# 3 Variable Equations | Algebra - Simple Equations

**3 Variable Equations | Algebra - Simple Equations**

Find the value of x + y + z, if x, y and z satisfies the given system of equations:

x - 2y + 3z = 9

4x + y - 5z = -6

x + 2y + 6z = 22

- A.
8

- B.
9

- C.
6

- D.
4

Answer: Option C

**Explanation** :

Given:

x - 2y + 3z = 9 …(1)

4x + y - 5z = -6 …(2)

x + 2y + 6z = 22 …(3)

Now, eliminating variable x.

(3) – (1)

⇒ 4y + 3z = 13 …(4)

(3) × 4 – (2)

⇒ 7y + 29z = 94 …(5)

Solving (4) and (5), we get y = 1, z = 3 and x = 2.

Hence, x + y + z = 2 + 1 + 3 = 6.

Hence, option (c).

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**3 Variable Equations | Algebra - Simple Equations**

The difference between a three-digit number and the number formed by reversing its digits is 396. Find the difference of its first and last digits.

Answer: 4

**Explanation** :

Let the number be abc with a > b

Difference of abc and cba = abc – cba = 100a + 10b + c - (100c + 10b + a) = 99(a - c)

∴ 99(a - c) = 396

⇒ a - c = 4

Hence, 4.

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh, Bharat, Chetan and Dinesh have a total of Rs. 300 with them. Ashutosh has one-fourth of the total amount with the others. Find the amount with Ashutosh (in Rs.).

- A.
40

- B.
50

- C.
60

- D.
75

Answer: Option C

**Explanation** :

Let the amount with Ashutosh be Rs. a.

∴ Total amount with the others = Rs. (300 – a)

⇒ a = ¼ × (300 -a)

⇒ 4a = 300 – a

⇒ 5a = 300

⇒ a = 60.

Hence, option (c).

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**3 Variable Equations | Algebra - Simple Equations**

In a stationery shop 5 notebooks, 7 pencils and 9 erasers cost Rs 134 and 8 notebooks, 11 pencils and 14 erasers cost Rs 210 then how much does one notebook, one pencil and one eraser cost?

- A.
20

- B.
24

- C.
18

- D.
Can't be determined

Answer: Option C

**Explanation** :

Let the cost of notebook, pencil and eraser be Rs. n, l and e respectively.

Hence,

⇒ 5n + 7l + 9e = 134 …(1)

⇒ 8n + 11l + 14e = 210 …(2)

Multiplying first equation with x and second equation with y and adding both the equations we get.

(5x + 8y)n + (7x + 11y)l + (9x + 14y)e = 134x + 210y … (3)

Now, we want the value of n + l + e …(4)

Hence, equating the coefficients, we get

5x + 8y = 1

7x + 11y = 1 &

9x + 14y = 1

Solving the above equations, we get x = -3 and m = 2

Putting these values in equation (3), we get

n + l + e = 134 × (-3) + 210 × 2 = 18

Hence, option (c).

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh goes to buy 10 notebooks, 5 pencils and three erasers which cost Rs 430. Had he bought just 2 notebooks and one pencil it would have cost him Rs 74. If he started with Rs 400 and tried to buy 6 notebooks, 3 pencils and 5 erasers, how much money would he be left with?

- A.
62

- B.
78

- C.
102

- D.
Can't be determined

Answer: Option B

**Explanation** :

Let n, l and e be the cost of notebook, pencil and eraser.

Hence, 10n + 5l + 3e = 430 …(1)

and 2n + l = 74

⇒ 10n + 5l = 74 × 5 = 370 …(2)

Solving (1) and (2)

⇒ 3e = 60 ⇒ e = 20.

We want the value of 6n + 3l + 5e = 3 × (2n + l) + 100 = 3 × 74 + 100 = 322

Hence, 400 – 322 = Rs. 78 would be left.

Hence, option (b).

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh goes to buy two mobiles, five binoculars and three PCs and realizes he has Rs. 14 less than what is required. He then removes three binoculars from his basket and can buy the remaining items with Rs. 4 left over. If he had tried buying a mobile and three PCs he would have Rs. 34 remaining, then what is the maximum number of PCs that could have been bought by Ashutosh?

- A.
4

- B.
3

- C.
5

- D.
Can't be determined

Answer: Option D

**Explanation** :

Let m, b and p be the cost of mobile, binocular and PC.

Let a be the amount with Ashutosh.

Hence,

2m + 5b + 3p = a + 14 …(1)

2m + 2b + 3p = a – 4 …(2)

(1) – (2)

⇒ b = 6.

Now, 2m + 3p = a – 16 (From (1) ∵ b = 6) …(3)

Also given, m + 3p = a – 34 …(4)

Solving (3) and (4) we get,

⇒ m = 18.

Hence, 3p = a - 52.

As there is no other equation given, the answer cannot be determined

Hence, option (d).

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh has 262 coins consisting of one rupee, 50 paisa and 25 paisa coins. The total value of the coins is Rs. 130. If the 50 paisa and 25 paisa coins are interchanged, the value comes down by Rs. 6. Find the number of 1-rupee coins he has?

- A.
46

- B.
62

- C.
38

- D.
54

Answer: Option A

**Explanation** :

Let X be the number of 1-rupee coin, Y be the 50 paise coin and Z be the 25 paise coin.

X + Y + Z = 262,

$X+\frac{Y}{2}+\frac{Z}{4}=130$,

$X+\frac{Y}{4}+\frac{Z}{2}=124$.

So, X = 23.

Hence, option (a).

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh has some Math, Physics and Chemistry books with him. He has more Math books than Physics books and more Physics books than Chemistry books, and he has 31 books in total. If he has at least 8 books of each subject, find the maximum number of Math books he can have.

- A.
11

- B.
12

- C.
13

- D.
14

Answer: Option D

**Explanation** :

Math > Physics > Chemistry.

The possibilities are (14, 9, 8), (13, 10, 8), (12, 11, 8), (11, 10, 9).

Therefore, the maximum number of Math books that he can have is 14.

Hence, option (d).

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**3 Variable Equations | Algebra - Simple Equations**

Ashutosh saved Rs. 100, Rs. 500 and Rs. 1,000 notes of value Rs. 54,400 in his house. In the following year, he accumulated black money and doubled the number of Rs. 500 and Rs. 1000 notes to get a balance of Rs. 107,400. One fine day, Rs. 500 and Rs. 1,000 are demonetized and he is able to deposit only the Rs. 100 notes. How much will his bank balance be if he starts a new bank account?

Answer: 1400

**Explanation** :

Let the number of hundred-rupee notes by h.

number of original five hundred-rupee notes by f.

and number of original thousand-rupee notes be t.

100h + 500f + 1000t = 54,400 ... (1)

After the black money, 100h + 1000f + 2000t = 107,400 .... (2)

Multiplying (1) by 2, we get 200h+ 1000f + 2000t = 108,800 ... (3)

Subtracting (2) from (3), we get 100 h = 1400

Hence, 1400.

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**3 Variable Equations | Algebra - Simple Equations**

The cost of 5 apples, 7 mangoes and 9 oranges is 184. The cost of 2 apples, 5 mangoes and 8 oranges is 144. What is the cost of 1 apple 1 orange and 1 mango?

- A.
32

- B.
16

- C.
20

- D.
24

- E.
Can't be determined

Answer: Option D

**Explanation** :

Let the cost of 1 apple be a, cost of 1 mango be m and the cost of 1 orange be k

Then,

5a + 7m + 9k = 184 ...(1)

2a + 5m + 8k = 144 ...(2)

We can observe the difference between the coefficients in equations (1) and (2) are 2 and 3 respectively.

So, multiplying equation (1) by 3 and equation (2) by 2, we get:

15a + 21m + 27k = 552

4a + 10m + 16k = 288

Now subtracting the two equations gives:

11a + 11m + 11k = 264

⇒ a + m + k = 264/11

⇒ a + m + k = 24

Hence, option (d).

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