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CDS I 2019 Elementary Mathematics with Solutions

Exam: CDS Year: 2019 (Session I) Questions: 100 Marks: 100 Negative Marking: 1/3

Q.34 [Number Theory]

A prime number contains the digit $X$ at unit's place. How many such digits of $X$ are possible?

(a) 3
(b) 4 ✓
(c) 5
(d) 6

Explanation: A prime number (greater than 5) can only end in 1, 3, 7, or 9. Including 2 and 5 themselves as primes, the possible unit digits for a prime are {1, 2, 3, 5, 7, 9}. However, the standard interpretation is the digits that can appear at the unit's place of ANY prime: 1, 2, 3, 5, 7, 9 — but 2 and 5 are single-digit primes. For multi-digit primes the unit digit must be 1, 3, 7, or 9. The question asks how many digits X are possible, and the answer is 4 (i.e., 1, 3, 7, 9 for primes > 5, plus 2 and 5 themselves). Standard answer: 4 possible digits (1, 3, 7, 9 for numbers >9, giving 4).

⚠ Answer needs review

Q.35 [Profit and Loss]

If an article is sold at a gain of 6% instead of a loss of 6%, the seller gets ₹6 more. What is the cost price of the article?

(a) ₹18
(b) ₹36
(c) ₹42
(d) ₹50 ✓

Explanation: Let CP = x. Selling at 6% gain gives 1.06x; selling at 6% loss gives 0.94x. Difference = 1.06x − 0.94x = 0.12x = 6, so x = 6/0.12 = ₹50.

Q.36 [Time and Work]

A field can be reaped by 12 men or 18 women in 14 days. In how many days can 8 men and 16 women reap it?

(a) 26 days
(b) 24 days
(c) 9 days ✓
(d) 8 days

Explanation: Total work = 12×14 = 168 man-days or 18×14 = 252 woman-days. So 1 man = 252/168 = 1.5 woman-days. 8 men + 16 women = 8×1.5 + 16 = 12 + 16 = 28 woman-equivalents per day. Total work in woman-days = 252. Days = 252/28 = 9 days.

Q.37 [Algebra / Exponents]

If $3^x = 4^y = 12^z$, then $z$ is equal to

(a) $xy$
(b) $x+y$
(c) $\frac{xy}{x+y}$ ✓
(d) $4x+3y$

Explanation: Let $3^x = 4^y = 12^z = k$. Then $3 = k^{1/x}$, $4 = k^{1/y}$, $12 = k^{1/z}$. Since $12 = 3 \times 4$, we get $k^{1/z} = k^{1/x} \cdot k^{1/y}$, so $1/z = 1/x + 1/y = (x+y)/(xy)$, thus $z = xy/(x+y)$.

Q.38 [Algebra / Proportions]

If $(4a+7b)(4c-7d) = (4a-7b)(4c+7d)$, then which one of the following is correct?

(a) $\frac{a}{b} = \frac{c}{d}$ ✓
(b) $\frac{a}{b} = \frac{c}{d}$
(c) $\frac{a}{d} = \frac{c}{b}$
(d) $\frac{a}{c} = \frac{7b}{d}$

Explanation: Expanding: $16ac - 28ad + 28bc - 49bd = 16ac + 28ad - 28bc - 49bd$. Simplifying: $-28ad + 28bc = 28ad - 28bc$, so $56bc = 56ad$, giving $bc = ad$, i.e., $a/b = c/d$ (or equivalently $a/c = b/d$).

Q.39 [Algebra / Polynomials]

Given that the polynomial $(x^2 + ax + b)$ leaves the same remainder when divided by $(x-1)$ or $(x+1)$. What are the values of $a$ and $b$ respectively?

(a) 4 and 0
(b) 0 and 3
(c) 3 and 0
(d) 0 and any integer ✓

Explanation: By the remainder theorem, remainder when divided by $(x-1)$ is $f(1) = 1 + a + b$, and by $(x+1)$ is $f(-1) = 1 - a + b$. Setting equal: $1+a+b = 1-a+b \Rightarrow 2a = 0 \Rightarrow a = 0$. Then $b$ can be any integer. So $a=0$ and $b$ = any integer.

Q.40 [Time and Work]

Tushar takes 6 hours to complete a piece of work, while Amar completes the same work in 10 hours. If both work together, what is the time required to complete the work?

(a) 3 hours
(b) 3 hours 15 minutes
(c) 3 hours 30 minutes
(d) 3 hours 45 minutes ✓

Explanation: Tushar's rate = 1/6 per hour, Amar's rate = 1/10 per hour. Combined rate = 1/6 + 1/10 = 5/30 + 3/30 = 8/30 = 4/15 per hour. Time = 15/4 = 3.75 hours = 3 hours 45 minutes.

Q.41 [Algebra / Nested Radicals]

What is the value of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$?

(a) 1
(b) 2 ✓
(c) 3
(d) 4

Explanation: Let $x = \sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$. Then $x = \sqrt{2+x}$, so $x^2 = 2+x$, giving $x^2 - x - 2 = 0$, $(x-2)(x+1)=0$. Since $x>0$, $x=2$.

Q.42 [Set Theory / Percentage]

In an examination, 52% candidates failed in English and 42% failed in Mathematics. If 17% failed in both subjects, then what percent passed in both subjects?

(a) 77
(b) 58
(c) 48
(d) 23 ✓

Explanation: Failed in at least one subject = 52 + 42 − 17 = 77%. Passed in both = 100 − 77 = 23%.

Q.43 [Ratio and Proportion]

A man left ₹3,90,000 to be divided among his wife, five sons and four daughters. Each son receives 3 times as much as each daughter, and each daughter receives twice as much as their mother. What was the wife's share?

(a) ₹14,000
(b) ₹12,000
(c) ₹10,000 ✓
(d) ₹9,000

Explanation: Let wife's share = x. Each daughter = 2x, each son = 6x. Total = x + 4(2x) + 5(6x) = x + 8x + 30x = 39x = 3,90,000. So x = 10,000. Wife's share = ₹10,000.

Q.44 [Compound Interest]

What is the least number of complete years in which a sum of money put out at 40% annual compound interest will be more than trebled?

(a) 3
(b) 4 ✓
(c) 5
(d) 6

Explanation: We need $(1.4)^n > 3$. $1.4^1=1.4$, $1.4^2=1.96$, $1.4^3=2.744$, $1.4^4=3.8416>3$. So minimum $n=4$ years.

Q.45 [Simple Interest]

A person divided ₹17,200 into three parts and invested at 5%, 6% and 9% per annum simple interest. At the end of two years, he got the same interest on each part. What is the money invested at 9%?

(a) ₹3,200
(b) ₹4,000 ✓
(c) ₹4,800
(d) ₹25,000

Explanation: Let parts be $P_1$, $P_2$, $P_3$ at 5%, 6%, 9%. Equal interest means $P_1 \times 5 = P_2 \times 6 = P_3 \times 9$ (rate × time same factor cancels). Let common value = k: $P_1=k/5$, $P_2=k/6$, $P_3=k/9$. Sum: $k(1/5+1/6+1/9) = 17200$. LCM=90: $k(18+15+10)/90 = k(43/90) = 17200$, so $k = 17200 \times 90/43 = 36000$. $P_3 = 36000/9 = 4000$. Answer: ₹4,000.

Q.46 [Algebra / Identities]

What is $\dfrac{(x-y)^3+(y-z)^3+(z-x)^3}{3(x-y)(y-z)(z-x)}$ equal to?

(a) 1 ✓
(b) 0
(c) $\frac{1}{3}$
(d) 3

Explanation: If $a+b+c=0$ then $a^3+b^3+c^3=3abc$. Here $a=(x-y)$, $b=(y-z)$, $c=(z-x)$ and $a+b+c=0$. So numerator $= 3(x-y)(y-z)(z-x)$. Dividing by $3(x-y)(y-z)(z-x)$ gives 1.

Q.47 [Algebra / Exponents]

If $a^x = b^y = c^z$ and $b^2 = ac$, then what is $\dfrac{2}{y}$ equal to?

(a) $\frac{1}{x}+\frac{1}{z}$ ✓
(b) $\frac{1}{x}-\frac{1}{z}$
(c) $\frac{1}{x+z}$
(d) $\frac{2}{x+z}$

Explanation: Let $a^x=b^y=c^z=k$. Then $a=k^{1/x}$, $b=k^{1/y}$, $c=k^{1/z}$. From $b^2=ac$: $k^{2/y}=k^{1/x+1/z}$, so $2/y = 1/x + 1/z$.

Q.48 [Algebra / Quadratics]

If $p$ and $q$ are the roots of the equation $x^2 - 15x + r = 0$ and $p - q = 1$, then what is the value of $r$?

(a) 55
(b) 56 ✓
(c) 60
(d) 64

Explanation: By Vieta's: $p+q=15$ and $pq=r$. Given $p-q=1$. From $p+q=15$ and $p-q=1$: $p=8$, $q=7$. So $r=pq=56$.

Q.49 [Algebra / Inequalities]

For the inequation $x^2 - 7x + 12 > 0$, which one of the following is correct?

(a) $3 < x < 4$
(b) $-\infty < x < 3$ only
(c) $4 < x < \infty$ only
(d) $-\infty < x < 3$ or $4 < x < \infty$ ✓

Explanation: $x^2-7x+12=(x-3)(x-4)>0$. This holds when both factors are positive ($x>4$) or both negative ($x<3$). So solution is $x<3$ or $x>4$.

Q.50 [Number Theory / Factors]

The expression $5^{2n} - 2^{5n}$ has a factor

(a) 3
(b) 7
(c) 17
(d) None of the above ✓

Explanation: $5^{2n}-2^{5n} = 25^n - 32^n$. For $n=1$: $25-32=-7$. So 7 divides it for $n=1$. Check $n=2$: $625-1024=-399=-3\times7\times19$; 7 divides it. In general $25^n-32^n$ is divisible by $25-32=-7$, i.e., by 7. So the answer is (b) 7.

⚠ Answer needs review

Q.51 [Trigonometry]

If $\tan x = 1$, $0 < x < 90°$, then what is the value of $2\sin x \cos x$?

(a) 3
(b) 1 ✓
(c) $\frac{1}{2}$
(d) $\sqrt{3}$

Explanation: $\tan x=1 \Rightarrow x=45°$. $2\sin 45°\cos 45° = 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 2 \cdot \frac{1}{2} = 1$.

Q.52 [Trigonometry]

What is the value of $\sin 46°\cos 44° + \cos 46°\sin 44°$?

(a) $\sin 2°$
(b) $\cos 2°$
(c) 1 ✓
(d) 2

Explanation: Using $\sin(A+B)=\sin A\cos B+\cos A\sin B$: this equals $\sin(46°+44°)=\sin 90°=1$.

Q.53 [Trigonometry / Inequalities]

Suppose $0 < \theta < 90°$. For every $\theta$, $4\sin^2\theta + 1$ is greater than or equal to

(a) 2
(b) $4\sin\theta$ ✓
(c) $4\cos\theta$
(d) $4\tan\theta$

Explanation: By AM-GM inequality: $4\sin^2\theta + 1 \geq 2\sqrt{4\sin^2\theta \cdot 1} = 2\cdot 2\sin\theta = 4\sin\theta$. Equality when $4\sin^2\theta=1$, i.e., $\sin\theta=1/2$, $\theta=30°$.

Q.54 [Trigonometry / Heights & Distances]

A person standing at point A observes the angle of elevation to the top of a tower at B as 30°, and the angle of elevation to the top of a tower at C is 45°. What is the ratio of the height of the towers at B and C?

(a) $1:\sqrt{3}$ ✓
(b) $1:3$
(c) $1:2$
(d) $1:2\sqrt{3}$

Explanation: If the person is at the same horizontal distance d from both towers, height of B = d·tan30° = d/√3 and height of C = d·tan45° = d. Ratio = (d/√3) : d = 1 : √3.

Q.55 [Trigonometry]

What is the value of $\tan 1° \cdot \tan 2° \cdot \tan 3° \cdots \tan 89°$?

(a) 0
(b) 1 ✓
(c) 2
(d) undefined

Explanation: Use the identity tan(90°−θ) = cotθ. Pair up: tan1°·tan89° = tan1°·cot1° = 1, tan2°·tan88° = 1, …, tan44°·tan46° = 1, and tan45° = 1. The entire product equals 1.

Q.56 [Trigonometry / Geometry]

There are two parallel streets each directed North to South. A person in the first street travelling from South to North wishes to take the second street which is on his right side. At some place, he makes a 150° turn to the right and travels for 15 minutes at 20 km/hr. After that he takes a left turn of 60° and travels for 20 minutes at 30 km/hr in order to meet the second street. What is the distance between the two streets?

(a) 7.5 km ✓
(b) 10.5 km
(c) 12.5 km
(d) 15 km

Explanation: A 150° right turn from North means the new direction is 30° west of south (heading at 210° bearing). Distance travelled = (15/60)×20 = 5 km. The horizontal (east-west) component after first leg: 5·sin30° = 2.5 km (to the right/east). Then a 60° left turn changes direction to due east (the 210°+60°=270° bearing, i.e. heading east). Distance = (20/60)×30 = 10 km. But we need the perpendicular (east-west) distance between the streets. The first leg contributes 5·sin30° = 2.5 km, and the second leg is at 0° to the east, so it is horizontal. However, re-examining: after 150° right turn, the person heads at bearing 150° (South-East direction, 30° east of south). Horizontal distance from first segment = 5·sin30° = 2.5 km. After a 60° left turn from bearing 150°, new bearing = 150°−60° = 90° (due east). Second leg is 10 km east. But the person meets the second street during the second leg, so the perpendicular distance between streets = 2.5 km + portion of second leg needed. The net east component from the first leg = 5·sin30° = 2.5 km. Since after the second leg the person meets the second street heading due east, and only the first leg contributes perpendicular distance, but that does not use the second leg. Re-check: bearing 150° means 30° east of due south. Horizontal (right/east) component = 5·sin30° = 2.5 km; vertical (south) = 5·cos30° = 4.33 km. Left turn of 60° from bearing 150°: new bearing = 90° (due east). He travels east, the second street is parallel and to his right (east). The total east distance to second street = 2.5 km + 10 km = 12.5 km — but answer (c) is 12.5 km. However, the person meets the second street during the 2nd leg implying only the perpendicular distance matters. The second street is 7.5 km away from first: at bearing 150° he moves east 5·sin30°=2.5 km; turning due east he travels until he meets the second street. To meet second street he needs total east = distance between streets. Second leg speed=30 km/hr for 20 min = 10 km east. So the distance = 2.5 + 10 = 12.5 km (option c). But CDS key says 7.5 km (a). Re-examine: 150° right turn from North = heading at 150° CW from north = SE direction. The east component = 5·sin(150°−90°)... Bearing 150° from north: east component = sin150°×5 = 0.5×5 = 2.5 km. Then 60° left from bearing 150°→bearing 90°(east): east= 10 km. Total = 12.5 km. Answer is (c) 12.5 km.

⚠ Answer needs review

Q.57 [Trigonometry]

If $3\tan\theta = \cot\theta$ where $0 < \theta < \frac{\pi}{2}$, then what is the value of $\theta$?

(a) $\frac{\pi}{6}$ ✓
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{2}$

Explanation: 3tanθ = cotθ = 1/tanθ → 3tan²θ = 1 → tan²θ = 1/3 → tanθ = 1/√3 → θ = π/6.

Q.58 [Trigonometry]

What is the value of $\sin^2 25° + \sin^2 65°$?

(a) 0
(b) 1 ✓
(c) 2
(d) 4

Explanation: sin65° = cos25°, so sin²25° + sin²65° = sin²25° + cos²25° = 1.

Q.59 [Trigonometry]

What is the value of $\sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta - 1$?

(a) 0 ✓
(b) 1
(c) 2
(d) 4

Explanation: Use the identity a³+b³ = (a+b)³ − 3ab(a+b) with a=sin²θ, b=cos²θ: sin⁶θ+cos⁶θ = (sin²θ+cos²θ)³ − 3sin²θcos²θ(sin²θ+cos²θ) = 1 − 3sin²θcos²θ. Therefore sin⁶θ+cos⁶θ+3sin²θcos²θ−1 = 1−3sin²θcos²θ+3sin²θcos²θ−1 = 0.

Q.60 [Trigonometry]

Consider the following for real numbers $\alpha, \beta, \gamma$ and $\delta$: 1. $\sec\alpha = 1/4$ 2. $\tan\beta = 20$ 3. $\cosec\gamma = 1/2$ 4. $\cos\delta = 2$. How many of the above statements are not possible?

(a) One
(b) Two
(c) Three ✓
(d) Four

Q.61 [Statistics]

Consider the following grouped frequency distribution: Class: 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 8, 12, 10, p, 5 respectively. If the mean of the above data is 25.2, then what is the value of p?

(a) 9 ✓
(b) 10
(c) 11
(d) 12

Explanation: Midpoints: 5, 15, 25, 35, 45. Sum of f = 8+12+10+p+5 = 35+p. Σfx = 8×5+12×15+10×25+p×35+5×45 = 40+180+250+35p+225 = 695+35p. Mean = (695+35p)/(35+p) = 25.2 → 695+35p = 25.2(35+p) = 882+25.2p → 9.8p = 187 → p = 19.08? Recheck: if 40-50 frequency is missing and total =50: 8+12+10+p+q=50. With mean 25.2 and assuming q = 50−30−p = 20−p: Σfx = 695+35p+45(20−p) = 695+35p+900−45p = 1595−10p. Mean = (1595−10p)/50 = 25.2 → 1595−10p = 1260 → 10p = 335 → p = 33.5, wrong. Try total = 8+12+10+p+5=35+p and p=9: mean = (695+315)/44 = 1010/44 = 22.95, not 25.2. Try frequencies 8,12,10,p,5 summing with 40-50 freq given as 5: total=35+p, mean=(695+35p)/(35+p)=25.2 → 695+35p=882+25.2p → 9.8p=187 → p≈19, not matching options. Given the OCR garbling of the 40-50 frequency, the standard CDS 2019 answer is p=9, option (a).

⚠ Answer needs review

Q.62 [Statistics]

Consider the following frequency distribution: x: 5, 6, 7, 8, 9 f: 4, 3, 2, 1, 1. What is the median for the distribution?

(a) 6 ✓
(b) 7
(c) 8
(d) 9

Explanation: Total frequency = 4+3+2+1+1 = 11. Median is the 6th value. Cumulative: up to x=5: 4, up to x=6: 7. The 6th value falls in x=6. Median = 6.

Q.63 [Statistics]

The average of 50 consecutive natural numbers is $x$. What will be the new average when the next four natural numbers are also included?

(a) $x+1$
(b) $x+2$ ✓
(c) $x+4$
(d) $x + \frac{x}{54}$

Explanation: If average of 50 consecutive numbers is x, then the numbers are x−24.5 to x+24.5, i.e., they are centered at x. When next 4 numbers (x+25.5, x+26.5, x+27.5, x+28.5) are added, the new average = (50x + 4x + 4×25.5 + (0.5+1.5+2.5+3.5))/(54)... Alternatively: sum of 50 numbers = 50x. Next 4 numbers sum = 4x + (25.5+26.5+27.5+28.5) relative terms. Since consecutive integers: let numbers be n, n+1,...,n+49 with mean = n+24.5 = x → n = x−24.5. Not integers. Let n be the first: mean = n+24.5 = x. Next 4: n+50, n+51, n+52, n+53. Their sum = 4n+206. New total sum = 50x + 4n+206 = 50x+4(x−24.5)+206 = 50x+4x−98+206 = 54x+108. New average = (54x+108)/54 = x+2.

Q.64 [Statistics]

Consider two-digit numbers which remain the same when the digits interchange their positions. What is the average of such two-digit numbers?

(a) 33
(b) 44
(c) 55 ✓
(d) 66

Explanation: A two-digit number that remains the same when digits are interchanged must have both digits equal: 11, 22, 33, 44, 55, 66, 77, 88, 99. Average = (11+22+…+99)/9 = 11(1+2+…+9)/9 = 11×45/9 = 11×5 = 55.

Q.65 [Statistics]

Diagrammatic representation of data includes which of the following? 1. Bar diagram 2. Pie diagram 3. Pictogram. Select the correct answer using the code given below:

(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3 ✓

Explanation: Bar diagrams, pie diagrams, and pictograms are all standard forms of diagrammatic representation of data. All three are valid.

Q.66 [Statistics]

The data collected from which one of the following methods is not a primary data?

(a) By direct personal interviews
(b) By indirect personal interviews
(c) By schedules sent through enumerators
(d) From published thesis ✓

Explanation: Primary data is collected first-hand by the researcher. Data from a published thesis is secondary data — it has already been collected and published by someone else.

Q.67 [Statistics]

The monthly expenditure of a person is ₹6,000. Distribution: Food ₹2,000, Clothing ₹660, Fuel and rent ₹1,200, Education ₹480, Miscellaneous ₹1,660. If the above data is represented by a percentage bar diagram of height 15 cm, what are the lengths of the segments corresponding to education and miscellaneous respectively?

(a) 1.25 cm and 5 cm
(b) 1.2 cm and 4.15 cm ✓
(c) 1.2 cm and 3.5 cm
(d) 4.15 cm and 6 cm

Explanation: Education % = (480/6000)×100 = 8%. Length = 8% of 15 cm = 1.2 cm. Miscellaneous % = (1660/6000)×100 = 27.67%. Length = 27.67% of 15 cm ≈ 4.15 cm. So education = 1.2 cm and miscellaneous ≈ 4.15 cm.

Q.68 [Statistics]

If the mean of $m$ observations out of $n$ observations is $n$ and the mean of the remaining observations is $m$, then what is the mean of all $n$ observations?

(a) $\frac{2m^2 - n}{n}$... wait, $2m - \frac{m^2}{n}$
(b) $2m + \frac{m^2}{n}$
(c) $m - \frac{m^2}{n}$
(d) $m + \frac{m^2}{n}$ ✓

Explanation: Sum of m observations = m×n = mn. Remaining observations = n−m, their sum = (n−m)×m = mn−m². Total sum = mn + mn − m² = 2mn − m². Mean of all n observations = (2mn−m²)/n = 2m − m²/n = m + (m − m²/n) — let me recompute: (2mn−m²)/n = 2m − m²/n. This matches option (a) if written as 2m − m²/n. But option (d) says m + m²/n. Re-examine: total sum = mn + (n−m)m = mn + mn − m² = 2mn − m². Mean = (2mn−m²)/n = 2m − m²/n. This is option (a): $2m - \frac{m^2}{n}$.

⚠ Answer needs review

Q.69 [Statistics]

Which one of the following pairs is correctly matched? (a) Median — Graphical location (b) Mean — Graphical location (c) Geometric mean — Ogive (d) (OCR cut off)

(a) Median — Graphical location ✓
(b) Mean — Graphical location
(c) Geometric mean — Ogive
(d) OCR unclear

Explanation: The median can be located graphically using an ogive (cumulative frequency curve). The median corresponds to the 50th percentile on the ogive, making 'Median — Graphical location' (via ogive) the correct match.

Q.70 [Statistics]

The following pairs relate to frequency distribution of a discrete variable and its frequency polygon. Which one of the following pairs is NOT correctly matched? (a) Baseline of the polygon — X-axis (b) Ordinates of the vertices of the polygon — Class frequencies (c) Abscissa of the vertices of the polygon — Class marks of the frequency distribution (d) Area of the polygon — Total frequency of the distribution

(a) Baseline of the polygon — X-axis
(b) Ordinates of the vertices of the polygon — Class frequencies
(c) Abscissa of the vertices of the polygon — Class marks of the frequency distribution
(d) Area of the polygon — Total frequency of the distribution ✓

Explanation: In a frequency polygon the area is NOT equal to the total frequency; rather the area equals (total frequency × class width). The total frequency equals the sum of the ordinates (heights). So option (d) is incorrectly matched.

Q.71 [Mensuration / Percentage]

In a rectangle, length is three times its breadth. If the length and the breadth of the rectangle are increased by 30% and 10% respectively, then its perimeter increases by what percentage?

(a) $\frac{40}{2}$% (i.e. 20%)
(b) 20%
(c) 25% ✓
(d) 27%

Explanation: Let breadth = b, length = 3b. Original perimeter = 2(3b + b) = 8b. New length = 1.3×3b = 3.9b, new breadth = 1.1b. New perimeter = 2(3.9b + 1.1b) = 2(5b) = 10b. Increase = (10b − 8b)/8b × 100 = 2/8 × 100 = 25%.

Q.72 [Mensuration / Percentage]

What is the percentage decrease in the area of a triangle if each side is halved?

(a) 75% ✓
(b) 50%
(c) 25%
(d) No change

Explanation: Area of a triangle scales as the square of its sides (by Heron's formula, all sides scaled by factor k gives area scaled by k²). Halving all sides: k = 1/2, new area = (1/4) original. Decrease = 75%.

Q.73 [Mensuration / Percentage]

The volume of a spherical balloon is increased by 700%. What is the percentage increase in its surface area?

(a) 300% ✓
(b) 400%
(c) 450%
(d) 500%

Explanation: Volume increases by 700% means new volume = 8 × original volume. Since V ∝ r³, new r = 2 × original r. Surface area ∝ r², so new SA = 4 × original SA. Increase = 300%.

Q.74 [Geometry / Circles]

If the lengths of two parallel chords in a circle of radius 10 cm are 12 cm and 16 cm, then what is the distance between these two chords?

(a) 1 cm or 7 cm
(b) 2 cm or 14 cm ✓
(c) 3 cm or 21 cm
(d) 4 cm or 28 cm

Explanation: For chord of length 12 cm: half-chord = 6, distance from centre = √(100−36) = 8. For chord of length 16 cm: half-chord = 8, distance from centre = √(100−64) = 6. Chords on same side: distance = 8−6 = 2 cm. Chords on opposite sides: distance = 8+6 = 14 cm. Answer: 2 cm or 14 cm.

Q.75 [Geometry / Areas]

Considering two opposite vertices of a square of side $a$ as centres, two circular arcs are drawn within the square joining the other two vertices, thus forming two sectors. What is the common area in these two sectors?

(a) $a^2\left(\frac{\pi}{2}+\frac{1}{2}\right)$
(b) $a^2\left(\frac{\pi}{2}-\frac{1}{2}\right)$
(c) $a^2\left(\frac{\pi}{2}-1\right)$ ✓
(d) $a^2\left(\frac{\pi}{2}+1\right)$

Explanation: Each arc has radius a (diagonal of square = a√2 is too large; the arc from one corner to the other two adjacent corners has radius a). Each sector is a quarter-circle of radius a, so area of each sector = (1/4)πa². The two sectors overlap. The common (lens-shaped) area = 2×(sector area) − area of square = 2×(πa²/4) − a² = a²(π/2 − 1).

Q.76 [Geometry / Regular Polygon]

The corners of a square of side $a$ are cut away so as to form a regular octagon. What is the side of the octagon?

(a) $a(\sqrt{2}-1)$ ✓
(b) $a(\sqrt{3}-1)$
(c) $\dfrac{a}{\sqrt{2}+2}$
(d) $\dfrac{a}{\sqrt{2}+2}$

Explanation: Let x be the length cut from each corner. The side of the octagon (the diagonal cut) = x√2, and the remaining part of each original side = a − 2x. For a regular octagon all sides are equal: x√2 = a − 2x → x(√2 + 2) = a → x = a/(√2+2) = a(√2−1)/((√2+2)(√2−1)) = a(√2−1)/(2−2+... Let's redo: x(√2+2)=a, so x = a/(2+√2). The octagon side = x√2 = a√2/(2+√2) = a√2(2−√2)/((2+√2)(2−√2)) = a√2(2−√2)/(4−2) = a√2(2−√2)/2 = a(2√2−2)/2 = a(√2−1). So side = a(√2−1).

Q.77 [Number Theory / Geometry]

Three consecutive integers form the lengths of a right-angled triangle. How many sets of such three consecutive integers are possible?

(a) Only one ✓
(b) Only two
(c) Only three
(d) Infinitely many

Explanation: Let the three consecutive integers be n, n+1, n+2 with hypotenuse n+2. Then (n+2)² = n² + (n+1)² → n²+4n+4 = n²+n²+2n+1 → 4n+4 = n²+2n+1 → n²−2n−3=0 → (n−3)(n+1)=0 → n=3. So only one set: 3, 4, 5.

Q.78 [Mensuration / Circles]

Two circles are drawn with the same centre. The circumference of the smaller circle is 44 cm and that of the bigger circle is double the smaller one. What is the area between these two circles?

(a) 154 square cm
(b) 308 square cm
(c) 462 square cm ✓
(d) 616 square cm

Explanation: Smaller circumference = 44 cm → r₁ = 44/(2π) = 7 cm. Larger circumference = 88 cm → r₂ = 14 cm. Area between = π(r₂²−r₁²) = π(196−49) = 147π = 147×(22/7) = 462 sq cm.

Q.79 [Mensuration / Area]

A rectangular red carpet of size 6 ft × 12 ft has a dark red border 6 inches (= 0.5 ft) wide. What is the area of the dark red border?

(a) 9 square feet
(b) 15 square feet
(c) 17 square feet ✓
(d) 18 square feet

Explanation: Total area = 6×12 = 72 sq ft. Inner area (excluding border of 0.5 ft on each side) = (6−1)×(12−1) = 5×11 = 55 sq ft. Border area = 72−55 = 17 sq ft.

Q.80 [Geometry / Right Triangle]

The perimeter of a right-angled triangle is $k$ times the shortest side. If the ratio of the other side to hypotenuse is 4:5, then what is the value of $k$?

(a) 2
(b) 3 ✓
(c) 4
(d) 5

Explanation: Let the sides be a (shortest), 4t, 5t. Since it's right-angled: (4t)²+(a)²=(5t)² → a²=9t² → a=3t. Perimeter = 3t+4t+5t = 12t = k×3t → k = 4. Wait, let's re-examine: ratio of 'other side' to hypotenuse = 4:5 means the two legs are in some ratio and hypotenuse. Let legs be a and 4m, hypotenuse = 5m. Then a²+(4m)²=(5m)² → a²=25m²−16m²=9m² → a=3m (shortest side). Perimeter = 3m+4m+5m=12m = k×3m → k=4.

⚠ Answer needs review

Q.81 [Mensuration / Algebra]

A 12 m long wire is cut into two pieces, one of which is bent into a circle and the other into a square enclosing the circle. What is the radius of the circle?

(a) $\dfrac{12}{\pi+4}$
(b) $\dfrac{6}{\pi+4}$ ✓
(c) $\dfrac{6}{\pi+4}$
(d) $\dfrac{6}{\pi+2\sqrt{2}}$

Explanation: Let the piece bent into a circle have length L. Then the circle has circumference L → radius r = L/(2π). The square must enclose the circle, so the square's side = 2r (diameter), and its perimeter = 8r = 12−L = 12−2πr → 8r+2πr=12 → r(8+2π)=12 → r=12/(2(π+4))=6/(π+4).

Q.82 [Geometry / Triangle]

The angles of a triangle are in the ratio 1:1:4. If the perimeter of the triangle is $k$ times its largest side, then what is the value of $k$?

(a) $1+\dfrac{2}{\sqrt{3}}$ ✓
(b) $1-\dfrac{2}{\sqrt{3}}$
(c) $1+\sqrt{3}$
(d) $2$

Explanation: Angles: 30°, 30°, 120°. By sine rule sides are proportional to sin30°, sin30°, sin120° = 1/2, 1/2, √3/2. Let sides be 1, 1, √3. Largest side = √3 (opposite 120°). Perimeter = 1+1+√3 = 2+√3. k = (2+√3)/√3 = 2/√3 + 1 = 1 + 2/√3.

Q.83 [Geometry / Right Triangle]

The hypotenuse of a right-angled triangle is 10 cm and its area is 24 cm². If the shorter side is halved and the longer side is doubled, the new hypotenuse becomes

(a) $\sqrt{245}$ cm
(b) $\sqrt{255}$ cm
(c) $\sqrt{265}$ cm ✓
(d) $\sqrt{275}$ cm

Explanation: Let legs be a and b. a²+b²=100 and (1/2)ab=24 → ab=48. New hypotenuse² = (a/2)²+(2b)² = a²/4+4b². We need a and b: a²+b²=100, ab=48. (a+b)²=100+96=196 → a+b=14; (a−b)²=100−96=4 → a−b=2. So a=8, b=6 (a is longer, b is shorter... actually b=6 is shorter). New hypotenuse² = (b/2)²+(2a)² = (3)²+(16)² = 9+256 = 265. New hypotenuse = √265 cm.

Q.84 [Geometry / Circles]

In a circle of radius 8 cm, AB and AC are two chords such that AB = AC = 12 cm. What is the length of chord BC?

(a) Not fully given in extracted text

Explanation: OCR unclear — needs manual review. The options for Q84 were not captured in the extracted text.

⚠ Answer needs review

Q.85 [Geometry – Cyclic Polygons]

Consider the following statements: 1. An isosceles trapezium is always cyclic. 2. Any cyclic parallelogram is a rectangle. Which of the above statements is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2 ✓
(d) Neither 1 nor 2

Explanation: Statement 1: An isosceles trapezium has equal base angles, so the sum of opposite angles is 180°, making it always cyclic. TRUE. Statement 2: A cyclic parallelogram has opposite angles supplementary (sum 180°), but opposite angles of a parallelogram are also equal, so each angle = 90°, making it a rectangle. TRUE. Both statements are correct.

Q.86 [Geometry – Pythagoras / Ladder Problem]

A ladder is resting against a vertical wall and its bottom is 2.5 m away from the wall. If it slips 0.8 m down the wall, then its bottom will move away from the wall by 1.4 m. What is the length of the ladder?

(a) 6.2 m
(b) 6.5 m ✓
(c) 6.8 m
(d) 7.5 m

Explanation: Let ladder length = L, initial height on wall = h. Initial: $h^2 + 2.5^2 = L^2$. After slip: height = h − 0.8, base = 2.5 + 1.4 = 3.9 m. So $(h-0.8)^2 + 3.9^2 = L^2$. Subtracting: $h^2 - (h-0.8)^2 = 3.9^2 - 2.5^2$. $1.6h - 0.64 = 15.21 - 6.25 = 8.96$. $1.6h = 9.6$, $h = 6$. $L^2 = 36 + 6.25 = 42.25$, $L = 6.5$ m.

Q.87 [Geometry – Circles]

Two equal circles intersect such that each passes through the centre of the other. If the length of the common chord of the circles is $10\sqrt{3}$ cm, then what is the diameter of the circle?

(a) 10 cm
(b) 15 cm
(c) 20 cm ✓
(d) 30 cm

Explanation: When each circle passes through the other's centre, the distance between centres equals the radius r. The common chord length = $r\sqrt{3}$. Given $r\sqrt{3} = 10\sqrt{3}$, so $r = 10$ cm, diameter = 20 cm.

Q.88 [Geometry – Circles]

Consider the following statements: 1. The number of circles that can be drawn through three non-collinear points is infinity. 2. The angle formed in a minor segment of a circle is acute. Which of the above statements is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2 ✓

Explanation: Statement 1: Exactly ONE circle can be drawn through three non-collinear points (not infinity). FALSE. Statement 2: The angle in a minor segment is obtuse (the arc is less than a semicircle, so the inscribed angle > 90°). FALSE. Neither statement is correct.

Q.89 [Geometry – Triangle Inequality]

Consider the following inequalities in respect of any triangle ABC: 1. $AC - AB < BC$, 2. $BC - AC < AB$, 3. $AB - BC < AC$. Which of the above are correct?

(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3 ✓

Q.90 [Geometry – Triangle Medians]

Consider the following statements: 1. The perimeter of a triangle is greater than the sum of its three medians. 2. In any triangle ABC, if D is any point on BC, then $AB + BC + CA > 2AD$. Which of the above statements is/are correct?

(a) 1 only
(b) 2 only
(c) Both 1 and 2 ✓
(d) Neither 1 nor 2

Explanation: Statement 1: It is a known result that the perimeter of a triangle > sum of medians (each median < sum of the two sides it connects, and combining gives perimeter > sum of medians). TRUE. Statement 2: By triangle inequality in triangles ABD and ACD: AB + BD > AD and AC + CD > AD, adding: AB + BC + CA > 2AD. TRUE. Both correct.

Q.91 [Mensuration – Sphere, Cone (Inscribed Solids)]

A cube is inscribed in a sphere. A right circular cylinder is within the cube touching all the vertical faces. A right circular cone is inside the cylinder with the same height and diameter as the cylinder. What is the ratio of the volume of the sphere to that of the cone?

(a) $6\sqrt{3} : 1$ ✓
(b) $7 : 2$
(c) $3\sqrt{3} : 1$
(d) $5\sqrt{3} : 1$

Explanation: Let cube side = $a$. Sphere radius $R = \frac{a\sqrt{3}}{2}$. Cylinder (touching all vertical faces) has radius $r = \frac{a}{2}$, height $h = a$. Cone: same radius $\frac{a}{2}$ and height $a$. $V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi\frac{3\sqrt{3}}{8}a^3 = \frac{\sqrt{3}}{2}\pi a^3$. $V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\frac{a^2}{4}\cdot a = \frac{\pi a^3}{12}$. Ratio $= \frac{\frac{\sqrt{3}}{2}\pi a^3}{\frac{\pi a^3}{12}} = \frac{\sqrt{3}}{2}\times 12 = 6\sqrt{3}$. So ratio is $6\sqrt{3}:1$.

Q.92 [Mensuration – Cube and Cylinder (Inscribed Solids)]

A cube is inscribed in a sphere. A right circular cylinder is within the cube touching all the vertical faces. What is the ratio of the volume of the cube to that of the cylinder?

(a) $4:3$
(b) $21:16$
(c) $14:11$ ✓
(d) $45:32$

Explanation: Let cube side = $a$. Cylinder radius = $\frac{a}{2}$, height = $a$. $V_{\text{cube}} = a^3$. $V_{\text{cylinder}} = \pi\left(\frac{a}{2}\right)^2 \cdot a = \frac{\pi a^3}{4}$. Ratio $= \frac{a^3}{\frac{\pi a^3}{4}} = \frac{4}{\pi}$. This equals approximately $1.273$, which matches $4:\pi$. Among the given options, the intended answer is (a) $4:3$ is closest as an approximation if $\pi\approx 3$, but the exact ratio is $4:\pi$. With $\pi$ taken as $\frac{22}{7}$: ratio $= \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}$, which matches option (c).

Q.93 [Mensuration – Surface Areas (Inscribed Solids)]

Consider the following statements: 1. The surface area of the sphere is $\sqrt{5}$ times the curved surface area of the cone. 2. The surface area of the cube is equal to the curved surface area of the cylinder. Which of the above statements is/are correct?

(a) 1 only ✓
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2

Explanation: Let cube side = $a$. Sphere: $SA = 4\pi R^2 = 4\pi\cdot\frac{3a^2}{4} = 3\pi a^2$. Cone: radius $= \frac{a}{2}$, height $= a$, slant $= \sqrt{\frac{a^2}{4}+a^2} = \frac{a\sqrt{5}}{2}$. Curved SA of cone $= \pi r l = \pi\cdot\frac{a}{2}\cdot\frac{a\sqrt{5}}{2} = \frac{\pi a^2\sqrt{5}}{4}$. Ratio $= \frac{3\pi a^2}{\frac{\pi a^2\sqrt{5}}{4}} = \frac{12}{\sqrt{5}} = \frac{12\sqrt{5}}{5}\neq\sqrt{5}$. Hmm — Statement 1 does not hold exactly. Statement 2: Cube SA $= 6a^2$. Cylinder curved SA $= 2\pi r h = 2\pi\cdot\frac{a}{2}\cdot a = \pi a^2\neq 6a^2$. Statement 2 is false. Revisiting statement 1 with the ratio $\frac{3\pi}{\frac{\pi\sqrt{5}}{4}} = \frac{12}{\sqrt{5}}\approx 5.37\neq\sqrt{5}$. Neither appears exactly correct; however, the intended answer per standard CDS 2019 key is (a) 1 only, likely using an approximate or differently stated ratio. Answer: (a) 1 only.

⚠ Answer needs review

Q.94 [Mensuration – Quadrilateral Area]

$ABCD$ is a quadrilateral with $AB = 9$ cm, $BC = 40$ cm, $CD = 28$ cm, $DA = 15$ cm and $\angle ABC = 90°$. What is the area of triangle $ADC$?

(a) $126$ cm$^2$ ✓
(b) $124$ cm$^2$
(c) $122$ cm$^2$
(d) $120$ cm$^2$

Explanation: First find diagonal $AC$: in right triangle $ABC$, $AC = \sqrt{9^2+40^2} = \sqrt{81+1600} = \sqrt{1681} = 41$ cm. Triangle $ADC$ has sides $AD=15$, $DC=28$, $AC=41$. Check: $15^2+28^2 = 225+784=1009\neq 1681$. Use Heron's formula: $s = \frac{15+28+41}{2}=42$. Area $= \sqrt{42\cdot27\cdot14\cdot1} = \sqrt{42\cdot27\cdot14} = \sqrt{15876} = 126$ cm$^2$.

Q.95 [Mensuration – Quadrilateral Area]

$ABCD$ is a quadrilateral with $AB = 9$ cm, $BC = 40$ cm, $CD = 28$ cm, $DA = 15$ cm and $\angle ABC = 90°$. What is the area of quadrilateral $ABCD$?

(a) $300$ cm$^2$
(b) $306$ cm$^2$ ✓
(c) $312$ cm$^2$
(d) $316$ cm$^2$

Explanation: Area of triangle $ABC = \frac{1}{2}\times 9\times 40 = 180$ cm$^2$. Area of triangle $ADC = 126$ cm$^2$ (from Q94). Total area $= 180 + 126 = 306$ cm$^2$.

Q.96 [Mensuration – Perimeter]

$ABCD$ is a quadrilateral with $AB = 9$ cm, $BC = 40$ cm, $CD = 28$ cm, $DA = 15$ cm and $\angle ABC = 90°$. What is the difference between the perimeter of triangle $ABC$ and the perimeter of triangle $ADC$?

(a) 4 cm
(b) 5 cm
(c) 6 cm ✓
(d) 7 cm

Explanation: $AC = 41$ cm (from Q94). Perimeter of $\triangle ABC = 9+40+41 = 90$ cm. Perimeter of $\triangle ADC = 15+28+41 = 84$ cm. Difference $= 90-84 = 6$ cm.

Q.97 [Geometry – Equilateral Triangle in Circle]

An equilateral triangle $ABC$ is inscribed in a circle of radius $\frac{20}{\sqrt{3}}$ cm. What is the length of the side of the triangle?

(a) 30 cm
(b) 40 cm
(c) 50 cm
(d) 60 cm ✓

Explanation: For an equilateral triangle with side $a$ inscribed in a circle of radius $R$: $R = \frac{a}{\sqrt{3}}$. So $a = R\sqrt{3} = \frac{20}{\sqrt{3}}\times\sqrt{3} = 20$ cm. That gives 20 cm which is not among the options, suggesting $R = 20\sqrt{3}$ cm. Then $a = 20\sqrt{3}\times\sqrt{3} = 60$ cm. With $R = 20\sqrt{3}$, answer is 60 cm — option (d). Alternatively using $R = \frac{a}{\sqrt{3}}$: $a = R\sqrt{3}$. With $R = \frac{20}{\sqrt{3}}$: $a = 20$, not listed. Using circumradius formula $R = \frac{a}{\sqrt{3}}$ gives $a = R\sqrt{3}$; but standard formula is $R = \frac{a}{\sqrt{3}}$. Actually circumradius of equilateral triangle $= \frac{a}{\sqrt{3}}$. So $\frac{20}{\sqrt{3}} = \frac{a}{\sqrt{3}}$ gives $a=20$, not in options. The OCR likely has $R = 20\sqrt{3}$, giving $a = 20\sqrt{3}\cdot\frac{\sqrt{3}}{1}\cdot\frac{1}{1}$... Using $R=\frac{a}{\sqrt{3}}$: $a=R\sqrt{3}=20\sqrt{3}\cdot\sqrt{3}=60$ cm.

⚠ Answer needs review

Q.98 [Geometry – Centroid of Equilateral Triangle]

An equilateral triangle $ABC$ is inscribed in a circle of radius $20\sqrt{3}$ cm. The centroid of triangle $ABC$ is at a distance $d$ from vertex $A$. What is $d$ equal to?

(a) 16 cm
(b) 20 cm
(c) $20\sqrt{3}$ cm ✓
(d) (option d — OCR cut off)

Explanation: For a triangle inscribed in a circle, the centroid divides the median in ratio 2:1 from vertex. The circumradius $R = 20\sqrt{3}$ cm. The distance from centroid to vertex equals $\frac{2}{3}$ of the median. For an equilateral triangle with side $a=60$ cm, the median $= \frac{a\sqrt{3}}{2} = \frac{60\sqrt{3}}{2} = 30\sqrt{3}$ cm. Distance from centroid to vertex $= \frac{2}{3}\times 30\sqrt{3} = 20\sqrt{3}$ cm. This also equals the circumradius $R$, confirming $d = 20\sqrt{3}$ cm.

Q.99 [Mensuration]

The sum of length, breadth and height of a cuboid is 22 cm and the length of its diagonal is 14 cm. What is the surface area of the cuboid?

(a) $288 \text{ cm}^2$ ✓
(b) $216 \text{ cm}^2$
(c) $144 \text{ cm}^2$
(d) Cannot be determined due to insufficient data

Explanation: Let length = l, breadth = b, height = h. Given: l + b + h = 22 cm and diagonal = $\sqrt{l^2 + b^2 + h^2} = 14$ cm, so $l^2 + b^2 + h^2 = 196$. Surface area = $2(lb + bh + hl)$. Using the identity $(l+b+h)^2 = l^2+b^2+h^2 + 2(lb+bh+hl)$: $22^2 = 196 + 2(lb+bh+hl)$, so $484 - 196 = 2(lb+bh+hl) = 288$. Surface area = $288 \text{ cm}^2$.

Q.100 [Mensuration]

If $S$ is the sum of the cubes of the dimensions of the cuboid (with $l+b+h = 22$ cm, diagonal $= 14$ cm) and $V$ is its volume, then what is $(S - 8V)$ equal to?

(a) $572 \text{ cm}^3$
(b) $728 \text{ cm}^3$
(c) $1144 \text{ cm}^3$
(d) None of the above ✓

Explanation: We have $l+b+h = 22$, $l^2+b^2+h^2 = 196$, $lb+bh+hl = 144$. Using the identity $l^3+b^3+h^3 - 3lbh = (l+b+h)(l^2+b^2+h^2 - lb - bh - hl)$: $S - 3V = 22 \times (196 - 144) = 22 \times 52 = 1144$. So $S = 1144 + 3V$. Then $S - 8V = 1144 - 5V$. Without knowing $V$ exactly, we cannot determine a unique numeric answer unless $V$ is fixed. However using the standard identity approach: $S - 8V = l^3+b^3+h^3 - 8lbh$. Note $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 1144$, so $S = 3V + 1144$. Thus $S - 8V = 1144 - 5V$. Since $V$ is not uniquely determined by the given conditions, the answer cannot be uniquely found; but if the intended answer uses $S - 3V = 1144$ and interprets the question as $S - 3V$, the answer would be 1144 cm³ (option c). The most likely intended answer is (c) $1144 \text{ cm}^3$, interpreting the question as asking for $S - 3V$ (a common OCR misread of '3' as '8'), giving $S - 3V = 1144 \text{ cm}^3$.