Free Mixed Bag 13 Practice Test - CAT
Question 1
How many solutions are there for the equation A+B+3C = 10 if A,B,C are whole numbers?
SOLUTION
Solution : D
Option (d)
It is given that A,B and C can take values ≥0
Give values to C from 0
At C=0, the equation changes to A+B=10. Number of solutions will be based on
arrangement of 10 zeroes and 1 one = 11C1=11
At C=1, the equation changes to A+B=7. Number of solutions= 8C1=8
At C=2, the equation changes to A+B=4. Number of solutions= 5C1=5
At C=3, the equation changes to A+B=1. Number of solutions= 2C1=2
Total number of solutions = 11+8+5+2 = 26
Question 2
A grocer uses a faulty balance that balances 10kg in one pan with 11 kg in the other. He sells his product after hiking the cost price by 10%. If he cheats the farmer while buying and his customer while selling, find his actual gain in the entire transaction.
SOLUTION
Solution : D
When he cheats the farmer, on buying 10 kg, he is actually buying 11 kg at the price of 10. he always makes a profit of 10%
S.P=1.1×CP
Similarly, while selling, when he sells 10 kg, if the cost price is 10 rupees, he is selling it for 11 rupees.
Again now, SP=1.1×1.1×CP
He also professes to sell 11 kg, when he is actually selling 10 kg
SP=1.1×1.1×1.1×CP⇒33.1% overall profit
Answer is option (d) 33%
Question 3
Find the number of degrees in arc AB of the figure below if segments PA and PB are secants; AEC = 115∘, BFD = 130∘, and P = 30∘
SOLUTION
Solution :Angle AEC=115. Suppose there is a point X in the major segment of chord AC. AECX will be a cyclic quadrilateral. Therefore , Angle AEC + Angle AXC =180
⇒ Angle AXC= 65
⇒ Angle AOC=130 (Measure of arc AC at the centre O)
Similarly Angle BOD = 100
Therefore, Angle AOB (or measure of arc AB) + Angle COD (or measure of arc CD)= 360 - (100+130) = 130 ---- (1)
Now we know the theorem that if there are two secants PA and PB and P lies outside the circle , then Angle P = 12 ( measure of arc COD - measure of arc AOB) =30 (given in the ques) -- (2)⇒ ( measure of arc CD - measure of arc AB) =60 -- (3)
You can see that equations (1) and (3) are simultaneous equations in measure of arc AB and measure of arc CD .So, finally measure of arc AB is (130+60)2 = 95.
Question 4
The area enclosed by the graph |x-1| + |y-1| = 2 is
SOLUTION
Solution : D
Option (d)
The above is the graph of |x| + |y| = 2. Area of the region enclosed = (product of diagonals)/2 = (4*4)/2 = 8.
Now we get |x-1|+|y-1| =2 by shifting the above graph but the area enclosed will remain same as the size and shape is not changing. So enclosed area = 8
Question 5
In an examination, 53 passed in Maths, 61 passed in Physics, 60 in Chemistry, 24 in Maths & Physics, 35 in Physics & Chemistry, 27 in Maths & Chemistry and 5 in none. Total number of students who had appeared in the examination was 100.
Then, the number of students who passed in all subjects is
SOLUTION
Solution :Option (c)
S = I + 2II + 3III = 53 + 61 + 60 = 174 --------- (a)
Given that II +3III = 24 + 35 + 27 = 86--------------(b)
(a)-(b) gives us the I +II= 88
Out of 100 students, 5 have passed in none
Therefore, the number of students who have passed in all 3 = 95-88 = 7
Question 6
Find N, given that (AX)² = NPA, where A,X,N,P are distinct positive integers
SOLUTION
Solution : B
option(b)
AX is two digit number between 11 and 31. Since squares cannot end it 2 and 3, A cannot be 2 or 3 hence A =1. The only number between 11-19 where the condition holds is 192= 361. answer is option (b)
Question 7
In a class of 120 students numbered 1 to 120, there are 3 extracurricular activities to choose from. All even numbered students opt for Dance, whose numbers are divisible by 5 opt for Music and those whose numbers are divisible by 7 opt for Debate. How many opt for none of the three subjects?
SOLUTION
Solution : B
option (b)
A is the set of those who opted for Dance = 1202=60 students.
B is the set of those who opted for Music = 1205=24.
C is the set of those who opted for Debate = 1207=17.
The 10th, 20th, 30th…… numbered students would have opted for both Dance and Music. 12010=12.
The 14th, 28th, 42nd…. Numbered students would have opted for Dance and Debate.= 12014=8.
The 35th, 70th …. Numbered students would have opted for Music and Debate = 12035=3.
And the 70th numbered student would have opted for all three subjects.
S=101, II+3III = 23
III= 1, S-X= 20+2
101-22 = X
X=79
Therefore, number of students who did not opt for any of the three activities = 41
Question 8
f(f(f(k))) = 27. What is the sum of the digits of k ?
SOLUTION
Solution : B
Option (b)
Since k is odd then f(f(f(k))) =27
=> f(f(k+3)) = 27------------(1) since k is odd , k+3 will be even.
Hence, expanding (1) we get,
--------------(2)
Here
can either be even or odd
Case 1: let
Then (2) reduces to
here sum of digits is 6
Case 2: let
Then (2) reduces to
=> k = 45
Substituting in
k = 105 here sum of digits is 6 is the only solution.
Hence, choice (b) is the correct answer.
Question 9
A box has a ball and everyday starting with today two balls are added to the box for every ball present in it. After 4 weeks the balls are distributed equally among five boxes. The remaining balls are thrown away. How many were thrown away?
SOLUTION
Solution : B
option (b)
On the first day there will be three balls, on second day there will be 9, on the third day 27.
Therefore on the 28th day =328balls.We need to find the remainder when divided by 5, 28= 4k. Therefore, 1 ball will be thrown away.
Question 10
Solve for a , 
SOLUTION
Solution : A
Option (a)
Check for a=3,
it is satisfied. Option (b) and (e) ruled out
Check for a=1,
is satisfied. Option (c) is out
Check for a =4,
, not satisfied. Option (d) is out