nयदि c, a और b का तृतीयानुपाती है, तो a:b=b:c , जिससे

c = \frac{b^{2}}{a}

प्राप्त होता है।\n

c = \frac{30^{2}}{18} = \frac{900}{18}

\nc=50\nअतः तृतीयानुपाती 50 है। इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":13552,"n":52,"subject":"Elementary Mathematics","stem":"Rupesh has two grandsons Mandar and Ketan. 13 year old Mandar gets some money from Rupesh's wealth and 14 year old Ketan gets rest of the money. But Mandar and Ketan will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate

5 %

compounded annually. When both turn 23 , they receive the same amount. How much had Rupesh given Ketan (in ₹) initially, if total money with Rupesh was ₹24600?","stemHi":"Rupesh के दो पोते Mandar और Ketan हैं। 13 वर्ष का Mandar, Rupesh की संपत्ति में से कुछ धन पाता है और 14 वर्ष का Ketan शेष धन पाता है। परंतु Mandar और Ketan को धन तभी मिलेगा जब वे 23 वर्ष के हो जाएँगे। तब तक धन बैंक में

5 %

वार्षिक की दर से चक्रवृद्धि ब्याज पर जमा रहता है। जब दोनों 23 वर्ष के होते हैं, तो उन्हें समान राशि मिलती है। यदि Rupesh के पास कुल धन ₹24600 था, तो Rupesh ने Ketan को आरंभ में कितना धन (₹ में) दिया था?","context":null,"contextHi":null,"options":[{"key":"a","text":"12950","textHi":null,"imageUrl":null},{"key":"b","text":"12000","textHi":null,"imageUrl":null},{"key":"c","text":"12600","textHi":null,"imageUrl":null},{"key":"d","text":"11750","textHi":null,"imageUrl":null}],"correct":"c","sol":"$31","solHi":"$32","pos":2,"neg":0.25},{"id":13553,"n":53,"subject":"Elementary Mathematics","stem":"If 4% of

x = 36

, then

x

is equal to:","stemHi":"यदि 4%

x = 36

है, तो

x

बराबर है:","context":null,"contextHi":null,"options":[{"key":"a","text":"900","textHi":null,"imageUrl":null},{"key":"b","text":"1000","textHi":null,"imageUrl":null},{"key":"c","text":"1800","textHi":null,"imageUrl":null},{"key":"d","text":"1900","textHi":null,"imageUrl":null}],"correct":"a","sol":"Percentage equation\n

4 %

of x=36 means

\frac{4}{100} \times x = 36

.\n

x = 36 \times \frac{100}{4}

x = 36 \times 25

\nx=900 , so the correct option is (a).","solHi":"प्रतिशत की समीकरण\nx का

4 % = 36

का अर्थ है

\frac{4}{100} \times x = 36

।\n

x = 36 \times \frac{100}{4}

x = 36 \times 25

\nx=900 । इसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":13554,"n":54,"subject":"Elementary Mathematics","stem":"Akshay invests a sum of ₹5400 and Atul invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 4 years, Atul gets ₹720 more interest than Akshay, then find the rate of interest per annum (in percentage).","stemHi":"Akshay ₹5400 की राशि और Atul ₹9400 की राशि समान वार्षिक साधारण ब्याज दर पर निवेश करते हैं। यदि 4 वर्ष के अंत में Atul को Akshay से ₹720 अधिक ब्याज मिलता है, तो वार्षिक ब्याज दर (प्रतिशत में) ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"6.5","textHi":null,"imageUrl":null},{"key":"b","text":"4.5","textHi":null,"imageUrl":null},{"key":"c","text":"3.5","textHi":null,"imageUrl":null},{"key":"d","text":"2.5","textHi":null,"imageUrl":null}],"correct":"b","sol":"Rate of simple interest\nFormula: simple interest

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

; here R is the same for both and T=4 for both.\nSo the extra ₹720 of interest comes only from the extra principal =9400-5400=4000 .\n

\frac{4000 \times R \times 4}{100} = 720

\n160R=720\n

R = \frac{720}{160} = 4.5

\nThe rate is

4.5 %

per annum, so the correct option is (b).","solHi":"साधारण ब्याज की दर\nसूत्र: साधारण ब्याज

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

; यहाँ दोनों के लिए R समान है और दोनों के लिए T=4 है।\nअतः ₹720 का अतिरिक्त ब्याज केवल अतिरिक्त मूलधन =9400-5400=4000 से आता है।\n

\frac{4000 \times R \times 4}{100} = 720

\n160R=720\n

R = \frac{720}{160} = 4.5

\nअतः वार्षिक दर

4.5 %

है। इसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":13555,"n":55,"subject":"Elementary Mathematics","stem":"The smallest natural number which is divisible by

12, 28, 36

and 27 is:","stemHi":"वह छोटी से छोटी प्राकृत संख्या जो

12, 28, 36

और 27 से विभाज्य है, है:","context":null,"contextHi":null,"options":[{"key":"a","text":"822","textHi":null,"imageUrl":null},{"key":"b","text":"688","textHi":null,"imageUrl":null},{"key":"c","text":"756","textHi":null,"imageUrl":null},{"key":"d","text":"691","textHi":null,"imageUrl":null}],"correct":"c","sol":"$33","solHi":"$34","pos":2,"neg":0.25},{"id":13556,"n":56,"subject":"Elementary Mathematics","stem":"The marked price of an article is

₹ 60, 000

. It is sold at successive discounts of

10 %, 15 %

and

20 %

. Find the selling price of the article.","stemHi":"एक वस्तु का अंकित मूल्य

₹ 60, 000

है। इसे

10 %, 15 %

और

20 %

की क्रमिक छूटों पर बेचा जाता है। वस्तु का विक्रय मूल्य ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"₹36,720","textHi":null,"imageUrl":null},{"key":"b","text":"₹36,710","textHi":null,"imageUrl":null},{"key":"c","text":"₹36,750","textHi":null,"imageUrl":null},{"key":"d","text":"₹36,700","textHi":null,"imageUrl":null}],"correct":"a","sol":"$35","solHi":"$36","pos":2,"neg":0.25},{"id":13557,"n":57,"subject":"Elementary Mathematics","stem":"If at same rate of interest, in 2 years, the simple interest is ₹40 and compound interest is ₹45, then what is the principal (in ₹)?","stemHi":"यदि समान ब्याज दर पर 2 वर्षों में साधारण ब्याज ₹40 और चक्रवृद्धि ब्याज ₹45 है, तो मूलधन (₹ में) क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"80","textHi":null,"imageUrl":null},{"key":"b","text":"84","textHi":null,"imageUrl":null},{"key":"c","text":"75","textHi":null,"imageUrl":null},{"key":"d","text":"73","textHi":null,"imageUrl":null}],"correct":"a","sol":"$37","solHi":"$38","pos":2,"neg":0.25},{"id":13558,"n":58,"subject":"Elementary Mathematics","stem":"Find the value of

((90 \div 3) \times (\frac{63}{7} + \frac{28}{3} \times (8 - 5)))

","stemHi":"

((90 \div 3) \times (\frac{63}{7} + \frac{28}{3} \times (8 - 5)))

का मान ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"1127","textHi":null,"imageUrl":null},{"key":"b","text":"1102","textHi":null,"imageUrl":null},{"key":"c","text":"1091","textHi":null,"imageUrl":null},{"key":"d","text":"1110","textHi":null,"imageUrl":null}],"correct":"d","sol":"BODMAS simplification\nBy BODMAS, clear the innermost bracket first: 8-5=3 .\nInside the braces, multiplication comes before addition:

\frac{28}{3} \times 3 = 28

.\nSo the braces give

\frac{63}{7} + 28 = 9 + 28 = 37

.\nThe first bracket gives

90 \div 3 = 30

.\n

30 \times 37 = 1110

\nSo the value is 1110 and the correct option is (d).","solHi":"BODMAS द्वारा सरलीकरण\nBODMAS के अनुसार सबसे भीतर के कोष्ठक को पहले हल कीजिए: 8-5=3 ।\nमझले कोष्ठक में गुणा, जोड़ से पहले आता है:

\frac{28}{3} \times 3 = 28

।\nअतः मझला कोष्ठक देता है

\frac{63}{7} + 28 = 9 + 28 = 37

।\nपहला कोष्ठक देता है

90 \div 3 = 30

।\n

30 \times 37 = 1110

\nअतः अभीष्ट मान 1110 है। इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":13559,"n":59,"subject":"Elementary Mathematics","stem":"A dishonest shopkeeper promises to sell his goods at cost price. However, he uses a weight that actually weighs

32 %

less than what is written on it. Find his profit percentage.","stemHi":"एक बेईमान दुकानदार अपना सामान क्रय मूल्य पर बेचने का वादा करता है। परंतु वह ऐसे बाट का प्रयोग करता है जिसका वास्तविक भार, उस पर लिखे भार से

32 %

कम है। उसका लाभ प्रतिशत ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"

48 \frac{2}{17} %

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

46 \frac{1}{17} %

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

47 \frac{1}{17} %

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

49 \frac{2}{17} %

","textHi":null,"imageUrl":null}],"correct":"c","sol":"$39","solHi":"$3a","pos":2,"neg":0.25},{"id":13560,"n":60,"subject":"Elementary Mathematics","stem":"Two numbers are in the ratio

1 : 3

. If the product of their HCF and LCM is 1200 , then the numbers are

____

and\n

____

.","stemHi":"दो संख्याएँ

1 : 3

के अनुपात में हैं। यदि उनके HCF और LCM का गुणनफल 1200 है, तो वे संख्याएँ

____

और\n

____

हैं।","context":null,"contextHi":null,"options":[{"key":"a","text":"20, 60","textHi":null,"imageUrl":null},{"key":"b","text":"30, 90","textHi":null,"imageUrl":null},{"key":"c","text":"15,45","textHi":null,"imageUrl":null},{"key":"d","text":"10, 30","textHi":null,"imageUrl":null}],"correct":"a","sol":"HCF, LCM and ratio\nLet the numbers be x and 3x . As 1:3 is already in lowest terms, their HCF is x and their LCM is 3x .\nRule: HCF

\times

LCM = product of the two numbers.\n

x \times 3 x = 1200

3 x^{2} = 1200

, so

x^{2} = 400

and x=20 .\nSo the numbers are 20 and

3 \times 20 = 60

, which is option (a).","solHi":"म.स., ल.स. और अनुपात\nमाना कि संख्याएँ x और 3x हैं। चूँकि 1:3 पहले से न्यूनतम रूप में है, उनका HCF x और LCM 3x है।\nनियम: HCF

\times

LCM = दोनों संख्याओं का गुणनफल।\n

x \times 3 x = 1200

3 x^{2} = 1200

, अतः

x^{2} = 400

और x=20 ।\nअतः संख्याएँ 20 और

3 \times 20 = 60

हैं। इसलिए सही विकल्प (a) है।","pos":2,"neg":0.25}],"test":{"slug":"ssc-gd-constable--18-feb-2025-shift-3-maths","title":"SSC GD Constable 18 February 2025 Shift-3 — Elementary Mathematics Sectional Test","examName":"SSC GD","paperLabel":"18 February 2025 · Shift 3","sectionSubject":"Elementary Mathematics","durationMinutes":15,"format":"sectional_test"}}]