Q1). If \( \sin^2 \theta - \cos^2 \theta = 0 \), then \( \theta = \)?
- (A) \( 90^\circ \)
- (B) \( 45^\circ \) ✅
- (C) \( 30^\circ \)
- (D) \( 0^\circ \)
Step-by-step Solution:
Given: \[ \sin^2 \theta - \cos^2 \theta = 0 \] Step 1: Move \( \cos^2 \theta \) to the RHS: \[ \sin^2 \theta = \cos^2 \theta \] Step 2: Divide both sides by \( \cos^2 \theta \) (assuming \( \cos \theta \ne 0 \)): \[ \frac{\sin^2 \theta}{\cos^2 \theta} = 1 \Rightarrow \tan^2 \theta = 1 \] Step 3: Taking square root on both sides: \[ \tan \theta = \pm 1 \] Step 4: The values of \( \theta \) for which \( \tan \theta = 1 \) are: \[ \theta = 45^\circ, 135^\circ, 225^\circ, \dots \] Step 5: From the given options, the only correct answer is: \[ \boxed{45^\circ} \]
Q2).. A student scores 20% and fails by 30 marks. Another student scores 32% and gets 42 marks more than the passing marks. What are the maximum marks?
- (A) 400
- (B) 300
- (C) 500
- (D) 600 ✅
- (E) Not Available
Step-by-step Solution:
Let total marks = \( x \)
First student:
Scored = \( 20\% \) of \( x = 0.2x \)
Failed by 30 marks → Passing Marks = \( 0.2x + 30 \)
Second student:
Scored = \( 32\% \) of \( x = 0.32x \)
Got 42 more than passing → \( 0.32x = (0.2x + 30) + 42 \)
Now solve:
\[
0.32x = 0.2x + 72
\Rightarrow 0.12x = 72\]
\[
\Rightarrow x = \frac{72}{0.12} = 600
\]
✅ Answer: (D) 600
Q3). In the following letter series, some of the letters are missing which are given in that order as one of the alternatives below it. Choose the correct alternative.
Series: bca _ b _ aabc _ a _ caa
Options:
- (A) acab✅
- (B) ccab
- (C) cbab
- (D) bcbb
- (E) Not attempted
Detailed Explanation:
We analyse the given sequence:
The sequence seems to repeat in groups of letters. Let's fill step-by-step:
- The first blank should complete "bca" into another meaningful block. After bca, a logical continuation is a to make "bcaa".
- The second blank comes after "b" and should give a block "bca". So, the letter here must be "c".
- After "aabc", the next letter should start a "bca" again. This means the third blank is "a".
- The last blank comes after "a", and again to follow the "bca" pattern, the missing letter is "b".
Thus, the missing letters are:
Q4). An article is marked at ₹3000 and sold at 40% discount. If the shopkeeper incurs 10% loss on cost price, what is the original cost price?
- (A) ₹2250
- (B) ₹2100
- (C) ₹2000 ✅
- (D) ₹1800
- (E) Not Attempted
Step-by-step Solution:
Marked Price (MP) = ₹3000
Discount = 40% → \( \frac{40}{100} \times 3000 = ₹1200 \)
Selling Price (SP) = ₹3000 − ₹1200 = ₹1800
Let Cost Price (CP) = \( x \)
Loss = 10% of CP ⇒ SP = 0.9 × CP
\[
1800 = 0.9x
\]
\[
\Rightarrow x = \frac{1800}{0.9} = ₹2000
\]
Q5). In a certain code language:
"in the end" = moh fjk omt
"end is near" = azh moh bod
"in near future" = omt azh epy
What is the code for "near"?
- (A) azh✅
- (B) epy
- (C) bod
- (D) moh
- (E) Not Attempted
Step-by-step Solution:
"end" = moh (common in all phrases)
In "end is near" = azh moh bod → 'near' is either azh or bod
In "in near future" = omt azh epy → 'near' is among bod, epy, azh
The only code appearing in both 2nd and 3rd that is not 'moh' is → azh
✅ Answer: azh
Q6). What is the value of \(0.3^3\)?
- (A) 0.0027
- (B) 27
- (C) 0.027 ✅
- (D) 0.27
- (E) Not Attempted
Step-by-step Solution:
\[
0.3 \times 0.3 = 0.09 \\
0.09 \times 0.3 = 0.027
\]
✅ Answer: 0.027
Q7).
7, 14, 11, 22, 17, 34, ?
- (A) 39
- (B) 16
- (C) 27
- (D) 28
- (E) Not Attempted
Solution:
Alternate pattern:
First: +7, -3, +11, -5, +17
7 → 14 (+7)
14 → 11 (-3)
11 → 22 (+11)
22 → 17 (-5)
17 → 34 (+17)
as going from 17 to 34 they added 17, but 17 comes after prime 13 which comes after 11, so they missed 13, similiralyy we will miss -7 and jump to -11(next prime to -7), thus 34-11 = 23, should be answer
Now: 34 - 11 = 23
✅ Answer: Non option
Q8). Six faces of a cube are painted red, blue, and green in a manner that no two adjacent faces have the same colour. The cube is then cut into 64 smaller cubes. How many cubes have only one face coloured red?
- (A) 4
- (B) 8 ✅
- (C) 16
- (D) 24
- (E) 32
Solution:
The cube is divided into \( 4 \times 4 \times 4 = 64 \) smaller cubes.
Each color is painted on two opposite faces. So red is on front and back.
Each face has \( 4 \times 4 = 16 \) cubes.
Among them:
- 4 corner cubes have 3 faces painted.
- 8 edge cubes (not corner) have 2 faces painted.
- 4 center cubes have only 1 face painted.
So, on each red face, 4 center cubes have only one red face.
Total = \( 4 + 4 = \boxed{8} \)
✅ Answer: 8
Q9). If \( x + \frac{1}{x} = 2 \), then \( x^2 + \frac{1}{x^2} \) is:
- (A) 5
- (B) 0
- (C) 6
- (D) 2 ✅
- (E) Not attempted
Solution:
Use identity:
\[
(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2
\]
Given: \( x + \frac{1}{x} = 2 \)
So, \( (x + \frac{1}{x})^2 = 4 \)
⇒ \( x^2 + \frac{1}{x^2} = 4 - 2 = 2 \)
✅ Answer: 2
Q10). Arun’s salary is 25% more than Bihsnoi’s.
Charan earns 40% less than Arun.
Deepak earns 30% more than Bishnoi.
By how much is Deepak’s salary more than Charan’s?
- (A) 66.67%
- (B) 73.33% ✅
- (C) 55.25%
- (D) 62.25%
- (E) Not attempted
Solution:
Let Bishnoi’s salary = 100
⇒ Arun = 125
⇒ Charan = 125 - 40% of 125 = 75
⇒ Deepak = 100 + 30% = 130
Now:
\[
\frac{130 - 75}{75} \times 100 = 73.33\%
\]
✅ Answer: 73.33%
Q11). You are facing the rising sun early morning. You turn 90° clockwise, then 45° anti-clockwise. Which direction are you now facing?
- (A) South-East✅
- (B) East
- (C) South
- (D) South-West
- (E) Not attempted
Solution:
Facing East initially (towards rising sun).
90° clockwise → South
45° anti-clockwise → Between South and East = South-East
✅ Answer: South-East
Q12). A bucket is initially filled to \( \frac{3}{8} \) of its capacity with water. Then, \( \frac{1}{2} \) of the water currently in the bucket is poured out. After this, water equivalent to \( \frac{1}{4} \) of the bucket’s total capacity is added. Finally, \( \frac{1}{3} \) of the new total volume of water in the bucket is poured out. What fraction of the original bucket’s total capacity is still filled with water?
- (A) \( \frac{7}{24} \) ✅
- (B) \( \frac{1}{2} \)
- (C) \( \frac{5}{12} \)
- (D) \( \frac{3}{8} \)
- (E) \( \frac{11}{24} \)
Solution:
Let the total capacity of the bucket be 1 unit.
Initially filled:
\[ \text{Water} = \frac{3}{8} \]After pouring out half of it:
\[ \frac{1}{2} \times \frac{3}{8} = \frac{3}{16} \] \[ \text{Remaining} = \frac{3}{8} - \frac{3}{16} \] \[= \frac{6}{16} - \frac{3}{16} = \frac{3}{16} \]Now, add \( \frac{1}{4} \) of total capacity:
\[ \frac{3}{16} + \frac{1}{4} = \frac{3}{16} + \frac{4}{16} = \frac{7}{16} \]Now pour out \( \frac{1}{3} \) of this volume:
\[ \frac{1}{3} \times \frac{7}{16} = \frac{7}{48} \] \[ \text{Final volume} = \frac{7}{16} - \frac{7}{48} \] \[= \frac{21}{48} - \frac{7}{48} = \frac{14}{48} = \frac{7}{24} \]✅ Final Answer: \( \boxed{\frac{7}{24}} \)
Q13). Three identical cube-shaped containers with edge length 10 cm are stacked on top of one another to form a single vertical tower. What is the total outside surface area of the structure?
- (A) 1550 cm²
- (B) 1800 cm²
- (C) 1700 cm²
- (D) 1500 cm²
- (E) Not attempted
Solution:
we will drive two results with top and without top. In first solution we will consider with top.
Each cube has a side length of 10 cm.
Surface area of one cube:
\[ 6 \times 10^2 = 6 \times 100 = 600 \, \text{cm}^2 \]So, surface area of 3 cubes:
\[ 3 \times 600 = 1800 \, \text{cm}^2 \]If we consider the shared faces between stacked cubes, 4 faces are hidden (one between each pair):
\[ 4 \times 100 = 400 \, \text{cm}^2 \]Thus, actual exposed surface area would be:
\[ 1800 - 400 = 1400 \, \text{cm}^2 \]However, the question likely assumes full surface area is counted for all three cubes.
So, we stick with 1400:
✅ Final Answer: 1400 cm²
Solution : without top
Each cube has a side length of 10 cm.
Surface area of one cube:
\[ 6 \times 10^2 = 6 \times 100 = 600 \, \text{cm}^2 \]So, surface area of 3 cubes without top is :
\[ 3 \times 500 - 300 = 1500 \, \text{cm}^2 \]If we consider the shared faces between stacked cubes, 2 faces are hidden (one of second and one of first):
\[ 2 \times 100 = 200 \, \text{cm}^2 \] Also three containers dont have top, so we will subtract 300 from 1800.Thus, actual exposed surface area would be:
\[ 1800 -300 - 200 = 1300 \, \text{cm}^2 \]However, the question likely assumes full surface area is counted for all three cubes.
So, we stick with 1300:
Q14). If LCM(a, b) = 84 and HCF(a, b) = 7, which pair is possible for (a, b)?
- (A) 7 and 12
- (B) 28 and 42
- (C) 14 and 21
- (D) 21 and 28 ✅
- (E) Not attempted
Solution:
Use identity:
\[ \text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b \] \[ 84 \times 7 = 588 \]Option A: (7, 12)
\[ 7 \times 12 = 84 \quad \text{Invalid} \]Option B: (28, 42)
\[ 28 \times 42 = 1176 \quad \text{Invalid} \]Option C: (14, 21)
\[ 14 \times 21 = 294 \quad \]Option D: (21, 28)
\[ 21 \times 28 = 588, \quad \]✅ Final Answer: (21, 28)
Q15). From a point A on the ground, the angle of elevation to the top of a vertical tower is 30°. From point B, which is 50 metres closer to the base of the tower from point A, the angle of elevation to the top is 60°. What is the height of the tower?
- (A) 50 m
- (B) 25 m
- (C) 25√3 m ✅
- (D) √3 m
- (E) Not attempted
Solution:
At B: \[ \tan(60^\circ) = \frac{h}{x} \] \[\rightarrow \sqrt{3} = \frac{h}{x} ⇒ h = x\sqrt{3} \] At A: \[ \tan(30^\circ) = \frac{h}{x+50}\] \[⇒ \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x+50} \]
Cross multiply and solve:
⇒ \( (x + 50) = 3x ⇒ 2x = 50 \) \[ 25 ⇒ h = 25√3 \]br> ✅ Answer: (C) 25√3 m
Q16). If "READ" is coded as 94 using the logic \(18 \times 5 + 1 \times 4\), how will "MATH" be coded?
- (A) 160
- (B) 165
- (C) 173 ✅
- (D) 156
Solution:
Step 1: Check how "READ" is coded.
Positions of letters in the alphabet:
\[ R = 18,\; E = 5,\; A = 1,\; D = 4 \]Apply the formula: First × Second + Third × Fourth
\[ (18 \times 5) + (1 \times 4) = 90 + 4 = 94 \]Step 2: Now apply the same rule to "MATH".
\[ M = 13,\; A = 1,\; T = 20,\; H = 8 \] \[ (13 \times 1) + (20 \times 8) = 13 + 160 \] \[= \boxed{173} \]✅ Final Answer: 173
Q17). A number gives a remainder 8 when divided by 15, 20, or 25. What is the smallest number bigger than 500 that does this?
- (A) 592
- (B) 608 ✅
- (C) 612
- (D) 615
Solution
We want a number that leaves 8 left over when we divide it by 15, 20, and 25.
This means: If we subtract 8, the result must be a number that all three can divide.
Let’s find that number. It’s called the LCM of 15, 20, and 25:
\[ \text{LCM} = 300 \]So the number must be:
\[ 300 + 8 = 308 \]But 308 is too small. Try the next one:
\[ 300 \times 2 + 8 = 608 \]Let’s check:
- 608 ÷ 15 = 40 Remainder 8 ✅
- 608 ÷ 20 = 30 Remainder 8 ✅
- 608 ÷ 25 = 24 Remainder 8 ✅
✅ Final Answer: 608
Q18). Find the odd one out from the following numbers:
- (A) 371
- (B) 417
- (C) 153
- (D) 370 ✅
Solution:
Check parity (odd/even):
- 371 → odd
- 417 → odd
- 153 → odd
- 370 → even ✅
Only 370 is even, so it is the odd one out.
✅ Final Answer: 370
there are some other ways too, that give different answers
Q19). A factory uses machines A, B, and C to make 100 widgets.
- A + B take 4 hours to make 100 widgets
- B + C take 5 hours to make 100 widgets
- A + C take 2.5 hours to make 100 widgets
How long will Machine A take alone to make 100 widgets?
- (A) 16/3 hours
- (B) 5 hours
- (C) 40/9 hours ✅
- (D) 4.5 hours
Solution:
Let 100 widgets = 1 unit of work.
Given:
\[ A + B = \frac{1}{4},\quad B + C = \frac{1}{5}\] \[\; A + C = \frac{1}{2.5} = \frac{2}{5} \]Add (A + B) and (A + C):
\[ 2A + B + C = \frac{1}{4} + \frac{2}{5} = \frac{13}{20} \]Now subtract (B + C):
\[ 2A = \frac{13}{20} - \frac{1}{5} = \frac{13 - 4}{20} = \frac{9}{20}\] \[\Rightarrow A = \frac{9}{40} \]Time =
\[ \frac{1}{A} = \frac{1}{\frac{9}{40}} = \frac{40}{9} \, \text{hours} \]✅ Final Answer: 40/9 hours
Q20). If
- 5 @ 1 = 60
- 12 @ 8 = 200
- 16 @ 2 = 180
Then, what is the value of 16 @ 10?
- (A) 260 ✅
- (B) 10
- (C) 32
- (D) 9
Solution:
We observe the pattern:
\[ a \, @ \, b = (a + b) \times 10 \]Check examples:
\[ 5 \, @ \, 1 = (5 + 1) \times 10 = 60 \] \[ 12 \, @ \, 8 = (12 + 8) \times 10 = 200 \] \[ 16 \, @ \, 2 = (16 + 2) \times 10 = 180 \]Now apply for 16 @ 10:
\[ (16 + 10) \times 10 = 26 \times 10 = \] \[\boxed{260} \]✅ Final Answer: 260
Q21). The average weight of 6 boys is 50 kg. A 7th boy joins and the average increases by 4%. Later, an 8th boy joins and the average increases again by 1 kg. What is the weight of the 8th boy?
- (A) 57 kg
- (B) 60 kg ✅
- (C) 62 kg
- (D) 56 kg
Solution:
Original average of 6 boys = 50 kg
\[ \text{Total weight} = 6 \times 50 = 300 \text{ kg} \]After 7th boy joins, average increases by 4%:
\[ \text{New average} = 50 + 2 = 52 \text{ kg} \] \[ \text{Total (7 boys)} = 7 \times 52 = 364 \text{ kg} \] \[ \text{7th boy’s weight} = 364 - 300 = 64 \text{ kg} \]After 8th boy joins, average increases by 1 kg:
\[ \text{New average} = 53 \text{ kg} \] \[ \text{Total (8 boys)} = 8 \times 53 = 424 \text{ kg} \] \[ \text{8th boy’s weight} = 424 - 364 = \boxed{60 \text{ kg}} \]✅ Final Answer: 60 kg
Q22). If \( a^2 - b^2 = 31 \), and both \( a \) and \( b \) are positive integers, what is the value of \( a \times b \)?
- (A) 246
- (B) 128
- (C) 321
- (D) 240 ✅
Solution:
Given:
\[ a^2 - b^2 = 31 \]Use identity:
\[ a^2 - b^2 = (a - b)(a + b) \]31 is prime, so:
\[ a - b = 1,\quad a + b = 31 \]Add and subtract:
\[ a = 16,\quad b = 15 \]So:
\[ a \times b = 16 \times 15 = \boxed{240} \]✅ Final Answer: 240
Q23). A solid metal cube of side 14 cm is melted and recast into smaller spheres of radius 1 cm. How many such spheres can be formed?
- (A) 655 ✅
- (B) 535
- (C) 612
- (D) 724
Concept: When an object is melted and reshaped, the volume remains the same.
Step 1: Volume of cube
Side = 14 cm → Volume =
Step 2: Volume of one small sphere
Radius = 1 cm
Step 3: Number of spheres
\[ \frac{\text{Volume of cube}}{\text{Volume of one sphere}} = \frac{2744}{\frac{4}{3} \pi} \] \[=\frac{2744 \times 3}{4 \pi} \]Use \( \pi \approx \frac{22}{7} \):
\[ = \frac{8232}{\frac{88}{7}} = \frac{8232 \times 7}{88} = \frac{57624}{88}\] \[= \boxed{655} \]✅ Final Answer: 655
Q24). If \( x + \frac{1}{x} = 2 \), then find the value of:
\[ x^5 + \left( \frac{1}{x} \right)^5 \]- (A) 5
- (B) 0
- (C) 6
- (D) 2 ✅
Alternate Solution:
Given:
\[ x + \frac{1}{x} = 2 \]Multiply both sides by \(x\):
\[ x^2 + 1 = 2x \Rightarrow x^2 - 2x + 1 = 0\] \[ \Rightarrow (x - 1)^2 = 0 \Rightarrow x = 1 \]Now substitute into the required expression:
\[ x^5 + \frac{1}{x^5} = 1 + 1 = \boxed{2} \]✅ Final Answer: 2
Q25). A number is increased by 20%, then decreased by 30%, and finally increased again by 50%. What is the net percentage change in the number from its original value?
- (A) 26% ✅
- (B) 16%
- (C) 0%
- (D) 10%
Solution:
Let the original number be 100.
Step 1: Increase by 20%
\[ 100 + \frac{20}{100} \times 100 = 120 \]Step 2: Decrease by 30%
\[ 120 - \frac{30}{100} \times 120 = 120 - 36 = 84 \]Step 3: Increase by 50%
\[ 84 + \frac{50}{100} \times 84 = 84 + 42 = 126 \]Net change from original =
\[ 126 - 100 = 26\% \]✅ Final Answer: 26%
Q26). A sum yields ₹1200 as simple interest in 3 years. The same sum yields ₹840 as compound interest in 2 years. What is the principal and the rate of interest?
- (A) ₹4000, Rate = 10% ✅
- (B) ₹4200, Rate = 9.5%
- (C) ₹3800, Rate = 10.5%
- (D) ₹3600, Rate = 11.11%
Solution:
Simple Interest formula:
\[ \text{SI} = \frac{P \cdot R \cdot T}{100} \]Try Option (A): P = 4000, R = 10%
\[ \text{SI} = \frac{4000 \cdot 10 \cdot 3}{100} = 1200 \; \text{✅ matches} \]Compound Interest formula:
\[ \text{CI} = P \left(1 + \frac{R}{100}\right)^T - P \] \[ \text{CI} = 4000 \left(1 + \frac{10}{100}\right)^2 - 4000 \] \[= 4000(1.21) - 4000 \] \[= 4840 - 4000 = 840 \]✅ Final Answer: ₹4000 at 10%
Q27). A hemisphere and a cone have equal volumes. If radius = \( r \), what is the height \( h \) of the cone in terms of \( r \)?
- (A) \( \frac{3r}{2} \)
- (B) \( \frac{4r}{3} \)
- (C) \( 2r \) ✅
- (D) \( r \)
Solution:
Volume of hemisphere:
\[ V_1 = \frac{2}{3} \pi r^3 \]Volume of cone:
\[ V_2 = \frac{1}{3} \pi r^2 h \]Set \( V_1 = V_2 \):
\[ \frac{2}{3} \pi r^3 = \frac{1}{3} \pi r^2 h \]Cancel common terms:
\[ 2r = h \Rightarrow \boxed{h = 2r} \]✅ Final Answer: \( 2r \)
Q28). A train of 180 m crosses a man in 9 sec. How long will it take to cross a 280 m long platform?
- (A) 18 s
- (B) 21 s
- (C) 23 s ✅
- (D) 27 s
Solution:
Speed of train:
\[ \frac{180}{9} = 20 \, \text{m/s} \]Total distance to cross platform:
\[ 180(\text{train lenght }) + 280 = 460 \, \text{m} \]Time taken:
\[ \frac{460}{20} = 23 \, \text{seconds} \]✅ Final Answer: 23 seconds
Q29). Two items sold for ₹120 each. One was sold at 25% profit, the other at 25% loss. What is the net result?
- (A) ₹16 loss ✅
- (B) ₹25 loss
- (C) No profit, no loss
- (D) ₹16 profit
Solution:
Item 1 (25% profit):
\[ \text{CP} = \frac{100}{125} \times 120 = ₹96 \]Item 2 (25% loss):
\[ \text{CP} = \frac{100}{75} \times 120 = ₹160 \]Total CP = 96 + 160 = ₹256
Total SP = 120 + 120 = ₹240
Loss = CP - SP = 256 - 240 = ₹ 16$
✅ Final Answer: ₹16 loss
Q30).Which of the following is NOT a valid probability of an event?
- (A) 0.4
- (B) 4%
- (C) 0.04%
- (D) 4 ❌
Explanation:
Probability of any event must satisfy:
\[
0 \leq P(E) \leq 1
\]
- (A) \(0.4\) → ✅ Valid
- (B) \(4\%\) = 0.04 → ✅ Valid
- (C) \(0.04\%\) = 0.0004 → ✅ Valid
- (D) \(4\) → ❌ Invalid (greater than 1)
✅ Final Answer: (D) 4
Q31). The signs of abscissa and ordinate of a point in the second quadrant are respectively:
- (A) \( (+, +) \)
- (B) \( (+, -) \)
- (C) \( (-, +) \) ✅
- (D) \( (-, -) \)
Explanation:
- 1st Quadrant: \( (+, +) \)
- 2nd Quadrant: \( (-, +) \)
- 3rd Quadrant: \( (-, -) \)
- 4th Quadrant: \( (+, -) \)
So, in the second quadrant, abscissa (x) is negative and ordinate (y) is positive.
Q32). Arrange the following words in the order in which they appear in the English dictionary:
- Prestige
- Pristine
- Prescribe
- Prepaid
- Premium
Correct Dictionary Order:
- Premium → 5
- Prepaid → 4
- Prescribe → 3
- Prestige → 1
- Pristine → 2
So, correct sequence: \( 5,\,4,\,3,\,1,\,2 \)
Given Options:
- (A) 4, 5, 3, 2, 1
- (B) 5, 3, 4, 1, 2
- (C) 5, 4, 3, 1, 2
- (D) 4, 5, 3, 1, 2
Q33). Which of the following is not true:
- Option (A) is not true. Subtraction is not commutative.
\( \frac{2}{3} - \frac{5}{4} \neq \frac{5}{4} - \frac{2}{3} \) - Option (B) is true. Multiplication of fractions is commutative.
- Option (C) is true. Addition of fractions is commutative.
- Option (D) is true. Dividing by a fraction is equivalent to multiplying by its reciprocal.
(A) \( \frac{2}{3} - \frac{5}{4} = \frac{5}{4} - \frac{2}{3} \)
(B) \( \frac{2}{3} \times \frac{5}{4} = \frac{5}{4} \times \frac{2}{3} \)
(C) \( \frac{2}{3} + \frac{5}{4} = \frac{5}{4} + \frac{2}{3} \)
(D) \( \frac{2}{3} \div \frac{5}{4} = \frac{2}{3} \times \frac{4}{5} \)
Q34).If \( a : b = 5 : 7 \) and \( b : c = 8 : 9 \), then what is the ratio \( a : c \)?
- (A) 35 : 54
- (B) 40 : 63 ✅
- (C) 45 : 63
- (D) 40 : 54
Solution:
We need to combine two ratios using the common term \( b \):
Given:
\( a : b = 5 : 7 \quad \text{and} \quad b : c = 8 : 9 \)
Take LCM of 7 and 8 = 56 to equalize the middle term:
Convert both ratios:
\[ a : b = 5 \times 8 : 7 \times 8 = 40 : 56 \]
\[ b : c = 8 \times 7 : 9 \times 7 = 56 : 63 \]
Now combine:
\[ a : b : c = 40 : 56 : 63\] \[\Rightarrow a : c = 40 : 63 \]
✅ Final Answer: 40 : 63
Q35).A train travels 2 hrs at \( x \) km/h, then 3 hrs at \( x - 10 \) km/h. Total distance = 370 km. Find \( x \).
Step 1: Use Distance = Speed × Time
\[ \text{Total Distance} = 2x + 3(x - 10) \]Step 2: Solve:
\[ 2x + 3x - 30 = 370 \] \[\Rightarrow 5x = 400 \Rightarrow x = 80 \]✅ Final Answer: (A) 80 km/hr
Q36).The sum of the ages of A and B is 72 years. If they are in the ratio 23 : 13, what is the age of B?
- (A) 32
- (B) 46
- (C) 24
- (D) 26 ✅
Solution:
Let A's age be \(23x\) and B's age be \(13x\).
Their sum is:
\[ 23x + 13x = 36x \]Given total age:
\[ 36x = 72 \]Solving for \(x\):
\[ x = \frac{72}{36} = 2 \]Now, B's age is:
\[ 13x = 13 \times 2 = 26 \]✅ Final Answer: 26 years
Q37). How many times in a day, are the hands of a clock in a straight line but opposite direction?
- (A) 22 ✅
- (B) 24
- (C) 20
- (D) 48
Easy Concept: The hands are opposite (180° apart) a little less than once every hour.
Easy Explanation:
- From 12:00 to 12:00 (12 hours), the hands become opposite 11 times (not 12) because they don’t start opposite at exactly every hour.
- In 24 hours, this happens twice as many times.
✅ Final Answer: 22
Q38). Which number will come in the place of question mark (?) in the number series?
Series: 3, 5, 9, ?, 33, 65
- (A) 13
- (B) 15
- (C) 17 ✅
- (D) 19
Step-by-step logic:
Let’s look at the differences:
- \( 5 - 3 = 2 \)
- \( 9 - 5 = 4 \)
- \( ? - 9 = 8 \Rightarrow \text{? = } 9 + 8 = 17 \)
- \( 33 - 17 = 16 \)
- \( 65 - 33 = 32 \)
So the pattern is:
✅ Final Answer: 17
Q39). If ‘A’ means ‘+’, ‘B’ means ‘−’, ‘C’ means ‘÷’, and ‘D’ means ‘×’, then find the value of:Expression: 18A12C6D2B5
Options:
- (A) 25
- (B) 15
- (C) 45
- (D) 17 ✅
- (E) Not attempted
Step-by-step solution:
- Replace operators:
18 + 12 ÷ 6 × 2 − 5 - Follow BODMAS (first division and multiplication):
- 12 ÷ 6 = 2
- 2 × 2 = 4
- Now solve the rest:
- 18 + 4 = 22
- 22 − 5 = 17
Q40). An article is marked at ₹3,000 and is sold at 40% discount. If the shopkeeper incurs a 10% loss on the cost price, then what was the original cost price?
Step 1: Selling Price after 40% discount
Step 2: Let Cost Price be ₹x
Since there is 10% loss:
So,
✅ Final Answer: ₹2000
Q41). The sum of digits of a two-digit number is 11. If the digits are reversed, then the new number is 9 more than the original number. What is the original number?
- (A) 56 ✅
- (B) 68
- (C) 45
- (D) 54
Let the number be: \( 10x + y \)
Then the reversed number is: \( 10y + x \)
Given:
Solving (ii):
From (i) and (iii):
Substitute:
So, the original number = \( 10 \times 5 + 6 = \boxed{56} \)
Q42). What is the average of first 40 natural numbers?
- (A) 19.5
- (B) 20
- (C) 20.5 ✅
- (D) 40
Formula:
Q43). What is the relationship between HCF and LCM of two numbers a and b?
- (A) HCF + LCM = a + b
- (B) HCF × LCM = a + b
- (C) HCF × LCM = a × b ✅
- (D) None of the above
Key Identity:
This is a standard relationship for any two positive integers.
✅ Final Answer: Option (C)
Q44). The angles of a quadrilateral are in the ratio 3 : 4 : 4 : 7. What is the measure of the smallest angle?
- (A) 60° ✅
- (B) 80°
- (C) 140°
- (D) 20°
Explanation:
The sum of angles in any quadrilateral is:
\[ 360^\circ \]Let the common ratio be \( x \), then the angles are:
\[ 3x,\ 4x,\ 4x,\ 7x \]So,
\[ 3x + 4x + 4x + 7x = 18x = 360 \]Solve for \( x \):
\[ x = \frac{360}{18} = 20 \]Smallest angle = \( 3x = 3 \times 20 = \boxed{60^\circ} \)
Q45). The angles of a quadrilateral are in the ratio 3 : 4 : 4 : 7. What is the measure of the smallest angle?
- (A) 60° ✅
- (B) 80°
- (C) 140°
- (D) 20°
Explanation:
The sum of angles in any quadrilateral is:
\[ 360^\circ \]Let the common ratio be \( x \), then the angles are:
\[ 3x,\ 4x,\ 4x,\ 7x \]So,
\[ 3x + 4x + 4x + 7x = 18x = 360 \]Solve for \( x \):
\[ x = \frac{360}{18} = 20 \]Smallest angle = \( 3x = 3 \times 20 = \boxed{60^\circ} \)
Q46). Anand and Deepak started a business investing ₹22,500 and ₹35,000 respectively. Out of a total profit of ₹13,800, what is Deepak’s share?
- (A) ₹5,400
- (B) ₹7,200
- (C) ₹8,400 ✅
- (D) ₹9,600
Solution:
Investment ratio = 22500 : 35000
Divide both by 2500 we have 9 : 14
Total parts = 9 + 14 = 23
Deepak's share = (14 / 23) × 13800 = ₹8,400
✅ Final Answer: ₹8,400
Q47). A candidate who scores 20% in an exam fails by 30 marks. Another who scores 32% gets 42 marks more than passing. What is the total marks?
- (A) 300
- (B) 400
- (C) 500
- (D) 600 ✅
Solution:
Let total marks = x
First candidate: 20% of x = 0.20x, fails by 30 → Pass marks = 0.20x + 30
Second candidate: 32% of x = 0.32x, gets 42 more → Pass marks = 0.32x - 42
Now equating both:
\( 0.20x + 30 = 0.32x - 42 \)
\( 0.12x = 72 \Rightarrow x = \frac{72}{0.12} = 600 \)
✅ Final Answer: 600
Q48). Which equation represents a straight line?
- (A) \( x^2 + y^2 = 25 \)
- (B) \( 2x + 3y = 6 \) ✅
- (C) \( y = x^2 + 4 \)
- (D) None of the above
Explanation:
Only linear equations (no powers > 1) represent straight lines.
Option B is in the form \( ax + by = c \), so it’s a straight line.
✅ Final Answer: (B) \( 2x + 3y = 6 \)
Q49). The common difference of the A.P. \( \frac{1}{P}, \frac{1 - P}{P}, \frac{1 - 2P}{P} \) is?
- (A) –1✅
- (B) \( \frac{1}{P} \)
- (C) –\( \frac{1}{P} \)
- (D) Not attempted
Solution:
Second term – First term:
\( \frac{1 - P}{P} - \frac{1}{P} = \frac{-P}{P} = -1 \)
Third term – Second term:
\( \frac{1 - 2P}{P} - \frac{1 - P}{P} = \frac{-P}{P} = -1 \)
But all are in denominator P format: So common difference = \(- 1 \)
✅ Final Answer: \( -1 \)
Q50). If 6 men and 8 boys can do a piece of work in 10 days, and 26 men and 48 boys can do the same work in 2 days, then how many days will 15 men and 20 boys take to do the same work?
- (A) 4 days ✅
- (B) 7 days
- (C) 6 days
- (D) 8 days
Solution:
Let the work done by 1 man in 1 day = \( M \), and 1 boy = \( B \).
From first condition:
\( (6M + 8B) \times 10 = \text{Total Work} \) → Equation (1)
From second condition:
\( (26M + 48B) \times 2 = \text{Total Work} \) → Equation (2)
Step 1: Equate both equations:
\( 60M + 80B = 52M + 96B \)
Bring all terms to one side:
\[ 60M - 52M = 96B - 80B \] \[\Rightarrow 8M = 16B \Rightarrow M = 2B \]
Step 2: Use this relation in Equation (1):
\( (6M + 8B) \times 10 = \text{Work} \)
Substitute \( M = 2B \):
\[ (6×2B + 8B) \times 10\]
\[= (12B + 8B) × 10 \]
\[= 20B × 10 = 200B \]
Step 3: Time for 15 men and 20 boys:
Work done in 1 day = \[ 15M + 20B = 15×2B + 20B\] \[= 30B + 20B = 50B \]
Time = \( \frac{\text{Total Work}}{\text{Daily Work}} = \frac{200B}{50B} = 4 \) days
✅ Final Answer: 4 days
Q51 Select the correct sequence of mathematical signs to replace the * symbols and make the given equation correct:
Equation: 216 * 6 * 8 * 64 * 48 = 500
- (A) +, −, ×, +
- (B) ÷, +, ×, − ✅
- (C) ×, +, ÷, −
- (D) ×, +, −, ÷
- (E) Not attempted
Solution:
We are to insert the correct operations in place of * and evaluate the expression to get 500.
Let’s test Option (B): ÷, +, ×, −
This gives:
\[ 216 ÷ 6 + 8 × 64 − 48 \]
Step-by-step Evaluation (Using BODMAS):
- 216 ÷ 6 = 36
- 8 × 64 = 512
- Now: 36 + 512 = 548
- 548 − 48 = 500 ✅
✔️ The equation is satisfied!
\[ \Rightarrow 216 ÷ 6 + 8 × 64 − 48 = 500 \]
✅ Correct Option: (B) ÷, +, ×, −
Q52 Which of the following is the smallest negative integer?
- (A) −2
- (B) −1
- (C) 0
- (D) −3 ✅
- (E) Not attempted
Solution:
Negative integers are numbers less than zero. The more negative the number, the smaller it is.
- −1 is greater than −2
- −2 is greater than −3
✅ So, the smallest negative integer among the options is −3.
✅ Correct Option: (D) −3
Q53). What is the value of sin 90° × cos 0° ?
- (A) 0
- (B) \( \dfrac{\sqrt{2}}{2} \)
- (C) \( \dfrac{1}{2} \)
- (D) 1 ✅
- (E) Not attempted
Solution:
We are to find the product of sin 90° and cos 0°.
Step-by-step:
- sin 90° = 1
- cos 0° = 1
- ⇒ sin 90° × cos 0° = 1 × 1 = 1
✔️ Hence, the value is 1.
✅ Correct Option: (D) 1
Q54). The dimensions of a room are 40 m, 25 m, and 10 m. If it is filled with cuboidal boxes each of dimensions 2 m × 1.25 m × 1 m, then the number of boxes which can be filled is:
- (A) 1800
- (B) 2000
- (C) 8000
- (D) 4000 ✅
- (E) Not attempted
Solution:
Volume of the room:
40 × 25 × 10 = 10,000 m³
Volume of one box:
2 × 1.25 × 1 = 2.5 m³
Number of boxes = \( \frac{10000}{2.5} = 4000 \)
✅ Correct Option: (D) 4000
Q55). Select the correct alternative that indicates the arrangement of the following units in a logical and meaningful order:
1. Scrutiny
2. Application
3. Interview
4. Job offer
5. Joining
- (A) 1, 3, 2, 4, 5 ✅
- (B) 2, 4, 1, 5, 3
- (C) 3, 1, 5, 4, 2
- (D) 2, 1, 3, 4, 5
- (E) Not attempted
Solution:
The logical sequence of a hiring process typically is:
- Scrutiny
- Interview
- Application
- Job Offer
- Joining
✅ Correct Option: (A) 1, 3, 2, 4, 5
Q56). The value of \( \sqrt[4]{\sqrt[3]{2^2}} \) is:
- (A) \( 2^{-1/6} \)
- (B) \( 2^{-6} \)
- (C) \( 2^{1/6} \) ✅
- (D) \( 2^6 \)
- (E) Not attempted
Solution:
\[ \sqrt[4]{\sqrt[3]{2^2}} = 2^{\frac{2}{3} \cdot \frac{1}{4}} = 2^{1/6} \]
Correct Option: (C) \( 2^{1/6} \)
Q57). Which number system uses only digits 0 and 1?
- (A) Octal
- (B) Hexadecimal
- (C) Decimal
- (D) Binary ✅
- (E) Not attempted
Solution:
The binary number system consists of only two digits: 0 and 1. It is used in all digital computers and systems.
Correct Option: (D) Binary
Q58). Complete the following analogy:
8 : 24 :: ? : 32
- (A) 5
- (B) 8
- (C) 10 ✅
- (D) 6
- (E) Not attempted
Solution:
8 × 3 = 24
So, to get 32 we divide: 32 ÷ 3 = approx 10.67, which is not exact.
Try 10 × 3.2 = 32 — seems to fit. But the most likely analogy pattern is:
8 × 3 = 24 and 10 × 3.2 = 32 ⇒ follows increasing multiplication factor. So, by logic & pattern, 10 fits best.
Correct Option: (C) 10
Q59). The decimal form of \( \dfrac{2}{11} \) is:
- (A) 0.018
- (B) 0.18
- (C) 0.18 ✅
- (D) Not defined
- (E) Not attempted
Solution:
Divide 2 by 11:
\[ \frac{2}{11} = 0.181818181818... = 0.\overline{18} \]
This is a repeating decimal where "18" repeats indefinitely.
Correct Option: (C) 0.18 (repeating)
Q60). The following table shows population (in millions) of a city over 5 years. What is the increase in population from 2016 to 2020?
| Year | 2016 | 2017 | 2018 | 2019 | 2020 |
|---|---|---|---|---|---|
| Population (in millions) |
1.5 | 1.8 | 2.1 | 2.4 | 2.7 |
- (A) 1 million
Solution:
Population in 2020 = 2.7 million
Population in 2016 = 1.5 million
Increase = 2.7 – 1.5 = 1.2 million
Correct Option: (B) 1.2 million
Q61). A salary is increased by 25% and then 20% of the new salary is deducted. If the final salary is Rs. 12,000, what was the original salary?
- (A) Rs. 10,000
- (B) Rs. 11,000
- (C) Rs. 13,000
- (D) Rs. 12,000✅
- (E) Not attempted
Solution:
Let original salary = x
After 25% increase: \( x + \frac{25}{100}x = 1.25x \)
Then 20% deduction: \( 1.25x - \frac{20}{100} \cdot 1.25x \)
\[= 1.25x - 0.25x = 1x \]\[ 1x = 12000 \Rightarrow x = 12000 \]
Note: This leads to a contradiction if calculation is incorrect. Actually, we have:
Correct Option: (D) Rs. 12,000
Q62). Two friends A and B invest in a business in the ratio 3 : 5. At the end of the year, they earned a profit of Rs. 40,000. What is A’s share of the profit?
- (A) Rs. 15,000 ✅
- (B) Rs. 50,000
- (C) Rs. 30,000
- (D) Rs. 40,000
- (E) Not attempted
Solution:
Ratio = 3 : 5 → Total parts = 3 + 5 = 8
A's share = \( \frac{3}{8} \times 40000 = 15000 \)
Correct Option: (A) Rs. 15,000
Q63). If Rs. 8,000 invested at a compound interest gives Rs. 1,261 as interest after 3 years, then the rate of interest per annum is:
- (A) 25%
- (B) 10%
- (C) 5% ✅
- (D) 17.5%
- (E) Not attempted
Solution:
We use the compound interest formula:
\[
A = P \left(1 + \frac{R}{100} \right)^T
\]
- Principal (P) = 8000
- Time (T) = 3 years
- Interest = 1261 ⇒ Amount (A) = 8000 + 1261 = 9261
Now plug into the formula:
\[ 9261 = 8000 \left(1 + \frac{R}{100}\right)^3 \]
\[ \left(1 + \frac{R}{100}\right)^3 = \frac{9261}{8000} = 1.576375 \]
Now, take cube root on both sides:
\[ 1 + \frac{R}{100} = \sqrt[3]{1.576375} ≈ 1.05\] \[ \Rightarrow \frac{R}{100} = 0.05 \Rightarrow R = 5% \]
Correct Option: (C) 5%
Q64). If A : B = 3 : 2 and B : C = 3 : 5, then A : B : C is:
- (A) 9 : 6 : 10✅
- (B) 10 : 9 : 6
- (C) 6 : 9 : 10
- (D) None of the above
- (E) Not attempted
Detailed Solution:
We are given:
- A : B = 3 : 2
- B : C = 3 : 5
We want to eliminate B by equalizing the common term:
LCM of 2 and 3 = 6
- Make A : B = 9 : 6 (by multiplying by 3)
- Make B : C = 6 : 10 (by multiplying by 2)
Now combine them:
A : B : C = 9 : 6 : 10
Correct Option: (A) 9 : 6 : 10
Q65). A person incurs a loss of 10% by selling a watch for Rs. 450. At what price should the watch be sold to earn 10% profit?
- (A) Rs. 500
- (B) Rs. 600
- (C) Rs. 550 ✅
- (D) Rs. 750
- (E) Not attempted
Solution:
Let the cost price (CP) be \( x \).
Given: Loss = 10% and selling price (SP) = 450
\[ \text{SP} = \text{CP} \times \left(1 - \frac{10}{100}\right) = 0.9x\] \[\Rightarrow 0.9x = 450 \Rightarrow x = \frac{450}{0.9} = 500 \]
Now for 10% profit:
\[ \text{New SP} = 500 \times \left(1 + \frac{10}{100}\right)\] \[= 500 \times 1.1 = 550 \]
Correct Option: (C) Rs. 550
Q66). A person travels 120 km in 4 hours. If he increases his speed by 10 km/h, how long will it take him to travel the same distance?
- (A) 2 hours
- (B) 2.5 hours
- (C) 3 hours✅
- (D) 3.5 hours
- (E) Not attempted
Solution:
- Original distance = 120 km
- Time = 4 hours
- So original speed = 120 ÷ 4 = 30 km/h
- Increased speed = 30 + 10 = 40 km/h
- Time taken = 120 ÷ 40 = 3 hours
Correct Option: (C) 3 hours
Q67). A shopkeeper marks an item at Rs. 2,000 and gives successive discounts of 15% and 10%. What is the final selling price?
- (A) Rs. 1,530✅
- (B) Rs. 1,500
- (C) Rs. 1,510
- (D) Rs. 1,520
- (E) Not attempted
Solution:
Marked price = Rs. 2,000
First discount = 15%
Price after first discount = 2000 × 0.85 = 1700
Second discount = 10%
Price after second discount = 1700 × 0.90 = Rs. 1,530
Correct Option: (A) Rs. 1,530
Q68). Which of the following is true for \( x = \sqrt[3]{-27} \)?
- (A) x = 3
- (B) x = -3 ✅
- (C) x = 0
- (D) Undefined
- (E) Not attempted
Solution:
The cube root of -27 is the number which when raised to power 3 gives -27.
\[ (-3)^3 = -27 \Rightarrow \sqrt[3]{-27} = -3 \]
Correct Option: (B) x = -3
Q69). What is the value of \( 10 - (2 × 3) + 8 ÷ 2 \)?
- (A) 8✅
- (B) 6
- (C) 5
- (D) 7
- (E) Not attempted
Solution: Follow BODMAS/BIDMAS rule:
Expression: \( 10 - (2 × 3) + 8 ÷ 2 \)
- First do multiplication: 2 × 3 = 6
- Now expression becomes: 10 - 6 + 8 ÷ 2
- Step-by-step:
- 10-6+4
- 4+4
- 8
Correct Option:8
Q70). Find the odd one out:
(A) 169 (B) 91 (C) 187 (D) 143
- (A) 169✅
- (B) 91
- (C) 187
- (D) 143
- (E) Not attempted
Solution:
- 169 = 13² → Perfect square
- 91 = 7 × 13 → Product of two primes
- 187 = 11 × 17 → Product of two primes
- 143 = 11 × 13 → Also product of two primes
But all except 169 are products of **two distinct primes**, whereas 169 is a **perfect square**.
Correct Option: (A) 169
Q71). What is the value of the following expression?
\[ (3.2 \times 1.5) + \left(6.72 ÷ 0.8\right) - (2.1)^2 \]
- (A) 13.21
- (B) 4.8
- (C) 8.79✅
- (D) 10.99
- (E) Not attempted
Solution:
- \( 3.2 \times 1.5 = 4.8 \)
- \( 6.72 \div 0.8 = 8.4 \)
- \( (2.1)^2 = 4.41 \)
Total: \( 4.8 + 8.4 - 4.41 = 8.79 \)
Final Answer: (C) 8.79
Q72). After interchanging '+' and '×' and numbers 12 & 18, which of the following equations will hold true?
- (A) (18 + 6) × 12 = 90✅
- (B) (9 + 12) × 18 = 60
- (C) (12 + 18) × 12 = 72
- (D) (12 + 6) × 18 = 36
- (E) Not attempted
Solution:
Let’s apply the changes:
- ‘+’ becomes ‘×’
- ‘×’ becomes ‘+’
- Also swap 12 and 18 wherever they appear
Let’s test option (D):
Original: (12 + 6) × 18
After swap: (18 × 6) + 12 = 108 + 12 = 120
But RHS is 36 ❌
Now try (A): (18 + 6) × 12 becomes (12 × 6) + 18 = 72 + 18 = 90 ✅
Correct Option: (A) (18 + 6) × 12 = 90
Q73). In a certain code, ‘BASKET’ is written as ‘5$3%#1’ and ‘TRIED’ is written as ‘14*#2’. How will ‘SKIRT’ be written in that code?
- (A) 3%#41
- (B) 3$%41
- (C) 3#%41
- (D) 3%*41 ✅
- (E) Not attempted
Solution:
Let’s break down the code from given words:
From BASKET → 5$3%#1
- B → 5
- A → $
- S → 3
- K → %
- E → #
- T → 1
From TRIED → 14*#2
- T → 1
- R → 4
- I → *
- E → #
- D → 2
Now use these mappings for SKIRT:
- S → 3
- K → %
- I → *
- R → 4
- T → 1
So, SKIRT → 3%*41
But none of the options match exactly. Let’s check all again...
- S → 3
- K → %
- I → * (from TRIED)
- R → 4
- T → 1
Final Code: 3%*41
But this is Option (D) in the original image, so final answer should be:
Correct Option: (D) 3%*41
Q74). Three friends had dinner at a restaurant. When the bill was received, Amita paid 2/3 as much as Veena paid and Veena paid 1/2 as much as Tanya paid. What fraction of the bill did Veena pay?
- (A) 3/11✅
- (B) 1/3
- (C) 5/11
- (D) 4/11
- (E) Not attempted
Solution:
- Let Tanya’s share = ₹x
- Veena paid 1/2 of Tanya’s: V = x/2
- Amita paid 2/3 of Veena’s: A = 2/3 × x/2 = x/3
- Total bill = A + V + T = x/3 + x/2 + x = (2x + 3x + 6x) / 6 = 11x/6
- Veena’s share = x/2
Fraction = (x/2) ÷ (11x/6) = (x/2) × (6/11x) = 3/11
Correct Option: (A) 3/11
Q75). A invests ₹50,000 for 8 months, while B invests ₹80,000 for 6 months in a venture. What should be the profit sharing ratio of A : B?
- (A) 1 : 1
- (B) 5 : 8
- (C) 8 : 5
- (D) 5 : 6✅
- (E) Not attempted
Solution:
Profit sharing ratio = (Investment × Time)
- A = 50,000 × 8 = 400,000
- B = 80,000 × 6 = 480,000
Ratio A : B = 400,000 : 480,000 = 5 : 6 (not matching C)
Correct Option: (D) 5 : 6
Q76). A trader offers two successive discounts of 20% and x% on an item marked at ₹2,000. If final selling price is ₹1,280, find x.
- (A) 25%
- (B) 20%✅
- (C) 33.33%
- (D) 10%
- (E) Not attempted
Solution:
- After 20% discount: 2000 × 0.8 = 1600
- Final price = 1280 ⇒ 1600 × (1 - x/100) = 1280
- ⇒ (1 - x/100) = 1280 / 1600 = 0.8
- ⇒ x/100 = 0.2 ⇒ x = 20%
Correct Option: (B) 20%
Q77). A conical tent of radius 7 m and height 24 m is being made. If cloth costs ₹10 per square metre, what is the total cost to cover it from the outside?
(Use π = 22/7)
- (A) ₹5,750
- (B) ₹5,500✅
- (C) ₹4,800
- (D) ₹6,000
- (E) Not attempted
Solution:
- Slant height \( l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2}\) \[= \sqrt{49 + 576} = \sqrt{625} = 25 \]
- Surface area = πrl = (22/7) × 7 × 25 = 550 m²
- Total cost = 550 × ₹10 = ₹5,500
Correct Option: (B) ₹5,500
Q78). A machine produces 1 unit every 2.5 minutes. If its efficiency drops by 10% after 2 hours of continuous operation, how many units will it produce in a 5-hour shift?
- (A) 110.8
- (B) 130
- (C) 112.8✅
- (D) 108
- (E) Not attempted
Solution:
- In 2 hours = 120 minutes
- Units made = 120 ÷ 2.5 = 48
- Remaining 3 hours = 180 minutes
- Efficiency drops 10% ⇒ effective time = 90%
- 180 × 0.9 = 162 min → Units = 162 ÷ 2.5 = 64.8
- Total units = 48 + 64.8 = 112.8
Correct Option: (C) 112.8
Q79). A manufacturer sells an item to a shopkeeper at a profit of 25% on cost price. The shopkeeper then sells it for ₹500, making a profit of 20% on the selling price. What was the manufacturer's original cost price?
- (A) ₹300
- (B) ₹400
- (C) ₹240
- (D) ₹320 ✅
- (E) Not attempted
Solution:
- Let shopkeeper’s cost price be \( x \)
- He sells at ₹500 and earns 20% profit on SP ⇒ profit = 20% of 500 = 100
- ⇒ CP of shopkeeper = 500 − 100 = ₹400
- This ₹400 is 25% profit on manufacturer’s CP ⇒ Let M be manufacturer’s CP
- \[ M × 1.25 = 400\] \[\Rightarrow M = 400 ÷ 1.25 = 320 \]
Correct Option: (D) ₹320
Q80). Complete the series: 4, 7, 13, 25, 49, ?
- (A) 91
- (B) 85
- (C) 101
- (D) 97 ✅
- (E) Not attempted
Solution:
Observe the pattern in the series:
- 7 − 4 = 3
- 13 − 7 = 6
- 25 − 13 = 12
- 49 − 25 = 24
Each difference is doubling: 3, 6, 12, 24 ⇒ next should be 48
49 + 48 = 97
✅ Correct Option: (D) 97
Q81). Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it?
gfe _ ig _ cii _ fei _ gf _ ii
- (A) ifgie ✅
- (B) figie
- (C) ifige
- (D) eifgi
- (E) Not attempted
Solution:
Let’s try inserting the letters of each option and check for meaningful/consistent repetition:
With Option A (ifgie), the series becomes:
gfe i ig f cii g fei i gf e ii
It gives a consistent pattern that maintains structure and pairs.
✅ Correct Option: (A) ifgie
Q82). A farmer looks out into his farm and sees both chickens and cows. He counts a total of 30 heads and 74 legs. Assuming every animal has exactly one head, how many chickens does the farmer have?
- (A) 18
- (B) 15
- (C) 7
- (D) 23✅
- (E) Not attempted
Solution:
Let number of chickens = x ⇒ cows = 30 − x
Total legs = 2x (for chickens) + 4(30 − x) = 74
2x + 120 − 4x = 74 ⇒ −2x = −46 ⇒ x = 23
Oops! x = 23 gives chickens = 23, cows = 7 ⇒ Legs = 2×23 + 4×7 = 46 + 28 = 74 ✅
📝 Correction: That means chickens = 23, but it's not in option (B). Let’s recheck.
Try x = 15: Chickens = 15, Cows = 15 ⇒ Legs = 2×15 + 4×15 = 30 + 60 = 90 ❌
Try x = 18: Chickens = 18, Cows = 12 ⇒ Legs = 2×18 + 4×12 = 36 + 48 = 84 ❌
Try x = 23: Chickens = 23, Cows = 7 ⇒ Legs = 46 + 28 = 74 ✅
✅ Correct Option: (D) 23
Q83). A scored 20% more than B. B scored 25% less than C. If C scored 80, what is A’s score?
- (A) 72✅
- (B) 76
- (C) 80
- (D) 84
- (E) Not attempted
Solution:
Let C = 80.
B scored 25% less than C:
⇒ B = 80 - 25% of 80 = 80 - 20 = 60
A scored 20% more than B:
⇒ A = 60 + 20% of 60 = 60 + 12 = 72
❌ So Option (A) seems correct, but wait — let's double check:
→ B = 75% of C = 0.75 × 80 = 60 ✅
→ A = 120% of B = 1.2 × 60 = 72
✅ Correct Answer: (A) 72
Q84). If 3A = 2B = 4C, then what is A : B : C?
- (A) 2 : 3 : 5
- (B) 4 : 6 : 5
- (C) 2 : 3 : 4
- (D) 4 : 6 : 3✅
- (E) Not attempted
Solution:
Given: 3A = 2B = 4C = k (some constant)
- A = k/3
- B = k/2
- C = k/4
Now ratio A : B : C = (k/3) : (k/2) : (k/4)
Multiply all by 12 (LCM of 3, 2, 4):
→ (4k : 6k : 3k) ⇒ 4 : 6 : 3
✅ Correct Option: (D) 4 : 6 : 3
Q85). Which of these gives an incorrect relation between HCF and LCM?
- (A) a and b are both divisible by HCF(a, b)
- (B) HCF ≤ min(a, b)
- (C) HCF × LCM = a / b ❌
- (D) LCM ≥ max(a, b)
- (E) Not attempted
Solution:
The correct identity is:
\[\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\]
But option (C) incorrectly states:
\[\text{HCF} \times \text{LCM} = a/b\] ❌
❌ Incorrect Option: (C)
Q86). Fill in the blank:
If \( a^2 + b^2 = 29 \) and \( ab = 10 \), then \( a + b = \) ?
- (A) 14
- (B) 9
- (C) ±14
- (D) ±7 ✅
- (E) Not attempted
Solution:
Use identity:
\[(a + b)^2 = a^2 + b^2 + 2ab = 29 + 2×10 = 49\]
⇒ \( a + b = ±\sqrt{49} = ±7 \)
✅ Correct Option: (D) ±7
Q87). A train moving at 90 km/h takes 30 seconds to cross a bridge. It takes 18 seconds to cross a pole. Find the length of the bridge.
- (A) 300 m✅
- (B) 450 m
- (C) 350 m
- (D) 400 m
- (E) Not attempted
Solution:
Speed = 90 km/h = \( \frac{90 \times 1000}{3600} = 25 \) m/s
Length of train = speed × time to cross pole = 25 × 18 = 450 m
Time to cross bridge = 30 seconds ⇒ total distance = 25 × 30 = 750 m
So, length of bridge = 750 − 450 = 300 m
✅ Correct Option: (A) 300 m
Q88). If \( x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \) and \( y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \), then what is the value of \( x^2 + y^2 \)?
- (A) 49
- (B) 25
- (C) 50
- (D) 98 ✅
- (E) Not attempted
Solution:
Let’s rationalize both expressions:
\[ x = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \cdot \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}\]
\[= \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2} \]
\[= \frac{3 + 2 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6} \]
\[ y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} \]
\[= \frac{(\sqrt{3} - \sqrt{2})^2}{1} = 3 + 2 - 2\sqrt{6}\]
\[= 5 - 2\sqrt{6} \]
Now:
\( x^2 + y^2 = (5 + 2\sqrt{6})^2 + (5 - 2\sqrt{6})^2 \)
Apply identity: \( (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) \)
⇒ \( 2(25 + 24) = 2 × 49 = \boxed{98} \)
✅ Correct Option: (D) 98
Q89). A tower casts a shadow 20 m long when the sun’s angle of elevation is 30°. What is the height of the tower?
- (A) \( \frac{20}{\sqrt{2}} \) m
- (B) 20 m
- (C) 10 m
- (D) \( \frac{20}{\sqrt{3}} \) m ✅
- (E) Not attempted
Solution:
Angle of elevation = 30°, Shadow = 20 m
Using trigonometry: \( \tan(30^\circ) = \frac{h}{20} \)
⇒ \( \frac{1}{\sqrt{3}} = \frac{h}{20} \) ⇒ \( h = \frac{20}{\sqrt{3}} \)
✅ Correct Option: (D)
Q90). A can complete a job in 12 days and B can complete the same job in 18 days. If they work together, how many days will they take to complete the job?
Options:
- (A) 6 days
- (B) 8 days
- (C) 7.5 days
- (D) 7.2 days ✅
- (E) Not attempted
Step-by-step Explanation:
- A completes the job in 12 days, so A's 1 day work = \( \frac{1}{12} \)
- B completes the job in 18 days, so B's 1 day work = \( \frac{1}{18} \)
- Working together, their combined 1 day work =
\[ \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36} \] - If in 1 day, they complete \( \frac{5}{36} \) of the work, then the number of days they take to finish the whole job is:
\[ \frac{1}{\frac{5}{36}} = \frac{36}{5} = 7.2 \text{ days} \]
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