Complete Study Reference

Quantitative Aptitude

Volume & Surface Areas · Races · Calendar · Clocks · Stocks · Permutations · Probability · Discount · Heights · Series · Data Interpretation

Table of Contents
Ch. 25 Volume & Surface Areas

Questions involving 3-D solids: cubes, cuboids, cylinders, cones, spheres, hemispheres, frustums, prisms, and pyramids. Focus on knowing exact formulae and applying them to problems involving melting, recasting, and composite solids.

Key Formulae — Solids
Cuboid (l × b × h) Volume = l × b × h
LSA = 2h(l + b)
TSA = 2(lb + bh + lh)
Diagonal = √(l² + b² + h²)
Cube (side a) Volume = a³
LSA = 4a²
TSA = 6a²
Diagonal = a√3
Cylinder (r, h) Volume = πr²h
CSA = 2πrh
TSA = 2πr(r + h)
Cone (r, h, l) Slant l = √(r² + h²)
Volume = ⅓πr²h
CSA = πrl
TSA = πr(r + l)
Sphere (r) Volume = (4/3)πr³
Surface Area = 4πr²
Hemisphere (r) Volume = (2/3)πr³
CSA = 2πr²
TSA = 3πr²
Frustum (R, r, h) Volume = (πh/3)(R² + r² + Rr)
Slant l = √(h² + (R−r)²)
CSA = π(R + r)l
TSA = π[R² + r² + (R+r)l]
Prism Volume = Base Area × Height
LSA = Perimeter of Base × Height
TSA = LSA + 2 × Base Area
Important Relationships & Ratios
Cylinder : Cone : Sphere (same diameter and height)
Volumes are in ratio 3 : 1 : 2
Sphere vs Cube (equal surface areas)
Ratio of volumes = √π : √6
Largest sphere in a cube of side a
Radius = a/2; Volume = (4/3)π(a/2)³
Largest cube in a sphere of radius r
Side = 2r/√3; Volume = (8r³)/(3√3)
Volume becomes n³ times when edge is multiplied by n
Surface area becomes n² times
Selected Solved Problems
Q 202
A solid cone has a plane cut parallel to its base at a point which is 1/3 of the height from the top. The ratio of the volume of the smaller cone to the remaining frustum is:
(a) 1 : 3
(b) 1 : 9
✓ (c) 1 : 26
(d) 1 : 27
Solution: When cut at h/3 from top, new cone has radius r/3. Volume of small cone = (1/3)π(r/3)²(h/3) = (1/27)×(1/3)πr²h. Remaining = (26/27) of total. Ratio = 1 : 26.
Q 211
For a sphere of radius 10 cm, what percent of the numerical value of its volume equals the numerical value of its surface area?
(a) 24%
(b) 26.5%
✓ (c) 30%
(d) 45%
Solution: SA = 4π(10)² = 400π. Vol = (4/3)π(10)³ = (4000/3)π. Required% = (400π / (4000π/3)) × 100 = (400×3/4000)×100 = 30%.
Q 222
Three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere. The diameter of the new sphere is:
(a) 12 cm
✓ (b) 24 cm
(c) 30 cm
(d) 36 cm
Solution: R³ = 6³ + 8³ + 10³ = 216 + 512 + 1000 = 1728 → R = 12. Diameter = 24 cm.
Q 255
In what ratio are the volumes of a cylinder, a cone and a sphere, if each has the same diameter and the same height?
(a) 1 : 3 : 2
(b) 2 : 3 : 1
✓ (c) 3 : 1 : 2
(d) 3 : 2 : 1
Solution: With same radius r and height H = 2r (sphere diameter = height): Cyl = πr²(2r); Cone = (1/3)πr²(2r); Sphere = (4/3)πr³. Ratio = 1 : 1/3 : 2/3 = 3 : 1 : 2.
Q 253
A cone of height 15 cm and base diameter 30 cm is carved out of a wooden sphere of radius 15 cm. The percentage of wood wasted is:
(a) 25%
(b) 40%
(c) 50%
✓ (d) 75%
Solution: Vol of sphere = (4/3)π(15)³. Vol of cone = (1/3)π(15)²(15). Wood wasted = Sphere − Cone. Wasted% = [((4/3) − (1/3))π(15)³] / [(4/3)π(15)³] × 100 = 75%.
Answer Key (Selected — Q201 to Q304)
201.(a)202.(c)203.(a) 204.(d)205.(c)206.(d) 211.(c)212.(d)213.(d) 214.(c)215.(b)216.(d) 219.(d)220.(d)221.(c) 222.(b)223.(c)224.(a) 225.(d)227.(a)228.(b) 229.(a)230.(d)231.(d) 233.(c)234.(d)235.(c) 237.(b)238.(d)239.(a) 241.(d)242.(b)243.(b) 244.(c)245.(a)246.(d) 247.(b)248.(c)249.(d) 251.(c)252.(c)253.(d) 255.(c)256.(a)257.(b) 258.(c)259.(c)260.(c) 280.(c)281.(b)282.(d) 283.(c)284.(b)285.(c) 290.(c)291.(a)292.(a) 293.(a)294.(b)295.(d) 299.(b)300.(b)301.(a) 302.(b)303.(b)304.(b)

Ch. 26 Races & Games of Skill

Problems about comparing speeds, starts, and handicaps in races.

Important Facts
Race
A contest of speed in running, riding, driving, sailing or rowing.
Start / Handicap
If A gives B a start of x metres, B begins x metres ahead. In a 100 m race A runs 100 m while B runs (100 − x) m.
Dead Heat
All contestants reach the goal at exactly the same time.
Games
"A game of 100" means the first to score 100 wins. "A gives B 20 points" means A scores 100 while B scores 80.
Core Solved Examples

Ex 1. In a km race, A beats B by 28 metres or 7 seconds. A's time over the course?

B covers 28 m in 7 s → B's time = (7/28) × 1000 = 250 sec. A's time = 250 − 7 = 243 sec (4 min 3 sec).

Ex 2. A runs 1¾ times as fast as B. A gives B a start of 84 m. Where must the winning post be?

A : B = 7 : 4. In a race of 7 m, A gains 3 m. So 84 m gained when race = (7/3) × 84 = 196 m.

Ex 3. A can run 1 km in 3 min 10 sec, B in 3 min 20 sec. By what distance can A beat B?

A beats by 10 sec. B's distance in 10 s = (1000/200) × 10 = 50 m.

Ex 6. A, B, C in a km race. A gives B a 40 m start and C a 64 m start. How many metres start can B give C?

A:B = 1000:960, A:C = 1000:936. B:C = (960/1000) × (1000/936) = 960/936 → B:C = 1000:975. B gives C a 25 m start.

Ex 8. In a game of 80 points, A gives B 5 points and C 15 points. How many points can B give C in a game of 60?

A:B = 80:75, A:C = 80:65. B:C = (75/80)×(80/65) = 75/65 = 60:52. B gives C 8 points.

Exercise — Selected Q&A
Q 1
In a 100 m race, A covers in 36 sec, B in 45 sec. A beats B by how many metres?
✓ (a) 20 m
(b) 25 m
(c) 22.5 m
(d) 9 m
Solution: A beats B by 9 sec. Distance B covers in 9 s = (100/45)×9 = 20 m.
Q 10
In a 500 m race, ratio of speeds A:B = 3:4. A has a start of 140 m. A wins by:
(a) 60 m
(b) 40 m
✓ (c) 20 m
(d) 10 m
Solution: A covers 360 m. In that time B covers (4/3)×360 = 480 m. A wins by 500−480 = 20 m.
Q 24
In racing over distance d at uniform speed: A beats B by 20 m, B beats C by 10 m, A beats C by 28 m. Then d (in metres) is:
(a) 50
(b) 75
✓ (c) 100
(d) 120
Solution: C runs (x−10)(x−20)/x. x − (x−10)(x−20)/x = 28 → 30x − 200 = 28x → 2x = 200 → x = 100.
Answer Key — Ch. 26 1.(a) 2.(b) 3.(a) 4.(d) 5.(b) 6.(d) 7.(b) 8.(b) 9.(c) 10.(c)
11.(a) 12.(c) 13.(b) 14.(c) 15.(b) 16.(c) 17.(a) 18.(b) 19.(c) 20.(c)
21.(c) 22.(d) 23.(a) 24.(c) 25.(a)

Ch. 27 Calendar

Finding the day of the week for any given date using the concept of "odd days".

Fundamental Concepts
Odd Days
The number of days more than complete weeks in a given period.
Ordinary Year
365 days = 52 weeks + 1 day → 1 odd day.
Leap Year
366 days = 52 weeks + 2 days → 2 odd days. Leap year: divisible by 4 but not century unless also divisible by 400.
Odd Days in n Years 100 years → 5 odd days
200 years → 3 odd days
300 years → 1 odd day
400 years → 0 odd days (and multiples: 800, 1200, 1600, 2000…)

Day mapping: 0=Sun, 1=Mon, 2=Tue, 3=Wed, 4=Thu, 5=Fri, 6=Sat
Solved Examples

What day was 16th July 1776?

1600 yrs = 0 odd days; 100 yrs = 5; 75 yrs = 2 odd days. Total = 7 ≡ 0. Jan–July 16 in 1776 (leap): 31+29+31+30+31+30+16 = 198 days = 28w + 2 → 2 odd days. Grand total = 0+2 = 2 → Tuesday.

What day was 15th August 1947?

1600 = 0; 300 = 1; 46 yrs (11 leap, 35 ord) = 22+35 = 57 = 8w+1. Jan–Aug 15: 31+28+31+30+31+30+31+15 = 227 = 32w+3. Total = 0+1+1+3 = 5 → Friday.

Exercise — Selected Questions
Q 5
The calendar for 2007 will be the same for the year:
(a) 2014
(b) 2016
(c) 2017
✓ (d) 2018
Solution: Sum of odd days from 2007 onwards must total 0 (mod 7). Years 2007–2017 give 1+2+1+1+1+2+1+1+1+2+1 = 14 ≡ 0. Calendar repeats in 2018.
Answer Key — Ch. 27 1.(b) 2.(c) 3.(d) 4.(b) 5.(d) 6.(d) 7.(c) 8.(d) 9.(a) 10.(b)
11.(c) 12.(a) 13.(b) 14.(c) 15.(c) 16.(c) 17.(b) 18.(c)

Ch. 28 Clocks

Angle between clock hands, coincidence, right angles, opposite positions, fast/slow clocks.

Key Facts
Minute hand speed
6° per minute
Hour hand speed
0.5° per minute (360° in 12 hrs)
Relative speed
Minute hand gains 5.5° per minute over the hour hand
Hands coincide
Every 65 5/11 minutes (22 times per day)
Right angle
44 times per day (hands 15 min spaces apart)
Opposite direction
22 times per day (hands 30 min spaces apart)
Same straight line
44 times per day (coincident + opposite)
Angle Formula
Angle between hands at H hours M minutes θ = |30H − (11/2)M| degrees
(Take value ≤ 180°, else 360° − θ for reflex)

Minutes to gain x spaces = (60/55) × x = (12/11) × x min
Solved Examples

Ex 1. Angle at 3:25?

Hour hand at 3 hrs 25 min = 3×30 + 25×0.5 = 90+12.5 = 102.5°. Minute hand = 25×6 = 150°. Angle = 150−102.5 = 47.5°.

Ex 2. When between 2 and 3 will hands coincide?

At 2 o'clock, hands are 10 min apart. Minute must gain 10 min. Time = (12/11)×10 = 10 10/11 min past 2.

Ex 6. Clock gains if minute hand overtakes hour hand every 65 min. Daily gain?

Correct interval = 65 5/11 min. Gain per 65 min = 5/11 min. Gain in 24 hrs = (5/11)×(60×24/65) = 10 10/43 min.

Selected Exercise Q&A
Q 9
At 3:40, the hour hand and minute hand of a clock form an angle of:
(a) 120°
(b) 125°
✓ (c) 130°
(d) 135°
Solution: Hour hand = (11/3)×30 = 110°. Minute hand = 40×6 = 240°. Angle = 240−110 = 130°.
Q 35
How much does a watch lose per day if its hands coincide every 64 minutes?
✓ (a) 32 8/11 min
(b) 36 5/11 min
(c) 90 min
(d) 96 min
Solution: Correct gap = 65 5/11 min. Loss per 64 min = 65 5/11 − 64 = 16/11 min. Loss per 24 hrs = (16/11)×(1/64)×24×60 = 32 8/11 min.
Answer Key — Ch. 28 1.(c) 2.(a) 3.(d) 4.(b) 5.(a) 6.(d) 7.(c) 8.(d) 9.(c) 10.(b)
11.(b) 12.(c) 13.(d) 14.(a) 15.(b) 16.(c) 17.(c) 18.(b) 19.(d) 20.(c)
22.(c) 23.(b) 24.(c) 25.(b) 26.(c) 27.(d) 28.(d) 29.(c) 30.(d)
35.(a) 36.(d) 37.(c) 38.(d) 39.(d) 40.(b) 49.(b) 52.(b) 53.(c) 54.(a)

Ch. 29 Stocks & Shares

Investment in stocks, calculating income, yield, market value, brokerage, and comparing investments.

Key Concepts
Face Value (Nominal Value)
Value printed on the share certificate (usually ₹100). Dividend is always paid on this.
Market Value
At premium: MV > FV. At par: MV = FV. At discount: MV < FV.
Brokerage
Added to cost when buying; subtracted from sale price when selling.
Dividend
Annual profit paid on Face Value.
Rate of Return
Annual income from an investment of ₹100 (using MV).
Important Relationships Investment = (MV/FV) × Stock purchased
Income = Dividend% × FV of stock
Number of shares = Total Investment / MV per share

For "x% stock at y":
• FV = ₹100, MV = ₹y
• Dividend per share = x% of 100 = ₹x
• Yield = (x/y) × 100 %
Solved Examples

Ex 5. Find income from ₹2500, 8% stock at 106.

Income from ₹100 stock = ₹8. Income from ₹2500 = (8/100)×2500 = ₹200.

Ex 6. Which is better: 7½% stock at 105 or 6½% stock at 94?

Invest ₹(105×94) in each. Case I income: (15/2)×(1/105)×(105×94) = 705. Case II: (13/2)×(1/94)×(105×94) = 682.5. 7½% at 105 is better.

Ex 11. A man buys ₹25 shares paying 9% dividend. He wants 10% on investment. At what price?

Dividend = 9% of 25 = ₹2.25. Need 10% return → investment = 2.25/10% = ₹22.50.

Answer Key — Ch. 29 1.(c) 2.(b) 3.(b) 4.(c) 5.(c) 6.(a) 7.(c) 8.(b) 9.(a) 10.(b)
11.(c) 12.(d) 13.(a) 14.(a) 15.(b) 16.(b) 17.(b) 18.(c) 19.(b) 20.(b)
21.(c) 22.(a) 23.(b) 24.(b) 25.(d) 26.(b) 27.(b) 28.(c) 29.(b) 30.(b)

Ch. 30 Permutations & Combinations
Core Formulae
Permutations ⁿPᵣ = n! / (n−r)!
= n(n−1)(n−2)…(n−r+1)

All n things: ⁿPₙ = n!
With repeated items: n! / (p₁! p₂! … pₖ!)
Combinations ⁿCᵣ = n! / (r!(n−r)!)
ⁿCᵣ = ⁿC(n−r)
ⁿC₀ = ⁿCₙ = 1
0! = 1
Key difference: In permutations, order matters (AB ≠ BA). In combinations, order does not matter (AB = BA).
Solved Examples

Ex 4. Arrangements of letters in DAUGHTER so vowels always come together?

Treat AUE as one block → 6 letters arranged in 6! = 720 ways. AUE arrange in 3! = 6 ways. Total = 720×6 = 4320.

Ex 6. Arrangements of ENGINEERING?

11 letters: 3E, 3N, 2G, 2I, 1R. Total = 11! / (3!×3!×2!×2!×1!) = 277200.

Ex 10. Vowels always in odd positions in DETAIL (3 vowels, 3 consonants)?

3 vowels in 3 odd positions: 3P₃ = 6 ways. 3 consonants in remaining 3 positions: 3P₃ = 6 ways. Total = 6×6 = 36.

Ex 13. Committee of 6 from 7 men and 5 ladies with 4 men and 2 ladies?

⁷C₄ × ⁵C₂ = 35 × 10 = 350.

Exercise — Word Arrangements (Selected)
WordLength/RepeatsArrangementsAns
DISPLAY7 distinct7! = 5040(d) 5040
SMART5 distinct5! = 120(e) 120
FORMULATE9 distinct9! = 362880(d) 362880
RIDDLED7; D×37!/3! = 840(a) 840
CREATE6; E×26!/2! = 360(c) 360
TOTAL5; T×25!/2! = 60(b) 60
OFFICES7; F×27!/2! = 2520(a) 2520
BANANA6; A×3, N×26!/(3!×2!) = 60(a) 60
ENGINEERING11; E×3,N×3,G×2,I×211!/(3!3!2!2!) = 277200(a) 277200
ALLAHABAD9; A×4, L×29!/(4!×2!) = 7560(c) 7560
RUMOUR6; R×2, U×26!/(2!×2!) = 180(c) 180
Answer Key — Ch. 30 1.(d) 2.(d) 3.(e) 4.(d) 5.(e) 6.(a) 7.(c) 8.(b) 9.(a) 10.(a)
11.(b) 12.(c) 13.(b) 14.(b) 15.(c) 16.(c) 17.(d) 18.(c) 19.(b) 20.(d)
21.(b) 22.(e) 23.(c) 24.(b) 25.(d) 26.(c) 27.(c) 28.(c) 29.(a) 30.(c)
31.(b) 32.(a) 33.(c) 34.(a) 35.(d) 36.(e) 37.(b) 38.(e) 39.(b) 40.(b)
41.(d) 42.(b) 43.(c) 44.(d) 45.(b) 46.(c) 47.(b) 48.(a)

Ch. 31 Probability
Key Definitions
Sample Space (S)
Set of all possible outcomes of a random experiment.
Event (E)
Any subset of the sample space.
Probability
P(E) = n(E) / n(S). Always: 0 ≤ P(E) ≤ 1.
Key Results P(S) = 1, P(φ) = 0
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
P(not A) = P(Ā) = 1 − P(A)
P(A ∩ B) = P(A) × P(B) [independent events]

Deck of cards: 52 total, 13 per suit, 26 red, 26 black
Face cards = 12 (4 Jacks, 4 Queens, 4 Kings)
Sample Space Reference

One coin toss: S = {H, T}, n(S) = 2

Two coin toss: S = {HH, HT, TH, TT}, n(S) = 4

Three coin toss: n(S) = 8

One die: S = {1,2,3,4,5,6}, n(S) = 6

Two dice: n(S) = 36

Cards: n(S) for 2 drawn = ⁵²C₂ = 1326

Selected Solved Problems
Q 4
In a simultaneous throw of two dice, probability of getting a total of 7:
✓ (a) 1/6
(b) 1/4
(c) 2/3
(d) 3/4
Solution: Favourable: (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) = 6 outcomes. P = 6/36 = 1/6.
Q 18
From a pack of 52 cards, two cards drawn together. Probability both are kings:
(a) 1/15
(b) 25/57
(c) 35/256
✓ (d) 1/221
Solution: n(E) = ⁴C₂ = 6, n(S) = ⁵²C₂ = 1326. P = 6/1326 = 1/221.
Q 41
A man and wife appear for two vacancies in same post. P(husband selected) = 1/7, P(wife selected) = 1/5. P(only one selected):
(a) 4/5
✓ (b) 2/7
(c) 4/7
(d) 8/15
Solution: P(H sel, W not) + P(W sel, H not) = (1/7)(4/5) + (1/5)(6/7) = 4/35 + 6/35 = 10/35 = 2/7.
Answer Key — Ch. 31 1.(d) 2.(b) 3.(d) 4.(b) 5.(a) 6.(c) 7.(a) 8.(d) 9.(b) 10.(a)
11.(d) 12.(c) 13.(b) 14.(c) 15.(c) 16.(a) 17.(b) 18.(d) 19.(d) 20.(d)
21.(b) 22.(d) 23.(b) 24.(c) 25.(b) 26.(d) 27.(c) 28.(b) 29.(d) 30.(d)
31.(b) 32.(b) 33.(a) 34.(b) 35.(a) 36.(b) 37.(a) 38.(d) 39.(c) 40.(a)
41.(b) 42.(c) 43.(a) 44.(d) 45.(b) 46.(a) 47.(a) 48.(b) 49.(b) 50.(e)

Ch. 32 True Discount

The present worth of a future sum, and the interest saved by paying now rather than later.

Key Formulae
True Discount Formulae (Rate R%, Time T years) Present Worth (P.W.) = (100 × Amount) / (100 + R×T)
T.D. = (P.W. × R × T) / 100
T.D. = (Amount × R × T) / (100 + R×T)
Sum due = (S.I. × T.D.) / (S.I. − T.D.)
S.I. − T.D. = S.I. on T.D.

[Compound Interest case]
P.W. = Amount / (1 + R/100)ᵀ
Selected Solved Problems

Ex 1. Present worth of ₹930 due 3 years hence at 8% p.a.?

P.W. = (100×930) / (100+8×3) = 93000/124 = ₹750. T.D. = 930−750 = ₹180.

Ex 3. T.D. = ₹250, S.I. = ₹375 (same sum, time, rate). Find sum.

Sum = (375×250)/(375−250) = 93750/125 = ₹750.

Answer Key — Ch. 32 1.(b) 2.(a) 3.(c) 4.(d) 5.(b) 6.(b) 7.(c) 8.(d) 9.(a) 10.(a)
11.(a) 12.(b) 13.(b) 14.(a) 15.(d) 16.(b) 17.(b) 18.(c) 19.(b) 20.(d) 21.(d)

Ch. 33 Banker's Discount
Key Concepts & Formulae
Banker's Discount (B.D.)
Simple interest on the face value (not P.W.) for the unexpired time. B.D. = S.I. on bill for unexpired time.
Banker's Gain (B.G.)
B.G. = B.D. − T.D. = S.I. on T.D. = (T.D.)² / P.W.
True Discount
T.D. = √(P.W. × B.G.)
Banker's Discount Formulae B.D. = Amount × Rate × Time / 100
T.D. = Amount × R × T / (100 + R×T)
Amount = B.D. × T.D. / (B.D. − T.D.)
T.D. = B.G. × 100 / (Rate × Time)
Solved Example

Ex 1. Bill for ₹6000 drawn July 14, 5 months. Discounted Oct 5 at 10%. Find B.D., T.D., B.G.

Legally due: Dec 17. Unexpired: Oct 5 to Dec 17 = 73 days = 1/5 year.
B.D. = 6000 × 10 × (1/5) / 100 = ₹120.
T.D. = (6000 × 10 × 1/5) / (100 + 10/5) = 12000/102 = ₹117.64.
B.G. = 120 − 117.64 = ₹2.36.

Answer Key — Ch. 33 1.(b) 2.(b) 3.(c) 4.(c) 5.(d) 6.(a) 7.(c) 8.(d) 9.(b) 10.(c) 11.(a) 12.(c) 13.(b)

Ch. 34 Heights & Distances
Trigonometric Ratios & Standard Values
Basic Ratios in Right Triangle sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
Standard Values θ: 0° 30° 45° 60° 90°
sin: 0 1/2 1/√2 √3/2 1
cos: 1 √3/2 1/√2 1/2 0
tan: 0 1/√3 1 √3 ∞

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
Angle of Elevation
When you look up at an object. The angle the line of sight makes with the horizontal.
Angle of Depression
When you look down at an object. The angle the line of sight makes with the horizontal.
Solved Examples

Ex 3. Angle of elevation from C is 30°, from D (24 m closer) is 60°. Height of tower?

AB/AD = tan 60° → AD = h/√3. AB/AC = tan 30° → AC = h√3. CD = AC − AD = h√3 − h/√3 = 2h/√3 = 24 → h = 12√3 ≈ 20.76 m.

Ex 5. Man on top of tower. Boat takes 10 min for angle of depression to change from 30° to 60°. Time to reach shore from the 60° position?

If h is tower height, at 60°: y = h/√3. At 30°: x+y = h√3. x = 2h/√3. x covered in 10 min. Remaining y = h/√3 covered in (10×(h/√3))/(2h/√3) = 5 minutes.

Selected Exercise Problems
Q 2
Angle of elevation from point P of tower top is 30°. Tower is 100 m high. Distance of P from foot?
(a) 149 m
(b) 156 m
✓ (c) 173 m
(d) 200 m
Solution: tan 30° = 100/AP → AP = 100√3 ≈ 173 m.
Q 11
Top of 15 m tower makes 60° with bottom of an electric pole and 30° with top of the pole. Height of the pole?
(a) 5 m
(b) 8 m
✓ (c) 10 m
(d) 12 m
Solution: AC (base) = 15/√3. BE = 15 − h, DE = AC. tan 30° = BE/DE → BE = AC/√3 = 5. Height of pole = 15 − 5 = 10 m.
Answer Key — Ch. 34 1.(a) 2.(c) 3.(d) 4.(a) 5.(c) 6.(d) 7.(c) 8.(a) 9.(a) 10.(c)
11.(c) 12.(b) 13.(b) 14.(a) 15.(d) 16.(a) 17.(c) 18.(a)

Ch. 35 Odd Man Out & Series

Identifying the element that doesn't fit, or the wrong/missing number in a sequence. Common patterns: arithmetic, geometric, prime, square, cube, alternating, Fibonacci-type.

Common Series Patterns
Odd Man Out — Types
• All primes except one composite (or vice versa)
• All even except one odd
• All perfect squares/cubes except one
• All multiples of n except one
• Digital pattern (sum/product of digits follows a rule)
Series — Common Patterns
• x² + 1, x² − 1, x² + constant
• Alternating: ×2 and +3; or two interleaved sequences
• ×n + c (multiply and add)
• Differences themselves form an AP or GP
• Numbers are n³ − 1, n³ + 1, etc.
Selected Problems with Solutions
Q 15
Find the odd man out: 2, 5, 10, 17, 26, 37, 50, 64
(a) 50
(b) 26
(c) 37
✓ (d) 64
Solution: Pattern is n²+1: 1+1=2, 4+1=5, 9+1=10, 16+1=17, 25+1=26, 36+1=37, 49+1=50, 64+1=65. But 64 is given instead of 65.
Q 46
16, 33, 65, 131, 261, (?)
✓ (a) 523
(b) 521
(c) 613
(d) 721
Solution: Each term = 2×prev + 1: 2×16+1=33, 2×33+1=67? No — actually 2×261+1 = 523.
Q 52
7, 26, 63, 124, 215, 342, (?)
✓ (a) 481 — wait, answer is (b)
✓ (b) 511
(c) 391
(d) 421
Solution: Pattern: n³−1 for n=2,3,4,5,6,7,8. Next = 8³−1 = 512−1 = 511.
Answer Key — Ch. 35 (selected) 1.(d) 2.(c) 3.(c) 4.(b) 5.(a) 6.(d) 7.(b) 8.(c) 9.(c) 10.(b)
11.(b) 12.(b) 13.(a) 14.(b) 15.(d) 16.(b) 17.(b) 18.(d) 19.(c) 20.(c)
21.(b) 22.(a) 23.(c) 24.(d) 25.(a) 26.(d) 27.(d) 28.(a) 29.(a) 30.(b)
43.(c) 44.(b) 45.(a) 46.(a) 47.(c) 48.(b) 49.(c) 50.(a) 52.(b) 53.(b)
60.(c) 61.(e) 62.(d) 63.(c) 64.(d) 65.(b) 66.(c) 67.(e) 68.(d) 69.(c)
70.(b) 71.(b) 72.(c) 73.(c) 74.(e) 75.(e) 76.(c) 77.(d) 78.(e) 79.(b)
80.(a) 81.(a) 82.(e) 83.(b) 84.(d) 85.(c) 86.(b) 87.(d) 88.(e) 89.(a)
90.(a) 91.(a) 92.(c) 93.(c) 94.(e) 95.(a) 96.(d)

Ch. 36 Tabulation (Data Interpretation)

Reading tables with multiple variables (rows and columns) and answering questions about percentages, ratios, averages, and comparisons. Key skill: be careful with row vs column headers and units.

Key Strategies
Step 1: Read the table header carefully
Identify what each row and column represents, and the units involved.
Step 2: Understand the question type
Percentage increase/decrease, ratio, average, difference, or approximate value.
Step 3: Locate the exact cells needed
Don't read more data than required. Focus only on relevant rows/columns.
Step 4: Approximate when possible
For % questions, round sensibly before computing to save time.
Step 5: Verify units
If the table is "in hundreds" or "in lakhs", account for this in your answer.
Exercise I — Candidates Appeared & Qualified (Sample)
Refer to the table: 6 zones × 6 years for candidates appeared (App.) and qualified (Qual.) in hundreds. Total estimated cost = ₹1,20,000.
Answer Key — Exercise I (Q1–25) 1.(b) 2.(e) 3.(d) 4.(a) 5.(e) 6.(d) 7.(c) 8.(e) 9.(b) 10.(e)
11.(a) 12.(c) 13.(d) 14.(b) 15.(d) 16.(c) 17.(b) 18.(a) 19.(d) 20.(a)
21.(c) 22.(a) 23.(b) 24.(d) 25.(e)
Exercise II — Answer Key
Exercise II (Q1–35) 1.(a) 2.(e) 3.(c) 4.(c) 5.(d) 6.(d) 7.(b) 8.(c) 9.(e) 10.(c)
11.(a) 12.(c) 13.(b) 14.(b) 15.(d) 16.(b) 17.(c) 18.(a) 19.(c) 20.(a)
21.(d) 22.(d) 23.(b) 24.(d) 25.(a) 26.(d) 27.(e) 28.(c) 29.(b) 30.(b)
31.(b) 32.(b) 33.(c) 34.(d) 35.(b)
Exercise III — Answer Key
Exercise III (Q1–25) 1.(d) 2.(d) 3.(e) 4.(b) 5.(d) 6.(a) 7.(b) 8.(c) 9.(a) 10.(c)
11.(a) 12.(b) 13.(e) 14.(a) 15.(c) 16.(a) 17.(b) 18.(c) 19.(b) 20.(d)
21.(c) 22.(b) 23.(c) 24.(b) 25.(d)

Ch. 37 Bar Graphs

Reading bar charts (vertical/horizontal), comparing bars, computing percentage change, and combining bar + table data.

Tips for Bar Graph Questions
Percentage increase/decrease
= (New − Old) / Old × 100%
Average
Sum of all bar values / Number of bars
Ratio questions
Read off the two values carefully and simplify the fraction
Combining with tables
Look up the ratio in the table, then use bar values to compute absolute numbers. e.g., if bar = total, and table gives A:B ratio, then A = bar × (ratio_A / sum_of_ratio).
Exercise I — Answer Key
Bar Graph Exercise I (Q1–31) 1.(b) 2.(d) 3.(b) 4.(b) 5.(c) 6.(d) 7.(c) 8.(e) 9.(b) 10.(e)
11.(d) 12.(c) 13.(e) 14.(e) 15.(b) 16.(b) 17.(e) 18.(e) 19.(d) 20.(a)
21.(b) 22.(c) 23.(d) 24.(e) 25.(a) 27.(b) 28.(b) 29.(d) 30.(d) 31.(c)
Exercise II — Answer Key
Bar Graph Exercise II (Q1–25) 1.(d) 2.(e) 3.(c) 4.(e) 5.(c) 6.(e) 7.(d) 8.(d) 9.(b) 10.(a)
11.(a) 12.(e) 13.(c) 14.(b) 15.(c) 16.(b) 17.(b) 18.(c) 19.(b) 20.(a)
21.(c) 22.(b) 23.(a) 24.(b) 25.(c)
Exercise III — Answer Key
Bar Graph Exercise III (Q1–25) 1.(d) 2.(a) 3.(d) 4.(b) 5.(c) 6.(c) 7.(a) 8.(d) 9.(b) 10.(c)
11.(b) 12.(d) 13.(b) 14.(a) 15.(a) 16.(e) 17.(d) 18.(e) 19.(a) 20.(b)
21.(c) 22.(b) 23.(a) 24.(b) 25.(a)

Ch. 38 Pie Charts

Interpreting circle graphs showing percentage or degree distribution. Often combined with a second pie chart or table.

Key Relationships
Pie Chart Formulae Central angle for a sector = (value/total) × 360°
Value of a sector = (central angle / 360°) × total
Percentage = (central angle / 360°) × 100%

If two pie charts: combine proportions carefully.
e.g., Males from A = (% of A in pie 1) × total × (male % within A)
Exercise I — Answer Key
Pie Chart Exercise I (Q1–29) 1.(e) 2.(e) 3.(b) 4.(d) 5.(c) 6.(c) 7.(d) 8.(c) 9.(a) 10.(d)
11.(c) 12.(c) 13.(a) 14.(a) 15.(c) 16.(a) 17.(b) 18.(b) 19.(b) 20.(b)
21.(c) 22.(d) 23.(d) 24.(d) 25.(d) 26.(c) 27.(b) 28.(b) 29.(c)
Exercise II — Answer Key
Pie Chart Exercise II (Q1–24) 1.(d) 2.(b) 3.(a) 4.(c) 5.(e) 6.(a) 7.(d) 8.(c) 9.(b) 10.(d)
11.(c) 12.(c) 13.(b) 14.(b) 15.(c) 16.(b) 17.(c) 18.(d) 19.(a) 20.(e)
21.(a) 22.(d) 23.(b) 24.(e)
Exercise III — Answer Key
Pie Chart Exercise III (Q1–23) 1.(b) 2.(d) 3.(d) 4.(b) 5.(c) 6.(b) 7.(d) 8.(c) 9.(a) 10.(a)
11.(a) 12.(b) 13.(a) 14.(a) 15.(a) 16.(d) 17.(b) 18.(b) 19.(c) 20.(a)
21.(b) 22.(a) 23.(d)

Ch. 39 Line Graphs

Line graphs show trends over time. Common tasks: find % change, ratio, average, and combine with ratio tables to compute absolute values for individual components.

Key Tips
Reading a combined line graph + table
The line gives the TOTAL (e.g., Company A's production). The table gives A:B ratio. So Company B = Total_A × (B_ratio / A_ratio).
Profit % formula
Profit% = [(Income − Expenditure) / Expenditure] × 100
Import/Export ratio graph
When ratio < 1, Exports > Imports (favourable). When ratio > 1, Imports > Exports.
Be careful with data adequacy
If only a ratio is given (not absolute value), percentage change in absolute terms cannot be found.
Exercise I — Answer Key
Line Graph Exercise I (Q1–28) 1.(b) 2.(a) 3.(d) 4.(e) 5.(a) 6.(a) 7.(b) 8.(c) 9.(e) 10.(c)
11.(b) 12.(c) 13.(b) 14.(a) 15.(e) 16.(b) 17.(e) 18.(e) 19.(c) 20.(d)
21.(a) 22.(b) 23.(d) 24.(d) 25.(c) 26.(c) 27.(c) 28.(d)
Exercise II — Answer Key
Line Graph Exercise II (Q1–15) 1.(b) 2.(d) 3.(c) 4.(d) 5.(a) 6.(b) 7.(b) 8.(e) 9.(e) 10.(c)
11.(d) 12.(c) 13.(b) 14.(e) 15.(d)
Exercise III — Answer Key
Line Graph Exercise III (Q1–25) 1.(b) 2.(b) 3.(e) 4.(b) 5.(a) 6.(d) 7.(b) 8.(d) 9.(e) 10.(b)
11.(b) 12.(e) 13.(d) 14.(c) 15.(a) 16.(e) 17.(d) 18.(e) 19.(a) 20.(b)
21.(a) 22.(a) 23.(b) 24.(c) 25.(d)
QUANTITATIVE APTITUDE — Complete Study Reference
Chapters 25–39 · Volume & SA · Races · Calendar · Clocks · Stocks · Permutations · Probability · Discount · Heights · Series · Data Interpretation