26:I[81166,["/_next/static/immutable/chunks/2corl6wyj4669.js","/_next/static/immutable/chunks/38n271cv4vmyv.js","/_next/static/immutable/chunks/2pf73jja2520w.js"],"default"] 27:Tbca,Compound interest, equal amounts Rupesh is 12, so his money grows for 23-12=11 years; Mandar is 13, so his grows for 23-13=10 years. Formula: Amount

A = P {(1 + \frac{r}{100})}^{n}

Both amounts are equal, so

P_{R} {(\frac{27}{25})}^{11} = P_{M} {(\frac{27}{25})}^{10}

(since

1 + \frac{8}{100} = \frac{27}{25}

). Cancelling

{(\frac{27}{25})}^{10}

gives

P_{R} \times \frac{27}{25} = P_{M}

, i.e.

\frac{P_{R}}{P_{M}} = \frac{25}{27}

P_{R} : P_{M} = 25 : 27

and 25+27=52 units = ₹24,700. 1 unit

= \frac{24700}{52} = 475

Mandar's share

= 27 \times 475 = 12825

Hence Shivam gave Mandar ₹12,825, so option (d) is correct.28:Td7d,चक्रवृद्धि ब्याज, समान राशि रूपेश 12 वर्ष का है, अतः उसका धन 23-12=11 वर्ष तक बढ़ता है; मंदार 13 वर्ष का है, अतः उसका धन 23-13=10 वर्ष तक बढ़ता है। सूत्र: मिश्रधन

A = P {(1 + \frac{r}{100})}^{n}

दोनों राशियाँ समान हैं, अतः

P_{R} {(\frac{27}{25})}^{11} = P_{M} {(\frac{27}{25})}^{10}

(क्योंकि

1 + \frac{8}{100} = \frac{27}{25}

)।

{(\frac{27}{25})}^{10}

को काटने पर

P_{R} \times \frac{27}{25} = P_{M}

, अर्थात्

\frac{P_{R}}{P_{M}} = \frac{25}{27}

।

अतः

P_{R} : P_{M} = 25 : 27

तथा 25+27=52 इकाई = ₹24,700। 1 इकाई

= \frac{24700}{52} = 475

मंदार का भाग

= 27 \times 475 = 12825

अतः शिवम ने मंदार को ₹12,825 दिए, इसलिए सही विकल्प (d) है।29:T867,Time and work, pairs Take one day’s work:

A + B = \frac{1}{18}

B + C = \frac{1}{32}

A + C = \frac{1}{52}

. Adding the three:

2 (A + B + C) = \frac{1}{18} + \frac{1}{32} + \frac{1}{52}

LCM of 18, 32 and 52 is 3744, so the sum

= \frac{208 + 117 + 72}{3744} = \frac{397}{3744}

A + B + C = \frac{397}{7488}

of the work per day Time

= \frac{7488}{397}

days

397 \times 18 = 7146

and 7488-7146=342, so the time

= 18 \frac{342}{397}

days Hence option (a) is correct.2a:T931,समय और कार्य, युग्म एक दिन का कार्य लीजिए:

A + B = \frac{1}{18}

B + C = \frac{1}{32}

A + C = \frac{1}{52}

। तीनों को जोड़ने पर:

2 (A + B + C) = \frac{1}{18} + \frac{1}{32} + \frac{1}{52}

18, 32 और 52 का ल.स. 3744 है, अतः योग

= \frac{208 + 117 + 72}{3744} = \frac{397}{3744}

A + B + C = \frac{397}{7488}

कार्य प्रतिदिन समय

= \frac{7488}{397}

दिन

397 \times 18 = 7146

तथा 7488-7146=342, अतः समय

= 18 \frac{342}{397}

दिन इसलिए सही विकल्प (a) है।2b:T817,Heron's formula First bring the sides to one unit: 0.08 m

= 0.08 \times 100 = 8

cm. So the sides are 6 cm, 8 cm and 4 cm.

Semiperimeter

s = \frac{6 + 8 + 4}{2} = \frac{18}{2} = 9

cm Formula (Heron's): Area

= \sqrt{s (s - a) (s - b) (s - c)}

Area

= \sqrt{9 (9 - 6) (9 - 8) (9 - 4)} = \sqrt{9 \times 3 \times 1 \times 5}

Area

= \sqrt{135} = \sqrt{9 \times 15} = 3 \sqrt{15}

​^{2}

Hence option (c) is correct.2c:T90f,हीरोन का सूत्र सबसे पहले भुजाओं को एक ही मात्रक में लाइए: 0.08 m

= 0.08 \times 100 = 8

cm। अतः भुजाएँ 6 cm, 8 cm और 4 cm हैं।

अर्धपरिमाप

s = \frac{6 + 8 + 4}{2} = \frac{18}{2} = 9

cm सूत्र (हीरोन): क्षेत्रफल

= \sqrt{s (s - a) (s - b) (s - c)}

क्षेत्रफल

= \sqrt{9 (9 - 6) (9 - 8) (9 - 4)} = \sqrt{9 \times 3 \times 1 \times 5}

क्षेत्रफल

= \sqrt{135} = \sqrt{9 \times 15} = 3 \sqrt{15}

​^{2}

इसलिए सही विकल्प (c) है।2d:T4d9,Proportion In a proportion a:b::c:d, the ratios are equal, so

\frac{a}{b} = \frac{c}{d}

\frac{0.3}{4.8} = \frac{0.2}{x}

Cross-multiplying:

0.3 \times x = 4.8 \times 0.2

0.3x = 0.96

x = \frac{0.96}{0.3} = 3.2

Shortcut:

4.8 = 16 \times 0.3

, so

x = 16 \times 0.2 = 3.2

; option (c) is correct.2e:T592,समानुपात समानुपात a:b::c:d में दोनों अनुपात बराबर होते हैं, अतः

\frac{a}{b} = \frac{c}{d}

।

\frac{0.3}{4.8} = \frac{0.2}{x}

तिर्यक गुणा करने पर:

0.3 \times x = 4.8 \times 0.2

0.3x = 0.96

x = \frac{0.96}{0.3} = 3.2

लघु विधि:

4.8 = 16 \times 0.3

, अतः

x = 16 \times 0.2 = 3.2

; इसलिए सही विकल्प (c) है।2f:T834,Successive discounts Formula: two successive discounts of

a %

and

b %

equal one discount of

(a + b - \frac{a b}{100}) %

. (a)

8 + 8 - \frac{8 \times 8}{100} = 16 - 0.64 = 15.36 %

(b)

10 + 10 - \frac{10 \times 10}{100} = 20 - 1 = 19 %

(c)

5 + 5 - \frac{5 \times 5}{100} = 10 - 0.25 = 9.75 %

(d)

8 + 11 - \frac{8 \times 11}{100} = 19 - 0.88 = 18.12 %

The largest single equivalent discount is

19 %

, so option (b) gives the highest discount.30:T8c8,क्रमिक छूट सूत्र:

a %

और

b %

की दो क्रमिक छूट,

(a + b - \frac{a b}{100}) %

की एक छूट के बराबर होती है। (a)

8 + 8 - \frac{8 \times 8}{100} = 16 - 0.64 = 15.36 %

(b)

10 + 10 - \frac{10 \times 10}{100} = 20 - 1 = 19 %

(c)

5 + 5 - \frac{5 \times 5}{100} = 10 - 0.25 = 9.75 %

(d)

8 + 11 - \frac{8 \times 11}{100} = 19 - 0.88 = 18.12 %

सबसे बड़ी तुल्य छूट

19 %

है, अतः विकल्प (b) सर्वाधिक छूट देता है।31:T4c5,Price, consumption and expenditure Expenditure = price

\times

consumption. Let the original price and consumption both be 100, so the expenditure

= 100 \times 100 = 10000

. New price

= 100 + 50 %

of 100=150 New expenditure

= 10000 + 41 %

of 10000=14100 New consumption

= \frac{14100}{150} = 94

Fall in consumption =100-94=6 Percentage reduction

= \frac{6}{100} \times 100 = 6 %

Hence option (a) is correct.32:T557,मूल्य, खपत और व्यय व्यय = मूल्य

\times

खपत। माना कि मूल मूल्य और मूल खपत दोनों 100 हैं, अतः व्यय

= 100 \times 100 = 10000

। नया मूल्य =100+100 का

50 % = 150

नया व्यय =10000+10000 का

41 % = 14100

नई खपत

= \frac{14100}{150} = 94

खपत में कमी =100-94=6 प्रतिशत कमी

= \frac{6}{100} \times 100 = 6 %

इसलिए सही विकल्प (a) है।33:T4a1,Cost price from gain percent Formula:

S P = C P (1 + \frac{gain %}{100})

219 = C P \times (1 + \frac{9.5}{100})

219 = C P \times \frac{109.5}{100} = C P \times \frac{219}{200}

C P = 219 \times \frac{200}{219} = 200

Hence the cost price is ₹200, so option (d) is correct.34:T50c,लाभ प्रतिशत से क्रय मूल्य सूत्र:

S P = C P (1 + \frac{लाभ %}{100})

219 = C P \times (1 + \frac{9.5}{100})

219 = C P \times \frac{109.5}{100} = C P \times \frac{219}{200}

C P = 219 \times \frac{200}{219} = 200

अतः क्रय मूल्य ₹200 है, इसलिए सही विकल्प (d) है।35:T9c0,Compound interest, equal amounts Rishi's money grows for 15 years and Shan's for 16 years at 4% per annum, and

1 + \frac{4}{100} = \frac{26}{25}

. Equal amounts:

P_{R} {(\frac{26}{25})}^{15} = P_{S} {(\frac{26}{25})}^{16}

Cancelling

{(\frac{26}{25})}^{15}

gives

P_{R} = P_{S} \times \frac{26}{25}

, so

\frac{P_{R}}{P_{S}} = \frac{26}{25}

. So

P_{R} : P_{S} = 26 : 25

and 26+25=51 units = ₹1,224. 1 unit

= \frac{1224}{51} = 24

Shan's share

= 25 \times 24 = 600

Hence Pravin gave Shan ₹600, so option (a) is correct.36:Tad3,चक्रवृद्धि ब्याज, समान राशि ऋषि का धन 15 वर्ष तक तथा शान का धन 16 वर्ष तक 4% वार्षिक दर से बढ़ता है, और

1 + \frac{4}{100} = \frac{26}{25}

। समान राशि:

P_{R} {(\frac{26}{25})}^{15} = P_{S} {(\frac{26}{25})}^{16}

{(\frac{26}{25})}^{15}

को काटने पर

P_{R} = P_{S} \times \frac{26}{25}

, अतः

\frac{P_{R}}{P_{S}} = \frac{26}{25}

। अतः

P_{R} : P_{S} = 26 : 25

तथा 26+25=51 इकाई = ₹1,224। 1 इकाई

= \frac{1224}{51} = 24

शान का भाग

= 25 \times 24 = 600

अतः प्रवीण ने शान को ₹600 दिए, इसलिए सही विकल्प (a) है।37:T52a,Inverse variation Inverse variation with

y^{2}

means

x \propto \frac{1}{y^{2}}

, i.e.

x y^{2} = k

(a constant). From the given values:

k = 78 \times 11^{2} = 78 \times 121 = 9438

When y=2.2:

y^{2} = 2.2 \times 2.2 = 4.84

x = \frac{9438}{4.84}

x=1950 Hence option (a) is correct.38:T5d4,व्युत्क्रम अनुपात

y^{2}

के व्युत्क्रमानुपाती होने का अर्थ है

x \propto \frac{1}{y^{2}}

, अर्थात्

x y^{2} = k

(अचर)। दिए गए मानों से:

k = 78 \times 11^{2} = 78 \times 121 = 9438

जब y=2.2:

y^{2} = 2.2 \times 2.2 = 4.84

x = \frac{9438}{4.84}

x=1950 इसलिए सही विकल्प (a) है।39:T479,Speed, time and delay For a fixed distance, time varies inversely with speed. Speed becomes

\frac{3}{7}

of normal, so time becomes

\frac{7}{3}

of the normal time T. Extra time

= \frac{7}{3} T - T = \frac{4}{3} T

\frac{4}{3} T = 28

T = 28 \times \frac{3}{4} = 21

minutes Hence option (c) is correct.3a:T586,चाल, समय और विलंब दूरी नियत होने पर समय, चाल के व्युत्क्रमानुपाती होता है। चाल सामान्य की

\frac{3}{7}

हो जाती है, अतः समय सामान्य समय T का

\frac{7}{3}

हो जाता है। अतिरिक्त समय

= \frac{7}{3} T - T = \frac{4}{3} T

\frac{4}{3} T = 28

T = 28 \times \frac{3}{4} = 21

मिनट इसलिए सही विकल्प (c) है।1c:["$","$L26",null,{"mode":"practice","dropped":0,"bilingual":true,"choiceSubjects":[],"questions":[{"id":8741,"n":41,"subject":"Elementary Mathematics","stem":"If 4% of x=108, then x is equal to:","stemHi":"यदि x का 4% = 108 है, तो x किसके बराबर है?","context":null,"contextHi":null,"options":[{"key":"a","text":"2700","textHi":null,"imageUrl":null},{"key":"b","text":"5500","textHi":null,"imageUrl":null},{"key":"c","text":"5400","textHi":null,"imageUrl":null},{"key":"d","text":"2800","textHi":null,"imageUrl":null}],"correct":"a","sol":"Percentage of a number\nLet the required number be x; the question says 4% of it is 108.\n

\frac{4}{100} \times x = 108

x = 108 \times \frac{100}{4}

x = 27 \times 100 = 2700

\nHence x = 2700, so option (a) is correct.","solHi":"किसी संख्या का प्रतिशत\nमाना कि अभीष्ट संख्या x है; प्रश्नानुसार इसका 4% = 108 है।\nअतः

\frac{4}{100} \times x = 108

x = 108 \times \frac{100}{4}

x = 27 \times 100 = 2700

\nअतः x = 2700, इसलिए सही विकल्प (a) है।","pos":2,"neg":0.25},{"id":8742,"n":42,"subject":"Elementary Mathematics","stem":"A soap company produces two kinds of soaps having weight 500 gm and 650 gm , respectively. If 200 soaps of the first kind and 300 soaps of second the kind were produced, then what is the weighted average weight (in gm ) of the soaps?","stemHi":"एक साबुन कंपनी दो प्रकार के साबुन बनाती है जिनका भार क्रमशः 500 gm और 650 gm है। यदि पहले प्रकार के 200 साबुन और दूसरे प्रकार के 300 साबुन बनाए गए, तो साबुनों का भारित औसत भार (gm में) क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"620","textHi":null,"imageUrl":null},{"key":"b","text":"590","textHi":null,"imageUrl":null},{"key":"c","text":"610","textHi":null,"imageUrl":null},{"key":"d","text":"600","textHi":null,"imageUrl":null}],"correct":"b","sol":"Weighted average\nFormula: weighted average = (total weight)

\div

(total number of items).\n $\"figure\"/$ \nTotal weight

= 200 \times 500 + 300 \times 650 = 100000 + 195000 = 295000

gm\nTotal number of soaps =200+300=500\nWeighted average

= \frac{295000}{500} = 590

gm\nHence the weighted average weight is 590 gm, so option (b) is correct.","solHi":"भारित औसत\nसूत्र: भारित औसत = कुल भार ÷ कुल संख्या।\n $\"figure\"/$ \nकुल भार

= 200 \times 500 + 300 \times 650 = 100000 + 195000 = 295000

gm\nसाबुनों की कुल संख्या =200+300=500\nभारित औसत

= \frac{295000}{500} = 590

gm\nअतः भारित औसत भार 590 gm है, इसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8743,"n":43,"subject":"Elementary Mathematics","stem":"LCM and HCF of two numbers are 140 and 5 respectively. If one of the numbers is 35 , then find the other number.","stemHi":"दो संख्याओं के ल.स. (LCM) और म.स. (HCF) क्रमशः 140 और 5 हैं। यदि उनमें से एक संख्या 35 है, तो दूसरी संख्या ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"23","textHi":null,"imageUrl":null},{"key":"b","text":"19","textHi":null,"imageUrl":null},{"key":"c","text":"22","textHi":null,"imageUrl":null},{"key":"d","text":"20","textHi":null,"imageUrl":null}],"correct":"d","sol":"LCM and HCF relation\nFormula: LCM

\times

HCF = product of the two numbers.\nLet the other number be N.\n

140 \times 5 = 35 \times N

\n700 = 35N\n

N = \frac{700}{35} = 20

\nCheck: HCF of 35 and 20 is 5 and their LCM is 140, so option (d) is correct.","solHi":"ल.स. और म.स. का संबंध\nसूत्र: ल.स.

\times

म.स. = दोनों संख्याओं का गुणनफल।\nमाना कि दूसरी संख्या N है।\n

140 \times 5 = 35 \times N

\n700 = 35N\n

N = \frac{700}{35} = 20

\nजाँच: 35 और 20 का म.स. 5 तथा ल.स. 140 है, इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":8744,"n":44,"subject":"Elementary Mathematics","stem":"Find the value of\n[(29 × 13) × {3 ÷ 3 × (18 − 14)/4}]","stemHi":"निम्नलिखित का मान ज्ञात कीजिए।\n[(29 × 13) × {3 ÷ 3 × (18 − 14)/4}]","context":null,"contextHi":null,"options":[{"key":"a","text":"380","textHi":null,"imageUrl":null},{"key":"b","text":"377","textHi":null,"imageUrl":null},{"key":"c","text":"364","textHi":null,"imageUrl":null},{"key":"d","text":"391","textHi":null,"imageUrl":null}],"correct":"b","sol":"BODMAS simplification\nBODMAS: clear the innermost bracket first, then the curly bracket, then multiply.\nInnermost bracket: 18-14=4, so

\frac{18 - 14}{4} = \frac{4}{4} = 1

.\nCurly bracket:

3 \div 3 \times 1 = 1 \times 1 = 1

.\nRound bracket:

29 \times 13 = 377

.\nNow

377 \times 1 = 377

.\nHence the value is 377, so option (b) is correct.","solHi":"BODMAS सरलीकरण\nBODMAS: पहले अंतरतम कोष्ठक, फिर मझले (कुंडलित) कोष्ठक, उसके बाद गुणा हल कीजिए।\nअंतरतम कोष्ठक: 18-14=4, अतः

\frac{18 - 14}{4} = \frac{4}{4} = 1

।\nकुंडलित कोष्ठक:

3 \div 3 \times 1 = 1 \times 1 = 1

।\nगोल कोष्ठक:

29 \times 13 = 377

।\nअब

377 \times 1 = 377

।\nअतः अभीष्ट मान 377 है, इसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8745,"n":45,"subject":"Elementary Mathematics","stem":"Evaluate:

16 + 12 \div 3 - 3 \times 3

","stemHi":"मान ज्ञात कीजिए: $16+12\\div 3-3\\times 3$","context":null,"contextHi":null,"options":[{"key":"a","text":"14","textHi":null,"imageUrl":null},{"key":"b","text":"10","textHi":null,"imageUrl":null},{"key":"c","text":"11","textHi":null,"imageUrl":null},{"key":"d","text":"13","textHi":null,"imageUrl":null}],"correct":"c","sol":"BODMAS simplification\nBy BODMAS, division and multiplication are done before addition and subtraction.\n

12 \div 3 = 4

and

3 \times 3 = 9

\n16+4-9\n= 20-9 = 11\nHence the value is 11, so option (c) is correct.","solHi":"BODMAS सरलीकरण\nBODMAS के अनुसार भाग और गुणा, जोड़ तथा घटाव से पहले किए जाते हैं।\n

12 \div 3 = 4

तथा

3 \times 3 = 9

\n16+4-9\n= 20-9 = 11\nअतः अभीष्ट मान 11 है, इसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":8746,"n":46,"subject":"Elementary Mathematics","stem":"The value of 69 − 4 × [4 + 3 × {20 − 5(7 − 5) × 2} ÷ 29] is:","stemHi":"69 − 4 × [4 + 3 × {20 − 5(7 − 5) × 2} ÷ 29] का मान है:","context":null,"contextHi":null,"options":[{"key":"a","text":"54","textHi":null,"imageUrl":null},{"key":"b","text":"45","textHi":null,"imageUrl":null},{"key":"c","text":"46","textHi":null,"imageUrl":null},{"key":"d","text":"53","textHi":null,"imageUrl":null}],"correct":"d","sol":"BODMAS simplification\nWork outward: round bracket, then curly bracket, then square bracket.\nRound bracket: 7-5=2, so

5 (7 - 5) \times 2 = 5 \times 2 \times 2 = 20

.\nCurly bracket: 20-20 = 0.\nNext,

3 \times 0 \div 29 = 0

.\nSquare bracket: 4+0 = 4.\n

69 - 4 \times 4 = 69 - 16 = 53

\nHence the value is 53, so option (d) is correct.","solHi":"BODMAS सरलीकरण\nअंदर से बाहर की ओर हल कीजिए: पहले गोल कोष्ठक, फिर कुंडलित कोष्ठक, फिर बड़ा कोष्ठक।\nगोल कोष्ठक: 7-5=2, अतः

5 (7 - 5) \times 2 = 5 \times 2 \times 2 = 20

।\nकुंडलित कोष्ठक: 20-20 = 0।\nआगे,

3 \times 0 \div 29 = 0

।\nबड़ा कोष्ठक: 4+0 = 4।\n

69 - 4 \times 4 = 69 - 16 = 53

\nअतः अभीष्ट मान 53 है, इसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":8747,"n":47,"subject":"Elementary Mathematics","stem":"Shivam has two grandsons Rupesh and Mandar 12 year old Rupesh gets some money from Shivam's wealth and 13 year old Mandar gets rest of the money. But Rupesh and Mandar will get money only when they turn 23 years old. Till then the money is in a bank getting interest at rate 8% compounded annually. When both turn 23 , they receive the same amount. How much had Shivam given Mandar (in ₹) initially, if total money with\nShivam was ₹24700 ?","stemHi":"शिवम के दो पोते रूपेश और मंदार हैं। 12 वर्ष का रूपेश शिवम की संपत्ति में से कुछ धन प्राप्त करता है और 13 वर्ष का मंदार शेष धन प्राप्त करता है। परंतु रूपेश और मंदार को यह धन तभी मिलेगा जब वे 23 वर्ष के हो जाएँगे। तब तक यह धन बैंक में 8% वार्षिक चक्रवृद्धि ब्याज दर पर रहता है। जब दोनों 23 वर्ष के होते हैं, तो उन्हें समान राशि प्राप्त होती है। यदि शिवम के पास कुल धन ₹24700 था, तो शिवम ने प्रारंभ में मंदार को कितना धन (₹ में) दिया था?","context":null,"contextHi":null,"options":[{"key":"a","text":"11875","textHi":null,"imageUrl":null},{"key":"b","text":"13175","textHi":null,"imageUrl":null},{"key":"c","text":"11625","textHi":null,"imageUrl":null},{"key":"d","text":"12825","textHi":null,"imageUrl":null}],"correct":"d","sol":"$27","solHi":"$28","pos":2,"neg":0.25},{"id":8748,"n":48,"subject":"Elementary Mathematics","stem":"A and B can do a piece of work in 18 days, B and C can do the same work in 32 days and

A

and

C

can do the same in 52 days. The time taken by A, B and C together to do the work is:","stemHi":"A और B एक कार्य को 18 दिनों में कर सकते हैं, B और C उसी कार्य को 32 दिनों में कर सकते हैं तथा A और C उसी कार्य को 52 दिनों में कर सकते हैं। A, B और C द्वारा मिलकर उस कार्य को करने में लिया गया समय है:","context":null,"contextHi":null,"options":[{"key":"a","text":"

18 \frac{342}{397}

days","textHi":"$$18\\frac{342}{397}$ दिन","imageUrl":null},{"key":"b","text":"

17 \frac{341}{397}

days","textHi":"$$17\\frac{341}{397}$ दिन","imageUrl":null},{"key":"c","text":"

15 \frac{339}{397}

days","textHi":"$$15\\frac{339}{397}$ दिन","imageUrl":null},{"key":"d","text":"

16 \frac{340}{397}

days","textHi":"$$16\\frac{340}{397}$ दिन","imageUrl":null}],"correct":"a","sol":"$29","solHi":"$2a","pos":2,"neg":0.25},{"id":8749,"n":49,"subject":"Elementary Mathematics","stem":"Nirmal invests a sum of ₹5400 and Rakesh invests a sum of ₹9400 at the same rate of simple interest per annum. If, at the end of 5 years, Rakesh gets ₹360 more interest than Nirmal, then find the rate of interest per annum (in percentage).","stemHi":"निर्मल ₹5400 की राशि निवेश करता है तथा राकेश ₹9400 की राशि समान वार्षिक साधारण ब्याज दर पर निवेश करता है। यदि 5 वर्ष के अंत में राकेश को निर्मल से ₹360 अधिक ब्याज मिलता है, तो वार्षिक ब्याज दर (प्रतिशत में) ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"3.2","textHi":null,"imageUrl":null},{"key":"b","text":"3.8","textHi":null,"imageUrl":null},{"key":"c","text":"1.8","textHi":null,"imageUrl":null},{"key":"d","text":"2.2","textHi":null,"imageUrl":null}],"correct":"c","sol":"Simple interest, difference\nFormula:

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

. Rate and time are the same for both, so the difference of interests comes only from the difference of principals.\nDifference of principals =9400-5400=4000\n

\frac{4000 \times R \times 5}{100} = 360

\n200R=360\n

R = \frac{360}{200} = 1.8

\nHence the rate is 1.8% per annum, so option (c) is correct.","solHi":"साधारण ब्याज, अंतर\nसूत्र:

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

। दर और समय दोनों के लिए समान हैं, अतः ब्याज का अंतर केवल मूलधन के अंतर से आता है।\nमूलधनों का अंतर =9400-5400=4000\n

\frac{4000 \times R \times 5}{100} = 360

\n200R=360\n

R = \frac{360}{200} = 1.8

\nअतः वार्षिक दर 1.8% है, इसलिए सही विकल्प (c) है।","pos":2,"neg":0.25},{"id":8750,"n":50,"subject":"Elementary Mathematics","stem":"Find the area of a triangle, whose sides are

6 cm, 0.08 m

and 4 cm .","stemHi":"एक त्रिभुज का क्षेत्रफल ज्ञात कीजिए, जिसकी भुजाएँ $6 cm,0.08 m$ और 4 cm हैं।","context":null,"contextHi":null,"options":[{"key":"a","text":"

5 \sqrt{15} {cm}^{2}

","textHi":"$$5\\sqrt{15}{ cm}^{2}$","imageUrl":null},{"key":"b","text":"

4 \sqrt{15} {cm}^{2}

","textHi":"$$4\\sqrt{15}{ cm}^{2}$","imageUrl":null},{"key":"c","text":"

3 \sqrt{15} {cm}^{2}

","textHi":"$$3\\sqrt{15}{ cm}^{2}$","imageUrl":null},{"key":"d","text":"

6 \sqrt{15} {cm}^{2}

","textHi":"$$6\\sqrt{15}{ cm}^{2}$","imageUrl":null}],"correct":"c","sol":"$2b","solHi":"$2c","pos":2,"neg":0.25},{"id":8751,"n":51,"subject":"Elementary Mathematics","stem":"Which of the following numbers will come in the place of x in the given equation?\n0.3 : 4.8 : : 0.2 : x","stemHi":"निम्नलिखित में से कौन-सी संख्या दिए गए समीकरण में x के स्थान पर आएगी?\n0.3 : 4.8 : : 0.2 : x","context":null,"contextHi":null,"options":[{"key":"a","text":"3.8","textHi":null,"imageUrl":null},{"key":"b","text":"2.1","textHi":null,"imageUrl":null},{"key":"c","text":"3.2","textHi":null,"imageUrl":null},{"key":"d","text":"4.2","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2d","solHi":"$2e","pos":2,"neg":0.25},{"id":8752,"n":52,"subject":"Elementary Mathematics","stem":"Which scheme provides the highest discount among the options given?","stemHi":"दिए गए विकल्पों में से कौन-सी योजना सर्वाधिक छूट प्रदान करती है?","context":null,"contextHi":null,"options":[{"key":"a","text":"Two successive discounts of 8% each","textHi":"8% की दो क्रमिक छूट","imageUrl":null},{"key":"b","text":"Two successive discounts of 10% each","textHi":"10% की दो क्रमिक छूट","imageUrl":null},{"key":"c","text":"Two successive discounts of 5% each","textHi":"5% की दो क्रमिक छूट","imageUrl":null},{"key":"d","text":"Two successive discounts of 8% and 11%","textHi":"8% और 11% की दो क्रमिक छूट","imageUrl":null}],"correct":"b","sol":"$2f","solHi":"$30","pos":2,"neg":0.25},{"id":8753,"n":53,"subject":"Elementary Mathematics","stem":"A train 268 m long is moving at a speed of

58 km / h

. How much time will it take to cross a man coming from the opposite direction at a speed of 14 km/h?","stemHi":"268 m लंबी एक रेलगाड़ी $58 km/h$ की चाल से चल रही है। विपरीत दिशा से 14 km/h की चाल से आ रहे एक व्यक्ति को पार करने में उसे कितना समय लगेगा?","context":null,"contextHi":null,"options":[{"key":"a","text":"12.8 seconds","textHi":"12.8 सेकंड","imageUrl":null},{"key":"b","text":"17.2 seconds","textHi":"17.2 सेकंड","imageUrl":null},{"key":"c","text":"15.5 seconds","textHi":"15.5 सेकंड","imageUrl":null},{"key":"d","text":"13.4 seconds","textHi":"13.4 सेकंड","imageUrl":null}],"correct":"d","sol":"Train crossing a man\nThey move in opposite directions, so the speeds add up.\nRelative speed =58+14=72 km/h\n72 km/h

= 72 \times \frac{5}{18} = 20

m/s\nA man is treated as a point, so the train covers only its own length, 268 m.\nTime

= \frac{distance}{relative speed} = \frac{268}{20} = 13.4

seconds\nHence option (d) is correct.","solHi":"रेलगाड़ी का व्यक्ति को पार करना\nदोनों विपरीत दिशाओं में चल रहे हैं, अतः चालें जुड़ जाती हैं।\nसापेक्ष चाल =58+14=72 km/h\n72 km/h

= 72 \times \frac{5}{18} = 20

m/s\nव्यक्ति को बिंदु माना जाता है, अतः रेलगाड़ी केवल अपनी लंबाई 268 m तय करती है।\nसमय

= \frac{दूरी}{सापेक्ष चाल} = \frac{268}{20} = 13.4

सेकंड\nइसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":8754,"n":54,"subject":"Elementary Mathematics","stem":"The price (per litre) of petrol increases by 50%. By what percent should its consumption be reduced such that the expenditure on it increases by 41% only ?","stemHi":"पेट्रोल की कीमत (प्रति लीटर) में 50% की वृद्धि होती है। इस पर होने वाला व्यय केवल 41% बढ़े, इसके लिए इसकी खपत में कितने प्रतिशत की कमी की जानी चाहिए?","context":null,"contextHi":null,"options":[{"key":"a","text":"6%","textHi":null,"imageUrl":null},{"key":"b","text":"8%","textHi":null,"imageUrl":null},{"key":"c","text":"94%","textHi":null,"imageUrl":null},{"key":"d","text":"93%","textHi":null,"imageUrl":null}],"correct":"a","sol":"$31","solHi":"$32","pos":2,"neg":0.25},{"id":8755,"n":55,"subject":"Elementary Mathematics","stem":"A man sold an article for ₹219, there by gaining 9.5%. The cost of the article (in ₹) was:","stemHi":"एक व्यक्ति ने एक वस्तु ₹219 में बेची, जिससे उसे 9.5% का लाभ हुआ। वस्तु का क्रय मूल्य (₹ में) था:","context":null,"contextHi":null,"options":[{"key":"a","text":"239","textHi":null,"imageUrl":null},{"key":"b","text":"198","textHi":null,"imageUrl":null},{"key":"c","text":"241","textHi":null,"imageUrl":null},{"key":"d","text":"200","textHi":null,"imageUrl":null}],"correct":"d","sol":"$33","solHi":"$34","pos":2,"neg":0.25},{"id":8756,"n":56,"subject":"Elementary Mathematics","stem":"Pravin has ₹1224 with him. He divided it amongst his sons Rishi and\nShan and asked them to invest it at 4% rate of interest compounded annually. It was seen that Rishi and Shan got same amount after 15 and 16 years respectively. How much (in ₹) did Pravin give to Shan ?","stemHi":"प्रवीण के पास ₹1224 हैं। उसने इसे अपने पुत्रों ऋषि और शान के बीच बाँटा तथा उनसे कहा कि वे इसे 4% वार्षिक चक्रवृद्धि ब्याज दर पर निवेश करें। यह देखा गया कि ऋषि और शान को क्रमशः 15 और 16 वर्ष बाद समान राशि प्राप्त हुई। प्रवीण ने शान को कितना धन (₹ में) दिया?","context":null,"contextHi":null,"options":[{"key":"a","text":"600","textHi":null,"imageUrl":null},{"key":"b","text":"724","textHi":null,"imageUrl":null},{"key":"c","text":"450","textHi":null,"imageUrl":null},{"key":"d","text":"624","textHi":null,"imageUrl":null}],"correct":"a","sol":"$35","solHi":"$36","pos":2,"neg":0.25},{"id":8757,"n":57,"subject":"Elementary Mathematics","stem":"The average of 21 numbers is 74 . The average of the first 11 numbers is 72.8 and that of the last 11 numbers is 77.2 . If the 10th number is five less than the 11th number and if 10th and 11th numbers are excluded, then what is the average of the remaining numbers ? (correct to nearest integer).","stemHi":"21 संख्याओं का औसत 74 है। पहली 11 संख्याओं का औसत 72.8 है तथा अंतिम 11 संख्याओं का औसत 77.2 है। यदि 10वीं संख्या, 11वीं संख्या से पाँच कम है और यदि 10वीं तथा 11वीं संख्याओं को हटा दिया जाए, तो शेष संख्याओं का औसत क्या है? (निकटतम पूर्णांक तक)","context":null,"contextHi":null,"options":[{"key":"a","text":"71","textHi":null,"imageUrl":null},{"key":"b","text":"70","textHi":null,"imageUrl":null},{"key":"c","text":"73","textHi":null,"imageUrl":null},{"key":"d","text":"72","textHi":null,"imageUrl":null}],"correct":"d","sol":"Average, overlapping groups\nSum of all 21 numbers

= 21 \times 74 = 1554

\nSum of the first 11 numbers

= 11 \times 72.8 = 800.8

\nSum of the last 11 numbers

= 11 \times 77.2 = 849.2

\nThe two blocks together cover 22 places, so the 11th number is counted twice.\n11th number =(800.8+849.2)-1554=1650-1554=96\n10th number =96-5=91\nSum of the remaining 19 numbers =1554-96-91=1367\nAverage

= \frac{1367}{19} = 71.9 \dots \approx 72

\nHence option (d) is correct.","solHi":"औसत, अतिव्यापी समूह\nसभी 21 संख्याओं का योग

= 21 \times 74 = 1554

\nपहली 11 संख्याओं का योग

= 11 \times 72.8 = 800.8

\nअंतिम 11 संख्याओं का योग

= 11 \times 77.2 = 849.2

\nदोनों समूह मिलकर 22 स्थान घेरते हैं, अतः 11वीं संख्या दो बार गिनी जाती है।\n11वीं संख्या =(800.8+849.2)-1554=1650-1554=96\n10वीं संख्या =96-5=91\nशेष 19 संख्याओं का योग =1554-96-91=1367\nऔसत

= \frac{1367}{19} = 71.9 \dots \approx 72

\nइसलिए सही विकल्प (d) है।","pos":2,"neg":0.25},{"id":8758,"n":58,"subject":"Elementary Mathematics","stem":"Baban sold 150 chairs and had a gain equal to the selling price of 90 chairs. What is his profit percentage ?","stemHi":"बबन ने 150 कुर्सियाँ बेचीं और उसे 90 कुर्सियों के विक्रय मूल्य के बराबर लाभ हुआ। उसका लाभ प्रतिशत क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"145%","textHi":null,"imageUrl":null},{"key":"b","text":"150%","textHi":null,"imageUrl":null},{"key":"c","text":"155%","textHi":null,"imageUrl":null},{"key":"d","text":"160%","textHi":null,"imageUrl":null}],"correct":"b","sol":"Profit percentage\nLet the cost price of one chair be c and the selling price of one chair be s.\nProfit on 150 chairs =150s-150c, and this equals the selling price of 90 chairs =90s.\n150s-150c=90s\n60s=150c, so

\frac{s}{c} = \frac{150}{60} = \frac{5}{2}

\nTake s=5 and c=2; profit per chair =5-2=3.\nProfit

% = \frac{3}{2} \times 100 = 150 %

\nHence option (b) is correct.","solHi":"लाभ प्रतिशत\nमाना कि एक कुर्सी का क्रय मूल्य c तथा एक कुर्सी का विक्रय मूल्य s है।\n150 कुर्सियों पर लाभ =150s-150c, और यह 90 कुर्सियों के विक्रय मूल्य =90s के बराबर है।\n150s-150c=90s\n60s=150c, अतः

\frac{s}{c} = \frac{150}{60} = \frac{5}{2}

\nमाना s=5 तथा c=2; प्रति कुर्सी लाभ =5-2=3।\nलाभ

% = \frac{3}{2} \times 100 = 150 %

\nइसलिए सही विकल्प (b) है।","pos":2,"neg":0.25},{"id":8759,"n":59,"subject":"Elementary Mathematics","stem":"If

x

varies indirectly to

y^{2}

and if

x =

78,

y = 11

, then find the value of x when y

= 2.2

.","stemHi":"यदि $x$, ${y}^{2}$ के व्युत्क्रमानुपाती है और यदि $x=$ 78, $y=11$ है, तो y $=2.2$ होने पर x का मान ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"1950","textHi":null,"imageUrl":null},{"key":"b","text":"1995","textHi":null,"imageUrl":null},{"key":"c","text":"1880","textHi":null,"imageUrl":null},{"key":"d","text":"1725","textHi":null,"imageUrl":null}],"correct":"a","sol":"$37","solHi":"$38","pos":2,"neg":0.25},{"id":8760,"n":60,"subject":"Elementary Mathematics","stem":"Bhanu travels from City A to City B. If Bhanu drives his car at

\frac{3}{7}

of his normal speed, then he reaches City B 28 minutes late. Find the time (in minutes) that Bhanu would have taken to travel from City A to City B if he drove at his normal speed.","stemHi":"भानु शहर A से शहर B तक यात्रा करता है। यदि भानु अपनी कार को अपनी सामान्य चाल के $\\frac{3}{7}$ भाग पर चलाता है, तो वह शहर B पर 28 मिनट देर से पहुँचता है। यदि वह सामान्य चाल से चलाता, तो शहर A से शहर B तक की यात्रा में भानु को लगने वाला समय (मिनटों में) ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"18","textHi":null,"imageUrl":null},{"key":"b","text":"11","textHi":null,"imageUrl":null},{"key":"c","text":"21","textHi":null,"imageUrl":null},{"key":"d","text":"19","textHi":null,"imageUrl":null}],"correct":"c","sol":"$39","solHi":"$3a","pos":2,"neg":0.25}],"test":{"slug":"ssc-gd-constable--17-feb-2025-shift-2-maths","title":"SSC GD Constable 17 February 2025 Shift-2 — Elementary Mathematics Sectional Test","examName":"SSC GD","paperLabel":"17 February 2025 · Shift 2","sectionSubject":"Elementary Mathematics","durationMinutes":15,"format":"sectional_test"}}]