Breakup G strategy | Percentages | Easy
A certain amount of money was divided among Pinu, Meena, Rinu and Seema. Pinu received 20% of the total amount and Meena received 40% of the remaining amount. If Seema received 20% less than Pinu, the ratio of the amounts received by Pinu and Rinu is
B) 2 : 1
C) 1 : 2
D) 8 : 5
Answer & Explanation
Correct answer A) 5 : 8
Suppose total = 100.
| Person | Basis of Calculation | Amount Received |
| Pinu | 20% of total | 20 |
| Remaining after Pinu | 100 − 20 | 80 |
| Meena | 40% of remaining (40% of 80) | 32 |
| Seema | 20% less than Pinu → 80% of 20 | 16 |
| Total given so far | 20 + 32 + 16 | 68 |
| Rinu | Remaining → 100 − 68 | 32 |
So Pinu got 20, Rinu got 32 thus ratio = 20 : 32 = 5 : 8
Breakup G strategy | Ratios + Table | Medium
The ratio of expenditures of Lakshmi and Meenakshi is 2 : 3, and the ratio of income of Lakshmi to expenditure of Meenakshi is 6 : 7. If excess of income over expenditure is saved by Lakshmi and Meenakshi, and the ratio of their savings is 4 : 9, then the ratio of their incomes is
A) 3 : 5
B) 5 : 6
C) 7 : 8
D) 2 : 1
Answer & Explanation
Correct answer A) 3 : 5
Let Lakshmi’s expenditure = 2x, Meenakshi’s expenditure = 3x (since their expenditures are in ratio 2:3).
Lakshmi’s income is given to be in ratio 6:7 with Meenakshi’s expenditure so Lakshmi’s income = (6/7) × (3x) = (18/7) x.
Then Lakshmi’s saving = income − expenditure = (18/7 x) − 2x = (4/7) x.
Let Meenakshi’s income = y, so her saving = y − 3x.
Given their savings ratio is 4:9 , (4/7 x) : (y − 3x) = 4 : 9.
So (4/7 x) / (y − 3x) = 4/9 , solving gives y = (30/7) x.
Hence incomes are (18/7 x) : (30/7 x) = 18 : 30 = 3 : 5
Visual Lens G Strategy 1.2/0.8 | Profit Loss | Easy
An item with a cost price of Rs. 1650 is sold at a certain discount on a fixed marked price to earn a profit of 20% on the cost price. If the discount was doubled, the profit would have been Rs. 110. The rate of discount, as a percentage, at which the profit percentage would be equal to the rate of discount, is nearest to
A) 12
B) 18
C) 16
D) 14
Answer & Explanation
Correct Answer D) 14
Given: CP = 1650, MP = 2200
Condition: profit % = discount %
| Discount % | SP = MP × (1 − d) | Profit (SP − 1650) | Profit % | Match? |
| A. 12% | 2200 × 0.88 = 1936 | 286 | ≈ 17% | ❌ |
| B. 18% | 2200 × 0.82 = 1804 | 154 | ≈ 9% | ❌ |
| C. 16% | 2200 × 0.84 = 1848 | 198 | ≈ 12% | ❌ |
| D. 14% | 2200 × 0.86 = 1892 | 242 | ≈ 14.7% | ✅ |
Visual Lens G strategy | 1.2 / 0.8 Percentages | Moderate
A loan of Rs 1000 is fully repaid by two instalments of Rs 530 and Rs 594, paid at the end of first and second year, respectively. If the interest is compounded annually, then the rate of interest, in percentage, is _____
A) 10
B) 9
C) 8
D) 11
Answer & Explanation
Correct Answer is c) 8%
Let the annual rate of interest be r%. The present value of the two instalments must equal the loan amount.
1000 = 530 / (1 + r/100) + 594 / (1 + r/100)²
| Rate (%) | 530 ÷ (1+r) | 594 ÷ (1+r)² | Total PV | Match |
| 10 | ≈ 481.8 | ≈ 491.0 | ≈ 972.8 | ❌ |
| 9 | ≈ 486.2 | ≈ 500.2 | ≈ 986.4 | ❌ |
| 8 | ≈ 490.7 | ≈ 508.9 | ≈ 999.6 | ✅ |
| 11 | ≈ 477.5 | ≈ 482.2 | ≈ 959.7 | ❌ |
Breakup G Strategy | Concepts of Time Speed Distance | Easy
Rita and Sneha can row a boat at 5 km/h and 6 km/h in still water, respectively. In a river flowing with a constant velocity, Sneha takes 48 minutes more to row 14 km upstream than to row the same distance downstream. If Rita starts from a certain location in the river, and returns downstream to the same location, taking a total of 100 minutes, then the total distance, in km, Rita will cover is ______
Answer & Explanation
Concept of upstream and Downstream:
Sneha’s speed = 6 km/h, let river speed = x km/h
Time difference: 14/(6-x) – 14/(6+x) = 0.8 , x = 1 km/h
Rita’s effective speed
Upstream = 5 – 1 = 4 km/h
Downstream = 5 + 1 = 6 km/h
Time to go and return = 100 min = 5/3 h
d/4 + d/6 = 5/3 , d = 4 km
Total distance:
Total distance = 4 + 4 = 8 km
Answer: 8 km
Breakup G Strategy | Mixtures + Equations | Moderate
A mixture of coffee and cocoa, 16% of which is coffee, costs Rs 240 per kg. Another mixture of coffee and cocoa, of which 36% is coffee, costs Rs 320 per kg. If a new mixture of coffee and cocoa costs Rs 376 per kg, then the quantity, in kg, of coffee in 10 kg of this new mixture is
A) 6
B) 2.5
C) 5
D) 4
Answer & Explanation
Correct Answer (c) 5 kg
Let the price of coffee be C and that of cocoa be K (per kg).
From the first mixture (16% coffee, price 240):
0.16C + 0.84K = 240
From the second mixture (36% coffee, price 320):
0.36C + 0.64K = 320
Multiply both equations by 100:
16C + 84K = 24000
36C + 64K = 32000
Solve these two:
From the pair, you get K = 176 and then C = 576.
Now for the new mixture with price 376 and coffee fraction x:
576x + 176(1 − x) = 376→576x + 176 − 176x = 376 → 400x = 200 → x = 0.5
So the new mixture has 50% coffee.
Therefore, in 10 kg of the new mixture, coffee = 0.5 × 10 = 5 kg.
Shortcut method:
Coffee % goes from 16% → 36%, Price goes from 240 → 320
So an increase of 20% coffee increases price by 80 ⇒ Each 1% coffee increases price by 4.
Now compare old price to new price → Price of new mixture = 376 → Difference from cheaper mixture = 376 − 240 = 136
So increase in coffee % = 136 ÷ 4 = 34%
Hence coffee % in new mixture = 16% + 34% = 50%
So in 10 kg, coffee = 50% of 10 = 5 kg
Breakup G strategy | Averages + Linear Equation | Easy
The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from second day to ninth day becomes 66. The number of copies sold on the first day of the book launch is ____
Answer & Explanation
Correct Answer: 49 copies
First 7 days avg = 60 ⇒ total = 7 × 60 = 420
First 8 days avg = 63 ⇒ total = 8 × 63 = 504
8th day sales = 504 − 420 = 84
9th day sales = 84 − 11 = 73
Avg from 2nd to 9th day = 66 ⇒ total (2nd–9th) = 8 × 66 = 528
Total (1st–9th) = 504 + 73 = 577
1st day sales = 577 − 528 = 49
LCM G Strategy | Time & Work | Moderate
Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone is
Answer & Explanation
Correct Answer 340
Let Chandan’s efficiency be 1 unit/day.
Then Bipin’s efficiency = 2 units/day and Ankita’s efficiency = 4 units/day.
For the first 20 days, all three work together, so work done = (4 + 2 + 1) × 20 = 140 units.
For the remaining 40 days, only Ankita and Chandan work, so work done = (4 + 1) × 40 = 200 units.
Total work = 140 + 200 = 340 units.
Since Chandan alone does 1 unit of work per day, he would complete the job in 340 days.
Correct Answer D) 14
Given: CP = 1650, MP = 2200
Condition: profit % = discount %
| Discount % | SP = MP × (1 − d) | Profit (SP − 1650) | Profit % | Match? |
| A. 12% | 2200 × 0.88 = 1936 | 286 | ≈ 17% | ❌ |
| B. 18% | 2200 × 0.82 = 1804 | 154 | ≈ 9% | ❌ |
| C. 16% | 2200 × 0.84 = 1848 | 198 | ≈ 12% | ❌ |
| D. 14% | 2200 × 0.86 = 1892 | 242 | ≈ 14.7% | ✅ |
Concept of Quadratic Equations | Hard
The equations 3x² − 5x + p = 0 and 2x² − 2x + q = 0 have one common root.
The sum of the other roots of these two equations is
A) 1/3 − 2p + 3/2 q
B) 2/3 + p + 3/4 q
C) 5/3 − p + 3/2 q
D) 4/3 − p + 2/3 q
Answer & Explanation
Correct Answer: Option C
Given equations:
3x² − 5x + p = 0
2x² − 2x + q = 0
For a quadratic ax² + bx + c = 0, sum of roots = −b/a.
Step 1: Use the fact that r is a common root
Since r satisfies both equations:
From (1): 3r² − 5r + p = 0 → p = 5r − 3r²
From (2): 2r² − 2r + q = 0 → q = 2r − 2r²
Step 2: Express sum of the “other” roots
Equation (1): Sum of roots = 5/3. So, other root of (1) = 5/3 − r
Equation (2): Sum of roots = 1. So, other root of (2) = 1 − r
Required sum = (5/3 − r) + (1 − r) = 8/3 − 2r
Step 3: Eliminate r using p and q
From p = 5r − 3r²; From q = 2r − 2r²
Rewrite both: p = r(5 − 3r); q = r(2 − 2r)
Solve for r in linear combination form:
Multiply second by 3/2: (3/2)q = 3r − 3r²
Now subtract from p: p − (3/2)q = (5r − 3r²) − (3r − 3r²) = 2r
So: r = (p − 3q/2)/2 = p/2 − 3q/4
Step 4: Substitute r into required sum Required sum = 8/3 − 2r = 8/3 − 2(p/2 − 3q/4) = 8/3 − p + 3q/2.
Concepts of Polynomial Equations + Quadratic | Hard
If 9(x² + 2x – 3) − 4(3(x² + 2x − 2)) + 27 = 0,
then the product of all possible values of x is _____
A) 20
B) 15
C) 30
D) 5
Answer & Explanation
Correct Answer: A) 20
Given equation:
9^(x² + 2x − 3) − 4·3^(x² + 2x − 2) + 27 = 0
Step 1: Write everything with base 3
9 = 3², so ⇒ 9^(x² + 2x − 3) = 3^(2x² + 4x − 6)
Also, 3^(x² + 2x − 2) = 3·3^(x² + 2x − 3)
Let t = 3^(x² + 2x − 3), where t > 0.
Then the equation becomes: t² − 4(3t) + 27 = 0 ⇒ t² − 12t + 27 = 0
Step 2: Solve the quadratic in t
t² − 12t + 27 = 0 ⇒ (t − 3)(t − 9) = 0 ⇒ So t = 3 or t = 9
Step 3: Convert back to x
Case 1: 3^(x² + 2x − 3) = 3 ⇒ x² + 2x − 3 = 1 ⇒ x² + 2x − 4 = 0
Roots: x = −1 ± √5 ⇒ Product of roots = −4
Case 2: 3^(x² + 2x − 3) = 9 = 3² ⇒ x² + 2x − 3 = 2 ⇒ x² + 2x − 5 = 0
Roots: x = −1 ± √6 ⇒ Product of roots = −5
Step 4: Product of all possible values of x
Total product = (product from case 1) × (product from case 2)
= (−4) × (−5) = 20
Vedic Patterns G Strategy | Quadratic Equations | Hard
If m and n are integers such that (m + 2n)(2m + n) = 27,
then the maximum possible value of 2m − 3n is ___
Answer & Explanation
Given: (m + 2n)(2m + n) = 27
Since m and n are integers, (m + 2n) and (2m + n) must be integer factors of 27.
Possible factor pairs (m + 2n, 2m + n) are:
(1, 27), (3, 9), (9, 3), (27, 1),
(−1, −27), (−3, −9), (−9, −3), (−27, −1)
We now solve each pair as a system: m + 2n = A; 2m + n = B
From these:
Multiply first equation by 2: 2m + 4n = 2A
Subtract from second: (2m + n) − (2m + 4n) = B − 2A
⇒ −3n = B − 2A ⇒ n = (2A − B) / 3
Then m = A − 2n
| m + 2n | 2m + n | n = (2A − B)/3 | m = A − 2n | 2m − 3n | Valid / Reject |
| 1 | 27 | (2 − 27)/3 = −25/3 | — | — | Reject |
| 3 | 9 | (6 − 9)/3 = −1 | 3 − 2(−1) = 5 | 13 | Valid |
| 9 | 3 | (18 − 3)/3 = 5 | 9 − 10 = −1 | −17 | Valid |
| 27 | 1 | (54 − 1)/3 = 53/3 | — | — | Reject |
| −1 | −27 | (−2 + 27)/3 = 25/3 | — | — | Reject |
| −3 | −9 | (−6 + 9)/3 = 1 | −3 − 2 = −5 | −13 | Valid |
| −9 | −3 | (−18 + 3)/3 = −5 | −9 + 10 = 1 | 17 | Valid |
| −27 | −1 | (−54 + 1)/3 = −53/3 | — | — | Reject |
Maximum possible valid value of 2m − 3n = 17
X maro G Strategy | Inequality | Easy
The set of all real values of x for which (x² − |x + 9| + x) > 0, is
A) (−∞, −3) ∪ (3, ∞)
B) (−9, −3) ∪ (3, ∞)
C) (−∞, −9) ∪ (3, ∞)
D) (−∞, −9) ∪ (9, ∞)
Answer & Explanation
Use X maro G Strategy put x = -5 and x = -10
| Option | Interval | Test x = −5 | Test x = −10 | Status |
| A | (−∞, −3) ∪ (3, ∞) | Included, LHS = 16 > 0 | Included, LHS = 89 > 0 | ✅ Possible |
| B | (−9, −3) ∪ (3, ∞) | Included, LHS = 16 > 0 | Not included | ❌ Eliminated |
| C | (−∞, −9) ∪ (3, ∞) | Not included | Included, LHS = 89 > 0 | ❌ Eliminated |
| D | (−∞, −9) ∪ (9, ∞) | Not included | Included, LHS = 89 > 0 | ❌ Eliminated |
Answer & Explanation
cc
Maxima Minima G Strategy | Easy
If a, b, c and d are integers such that their sum is 46, then the minimum possible value of
(a − b)² + (a − c)² + (a − d)² is ______
Answer & Explanation
Minimum possible value is 2
Since a, b, c, d must be integers, they must be the integers closest to 11.5, which are 11 and 12.
To satisfy the sum constraint a + b + c + d = 46, we can take values 11, 11, 12, 12.
Expression becomes: (11 − 12)² + (11 − 12)² + (11 − 11)² = 1 + 1 + 0 = 2
Concept of Factors | Hard
The number of divisors of 2⁶ × 3⁵ × 5³ × 7², which are of the form (3r + 1), where r is a non-negative integer, is
A) 24
B) 56
C) 36
D) 42
Answer & Explanation
Correct Answer: 42
N = 2⁶ × 3⁵ × 5³ × 7²
We want the number of divisors of N that are of the form (3r + 1), i.e.
divisors ≡ 1 (mod 3).
Step 1: Kill the 3-factor
Any divisor containing 3¹ or higher is divisible by 3 → it cannot be 3r + 1.
So exponent of 3 must be 0.
We only use: 2ᵃ × 5ᶜ × 7ᵈ
a: 0 to 6 → 7 values; c: 0 to 3 → 4 values; d: 0 to 2 → 3 values
Step 2: Work mod 3
2 ≡ -1 (mod 3); 5 ≡ -1 (mod 3); 7 ≡ 1 (mod 3)
So divisor ≡ (-1)^(a + c) (mod 3).
For divisor ≡ 1 (mod 3):
(-1)^(a + c) = 1 → a + c must be even.
Step 3: Count (a, c) with a + c even
a = 0,1,2,3,4,5,6; even: 0,2,4,6 → 4 values; odd: 1,3,5 → 3 values
c = 0,1,2,3; even: 0,2 → 2 values; odd: 1,3 → 2 values
a + c even when: a even, c even → 4 × 2 = 8; a odd, c odd → 3 × 2 = 6
Total (a, c) good pairs = 8 + 6 = 14. d has 3 values → 0,1,2.
Total divisors = 14 × 3 = 42.
Vedic Patterns G Strategy | Moderate
Suppose a, b, c are three distinct natural numbers, such that 3ac = 8(a + b).
Then, the smallest possible value of 3a + 2b + c is _______
Answer & Explanation
Final Answer: 12
Given: 3ac = 8(a + b) ⇒ 8b = 3ac − 8a ⇒ b = (3ac/8) − a
Since a, b, c are natural numbers, 3ac must be divisible by 8.
Now try small values of a (since we want the minimum value of 3a + 2b + c).
Try a = 1
Then 3c must be divisible by 8 ⇒ c = 8
b = (3×1×8)/8 − 1 = 3 − 1 = 2
Value = 3a + 2b + c = 3 + 4 + 8 = 15
Try a = 2
Then 6c divisible by 8 ⇒ c = 4 is the smallest possible
b = (3×2×4)/8 − 2 = 3 − 2 = 1
Here a, b, c = 2, 1, 4 are distinct natural numbers
Value = 3a + 2b + c = 6 + 2 + 4 = 12
Try a = 3
Then 9c divisible by 8 ⇒ c = 8
b = (3×3×8)/8 − 3 = 9 − 3 = 6
Value = 9 + 12 + 8 = 29 (larger)
Hence the smallest possible value occurs at a = 2, b = 1, c = 4
Visual Lens G Strategy | Hexagon Area | Moderate
Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively.
Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is _______
Answer & Explanation
Correct answer is B) 5 : 24
Let area of the hexagon is 6E, where E represents the area of any one of the six congruent central triangles, such as triangle OBC.
The area of trapezium PBCQ is obtained by decomposing it into simpler regions. It consists of the full area of triangle OBC along with suitable fractional parts of the triangles adjacent to the base BC.
This decomposition gives: Area(PBCQ) = 5E/4.
Therefore, the required ratio of areas is: (5E/4) ÷ 6E = 5/24.
Visual Lens G Strategy | Triangle
In a triangle ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is
A) 9 : 4
B) 7 : 4
C) 11 : 4
D) 15 : 4
Answer & Explanation
Correct answer: C) 11:4
Given, AD : AT = 4 : 3 ⇒ AT : TD = 3 : 1 and BE : BT = 5 : 4 ⇒ BT : TE = 4 : 1
By the mass points method, masses are assigned inversely to the segments.
From AT : TD = 3 : 1 ⇒ Mass at A : Mass at D = 1 : 3.
From BT : TE = 4 : 1, ⇒ Mass at B : Mass at E = 1 : 4.
Choosing consistent masses,
A = 5/4, D = 15/4 and B = 1, E = 4, so mass at T = 5.
Since D lies on BC, Mass at D = Mass at B + Mass at C
15/4 = 1 + Mass at C ⇒ Mass at C = 11/4.
Therefore, BD : DC = Mass at C : Mass at B ⇒ (11/4) : 1 ⇒ 11 : 4.
Concept of Angles of Circles | Moderate
Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR, respectively, such that AB is also a tangent drawn to the same circle. If ∠AOB = 50°, then ∠APB is _____
Answer & Explanation
Correct Answer 80
From P, PQ and PR are tangents to the circle with centre O, so OQ ⟂ PQ and OR ⟂ PR.
AB is also a tangent, touching the circle at S, hence OS ⟂ AB.
From an external point, the line joining the centre bisects the angle between tangents.
Thus OA bisects ∠QOS and OB bisects ∠ROS.
Let AOQ = AOS = x and BOR = BOS = y.
Given ∠AOB = x + y = 50°.
Then ∠QOS = 2x, ∠ROS = 2y and hence ∠QOR = 2(x + y) = 100°.
In quadrilateral ORPQ, angles at Q and R are 90°, so
∠QOR + ∠APB = 180°.
Therefore, ∠APB = 180° − 100° = 80°.
Concepts of Indices | Moderate
The sum of digits of the number (625)⁶⁵ × (128)³⁶ is ______
Answer & Explanation
Final answer 25
Given: (625)⁶⁵ × (128)³⁶
Express the bases as powers of 5 and 2: 625 = 5⁴ and 128 = 2⁷
(625)⁶⁵ × (128)³⁶
= 5^(4×65) × 2^(7×36)
= 5^260 × 2^252
= 5^8 × 5^252 × 2^252
= 5^8 × 10^252
Now 5⁸ = 390625
So the full number is:
390625 followed by 252 zeros.
Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 = 25
Answer: 25
Concepts of Logs + Maxima Minima | Hard
If log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is
A) 36
B) 48
C) 96
D) 24
Answer & Explanation
Correct Answer B (48)
Given log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4 x, y, z > 0
Step 1: Convert all logs to the same base (base 8)
64 = 8²; 512 = 8³
Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x
Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z)
Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4
Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096
Apply the Arithmetic Mean – Geometric Mean (AM-GM) inequality.
For positive real numbers x, y, and z, the minimum value of their sum occurs when x = y = z.
(x + y + z) / 3 ≥ ³√(xyz)
Substitute the value of xyz: (x + y + z) / 3 ≥ ³√(4096)
Calculate the cube root: Since 16³ = 4096, we have: (x + y + z) / 3 ≥ 16
Solve for the minimum value of (x + y + z): x + y + z ≥ 3 × 16 x + y + z ≥ 48
Concepts of AP GP + Quadratic Equations | Hard
Let an be the nth term of a decreasing infinite geometric progression.
If a₁ + a₂ + a₃ = 52 and a₁a₂ + a₂a₃ + a₃a₁ = 624, then the sum of this geometric progression is
A) 54
B) 60
C) 63
D) 57
Answer & Explanation
Correct Answer: 54
Let the GP be a, ar, ar² with common ratio r (0 < r < 1 since it is decreasing and infinite).
Given:
a + ar + ar² = 52 → a(1 + r + r²) = 52
a·ar + ar·ar² + ar²·a = 624 → a²(r + r² + r³) = 624
From these, form a relation involving only r.
Divide equation (2) by [equation (1)]²:
[a²(r + r² + r³)] / [a²(1 + r + r²)²] = 624 / 52²
So:
(r + r² + r³) / (1 + r + r²)² = 624 / 2704 = 39 / 169
Now just try simple values of r between 0 and 1.
Try r = 1/2:
r + r² + r³ = 1/2 + 1/4 + 1/8 = 7/8
1 + r + r² = 1 + 1/2 + 1/4 = 7/4 → squared = 49/16
Ratio = (7/8) / (49/16) = 2/7 (not 39/169).
Try r = 1/3:
r + r² + r³ = 1/3 + 1/9 + 1/27 = 13/27
1 + r + r² = 1 + 1/3 + 1/9 = 13/9 → squared = 169/81
Ratio = (13/27) / (169/81) = 39/169 (matches).
So r = 1/3.
From a(1 + r + r²) = 52 → a(1 + 1/3 + 1/9) = 52 → a(13/9) = 52 → a = 36.
Sum of infinite GP = a / (1 − r) = 36 / (1 − 1/3) = 36 / (2/3) = 54.
