IQ (Intelligence Quotient)

Replacing the symbols as instructed ($\times\to\div$, $-\to\times$, $+\to+$, $\div\to-$):

$252 \div 9 \times 5 + 32 - 92 = 28\times 5 + 32 - 92 = 140 + 32 - 92 = 80$

The value is $80$, which does not match any of the options (95, 168, 192, 200). The question or its options appear to be misprinted, so none of the given choices is correct.

(b) 28

Using BODMAS:
$30 \div 2 + 3 \times 6 - 5 = 15 + 18 - 5 = 28$

(d) 5,4,2,3,1

This follows the order from the outermost part to the shoulder: Finger → Palm → Wrist → Elbow → Shoulder.

(c) 2, 3, 1, 4

The natural building order of language is: Letter → Word → Phrase → Sentence. Letters combine to form words, words combine to form a phrase, and phrases combine to form a sentence. So the sequence is 2, 3, 1, 4.

(Note: options (a) and (b) are printed identically in the question, but neither matches the correct order — the answer is (c).)

(b) Rs. 17.50

Let the amount spent by each engineer be Rs. $E$.

Given:

• $5T = 4E \Rightarrow T = \frac{4E}{5}$
• $12E = 9D \Rightarrow D = \frac{4E}{3}$
• $6D = 8S \Rightarrow S = \frac{3D}{4} = E$

Total spent: $25T + 20E + 18D + 12S = 1330$ $25\left(\frac{4E}{5}\right) + 20E + 18\left(\frac{4E}{3}\right) + 12E = 1330$ $20E + 20E + 24E + 12E = 1330$ $76E = 1330$ $E = 17.50$

So, each engineer spent Rs. 17.50.

(a) 40

Let A, B, C, D be the number of cards each has.

From the first condition:

• If A gives B 8 cards, B has as many as C:
  (B+8=C \Rightarrow B=C-8)
• A will have 3 less than C:
  (A-8=C-3 \Rightarrow A=C+5)

From the second condition:

• If A takes 6 cards from C, A has twice D:
  (A+6=2D \Rightarrow D=\frac{A+6}{2})

Given (B+D=50): [ (C-8)+\frac{C+5+6}{2}=50 ] [ C-8+\frac{C+11}{2}=50 ] [ 2C-16+C+11=100 ] [ 3C=105 \Rightarrow C=35 ] So, [ A=C+5=35+5=40 ]

Therefore, A has 40 cards.

(d) 50 years

Let the elder son’s age be $E$ and the younger son’s age be $Y$.

$F=2E,\qquad E-Y=15$ $F+10=3(Y+10)$

Since $E=Y+15$, we get $F=2(Y+15)=2Y+30$.

Also, $2Y+30+10=3Y+30$ $2Y+40=3Y+30$ $Y=10$

So, $F=2(10+15)=50$

Thus, the father’s age is 50 years.

(c) 23

Total language-speakers counted with repetition:

$6 + 15 + 6 = 27$

Let the number speaking exactly two languages be 2, and all three languages be 1.

So,
$\text{Total} = 27 - 2 - 2 \times 1 = 23$

Hence, the number of persons in the group is 23.

(b) 990

Pages 1–9 use $9$ digits, pages 10–99 use $90 \times 2 = 180$ digits, and pages 100–366 use $267 \times 3 = 801$ digits.
Total $= 9 + 180 + 801 = 990$.

(b) 45

Each handshake involves a pair of people, so the number of handshakes is the number of ways to choose 2 people from 10:

$\binom{10}{2} = \frac{10 \times 9}{2} = 45$

(d) None of these

The monkey climbs 30 ft at the start of each hour: 30 ft at 8:00 a.m., 60 ft at 9:00, 90 ft at 10:00 and 120 ft at 11:00 a.m. So he first reaches the flag at 120 ft at about 11:00 a.m. (the fourth climb), which is none of 4:00, 5:00 or 6:00 p.m.

(c) 51

Pattern: in each column, multiply the 2nd and 3rd numbers, then subtract the 1st number.

• Column 1: (4 \times 7 - 1 = 28 - 1 = 27)
• Column 2: (5 \times 8 - 2 = 40 - 2 = 38)
• Column 3: (6 \times 9 - 3 = 54 - 3 = 51)

(c) 68.3 meters

Let the height of the tower be $h$ m and the first point be $x$ m from its base.

At $45^\circ$: $\tan 45^\circ=\frac{h}{x}=1 \Rightarrow x=h$

At $30^\circ$, after going 50 m further: $\tan 30^\circ=\frac{h}{x+50}=\frac{1}{\sqrt3}$ $h=\frac{h+50}{\sqrt3}$ $\sqrt3h=h+50$ $h(\sqrt3-1)=50$ $h=\frac{50}{\sqrt3-1}=25(\sqrt3+1)\approx 68.3\text{ m}$

So the correct option is (c) 68.3 meters.

The table gives rows $(4,5,6)$, $(2,3,4)$, $(3,4,5)$ and $(10,17,?)$.

In the first three rows the pattern is $c=b+1$ and $b=a+1$, but the row $(10,17,?)$ does not follow $b=a+1$. Applying $c=b+1$ would give $18$, which is not among the options (7, 5, 3, 0). No single rule fits all four rows and yields one of the listed options, so the question appears to be misprinted or incomplete and has no reliably correct choice.

(d) None of these

The given values are $3, 49, 2, 7, 40, 5, 6, 55, 4$. Each value occurs only once, so there is no repeated value. Hence the distribution has no mode.

(a) Diagonals are equal and bisect each other at right angles

In a square, both diagonals are equal in length, and they bisect each other perpendicularly (at 90°).

(d) has infinitely many solutions

The second equation is just twice the first: $6x+2y-2=2(3x+y-1)=0$ So both equations represent the same straight line, and every point on it satisfies the system. Hence there are infinitely many solutions.

(c) 0

Since $16x^2+24x+9=(4x+3)^2$, putting $x=-\frac34$ gives $(4(-\frac34)+3)^2=(-3+3)^2=0$.

(d) 3

Putting $A=3$: $1+3+3^2+3^3=1+3+9+27=40$, so $A=3$.

(d) 11:16

Let (a=3k) and (b=4k). Then [ 6a+b=6(3k)+4k=22k,\qquad 4a+5b=4(3k)+5(4k)=32k ] So the ratio is (22k:32k=11:16).

(b) idempotent matrix

Given $AB=A$ and $BA=B$, substitute $A=AB$ for the left factor and then $BA=B$:

$A^2=(AB)A=A(BA)=AB=A$

Here the associative law gives $(AB)A=A(BA)$, then $BA=B$ gives $AB$, and $AB=A$. So $A^2=A$, i.e. $A$ is idempotent.

(d) $A^2 - BA - AB + B^2$

The question is a misprint for $(A-B)^2$ (as printed, $(A-2B)$ is just a matrix and matches no option). Expanding:

$(A-B)^2=(A-B)(A-B)=A^2-AB-BA+B^2$

Matrix multiplication is not commutative, so $AB$ and $BA$ must be kept separate. Option (d) matches this when its last term is read as $+B^2$.

(c) Rs. 500

Let the cost price be Rs. $C$.

Loss at SP Rs. 400: $\frac{C-400}{C}\times 100 = x$

Loss at SP Rs. 380: $\frac{C-380}{C}\times 100 = x+4$

Subtracting: $\frac{20}{C}\times 100 = 4$ $C = \frac{2000}{4} = 500$

So, the cost price of the fan is Rs. 500.

(a) 1

A single coin flip has only two possible outcomes, head or tail, so getting either head or tail is a certain event. Hence the probability is $\frac{2}{2}=1$.

(b) 360

Boys $=\frac{5}{6}\times 36=30$, girls $=6$.
Average age of boys $=2\times 6=12$ years.
Total age of boys $=30\times 12=360$.

(c) 336

The pattern is $n^3-n$:
$1^3-1=0,\ 2^3-2=6,\ 3^3-3=24,\ 4^3-4=60,\ 5^3-5=120,\ 6^3-6=210$
So next: $7^3-7=343-7=336$.

(b) 81

The terms are squares of consecutive odd numbers: $1^2, 3^2, 5^2, 7^2, 9^2, 11^2$. So the missing term is $9^2 = 81$.

(a) 10, 10

Split into two alternating series:

• Odd positions: 20, 19, 17, 14, ? → subtract 1, 2, 3, 4 → next is 10
• Even positions: 20, 16, 13, 11, ? → subtract 4, 3, 2, 1 → next is 10

So the missing terms are 10, 10.

(b) G

Positions: R(18) → U(21) → X(24) → A(1) → D(4), adding 3 each time (wrapping after Z). So next is G(7).

(c) MNOP

The groups have 1, 2, 3, 4, 5 letters respectively, and the gaps between groups increase by one letter:
A | skip B | CD | skip EF | GHI | skip JKL | MNOP | skip QRST | UVWXY.

(a) 12

All the other numbers (1, 3, 9, 19, 29) are odd; 12 is the only even number, so it is the odd one out.

(d) 18

All the other numbers are perfect cubes:
$27 = 3^3,\ 64 = 4^3,\ 125 = 5^3,\ 216 = 6^3,\ 343 = 7^3$.
But $18$ is not a perfect cube, so it is the odd one out.

(c) 14

All the other numbers are odd, while 14 is the only even number in the list.

The question stem is incomplete (only the series 2, 4, 16, 23, 25, 26 and the options 2, 8, 16, 30 survive), so no single option can be justified from the text. The series gives no pattern that selects any of these options, and 8 and 30 are not in the series at all. The original stem should be restored from the source paper before a key is assigned.

(a) 7

The average of the given numbers is
[ \frac{2+4+6+8+10+12}{6}=\frac{42}{6}=7. ]

(b) Often : Rarely

"Always" and "Never" are opposite in meaning. Similarly, "Often" and "Rarely" are opposites.

(a) Ice : Cold

Sugar is known for its quality of being sweet, just as ice is known for its quality of being cold. The other options do not show this "thing : characteristic quality" relationship.

(d) Road : Car

A train runs on a track in the same way a car runs on a road — both are man-made paths built for the vehicle to move along.

(c) Jhapa : Bhadrapur

Pokhara is the district headquarters of Kaski district. Similarly, Bhadrapur is the district headquarters of Jhapa district.

(b) Interest : Capital

Rent is the reward (return) earned from the factor of production land, just as interest is the reward earned from capital.

(a) January : February

Monday and Tuesday are consecutive days of the week, just as January and February are consecutive months of the year.

(c) Chalk : Blackboard

Ink is used to write on paper, just as chalk is used to write on a blackboard.

(d) Patient : Client

Doctor and Lawyer are both professionals; Patient and Client are the persons they respectively serve. A doctor serves a patient, and a lawyer serves a client. Hence, Doctor : Lawyer :: Patient : Client.

(b) Cabin : Ship

A room is a structural compartment/part of a house. Similarly, a cabin is a structural compartment/part of a ship. Option (c) is also a part–whole relation, but a wheel is not an essential/structural compartment of a typical chair, so (b) is the best match. Options (a) and (d) are merely objects located in a place.

(c) Mouse : Hole

A bee lives in a hive, just as a mouse lives in a hole — the second is the dwelling or shelter of the first.

(a) Calendar

MS Excel has built-in date functions DATE(), DAY() and MONTH() — DATE(year, month, day) builds a date, DAY(serial) returns the day of the month (1–31) and MONTH(serial) returns the month number (1–12). "Calendar" is not a worksheet function in Excel, so it is the odd one out.

(b) Vulture

Odd one out: the vulture is a scavenging bird of prey that feeds on carrion, while the parrot, swan and sparrow are not.

(c) Fifth

The fifth generation of computers is based on artificial intelligence (AI) and parallel processing, using technologies such as voice recognition and natural-language understanding.

(d) Nagarkot

Odd one out: Tansen (Palpa), Dhangadi (Kailali) and Nepalgunj (Banke) are all district headquarters and towns, while Nagarkot is a hill-top viewpoint village near Bhaktapur, known for Himalayan sunrise views.

(c) Bat

Bat is the only mammal among the four — it is a flying mammal (order Chiroptera) that gives birth to live young and feeds them milk. Parrot and Crow are birds, and Shark is a cartilaginous fish.

(c) One-Third

A cone (or pyramid) and a cylinder (or prism) with the same base area and the same height are related by the factor $\frac{1}{3}$:

$V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\,V_{\text{cylinder}}$

So the volume of the cone is one-third of the volume of the cylinder having the same base and height.

(b) New York

The United Nations headquarters is located in New York City, USA. (Paris hosts UNESCO, Kathmandu is Nepal's capital, and Tokyo is Japan's capital — none of these is the UN headquarters.)

(c) 135

Odd one out: 91 = 7 × 13, 112 = 7 × 16 and 154 = 7 × 22 are all multiples of 7, but 135 is not (135 = 7 × 19 + 2).

(a) Bagful

"Bagful" is a noun (the amount a bag can hold), whereas "careful", "skilful" and "hopeful" are adjectives formed with the suffix -ful. So "Bagful" is the odd one out.

(d) Grass

Lamb, dog and cat are all animals, while grass is a plant — so grass is the odd one out.

(a) f, a, b, c, e, d

Going upward from the foot, the body parts come in the order: foot → waist → belly → chest → neck → head.

(c) HIGH

Each digit group is the alphabet position of the letters: $819$ is H-A-I ($8,1,9$) and $945$ is I-D-E ($9,4,5$). Decoding $8978$ the same way gives $8=H$, $9=I$, $7=G$, $8=H$, so the word is HIGH.

(c) TRUTH

In the example, each pair of letters surrounds the letter it stands for: LN gives M, NP gives O, QS gives R, and the last pair gives E. Applying this to SU, QS, TV, SU gives T, R, U, T. These are the first four letters of TRUTH, the only option that begins TRUT. The fifth pair (for H) appears to be missing from the question as printed.

(d) SRTS

Using the reverse alphabet code: A↔Z, B↔Y, C↔X, D↔W, etc.
So H→S, I→R, G→T, H→S, giving HIGH → SRTS.

(c) 2428

Pattern: multiply each number by 4 and write the results side by side. $2 \times 4 = 8$, $3 \times 4 = 12 \Rightarrow 812$; $4 \times 4 = 16$, $5 \times 4 = 20 \Rightarrow 1620$. So, $6 \times 4 = 24$, $7 \times 4 = 28 \Rightarrow 2428$.

(a) 12

All the other numbers — 1, 3, 9, 13 and 29 — are odd, while 12 is the only even number in the list.

(b) 150

Previous height $= \frac{159}{1.06} = 150$ cm.

(c) 166.66

He spent 25%, so 75% of his money is left.
$75\% \text{ of } x = 125 \Rightarrow x = \frac{125 \times 100}{75} = 166.66$

(a) 23.07%

If the cost increases by 30%, the new cost becomes $1.3$ times the original. To keep total spending the same, consumption must be reduced by
$1-\frac{1}{1.3}=\frac{0.3}{1.3}=0.2307=23.07\%$

(b) 8%

Let original price = $P$ and quantity = $Q$. New price = $1.25P$, and new expenditure = $1.15PQ$.
New quantity $=\dfrac{1.15PQ}{1.25P}=0.92Q$, so reduction $=Q-0.92Q=0.08Q=8\%$.

28,000 votes (this value is not among the listed options)

Let the total votes be $T$. Congress got 10% of $T$ more than Non-Congress, so Congress has $55\%$ and Non-Congress $45\%$ of $T$.

$0.45\,T = 126{,}000 \Rightarrow T = 280{,}000$

Congress got $154{,}000$ votes, so the margin is $154{,}000 - 126{,}000 = 28{,}000$. None of (a)-(d) equals this, so the question or its options are misprinted.

(a) 285,000

Net growth rate = birth rate − death rate = 50% − 31% = 19%.
Net change = 1,500,000 × 19% = 1,500,000 × 0.19 = 285,000.

(d) 1%

Let the original side of the square be $s$.
New length $=1.1s$, new width $=0.9s$.

New area: $1.1s \times 0.9s = 0.99s^2$

Original area $=s^2$, so change: $\frac{s^2-0.99s^2}{s^2}\times 100 = 1\%$

So the area decreases by 1%.

(a) 112.5

Ram’s marks = $40\%$ of 200 = 80
Shyam’s marks = $80 + 25\%$ of 80 = $80+20=100$

Let Bhuwan’s marks be $x$.
Bhuwan is ahead of Shyam by one-ninth of his own marks:

$x = 100 + \frac{x}{9}$

$x-\frac{x}{9}=100 \Rightarrow \frac{8x}{9}=100 \Rightarrow x=112.5$

So Bhuwan gets 112.5 marks.

(b) 10

Only one Englishman (E1) can speak to both groups, so every secret must pass through E1.

• 4 calls: the three Frenchmen and E1 share all four secrets (each Frenchman calls E1, and the French exchange among themselves in French).
• 3 calls: E1-E2, E2-E3, E1-E3 give all three Englishmen every secret.
• 3 calls: E1 (now knowing all six) calls each Frenchman once more so that they learn the English secrets.

Total $4+3+3 = 10$ calls.

(d) 18

Let the number of 1 Miso, 10 Miso, and 50 Miso coins be $x$, $y$, and $z$ respectively.

$x + 10y + 50z = 107$

Possible values of $z$:

• $z=0$: $x+10y=107$ → $y=0$ to $10$ → 11 ways
• $z=1$: $x+10y=57$ → $y=0$ to $5$ → 6 ways
• $z=2$: $x+10y=7$ → $y=0$ → 1 way

Total ways:

$11+6+1=18$

So the correct answer is (d) 18.

Let the cheque amount be Rs $x$ and $y$ paisa, so its value in paisa is:

$100x + y$

The teller transposed rupees and paisa, so Shailaja received:

$100y + x \text{ paisa}$

After spending 50 paisa, she had three times the cheque amount:

$100y + x - 50 = 3(100x + y)$

$100y + x - 50 = 300x + 3y$

$97y - 299x = 50$

Solving gives:

$x = 18,\quad y = 56$

So the cheque amount is Rs 18.56, which is over Rs 18 but less than Rs 19.

Answer: (d) Over Rupees 18 but less than Rupees 19

(a) 24 years

At present, the 6 surviving original members have total age
$231 - (60-3) - (60-6) + 6 \times 10 = 231 - 57 - 54 + 60 = 180$

The two children born 7 years ago and 4 years ago have current ages $7$ and $4$ years.

Total current age of 8 members: $180 + 7 + 4 = 191$

Average: $\frac{191}{8} = 23.875 \approx 24 \text{ years}$

So the nearest average age is 24 years.

(d) None of the above

By inclusion-exclusion, the number of animals before the transfer is

$42+34+20-24-10-12+4 = 54$

The four animals grazing in all three sectors are then transported to another zoo, leaving $54-4 = 50$ animals. This matches none of (a)-(c), so the answer is (d).

(a) 25,000

Let total enrolled voters be $x$.

Votes polled $=90\%$ of $x=0.9x$
Valid votes $=90\%$ of polled $=0.9\times 0.9x=0.81x$

Majority $=54\%-46\%=8\%$ of valid votes.

$8\% \text{ of } 0.81x=1620$ $0.08\times 0.81x=1620$ $x=\frac{1620}{0.0648}=25000$

So, the number of voters enrolled was 25,000.

(a) Rs 15,600 (approximately)

Bring each withdrawal back to the opening date using simple interest at 10% per year:

$\frac{10000}{1+0.1\times4}+\frac{6000}{1+0.1\times3}+\frac{5000}{1+0.1\times2}\approx 7143+4615+4167 \approx 15{,}925$

The deposit is about Rs 15,900, and the nearest option is Rs 15,600.

(d) 8:00 pm

Total work = (6 \times 10 = 60) technician-hours.

From 11:00 AM to 5:00 PM = 6 hours.
Work done by 6 technicians = (6 \times 6 = 36) technician-hours.
Remaining work = (60 - 36 = 24) technician-hours.

From 5:00 PM onward, one technician is added each hour:

• 5:00–6:00 PM: 7 technicians → (7) hours of work, remaining (24 - 7 = 17)
• 6:00–7:00 PM: 8 technicians → (8) hours of work, remaining (17 - 8 = 9)
• 7:00–8:00 PM: 9 technicians → (9) hours of work, remaining (9 - 9 = 0)

So the server is completed at 8:00 PM.

(c) 10 minute

Let the slowest runner’s speed be $v$ m/min.
Then the fastest runner’s speed is $2v$ m/min.

They meet at the same point for the first time after 5 minutes, so their relative distance covered must equal one circumference:

$(2v - v)\times 5 = 100$

$5v = 100 \Rightarrow v = 20$

So the fastest runner’s speed is:

$2v = 40 \text{ m/min}$

Time taken by the fastest runner to finish 400 m:

$\frac{400}{40} = 10 \text{ minutes}$

So the answer is (c) 10 minute.

(c) 52 seconds

Speed = $45$ km/h $=45\times\frac{5}{18}=12.5$ m/s.
In 60 seconds, the train covers $12.5\times60=750$ m.
So, train length = $750-100=650$ m.
Time to cross an electric pole = $\frac{650}{12.5}=52$ seconds.

(c) 9

Let the consecutive odd numbers be (P=x,\ Q=x+2,\ R=x+4).

Given: [ 3P = 2R - 3 ] [ 3x = 2(x+4)-3 = 2x+5 ] [ x=5 ] So, [ R=x+4=5+4=9 ]

Hence, R is 9.

(c) Rs. 528,000

Perimeter of square hall = 400 m
Side = $400/4 = 100$ m
Area = $100^2 = 10000$ m²

Damaged tiles = 10% of 10000 = 1000 m²
Total tiles to be paid for = $10000 + 1000 = 11000$ m²

Amount = $11000 \times 48 = \text{Rs. } 528000$

So the correct option is (c) Rs. 528,000.

(a) 39

Let the average after 16 innings be $x$.
After scoring 87 in the 17th innings, the new average is $x+3$.

$17(x+3)=16x+87$ $17x+51=16x+87$ $x=36$

So the average after 17 innings is $36+3=39$.

(b) 21

First three sum: $a+b+c=16\times3=48$.
Last three sum: $b+c+d=15\times3=45$.
Given $d=18$, so $b+c=45-18=27$.
Thus $a=48-(b+c)=48-27=21$.

(c) Rs. 21,000

Investment ratio: [ 120000:135000:150000 = 8:9:10 ] Total parts (=8+9+10=27)

Pavan’s share: [ \frac{10}{27}\times 56700 = 21000 ]

So Pavan’s share is Rs. 21,000.

(b) 5:12

Rita and her mother's ages are $3x$ and $8x$. Given $8x-3x=35$, so $x=7$. Thus Rita is 21 and her mother is 56. After 4 years, the ratio is $25:60=5:12$.

(d) Rs 9,000

This is a standard problem whose original data is A's capital of Rs 3,500. With the printed Rs 45,000, B's capital would come to about Rs 1,15,714, which is not an option, so the figure is evidently a misprint.

A invests for 12 months and B for $12-5=7$ months. Profits are shared in proportion to capital $\times$ time:

$3500\times12 : 7x = 2:3 \Rightarrow 14x = 126000 \Rightarrow x = 9000$

So B's contribution is Rs 9,000.