APPSC Group-1 2017 Mental Ability Previous Year Questions

Ans: (1) Anusha, Pradip, Rajesh, Samir (as per APPSC official final key)

The conditions are:

• P and S must be together.
• Q cannot be with B or C.
• R and A must be together.
• T cannot be with A or D.

Q is in the team, so B and C are out. If T were in, A and D would be out, so R (who needs A) would also be out. That leaves only Q, T, P and S, which is four people, not five. So T is out. The remaining four members must come from P, S, R, A and D. Since P–S and R–A go in pairs, the team is Q, A, P, R, S, which is option (1).

Note:

• In such puzzles, first apply the ‘must be together’ pairs, then remove the forbidden combinations.

Ans: (3) Bhagya, Chinmayi, Durga, Pradip, Samir (as per APPSC official final key)

The conditions are:

• P and S must be together.
• Q cannot be with B or C.
• R and A must be together.
• T cannot be with A or D.

Check each option:

• (1) A, B, C, R, T – T is with A, which is not allowed.
• (2) A, C, D, R, S – S is without P, which is not allowed.
• (4) A, B, C, D, R – this has four girls, not three.
• (3) B, C, D, P, S – three girls, with P and S together, and no Q, T, R or A. All the rules are satisfied.

So the team is B, C, D, P, S.

Note:

• Check every condition for each option; one broken rule rules it out.

Ans: (2) Pradip, Quddus, Rajesh, Samir (as per APPSC official final key)

The conditions are:

• P and S must be together.
• Q cannot be with B or C.
• R and A must be together.
• T cannot be with A or D.

A is in, so R must be in. T cannot be with A, so T is out. The four boys must then be chosen from P, Q, R and S, and they are exactly four: P, Q, R, S. Q has no conflict, because B and C are absent, and P–S are together.

Ans: (4) Chinmayi, Durga, Pradip, Rajesh, Samir (as per APPSC official final key)

The conditions are:

• P and S must be together.
• Q cannot be with B or C.
• R and A must be together.
• T cannot be with A or D.

Option (4) has R without A. Since R and A must be together, the team C, D, P, R, S is not possible. The other options satisfy all the conditions.

Ans: (4) 7·33, 8, 8 (as per APPSC official final key)

Arrange the numbers in order: 2, 3, 5, 7, 8, 8, 9, 10, 14 (9 numbers).

• Mean = sum ÷ count = 66 ÷ 9 = 7.33
• Median = the middle (5th) value = 8
• Mode = the most frequent value = 8

Note:

• For an even count, the median is the average of the two middle values.
• Empirical relation: Mode ≈ 3 Median − 2 Mean.

Ans: (2) 7·7 kg (as per APPSC official final key)

For grouped data, mean = Σ(f × x) ÷ Σf, where x is the class mid-point.

• 3–5: mid 4 × 8 = 32
• 5–7: mid 6 × 25 = 150
• 7–9: mid 8 × 45 = 360
• 9–11: mid 10 × 18 = 180
• 11–13: mid 12 × 4 = 48

Σfx = 770, Σf = 100. So mean = 770 ÷ 100 = 7.7 kg.

Note:

• The assumed-mean (step deviation) method gives the same answer faster.

Ans: (1) 13, 9, 12, 11 (as per APPSC official final key)

Each letter is replaced by its position in the alphabet: C = 3, O = 15, F = 6, F = 6, E = 5, E = 5.

So for MILK: M = 13, I = 9, L = 12, K = 11, which gives 13, 9, 12, 11.

Note:

• Useful trick: EJOTY = 5, 10, 15, 20, 25.
• Reverse position = 27 − position (A = 26 … Z = 1).

Ans: (3) Cube (3) in the figure (as per APPSC official final key)

In this question the net of a cube is shown, and you must find which of the four cubes can be formed from it. When a net is folded, faces that are next to each other in the net become adjacent on the cube, and faces separated by one face in a straight line become opposite faces. Opposite faces can never be seen together on a cube. Applying these rules to the figure, cube (3) is the one that can be formed, as per the official key.

Note:

• Opposite faces rule: in a row of faces in the net, alternate faces (1st and 3rd) are opposite.
• Two opposite faces never appear together in a cube view.

Ans: (2) 3 (as per APPSC official final key)

In each square, the centre number = (top-left × bottom-right) − (top-right × bottom-left).

• First square: 9 × 4 − 8 × 4 = 36 − 32 = 4 ✓
• Second square: 7 × 8 − 8 × 2 = 56 − 16 = 40 ✓
• Third square: 3 × 9 − 3 × 8 = 27 − 24 = 3

Note:

• Try sums, differences and products of the diagonal pairs first in such figure puzzles.

Ans: (4) 216 (as per APPSC official final key)

The terms are multiplied alternately by 2 and 3: 1 ×2 = 2, ×3 = 6, ×2 = 12, ×3 = 36, ×2 = 72, ×3 = 216.

Note:

• Common series patterns: alternate operations, squares/cubes, differences in AP.

Ans: (2) Grief (as per APPSC official final key)

A sedative is a medicine that calms the body and relieves pain. In the same way, solace (comfort) relieves grief (sorrow). So the relation is “remedy and what it relieves”.

Note:

• Similar pair: Antidote : Poison.

Ans: (2) No (as per APPSC official final key)

From the chart, the heights are: Bio-fertilizer 21 cm, DAP 17, Super phosphate 7, Urea 11, and No fertilizer 12.

With super phosphate (7) and urea (11), the plants grew less than with no fertilizer (12). So fertilizer is not always beneficial, and the answer is No.

Note:

• In data-interpretation questions, look for the ‘control’ value (no fertilizer) to compare.

Ans: (4) 75% (as per APPSC official final key)

Bio-fertilizer gave 21 cm and no fertilizer gave 12 cm.

Increase = 21 − 12 = 9 cm. Percentage increase = 9 ÷ 12 × 100 = 75%.

Note:

• Percentage increase is always calculated on the original (base) value.

Ans: (4) Data is insufficient (as per APPSC official final key)

The chart shows only the effect of each fertilizer used separately. There is no information about using urea and super phosphate together. So no conclusion can be drawn: the data is insufficient.

Note:

• Do not assume information that the data does not show.

Ans: (1) Yes (as per APPSC official final key)

Super phosphate (7 cm) and urea (11 cm) gave less growth than no fertilizer at all (12 cm). So we can safely conclude that certain fertilizers may not be suitable for bean plants. The answer is Yes.

Note:

• Beans are legumes; they fix nitrogen with Rhizobium bacteria, so extra urea is often not needed.