Algebra Practice Questions for CAT with Solutions

Question 1:
A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?
[1] 2 ≤ x ≤ 6
[2] 5 ≤ x ≤ 8
[3] 9 ≤ x ≤ 12
[4] 11 ≤ x ≤ 14
After first customer, amount of rice left is 0.5x – 0.5

After second customer, amount of rice left is 0.5(0.5x -0.5) – 0.5

After third customer, amount of rice left is 0.5(0.5(0.5x -0.5) – 0.5) – 0.5 = 0

0.5(0.5(0.5x -0.5) – 0.5) = 0.5
0.5(0.5x -0.5) – 0.5 = 1
0.5x -0.5 = 3
x = 7. Option B

Verification for better understanding:

Originally there were 7 kgs of rice.

First customer purchased 3.5kgs + 0.5kgs = 4 kgs.

After first customers, amount of rice left is 3 kgs.

Second customer purchased 1.5kgs + 0.5 kgs = 2 kgs.

After second customer, amount of rice left is 1 kg.

Third customer purchased 0.5kgs + 0.5kgs = 1 kg.

No rice is left after the third customer.


Question 2:
If p and Q are integers such that \(\frac{7}{10}<\frac{p}{q}<\frac{11}{15} \) , find the smallest possible value of q.
[1] 13
[2] 60
[3] 30
[4] 7
The fraction lies in the range (0.7,0.733333)

We know that \(\frac{8}{{11}}\) = 0.727272.. is valid value.

The smallest value of q has to be less than or equal to 11. Only 7 fits in the range.

With a little hit and trial we get a valid value of \(\frac{p}{q}\) as \(\frac{5}{7}\)

Smallest value of q = 7. Option D


Question 3:
Given the system of equations \(\left\{ {\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z = - 1}\\{3x + 2y + z = 9}\end{array}} \right. \), find the value of x+y+z.
[1] -1
[2] 3.5
[3] 2
[4] 1
The given equations are

\(2x + y + 2z = 4 \ldots \left( 1 \right)\)

\(x + 2y + 3z =  - 1 \ldots \left( 2 \right)\)

\(3x + 2y + z = 9 \ldots \left( 3 \right)\)

Take the first and the second equation :

\(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{x + 2y + 3z =  - 1}\\{}\end{array}\)  multiply equation 2 by -2 , thus the 2 equations we get after multiplying are \(\begin{array}{*{20}{c}}{2x + y + 2z = 4}\\{ - 2x - 4y - 6z = 2}\\{}\end{array}\), on solving this we get \( - 3y - 4z = 6\) … (4)

Now take equation (2) and (3)

\(\begin{array}{*{20}{c}}{}\\{x + 2y + 3z =  - 1}\\{3x + 2y + z = 9}\end{array}\)    multiply equation (2) by -3, thus the equations will be

\(\begin{array}{*{20}{c}}{}\\{ - 3x - 6y - 9z = 3}\\{3x + 2y + z = 9}\end{array}\)

On solving the above 2 equations we get  -4y-8z=12  ie.  \( - y - 2z = 3\) (5)

Again on multiplying equation (5) by -3 we get   \( - 3y + 6z =  - 9\).

Adding equations (4)(5) :

We get z=-1.5 and y=0 , substituting these values in any of the 3 main equations , we get x = ½ or 0.5

Adding x + y + z = 0.5+0-1.5 = -1, Option A


Question 4:
If x and y are positive integers and x+y+xy=54, find x+y
[1] 12
[2] 14
[3] 15
[4] 16
With x + y = 12, maximum value possible is 6 + 6 + 6×6 = 48

With x + y = 14, maximum value possible is 7 + 7 + 7×7 = 63

6 + 8 + 6×8 = 62

5 + 9 + 5×9 = 59

4 + 10 + 4×10 = 54

So, x + y = 14. Option B

Alternatively,

x + y + xy = 54

1 + x + y + xy = 55
(1 + x)(1 + y) = 55
55 can be split as 5 and 11
So x and y can be 4 and 10
x + y = 14. Option B

Question 5:
How many pairs of integers (x, y) exist such that x2 + 4y2 < 100?
[1] 95
[2] 90
[3] 147
[4] 180
y will lie in the range [-4, 4]

When y = 4 or – 4, x will lie in the range [-5, 5] = 11 values. Total values = 22

When y = 3 or – 3, x will lie in the range [-7, 7] = 15 values. Total values = 30

When y = 2 or – 2, x will lie in the range [-9, 9] = 19 values. Total values = 38

When y = 1 or – 1, x will lie in the range [-9, 9] = 19 values. Total values = 38

When y = 0, x will lie in the range [-9, 9] = 19 values.

Total values = 22 + 30 + 38 + 38 + 19 = 147.


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Algebra Practice Questions for CAT with Solutions
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