Special Equations Problems for CAT PDF

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Special Equations Problems for CAT PDF
Special Equations Problems for CAT PDF

Special Equations Problems for CAT PDF

Download important Special Equations Questions for CAT PDF based on previously asked questions in CAT exam. Practice Special Equations Questions PDF for CAT exam.

Download Special Equations Problems for CAT PDF

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Question 1: Ravi and Hari went to a market to buy some fruits. Ravi purchased 23 more apples than those purchased by Hari and they both purchased some number of oranges as well. Ravi and Hari purchased a total of 30 and 40 fruits respectively by spending equal amount of money. What is the price of an apple given that it is 10 rupees costlier than an orange?

a) Rs. 20

b) Rs. 23

c) Rs. 28

d) Rs. 33

Question 2: A man buys 3 bats, 2 gloves, 5 stumps and 7 helmet which costs him 86 Rs. Other person buys 12 bats, 9 gloves, 19 stumps and 28 helmets at 339 Rs from the same shop. The third person buys 18 bats, 10 gloves, 33 stumps and 42 helmets at 533 Rs also from the same shop. How much does one stump cost?

a) 7 Rs

b) 6 Rs

c) 8 Rs

d) Cannot be determined

Question 3: Sam had 16 francs and 9 dinars. Ram had 14 francs and 12 dinars. Esha, who had no money, received money equivalent to 26 rupees and 32 rupees from Sam and Ram, respectively. Now, each of them had equal value of money with them. Find the sum of the values of one franc and one dinar in rupees.

Question 4: A chocolate dealer has to send chocolates of three brands to a shopkeeper. All the brands are packed in boxes of same size. The number of boxes to be sent is 96 of brand A, 240 of brand B and 336 of brand C. These boxes are to be packed in cartons of same size containing equal number of boxes. Each carton should contain boxes of same brand of chocolates. What could be the minimum number of cartons that the dealer has to send?

a) 20

b) 14

c) 42

d) 38

e) 16

Question 5: The cost of 4 pens, 3 papers and 2 crayons is Rs. 29. The cost of 9 pens, 7 papers and 5 crayons is Rs. 68. Find the cost of 4 crayons and 7 papers, if the cost of each item is a distinct natural number. Also, the cost of the paper is the least among the three.

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Question 6: Find the number of positive integral solutions of the equation 29x + 21y + 8z = 256.

Question 7: Mrs. Sonia buys Rs. 249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavoured candy priced at Rs. 3.30 per candy; each boy receives a chocolate flavoured candy priced at Rs. 2.90 per candy. How many candies of each type did she buy?

a) 21, 57

b) 57, 21

c) 37, 51

d) 27, 51

Question 8: Mother Dairy sells milk packets in boxes of different sizes to its vendors. The vendors are charged Rs. 20 per packet up to 2000 packets in a box. Additions can be made only in a lot size of 200 packets. Each addition of one lot to the box results in a discount of one rupee an all the packets in the box. What should be the maximum size of the box that would maximize the revenue per box for Mother Dairy?

a) 2400 packets

b) 3000 packets

c) 4000 packets

d) None of the above

Question 9: Find $x +y$ if $\frac{90}{3x – y} + \frac{46}{2x + 3y} = 7$ and $\frac{92}{2x + 3y} – \frac{36}{3x – y} = 2$.

Question 10: How many integral solutions exist for the equation 5x + 7|y| = 405 such that x > 0?

a) 22

b) 23

c) 24

d) 25

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Answers & Solutions:

1) Answer (D)

Let,
The number of apples bought by Ravi = x+23
The number of oranges bought by Ravi = 7-x
The number of apples bought by Hari = x
The number of oranges bought by Hari = 40-x
The price of an apple = y+10
The price of an orange = y
It is given that they both paid the same amount. So we get,
(y+10)(23+x)+y(7-x) = (y+10)x+y(40-x)
=> y = 23
Therefore the price of an apple = 23+10 = Rs. 33

2) Answer (A)

Let cost of 1 bat be ‘b’ Rs , cost of 1 gloves be ‘g’ Rs,
cost of 1 stumps be ‘s’ Rs , cost of 1 helmet be ‘h’ Rs
From question we get following equations,
3b+2g+5s+7h = 86 …(1)
12b+9g+19s+28h = 339 …(2)
18b+10g+33s+42h = 533 …(3)
Multiplying eqn (1) by 4 and subtracting it from eqn (2) we get
g – s = -5 …(4)
Multiplying eqn (1) by 6 and subtracting it from eqn (3) we get
-2g + 3s = 17 …(5)
Solving equation 4 and 5 we get,
g = 2 and s = 7
Hence, one stump costs 7 Rs

3) Answer: 7

Let one franc be f rupees, one dinar be d rupees.
Let’s calculate all the values in rupees.

Sam now has 58 rupees.

This means that Sam, after giving 26 rupees had 58 rupees with him.

16f + 9d – 26 = 58
16f + 9d = 84 …… i

Similarly, Ram, after giving 32 rupees had 58 rupees with him.

14f + 12d – 32 = 58
14f + 12d = 90 …… ii

Multiplying i by 4 and ii by 3 and subtracting, we get

22f =66
f = 3

Substituting the value of f in I, we get d = 4

Thus, the value of one franc and one dinar is 7 rupees.

4) Answer (B)

Since each carton should contain boxes of the same brand of chocolates and all boxes being of equal size, to get the minimum number of cartons, we should have the maximum number of boxes in each carton.

Thus, the number of boxes in each carton = H.C.F (96,240,336)

= 48

So, we will get minimum number of cartons if there are 48 boxes in each carton.

$\therefore$ Number of cartons = $\frac{96}{48} + \frac{240}{48} + \frac{336}{48}$

= $2 + 5 + 7 = 14$

5) Answer: 27

Let the cost of one pen, one paper and one crayon be x, y, z respectively.
Given,
4x + 3y + 2z = 29 ….. i
And
9x + 7y + 5z = 68 …. Ii

Multiplying I by 2 we get,
8x + 6y + 4z = 32

Subtracting from ii

x + y + z = 10

Since, x, y and z are distinct natural numbers, the possible cases are:

(1, 3, 6); (1, 4, 5); (1, 2, 7); (2, 3, 5)

Given, the cost of the paper is least. Hence, y can be 1 or 2.

Checking the values by substituting in i, we get

x = 4, y = 1 and z = 5.

Cost of 4 crayons and 7 papers = 4 * 5 + 7 * 1 = 27

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6) Answer: 7

The given equation is
$29x + 21y + 8z = 256$

Dividing the equation by 8,

$3x + 2y + \frac{5(x + y)}{8} + z = 32$

We know that x, y and z are positive integers.
The above equation will be true only if x + y = 8k, where k is a natural number.

Let’s start with x + y = 8

$3x + 2y + z = 27$
Let’s check for all x + y = 8,

x = 1, y = 7, z = 10
x = 2, y = 6, z = 9
x = 3, y = 5, z = 8
x =4, y = 4, z = 7
x = 5, y = 3, z = 6
x = 6, y = 2, z = 5
x = 7, y = 1, z = 4

Now, let x+y = 16

3x + 2y + z = 22

There is no value of x, y such that they satisfy both the equations. (x, y and z are positive integers)

Hence, there are 7 solutions.

7) Answer (B)

Let the number of strawberry flavoured candy = x and the number of chocolate flavoured candy = y
Thus, as given in the question:-
3.3x + 2.9y = 249
Using options,
Option A = 3.3*21 + 2.9*57 = 69.3+165.3 = 234.6 Rs $\neq$ 249
Hence, option A cannot be the answer.
Option B = 3.3*57 + 2.9*21 = 188.1+60.9 = 249
Thus, option B can be the answer.
Option C = 3.3*37 + 2.9*57 = 241.8 $\neq$ 249
Hence, option C cannot be the answer.
Option D = 3.3*27 + 2.9*51 = 237 $\neq$ 249
Thus, option D cannot be the answer.

Hence, option B is the correct answer.

8) Answer (B)

We are given that, The vendors are charged Rs. 20 per packet up to 2000 packets in a box. Additions can be made only in a lot size of 200 packets. Each addition of one lot to the box results in a discount of one rupee an all the packets in the box.
Let x be number of additional lots.
Thus,
$(20-x)(2000+200x) = 40000 – 2000x + 4000x -200x^2$
=> $-200x^2+2000x+40000$
We need to maximize this
The minimum/maximum value of a quadratic equation is when $x = -b/2a$
Thus, the maximum value = $-2000/400 = 5$
Thus, the maximum size of the box that would maximize the revenue per box for Mother Dairy = 2000+200*5 = 3000

9) Answer: 10

Let
$\frac{1}{3x – y} = a$ and $\frac{1}{2x + 3y} = b$

Therefore,

$92b – 36a = 2$ and

$46b + 90a = 7$

Multiplying ii by 2 and subtracting i from it,

$216a = 12$
$a = \frac{1}{18}$
Substituting in ii,
$b = \frac{1}{23}$

Therefore,
$3x – y = 18$ and $2x + 3y = 23$

Multiplying the first equation by 3 and adding both the equations.

$11x = 77$, $ x = 7$

Hence, $y = 3$

$x + y = 10$

10) Answer (B)

Given equation 5x + 7|y| = 405?
If x = 4, then y can be 55 or -55. Now the next solution will be obtained when x increased by 7 and y decreases by 5. Thus the next solution will be
x = 11 and y = 50, -55. Thus we can see that for every value of x, there are two corresponding values of y.
Hence we can just find the possible values of x.
The values of x will be given by 4, 11, 18 . . . . . . 81. If x is greater than 81, then |7y| will have to be less than 0 which is not possible. At x =81, only one value of ‘y’ will be possible (0). For, all other values of x, y can take two values. Hence, the required number of solutions will be
11*2 + 1 = 23.

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