TS ICET 2020 Question Paper Shift-1 (1st Oct)

Instructions

For the following questions answer them individually

TS ICET 2020 Shift-1 (1st Oct) - Question 71


A person starts from his house and walks 80 meters towards West, then he walks 60 meter meters towards his left, then 80 meters towards West and then 40 meters towards South, finally, he turns right and walks 50 meters. What is the horizontal distance covered by him from his house? (in meters)

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TS ICET 2020 Shift-1 (1st Oct) - Question 72


Four persons A, B, C and D live in different towns which are located as
i) B is to the South West of A;
ii) C is to the East of B and is also to the South East of A:
iii) D is to the North of C in line with B and A;
What is the direction of A relative to D?

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TS ICET 2020 Shift-1 (1st Oct) - Question 73


If $$a* b = ab +\frac{1}{ab} + a + \frac{1}{b}$$ then $$(-1) * (2 * 1) = $$

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TS ICET 2020 Shift-1 (1st Oct) - Question 74


If $$a \triangle b = \frac{a}{b} + \frac{b}{a}$$ and $$a * b = ab + \frac{1}{ab}$$ then $$(1 \triangle 2) * 4 =$$

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TS ICET 2020 Shift-1 (1st Oct) - Question 75


If $$a \oplus b = a^{2} + b^{2}, a \ominus b = a^{2} - b^{2}$$ then $$(2 \oplus 3) \ominus 13 = $$

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TS ICET 2020 Shift-1 (1st Oct) - Question 76


If $$a (\sqrt[3]{x^{2}}) + a(\sqrt[3]{x}) + c = 0$$ then $$a^{3}x^{2} + b^{3}x + c^{3} = $$

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TS ICET 2020 Shift-1 (1st Oct) - Question 77


If s, t are rational numbers such that $$(3^{t-2})$$ X $$(2^{s+3})^{3} = 432$$, then 4t - 5s=

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TS ICET 2020 Shift-1 (1st Oct) - Question 78


A bag contains the coins of denomination of Rs.5. Rs.2, and Rs.1 in the ratio of 2 : 3 : 5 respectively. If total value of the money in the bag is Rs.2100, then how many Rs.5 coins are there in the bag?

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TS ICET 2020 Shift-1 (1st Oct) - Question 79


The shares of two persons P and Q in a company are in the ratio 3 : 7. The company have each of them 20 additional shares, as a result of which, the ratio got altered to 1 : 2. How many shares of company does Q now hold?

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TS ICET 2020 Shift-1 (1st Oct) - Question 80


$$x = \sqrt[3]{p + \sqrt{p^{2} + q^{3}}} + \sqrt[3]{p - \sqrt{p^{2} + q^{3}}}$$, then $$x^{3} + 3qx - 2p = $$

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