CAT 1991 QA
Paper year paper questions for CAT 1991 QA
Read the following information and answer the questions that follows:
Ghosh Babu deposited a certain sum of money in a bank in 1986. The bank calculated interest on the principal at 10 percent simple interest, and credited it to the account once a year. After the 1st year, Ghosh Babu withdrew the entire interest and 20% of the initial amount. After the 2nd year, he withdrew the interest and 50% of the remaining amount. After the 3rd year, he withdrew the interest and 50% of the remaining amount. Finally after the 4th year, Ghosh Babu closed the account and collected the entire balance of Rs. 11,000.
The initial amount in rupees, deposited by Ghosh Babu was:
 A.
25,000
 B.
75,000
 C.
50,000
 D.
None of these
Answer: Option C
Explanation :
Let us assume that Ghosh Babu had deposited Rs.100 initially.
Hence, had he deposited Rs.100 initially, he should have withdrawn Rs.22 at the end to close the account.
Since he withdrew Rs.11000, at the end, he should have initially deposited = 100/22 × 11,000 = Rs.50,000.
Hence, option (c).
Workspace:
The year, at the end of which, Ghosh Babu withdrew the smallest amount was:
 A.
First
 B.
Second
 C.
Third
 D.
Fourth
Answer: Option D
Explanation :
Consider the solution for first question of this set.
He withdrew the smallest amount after the 4^{th} year.
Hence, option (d).
Workspace:
The year, at the end of which, Ghosh Babu collected the maximum interest was:
 A.
First
 B.
Second
 C.
Third
 D.
Fourth
Answer: Option A
Explanation :
Consider the solution for first question of this set.
He collected the maximum interest after the 1^{st} year.
Hence, option (a).
Workspace:
The year, at the end of which, Ghosh Babu withdrew the maximum amount was:
 A.
First
 B.
Second
 C.
Third
 D.
Fourth
Answer: Option B
Explanation :
Ghosh Babu withdrew the maximum amount after the 2nd year.
Hence, option (b).
Workspace:
The total interest, in rupees, collected by Ghosh Babu was:
 A.
12,000
 B.
20,000
 C.
4,000
 D.
11,000
Answer: Option A
Explanation :
Consider the solution for first question of this set.
As seen from the table, the total interest collected by Ghosh Babu is Rs.24 on Rs.100.
Hence on Rs.50000, it would be 24/100 × Rs. 50,000 = Rs. 12,000
Hence, option (a).
Workspace:
Use the following information:
Prakash has to decide whether or not to test a batch of 1000 widgets before sending them to the buyer. In case he decides to test, he has two options: (a) Use test I ; (b) Use test II. Test I cost Rs. 2 per widget. However, the test is not perfect. It sends 20% of the bad ones to the buyer as good. Test II costs Rs. 3 per widget. It brings out all the bad ones. A defective widget identified before sending can be corrected at a cost of Rs. 25 per widget. All defective widgets are identified at the buyer’s end and penalty of Rs. 50 per defective widget has to be paid by Prakash.
Prakash should not test if the number of bad widgets in the lot is:
 A.
less than 100
 B.
more than 200
 C.
between 120 & 190
 D.
Cannot be found out.
Answer: Option A
Explanation :
Let the number of defective widgets be ‘x’.
Cost to Prakash if he does not use any test = 50x
Cost to Prakash if he uses test 1 = 2 × 1000 + $\frac{4x}{5}$× 25 + $\frac{x}{5}$× 50 = 2000 + 30x
Cost to Prakash if he uses test 2 = 3000 + 25x
Prakash should not test when 50x ≤ (2000 + 30x) and (3000 + 25x)
50x ≤ (2000 + 30x) and 50x ≤ (3000 + 25x)
20x ≤ 2000 and 25x ≤ 3000
x ≤ 100 and x ≤ 120
x ≤ 100
Prakash should use test 1 when test 1 is cheaper than test 2
3000 + 25x ≥ 2000 + 30x
5x ≤ 1000
x ≤ 200
For x ≥ 200 he can use test 2.
Below 100, no test would be cheaper.
Hence, option (a).
Workspace:
If there are 120 defective widgets in the lot, Prakash:
 A.
should either use Test I or not test.
 B.
should either use Test II or not test
 C.
should use Test I or Test II.
 D.
should use Test I only.
Answer: Option D
Explanation :
Consider the solution to first question of this set.
For widgets ≤ 100, no test is required.
For 100 ≥ widgets ≤ 200 test 1 is preferred.
For widgets ≥ 200 test 2 is preferred.
If there are 120 widgets, he should go for test I as it is cheaper.
Hence, option (d).
Workspace:
If the number of defective widgets in the lot is between 200 and 400, Prakash:
 A.
may use Test I or Test II
 B.
should use Test I only
 C.
should use Test II only
 D.
cannot decide
Answer: Option C
Explanation :
Consider the solution to first question of this set.
For widgets ≤ 100, no test is required.
For 100 ≥ widgets ≤ 200 test 1 is preferred.
For widgets ≥ 200 test 2 is preferred.
Hence, option (c).
Workspace:
If Prakash is told that the lot has 160 defective widgets, he should:
 A.
use Test I only
 B.
use Test II only
 C.
do no testing
 D.
either use Test I or do not test
Answer: Option A
Explanation :
Consider the solution to first question of this set.
For widgets ≤ 100, no test is required.
For 100 ≥ widgets ≤ 200 test 1 is preferred.
For widgets ≥ 200 test 2 is preferred.
In case of 160 defectives he should use test I as it is cheaper.
Hence, option (a).
Workspace:
If there are 200 defective widgets in the lot, Prakash:
 A.
may use either Test I or Test II
 B.
should use Test I or not use any test
 C.
should use Test II or not use any test
 D.
cannot decide
Answer: Option A
Explanation :
Consider the solution to first question of this set.
For widgets ≤ 100, no test is required.
For 100 ≥ widgets ≤ 200 test 1 is preferred.
For widgets ≥ 200 test 2 is preferred.
If there are 200 defective widgets in the lot, Prakash may use either Test I or Test II as the cost of both the Tests is same = Rs.8000.
Hence, option 1.
Workspace:
A function can sometimes reflect on itself, i.e. if y = f(x), then x = f(y). Both of them retain the same structure and form. Which of the following functions has this property?
 A.
y = $\frac{2x+3}{3x+4}$
 B.
y = $\frac{2x+3}{3x2}$
 C.
y = $\frac{3x+4}{4x5}$
 D.
None of the above
Answer: Option B
Explanation :
For each of the given expressions, you may have to simplify and express x in terms of y and hence verify for which one does the form & structure remain the same.
In general, any function of the form y = $\frac{ax+b}{bxa}$ reflects on to itself as we can rearrange it as
x = $\frac{ay+b}{bya}$
Let us check option (b).
y = f(x)
⇒ y = $\frac{2x+3}{3x2}$
⇒ 3xy  2y = 2x + 3
⇒ 3xy  2x = 2y + 3
⇒ x(3y  2) = (2y + 3)
⇒ x = $\frac{2y+3}{3y2}$
∴ x = f(y)
Hence, the answer is (b).
Workspace:
What is the value of k for which the following system of equations has no solution:
2x – 8y = 3 and kx +4y = 10
 A.
2
 B.
1
 C.
1
 D.
2
Answer: Option C
Explanation :
For two lines a_{1}x + b_{1}y = c_{1} and a_{2}x + b_{2}y = c_{2} to be parallel, the condition is
$\frac{{a}_{1}}{{a}_{2}}$ = $\frac{{b}_{1}}{{b}_{2}}$ ≠ $\frac{{c}_{1}}{{c}_{2}}$
For no solution, lines must be parallel and not over lapping
⇒ $\frac{2}{k}=\frac{8}{4}\ne \frac{3}{10}$
∴ k = − 1.
Hence, option 3.
Workspace:
How many 3digit even numbers can you form such that if one of the digits is 5 then the following digit must be 7?
 A.
5
 B.
405
 C.
365
 D.
495
Answer: Option C
Explanation :
Case 1: The number has 5 as one of the digits
Three digit number such that 7 follows 5 could be of the form 57_ or _57
Since, the number is an even number
Therefore possible numbers are 570,572,574,576 or 578.
Hence, 5 such numbers are possible.
Case 2: The number does not have 5 as one of the digits
First digit can be fille din 8 ways, second in 9 ways and third in 5 ways (even digits only).
∴ Number of ways = 8 × 9 × 5 = 360
∴ Total number of ways = 5 + 360 = 365
Hence, option (c).
Workspace:
Alord got an order from a garment manufacturer for 480 Denim Shirts. He brought 12 sewing machines and appointed some expert tailors to do the job. However, many didn’t report to duty. As a result, each of those who did, had to stitch 32 more shirts than originally planned by Alord, with equal distribution of work. How many tailors had been appointed earlier and how many had not reported for work?
 A.
12, 4
 B.
10, 3
 C.
10, 4
 D.
None of these
Answer: Option C
Explanation :
Let x be the number of tailors initially appointed.
Let n be the number of shirts that had to be stitched by each tailor initially.
Let y be the number of tailors who did not come.
x × n = 480.
(x – y)(n + 32) = 480
Only option (c) satisfies the equation.
Hence, option (c).
Workspace:
Iqbal dealt some cards to Mushtaq and himself from a full pack of playing cards and laid the rest aside. Iqbal then said to Mushtaq. “If you give me a certain number of your cards, I will have four times as many cards as you will have. If I give you the same number of cards, I will have thrice as many cards as you will have “. Of the given choices, which could represent the number of cards with Iqbal?
 A.
9
 B.
31
 C.
12
 D.
35
Answer: Option B
Explanation :
Let Mushtaq has M cards while Iqbal has got I cards with him.
Let number of cards exchanged be x.
Case 1: I + x = 4(M – x) ... (i)
Case 2: I – x = 3(M + x) ... (ii)
From (i) and (ii)
I = 31x
Since I + M has to be less than or equal to 52, the only possible value for I could be 31.
Hence, option (b).
Workspace:
Fifty college teachers are surveyed as to their possession of colour TV, VCR and tape recorder. Of them, 22 own colour TV, 15 own VCR and 14 own tape recorders. Nine of these college teachers own exactly two items out of colour TV, VCR and tape recorder; and, one college teacher owns all three. How many of the 50 teachers own none of the three, colour TV, VCR or tape recorder?
 A.
4
 B.
9
 C.
10
 D.
11
Answer: Option C
Explanation :
Let us find out how many teachers own at least one of the three items.
We konw, A∪B∪C = A + B + C – (exactly two of A, B and C)  2(all three)
∴ TV∪VCR∪TR = 22 + 15 + 14  (9)  2(1) = 40
∴ 40 people own at least one of TV, VCR and Tape Recorder.
Hence the number of teachers owing none = 50 – 40 = 10.
Hence, option (c).
Workspace:
Three times the first of three consecutive odd integers is 3 more than twice the third. What is the third integer?
 A.
15
 B.
9
 C.
11
 D.
5
Answer: Option A
Explanation :
Let the 3 odd numbers be (x – 2), x and (x + 2).
It is given that 3(x – 2) = 3 + 2(x + 2).
Hence x = 13.
So the third integer is (x + 2) = 15.
Hence, option (a).
Workspace:
What is the total number of ways to reach A to B in the network given?
 A.
12
 B.
16
 C.
20
 D.
22
Answer: Option B
Explanation :
To go from A to B, we need to pass through 3 levels of nodes.
Number of ways of goind from A to any of the level 1 node = 4
Number of ways of goind from any of the level 1 node to any of the level 2 node = 2
Number of ways of goind from any of the level 2 node to B = 2
Total number of ways to reach A to B = 4 × 2 × 2 × 1 = 16.
Hence, option (b).
Workspace:
Let the consecutive vertices of a square S be A, B, C & D. Let E, F & G be the midpoints of the sides AB, BC & AD respectively of the square. Then the ratio of the area of the quadrilateral EFDG to that of the square S is nearest to
 A.
1/2
 B.
1/3
 C.
1/4
 D.
1/8
Answer: Option A
Explanation :
Let the side of square be a.
Area of quadrilateral EFDG = Ar(△DGF) + Ar(△GEF) = 1/2 × a/2 × a + 1/2 × a × a/2 = 1/2 × a^{2}
Ar(EFDG) : Ar(S) = 1/2 × a^{2} : a^{2} = 1 : 2
Hence, option (a).
Workspace:
2^{73} – 2^{72} – 2^{71} is the same as
 A.
2^{69}
 B.
2^{70}
 C.
2^{71}
 D.
2^{72}
Answer: Option C
Explanation :
2^{73} – 2^{72} – 2^{71} = 2^{71}(2^{2} – 2^{1}  1) = 2^{71}(4  2  1) = 2^{71}(1) = 2^{71}
Hence, option (c).
Workspace:
The number of integers n satisfying –n + 2 ≥ 0 and 2n ≥ 4 is
 A.
0
 B.
1
 C.
2
 D.
3
Answer: Option B
Explanation :
The two equations can be simplified into n ≤ 2 and n ≥ 2.
The only value that satisfies both these conditions is n = 2.
Hence, option (b).
Workspace:
The sum of two integers is 10 and the sum of their reciprocals is 5/12. Then the larger of these integers is
 A.
2
 B.
4
 C.
6
 D.
8
Answer: Option C
Explanation :
Since the sum of reciprocals is 5/12.
The two numbers must have their LCM as 12 and their sum as given will be 10.
Possible numbers are 4 and 6 only.
Alternately,
Let the numbers be a and b.
a + b = 10 ...(1)
1/a + 1/b = 5/12 ...(2)
Solving (1) and (2), we get (a, b) = (4, 6)
Hence, option (c).
Workspace:
A circle is inscribed in a given square and another circle is circumscribed about the square. What is the ratio of the area of the inscribed circle to that of the circumscribed circle?
 A.
2 : 3
 B.
3 : 4
 C.
1 : 4
 D.
1 : 2
Answer: Option D
Explanation :
As it is apparent from the following diagram, the diameter of the inscribed circle is equal to the side of the square, while the diameter of the circumscribed square is equal to the diagonal of the square.
Since the ratio of any two circles is equal to the ratio of the squares of their diameters, in this case the required ratio is equal to (side)^{2} : (diagonal)^{2}
If the side of the square is a, its diagonal will be √2a
Now, the ratio of the side to the diagonal of a square = 1 : √2, the ratio of their squares will be 1 : 2.
Hence, option (d).
Workspace:
If y = f(x) and f(x) = $\frac{(1\u2013x)}{(1+x)}$, which of the following is true?
 A.
f(2x) = f(x)  1
 B.
x = f(2y)  1
 C.
f(1/x) = f(x)
 D.
x = f(y)
Answer: Option D
Explanation :
y = f(x) = $\frac{1x}{1+x}$
⇒ y(1 + x) = 1 – x
⇒ y + xy = 1 – x
⇒ x + xy = 1 – y
⇒ x(1 + y) = (1 – y)
⇒ x = $\frac{1y}{1+y}$ = f(y)
Since here, y = f(x) and x = f(y), the given function refleccts on itself.
Hence, option (d).
Workspace:
Answer the following questions based on the information given below:
There were a hundred schools in a town. Of these, the number of schools having a play – ground was 30, and these schools had neither a library nor a laboratory. The number of schools having a laboratory alone was twice the number of those having a library only. The number of schools having a laboratory as well as a library was one fourth the number of those having a laboratory alone. The number of schools having either a laboratory or a library or both was 35.
How many schools had none of the three viz., laboratory, library or play – ground?
 A.
20
 B.
5
 C.
30
 D.
35
Answer: Option D
Explanation :
It is given that x + 2x + x/2 = 35.
Hence x = 10.
Total number of schools that had at least one of the three = 30 + 10 + 20 + 5 = 65.
Hence the number of schools having none of them = 100 – 65 = 35.
Hence, option 4.
Workspace:
What was the ratio of schools having laboratory to those having library?
 A.
1 : 2
 B.
5 : 3
 C.
2 : 1
 D.
2 : 3
Answer: Option B
Explanation :
Number of schools having library = 15. And number of schools having laboratory = 25.
Hence the ratio = 25 : 15 = 5 : 3.
Hence, option 2.
Workspace:
A player rolls a die and receives the same number of rupees as the number of dots on the face that turns up. What should the player pay for each roll if he wants to make a profit of one rupee per throw of the die in the long run?
 A.
Rs. 2.50
 B.
Rs. 2
 C.
Rs. 3.50
 D.
Rs. 4
Answer: Option A
Explanation :
Since in the long run the probability of each number appearing is the same, we can say in ‘n’ throws one can get 1, 2, 3, 4, 5 and 6, n/6 times each.
Hence he would earn (1 + 2 + 3 + 4 + 5 + 6)n/6 = Rs. 7n/2 = Rs. 3.5n.
∴ His average earning for each through is 3.5n/n = Rs. 3.5
In order to make a profit of 1 Re. per throw his cost for the each through should be = 3.5 – 1 = Rs. 2.50
Hence, option (a).
Workspace:
Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously? (in hous)
 A.
12
 B.
10
 C.
8
 D.
6
Answer: Option B
Explanation :
B being twice as fast as A, will take half of time of A to produce a million units i.e. = 30 hrs.
Since, Machine C takes the same amount of time as A & B running together,
∴ $\frac{1}{C}$ = $\frac{1}{A}$ + $\frac{1}{B}$ or
$\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{C}$ = $\frac{2}{A}$ + $\frac{2}{B}$
⇒ $\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{C}$ = $2\left(\frac{1}{A}+\frac{1}{B}\right)$ = $2\left(\frac{1}{60}+\frac{1}{30}\right)$ = $\frac{1}{10}$
Hence, it will take 10 hours for all three of them to produce 1 million units.
Hence, option (b).
Workspace:
Let Y = minimum of {(x + 2), (3 – x)}. What is the maximum value of Y for 0 ≤ x ≤ 1?
 A.
1.0
 B.
1.5
 C.
3.1
 D.
2.5
Answer: Option D
Explanation :
For Y to be maximum,
x + 2 = 3  x
∴ x = 0.5
Maximum value of Y = x + 2 = 2.5
Hence, option (d).
Workspace:
There are 3 clubs A, B & C in a town with 40, 50 & 60 members respectively. While 10 people are members of all 3 clubs, 70 are members in only one club. How many belong to exactly two clubs?
 A.
20
 B.
25
 C.
50
 D.
70
Answer: Option B
Explanation :
Given, x + y + z = 70
x + a + b = 40 – 10 = 30 ... (i)
y + b + c = 50 – 10 = 40 ... (ii)
z + a + c = 60 – 10 = 50 ... (iii)
From (i), (ii) and (iii), we get
(x + y + z) + 2(a + b + c) = 120
⇒ a + b + c = 25.
Hence, option (b).
Workspace:
A square piece of cardboard of sides ten inches is taken and four equal squares pieces are removed at the corners, such that the side of this square piece is also an integer value. The sides are then turned up to form an open box. Then the maximum volume such a box can have is
 A.
72 cubic inches.
 B.
24.074 cubic inches.
 C.
2000/27 cubic inches.
 D.
64 cubic inches.
Answer: Option A
Explanation :
If x = 1, then for the rectangular box : l = 8, b = 8 and h = 1, so volume = 64.
If x = 2, l = 6, b = 6 and h = 2 and volume = 72.
If x = 3, l = 4, b = 4 and h = 3 and volume = 48.
Therefore, we can see that for a value of x between 2 and 3, the volume of box decreases and will go on decreasing further as x increases.
Hence, the maximum volume that the box can have is 72 cubic inches.
Hence, option (a).
Workspace:
x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?
 A.
(x  1)yz
 B.
x(y  1)z
 C.
xy(z  1)
 D.
x(y + 1)z
Answer: Option A
Explanation :
Going by the options:
(a) (x  1)yz = xyz − yz
(b) x(y  1)z = xyz  xz
(c) xy(z  1) = xyz  xy
(d) x(y + 1)z = xyz + xz
Here, yz will be minimum out of yz, xz, xy, xz as x > y > z
Hence, option (1).
Workspace:
What is the greatest power of 5 which can divide 80! exactly.
 A.
16
 B.
20
 C.
19
 D.
None of these
Answer: Option C
Explanation :
Number of powers of 5 in 80! = $\left(\frac{80}{5}=16\right)$ + $\left(\frac{16}{5}=3\right)$ = 19.
Hence, option (c).
Workspace:
A third standard teacher gave a simple multiplication exercise to the kids. But one kid reversed the digits of both the numbers and carried out the multiplication and found that the product was exactly the same as the one expected by the teacher. Only one of the following pairs of numbers will fit in the description of the exercise. Which one is that?
 A.
14, 22
 B.
13, 62
 C.
19, 33
 D.
42, 28
Answer: Option B
Explanation :
The last digits obtained by multiplying the units place digits should be the same as that obtained by multiplying the tens place digits.
Hence, option (b).
Workspace:
Find the minimum integral value of n such that the division 55n/124 leaves no remainder.
 A.
124
 B.
123
 C.
31
 D.
62
Answer: Option A
Explanation :
As 55 does not have factor common with 124, for 55n to be exactly divisible by 124, n should be a multiple of 124.
Hence the minimum value that n can have is 124 itself.
Hence, option (a).
Workspace:
Let k be a positive integer such that k + 4 is divisible by 7. Then the smallest positive integer n, greater than 2, such that k + 2n is divisible by 7 equals
 A.
9
 B.
7
 C.
5
 D.
3
Answer: Option A
Explanation :
a = k + 4 is divisible by 7
b = k + 2n is divisible by 7
⇒ b – a = 2n – 4 is divisible by 7
⇒ 2n – 4 = 0 or 7 or 14 ...
For minimum integral value of n, greater than 2
2n – 4 = 14
⇒ n = 9
Hence, option (a).
Workspace:
A calculator has two memory buttons, A and B. Value 1 is initially stored in both memory locations.
The following sequence of steps is carried out five times:
 add 1 to B
 multiply A to B
 store the result in A
What is the value stored in memory location A after this procedure?
 A.
120
 B.
450
 C.
720
 D.
250
Answer: Option C
Explanation :
A  B  Step I (B + 1) 
Step II (A × B) 
Step III A 
Step III B 


Beginning  1  1  
1^{st} time  1  1  B = 2  (1 × 2) = 2  2  2 
2^{nd} time  2  2  B = 3  (2 × 3) = 6  6  3 
3^{rd} time  6  3  B = 4  (6 × 4) = 24  24  4 
4^{th} time  24  4  B = 5  (24 × 5) = 120  120  5 
5^{th} time  120  5  B = 6  (120 × 6) = 720  720  6 
Hence, option 3.
Workspace:
A one rupee coin is placed on a table. The maximum number of similar one rupee coins which can be placed on the table, around it, with each one of them touching it and only two others is
 A.
8
 B.
6
 C.
10
 D.
4
Answer: Option B
Explanation :
It can be seen that if we place 3 coins touching each other, their centers form an equilateral triangle.
Hence the angle made by the centers of the coins around the central coin is 60°.
Since the total angle to be covered is 360°, there has to be 6 coins surrounding the central coin.
Hence, option (b).
Workspace:
Gopal went to a fruit market with certain amount of money. With this money he can buy either 50 oranges or 40 mangoes. He retains 10% of the money for taxi fare. If he buys 20 mangoes, then the number of oranges he can buy is
 A.
25
 B.
20
 C.
18
 D.
6
Answer: Option B
Explanation :
Let the amount with Gopal be Rs. 400.
Therefore price of an orange is then Rs. 8 and that of a mango is Rs.10.
If he keeps 10% of the money for taxi fare, he is left with Rs.360.
Now if he buys 20 mangoes, then he spends on mangoes Rs. 200.
Now he is left with Rs.160, in which he can buy = 160/8 = 20 oranges.
Hence, option (b).
Workspace:
Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
 A.
1 hour
 B.
50 minutes
 C.
1/2 hour
 D.
55 minutes
Answer: Option D
Explanation :
Since her husband meets her mid way, the total time saved by him can be equally divided into time saved while going to station and that saved while returning home.
In other words, he saved 5 mins while going and 5 mins while coming.
So instead of usual time of 6:00 pm he must have met her at 5:55 pm.
So she must have walked for 55 mins.
Hence, option (d).
Workspace:
In Sivakasi, each boy’s quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?
 A.
200
 B.
150
 C.
125
 D.
175
Answer: Option B
Explanation :
Let x be the total number of sticks assigned to each boy and y be the number of boxes in which he has to fill them. If he reduces number of sticks per box by 25, he would fill (x/y  25) in each box and hence he would now fill (y + 3) boxes.
∴ x = (x/y  25)(y + 3) = x + 3x/y  25y  75.
⇒ 3x = (25y + 75)y
⇒ x = $\frac{(25{y}^{2}+75y)}{3}=\frac{25y(y+3)}{3}$
For x to have an integer value, y(y + 3) has to be a multiple of 3. This is possible only when y is a multiple of 3.
Subsequently, (y + 3) will be a multiple of 3.
Therefore, 25y(y + 3)/3 = x will be a multiple of 3.
Hence, option (b).
Workspace:
A sum of money compounded annually becomes Rs. 625 in two years and Rs. 675 in three years. The rate of interest per annum is
 A.
7%
 B.
8%
 C.
6%
 D.
5%
Answer: Option B
Explanation :
For a difference of 1 year, CI can be computed as SI.
Hence, from the 2nd year to the 3rd year interest earned = (675 – 625) = Rs. 50 on Rs. 625.
Hence the Rate of interest 50/625 × 100 = 8% p.a.
Hence, option (b).
Workspace:
In a sixnode network, two nodes are connected to all the other nodes. Of the remaining four, each is connected to four nodes. What is the total number of links in the network?
 A.
13
 B.
15
 C.
7
 D.
26
Answer: Option A
Explanation :
Nodes 1 & 2 will be connected with 5 links each hence, we get 10 links
Nodes 3, 4, 5, and 6 will be connected witih 4 links each hence, we get 16 links.
Total links = 10 + 16 = 26.
Now since each link connects 2 nodes, unique number of links = 26/2 = 13.
Alternately,
(Note : In the diagram given below, the top two nodes are connected to all the other nodes, while the remaining four are connected to only four other nodes).
You find that the total number of links in the network is 13.
Hence, option (a).
Workspace:
If x is a positive integer such that 2x + 12 is perfectly divisible by x, then the number of possible values of x is
 A.
2
 B.
5
 C.
6
 D.
12
Answer: Option C
Explanation :
If (2x + 12) is perfectly divisible by x, then (2x + 12)/x has to be an integer.
Now if we divide, the expression simplifies to (2 + 12/x).
The only way in which this expression would be an integer is when 12/x is an integer or if 12 is perfectly divisible by x. [i.e., x is a factor of 12]
This is possible if x takes either of these values : 1, 2, 3, 4, 6, 12. Hence the answer is 6 values.
Hence, option (c).
Workspace:
An outgoing batch of students wants to gift PA system worth Rs. 4200 to their school. If the teachers offer to pay 50% more than the students, and an external benefactor gives three times teachers’ contribution, how much should the teachers donate?
 A.
600
 B.
840
 C.
900
 D.
1200
Answer: Option C
Explanation :
The ratios of the share of students, teachers and benefactor is 1 : 1.5 : 4.5.
So, the proportion of teachers share is 1.5/7.
So teachers would donate: (1.5 × 4200)/7 = Rs. 900
Hence, option (c).
Workspace:
A positive integer is said to be a prime number if it is not divisible by any positive integer other than itself and 1. Let p be a prime number greater than 5. Then (p^{2} – 1) is
 A.
never divisible by 6
 B.
always divisible by 6, and may or may not be divisible by 12
 C.
always divisible by 12, and may or may not be divisible by 24
 D.
always divisible by 24
Answer: Option D
Explanation :
As each prime number greater than or equal 3 is of from 6k ± 1, here k is a natural number, p = 6k ± 1.
p^{2}  1 = (6k ± 1)^{2}  1 = 12k(3k ± 1)
Since k(3k ± 1) is an even number, above number is always divisible by 24 irrespective of value of k.
Hence, option (d).
Workspace:
To decide whether a n digit number is divisible by 7, we can define a process by which its magnitude is reduced as follows: (i_{1}, i_{2}, i_{3}, … , are the digits of the number, starting from the most significant digit).
i_{1 }i_{2} ……. i_{n} ⇒ i_{1 }. 3^{n1} + i_{2} . 3^{n2} + ……… + i_{n} . 3^{0}.
e.g. 259 ⇒ 2.3^{2} + 5.3^{1} + 9.3^{0} = 18 + 15 + 9 = 42
Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?
 A.
2
 B.
3
 C.
4
 D.
1
Answer: Option A
Explanation :
203 = 2.3^{2} + 0.3^{1} + 3.3^{0} = 18 + 0 + 1 = 21
21 = 2.3^{1} + 1.3^{0} = 6 + 1 = 7.
Therefore we can reduce 203 to 7 in 2 steps.
Hence, option (a).
Workspace:
What is the distance between the points A(3, 8) and B(–2, –7)?
 A.
5√2
 B.
5
 C.
5√10
 D.
10√2
Answer: Option A
Explanation :
The distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given as $\sqrt{{({x}_{2}{x}_{1})}^{2}+{({y}_{2}{y}_{1})}^{2}}$
Hence, required distance = $\sqrt{{(23)}^{2}+{(78)}^{2}}$ = 5√10.
Hence, option (c).
Workspace:
Each of these items has a question followed by two statements. As the answer,
Type 1, If the question can be answered with the help of statement I alone,
Type 2, If the question can be answered with the help of statement II alone,
Type 3, If both the statement I and statement II are needed to answer the question, and
Type 4, If the question cannot be answered even with the help of both the statements.
Is it more profitable for Company M to produce Q?
 Product R is sold at a price four times that of Q.
 One unit of Q requires 2 units of labour, while one unit of R requires 5 units of labour. There is no other constraint on production.
Answer: 3
Explanation :
Statement I only gives the comparison of the selling prices. This information alone is not enough to answer the question as the profit also depends on cost.
Statement II only gives the infomation about cost and not selling price.
Hence, the 2 statements alone are not enought to answer the question.
So we also need to analyze both the statements.
Since there is no other constraint on production, we can solely compare the profitability of two products on the basis of labour.
According to it, if 10 units of labour is available, it can produce 5 units of Q and 2 units of R. So, from 10 units of labour, I can earn (5 x 1) = 5 units of sales revenue from Q and (2 x 4) = 8 units of sales revenue from R.
So by taking both statements together we can determine which would be more profitable.
Hence, 3.
Workspace:
A train started from Station A, developed engine trouble and reached Station B, 40 minutes late. What is the distance between Stations A and B?
 The engine trouble developed after travelling 40 km from Station A and the speed reduced to 1/4^{th} of the original speed.
 The engine trouble developed after travelling 40 km from station A in two hours and the speed reduced to ¼th of the original speed.
Answer: 2
Explanation :
In order to solve the question, we need to know two things :
(a) the original speed of the train or the new speed of the train and
(b) at what distance from A or after how much time after leaving A the train brokedown.
The statement II provides both of these data viz. original speed = 20 kmph and distance from A = 40 kms. and hence only this is required to answer the question.
For e.g: If the distance between A & B is considered to be x, then time taken had it not broken down is x/20 hours.
The new time taken is [2 + (x – 40)/5] hours and we know that this time is 40 min. more than the original time.
The equation becomes : x/20 + 40/60 = [2 + (x – 40)/5], which can be easily solved to get value of x.
Hence, 2.
Workspace:
What is the value of prime number x?
 x^{2} + x is a two digit number greater than 50.
 x^{3} is a three digit number.
Answer: 1
Explanation :
From statement I : x^{2} + x > 50 and x^{2} + x < 100
⇒ x(x + 1) > 50
Only prime number 7 satisfies the above equation.
So the question can be answered from statement 1 alone.
From statement II : x can be 5 or 7.
Hence, 1.
Workspace:
The average of three unequal quotations for a particular share is Rs.110. If all are quoted in integral values of rupee, does the highest quotation exceed Rs.129?
 The lowest quotation is Rs.100.
 One of the quotations is Rs.115.
Answer: 1
Explanation :
Sum of all the three quotations = 110 × 3 = 330
From statement I, if the lowest quotation is 110 then the sum of other two quotations = 330 – 100 = Rs. 230.
Now, minimum value of second quotation = Rs. 101
Therefore, maximum value of third quotation = 230 – 101 = Rs. 129.
Hence, 1.
Workspace:
How many people (from the group surveyed) read both Indian Express and Times of India?
 Out of total of 200 readers, 100 read Indian Express, 120 read Times of India and 50 read Hindu.
 Out of a total of 200 readers, 100 read Indian Express, 120 read Times of India and 50 read neither
Answer: 2
Explanation :
Using this we cannot find the answer.
From the statement II however we can find the answer, as we get the following Venn diagram.
Hence, 2.
Workspace:
X says to Y, “I am 3 times as old as you were 3 years ago”. How old is X?
 Y’s age 17 years from now will be same as X’s present age.
 X’s age nine years from now is 3 times Y’s present age.
Answer: 1
Explanation :
Let present age of X and Y be x and y respectively.
We have, x = 3(y  3) ... (i)
From statement I, x = y + 17 ... (ii)
From statement II, x + 9 = 3y ... (iii)
From (i) and (ii) x = 37.5 years and y = 12.5 years
But, equations (i) and (iii) are same so we cannot find values of x and y.
Hence, 1.
Workspace:
What is the area under the line GHI – JKL in the given quadrilateral OPQR, knowing that all the small spaces are squares of the same area?
 Length ABCDEQ is greater than or equal to 60.
 Area OPQR is less than or equal to 1512.
Answer: 3
Explanation :
Let length of the smaller square be x.
From statement I,Length of ABCDEQ = 10x ≥ 60
Therefore, x ≥ 6
From statement II, Area of rectangle OPQR = 42x^{2} ≤ 1512
⇒ x^{2} ≤ 36
∴ x ≤ 6
By combining results from statement I and II we have, x = 6
Now, area under the ling GHIJKL can be found out using x.
Hence, 3.
Workspace:
What is the radius of the circle?
 Ratio of its area to circumference is > 7.
 Diameter of the circle is ≤ 32.
Answer: 4
Explanation :
Let the radius of the circle be r.
From statement I : πr^{2}/2πr > 7
⇒ r > 14 ...(i)
From statement II : 2r ≤ 32
⇒ r ≤ 16 ...(ii)
Combining (i) and (ii) 14 < r ≤ 16
So r cannot be uniquely determined.
Hence, 4.
Workspace:
What is the time difference between New York and London?
 The departure time at New York is exactly 9:00 a.m. local time and the arrival time at London is at 10:00 a.m. local time.
 The flight time is 5 hours.
Answer: 4
Explanation :
From both statements I and II : Arrival time of flight by New York local time will be 2 P.M.
Since we don’t know whether the flight landed on same date or other we can not find the answer.
Hence, 4.
Workspace:
Mr. Murthy takes the morning train to his office from station A to station B, and his colleague Mr.Rahman joins him on the way. There are three stations C, D and E on the way not necessarily in that sequence. What is the sequence of stations?
 Mr. Rahman boards the train at D.
 Mr. Thomas, who travels between C & D has two segments of journey in common with Mr. Murthy but none with Mr. Rahman.
Answer: 3
Explanation :
Statement I alone no conclusion can be drawn.
Statement II following sequence of stations is possible
A  C/D  E  D/C  B 

But, From I Mr. Rahman boards the station at D which is possible at fourth position only because Mr. Thomas and Mr. Rahman have no common station.
Hence, 3.
Workspace:
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