Practice Questions for Algebra - Simple Equations

1. 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

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Explanation :

Two linear equations:

a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0

have unique solution when $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

∴ For the given system of equations,

$\frac{5}{p} \neq \frac{p}{5}$

⇒ p² ≠ 25

⇒ p ≠ ±5

Hence, option (c).

2. 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

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Explanation :

Two linear equations:

a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0

have no solution when $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

∴ For the given system of equations,

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

⇒ p = 9

Hence, option (d).

3. 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

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Explanation :

Two linear equations:

a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 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,

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

⇒ p = 12

Hence, (d).

4. 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

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Explanation :

Two linear equations:

a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0

have unique solution when $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

∴ For the given system of equations,

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

⇒ k ≠ 161

Hence, option (d).

5. 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 B

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Explanation :

The numbers can be arranged in ascending order thus:

Min < Median < Max

According to the question,

Min + Max = 26

$M e d i a n - \frac{M i n + M e d i a n + M a x}{3}$ = $M e d i a n - \frac{26 + M e d i a n}{3}$ = 4

Solving, we get Median = 19

Hence, option (b).

6. 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

7. 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

8. 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

9. 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?

Backspace

7 8 9 4 5 6 1 2 3 0 . -

Clear All Submit

Answer: 25

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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 $\frac{\frac{x - 1}{2} - 2}{2}$ .

To the third guard he gives 3 chocolate and half of the remaining. Hence, no. of chocolates left with the man is $\frac{[\frac{\frac{x - 1}{2} - 2}{2} - 3]}{2}$ .

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

⇒ $\frac{\frac{x - 1}{2} - 2}{2} - 3 = 2$

⇒ $\frac{\frac{x - 1}{2} - 2}{2} = 5$

⇒ $\frac{x - 1}{2} - 2 = 10$

⇒ $\frac{x - 1}{2} = 12$

⇒ x = 25

Hence, 25.

10. 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

Simple Equations | Algebra - Simple Equations

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