# Algebra Questions for IIFT PDF

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## Algebra Questions for IIFT PDF

Download important IIFT Algebra Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Algebra Questions and answers for IIFT and other  exams.

Question 1: How many different pairs(a,b) of positive integers are there such that $a\geq b$ and $\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$?

Question 2: 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 3: Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindrical tank holds 500 litres more than the conical tank. After 200 litres of fuel has been pumped out from each tank the cylindrical tank contains twice the amount of fuel in the conical tank. How many litres of fuel did the cylindrical tank have when it was full?

a) 700

b) 1000

c) 1100

d) 1200

Question 4: Which one of the following conditions must p, q and r satisfy so that the following system of linear simultaneous equations has at least one solution, such that p + q + r $\neq$ 0?
x+ 2y – 3z = p
2x + 6y – 11z = q
x – 2y + 7z = r

a) 5p -2q – r = 0

b) 5p + 2q + r = 0

c) 5p + 2q – r = 0

d) 5p – 2q + r = 0

Question 5: Consider the following steps :
1. Put x = 1, y = 2
2. Replace x by xy
3. Replace y by y +1
4. If y = 5 then go to step 6 otherwise go to step 5.
5. Go to step 2
6. Stop Then the final value of x equals

a) 1

b) 24

c) 120

d) 720

Question 6: 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

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: If $12x^2- ax + 7 = ax^2 + 9x + 3$ has only one (repeated) solution, then the positive integral solution of a is:

a) 2

b) 4

c) 3

d) 5

Question 9: If $13x^2 = 17^2 – 9^2$, find the value of x?

a) 16

b) 12

c) 8

d) 4

Question 10: If 5x + 5 > 2 + 2x and 5x + 3 ≤ 4x + 5; then x can take which of the following values?

a) 3

b) -2

c) -3

d) 1

$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$
=> $ab = 9(a + b)$
=> $ab – 9(a+b) = 0$
=> $ab – 9(a+b) + 81 = 81$
=> $(a – 9)(b – 9) = 81, a > b$
Hence we have the following cases,
$a – 9 = 81, b – 9 = 1$ => $(a,b) = (90,10)$
$a – 9 = 27, b – 9 = 3$ => $(a,b) = (36,12)$
$a – 9 = 9, b – 9 = 9$ => $(a,b) = (18,18)$
Hence there are three possible positive integral values of (a,b)

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)

Let the current capacity of conical flask be C. So, cylinder = C+500.

After pumping out 200 liters, C+300 = 2(C-200) => C = 700

So, full capacity of cylinder = 700+500 = 1200

Substitute value of p,q,r in the options only option A satisfies .

5(x+2y-3z)-2(2x+6y-11z)-(x-2y+7z) = 5x+10y-15z-4x-12y+22z-x+2y-7z  = 0

1. x=1 ; y=2

2. x=2 ; y=3

3. x=6 ; y=4

4. x=24 ; y=5

Hence when y=5 , x will be 24

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 $x = 8.5 – 7.5 = 1$

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

Given, $12x^2- ax + 7 = ax^2 + 9x + 3$
$(a-12)x^2 + (a+9)x-4 = 0$
If $ax^2+bx+c=0$ has equal roots, then $b^2 = 4ac$
$(a+9)^2 = 4(a-12)(-4)$
$a^2+81+18a = 192-16a$
$a^2+34a-111 = 0$
On solving above equation, we get a = 3 and a = -37.
Here, The positive integral solution will be 3.

Expression : $13x^2 = 17^2 – 9^2$

=> $13 x^2 = (17 – 9) (17 + 9)$

=> $13 x^2 = 8 \times 26$

=> $x^2 = 8 \times 2 = 16$

=> $x = \sqrt{16} = 4$

Expression 1 : 5x + 5 > 2 + 2x

=> $5x – 2x$ > $2 – 5$

=> $3x$ > $-3$

=> $x$ > $-1$ ————-(i)

Expression 2 : 5x + 3 ≤ 4x + 5

=> $5x – 4x \leq 5 – 3$

=> $x \leq 2$ ————(ii)

Combining inequalities (i) and (ii), => $-1$ < $x \leq 2$

Thus, the values that $x$ can take = 0 , 1 , 2

=> Ans – (D)

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