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CAT 2018 quants section lesson 1
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Lesson 1 of the QA CAT 2018 course is about a question from mensuration and another interesting one from the chapter of pipes and cisterns.

## Arun Sharma is teaching live on Unacademy Plus

Arun Sharma
An IIM B pass out, an author and trainer for CAT, CSAT & other aptitude exams having trained lacs of students over the past twenty years.

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Ts
Sir can you please show us the solution of 3rd question again. It is not clear.
the 2nd ques ans will be 66mins
Sir Publish the updated LR-DI books asap plz ; Much needed for CAT 2019 .TIA
1. CAT 2018 Slot 2 Lesson 1

2. Arun Sharma IIM Bangalore Alumnus CAT 2018 99.99 percentiler, Cracked CAT 19 times Author McGraw Hill

3. Q.1 From a rectangle ABCD of area 768 sq cm, a semicircular part with diameter AB and area 72 sq cm is removed. The perimeter of the leftover portion, in cm, is 1, 82 + 24 2. 88 12 3. 80 16n 4. 86 + 8

4. Area of semicircular portion AB= 72 AB 2 AB = 24 cm AB"AD= 768 768 AD-768/AB-24-32 CM perimeter of semicircular portion AB= (AB) = (24) 12 cm Perimeter of the leftover portion AD+ DC+ Bc+ perimeter of semicircular portion AB 32 24 32+ 12 = 88 + 12 cm. Hence, option b is correct.

5. Q.2 A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are

6. Let the efficiency of inlet of type A and inlet of type B be 'a' and 'b' respectively and the capacity of tank be 'c'. According to the question: (10a+45b)1/2- c & (8a+18b)c (10a+45b)1/2- (8a+18b) 10a +45b-16a+36b 6a-9b or 2a 3b or b- 2a/3 Capacity of the tank- 8a+ 18b- 8a+ 18(2a/3)-20a

7. Net efficiency of 7 inlets of type A and 27 inlets of type B-7a+ 27b- 7a+27(2a/3)-7a+ 18a- 25a. Required time to fill the tank- 20a/25a- 0.8 hours- 48 minutes.

8. Q.3 The smallest integer n such that 3. n3-11n2 +32n -28> 0 (n -7) (n 2)2> 0 (n -2)2 is always greater than 0. Hence, n- 7> 0 orn>7 So the smallest possible integer must be 8.