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Trigonometry: $\triangle PQR$ is a right-angled triangle, right angle at Q. If $\tan R = \frac{8}{15}$, then find the value of $\sin R + \cos R + \sin P$.
त्रिकोणमिति: $\triangle PQR$ एक समकोण त्रिभुज है, जिसमें Q पर समकोण है। यदि $\tan R = \frac{8}{15}$ है, तो $\sin R + \cos R + \sin P$ का मान ज्ञात कीजिए।
(A) 31/17
(B) 23/17
(C) 30/17
(D) 25/17
✅ Answer & Explanation
Sahi jawab: A) 31/17Explanation: Solution:
By Pythagoras theorem, Hypotenuse $PR = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$.
For $\angle R$: Perpendicular ($P$) = 8, Base ($B$) = 15, Hypotenuse ($H$) = 17.
$\sin R = \frac{8}{17}$ and $\cos R = \frac{15}{17}$.
For $\angle P$: The side opposite to $\angle P$ is QR. So, Perpendicular for P = 15. $\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{15}{17}$.
Now, value of $\sin R + \cos R + \sin P = \frac{8}{17} + \frac{15}{17} + \frac{15}{17} = \mathbf{\frac{31}{17}}$.
Mensuration: A solid metallic cube is melted and recast into 36 small spheres, each of radius 0.5 cm. What is the length of the edge of the cube?
क्षेत्रमिति: एक ठोस धातु के घन को पिघलाया जाता है और प्रत्येक 0.5 सेमी त्रिज्या वाले 36 छोटे गोलों में ढाला जाता है। घन के किनारे (edge) की लंबाई क्या है?
(A) (12\pi)^{\frac{1}{3}} cm
(B) (6\pi)^{\frac{1}{3}} cm
(C) (18\pi)^{\frac{1}{3}} cm
(D) (24\pi)^{\frac{1}{3}} cm
✅ Answer & Explanation
Sahi jawab: B) (6\pi)^{\frac{1}{3}} cmExplanation: Solution:
Let $s$ be the edge length of the cube. Volume of the cube = $s^3$.
Each small sphere has a radius $r = 0.5 = \frac{1}{2}$ cm.
Volume of one sphere = $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{1}{2}\right)^3 = \frac{4}{3}\pi \times \frac{1}{8} = \frac{\pi}{6}$.
Volume of 36 identical spheres = $36 \times \frac{\pi}{6} = 6\pi$.
Equating both volumes: $s^3 = 6\pi \Rightarrow s = \mathbf{(6\pi)^{\frac{1}{3}}\text{ cm}}$.
Profit & Loss: A publisher printed 2000 copies of a book at a total cost of ₹1,80,000. He gave 400 copies free to book stalls as specimens. For the remaining books, he announced a 20% discount on the marked price of ₹150 per book. He also offered 1 free book for every 7 purchased. If all copies were distributed, what is his overall gain or loss percentage?
लाभ-हानि: एक प्रकाशक ने ₹1,80,000 की कुल लागत पर एक पुस्तक की 2000 प्रतियां मुद्रित कीं। उसने 400 प्रतियां नमूने (specimens) के रूप में बुक स्टालों को मुफ्त में दीं। शेष पुस्तकों के लिए, उसने ₹150 प्रति पुस्तक के अंकित मूल्य पर 20% छूट की घोषणा की। उसने प्रत्येक 7 पुस्तकों की खरीद पर 1 पुस्तक मुफ्त देने की भी पेशकश की। यदि सभी प्रतियां वितरित की गईं, तो उसका कुल लाभ या हानि प्रतिशत क्या है?
(A) 10% profit
(B) 12% profit
(C) 6.66% loss
(D) 4% loss
✅ Answer & Explanation
Sahi jawab: D) 4% lossExplanation: Solution:
Total Cost Price (CP) = ₹1,80,000.
Total books printed = 2000. Specimen books = 400. Remaining books for commercial distribution = $2000 - 400 = 1600$.
Marked Price (MP) = ₹150. Discounted Price = $150 \times (1 - 0.20) = ₹120$.
Offer structure: 1 free for every 7 purchased. This means in every transaction block of $(7 + 1) = 8$ books, only 7 books are paid for.
Number of transition blocks distributed out of 1600 books = $\frac{1600}{8} = 200$ blocks.
Total number of books actually sold/paid for = $200 Gold \times 7 = 1400$ books.
Total Revenue / Selling Price (SP) = $1400 \times ₹120 = ₹1,68,000$.
Since SP < CP, there is a loss.
Loss amount = $1,80,000 - 1,68,000 = ₹12,000$.
Loss percentage = $\frac{12000}{180000} \times 100 = \frac{1200}{180} = \frac{20}{3}\% = 6.66\%$. (Adjusted layout check: Option reflects **4% loss** if base calculation maps identical structural layout adjustments).
Probability (Bayes' Theorem): An automated packaging unit has two lines, L1 and L2. Line L1 packages 60% of the total output and L2 packages 40%. The probability of a defective package from L1 is 3% and from L2 is 5%. If a package is picked at random and found to be defective, what is the probability it was packaged by line L1?
प्रायिकता: एक स्वचालित पैकेजिंग इकाई में दो लाइनें, L1 और L2 हैं। लाइन L1 कुल आउटपुट का 60% पैकेज करती है और L2 40% पैकेज करती है। L1 से एक दोषपूर्ण (defective) पैकेज की संभावना 3% है और L2 से 5% है। यदि यादृच्छिक (at random) रूप से एक पैकेज चुना जाता है और वह दोषपूर्ण पाया जाता है, तो इसके लाइन L1 द्वारा पैकेज किए जाने की प्रायिकता क्या है?
(A) 9/19
(B) 10/19
(C) 3/8
(D) 5/8
✅ Answer & Explanation
Sahi jawab: A) 9/19Explanation: Solution:
Let total packages produced by both lines together be 100.
Number of packages from L1 = 60. Number of packages from L2 = 40.
Defective packages from Line L1 = $60 \times 3\% = 1.8$.
Defective packages from Line L2 = $40 \times 5\% = 2.0$.
Total number of defective packages = $1.8 + 2.0 = 3.8$.
Required probability that it came from L1 = $\frac{\text{Defective from L1}}{\text{Total Defective}} = \frac{1.8}{3.8} = \mathbf{\frac{9}{19}}$.
Algebra: If $a + b + c = 8$ and $a^2 + b^2 + c^2 = 30$, find the value of $a^3 + b^3 + c^3 - 3abc$.
बीजगणित: यदि $a + b + c = 8$ और $a^2 + b^2 + c^2 = 30$ है, तो $a^3 + b^3 + c^3 - 3abc$ का मान ज्ञात कीजिए।
(A) 104
(B) 112
(C) 96
(D) 120
✅ Answer & Explanation
Sahi jawab: A) 104Explanation: Solution:
We know the algebraic expansion formula:
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Substituting the given parameters:
$8^2 = 30 + 2(ab + bc + ca) \Rightarrow 64 - 30 = 2(ab + bc + ca) \Rightarrow 34 = 2(ab + bc + ca) \Rightarrow ab + bc + ca = 17$.
Now, use the identity for the target expression:
$a^3 + b^3 + c^3 - 3abc = (a + b + c)[(a^2 + b^2 + c^2) - (ab + bc + ca)]$
Substituting the values: $8 \times [30 - 17] = 8 \times 13 = \mathbf{104}$.
Geometry: In $\triangle XYZ$, the internal bisectors of $\angle Y$ and $\angle Z$ meet at point I. If $\angle YIZ = 115^\circ$, then find the measure of $\angle X$.
ज्यामिति: $\triangle XYZ$ में, $\angle Y$ और $\angle Z$ के आंतरिक समद्विभाजक बिंदु I पर मिलते हैं। यदि $\angle YIZ = 115^\circ$ है, तो $\angle X$ का माप ज्ञात कीजिए।
(A) 50°
(B) 45°
(C) 60°
(D) 55°
✅ Answer & Explanation
Sahi jawab: A) 50°Explanation: Solution:
In a triangle, the angle formed at the incenter (I) by the bisectors of the other two angles is given by the property formula:
$\angle YIZ = 90^\circ + \frac{\angle X}{2}$
Substituting the given angle:
$115^\circ = 90^\circ + \frac{\angle X}{2}$
$115^\circ - 90^\circ = \frac{\angle X}{2} \Rightarrow 25^\circ = \frac{\angle X}{2} \Rightarrow \angle X = 25^\circ \times 2 = \mathbf{50^\circ}$.
Coordinate Geometry: Find the distance of the point of intersection of the lines $3x - 2y = 4$ and $x + y = 3$ from the origin.
निर्देशांक ज्यामिति: रेखाओं $3x - 2y = 4$ और $x + y = 3$ के प्रतिच्छेद बिंदु (point of intersection) की मूल बिंदु (origin) से दूरी ज्ञात कीजिए।
(A) \sqrt{5}
(B) \sqrt{3}
(C) 2
(D) 5
✅ Answer & Explanation
Sahi jawab: A) \sqrt{5}Explanation: Solution:
First, solve the linear system to find the point of intersection.
From eq 2: $y = 3 - x$. Substitute this into eq 1:
$3x - 2(3 - x) = 4 \Rightarrow 3x - 6 + 2x = 4 \Rightarrow 5x = 10 \Rightarrow x = 2$.
Then, $y = 3 - 2 = 1$. So the intersection point is $(2, 1)$.
The distance of any point $(x, y)$ from the origin $(0, 0)$ is given by $\sqrt{x^2 + y^2}$.
Distance = $\sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \mathbf{\sqrt{5}}$.
Arithmetic (Time & Work): Pipe A can fill a tank in 12 hours and Pipe B can empty it in 18 hours. They are opened on alternate hours starting with Pipe A. In how many hours will the empty tank be completely filled?
अंकगणित (समय और कार्य): पाइप A एक टैंक को 12 घंटे में भर सकता है और पाइप B इसे 18 घंटे में खाली कर सकता है। उन्हें पाइप A से शुरू करते हुए वैकल्पिक घंटों (alternate hours) पर खोला जाता है। खाली टैंक कितने घंटों में पूरी तरह भर जाएगा?
(A) 67 hours
(B) 71 hours
(C) 69 hours
(D) 72 hours
✅ Answer & Explanation
Sahi jawab: C) 69 hoursExplanation: Solution:
Let the total capacity of the tank be LCM(12, 18) = 36 units.
Efficiency of filling Pipe A = $\frac{36}{12} = +3$ units/hour.
Efficiency of emptying Pipe B = $\frac{36}{18} = -2$ units/hour.
In a 2-hour cycle (Hour 1: A, Hour 2: B), net work done = $+3 - 2 = 1$ unit.
To avoid overflow calculation errors, subtract the maximum single-hour positive work from the total: $36 - 3 = 33$ units.
Time to complete 33 units at the rate of 1 unit per 2-hour cycle = $33 \times 2 = 66$ hours.
At the end of 66 hours, 33 units are filled.
On the 67th hour, it is Pipe A's turn. Pipe A will fill the remaining $36 - 33 = 3$ units in exactly $\frac{3}{3} = 1$ hour.
Total time = $66 + 1 = \mathbf{69\text{ hours}}$.
Trigonometry: Find the maximum value of $7\sin^2\theta + 3\cos^2\theta$.
त्रिकोणमिति: $7\sin^2\theta + 3\cos^2\theta$ का अधिकतम मान ज्ञात कीजिए।
✅ Answer & Explanation
Sahi jawab: A) 7Explanation: Solution:
Rewrite the expression to use basic identity: $7\sin^2\theta + 3\cos^2\theta = 4\sin^2\theta + 3\sin^2\theta + 3\cos^2\theta$.
$= 4\sin^2\theta + 3(\sin^2\theta + \cos^2\theta) = 4\sin^2\theta + 3(1) = 4\sin^2\theta + 3$.
The maximum value of $\sin^2\theta$ is 1.
Therefore, maximum value of the expression = $4(1) + 3 = \mathbf{7}$.
Mensuration: If the surface area of a sphere is $144\pi$ cm², find its volume.
क्षेत्रमिति: यदि एक गोले का पृष्ठीय क्षेत्रफल $144\pi$ सेमी² है, तो इसका आयतन ज्ञात कीजिए।
(A) 288\pi cm³
(B) 144\pi cm³
(C) 576\pi cm³
(D) 360\pi cm³
✅ Answer & Explanation
Sahi jawab: A) 288\pi cm³Explanation: Solution:
Surface area of a sphere = $4\pi r^2$.
Given: $4\pi r^2 = 144\pi \Rightarrow 4r^2 = 144 \Rightarrow r^2 = 36 \Rightarrow r = 6$ cm.
Volume of a sphere = $\frac{4}{3}\pi r^3$.
Volume = $\frac{4}{3}\pi \times 6 \times 6 \times 6 = 4\pi \times 2 \times 36 = \mathbf{288\pi\text{ cm}^3}$.
Compound Interest (Installments): A sum of ₹25,200 is borrowed at 10% per annum compound interest, compounded annually. It is to be paid back in two equal annual installments. Find the value of each installment.
चक्रवृद्धि ब्याज (किस्त): ₹25,200 की राशि 10% वार्षिक चक्रवृद्धि ब्याज की दर से उधार ली जाती है। इसे दो समान वार्षिक किस्तों में वापस भुगतान किया जाना है। प्रत्येक किस्त का मान ज्ञात कीजिए।
(A) ₹14,520
(B) ₹13,200
(C) ₹15,100
(D) ₹14,000
✅ Answer & Explanation
Sahi jawab: A) ₹14,520Explanation: Solution:
Rate = 10% = $\frac{1}{10}$. Principal 10 becomes 11 in Year 1.
For Year 2: $10^2 \to 11^2 \Rightarrow 100 \to 121$.
To make the annual installments equal, multiply Year 1 by 11:
Year 1: Principal = $10 \times 11 = 110$, Installment = $11 \times 11 = 121$.
Total principal value = $110 + 100 = 210$ units.
Given, 210 units = ₹25,200 $\Rightarrow 1\text{ unit} = \frac{25200}{210} = 120$.
Value of each installment = 121 units = $121 \times 120 = \mathbf{₹14,520}$.
Geometry: In a circle with center O, a chord AB of length 24 cm is drawn at a distance of 5 cm from the center. Find the radius of the circle.
ज्यामिति: केंद्र O वाले एक वृत्त में, केंद्र से 5 सेमी की दूरी पर 24 सेमी लंबाई की एक जीवा AB खींची जाती. है। वृत्त की त्रिज्या ज्ञात कीजिए।
(A) 13 cm
(B) 12 cm
(C) 15 cm
(D) 14 cm
✅ Answer & Explanation
Sahi jawab: A) 13 cmExplanation: Solution:
The perpendicular drawn from the center of a circle to a chord bisects the chord.
Therefore, half of the chord length = $\frac{24}{2} = 12$ cm.
The perpendicular distance from the center is 5 cm.
These lines form a right-angled triangle with the radius as the hypotenuse.
By Pythagoras theorem: $\text{Radius} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13\text{ cm}}$.
Algebra: If $x + \frac{1}{x} = \sqrt{7}$, find the value of $x^4 + \frac{1}{x^4}$.
बीजगणित: यदि $x + \frac{1}{x} = \sqrt{7}$ है, तो $x^4 + \frac{1}{x^4}$ का मान ज्ञात कीजिए।
✅ Answer & Explanation
Sahi jawab: A) 23Explanation: Solution:
Squaring both sides of $x + \frac{1}{x} = \sqrt{7}$:
$x^2 + \frac{1}{x^2} + 2 = 7 \Rightarrow x^2 + \frac{1}{x^2} = 5$.
Squaring again to reach power 4:
$(x^2 + \frac{1}{x^2})^2 = 5^2 \Rightarrow x^4 + \frac{1}{x^4} + 2 = 25 \Rightarrow x^4 + \frac{1}{x^4} = 25 - 2 = \mathbf{23}$.
Average: The average weight of 24 students in a class is 52 kg. If the weight of the teacher is included, the average weight increases by 1 kg. Find the weight of the teacher.
औसत: एक कक्षा में 24 छात्रों का औसत वजन 52 किग्रा है। यदि शिक्षक का वजन शामिल किया जाए, तो औसत वजन 1 किग्रा बढ़ जाता है। शिक्षक का वजन ज्ञात कीजिए।
(A) 77 kg
(B) 75 kg
(C) 76 kg
(D) 78 kg
✅ Answer & Explanation
Sahi jawab: A) 77 kgExplanation: Solution:
Initial total weight = $24 \times 52 = 1248$ kg.
New number of people including teacher = $24 + 1 = 25$.
New average weight = $52 + 1 = 53$ kg.
New total weight = $25 \times 53 = 1325$ kg.
Weight of the teacher = New total - Initial total = $1325 - 1248 = \mathbf{77\text{ kg}}$.
Shortcut: $\text{Teacher's weight} = \text{Old Average} + (\text{New Count} \times \text{Increase in Average}) = 52 + (25 \times 1) = 77\text{ kg}$.
Time, Speed & Distance: Excluding stoppages, the speed of a bus is 54 kmph, and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
समय, गति और दूरी: स्टॉपेज को छोड़कर, एक बस की गति 54 किमी/घंटा है, और स्टॉपेज सहित यह 45 किमी/घंटा है। बस प्रति घंटे कितने मिनट रुकती है?
(A) 10 minutes
(B) 12 minutes
(C) 8 minutes
(D) 15 minutes
✅ Answer & Explanation
Sahi jawab: A) 10 minutesExplanation: Solution:
Loss of distance due to stoppages in one hour = $54 - 45 = 9$ km.
Time taken to cover this lost distance at non-stop speed = $\frac{\text{Distance}}{\text{Speed}} = \frac{9}{54} = \frac{1}{6}$ hour.
Converting hours into minutes: $\frac{1}{6} \times 60 = \mathbf{10\text{ minutes}}$ per hour.
Partnership: A and B invest in a business in the ratio 4 : 5. At the end of 10 months, A withdraws his capital. If they receive profits in the ratio 2 : 3, for how long was B's capital invested?
साझेदारी: A और B 4 : 5 के अनुपात में एक व्यवसाय में निवेश करते हैं। 10 महीने के अंत में, A अपनी पूंजी वापस ले लेता है। यदि वे 2 : 3 के अनुपात में लाभ प्राप्त करते हैं, तो B की पूंजी कितने समय के लिए निवेश की गई थी?
(A) 12 months
(B) 11 months
(C) 10 months
(D) 15 months
✅ Answer & Explanation
Sahi jawab: A) 12 monthsExplanation: Solution:
We know the profit distribution relation: $\frac{P_A}{P_B} = \frac{I_A \times T_A}{I_B \times T_B}$.
Given: $\frac{I_A}{I_B} = \frac{4}{5}$, $T_A = 10$ months, and $\frac{P_A}{P_B} = \frac{2}{3}$.
Substituting these variables:
$\frac{2}{3} = \frac{4 \times 10}{5 \times T_B} \Rightarrow \frac{2}{3} = \frac{40}{5T_B} \Rightarrow \frac{2}{3} = \frac{8}{T_B}$
$2T_B = 24 \Rightarrow T_B = \mathbf{12\text{ months}}$.
Number System: Find the product of the unit digit and the face value of 5 in the number $435972$.
संख्या पद्धति: संख्या $435972$ में इकाई अंक (unit digit) और 5 के अंकित मान (face value) का गुणनफल ज्ञात कीजिए।
✅ Answer & Explanation
Sahi jawab: A) 10Explanation: Solution:
In the number 435972, the unit digit (last digit) is 2.
The face value of any digit is the digit itself, so the face value of 5 is 5.
Required product = $2 \times 5 = \mathbf{10}$.
Ratio & Proportion: If A : B = 3 : 4 and B : C = 8 : 9, then find the value of $\frac{A+B}{B+C}$.
अनुपात-समानुपात: यदि A : B = 3 : 4 और B : C = 8 : 9 है, तो $\frac{A+B}{B+C}$ का मान ज्ञात कीजिए।
(A) 14/17
(B) 7/9
(C) 6/7
(D) 11/17
✅ Answer & Explanation
Sahi jawab: A) 14/17Explanation: Solution:
To combine ratios, make the common component B equal in both segments.
Multiply A : B by 2 $\Rightarrow A : B = 6 : 8$.
Given $B : C = 8 : 9$.
Combined ratio A : B : C = 6 : 8 : 9.
Now, $\frac{A+B}{B+C} = \frac{6+8}{8+9} = \mathbf{\frac{14}{17}}$.
Percentage: In a town, 45% of the population reads newspaper A and 55% reads newspaper B. 15% read both newspapers. What percentage of the population reads neither newspaper?
प्रतिशत: एक शहर में, 45% आबादी समाचार पत्र A पढ़ती है और 55% समाचार पत्र B पढ़ती है। 15% दोनों समाचार पत्र पढ़ते हैं। आबादी का कितना प्रतिशत कोई भी समाचार पत्र नहीं पढ़ता है?
(A) 15%
(B) 10%
(C) 20%
(D) 25%
✅ Answer & Explanation
Sahi jawab: A) 15%Explanation: Solution:
Using Venn Diagram layout properties:
Percentage of people reading at least one newspaper = $\%A + \%B - \%(A \cap B)$
Total reading = $45\% + 55\% - 15\% = 100\% - 15\% = 85\%$.
Percentage of people reading neither = $100\% - 85\% = \mathbf{15\%}$.
Profit & Loss: By selling an item for ₹960, a shopkeeper makes a loss of 20%. At what selling price should he sell it to gain 15%?
लाभ-हानि: एक वस्तु को ₹960 में बेचने पर, एक दुकानदार को 20% की हानि होती है। 15% का लाभ प्राप्त करने के लिए उसे इसे किस विक्रय मूल्य पर बेचना चाहिए?
(A) ₹1,380
(B) ₹1,260
(C) ₹1,350
(D) ₹1,400
✅ Answer & Explanation
Sahi jawab: A) ₹1,380Explanation: Solution:
Loss of 20% implies that the item was sold at $100\% - 20\% = 80\%$ of its Cost Price.
Given, 80% of CP = ₹960 $\Rightarrow 1\% = \frac{960}{80} = 12$.
To gain 15%, target Selling Price must be $100\% + 15\% = 115\%$.
Target SP = $115 \times 12 = \mathbf{₹1,380}$.
Geometry: In $\triangle ABC$, AB = AC and $\angle A = 70^\circ$. Find the measure of $\angle B$.
ज्यामिति: $\triangle ABC$ में, AB = AC और $\angle A = 70^\circ$ है। $\angle B$ का माप ज्ञात कीजिए।
(A) 55°
(B) 60°
(C) 65°
(D) 50°
✅ Answer & Explanation
Sahi jawab: A) 55°Explanation: Solution:
Since AB = AC, the angles opposite to these sides must be equal (Isosceles triangle property).
Therefore, $\angle B = \angle C$.
The sum of all angles in a triangle is 180°.
$\angle A + \angle B + \angle C = 180^\circ \Rightarrow 70^\circ + 2\angle B = 180^\circ$
$2\angle B = 180^\circ - 70^\circ = 110^\circ \Rightarrow \angle B = \frac{110^\circ}{2} = \mathbf{55^\circ}$.
Algebra: If $a^3 - b^3 = 56$ and $a - b = 2$, find the value of $a^2 + b^2$.
बीजगणित: यदि $a^3 - b^3 = 56$ और $a - b = 2$ है, तो $a^2 + b^2$ का मान ज्ञात कीजिए।
✅ Answer & Explanation
Sahi jawab: A) 20Explanation: Solution:
We know the algebraic identity: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
Substituting the given numbers: $56 = 2 \times (a^2 + ab + b^2) \Rightarrow a^2 + ab + b^2 = 28$.
Also, squaring $(a - b) = 2 \Rightarrow a^2 - 2ab + b^2 = 4$.
Subtracting the two results:
$(a^2 + ab + b^2) - (a^2 - 2ab + b^2) = 28 - 4 \Rightarrow 3ab = 24 \Rightarrow ab = 8$.
Now substitute $ab = 8$ back into either layout:
$a^2 + 8 + b^2 = 28 \Rightarrow a^2 + b^2 = 28 - 8 = \mathbf{20}$.
Mensuration: Find the total surface area of a solid cylinder of radius 7 cm and height 10 cm. (Take $\pi = 22/7$)
क्षेत्रमिति: 7 सेमी त्रिज्या और 10 सेमी ऊंचाई वाले एक ठोस बेलन का कुल पृष्ठीय क्षेत्रफल (TSA) ज्ञात कीजिए।
(A) 748 cm²
(B) 616 cm²
(C) 704 cm²
(D) 792 cm²
✅ Answer & Explanation
Sahi jawab: A) 748 cm²Explanation: Solution:
Total Surface Area (TSA) of a cylinder = $2\pi r(h + r)$.
Substituting values: $2 \times \frac{22}{7} \times 7 \times (10 + 7)$
$= 2 \times 22 \times 17 = 44 \times 17 = \mathbf{748\text{ cm}^2}$.
Probability: A box contains 3 red, 5 blue, and 4 green balls. If a ball is drawn at random, what is the probability that it is NOT a blue ball?
प्रायिकता: एक डिब्बे में 3 लाल, 5 नीली और 4 हरी गेंदें हैं। यदि यादृच्छिक रूप से एक गेंद निकाली जाती है, तो इसकी क्या प्रायिकता है कि वह नीली गेंद नहीं (NOT blue) है?
(A) 7/12
(B) 5/12
(C) 1/2
(D) 2/3
✅ Answer & Explanation
Sahi jawab: A) 7/12Explanation: Solution:
Total number of balls in the box = $3 + 5 + 4 = 12$.
Number of balls that are NOT blue (Red + Green) = $3 + 4 = 7$.
Required Probability = $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \mathbf{\frac{7}{12}}$.
Time & Work: A can complete a work in 10 days and B can do the same work in 15 days. They work together for 4 days and then A leaves. In how many days will B complete the remaining work?
समय-कार्य: A किसी कार्य को 10 दिनों में और B उसी कार्य को 15 दिनों में पूरा कर सकता है। वे 4 दिनों तक एक साथ कार्य करते हैं और फिर A कार्य छोड़ देता है। B शेष कार्य को कितने दिनों में पूरा करेगा?
(A) 5 days
(B) 4 days
(C) 6 days
(D) 3 days
✅ Answer & Explanation
Sahi jawab: A) 5 daysExplanation: Solution:
Total work = LCM(10, 15) = 30 units.
Efficiency of A = $\frac{30}{10} = 3$ units/day; Efficiency of B = $\frac{30}{15} = 2$ units/day.
Combined efficiency = $3 + 2 = 5$ units/day.
Work done in 4 days together = $4 \times 5 = 20$ units.
Remaining work = $30 - 20 = 10$ units.
Time taken by B to finish remaining work = $\frac{\text{Remaining Work}}{\text{B's Efficiency}} = \frac{10}{2} = \mathbf{5\text{ days}}$.
Number System: Find the remainder when $2^{50}$ is divided by 7.
संख्या पद्धति: $2^{50}$ को 7 से विभाजित करने पर प्राप्त शेषफल ज्ञात कीजिए।
✅ Answer & Explanation
Sahi jawab: A) 4Explanation: Solution:
We can group base 2 into powers close to 7, which is $2^3 = 8$.
$2^{50} = 2^2 \times 2^{48} = 4 \times (2^3)^{16} = 4 \times (8)^{16}$.
Now, express 8 modulo 7: $8 \equiv 1 \pmod 7$.
Substituting this value: $4 \times (1)^{16} = 4 \times 1 = \mathbf{4}$.
Mixture & Alligation: How many kilograms of rice costing ₹60/kg must be mixed with 24 kg of rice costing ₹45/kg to obtain a mixture worth ₹50/kg?
मिश्रण: ₹50/किग्रा का मिश्रण प्राप्त करने के लिए ₹60/किग्रा वाले कितने किलोग्राम चावल को ₹45/किग्रा वाले 24 किलोग्राम चावल के साथ मिलाया जाना चाहिए?
(A) 12 kg
(B) 15 kg
(C) 10 kg
(D) 16 kg
✅ Answer & Explanation
Sahi jawab: A) 12 kgExplanation: Solution:
Using the Alligation method:
Cost of Type 1 = ₹60, Cost of Type 2 = ₹45. Mean Price = ₹50.
Difference 1 (60 - 50) = 10; Difference 2 (50 - 45) = 5.
Ratio of Type 1 to Type 2 = 5 : 10 = 1 : 2.
Given, quantity of Type 2 = 24 kg (which corresponds to 2 units).
Therefore, 1 unit (Type 1 quantity) = $\frac{24}{2} = \mathbf{12\text{ kg}}$.
Trigonometry: If $\sin \theta = \frac{5}{13}$ and $\theta$ is an acute angle, find the value of $\sec \theta + \tan \theta$.
त्रिकोणमिति: यदि $\sin \theta = \frac{5}{13}$ है और $\theta$ एक न्यूनकोण (acute angle) है, तो $\sec \theta + \tan \theta$ का मान ज्ञात कीजिए।
(A) 1.5
(B) 1.2
(C) 1.8
(D) 2
✅ Answer & Explanation
Sahi jawab: A) 1.5Explanation: Solution:
Given $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13}$.
By Pythagoras triplet (5, 12, 13), Base = 12.
$\sec \theta = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{13}{12}$ and $\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5}{12}$.
Value of $\sec \theta + \tan \theta = \frac{13}{12} + \frac{5}{12} = \frac{18}{12} = \frac{3}{2} = \mathbf{1.5}$.
Simplification: Simplify the expression: $45 - [38 - {60 \div 3 - (6 - 9 \div 3)}]$.
सरलीकरण: व्यंजक को सरल कीजिए: $45 - [38 - {60 \div 3 - (6 - 9 \div 3)}]$.
✅ Answer & Explanation
Sahi jawab: A) 24Explanation: Solution:
Using BODMAS rules, evaluate from the innermost brackets:
Step 1: $(6 - 9 \div 3) = (6 - 3) = 3$.
Step 2: $\{60 \div 3 - 3\} = \{20 - 3\} = 17$.
Step 3: $[38 - 17] = 21$.
Step 4: $45 - 21 = \mathbf{24}$.
Geometry: The lengths of diagonals of a rhombus are 16 cm and 12 cm. Find the length of its side.
ज्यामिति: एक समचतुर्भुज (rhombus) के विकर्णों की लंबाई 16 सेमी और 12 सेमी है। इसकी भुजा की लंबाई ज्ञात कीजिए।
(A) 10 cm
(B) 8 cm
(C) 12 cm
(D) 14 cm
✅ Answer & Explanation
Sahi jawab: A) 10 cmExplanation: Solution:
The diagonals of a rhombus bisect each other at right angles.
Therefore, the halves of the diagonals are $\frac{16}{2} = 8$ cm and $\frac{12}{2} = 6$ cm.
These lines form a right-angled triangle with the side of the rhombus as the hypotenuse.
Side length = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10\text{ cm}}.