100+ TNPSC Simplification Aptitude Questions and Answers - 1

Question: 1

- 76 × 33 + 221 = ?

(A) 2287

(B) -2287

(C) -19304

(D) 19304

Ans: B

Given expression = -2508 + 221 = -2287.

Question: 2

74 is divided into two parts so that 5 times one part and 11 times the other part are together equal to 454. The parts are

(A) 14, 60

(B) 30, 44

(C) 44, 30

(D) 60, 14

Ans: D

Let the two parts be x and (74 - x).

Then, 5x + 11(74 – x) = 454 ⇔ 6x = 360 ⇔ x = 60.

So, the two parts are 60 and 14.

Question: 3

$$[{125 × 125} / {50} × 20]$$ ÷ 25 = ?

(A) 100

(B) 120

(C) 150

(D) 250

Ans: D

Given expression = $$[{125 × 125} / {50} × 20]$$ ÷ 25 = 6250 ÷ 25 = 250.

Question: 4

A class starts at 10 a.m. and lasts till 1.27 p.m. Four periods are held during this interval. After every period, 5 minutes are given free to the students. The exact duration of each period is

(A) 42 minutes

(B) 48 minutes

(C) 51 minutes

(D) 53 minutes

Ans: B

Time between 10 a.m. and 13.27 hours = 3 hrs. 27 min. = 207 min.

Total duration of free time = (5 × 3) min = 15 min.

Remaining time = (207 – 15) min. = 192 min.

∴ Duration of each of the 4 periods = $$({192}/{4})$$ min. = 48 min.

Question: 5

The marks scored in an examination are converted from 50 to 10 for the purpose of internal assessment. The highest marks were 47 and the lowest were 14. The difference between the maximum and minimum internal assessment scores is

(A) 3.3

(B) 4.8

(C) 6.6

(D) 7.4

Ans: C

Maximum internal assessment score = $$({47}/{50} × 10)$$ = 9.4

Minimum internal assessment score = $$({14}/{50} × 10)$$ = 2.8

∴ Required difference = (9.4 – 2.8) = 6.6.

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