Practice Set — TITA (Type In The Answer)

Eight questions with no options. You type the answer, and a wrong typed answer costs you nothing — which makes the arithmetic of attempting completely different from the MCQ half of the paper. On an MCQ at +3/−1 a blind guess has negative expected value; on a TITA question every attempt has non-negative expected value, so the only question worth asking is whether the minute is better spent elsewhere. Two habits are worth building here. First, never leave a TITA question blank if you have a number — even a half-finished computation is free to submit. Second, type the form the question asked for: a question that says "in ₹ lakh" is not answered by the figure in crore, and an answer of 5 typed into a question that asked for a percentage is right while 0.05 is wrong. Nothing below needs a calculator, and the on-screen calculator the real exam gives you would slow most of them down.

How a TITA Question Is Attempted

  1. Read what form the answer must take before starting — a count, a percentage, an amount, a number of solutions. Most TITA marks are lost to a correctly computed number typed in the wrong unit.
  2. With no options there is nothing to work backwards from, so estimation and substitution stop being shortcuts. Expect to actually solve, and budget the minute accordingly.
  3. Type an integer as an integer. Both 12 and 12.00 are accepted here and by the real exam, but a stray unit or word in the box is not a number and will not be read as one.
  4. Answer every one you have touched. There is no penalty, so a blank box is a mark you decided not to have.

Key takeaways

  • A wrong TITA answer costs nothing, so any number beats a blank. This single asymmetry is the most valuable thing to know about the CAT paper’s structure.
  • With no options, backsolving is unavailable — a TITA question generally takes longer than an MCQ of the same topic, and should be triaged on that basis.
  • x + 1/x = k gives x² + 1/x² = k² − 2 directly. Squaring the given expression is faster than finding x.
  • |x − a| + |x − b| equals the distance between a and b for every point between them, so such an equation has infinitely many real solutions and a countable number of integer ones.
  • For a sum of terms in AP, find the first and last member of the range first; the count follows from the difference, and the sum from the average of the ends.
  • A two-set overlap question is one equation: |A ∪ B| = |A| + |B| − |A ∩ B|, with the "neither" count taken out of the universe first.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. If x + 1/x = 4, what is the value of x² + 1/x²?

    Numerical answer — type the value.

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    Answer: 14

    Square the given expression: (x + 1/x)² = x² + 2 + 1/x², so x² + 1/x² = 4² − 2 = 14. The wasted route is solving x² − 4x + 1 = 0 for x = 2 ± √3 and then squaring an irrational number — it gives 14 as well, but takes four times as long and invites an arithmetic slip. Note that the same identity run once more gives x⁴ + 1/x⁴ = 14² − 2 = 194, which is the usual follow-up.
  2. How many integer values of x satisfy |x − 3| + |x + 2| = 5?

    Numerical answer — type the value.

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    Answer: 6

    Read the expression as a distance: |x − 3| + |x + 2| is the distance from x to 3 plus the distance from x to −2. Those two points are 5 apart, so the sum equals 5 for every x between them (inclusive) and exceeds 5 outside. The integers in [−2, 3] are −2, −1, 0, 1, 2, 3 — 6 of them. The trap is to solve it as an equation in three cases and report the two endpoints only, giving 2; the trap in the other direction is to answer "infinitely many", which is true of the real solutions but not of the integers the question asked for.
  3. What is the sum of all three-digit natural numbers that are divisible by 17?

    Numerical answer — type the value.

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    Answer: 28832

    Find the ends first. The smallest three-digit multiple of 17 is 17 × 6 = 102; the largest is 17 × 58 = 986. So the terms are 17 × 6 through 17 × 58 — that is 58 − 6 + 1 = 53 terms in AP. Their sum is 53 × (102 + 986)/2 = 53 × 544 = 28,832. Two slips are common: starting at 17 × 5 = 85, which is not three-digit, and taking the count as 58 − 6 = 52, which forgets that both ends are included.
  4. How many four-digit numbers can be formed in which every digit is odd and no digit repeats?

    Numerical answer — type the value.

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    Answer: 120

    The odd digits are 1, 3, 5, 7, 9 — five of them, and none is 0, so the leading-digit restriction that usually complicates these questions does not apply here. Choosing four of the five in order is 5 × 4 × 3 × 2 = 120, which is ⁵P₄. The common error is 5⁴ = 625, which allows repetition; the other is to subtract a case for a leading zero, of which there are none because 0 is even.
  5. If log₂ x + log₄ x = 6, what is the value of x?

    Numerical answer — type the value.

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    Answer: 16

    Put both logs on the same base. log₄ x = log₂ x / log₂ 4 = (1/2)·log₂ x, so the equation is log₂ x + (1/2)log₂ x = 6, that is (3/2)·log₂ x = 6, so log₂ x = 4 and x = 2⁴ = 16. Check: log₂ 16 = 4 and log₄ 16 = 2, and 4 + 2 = 6. The usual mistake is adding the arguments rather than converting the bases, which turns the left side into log of a product and gives a number that satisfies nothing.
  6. What is the area, in square units, of a triangle whose sides measure 13, 14 and 15 units?

    Numerical answer — type the value.

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    Answer: 84

    Heron’s formula: s = (13 + 14 + 15)/2 = 21, and the area is √(21 × 8 × 7 × 6) = √7056 = 84. The arithmetic is friendlier than it looks — 21 × 8 = 168, 7 × 6 = 42, and 168 × 42 = 7056, whose square root is exactly 84. This triangle is worth memorising, because CAT reuses it: sides 13, 14, 15 give area 84, which makes the altitude to the side of length 14 exactly 2 × 84 / 14 = 12. Checking that the sides satisfy the triangle inequality first is a habit worth keeping, because a "triangle" with sides 3, 4 and 9 has no area and Heron will hand you the square root of a negative number rather than an error.
  7. In a group of 100 people, 60 drink tea, 50 drink coffee and 20 drink neither. How many drink both?

    Numerical answer — type the value.

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    Answer: 30

    Take the "neither" group out of the universe first: 100 − 20 = 80 people drink at least one of the two, so |T ∪ C| = 80. Then 80 = 60 + 50 − |T ∩ C|, giving |T ∩ C| = 110 − 80 = 30. The commonest wrong answer is 110 − 100 = 10, which forgets that 20 of the hundred are outside both circles and so uses the wrong union. A quick sanity check: with 30 in both, 30 drink only tea and 20 only coffee, and 30 + 20 + 30 + 20 = 100.
  8. The line 3x + 4y = 24 cuts the x-axis at A and the y-axis at B. What is the area, in square units, of triangle OAB, where O is the origin?

    Numerical answer — type the value.

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    Answer: 24

    Set y = 0 for the x-intercept: 3x = 24, so A = (8, 0). Set x = 0 for the y-intercept: 4y = 24, so B = (0, 6). The triangle is right-angled at the origin, so its area is (1/2) × 8 × 6 = 24. The general result is worth carrying: a line x/a + y/b = 1 cuts off a triangle of area |ab|/2 with the axes, and 3x + 4y = 24 is x/8 + y/6 = 1. The trap is to read the coefficients 3 and 4 as the intercepts and answer 6.