Question 1
At what time between 9 o’clock and 10 o’clock will the hands of a clock be in the same straight line but not together?
(A) 16.36 minutes after 9
(B) 16.18 minutes after 9
(C) 15.18 minutes after 9
(D) 15.27 minutes after 9
Question 2
If MK = 169/121, then JH = ?
(A) 100/64
(B) 100/81
(C) 64/100
(D) 81/100
Question 3
Six children A, B, C, P, Q and R are standing in two rows in such a way that each child in one row is facing a child in another row. P is not at the end of any row and is to the right of Q, who is facing C. R is to the left of B, who is facing P. Which of the following groups of children is in the same row?
(A) ACP
(B) CBQ
(C) RBP
(D) None of these
Question 4
What is the face opposite to A?
(A) H
(B) P
(C) B
(D) M
Question 5
If it was Saturday on December 17, 1899, then what will be the day on December 22, 1901?
(A) Thursday
(B) Friday
(C) Saturday
(D) Sunday
N.B. Please note that the official timings for publication of these questions is 11 am daily.
SOLUTIONS TO DAILY CSAT MISSION # 80
1. (D) 2. (C) 3. (B) 4. (D) 5. (A)
Explanations
1. If the HCF is 21, both the numbers have to be multiples of 21.
Now, one of the numbers lies between 200 and 300. But the only numbers that are divisible by 21 and lie between 200 and 300 are 210, 231, 252, 273 and 294. So, our first number is one of these five.
We also know that LCM x HCF = (First number) x (Second number). Since LCM x HCF is an odd number, both the first and second numbers have to be odd numbers too. This rules out 210, 252 and 294, leaving our first number to be either 231 or 273.
4641 is not divisible by 231 (you manually find this out). Thus our first number is 273 and the second number is 4641*21/273 = 357.
OR
You can take the elimination route to solving this problem. Here also, the things to remember are that both numbers have to be divisible by 21, and that LCM x HCF = (First number) x (Second number).
In option (A), 351 is not divisible by 21. So this option is discarded.
In option (B), both are divisible by 21. But the product of the two numbers will end in 7 while LCM x HCF has last digit as 1. So this option is discarded.
In option (C), both are divisible by 21. And, the product of the two numbers will also end in 1. So this can be the solution.
In option (D), both are divisible by 21. And, the product of the two numbers will also end in 1. So this can also be the solution.
Since there are two different options both of which can be the answer, manually multiply them to find out in which case, LCM x HCF = (First number) x (Second number).
2. Let the cost price be ‘x’. Then, selling price in case of 25% loss = 0.75x and selling price in case of 10% profit = 1.1x. Since 1.1x – 0.75x = 21, x = 60.
3. There are two intervals between strikes when the clock strikes 3 o’clock. Time taken for these two intervals = 4 seconds, so each interval is 2 seconds long.
When it will strike 9 o’clock, there will be 8 intervals. So the time taken for striking 9 o’clock will be 16 seconds.
4. The statements in case of option D can be summarised as:
None of the other options give such a comprehensive description.
5. The first statement gives Priya > Vandana. So till now, Vandana is the shortest. The second statement gives Shilpa > Vandana. So Vandana is still the shortest. The third statement gives no indication to anything that is known, so we put it aside for the time being. The fourth statement gives Vandana > Tista. So now, Tista becomes the shortest. We come back to the third statement. It gives Archana > Tista which means Tista stays the shorest. The fifth statement indicates Shilpa > Archana and since Archana > Tista, so Shilpa > Tista. Tista will continue to remain the shortest.
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