28:I[81166,["/_next/static/immutable/chunks/2corl6wyj4669.js","/_next/static/immutable/chunks/38n271cv4vmyv.js","/_next/static/immutable/chunks/2pf73jja2520w.js"],"default"] 29:T52b,Algebraic identity Use the identity

{(A + B + C)}^{2} = A^{2} + B^{2} + C^{2} + 2 (A B + B C + C A)

. Substituting A+B+C=0 and

A^{2} + B^{2} + C^{2} = 64

0^{2} = 64 + 2 (A B + B C + C A)

2(AB+BC+CA)=-64 AB+BC+CA=-32 Hence option (a) is correct.2a:T5a2,बीजीय सर्वसमिका सर्वसमिका का प्रयोग कीजिए:

{(A + B + C)}^{2} = A^{2} + B^{2} + C^{2} + 2 (A B + B C + C A)

। A+B+C=0 तथा

A^{2} + B^{2} + C^{2} = 64

रखने पर -

0^{2} = 64 + 2 (A B + B C + C A)

2(AB+BC+CA)=-64 AB+BC+CA=-32 अतः सही विकल्प (a) है।2b:T5b3,Partnership profit share Take the capitals as 4, 5 and 6 units; the whole period is 2 years, i.e. 24 months. P keeps 4 units for the first 8 months and then

4 \times 1.5 = 6

units for the remaining 16 months. P's capital-months

= 4 \times 8 + 6 \times 16 = 32 + 96 = 128

. Q's capital-months

= 5 \times 24 = 120

and R's

= 6 \times 24 = 144

So the profit is shared in the ratio 128:120:144, whose parts total 128+120+144=392. P's share

= \frac{128}{392} \times 140000 = 45714.28

, i.e. Rs. 45,714.28. Hence option (d) is correct.2c:T6ef,साझेदारी में लाभ-विभाजन पूँजियाँ 4, 5 तथा 6 इकाई मानिए; पूरा समय 2 वर्ष अर्थात् 24 महीने है। P पहले 8 महीने 4 इकाई तथा शेष 16 महीने

4 \times 1.5 = 6

इकाई लगाए रखता है। P का पूँजी-मास

= 4 \times 8 + 6 \times 16 = 32 + 96 = 128

। Q का पूँजी-मास

= 5 \times 24 = 120

तथा R का

= 6 \times 24 = 144

।

अतः लाभ 128:120:144 के अनुपात में बँटेगा, जिसके कुल भाग 128+120+144=392 हैं। P का भाग

= \frac{128}{392} \times 140000 = 45714.28

, अर्थात् Rs. 45,714.28। अतः सही विकल्प (d) है।2d:T83f,Trigonometric identity Since

\sin A = \frac{15}{17}

, we get

\cos A = \sqrt{1 - \frac{225}{289}} = \frac{8}{17}

. Since

\cos B = \frac{8}{17}

, we get

\sin B = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}

. The expression

\cos A \cos B + \sin A \sin B

is the expansion of

\cos (A - B)

= \frac{8}{17} \times \frac{8}{17} + \frac{15}{17} \times \frac{15}{17}

= \frac{64 + 225}{289} = \frac{289}{289} = 1

Hence option (c) is correct.2e:T8dd,त्रिकोणमितीय सर्वसमिका चूँकि

\sin A = \frac{15}{17}

है, अतः

\cos A = \sqrt{1 - \frac{225}{289}} = \frac{8}{17}

। चूँकि

\cos B = \frac{8}{17}

है, अतः

\sin B = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}

। व्यंजक

\cos A \cos B + \sin A \sin B

वास्तव में

\cos (A - B)

का प्रसार है।

= \frac{8}{17} \times \frac{8}{17} + \frac{15}{17} \times \frac{15}{17}

= \frac{64 + 225}{289} = \frac{289}{289} = 1

अतः सही विकल्प (c) है।2f:T5a0,Profit percentage Cost price of one apple

= \frac{80}{6} = \frac{40}{3}

. Selling price of one apple

= \frac{90}{4} = \frac{45}{2}

. Profit on one apple

= \frac{45}{2} - \frac{40}{3} = \frac{135 - 80}{6} = \frac{55}{6}

. Profit %

= \frac{55}{6} \times \frac{3}{40} \times 100 = \frac{55}{80} \times 100 = 68.75

. The selling price is above the cost price, so it is a profit of 68.75%. Hence option (d) is correct.30:T677,लाभ प्रतिशत एक सेब का क्रय मूल्य

= \frac{80}{6} = \frac{40}{3}

। एक सेब का विक्रय मूल्य

= \frac{90}{4} = \frac{45}{2}

। एक सेब पर लाभ

= \frac{45}{2} - \frac{40}{3} = \frac{135 - 80}{6} = \frac{55}{6}

। लाभ प्रतिशत

= \frac{55}{6} \times \frac{3}{40} \times 100 = \frac{55}{80} \times 100 = 68.75

। विक्रय मूल्य क्रय मूल्य से अधिक है, अतः यह 68.75% का लाभ है। अतः सही विकल्प (d) है।31:T45e,Ratio and division of money Let P have P and Q have Q, so that P+Q=12000. It is given that

\frac{5}{11} P = \frac{4}{7} Q

\frac{P}{Q} = \frac{4}{7} \times \frac{11}{5} = \frac{44}{35}

So P : Q = 44 : 35 and the total number of parts is 44+35=79. P's amount

= \frac{44}{79} \times 12000 = \frac{528000}{79} = 6683.5

(approx.), i.e. Rs. 6683.5. Hence option (a) is correct.32:T528,अनुपात एवं धन-विभाजन माना कि P के पास P तथा Q के पास Q राशि है, अतः P+Q=12000। दिया है कि

\frac{5}{11} P = \frac{4}{7} Q

।

\frac{P}{Q} = \frac{4}{7} \times \frac{11}{5} = \frac{44}{35}

अतः P : Q = 44 : 35 तथा कुल भागों की संख्या =44+35=79। P की राशि

= \frac{44}{79} \times 12000 = \frac{528000}{79} = 6683.5

(लगभग), अर्थात् Rs. 6683.5। अतः सही विकल्प (a) है।33:T4e7,LCM with a remainder A number that leaves the same remainder 5 with each of the divisors must be a common multiple of them increased by 5. 11=11,

15 = 3 \times 5

24 = 2^{3} \times 3

, so LCM

= 2^{3} \times 3 \times 5 \times 11 = 1320

The smallest such number =1320+5=1325. Check:

1325 = 11 \times 120 + 5 = 15 \times 88 + 5 = 24 \times 55 + 5

. Hence option (b) is correct.34:T600,ल.स. एवं शेषफल जो संख्या प्रत्येक भाजक से भाग देने पर समान शेषफल 5 छोड़े, वह उन भाजकों के किसी उभयनिष्ठ गुणज में 5 जोड़ने से बनती है। 11=11,

15 = 3 \times 5

24 = 2^{3} \times 3

, अतः ल.स.

= 2^{3} \times 3 \times 5 \times 11 = 1320

।

अभीष्ट न्यूनतम संख्या =1320+5=1325। जाँच:

1325 = 11 \times 120 + 5 = 15 \times 88 + 5 = 24 \times 55 + 5

। अतः सही विकल्प (b) है।35:T48f,Time and work Take the total work as the LCM of 20 and 30, i.e. 60 units. One day's work of

P = \frac{60}{20} = 3

units and of

Q = \frac{60}{30} = 2

units.

Work done by P alone in 8 days

= 8 \times 3 = 24

units. Remaining work =60-24=36 units. After Q joins, they together do 3+2=5 units in a day. Required time

= \frac{36}{5} = 7.2

days. Hence the correct option is (a).36:T659,समय और कार्य कुल कार्य को 20 और 30 का लघुत्तम समापवर्त्य अर्थात् 60 इकाई मानिए। P का एक दिन का कार्य

= \frac{60}{20} = 3

इकाई तथा Q का एक दिन का कार्य

= \frac{60}{30} = 2

इकाई है।

P द्वारा अकेले 8 दिनों में किया गया कार्य

= 8 \times 3 = 24

इकाई। शेष कार्य =60-24=36 इकाई। Q के जुड़ने के बाद दोनों मिलकर एक दिन में 3+2=5 इकाई कार्य करते हैं। अभीष्ट समय

= \frac{36}{5} = 7.2

दिन। अतः सही विकल्प (a) है।37:T573,Discount and profit Selling price = marked price

\times (1 - \frac{25}{100})

= 8000 \times \frac{3}{4} = 6000

, i.e. the fan is sold for ₹6,000.

A profit of 20% means selling price = cost price

\times \frac{120}{100}

6000 = C P \times \frac{6}{5}

C P = 6000 \times \frac{5}{6} = 5000

So the fan cost the retailer ₹5,000. Hence the correct option is (d).38:T68a,छूट और लाभ विक्रय मूल्य = अंकित मूल्य

\times (1 - \frac{25}{100})

= 8000 \times \frac{3}{4} = 6000

, अर्थात् पंखा ₹6,000 में बेचा जाता है।

20% लाभ का अर्थ है विक्रय मूल्य = क्रय मूल्य

\times \frac{120}{100}

।

6000 = C P \times \frac{6}{5}

C P = 6000 \times \frac{5}{6} = 5000

अतः फुटकर विक्रेता के लिए पंखे का क्रय मूल्य ₹5,000 था। अतः सही विकल्प (d) है।39:T49c,Simple interest In simple interest the same amount of interest is added every year, so the rise in the amount is proportional to the time. Interest of 9-7=2 years =9880-8840=1040. Interest of 1 year

= \frac{1040}{2} = 520

, i.e. ₹520.

Interest of 7 years

= 7 \times 520 = 3640

, so principal =8840-3640=5200, i.e. ₹5,200. Rate

= \frac{S I \times 100}{P \times T} = \frac{520 \times 100}{5200 \times 1} = 10 %

. Hence the correct option is (c).3a:T5dc,साधारण ब्याज साधारण ब्याज में प्रत्येक वर्ष समान ब्याज जुड़ता है, अतः मिश्रधन की वृद्धि समय के अनुक्रमानुपाती होती है। 9-7=2 वर्षों का ब्याज =9880-8840=1040। 1 वर्ष का ब्याज

= \frac{1040}{2} = 520

, अर्थात् ₹520।

7 वर्षों का ब्याज

= 7 \times 520 = 3640

, अतः मूलधन =8840-3640=5200, अर्थात् ₹5,200। दर

= \frac{S I \times 100}{P \times T} = \frac{520 \times 100}{5200 \times 1} = 10 %

। अतः सही विकल्प (c) है।3b:T7b0,Infinite nested radical Let

x = \sqrt{15 + \sqrt{15 + \sqrt{15 + \dots}}}

. The expression sitting under the first root is the very same infinite radical, so it is again x.

x = \sqrt{15 + x}

Squaring:

x^{2} = 15 + x

, i.e.

x^{2} - x - 15 = 0

x = \frac{1 + \sqrt{1 + 60}}{2} = \frac{1 + \sqrt{61}}{2}

(the negative root is rejected because a square root is non-negative).

\sqrt{61} \approx 7.81

, so

x \approx \frac{1 + 7.81}{2} \approx 4.4

, a value lying between 4 and 5. Of the four choices only 5 is near this value (3 is too small, 8 and 15 far too large), so the value of the radical is taken as 5. Hence the correct option is (a).3c:T99c,अनंत करणी शृंखला माना कि

x = \sqrt{15 + \sqrt{15 + \sqrt{15 + \dots}}}

है। पहले करणी-चिह्न के नीचे वही अनंत करणी शृंखला है, अतः वह पुनः x के बराबर है।

x = \sqrt{15 + x}

वर्ग करने पर:

x^{2} = 15 + x

, अर्थात्

x^{2} - x - 15 = 0

।

x = \frac{1 + \sqrt{1 + 60}}{2} = \frac{1 + \sqrt{61}}{2}

(ऋणात्मक मूल अस्वीकार्य है क्योंकि वर्गमूल ऋणेतर होता है)।

\sqrt{61} \approx 7.81

, अतः

x \approx \frac{1 + 7.81}{2} \approx 4.4

, जो 4 और 5 के बीच का मान है। दिए गए चार विकल्पों में केवल 5 ही इस मान के निकट है (3 बहुत छोटा तथा 8 व 15 बहुत बड़े हैं), अतः इस करणी का मान 5 लिया जाता है। अतः सही विकल्प (a) है।3d:T45f,Mixture - replacement Acid at the start

= \frac{80}{100} \times 120 = 96

litres. The 16 litres taken out is ordinary mixture, so it carries away acid in the same ratio:

\frac{80}{100} \times 16 = 12.8

litres. Acid left =96-12.8=83.2 litres. The 16 litres put back is pure water, so it adds no acid, and the drum is full again with 120 litres.

New percentage of acid

= \frac{83.2}{120} \times 100 = 69.33 %

. Hence the correct option is (c).3e:T5bf,मिश्रण - प्रतिस्थापन आरंभ में अम्ल

= \frac{80}{100} \times 120 = 96

लीटर। निकाले गए 16 लीटर मिश्रण ही है, अतः उसमें अम्ल उसी अनुपात में जाता है:

\frac{80}{100} \times 16 = 12.8

लीटर। शेष अम्ल =96-12.8=83.2 लीटर। वापस डाला गया 16 लीटर शुद्ध जल है, अतः उससे अम्ल नहीं बढ़ता और ड्रम पुनः 120 लीटर भर जाता है।

अम्ल का नया प्रतिशत

= \frac{83.2}{120} \times 100 = 69.33 %

। अतः सही विकल्प (c) है।3f:T598,Chord of a circle The perpendicular drawn from the centre of a circle to a chord bisects that chord.

A M = M B = \frac{24}{2} = 12

cm and OM=9 cm. In the right triangle OMB, by Pythagoras theorem

O B^{2} = O M^{2} + M B^{2}

r^{2} = 9^{2} + 12^{2} = 81 + 144 = 225

r = \sqrt{225} = 15

cm. Hence the correct option is (a).40:T672,वृत्त की जीवा वृत्त के केंद्र से जीवा पर डाला गया लंब उस जीवा को समद्विभाजित करता है।

अतः

A M = M B = \frac{24}{2} = 12

cm तथा OM=9 cm। समकोण त्रिभुज OMB में, पाइथागोरस प्रमेय से

O B^{2} = O M^{2} + M B^{2}

।

r^{2} = 9^{2} + 12^{2} = 81 + 144 = 225

r = \sqrt{225} = 15

cm। अतः सही विकल्प (a) है।41:T4fe,Volume of spheres Melting and recasting does not change the volume of metal, so the big sphere's volume is shared equally by the small spheres.

Number of small spheres

= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} π r^{3}} = {(\frac{R}{r})}^{3}

= {(\frac{8}{2})}^{3} = 4^{3} = 64

Hence the correct option is (c).42:T5bd,गोले का आयतन पिघलाकर पुनः ढालने से धातु का आयतन नहीं बदलता, अतः बड़े गोले का आयतन ही छोटे गोलों में बँट जाता है।

छोटे गोलों की संख्या

= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} π r^{3}} = {(\frac{R}{r})}^{3}

= {(\frac{8}{2})}^{3} = 4^{3} = 64

अतः सही विकल्प (c) है।43:T539,Sphere recast into cylinder Let the radius of the sphere be r; the cylinder is given the same radius r and height h.

Recasting keeps the volume unchanged:

\frac{4}{3} π r^{3} = π r^{2} h

h = \frac{\frac{4}{3} π r^{3}}{π r^{2}} = \frac{4 r}{3}

Required fraction

= \frac{h}{r} = \frac{4}{3}

. Hence the correct option is (a).44:T5fc,गोले से बेलन बनाना माना कि गोले की त्रिज्या r है; बेलन की त्रिज्या भी r तथा ऊँचाई h है।

पुनः ढालने पर आयतन अपरिवर्तित रहता है:

\frac{4}{3} π r^{3} = π r^{2} h

h = \frac{\frac{4}{3} π r^{3}}{π r^{2}} = \frac{4 r}{3}

अभीष्ट भाग

= \frac{h}{r} = \frac{4}{3}

। अतः सही विकल्प (a) है।45:T7a5,Rationalising surds Rationalise each term by multiplying with the conjugate of its denominator.

\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{{(\sqrt{7})}^{2} - {(\sqrt{6})}^{2}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}

\frac{1}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} - \sqrt{6}}{7 - 6} = \sqrt{7} - \sqrt{6}

Adding:

(\sqrt{7} + \sqrt{6}) + (\sqrt{7} - \sqrt{6}) = 2 \sqrt{7}

. Hence the correct option is (a).46:T840,करणी का परिमेयकरण प्रत्येक पद को उसके हर के संयुग्मी से गुणा करके परिमेय बनाइए।

\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{{(\sqrt{7})}^{2} - {(\sqrt{6})}^{2}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}

\frac{1}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} - \sqrt{6}}{7 - 6} = \sqrt{7} - \sqrt{6}

जोड़ने पर:

(\sqrt{7} + \sqrt{6}) + (\sqrt{7} - \sqrt{6}) = 2 \sqrt{7}

। अतः सही विकल्प (a) है।47:T542,Discriminant of a quadratic Roots are real and distinct only when the discriminant

D = b^{2} - 4 a c > 0

. Comparing

y^{2} + k y + 36 = 0

with

a y^{2} + b y + c = 0

: a=1, b=k, c=36.

k^{2} - 4 \times 1 \times 36 > 0

k^{2} > 144

k>12 or k<-12. Among the options only k>12 satisfies this (k<-11 fails, for example k=-11.5 gives D<0). Hence the correct option is (a).48:T66b,द्विघात समीकरण का विविक्तकर मूल वास्तविक और भिन्न तभी होते हैं जब विविक्तकर

D = b^{2} - 4 a c > 0

हो।

y^{2} + k y + 36 = 0

की तुलना

a y^{2} + b y + c = 0

से करने पर: a=1, b=k, c=36।

k^{2} - 4 \times 1 \times 36 > 0

k^{2} > 144

k>12 अथवा k<-12। विकल्पों में केवल k>12 ही इसे संतुष्ट करता है (k<-11 नहीं, जैसे k=-11.5 पर D<0 होता है)। अतः सही विकल्प (a) है।49:Tea1,Trigonometric identities Use

1 + \cot^{2} x = {cosec}^{2} x

and

1 + \tan^{2} x = \sec^{2} x

. First factor

= \frac{{cosec}^{2} x}{\sec^{2} x} = \frac{1 / \sin^{2} x}{1 / \cos^{2} x} = \frac{\cos^{2} x}{\sin^{2} x} = \cot^{2} x

. For the second factor multiply numerator and denominator by

(cosec x - \cot x)

\frac{{(cosec x - \cot x)}^{2}}{{cosec}^{2} x - \cot^{2} x} = {(cosec x - \cot x)}^{2}

, since

{cosec}^{2} x - \cot^{2} x = 1

{(cosec x - \cot x)}^{2} = {(\frac{1 - \cos x}{\sin x})}^{2} = \frac{{(1 - \cos x)}^{2}}{1 - \cos^{2} x} = \frac{1 - \cos x}{1 + \cos x}

Multiplying the two factors:

\frac{\cot^{2} x (1 - \cos x)}{1 + \cos x}

. Hence the correct option is (d).4a:Tfb0,त्रिकोणमितीय सर्वसमिकाएँ

1 + \cot^{2} x = {cosec}^{2} x

तथा

1 + \tan^{2} x = \sec^{2} x

का प्रयोग कीजिए। पहला गुणनखंड

= \frac{{cosec}^{2} x}{\sec^{2} x} = \frac{1 / \sin^{2} x}{1 / \cos^{2} x} = \frac{\cos^{2} x}{\sin^{2} x} = \cot^{2} x

। दूसरे गुणनखंड के अंश और हर को

(cosec x - \cot x)

से गुणा कीजिए:

\frac{{(cosec x - \cot x)}^{2}}{{cosec}^{2} x - \cot^{2} x} = {(cosec x - \cot x)}^{2}

, क्योंकि

{cosec}^{2} x - \cot^{2} x = 1

है।

{(cosec x - \cot x)}^{2} = {(\frac{1 - \cos x}{\sin x})}^{2} = \frac{{(1 - \cos x)}^{2}}{1 - \cos^{2} x} = \frac{1 - \cos x}{1 + \cos x}

दोनों गुणनखंडों को गुणा करने पर:

\frac{\cot^{2} x (1 - \cos x)}{1 + \cos x}

। अतः सही विकल्प (d) है।4b:T63a,Fundamental trigonometric identity Identity:

\sin^{2} θ + \cos^{2} θ = 1

, so

\sin^{2} θ = 1 - \cos^{2} θ

\cos θ = \frac{8}{27} \Rightarrow \cos^{2} θ = \frac{64}{729}

\sin^{2} θ = 1 - \frac{64}{729} = \frac{729 - 64}{729} = \frac{665}{729}

(Option (b) is impossible anyway, since

\sin^{2} θ

can never exceed 1.) Hence the correct option is (a).4c:T6f4,मूल त्रिकोणमितीय सर्वसमिका सर्वसमिका:

\sin^{2} θ + \cos^{2} θ = 1

, अतः

\sin^{2} θ = 1 - \cos^{2} θ

।

\cos θ = \frac{8}{27} \Rightarrow \cos^{2} θ = \frac{64}{729}

\sin^{2} θ = 1 - \frac{64}{729} = \frac{729 - 64}{729} = \frac{665}{729}

(विकल्प (b) वैसे भी असंभव है, क्योंकि

\sin^{2} θ

कभी 1 से अधिक नहीं हो सकता।) अतः सही विकल्प (a) है।4d:T5d5,Trigonometric ratios of 45

\cot B = 1 \Rightarrow \tan B = 1 \Rightarrow B = 45^{\circ}

\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}

2 \sin B \cos B = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{2} = 1

The same result follows from the double-angle form

2 \sin B \cos B = \sin 2 B = \sin 90^{\circ} = 1

. Hence the correct option is (c).4e:T66a,45 अंश के त्रिकोणमितीय अनुपात

\cot B = 1 \Rightarrow \tan B = 1 \Rightarrow B = 45^{\circ}

।

\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}

2 \sin B \cos B = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{2} = 1

यही परिणाम द्विगुण-कोण रूप

2 \sin B \cos B = \sin 2 B = \sin 90^{\circ} = 1

से भी प्राप्त होता है। अतः सही विकल्प (c) है।1e:["$","$L28",null,{"mode":"exam","dropped":0,"bilingual":true,"choiceSubjects":[],"questions":[{"id":6251,"n":51,"subject":"Quantitative Aptitude","stem":"If

A + B + C = 0

, and

A^{2} + B^{2} + C^{2} =

64 , then find the value of

A B + B C + C A

","stemHi":"यदि

A + B + C = 0

है, तथा

A^{2} + B^{2} + C^{2} =

64 है, तो

A B + B C + C A

का मान ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"-32","textHi":null,"imageUrl":null},{"key":"b","text":"32","textHi":null,"imageUrl":null},{"key":"c","text":"-64","textHi":null,"imageUrl":null},{"key":"d","text":"64","textHi":null,"imageUrl":null}],"correct":"a","sol":"$29","solHi":"$2a","pos":2,"neg":0.5},{"id":6252,"n":52,"subject":"Quantitative Aptitude","stem":"P, Q, and R invest in a business with their capitals in the ratio 4 : 5 : 6. After 8 months, P increases his capital by

50 %

. At the end of 2 years, the total profit earned is ₹1,40,000. How much profit does P receive?","stemHi":"P, Q तथा R किसी व्यवसाय में 4 : 5 : 6 के अनुपात में पूँजी निवेश करते हैं। 8 महीने बाद P अपनी पूँजी

50 %

बढ़ा देता है। 2 वर्ष के अंत में अर्जित कुल लाभ ₹1,40,000 है। P को कितना लाभ प्राप्त होता है?","context":null,"contextHi":null,"options":[{"key":"a","text":"Rs. 58,000","textHi":null,"imageUrl":null},{"key":"b","text":"Rs. 40,456.78","textHi":null,"imageUrl":null},{"key":"c","text":"Rs. 42,000","textHi":null,"imageUrl":null},{"key":"d","text":"Rs. 45,714.28","textHi":null,"imageUrl":null}],"correct":"d","sol":"$2b","solHi":"$2c","pos":2,"neg":0.5},{"id":6253,"n":53,"subject":"Quantitative Aptitude","stem":"If

\sin A = \frac{15}{17}

and

\cos B = \frac{8}{17}

, find

\cos A \cos B + \sin A \sin B

","stemHi":"यदि

\sin A = \frac{15}{17}

तथा

\cos B = \frac{8}{17}

है, तो

\cos A \cos B + \sin A \sin B

का मान ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"

\frac{120}{289}

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

\frac{17}{289}

","textHi":null,"imageUrl":null},{"key":"c","text":"1","textHi":null,"imageUrl":null},{"key":"d","text":"

\frac{8}{17}

","textHi":null,"imageUrl":null}],"correct":"c","sol":"$2d","solHi":"$2e","pos":2,"neg":0.5},{"id":6254,"n":54,"subject":"Quantitative Aptitude","stem":"A vendor buys apples at 6 for ₹80 and sells them at 4 for ₹90. What is his profit or loss percentage?","stemHi":"एक विक्रेता 6 सेब ₹80 में खरीदता है तथा 4 सेब ₹90 में बेचता है। उसका लाभ अथवा हानि प्रतिशत क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"64.75% profit","textHi":"64.75% लाभ","imageUrl":null},{"key":"b","text":"65.65% loss","textHi":"65.65% हानि","imageUrl":null},{"key":"c","text":"67.50% loss","textHi":"67.50% हानि","imageUrl":null},{"key":"d","text":"68.75% profit","textHi":"68.75% लाभ","imageUrl":null}],"correct":"d","sol":"$2f","solHi":"$30","pos":2,"neg":0.5},{"id":6255,"n":55,"subject":"Quantitative Aptitude","stem":"P and Q together have Rs. 12000. If

\frac{5}{11}

P^{'}

s amount is equal to

\frac{4}{7}

Q^{'}

s amount, how much amount does P have?","stemHi":"P तथा Q के पास मिलाकर Rs. 12000 हैं। यदि

\frac{5}{11}

गुना

P^{'}

की राशि,

\frac{4}{7}

गुना

Q^{'}

की राशि के बराबर है, तो P के पास कितनी राशि है?","context":null,"contextHi":null,"options":[{"key":"a","text":"Rs. 6683.5","textHi":null,"imageUrl":null},{"key":"b","text":"Rs. 7000","textHi":null,"imageUrl":null},{"key":"c","text":"Rs. 5016","textHi":null,"imageUrl":null},{"key":"d","text":"Rs. 5498.5","textHi":null,"imageUrl":null}],"correct":"a","sol":"$31","solHi":"$32","pos":2,"neg":0.5},{"id":6256,"n":56,"subject":"Quantitative Aptitude","stem":"An amount is said to double in 8 years with compound interest. How many years will it take for the amount to grow to 16 times its original value ?","stemHi":"चक्रवृद्धि ब्याज पर कोई धनराशि 8 वर्षों में दोगुनी हो जाती है। उसी धनराशि को अपने मूल मान का 16 गुना होने में कितने वर्ष लगेंगे?","context":null,"contextHi":null,"options":[{"key":"a","text":"35","textHi":null,"imageUrl":null},{"key":"b","text":"30","textHi":null,"imageUrl":null},{"key":"c","text":"32","textHi":null,"imageUrl":null},{"key":"d","text":"20","textHi":null,"imageUrl":null}],"correct":"c","sol":"Compound interest doubling\nUnder compound interest the amount is multiplied by the same factor in every equal interval of time.\nThe amount becomes 2 times in 8 years, so in 8n years it becomes

2^{n}

times.\n $\"figure\"/$ \nWe need 16 times, and

16 = 2^{4}

, so n=4.\nRequired time

= 4 \times 8 = 32

years.\nHence option (c) is correct.","solHi":"चक्रवृद्धि ब्याज - दोगुना होना\nचक्रवृद्धि ब्याज में प्रत्येक समान समय-अंतराल में मिश्रधन उसी गुणक से गुणित होता रहता है।\nधन 8 वर्ष में 2 गुना हो जाता है, अतः 8n वर्ष में यह

2^{n}

गुना हो जाएगा।\n $\"figure\"/$ \nहमें 16 गुना चाहिए और

16 = 2^{4}

है, अतः n=4।\nअभीष्ट समय

= 4 \times 8 = 32

वर्ष।\nअतः सही विकल्प (c) है।","pos":2,"neg":0.5},{"id":6257,"n":57,"subject":"Quantitative Aptitude","stem":"The ratio of the speeds of a bike and a car is

5 : 7

. If the bike travels 180 km in 6 hours, how far will the car travel in 8 hours ?","stemHi":"एक बाइक तथा एक कार की चालों का अनुपात

5 : 7

है। यदि बाइक 6 घंटे में 180 km की दूरी तय करती है, तो कार 8 घंटे में कितनी दूरी तय करेगी?","context":null,"contextHi":null,"options":[{"key":"a","text":"300 km","textHi":null,"imageUrl":null},{"key":"b","text":"336 km","textHi":null,"imageUrl":null},{"key":"c","text":"350 km","textHi":null,"imageUrl":null},{"key":"d","text":"380 km","textHi":null,"imageUrl":null}],"correct":"b","sol":"Ratio of speeds\nSpeed of the bike

= \frac{180}{6} = 30

km/h.\nLet the speeds be 5x and 7x; then 5x=30, so x=6.\nSpeed of the car

= 7 \times 6 = 42

km/h.\nDistance covered by the car in 8 hours

= 42 \times 8 = 336

km.\nHence option (b) is correct.","solHi":"चालों का अनुपात\nबाइक की चाल

= \frac{180}{6} = 30

km/h।\nमाना कि चालें 5x तथा 7x हैं; तब 5x=30, अर्थात् x=6।\nकार की चाल

= 7 \times 6 = 42

km/h।\nकार द्वारा 8 घंटे में तय की गई दूरी

= 42 \times 8 = 336

km।\nअतः सही विकल्प (b) है।","pos":2,"neg":0.5},{"id":6258,"n":58,"subject":"Quantitative Aptitude","stem":"The average marks of 40 students is 66 . If one student who scored 50 is replaced by another who scored 90 , what will be the new average ?","stemHi":"40 छात्रों के औसत अंक 66 हैं। यदि 50 अंक प्राप्त करने वाले एक छात्र के स्थान पर 90 अंक प्राप्त करने वाला दूसरा छात्र आ जाए, तो नया औसत क्या होगा?","context":null,"contextHi":null,"options":[{"key":"a","text":"67","textHi":null,"imageUrl":null},{"key":"b","text":"68","textHi":null,"imageUrl":null},{"key":"c","text":"60","textHi":null,"imageUrl":null},{"key":"d","text":"65","textHi":null,"imageUrl":null}],"correct":"a","sol":"Average after replacement\nTotal marks of the 40 students

= 40 \times 66 = 2640

.\nReplacing the student who scored 50 by one who scored 90 changes the total by 90-50=40.\nNew total =2640+40=2680.\nNew average

= \frac{2680}{40} = 67

.\nHence option (a) is correct.","solHi":"प्रतिस्थापन के बाद औसत\n40 छात्रों के कुल अंक

= 40 \times 66 = 2640

।\n50 अंक पाने वाले छात्र के स्थान पर 90 अंक पाने वाला छात्र आने से कुल अंकों में 90-50=40 का परिवर्तन होता है।\nनया योग =2640+40=2680।\nनया औसत

= \frac{2680}{40} = 67

।\nअतः सही विकल्प (a) है।","pos":2,"neg":0.5},{"id":6259,"n":59,"subject":"Quantitative Aptitude","stem":"Which of the following is the smallest number divisible by 11,15 , and 24 and leaves a remainder of 5 in each case ?","stemHi":"निम्नलिखित में से कौन-सी सबसे छोटी संख्या है, जो 11, 15 तथा 24 से भाग देने पर प्रत्येक स्थिति में शेषफल 5 छोड़ती है?","context":null,"contextHi":null,"options":[{"key":"a","text":"1320","textHi":null,"imageUrl":null},{"key":"b","text":"1325","textHi":null,"imageUrl":null},{"key":"c","text":"1250","textHi":null,"imageUrl":null},{"key":"d","text":"1300","textHi":null,"imageUrl":null}],"correct":"b","sol":"$33","solHi":"$34","pos":2,"neg":0.5},{"id":6260,"n":60,"subject":"Quantitative Aptitude","stem":"The price of a shirt is increased by

30 %

and then again increased by

20 %

. What is the net percentage increase in price?","stemHi":"एक शर्ट की कीमत

30 %

बढ़ा दी जाती है और फिर उसे पुनः

20 %

बढ़ा दिया जाता है। कीमत में शुद्ध प्रतिशत वृद्धि क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"50%","textHi":null,"imageUrl":null},{"key":"b","text":"56%","textHi":null,"imageUrl":null},{"key":"c","text":"51%","textHi":null,"imageUrl":null},{"key":"d","text":"53%","textHi":null,"imageUrl":null}],"correct":"b","sol":"Successive percentage increase\nTake the original price as ₹100.\nAfter the first rise the price is

100 \times \frac{130}{100} = 130

.\nAfter the second rise it is

130 \times \frac{120}{100} = 156

.\nNet increase =156-100=56 on 100, that is 56%.\nShortcut:

30 + 20 + \frac{30 \times 20}{100} = 56

.\nHence option (b) is correct.","solHi":"क्रमिक प्रतिशत वृद्धि\nमूल कीमत ₹100 मानिए।\nपहली वृद्धि के बाद कीमत

= 100 \times \frac{130}{100} = 130

।\nदूसरी वृद्धि के बाद कीमत

= 130 \times \frac{120}{100} = 156

।\nशुद्ध वृद्धि =156-100=56 प्रति 100, अर्थात् 56%।\nसंक्षिप्त विधि:

30 + 20 + \frac{30 \times 20}{100} = 56

।\nअतः सही विकल्प (b) है।","pos":2,"neg":0.5},{"id":6261,"n":61,"subject":"Quantitative Aptitude","stem":"P can complete a task in 20 days and ,

Q

can finish a task in 30 days respectively.

P

starts the work and works for 8 days, after which

Q

joins P. How many more days will they take to complete the task ?","stemHi":"P एक कार्य को 20 दिनों में और

Q

एक कार्य को क्रमशः 30 दिनों में पूरा कर सकता है।

P

कार्य आरंभ करता है और 8 दिनों तक कार्य करता है, जिसके बाद

Q

भी P के साथ कार्य में जुड़ जाता है। शेष कार्य पूरा करने में उन्हें और कितने दिन लगेंगे?","context":null,"contextHi":null,"options":[{"key":"a","text":"7.2 days","textHi":"7.2 दिन","imageUrl":null},{"key":"b","text":"7.5 days","textHi":"7.5 दिन","imageUrl":null},{"key":"c","text":"7.8 days","textHi":"7.8 दिन","imageUrl":null},{"key":"d","text":"7.4 days","textHi":"7.4 दिन","imageUrl":null}],"correct":"a","sol":"$35","solHi":"$36","pos":2,"neg":0.5},{"id":6262,"n":62,"subject":"Quantitative Aptitude","stem":"A fan's cost price is unknown. A retailer marks it up so that its marked price is ₹8,000. They offer a 25% discount on this marked price. Even after the discount, they make a profit of 20%. What was the original cost price of the fan for the retailer ?","stemHi":"एक पंखे का क्रय मूल्य ज्ञात नहीं है। एक फुटकर विक्रेता उस पर इस प्रकार मूल्य अंकित करता है कि उसका अंकित मूल्य ₹8,000 हो जाता है। वह इस अंकित मूल्य पर 25% की छूट देता है। छूट देने के बाद भी उसे 20% का लाभ होता है। फुटकर विक्रेता के लिए पंखे का मूल क्रय मूल्य कितना था?","context":null,"contextHi":null,"options":[{"key":"a","text":"₹5,000","textHi":null,"imageUrl":null},{"key":"b","text":"₹5,500","textHi":null,"imageUrl":null},{"key":"c","text":"₹5,200","textHi":null,"imageUrl":null},{"key":"d","text":"₹5,000","textHi":null,"imageUrl":null}],"correct":"d","sol":"$37","solHi":"$38","pos":2,"neg":0.5},{"id":6263,"n":63,"subject":"Quantitative Aptitude","stem":"A sum of money amounts to ₹8,840 in 7 years and to ₹9,880 in 9 years at a certain rate of simple interest. What is the annual rate of interest?","stemHi":"एक धनराशि साधारण ब्याज की किसी निश्चित दर पर 7 वर्षों में ₹8,840 और 9 वर्षों में ₹9,880 हो जाती है। ब्याज की वार्षिक दर कितनी है?","context":null,"contextHi":null,"options":[{"key":"a","text":"12%","textHi":null,"imageUrl":null},{"key":"b","text":"15%","textHi":null,"imageUrl":null},{"key":"c","text":"10%","textHi":null,"imageUrl":null},{"key":"d","text":"13%","textHi":null,"imageUrl":null}],"correct":"c","sol":"$39","solHi":"$3a","pos":2,"neg":0.5},{"id":6264,"n":64,"subject":"Quantitative Aptitude","stem":"Find the value of\n

\sqrt{15 + \sqrt{15 + \sqrt{15 + \dots}}}

","stemHi":"निम्नलिखित का मान ज्ञात कीजिए:\n

\sqrt{15 + \sqrt{15 + \sqrt{15 + \dots}}}

","context":null,"contextHi":null,"options":[{"key":"a","text":"5","textHi":null,"imageUrl":null},{"key":"b","text":"3","textHi":null,"imageUrl":null},{"key":"c","text":"8","textHi":null,"imageUrl":null},{"key":"d","text":"15","textHi":null,"imageUrl":null}],"correct":"a","sol":"$3b","solHi":"$3c","pos":2,"neg":0.5},{"id":6265,"n":65,"subject":"Quantitative Aptitude","stem":"A drum contains 120 litres of a

80 %

pure acid solution. If 16 litres of this solution are removed and replaced with 16 litres of pure water, what is the new percentage of acid in the solution ?","stemHi":"एक ड्रम में

80 %

शुद्ध अम्ल वाले विलयन के 120 लीटर हैं। यदि इस विलयन में से 16 लीटर निकालकर उसके स्थान पर 16 लीटर शुद्ध जल मिला दिया जाए, तो विलयन में अम्ल का नया प्रतिशत क्या होगा?","context":null,"contextHi":null,"options":[{"key":"a","text":"60.33%","textHi":null,"imageUrl":null},{"key":"b","text":"61 %","textHi":null,"imageUrl":null},{"key":"c","text":"69.33%","textHi":null,"imageUrl":null},{"key":"d","text":"65%","textHi":null,"imageUrl":null}],"correct":"c","sol":"$3d","solHi":"$3e","pos":2,"neg":0.5},{"id":6266,"n":66,"subject":"Quantitative Aptitude","stem":"A chord of a circle measures 24 cm in length, and the perpendicular distance from the center of the circle to the chord is 9 cm . What is the radius of the circle ?","stemHi":"एक वृत्त की जीवा की लंबाई 24 cm है तथा वृत्त के केंद्र से उस जीवा पर डाली गई लंब दूरी 9 cm है। वृत्त की त्रिज्या कितनी है?","context":null,"contextHi":null,"options":[{"key":"a","text":"15 cm","textHi":null,"imageUrl":null},{"key":"b","text":"16 cm","textHi":null,"imageUrl":null},{"key":"c","text":"20 cm","textHi":null,"imageUrl":null},{"key":"d","text":"24 cm","textHi":null,"imageUrl":null}],"correct":"a","sol":"$3f","solHi":"$40","pos":2,"neg":0.5},{"id":6267,"n":67,"subject":"Quantitative Aptitude","stem":"A 8 cm-radius solid metal sphere is melted and reshaped into smaller spheres, each with a 2 cm radius. From the original sphere, how many smaller spheres can be created ?","stemHi":"8 cm त्रिज्या वाले एक ठोस धातु के गोले को पिघलाकर 2 cm त्रिज्या वाले छोटे गोले बनाए जाते हैं। मूल गोले से ऐसे कितने छोटे गोले बनाए जा सकते हैं?","context":null,"contextHi":null,"options":[{"key":"a","text":"63","textHi":null,"imageUrl":null},{"key":"b","text":"66","textHi":null,"imageUrl":null},{"key":"c","text":"64","textHi":null,"imageUrl":null},{"key":"d","text":"65","textHi":null,"imageUrl":null}],"correct":"c","sol":"$41","solHi":"$42","pos":2,"neg":0.5},{"id":6268,"n":68,"subject":"Quantitative Aptitude","stem":"A solid spherical block is melted and recast into a cylinder of the same radius. The height of the cylinder is what fraction of the radius of the sphere ?","stemHi":"एक ठोस गोलाकार ब्लॉक को पिघलाकर उसी त्रिज्या का एक बेलन बनाया जाता है। बेलन की ऊँचाई, गोले की त्रिज्या का कितना भाग है?","context":null,"contextHi":null,"options":[{"key":"a","text":"

\frac{4}{3}

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

\frac{3}{4}

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

\frac{5}{4}

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

\frac{3}{5}

","textHi":null,"imageUrl":null}],"correct":"a","sol":"$43","solHi":"$44","pos":2,"neg":0.5},{"id":6269,"n":69,"subject":"Quantitative Aptitude","stem":"What is the value of

(4.5)^{2} + \sqrt{2.25} - (6.5 \div 1.3)

?","stemHi":"

(4.5)^{2} + \sqrt{2.25} - (6.5 \div 1.3)

का मान क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"20.45","textHi":null,"imageUrl":null},{"key":"b","text":"21.25","textHi":null,"imageUrl":null},{"key":"c","text":"20","textHi":null,"imageUrl":null},{"key":"d","text":"16.75","textHi":null,"imageUrl":null}],"correct":"d","sol":"Decimal simplification\nWork out each part separately and then apply BODMAS.\n

{(4.5)}^{2} = 4.5 \times 4.5 = 20.25

\sqrt{2.25} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5

6.5 \div 1.3 = \frac{65}{13} = 5

\nSo the expression =20.25+1.5-5=16.75.\nHence the correct option is (d).","solHi":"दशमलव सरलीकरण\nप्रत्येक भाग का मान अलग-अलग निकालिए और फिर BODMAS लागू कीजिए।\n

{(4.5)}^{2} = 4.5 \times 4.5 = 20.25

\sqrt{2.25} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5

6.5 \div 1.3 = \frac{65}{13} = 5

\nअतः व्यंजक =20.25+1.5-5=16.75।\nअतः सही विकल्प (d) है।","pos":2,"neg":0.5},{"id":6270,"n":70,"subject":"Quantitative Aptitude","stem":"Simplify:

\frac{1}{\sqrt{7} - \sqrt{6}} + \frac{1}{\sqrt{7} + \sqrt{6}}

","stemHi":"सरल कीजिए:

\frac{1}{\sqrt{7} - \sqrt{6}} + \frac{1}{\sqrt{7} + \sqrt{6}}

","context":null,"contextHi":null,"options":[{"key":"a","text":"

2 \sqrt{7}

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

\sqrt{6}

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

\sqrt{7}

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

2 \sqrt{6}

","textHi":null,"imageUrl":null}],"correct":"a","sol":"$45","solHi":"$46","pos":2,"neg":0.5},{"id":6271,"n":71,"subject":"Quantitative Aptitude","stem":"A line passes through the point

(6, 8)

and has a slope of -3 . What is the equation of the line ?","stemHi":"एक रेखा बिंदु

(6, 8)

से होकर जाती है और उसकी प्रवणता (ढाल) -3 है। उस रेखा का समीकरण क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"

y = - 3 x + 26

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

y = - 3 x + 24

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

y = - 3 x - 26

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

y = - x + 26

","textHi":null,"imageUrl":null}],"correct":"a","sol":"Equation of a line\nPoint-slope form:

y - y_{1} = m (x - x_{1})

.\nHere m=-3,

x_{1} = 6

and

y_{1} = 8

.\ny-8=-3(x-6)\ny-8=-3x+18\ny=-3x+26\nCheck: putting x=6 gives y=-18+26=8, so the point (6,8) does lie on it.\nHence the correct option is (a).","solHi":"रेखा का समीकरण\nबिंदु-प्रवणता रूप:

y - y_{1} = m (x - x_{1})

।\nयहाँ m=-3,

x_{1} = 6

तथा

y_{1} = 8

है।\ny-8=-3(x-6)\ny-8=-3x+18\ny=-3x+26\nजाँच: x=6 रखने पर y=-18+26=8 आता है, अतः बिंदु (6,8) इस रेखा पर स्थित है।\nअतः सही विकल्प (a) है।","pos":2,"neg":0.5},{"id":6272,"n":72,"subject":"Quantitative Aptitude","stem":"Find the value of

k

such that the quadratic equation

y^{2} + k y + 36 = 0

has real and distinct roots.","stemHi":"

k

का वह मान ज्ञात कीजिए जिसके लिए द्विघात समीकरण

y^{2} + k y + 36 = 0

के मूल वास्तविक और भिन्न हों।","context":null,"contextHi":null,"options":[{"key":"a","text":"

k > 12

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

k > 10

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

k < - 15

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

k < - 11

","textHi":null,"imageUrl":null}],"correct":"a","sol":"$47","solHi":"$48","pos":2,"neg":0.5},{"id":6273,"n":73,"subject":"Quantitative Aptitude","stem":"Simplify:\n

\frac{1 + \cot^{2} (x)}{1 + \tan^{2} (x)} \times \frac{cosec (x) - \cot (x)}{cosec (x) + \cot (x)}

","stemHi":"सरल कीजिए:\n

\frac{1 + \cot^{2} (x)}{1 + \tan^{2} (x)} \times \frac{cosec (x) - \cot (x)}{cosec (x) + \cot (x)}

","context":null,"contextHi":null,"options":[{"key":"a","text":"

\frac{1}{1 + \cos (x)}

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

\frac{\cot^{2} (x)}{1 + \cos (x)}

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

\frac{1 - \cos (x)}{1 + \cos (x)}

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

\frac{\cot^{2} (x) [1 - \cos (x)]}{1 + \cos (x)}

","textHi":null,"imageUrl":null}],"correct":"d","sol":"$49","solHi":"$4a","pos":2,"neg":0.5},{"id":6274,"n":74,"subject":"Quantitative Aptitude","stem":"If

\cos θ = \frac{8}{27}

find

\sin^{2} θ

.","stemHi":"यदि

\cos θ = \frac{8}{27}

है, तो

\sin^{2} θ

ज्ञात कीजिए।","context":null,"contextHi":null,"options":[{"key":"a","text":"

\frac{665}{729}

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

\frac{793}{729}

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

\frac{625}{729}

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

\frac{650}{729}

","textHi":null,"imageUrl":null}],"correct":"a","sol":"$4b","solHi":"$4c","pos":2,"neg":0.5},{"id":6275,"n":75,"subject":"Quantitative Aptitude","stem":"In a right triangle

A B C

, right-angled at

B

, if

\cot B = 1

, what is the value of

2 \sin B \cos B

?","stemHi":"एक समकोण त्रिभुज

A B C

में, जिसका समकोण

B

पर है, यदि

\cot B = 1

है, तो

2 \sin B \cos B

का मान क्या है?","context":null,"contextHi":null,"options":[{"key":"a","text":"3","textHi":null,"imageUrl":null},{"key":"b","text":"4","textHi":null,"imageUrl":null},{"key":"c","text":"1","textHi":null,"imageUrl":null},{"key":"d","text":"6","textHi":null,"imageUrl":null}],"correct":"c","sol":"$4d","solHi":"$4e","pos":2,"neg":0.5}],"test":{"slug":"ssc-chsl-tier-1--12-nov-2025-shift-2-quant","title":"SSC CHSL Tier-1 12 November 2025 Shift-2 — Quantitative Aptitude Sectional Test","examName":"CHSL Tier-1","paperLabel":"12 November 2025 · Shift 2","sectionSubject":"Quantitative Aptitude","durationMinutes":15,"format":"sectional_test"}}]