CAT 50 Day Streak Day 16

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CAT 50 Day Streak – Day 16

30 questions across VARC, DILR & QA · +3 for a correct answer · −1 for a wrong MCQ · no negative marking for TITA

0/30Attempted

0Correct

0Wrong

0Score / 90

VARC · Verbal Ability & Reading Comprehension

Q1 – Q10 0/10

Passage · Q1 – Q4

Dust controls numerous atmospheric and surface processes on Mars. In the atmosphere, suspended dust can lead to substantial temperature variations, can perturb atmospheric dynamics, and can lead to local, regional, and even global dust storms. On the surface, seasonal and post–storm dust settling has been tied to the occurrence and modification of surface features such as slope streaks and recurring slope lineae. The role of dust in controlling the temperature of the atmosphere and surface makes it a central part of the weather and climate system. In addition, dust might affect Mars’ past and current habitability through the exchange and distribution of nutrients and shielding of ultraviolet radiation. Dust has a direct effect on the exploration of Mars as well, by affecting remote sensing observations, surface hardware, crew operations, and the power efficiency of solar panels; in particular, large-scale dust storms and planet-encircling dust events can severely affect operations on Mars. The National Aeronautics and Space Administration’s most recent Mars Exploration Program specifically calls out the need for a better understanding of the dynamics and interactions of Mars’ dusty surface and atmosphere in the context of sustainable, future exploration.

Most likely, martian dust is the result of aeolian processes eroding exposed bedrock over geologic time frames. Past work established that wind stress exerted by (i) “vortical” processes, such as convective vortices and dust devils, and (ii) “nonvortical” processes, such as (straight-line) wind gusts, convection cell fronts, and meridional circulation, is responsible for the bulk injection of dust into the atmosphere. Dust particles are difficult to lift directly into the thin Martian atmosphere due to cohesive forces, especially if they do not form dust aggregates. However, the saltation of larger grains is easier to initiate on Mars at wind shear velocities beyond the “static threshold”. In turn, saltating sand particles can directly and efficiently inject finer-sized particles into the atmosphere upon reimpact on the surface.

Early modeling efforts and wind tunnel experiments suggested that the static threshold of ~100- to ~200-μm-sized martian grains is located at wind shear velocities between ~1.5 and ~2 m/s, which can be converted to wind velocities between ~15 and ~22 m/s at an altitude of ~1.5 m above the surface, assuming an average aerodynamic surface roughness and a convectively stable atmosphere. However, the atmospheric data collected by missions on Mars seemed to suggest that winds do not—or only rarely—exceed this threshold, despite frequent observations of active aeolian surface processes from orbit, a discrepancy termed the “martian sand transport puzzle”. More recent in situ measurements on Mars—most prominently by the InSight lander (Interior Exploration using Seismic Investigations, Geodesy and Heat Transport) and Perseverance rover—showed that wind velocities can be substantially higher and that sand and dust transport can be initiated at lower wind shear velocities than anticipated earlier.

Q1MCQ

The “martian sand transport puzzle” mentioned in the passage represents a conflict between:

Aobservations of frequent dust storms from orbit and the lack of dust measured on the surface by landers.

Btheoretical models predicting high wind velocity thresholds for sand movement and the lower wind speeds previously measured on Mars.

Cthe cohesive forces affecting fine dust particles and the shear forces required to initiate saltation of larger grains.

Dwind tunnel experiments conducted on Earth and the aerodynamic principles governing the thin Martian atmosphere.

Solution Answer: B

Models and wind-tunnel work put the threshold at about 15–22 m/s. Mission data suggested winds “do not—or only rarely—exceed this threshold”, yet sand was clearly moving. That mismatch between the predicted high threshold and the lower measured winds is the puzzle.

A: the puzzle concerns sand transport, not a lack of surface dust. C describes a mechanism, not the discrepancy. D: the passage never sets wind tunnels against aerodynamic principles.

Q2MCQ

Why does the author mention “cohesive forces” in the second paragraph?

ATo explain why the saltation of larger grains is easier to initiate than the lifting of fine dust.

BTo argue that Martian dust is fundamentally different from sand found on Earth.

CTo introduce the concept of dust aggregates as the primary form of Martian surface material.

DTo illustrate the primary mechanism by which wind gusts inject dust into the atmosphere.

Solution Answer: A

Cohesive forces make fine dust “difficult to lift directly”. The next sentence then says “However, the saltation of larger grains is easier to initiate”. The mention sets up the contrast between hard-to-lift dust and easier-to-move grains.

B: there is no comparison with Earth sand. C: aggregates are mentioned only in passing. D: cohesive forces hinder lifting; they are not an injection mechanism.

Q3MCQ

Which of the following scenarios is most analogous to the process of dust injection via saltation as described in the passage?

AA strong wind blowing fine powder off a completely flat and smooth surface.

BA swift river current washing away silt from the riverbed.

CA billiard player using the cue ball to scatter a tightly packed rack of other balls.

DA large wave crashing on a beach, pulling fine sand back into the ocean as it recedes.

Solution Answer: C

In saltation, wind moves larger grains, and these grains then strike the surface and knock finer particles loose. A cue ball (the moving grain) striking a tightly packed rack (the cohesive fine particles) and scattering it matches this impact-driven transfer of energy.

A and B describe direct lifting or washing by the fluid itself, which is exactly what the passage says is hard for fine dust. D involves no impact that ejects other particles.

Q4MCQ

Which of the following, if discovered to be true, would most seriously weaken the proposed resolution to the “martian sand transport puzzle”?

AIt is determined that the aerodynamic surface roughness on Mars is significantly lower than the average value used in models.

BScientists find that dust devils are far more effective at lifting sand than previously believed, even at low wind speeds.

CNew orbital imagery shows that active sand transport occurs even in regions where rovers have measured consistently calm conditions.

DFurther analysis reveals that the wind sensors on the InSight and Perseverance missions were miscalibrated, consistently over-reporting wind speeds.

Solution Answer: D

The resolution rests on in situ measurements by InSight and Perseverance showing that winds can be “substantially higher”. If those sensors over-reported wind speeds, the evidence for the resolution falls away and the puzzle returns.

A changes model assumptions, not the new measurements. B offers a different way sand might move, so it doesn’t undercut the measured data. C is local and leaves the measured high winds intact.

Q5TITA

Five jumbled-up sentences (labelled 1, 2, 3, 4 and 5), related to a topic, are given below. Four of them can be put together to form a coherent paragraph. Identify the odd sentence and key in the number of that sentence as your answer.

Many of us still know something about these treatments: studies estimate that as much as 80 per cent of the world’s population still use traditional plant- and animal-based medicines.
We treated insomnia with lettuce and halitosis with parsley – according to a 10th-century cookbook from Baghdad, which devotes an entire chapter to the humoral properties of different vegetables.
We treated pain with willow bark and terminated pregnancies with abortifacient herbs.
At one point or another, we all knew something about how to heal ourselves using the plants and animals that surround us, at least somewhat.
Lately, we often relegate this kind of intelligence to laboratories.

Solution Answer: 5

The coherent paragraph is 4 → 3 → 2 → 1:

4: We all once knew how to heal ourselves with plants and animals.
3 and 2: Examples (“We treated…”): willow bark, abortifacient herbs, lettuce and parsley.
1: “Many of us still know something about these treatments”, since about 80% still use them.

Sentence 5 shifts to modern laboratories. That contrast is never developed, and it cuts against sentence 1’s point that such knowledge is still widely used.

Q6MCQ

The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.

Passage There was a moment during Scotland’s match against South Africa, as tempers frayed and testosterone flowed, that harked back to rugby union’s bygone era. Damian de Allende pinned an elbow to Ben White’s head. Darcy Graham and Cheslin Kolbe grabbed each other’s necks. The big men came steaming in. And while what the Daily Record called a “furious World Cup rammy” soon fizzled out, it reflected a basic tenet of collision sports: win the physical battle, win the match. Some players, however, believe this goes even further. And that when a fight breaks out on the pitch, winning it can change the momentum of a match. Perhaps even a series.

AA tense World Cup rugby clash between Scotland and South Africa saw tempers flare, highlighting the enduring belief that physical dominance in contact sports secures victory.

BIndividual scuffles in rugby symbolize outdated aggression that rarely affects match outcomes, despite occasional nostalgia for the sport’s past.

CMoments of on-field fighting in rugby are believed not only to influence immediate physical battles but also to shift match or series momentum.

DThe brief fight in a World Cup rugby match reflected a decline in sportsmanship, with physical contests seen as detrimental to team performance.

Solution Answer: C

The passage builds from a basic tenet (win the physical battle, win the match) to its real point: some players believe that winning a fight can change the momentum of a match, “perhaps even a series”. C captures both layers.

A stops at the basic tenet and misses the momentum claim. B says fights rarely affect outcomes, which is the opposite of the claim. D introduces a decline in sportsmanship that the passage never mentions.

Q7MCQ

There is a sentence that is missing in the paragraph below. Look at the paragraph and decide where (option 1, 2, 3, or 4) the following sentence would best fit.

Sentence The juxtaposition between his supposed arrested development and his numerical facility made his mathematical feats even more impressive.

Paragraph On 25 November 1915, the American newspaper The Review published the extraordinary case of an 11-year-old boy with prodigious mathematical abilities. ___(1)___. Perched on a hill close to a set of railroad tracks, he could memorise all the numbers of the train carriages that sped by at 30 mph, add them up, and provide the correct total sum. ___(2)___. What was remarkable about the case was not just his ability to calculate large numbers, but the fact that he could barely eat unassisted or recognise faces of people he met. ___(3)___. ‘How can you account for it?’ asked the article’s author. ___(4)___. The answer took the form of a medical label: the boy was what 19th-century medicine termed an ‘idiot savant’.

AOption 1

BOption 2

COption 3

DOption 4

Solution Answer: C (Option 3)

The sentence refers to a “juxtaposition” of “arrested development” and “numerical facility”. Both halves are first laid out in the sentence before blank 3: he could calculate large numbers, but he could barely eat unassisted or recognise faces. The new sentence then sums up why that made the feats impressive, which leads naturally into “How can you account for it?”

At blanks 1 and 2 his developmental limits haven’t been mentioned yet. Blank 4 would wedge the sentence between the question and its answer.

Q8TITA

The four sentences (labelled 1, 2, 3 and 4) given below, when properly sequenced, would yield a coherent paragraph. Decide on the proper sequencing of the order of the sentences and key in the sequence of the four numbers as your answer.

While a new Alzheimer’s drug, Leqembi, has recently been approved by the FDA, it is available only for a small portion of people living with dementia.
People living with dementia also require significantly more hours of care than older adults without dementia.
Direct medical spending for dementia is relatively small because there aren’t many treatments available.
While many elderly adults eventually need help with activities of daily living, such as eating and bathing, people living with dementia often need constant or near-constant supervision.

Solution Answer: 3124

The order runs like this:

3: Medical spending on dementia is small because few treatments exist.
1: Qualifies this: a new drug exists, but only for a small portion of patients.
2: “Also” moves to a second cost driver: many more hours of care.
4: Explains why: constant or near-constant supervision.

Q9TITA

For the first couple of centuries after the invention of the printing press, reading remained largely an elite pursuit.
It was an age of monumental works of thought and knowledge.
People alive at the time understood that something momentous was going on.
But by the beginning of the 1700s, the expansion of education and an explosion of cheap books began to diffuse reading rapidly down through the middle classes and even into the lower ranks of society.
Suddenly, it seemed that everyone was reading everywhere: men, women, children, the rich, the poor.

Solution Answer: 2

The coherent paragraph is 1 → 4 → 5 → 3:

1: Reading was an elite pursuit for two centuries after the printing press.
4: “But” by the 1700s it spread to the middle and lower classes.
5: “Suddenly” everyone was reading everywhere.
3: People sensed “something momentous was going on”.

Sentence 2 is about the content of the age (monumental works), not the spread of reading, so it breaks the thread.

Q10MCQ

There is a sentence that is missing in the paragraph below. Look at the paragraph and decide where (option 1, 2, 3, or 4) the following sentence would best fit.

Sentence Recovering alcoholics notwithstanding, I enjoyed the collective sense of what was meant by a night out.

Paragraph I miss the ubiquity of a rambunctious drinking culture, whether I participated in it or not, at whatever level. ___(1)___. These are not the rantings of someone in her forties assuming that there’s no party because she wasn’t invited. This is a wave of moderation. ___(2)___. The new language of sobriety is particularly ill-suited to writers. ___(3)___. Mocktails are etymologically and financially silly. ___(4)___. And there’s nothing so tedious as a conversation about abstinence.

AOption 1

BOption 2

COption 3

DOption 4

Solution Answer: A (Option 1)

The sentence extends the opening line. “The ubiquity of a rambunctious drinking culture” becomes “the collective sense of what was meant by a night out”. “Recovering alcoholics notwithstanding” adds the obvious caveat to her nostalgia. The next sentence (“These are not the rantings…”) then defends that nostalgia.

Blanks 2–4 fall inside her complaints about the new sobriety (its language, mocktails, abstinence talk), where a fond memory of nights out would interrupt the flow.

DILR · Data Interpretation & Logical Reasoning

Q11 – Q20 0/10

Set 1 · Q11 – Q15

In an Avenger fan club which has 3000 kids, a survey was conducted for their liking of Captain America, Thor or Iron Man. The ratio of boys to girls in the club is 5:3. On the day of the survey, 75 boys and 75 girls were absent. The results obtained from the remaining kids were:

Each kid liked at least one superhero.
50% of the boys liked Captain America, while 50% of the girls liked Thor.
3% of the boys liked all three superheroes, while this percentage was 4% among girls.
The number of boys liking only Thor was 50% more than the number of boys liking only Iron Man.
The number of girls liking only Captain America was the same as the number of girls liking only Iron Man.
The percentage of boys liking only Captain America and Thor was 7%, only Thor and Iron Man was 5%, and only Iron Man and Captain America was 9%. The respective percentages in girls were 8%, 6% and 6%.

Q11MCQ

What is the ratio of boys liking Iron Man to those liking Thor?

A2:3

B5:6

C7:10

D21:25

Solution Answer: B

Surveyed. Boys = 1875 − 75 = 1800; girls = 1125 − 75 = 1050; total 2850.
Boys. Captain America (CA) = 900. All three = 54, CA&T only = 126, T&IM only = 90, IM&CA only = 162. Only CA = 900 − 54 − 126 − 162 = 558. Only T + only IM = 1800 − 558 − 432 = 810 with only T = 1.5 × only IM, so only IM = 324 and only T = 486.
Girls. Thor = 525. All three = 42, CA&T only = 84, T&IM only = 63, IM&CA only = 63. Only T = 525 − 42 − 84 − 63 = 336. Only CA + only IM = 1050 − 336 − 252 = 462, so each is 231.

Category	Boys	Girls	Total
Only CA	558	231	789
Only Thor	486	336	822
Only Iron Man	324	231	555
CA & Thor only	126	84	210
Thor & IM only	90	63	153
IM & CA only	162	63	225
All three	54	42	96
Total	1800	1050	2850

Boys liking Iron Man = 324 + 90 + 162 + 54 = 630. Boys liking Thor = 486 + 126 + 90 + 54 = 756. Ratio = 630 : 756 = 5 : 6.

Q12TITA

What is the number of girls who like Iron Man?

Solution Answer: 399

From the table in Q11: 231 (only IM) + 63 (T&IM) + 63 (IM&CA) + 42 (all three) = 399.

Q13TITA

What is the difference between the number of only Captain America fans and only Thor fans?

Solution Answer: 33

Only CA = 558 + 231 = 789. Only Thor = 486 + 336 = 822. Difference = 33.

Q14MCQ

What percentage of the users surveyed like all three superheroes?

A3.37%

B3.20%

C3.50%

D3.44%

Solution Answer: A

All three = 54 + 42 = 96 out of 2850 surveyed. 96/2850 × 100 ≈ 3.37%.

Q15MCQ

How many of the surveyed kids like Captain America?

A1245

B1270

C1310

D1320

Solution Answer: D

Boys liking CA = 900. Girls liking CA = 231 + 84 + 63 + 42 = 420. Total = 1320.

Set 2 · Q16 – Q20

Six friends, Parag, Harsh, Roopa, Azam, Siddhi, and Edwin, are natives of different cities from among Patiala, Hisar, Rourkela, Agra, Siliguri and Erode. Each of them works in a different city from among Rome, Auckland, Istanbul, Seoul, Edinburgh and Delhi, with a different annual CTC (in Rs. lakhs) out of 20, 25, 30, 35, 40 and 45. It is also known that:

For each person, the initials of name, native place and city in which they work are different.

Azam works in Seoul.
Neither of the persons who work in Auckland and Istanbul have the highest or the lowest CTC. The sum of the CTC of these two persons is equal to the sum of the CTC of the persons who are natives of Hisar and Patiala.
The CTC of Edwin, who works in Rome, is less than the CTC of the person from Agra.
The person with the lowest CTC is from Siliguri and works in Delhi.
The CTC of the person working in Istanbul is Rs. 35 lakhs.
The CTC of each of Siddhi, Parag, Azam and the person who is from Agra is less than the CTC of Roopa.

Q16MCQ

What is the sum of the CTC of the persons who are natives of Erode and Rourkela?

ARs. 65 Lakhs

BRs. 70 Lakhs

CRs. 75 Lakhs

DCannot be determined

Solution Answer: A

Delhi / Siliguri / 20. This person can’t be Siddhi (initial S), Azam (Seoul), Edwin (Rome) or Roopa (she out-earns four people). So it is Parag or Harsh.
Edwin’s native place. It can’t be Erode (E), Rourkela (he works in Rome), Agra (he earns less than the Agra person) or Siliguri. So Edwin is from Patiala or Hisar.
Roopa. She earns more than Siddhi, Parag, Azam, the Agra person and therefore Edwin, so her CTC is 40 or 45. She can’t be in Istanbul (35).
If Roopa = 45: she can’t work in Auckland (highest CTC), so she works in Edinburgh, and her native place must be Patiala or Hisar. Hisar + Patiala = Edwin + 45 = Auckland + 35, so Auckland = Edwin + 10. That forces Auckland = 40 and Edwin = 30, leaving 25 for Azam.
If Roopa = 40: only Harsh can earn 45, so Harsh works in Edinburgh and Roopa in Auckland (40). Then Hisar + Patiala = 75 = Edwin + 45, so Harsh is from Patiala/Hisar, Edwin = 30 and Azam = 25.
Agra. The Agra person earns more than 30 and less than Roopa, and can’t work in Auckland (initial A). In both cases that makes the Agra CTC 35.

Total CTC = 195, so Erode + Rourkela = 195 − 20 (Siliguri) − 75 (Hisar + Patiala) − 35 (Agra) = Rs. 65 lakhs.

Q17MCQ

Who works in Istanbul?

ARoopa

BSiddhi

CParag

DHarsh

Solution Answer: D

Detailed solution coming soon.

Q18MCQ

What is the CTC of the person who works in Seoul?

ARs. 35 Lakhs

BRs. 30 Lakhs

CRs. 25 Lakhs

DRs. 20 Lakhs

Solution Answer: C

Azam works in Seoul. As shown in Q16, both possible cases for Roopa give Edwin = 30 and leave Rs. 25 lakhs for Azam.

Q19TITA

What is the CTC (in Rs. Lakhs) of Siddhi?

Solution Answer: 40

Detailed solution coming soon.

Q20MCQ

Which of the following additional information will enable us to uniquely determine the hometown, working city and CTC of each of them?

AThe CTC of the person working in Rome is less than that of the person who is from Rourkela.

BThe person who is from Rourkela earns more than the person who is from Patiala.

CThe person who is from Erode earns less than the person who is from Rourkela.

DThe person who is from Hisar earns more than the person who is from Erode.

Solution Answer: B

Detailed solution coming soon.

QA · Quantitative Ability

Q21 – Q30 0/10

Q21TITA

Find the sum of all the possible values of p such that p⁴⁸ − p³³ has the unit’s digit as 2, where 20 ≤ p ≤ 30.

Solution Answer: 53

Unit digits repeat in cycles of 4. Since 48 ≡ 0 and 33 ≡ 1 (mod 4), compare the unit digit of d⁴ with d itself, where d is the unit digit of p.

d = 0, 1, 5, 6: difference ends in 0.
d = 2: 6 − 2 → 4. d = 3: 1 − 3 → 8. d = 7: 1 − 7 → 4. d = 8: 6 − 8 → 8.
d = 4: 6 − 4 → 2 ✓. d = 9: 1 − 9 → 2 ✓.

So p = 24 or 29, and the sum is 53.

Q22TITA

How many values of x satisfy the system of equations |2 − y| = 3 and 2|(|x| − 3)| = y + 1?

Solution Answer: 5

|2 − y| = 3 gives y = 5 or y = −1.

y = 5: ||x| − 3| = 3, so |x| = 6 or 0, giving x = 6, −6, 0.
y = −1: ||x| − 3| = 0, so |x| = 3, giving x = 3, −3.

Total = 5 values of x.

Q23MCQ

Sarika takes 27 seconds to ride an escalator when walking at her normal speed in the same direction. When she doubles her speed in the same direction, it takes her 18 seconds. If Sarika stands still on the escalator, how long will it take her to complete the ride?

A36 seconds

B48 seconds

C45 seconds

D54 seconds

Solution Answer: D

Let the escalator length be L, her speed v and the escalator’s speed e. Then v + e = L/27 and 2v + e = L/18.

Subtracting: v = L/18 − L/27 = L/54, so e = L/27 − L/54 = L/54.

Standing still, time = L/e = 54 seconds.

Q24MCQ

In the given figure, the area of each regular hexagon is 12 sq cm. What is the area of the square?

Three identical regular hexagons meeting at a common point, enclosed in a square that touches their outer vertices and edges

A56 sq cm

B54 sq cm

C48 sq cm

D45 sq cm

Solution Answer: A

Let each hexagon have side a. Area = 3√32a² = 12, so a² = 8√3.

The three hexagons meet at one point. The left hexagon spans 2a horizontally, and the two right hexagons reach 1.5a beyond the meeting point. So the side of the square = 3.5a.

Area = 12.25a² = 12.25 × 8√3 = 98√3 ≈ 56.6, so the closest option is 56 sq cm.

Q25MCQ

A person had borrowed some money at the rate of 10% interest. At the end of the first year, he paid Rs 10000, and the rest of the amount was paid in the second year. The rate of interest in the second year was 8%. The ratio of the interest which he paid in the first year to that in the second was 3:2. Find out the amount that was borrowed.

ARs 45000

BRs 40000

CRs 37500

DRs 52500

Solution Answer: C

Year 1 interest = 0.1P. Amount after year 1 = 1.1P, and after paying 10000, 1.1P − 10000 remains.

Year 2 interest = 0.08(1.1P − 10000). So 0.1P : 0.08(1.1P − 10000) = 3 : 2, which gives 0.2P = 0.264P − 2400.

0.064P = 2400, so P = Rs 37500.

Q26TITA

For natural numbers a, b, and c, let f(x) = ax − 3 and g(x) = x^b + c, where x ∈ R. If (fog)⁻¹(x) = (x − 72)^1/3, then (gof)(b) is equal to

Solution Answer: 32

Inverting (fog)⁻¹: (fog)(x) = 2x³ + 7.

Also (fog)(x) = a(x^b + c) − 3 = ax^b + ac − 3. Comparing coefficients: a = 2, b = 3, and 2c − 3 = 7, so c = 5.

(gof)(3) = g(f(3)) = g(2·3 − 3) = g(3) = 3³ + 5 = 32.

Q27MCQ

The probability of success in an event is 40%. What is the probability of at least 1 success in 4 such events? (Rounded off to the nearest integer)

A80%

B82%

C85%

D87%

Solution Answer: D

P(no success) = 0.6⁴ = 0.1296.

P(at least one) = 1 − 0.1296 = 0.8704 ≈ 87%.

Q28MCQ

Ramya had kept three jugs, A, B and C, in her kitchen that contained 400 ml each of carrot, papaya and orange juice respectively. Her son poured 100 ml of juice from jug A to B, mixed the contents and poured 100 ml of the mixture into jug C. He then poured 100 ml of the mixture from jug C to jug A. What is the ratio of the final quantity of orange juice in jug A to that of papaya juice in jug C?

A6:5

B5:4

C4:3

D3:2

Solution Answer: B

A → B: B = 400 papaya + 100 carrot (500 ml). 100 ml of this = 80 papaya + 20 carrot.

B → C: C = 400 orange + 80 papaya + 20 carrot (500 ml). 100 ml of this = 80 orange + 16 papaya + 4 carrot.

C → A: A gains 80 ml orange. C keeps 80 − 16 = 64 ml papaya.

Ratio = 80 : 64 = 5 : 4.

Q29MCQ

If 2^x = 4^y = 8^z and xyz = 288, evaluate 12x + 14y + 18z

A1196