## Arun Sharma is teaching live on Unacademy Plus

CSAT PROBLEM SOLVING CSAT 2017 (Quantitative Aptitude) LESSON 1

ARUN SHARMA EX. FACULTY VAJIRAM AND RAVI AUTHOR McGRAW-HILL IIM BANGALORE ALUMNUS TRAINING FOR APTITUDE FOR OVER 2 DECADES

Back to Plus Complete CSAT Paper 2 for UPSC CSE Prelims 2019 Arun Sharma 198 followers unacademy 18th January to 22nd March 2019 This course by Arun Sharma covers the entire portion of CSAT-Basic Numeracy, Data Interpretation, Reasoning & Reading Comprehension. It would be very helpful for all UPSC CSE Prelims aspirants. The course will also be useful for other government exams involving Aptitude. This course would be conducted in English and notes will also be provided in English. 45 hours of live classes Private Discussion Forum Doubt clearing sessions and Live quizzes 5,500 includes 18% GST Apply for this Plus Course About this course O Last date to apply is 18th Jan In HIGH DEMAND. Cost increases on 22nd Dec. Live Sessions The course consists of 30 sessions of 1 hour 30 mins duration each. Session recording will be provided for all the sessions

1. The age of Mr. X last year was the square of a number and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again? (a) 42 (b) 38 (c) 25 (d) 16

If the age of Mr. X is 26 years Last year his age was 25 years which is square of 5. Next year his age will be 27 years, which is cube of 3 Next cubic number is 64. So, he has to wait for 64- 26 38 years. Hence, option (b) is correct.

2.How many numbers are there between 99 and 1000 such that the digit 8 occupies the units place? (a) 64 (b) 80 (c) 90 (d) 104

Between 99 to 1000 following numbers has 8 as its unit digit. 108, 118, 128, 138, ....998 The above series is an Arithmetic progression with common difference 10 Last term- First term+ (number of terms 1)common difference. 998 108+ (number of terms - 1) 10 89 Number of terms -1 Number of terms- 90. Hence, option (c) is correct.

3. If for a sample data Mean < Median < Mode then the distribution is (a) symmetric (b) skewed to the right (c) neither symmetric nor skewed (d) skewed to the left

If the distribution of data is skewed to the left, the mean is less than the median, which is often less than the mode. (mean < median <mode) Hence, option (d) is correct.

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