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CDS II 2023 Elementary Mathematics with Solutions

Exam: CDS Year: 2023 (Session II) Questions: 100 Marks: 100 Negative Marking: 1/3

Q.79 [Profit and Loss]

A person bought an article and sold it at a profit of 20%. Had he bought it at 20% less and the selling price remained the same, what would have been the profit percentage?

(a) 25%
(b) 40%
(c) 50% ✓
(d) 60%

Explanation: Let CP = 100, SP = 120 (20% profit). New CP = 80 (20% less). New profit % = (120 − 80)/80 × 100 = 40/80 × 100 = 50%.

Q.80 [Algebra]

If $2s = a + b + c$, then what is $s^2 + (s-a)(s-b) + (s-b)(s-c) + (s-c)(s-a)$ equal to?

(a) $(a+b+c)^2$
(b) $ab+bc+ca$ ✓
(c) $2(ab+bc+ca)$
(d) $3(ab+bc+ca)$

Explanation: With 2s = a+b+c, let x = s−a, y = s−b, z = s−c so x+y+z = 3s−(a+b+c) = 3s−2s = s. The expression = s² + xy + yz + zx. Now (x+y+z)² = x²+y²+z²+2(xy+yz+zx), and one can expand directly: s² + (s−a)(s−b)+(s−b)(s−c)+(s−c)(s−a). Expand each: (s−a)(s−b) = s²−s(a+b)+ab, etc. Sum of three products = 3s²−2s(a+b+c)+ab+bc+ca = 3s²−2s(2s)+ab+bc+ca = 3s²−4s²+ab+bc+ca = −s²+ab+bc+ca. Total = s²+(−s²+ab+bc+ca) = ab+bc+ca.

Q.81 [Mensuration]

A sphere of radius 5 cm is dropped into a right circular cylindrical vessel partly filled with water. The radius of the cylindrical vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise?

(a) $\frac{1}{3}$ cm
(b) $\frac{5}{3}$ cm ✓
(c) 1 cm
(d) 3 cm

Explanation: Volume of sphere = $\frac{4}{3}\pi(5)^3 = \frac{500\pi}{3}$ cm³. Rise in water level h satisfies $\pi(10)^2 h = \frac{500\pi}{3}$, so $100h = \frac{500}{3}$, giving $h = \frac{5}{3}$ cm.

Q.82 [Geometry]

Consider the following statements: 1. The angle in a segment greater than a semicircle is less than a right angle. 2. If two sides of a pair of opposite sides of a cyclic quadrilateral are equal, then its diagonals are also equal. Which of the statements given above is/are correct?

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

Explanation: Statement 1: An angle inscribed in a major segment (segment greater than semicircle) subtends a minor arc, so it is less than 90°. True. Statement 2: In a cyclic quadrilateral, if one pair of opposite sides are equal (i.e., it is an isosceles trapezium), then its diagonals are equal. True. Both statements are correct.

Q.83 [Algebra]

If $a, b, c, x, y, z$ are real numbers such that $(a+b+c)^2 - 3(ab+bc+ca) + 3(x^2+y^2+z^2) = 0$, then which one of the following is correct?

(a) $a=b=c,\ x=y=z\neq 0$
(b) $a=b=c=0,\ x=y=z=1$
(c) $a=b=c,\ x=y=z=0$ ✓
(d) $a\neq b\neq c,\ x=y=z=0$

Explanation: $(a+b+c)^2 - 3(ab+bc+ca) = a^2+b^2+c^2 - ab - bc - ca = \frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2] \geq 0$. Also $3(x^2+y^2+z^2) \geq 0$. For the sum to be 0, both parts must be 0. So $(a-b)^2+(b-c)^2+(c-a)^2=0 \Rightarrow a=b=c$, and $x^2+y^2+z^2=0 \Rightarrow x=y=z=0$.

Q.84 [Geometry / Trigonometry]

In a triangle $ABC$, $\angle B = 90°$ and $p$ is the length of the perpendicular from $B$ to $AC$. If $BC = 10$ cm and $AC = 12$ cm, what is the value of $p$?

(a) $\frac{5\sqrt{11}}{3}$ cm ✓
(b) $\frac{10\sqrt{11}}{3}$ cm
(c) $\frac{40}{\sqrt{61}}$ cm
(d) $\frac{12}{\sqrt{61}}$ cm — (reconstructed as $\frac{5\sqrt{11}}{\sqrt{something}}$)

Explanation: With ∠B = 90°, BC = 10, AC = 12 (hypotenuse). Then AB = √(AC²−BC²) = √(144−100) = √44 = 2√11. Area of triangle = ½·AB·BC = ½·2√11·10 = 10√11. Also Area = ½·AC·p = ½·12·p = 6p. So 6p = 10√11 → p = 10√11/6 = 5√11/3.

Q.85 [Statistics]

The mean of $p, q, r, s$ and $t$ is 280. If the mean of $p, r$ and $t$ is 240, what is the mean of $q$ and $s$?

(a) 310
(b) 320 ✓
(c) 330
(d) 340

Explanation: Sum of all five = 5×280 = 1400. Sum of p, r, t = 3×240 = 720. Sum of q and s = 1400−720 = 680. Mean of q and s = 680/2 = 340. Wait — that gives 340, which is option (d). Let me recheck: 1400−720 = 680, 680/2 = 340. Answer is (d).

⚠ Answer needs review

Q.86 [Data Interpretation (Age-Weight)]

[Data set: A, B, C, D, E, F, G are cousins. D is thrice as old as A. C is as many years younger to B as G is to E and E is to D. Average age of D and G is 16; average age of A and E is 11; average age of B and C is 11. D has age:weight ratio 9:20; A has age:weight ratio 2:5; A's weight is 10 kg less than B's; D is 4 kg heavier than E; E is 4 kg heavier than F; F is 4 kg heavier than G; none exceeds 40 kg.] What is D's age (in years)?

(a) 15
(b) 16
(c) 17
(d) 18 ✓

Explanation: D = 3A. Average of A and E = 11 → A+E = 22. Average of D and G = 16 → D+G = 32. Let D = 3A. D's age:weight = 9:20, A's age:weight = 2:5. Let D's age = 9k, weight = 20k. Since none exceeds 40 kg: 20k ≤ 40 → k ≤ 2. For A: age = (2/5)·weight_A. Also D = 3A (ages), so 9k = 3·A_age → A_age = 3k. A's age:weight = 2:5 → weight_A = (5/2)·3k = 7.5k. A+E = 22 → E_age = 22−3k. D+G = 32 → G = 32−9k. E to D difference = D−E = 9k−(22−3k) = 12k−22. G to E difference = E−G = (22−3k)−(32−9k) = 6k−10. C to B difference = same as G to E = 6k−10. These must all be equal: 12k−22 = 6k−10 → 6k = 12 → k = 2. So D's age = 9×2 = 18 years.

Q.87 [Data Interpretation (Age-Weight)]

Using the same data set as Q86, what is the average age (in years) of B, C, D, E and G?

(a) 12
(b) 13
(c) 14 ✓
(d) 15

Explanation: From Q86: k=2. D=18, A=6, E=22−6=16, G=32−18=14. D−E=2 (common difference). B+C=22 (average 11). Average of B,C,D,E,G = (B+C+D+E+G)/5 = (22+18+16+14)/5 = 70/5 = 14.

Q.88 [Data Interpretation (Age-Weight)]

Using the same data set as Q86, what is the difference between the weights (in kg) of G and C?

(a) 4
(b) 3
(c) 2
(d) 1 ✓

Explanation: k=2: D weight=40 kg, E=40−4=36, F=36−4=32, G=32−4=28. A weight=7.5×2=15 kg. B weight=15+10=25 kg. B and C have equal weight, so C=25 kg. Difference G−C = 28−25 = 3, or C−G = 25−28 = −3. |difference| = 3. Answer is (b) 3.

⚠ Answer needs review

Q.89 [Data Interpretation (Age-Weight)]

Using the same data set as Q86, what is the average weight (in kg) of A, B, C, D, E, F and G?

(a) $\frac{201}{7}$ ✓
(b) $\frac{197}{7}$
(c) 30
(d) 32

Explanation: Weights: D=40, E=36, F=32, G=28, A=15, B=25, C=25. Sum = 40+36+32+28+15+25+25 = 201. Average = 201/7.

Q.90 [Data Interpretation (Age-Weight)]

Using the same data set as Q86, consider the following statements: 1. The age of F cannot be determined due to insufficient data. 2. The average weight of D and F is equal to the weight of E. 3. The weight difference is maximum for D and A. Which of the statements given above are correct?

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

Explanation: Statement 1: F's age — we know D−E = E−F = G−? (common difference in ages among D,E,F,G). D=18, E=16, so D−E=2. E−F=2 → F=14, G=12. F's age IS determinable. Statement 1 is FALSE. Statement 2: Average weight of D and F = (40+32)/2 = 36 = weight of E. TRUE. Statement 3: Weights: A=15, B=25, C=25, D=40, E=36, F=32, G=28. Maximum difference = D−A = 40−15 = 25. D−A is indeed the largest difference. TRUE. Statements 2 and 3 are correct.

⚠ Answer needs review