Linear Equations(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

LINEAR EQUATIONS MCQs

Linear Equations In One Variable, Linear Equations In Two Variables, Pair Of Linear Equations In Two Variables (8th, 9th And 10th Grade)

Total Questions : 74 | Page 7 of 8 pages

Question 61. Pawan is a cashier in a bank. He has currency notes of denominations 1000, 500 and 100 respectively. The ratio of the number of these notes is 2:3:5. If the total cash with Pawan is ₹5,00,000, then how many notes of ₹1000 denomination does he have?

Discuss Question

Answer: Option A. -> 250
:
A
Given that the denomination notes of 1000, 500, 100 are in the ratio 2:3:5.
Numberof 1000 denomination notes with Pawan=

2 x

Therefore, amount =

₹ 2000 x

Numberof 500 denomination notes with Pawan=

3 x

Therefore, amount =

₹ 1500 x

Numberof 100 denomination notes with Pawan =

5 x

Therefore, amount =

₹ 500 x

Total amount =

₹ 5, 00, 000

(Given)

2000 x + 1500 x + 500 x = ₹ 5, 00, 000

4000 x = 5, 00, 000

x = \frac{5, 00, 000}{4, 000} = 125

Therefore, number of 1000 denomination notes =

2 x = 2 \times 125

250

notes.

Question 62. Solve for x:

\frac{x + 5}{6} - \frac{x + 1}{9} = \frac{x + 3}{4}

−27
−17
7
−57

Discuss Question

Answer: Option B. -> −17
:
B
Given,

\frac{x + 5}{6} - \frac{x + 1}{9} = \frac{x + 3}{4}

Multiplying both sides by 36 (LCM of 4, 6 and 9), we get,

(\frac{x + 5}{6} \times 36) - (\frac{x + 1}{9} \times 36) = (\frac{x + 3}{4} \times 36)

\Rightarrow 6 (x + 5) - 4 (x + 1) = 9 (x + 3)

\Rightarrow 6 x + 30 - 4 x - 4 = 9 x + 27

\Rightarrow 2 x + 26 = 9 x + 27

Transposing 9x from RHS to LHS, and 26 from LHS to RHS, we get,

2 x - 9 x = 27 - 26

\Rightarrow - 7 x = 1

Dividing both sides by -7, we get,

x = - \frac{1}{7}

Question 63. The perimeter of a rectangular swimming pool is

64 m

. If its length and breadth are in the ratio of

5 : 3

, then find the length of the pool (in metre).

Discuss Question

Answer: Option B. -> 20
:
B
Perimeter of swimming pool

= 64 (G i v e n)

Perimeter of rectangle

= 2 (l e n g t h + b r e a d t h)

Since length and breadth are given to be in the ratio

5 : 3

, we can assume them to be

5 x

and

3 x

respectively.

\Rightarrow

Length =

5 x

andbreadth =

3 x

Substituting these in the formula of perimeter, we get

64 = 2 (5 x + 3 x) \Rightarrow 8 x = 32 \Rightarrow x = 4

∴

Length of the swimming pool

= 5 x = 5 \times 4 = 20 m

Question 64. If a sequence follows the pattern

2.5 n - 10

, then find the

10^{t h}

term of the sequence.

Discuss Question

Answer: Option C. -> 15
:
C
By substituting n= 10 in the given equation, we can find out the

10^{t h}

term i.e.,

2.5 x - 10

=(2.5

\times

10) - 10
= 25- 10
= 15

Question 65. The present age of Vinay’s mother is three times the present age of Vinay. After 5 years, sum of their ages will be 70 years. Find their present ages.

10 years, 30 years
15 years, 45 years
20 years, 60 years
18 years, 54 years

Discuss Question

Answer: Option B. -> 15 years, 45 years
:
B

Let Vinay’s present age be x Then, his mother’s present age = 3 x

After 5 years, Vinay’s age = x + 5 Vinay's mother’s age = 3 x + 5

Sum of Vinay and Vinay's Mother's age after 5 years = 70

\Rightarrow x + 5 + 3 x + 5 = 70 \Rightarrow 4 x + 10 = 70 \Rightarrow 4 x = 60 \Rightarrow x = 15 \Rightarrow 3 x = 45

∴ Vinay’s age is 15 years and his

mother’s age is 45 years.

Question 66. The altitude of a triangle is three-fifth the length of its corresponding base. If the altitude is increased by 4 cm and the base is decreased by 2 cm, then the area of the triangle remains the same. Find the base of the triangle in cm.

Discuss Question

Answer: Option C. -> 267
:
C

Let the base of the triangle be x c m .

Then, the corresponding altitude of the triangle would be \frac{3 x}{5} c m .

If altitude is increased by 4 cm, the new altitude = \frac{3 x}{5} + 4 c m

New base = x - 2 c m

Area of triangle = \frac{1}{2} \times base \times height

Since the area remains the same, \frac{1}{2} \times x \times \frac{3 x}{5} = \frac{1}{2} \times (\frac{3 x}{5} + 4) \times (x - 2)

\Rightarrow \frac{3 x^{2}}{5} = \frac{3 x^{2}}{5} + 4 x - \frac{6 x}{5} - 8

\Rightarrow 4 x - \frac{6 x}{5} = 8 \Rightarrow \frac{20 x - 6 x}{5} = 8

\Rightarrow 14 x = 40

\Rightarrow x = \frac{40}{14} = \frac{20}{7} \Rightarrow x = 2 \frac{6}{7} cm

Hence, base of the triangle is 2 \frac{6}{7} c m long.

Question 67. Half the students of a class are dancing and three-fourths of the remaining are playing cricket. If the remaining 9 students are studying in the library, then find the total number of students in the class.

Discuss Question

Answer: Option A. -> 72
:
A
Let the total number of students be

x

According to the question,
Numberof students dancing

= \frac{x}{2}

Numberof students remaining

= x - \frac{x}{2}

= \frac{x}{2}

Students playing cricket

= (\frac{3}{4}) (\frac{x}{2})

= \frac{3 x}{8}

Remaining students

= 9

\Rightarrow (x - \frac{x}{2}) - \frac{3 x}{8} = 9 \Rightarrow \frac{x}{8} = 9 \Rightarrow x = 8 \times 9 \Rightarrow x = 72

Hence, the total number of students in theclass = 72

Question 68. The digit at tens place of a two digit number is two times the digit at its units place. If the digits are interchanged and added to the original number, then the result will be 66. Find the original number.

Discuss Question

Answer: Option A. -> 42
:
A
Let the digit in units placebe

x

.
Then, the digit in tens placewill be

2 x

.
So, the number will be

(10 \times 2 x) + (1 \times x) = 20 x + x = 21 x

When we interchange the digits, the number will be

(10 \times x) + (1 \times 2 x) = 10 x + 2 x = 12 x

It is given that,

21 x + 12 x = 66

\Rightarrow 33 x = 66

\Rightarrow x = \frac{66}{33}

\Rightarrow x = 2

∴

The digit in units place

= 2

The digit in tens place

= 4

The original number is

42

.
Verification:
Original number

= 42

Interchanging the digits of

42

, we get

24

.
Sum of the original number and the interchanged number is

42 + 24 = 66

Question 69. A total of ₹10,000 is distributed among 150 persons as gifts. If a gift is either of ₹50 or ₹100 denominations, then the total number of gifts worth ₹50 is 50.

True
False
2x = 3x + 4
2x + 4z = 5

Discuss Question

Answer: Option B. -> False
:
B
Total number of gifts = 150
Let the number of gifts worth ₹50bex
Then the number of gifts worth ₹100 would be (150 - x)
Amount spent on x gifts of ₹50 =₹50x
Amount spent on (150 - x) gifts of ₹100 = ₹100(150 - x)
So, 50x + 100 (150 - x) = 10000
50x + 15000 - 100x = 10000
50x = 15000 - 10000 = 5000
x = 100
The total number of ₹50 gifts =x = 100 and hence the given statement is False.

Question 70.