**Algebra Questions for CMAT**

Download important Algebra Questions for CMAT PDF based on previously asked questions in the CMAT exam. Practice Algebra Questions PDF for the CMAT exam.

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**Question 1: **The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the nth day of 2007 (n=1, 2, …, 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the nth day of 2007 (n = 1, 2, …, 365). On which date in 2007 will the prices of these two varieties of tea be equal?

a) May 21

b) April 11

c) May 20

d) April 10

e) June 30

**Question 2: **When you reverse the digits of the number 13, the number increases by 18. How many other two-digit numbers increase by 18 when their digits are reversed?

a) 5

b) 6

c) 7

d) 8

e) 10

**Question 3: **At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes, and one order of fries for Rs. 120 exactly. At the same place it would cost Rs. 164.5 for 4 burgers, 10 shakes, and one order of fries. How much would it cost for an ordinary meal of one burger, one shake, and one order of fries?

a) Rs. 31

b) Rs. 41

c) Rs. 21

d) Cannot be determined

**Question 4: **Three travellers are sitting around a fire, and are about to eat a meal. One of them has 5 small loaves of bread, the second has 3 small loaves of bread. The third has no food, but has 8 coins. He offers to pay for some bread. They agree to share the 8 loaves equally among the three travellers, and the third traveller will pay 8 coins for his share of the 8 loaves. All loaves were the same size. The second traveller (who had 3 loaves) suggests that he will be paid 3 coins, and that the first traveller be paid 5 coins. The first traveller says that he should get more than 5 coins. How much should the first traveller get?

a) 5

b) 7

c) 1

d) None of these

**Question 5: **The sum of the possible values of X in the equation |X + 7| + |X – 8| = 16 is:

a) 0

b) 1

c) 2

d) 3

e) None of the above

CMAT Free Solved Previous Papers.

**Question 6: **Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs 16 kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

a) 3.33 kg

b) 4.67 kg

c) 5.33 kg

d) None of the above

**Question 7: **The number of solutions $(x, y, z)$ to the equation $x – y – z = 25$, where x, y, and z are positive integers such that $x\leq40,y\leq12$, and $z\leq12$ is

a) 101

b) 99

c) 87

d) 105

**Question 8: **Let a, b, x, y be real numbers such that $a^2 + b^2 = 25, x^2 + y^2 = 169$, and $ax + by = 65$. If $k = ay – bx$, then

a) $0 < k \leq \frac{5}{13}$

b) $k > \frac{5}{13}$

c) $k = \frac{5}{13}$

d) k = 0

**Question 9: **David is trying to solve the expression :-

$\frac{(4)^2 \times 2^{n + 1} – 4 \times 2^n}{(4)^2 \times 2^{n + 2} – 2 \times 2^{n + 2}}$

And you help him to do the same and ﬁnally arrive at the answer with correct to one decimal which would be –

(Note:- DO NOT include spaces in your answer)

**Question 10: **Ramesh is trying to Simplify the expression

$(p + q)^3 – (p -q)^3 – 6q(p^2 – q^2)$ and if $q = 1$-

You helped him and the solution arrived was……….

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**Answers & Solutions:**

**1) Answer (C)**

Price of Darjeeling tea on 100th day= 100+(0.1*100)=110

Price of Ooty tea on nth day= 89+0.15n

Let us assume that the price of both varieties of tea would become equal on nth day where n<=100

So

89+0.15n=100+0.1n

n=220 which does not satisfy the condition of n<=100

So the price of two varieties would become equal after 100th day.

89+0.15n=110

n=140

140th day of 2007 is May 20 (Jan=31,Feb=28,March=31,April=30,May=20)

**2) Answer (B)**

Let the number be xy

10y + x = 10x + y + 18

=> 9y – 9x = 18

=> y – x = 2

So, y can take values from 9 to 4 (since 3 is already counted in 13)

Number of possible values = 6

**3) Answer (A)**

Let the price of 1 burger be x and the price of 1 shake be y and the prize of 1 french fries be z

3x + 7y + z = 120

4x + 10y + z = 164.5

=> x + 3y = 44.5

=> x = 44.5 – 3y

=> 3(44.5 – 3y) + 7y + z = 120 => z = 120 – 133.5 + 2y

So, x+y+z = 44.5 – 3y + y -13.5 + 2y = 31

So, the cost of a meal consisting of 1 burger, 1 shake and 1 french fries = Rs 31

**4) Answer (B)**

Suppose A, B and C have 5 pieces of bread, 3 pieces of bread and 8 coins respectively. Since in total there are 8 pieces of bread, each one should get around 2.66 bread. So A must give 2.33 part of his bread to C and B must give 0.33. Distributing the amount in the same ratio of bread contribution, A must get 7 coins and B must get 1 coin.

**5) Answer (B)**

Expression : $|x + 7| + |x – 8| = 16$

Case 1 : $x < -7$

=> $-(x + 7) – (x – 8) = 16$

=> $-2x + 1 = 16$

=> $x = \frac{-15}{2} = – 7.5$

Case 2 : $-7 \leq x < 8$

=> $(x + 7) – (x – 8) = 16$

=> $15 = 16$, which is not possible.

Case 3 : $x \geq 8$

=> $(x + 7) + (x – 8) = 16$

=> $2x – 1 = 16$

=> $x = \frac{17}{2} = 8.5$

$\therefore$ Sum of all possible values of $x = 8.5 – 7.5 = 1$

CMAT Free Solved Previous Papers.

**6) Answer (C)**

Let x and y be the percentage of aluminum in the 8 kg and 16 kg piece.

Let us take a kg away from both.

Thus, we will have xa and ya amount of aluminum in both the pieces.

In the remaining 8-a and 16-a pieces we will have (8-a)x and (16-a)y amount of aluminum.

Given that, $\dfrac{(8-a)x+ay}{8} = \dfrac{(16-a)y + ax}{16}$

Thus, $2*(8x-ax+ay) = 16y-ay+ax$

=> $16x-2ax+2ay = 16y-ay+ax$

Thus, $16(x-y) = 3a(x-y)$

Thus, $a = \dfrac{16}{3} = 5.33$

Hence, option C is the correct answer.

**7) Answer (B)**

x – y – z = 25 and $x\leq40,y\leq12$, $z\leq12$

If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions.Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.

If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.

If x = 37 then y + z = 12 which will give 11 solutions.

Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.

Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11

= 12*13/2 + 21

78 + 21 = 99

**8) Answer (D)**

$\left(ax+by\right)^2=65^2$

$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$

$k = ay – bx$

$k^2\ =\ a^2y^2+b^2x^2-2abxy$

$(a^2 + b^2)(x^2 + y^2 )= 25* 169$

$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$

$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$

k = 0

D is the correct answer.

**9) Answer: 0.5**

$\frac{(4)^2 \times 2^{n + 1} – 4 \times 2^n}{(4)^2 \times 2^{n + 2} – 2 \times 2^{n + 2}}$

$\frac{(2)^4 \times 2^{n + 1} – 2^2 \times 2^n}{(2)^4 \times 2^{n + 2} – 2 \times 2^{n + 2}}$

$\frac{(2)^3 \times 2^{n}*2 -2 \times 2^n}{(2)^3 \times 2^{n}*2^2 – 2^{n}*2^2}$

$\ \frac{\ 2^3\times\ 2-2}{2^3\cdot2^2-2^2}$

$\ \frac{\ 14}{28}$

=0.5

**10) Answer: 8**

($p^3+3p^2q+3pq^2+q^3$) – ($p^3-3p^2q+3pq^2-q^3$) – ($6q(p^2 – q^2)$)

=$6p^2q+2q^3-6q(p^2-q^2)$

= 8$q^3$

=8 (Since q = 1)

8 is the correct answer.

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