# Simple Equations | Algebra - Simple Equations

**Simple Equations | Algebra - Simple Equations**

Find the value of p if the equations 5x + py = 1 and px + 5y = 4 have unique solution.

- A.
p ≠ 5

- B.
p ≠ -5

- C.
Both (a) and (b)

- D.
None of these

Answer: Option C

**Explanation** :

Two linear equations:

a_{1}x + b_{1}y + c_{1} = 0 and

a_{2}x + b_{2}y + c_{2} = 0

have unique solution when $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$

∴ For the given system of equations,

$\frac{5}{p}\ne \frac{p}{5}$

⇒ p^{2} ≠ 25

⇒ p ≠ ±5

Hence, option (c).

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**Simple Equations | Algebra - Simple Equations**

Find the value of p if the equations 4x + 6y = 10 and 6x + py = 20 have no solution.

- A.
p = 12

- B.
p ≠ 9

- C.
p ≠ 12

- D.
9

Answer: Option D

**Explanation** :

Two linear equations:

a_{1}x + b_{1}y + c_{1} = 0 and

a_{2}x + b_{2}y + c_{2} = 0

have no solution when $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}\ne \frac{{c}_{1}}{{c}_{2}}$

∴ For the given system of equations,

$\frac{4}{6}=\frac{6}{p}\ne \frac{10}{20}$

⇒ p = 9

Hence, option (d).

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**Simple Equations | Algebra - Simple Equations**

Find the value of p if the equations 3x + (p/3 + 2)y = 1 and px + 2py = 4 have infinite solutions.

- A.
9

- B.
6

- C.
18

- D.
12

Answer: Option D

**Explanation** :

Two linear equations:

a_{1}x + b_{1}y + c_{1} = 0 and

a_{2}x + b_{2}y + c_{2} = 0

have infinite solutions when $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}$

∴ For the given system of equations,

$\frac{3}{p}=\frac{\frac{p}{3}+2}{2p}=\frac{1}{4}$

⇒ p = 12

Hence, (d).

Workspace:

**Simple Equations | Algebra - Simple Equations**

If the following system of equations have a unique solution what can be said about the value of k?

13x + 91y = 79

23x + ky = 89

- A.
k ≠ 151

- B.
k = 151

- C.
k = 161

- D.
k ≠ 161

- E.
k ≠ 138

Answer: Option D

**Explanation** :

Two linear equations:

a_{1}x + b_{1}y + c_{1} = 0 and

a_{2}x + b_{2}y + c_{2} = 0

have unique solution when $\frac{{a}_{1}}{{a}_{2}}\ne \frac{{b}_{1}}{{b}_{2}}$

∴ For the given system of equations,

$\frac{13}{23}\ne \frac{91}{k}$

⇒ k ≠ 161

Hence, option (d).

Workspace:

**Simple Equations | Algebra - Simple Equations**

Consider three numbers a, b and c. Max (a, b, c) + Min (a, b, c) = 26. Median (a, b, c) - Mean (a, b, c) = 4. Find the median of a, b, and c.?

- A.
13

- B.
19

- C.
18

- D.
24

Answer: Option C

**Explanation** :

The numbers can be arranged in ascending order thus:

Min < Median < Max

According to the question,

Min + Max = 26

$Median\u2013\frac{Min+Median+Max}{3}$ = $Median\u2013\frac{26+Median}{3}$ = 4

Solving, we get Median = 19

Hence, option (c).

Workspace:

**Simple Equations | Algebra - Simple Equations**

Suresh, Harish and Jayant had some candies each. Together Suresh and Harish had 38 candies. Even after giving six candies to Jayant, Suresh had four more candies than him. Then Harish gave four of his candies to Jayant and was also left with four more candies than him. How many candies does Jayant have now?

- A.
2

- B.
8

- C.
10

- D.
12

- E.
16

Answer: Option D

**Explanation** :

Let Suresh, Harish and Jayant have s, h and j number of candies with them respectively.

If Suresh gives six candies to Jayant, then he is left with s − 6 candies and Jayant now has j + 6 candies.

∴ s - 6 = j + 6 + 4

⇒ s = j + 16 …(1)

If Harish gives four candies to Jayant, he is now left with h − 4 candies whereas Jayant has j + 6 + 4 = j + 10 candies.

∴ h – 4 = j + 10 + 4

⇒ h = j + 18 …(2)

Adding (1) and (2),

s + h = 2j + 34

∴ 38 = 2j + 34

∴ j = 2

⇒ Jayant has now j + 10 = 12 candies

Hence, option (d).

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**Simple Equations | Algebra - Simple Equations**

There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is

- A.
20

- B.
80

- C.
100

- D.
200

- E.
None of these

Answer: Option C

**Explanation** :

Let the number of students in rooms A and B be x and y respectively. Then, x - 10 = y + 10

⇒ x – y = 20 … (i) and

x + 20 = 2(y – 20) ⇒ x – 2y = -60… (ii)

Solving (i) and (ii), we get x = 100 and y = 80.

∴ Required answer A = 100

Hence, option (c).

Workspace:

**Simple Equations | Algebra - Simple Equations**

A sum of Rs. 1360 has been divided among A, B and C such that A gets 2/3 of what B gets and B gets 1/4 of what C gets. B's share is

- A.
Rs. 120

- B.
Rs. 160

- C.
Rs. 240

- D.
Rs. 300

- E.
None of these

Answer: Option C

**Explanation** :

Let C’s share = Rs. x

Then, B’s share = Rs. x/4, A’s share = Rs. (2/3) × x/4 = x/6

∴ x/6 + x/4 + x = 1360.

Hence, x = 960

∴ B’s share = 960/4 = Rs. 240

Hence, option (c).

Workspace:

**Simple Equations | Algebra - Simple Equations**

A thief somehow managed to steal some chocolates from a Mall but while coming out of it at the first door he was caught by the guard and he successfully dealt him by giving 1 chocolate plus half of the rest chocolates. Further he had to pay 2 chocolates and half of the rest to the second guard. Once again at the third gate he gave 3 chocolates and then half of the rest. After that he was left with only one chocolate. How many chocolates had he stolen?

Answer: 25

**Explanation** :

Let the number of chocolates initially with him = x.

To the first guard he gives 1 chocolate and half of the remaining. Hence, no. of chocolates left with the man is (x – 1)/2.

To the second guard he gives 2 chocolate and half of the remaining. Hence, no. of chocolates left with the man is $\frac{\frac{x\u20131}{2}-2}{2}$.

To the third guard he gives 3 chocolate and half of the remaining. Hence, no. of chocolates left with the man is $\frac{\left[\frac{\frac{x\u20131}{2}-2}{2}-3\right]}{2}$.

Finally, he is left with only 1 chocolate, hence $\frac{\left[\frac{\frac{x\u20131}{2}-2}{2}-3\right]}{2}=1$

⇒ $\frac{\frac{x\u20131}{2}-2}{2}-3=2$

⇒ $\frac{\frac{x\u20131}{2}-2}{2}=5$

⇒ $\frac{x\u20131}{2}-2=10$

⇒ $\frac{x-1}{2}=12$

⇒ x = 25

Hence, 25.

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**Simple Equations | Algebra - Simple Equations**

A zookeeper was asked how many animals are there in the zoo. He replied that there are all lions but 5, all monkeys but 7, and all elephants but 4. How many animals are there?

- A.
10

- B.
8

- C.
16

- D.
14

Answer: Option B

**Explanation** :

Let total number of animals is T.

There are all lions but 5 - means that except for 5 animals all other are lions.

∴ Lions = T – 5 …(1)

All monkeys but 7 – means except for 7 animals all other are monkeys.

∴ Monkey = T - 7 …(2)

All elephant but 4 – means except for 4 animals all other are elephants.

∴ Elephants = T – 4 …(3)

Adding all three equations, we get

Lions + Monkeys + Elephants = 3T – 16

⇒ T = 3T – 16

⇒ T = 8.

Hence, option (b).

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