May 28, 2021 - CSAT Quiz
1. A rabbit on a controlled diet is fed daily 300 grams of a mixture of two foods, food X and food Y. Food X contains 10 percent protein and food Y contains 15 percent protein. If the rabbits diet provides exactly 38 grams of protein daily, how many grams of food X are in the mixture?.
(a) 100
(b) 140
(c) 150
(d) 160
2. In how many different ways can the letters of the word "CORPORATION" be arranged in such a way that no two vowels are together?
(a) 50400
(b) 5040
(c) 144800
(d) 1080
3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?.
(a) 256
(b) 756
(c) 956
(d) 456
4. In how many ways can 5 examination papers be arranged so that the best and the worst papers never come together?
(a) 72
(b) 144
(c) 20
(d) 15
5. If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?
(a) 1 : 6
(b) 6 : 1
(c) 5 : 1
(d) 1 : 5
Answers
1 – (b)
x= mixture A ,, then 300-x is Mixture B
0.1 x + (300-x)0.15=38
x=140
2 – (a)
Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 letters, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400
3 – (b)
Since at least 3 men must be chosen, we consider all committees which include 3, 4, and 5 men, with 2, 1, and 0 women, respectively. That is, we want to add the number of ways to:
a. Choose 3 from 7 men and 2 from 6 women = (7C3)*(6C2) = 35*15 = 525
b. Choose 4 from 7 men and 1 from 6 women = (7C4)*(6C1) = 35*6 = 210
c. Choose 5 from 7 men and 0 from 6 women = (7C5)*(6C0) = 21*1 = 21
a+b+c= 756.
4 – (a)
No. of ways in which 5 paper can arranged is 5! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 4 papers.
These 4 papers can be arranged in 4! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 5!- 4!.2! = 4!(5-2) = 24 x 3 = 72.
(n-2)(n-1)! = (5-2)(5-1) = 72
5 – (b)
Numbers of Options applicable for 5 letter digit -> 10∗9∗8∗7∗610∗9∗8∗7∗6
( as option pool for first digit is 10, for second 9 because one is removed and so on)
Numbers of Options applicable for 5 letter digit -> 10∗9∗8∗710∗9∗8∗7
Required Ratio -> (10∗9∗8∗7∗6)/(10∗9∗8∗7)(10∗9∗8∗7∗6)/(10∗9∗8∗7) = 6:1
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