3 Mark Question

Question 13 Marks

Solve the following equations by trial and error method:
4x = 28

Answer

4x = 28 Here, L.H.S. = 4x and R.H.S. = 28

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
1	4 × 1 = 4	28	No
2	4 × 2 = 8	28	No
3	4 × 3 = 12	28	No
4	4 × 4 = 16	28	No
5	4 × 5 = 20	28	No
6	4 × 6 = 24	28	No
7	4 × 7 = 28	28	Yes

Therefore, if x = 7, L.H.S. = R.H.S.
Hence, x = 7 is the solution to this equation.

View full question & answer→

Question 23 Marks

$\frac{\text{x}}{4}=\frac78$

Answer

$\frac{\text{x}}{4}=\frac78$
Multiplying both sides by 4, we get
$\frac{\text{x}}{4}\times4=\frac{7}{8}\times4$
$\text{x}=\frac72$
Verification:
Substituting $\text{x}=\frac72$ in L.H.S., we get
$\text{L.H.S.}=\frac{\frac72}{4}=\frac78,$ and $\text{R.H.S.}=\frac78$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.

View full question & answer→

Question 33 Marks

One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?

Answer

Let the number of sea shells found by Sandy = 'x'.
So, the number of sea shells found by Anita = (x + 5).
The number of sea shells found by Shella = 2 (x + 5).
According to the question:
⇒ x + 2 (x + 5) = 16
⇒ x + 2x + 10 = 16
⇒ 3x + 10 = 16
Subtracting 10 from both sides, we get
⇒ 3x + 10 - 10 = 16 - 10
⇒ 3x = 6
Dividing both sides by 3, we get
$\Rightarrow\frac{\text{3x}}{3}=\frac{6}{3}$
$\Rightarrow\text{x}=2$
Thus, the number of sea shells found by Sandy = x = 2, the number of sea shells found by Anita = x + 5 = 2 + 5 = 7, and the number of sea shells found by Shella = 2(x + 5) = 2(2 + 5) = 2(7) = 14.

View full question & answer→

Question 43 Marks

The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres, find the dimensions of the field.

Answer

Let the breadth of the rectangle = ‘x’ metres.
According to the question:
Length of the rectangle = ‘2x’ metres
Perimeter of a rectangle = 2 (length + breadth)
So, 2(2x + x) = 228
⇒ 2(3x) = 228
⇒ 6x = 228
Dividing both sides by 6, we get
$\Rightarrow\frac{6\text{x}}{6}=\frac{228}{6}$
$\Rightarrow\text{x}=38$
So, the breadth of the rectangle = x = 38 metres,
and the length of the rectangle = 2x = 2(38) = 76 metres.

View full question & answer→

Question 53 Marks

$\frac{3}{\text{x}}=0$

Answer

$\frac{3}{\text{x}}=0$
Dividing both sides by 3, we get
$\frac{3\text{x}}{3}=\frac{0}{3}$
$\text{x}=0$
Verification:
Substituting x = 0 in L.H.S. = 3x, we get
L.H.S. = 3 × 0 = 0 and R.H.S. = 0
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 63 Marks

x - 3 = 5

Answer

x - 3 = 5
Adding 3 to both sides, we get
x - 3 + 3 = 5 + 3
x = 8
Verification:
Substituting x = 8 in L.H.S., we get
L.H.S. = x - 3 and R.H.S. = 5
L.H.S. = 8 - 3 = 5 and R.H.S. = 5
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 73 Marks

In a hostel mess, 50kg rice are consumed everyday. If each student gets 400gm of rice per day, find the number of students who take meals in the hostel mess.

Answer

Let the number of students in the hostel be ‘x’.
Quantity of rice consumed by each student = 400gm.
So, daily rice consumption in the hostel mess = 400(x).
But, daily rice consumption = 50kg = 50 × 1000 = 50000gm [since 1kg = 1000gm].
According to the question:
400x = 50000
Dividing both sides by 400, we get
$\Rightarrow\frac{400\text{x}}{400}=\frac{50000}{400}$
$\Rightarrow\text{x}=125$
Thus, 125 students have their meals in the hostel mess.

View full question & answer→

Question 83 Marks

Solve the following equations by trial and error method:
$\frac{\text{x}}{4}= 12$

Answer

$\frac{\text{x}}{4}=12$
Here, $\text{L.H.S.}=\frac{\text{x}}{4}$ and R.H.S. = 12
Since R.H.S. is a natural number, $\frac{\text{x}}{4}$ must also be a natural number, so we must substitute values of x that are multiples of 4.

X	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
$16$	$\frac{16}{4}=4$	$12$	No
$20$	$\frac{20}{4}=5$	$12$	No
$24$	$\frac{24}{4}=6$	$12$	No
$28$	$\frac{28}{4}=7$	$12$	No
$32$	$\frac{32}{4}=8$	$12$	No
$36$	$\frac{36}{4}=9$	$12$	No
$40$	$\frac{40}{4}=10$	$12$	No
$44$	$\frac{44}{4}=11$	$12$	No
$48$	$\frac{48}{4}=12$	$12$	Yes

Therefore, if x = 48, L.H.S. = R.H.S.
Hence, x = 48 is the solution to this equation.

View full question & answer→

Question 93 Marks

A man is 4 times as old as his son. After 16 years, he will be only twice as old as his son. Find the their present ages.

Answer

Let the present age of the son = ‘x’ years.Therefore, the present age of his father = ‘4x’ years.
So, after 16 years,
Son’s age = (x + 16) and father’s age = (4x + 16) years
According to question:
⇒ 4x + 16 = 2(x + 16)
⇒ 4x + 16 = 2x + 32
Transposing 2x to L.H.S. and 16 to R.H.S., we get
⇒ 4x - 2x = 32 - 16
⇒ 2x = 16
Dividing both sides by 2, we get
$\frac{2\text{x}}{2}=\frac{16}{2}$
⇒ x = 8
So, the present age of the son = x = 8 years, and the present age of the father = 4x = 4(8) = 32 years.

View full question & answer→

Question 103 Marks

Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. If Andy has 75 marbles more than Sandy. How many does each of them have?

Answer

Let the number of marbles with Pandy = ‘x’.So, the number of marbles with Andy = ‘2x’.
Thus, the number of marbles with Sandy $=\frac{\text{x}}{2}+\frac{2\text{x}}{2}=\frac{3\text{x}}{2}$
According to the question:
$2\text{x}+75=\frac{3\text{x}}{2}$
$2\text{x}-\frac{3\text{x}}{2}=-75$
$\frac{4\text{x}-3\text{x}}{2}=-75$
$\frac{\text{x}}{2}=-75$
$\text{x}=-150$
Since, no. of marbles can not be negative.
Therefore, x = 150
So, Pandy has 150 marbles, Andy has 2x = 2(150) = 300 marbles, and Sandy has $\frac{3\text{x}}{2}=225$ marbles.

View full question & answer→

Question 113 Marks

x + 9 = 13

Answer

x + 9 = 13
Subtracting 9 from both sides, we get
⇒ x + 9 - 9 = 13 – 9
⇒ x = 4
Verification:
Substituting x = 4 on L.H.S., we get
L.H.S. = 4 + 9 = 13 = R.H.S.
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 123 Marks

Solve the following equations by trial and error method:
x + 3 = 12

Answer

x + 3 = 12
Here, L.H.S. = x + 3 and R.H.S. = 12

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
1	1 + 3 = 4	12	No
2	2 + 3 = 5	12	No
3	3 + 3 = 6	12	No
4	4 + 3 = 7	12	No
5	5 + 3 = 8	12	No
6	6 + 3 = 9	12	No
7	7 + 3 = 10	12	No
8	8 + 3 = 11	12	No
9	9 + 3 = 12	12	Yes

Therefore, if x = 9, L.H.S. = R.H.S.
Hence, x = 9 is the solution to this equation.

View full question & answer→

Question 133 Marks

There are only 25 paise coins in a purse. The value of money in the purse is Rs. 17.50. Find the number of coins in the purse.

Answer

Let the number of 25-paise coins in the purse be ‘x’.
So, the value of money in the purse = 0.25x.
But 0.25x = 17.5.
Dividing both sides by 0.25, we get
$\Rightarrow\frac{0.25\text{x}}{0.25}=\frac{17.5}{0.25}$
$\Rightarrow\text{x}=70$
Thus, the number of 25-paise coins in the purse = 70.

View full question & answer→

Question 143 Marks

$\text{x}-\frac35=\frac75$

Answer

$\text{x}-\frac35=\frac75$
Adding $\frac35$ to both sides, we get
$\Rightarrow\text{x}-\frac{3}{5}+\frac35=\frac{7}{5}+\frac35$
$\Rightarrow\text{x}=\frac{7}{5}+\frac{3}{5}$
$\Rightarrow\text{x}=\frac{10}{5}$
$\Rightarrow\text{x}=2$
Verification:
Substituting x = 2 in L.H.S., we get
$\text{L.H.S.}=2-\frac{3}{5}=10-\frac35=\frac75$ and $\text{R.H.S.}=\frac75$
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 153 Marks

A man says, "I am thinking of a number. When I divide it by 3 and then add 5, my answer is twice the number I thought of". Find the number.

Answer

Let the number thought of by the man be ‘x’.
So, According to question:
$\frac{\text{x}}{3}+5=2\text{x}$
Transposing $\frac{\text{x}}{3}$ to R.H.S., we get
$5=\text{2x}-\frac{\text{x}}{3}$
$5=6-\frac{\text{x}}{3}$
$5=\frac{5\text{x}}{3}$
Multiplying both sides by 3, we get
$5\times3=\frac{5\text{x}}{3}\times3$
$15=5\text{x}$
Dividing both sides by 5, we get
$\frac{15}{5}=\frac{\text{5x}}{5}$
$\text{x}=3$
Thus, the number thought of by the man is 3.

View full question & answer→

Question 163 Marks

Solve the following equations by trial and error method:
$\frac{\text{x}}{2}+7=11$

Answer

$\frac{\text{x}}{2}+7=11$
Here, $\text{L.H.S.} = \frac{\text{x}}{2}+7$ and R.H.S. = 11.
Since R.H.S. is a natural number, $\frac{\text{x}}{2}$ must also be a natural number, so we must substitute values of x that are multiples of 2.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
$2$	$\frac{2}{2}+7=8$	$11$	No
$4$	$\frac{4}{2}+7=9$	$11$	No
$6$	$\frac{6}{2}+7=10$	$11$	No
$8$	$\frac{8}{2}+7=11$	$11$	Yes

Therefore, if x = 8, L.H.S. = R.H.S.
Hence, x = 8 is the solution to this equation.

View full question & answer→

Question 173 Marks

If a number is tripled and the result is increased by 5, we get 50. Find the number.

Answer

Let the required number be ‘x’.
So, According to question:
⇒ 3x + 5 = 50
Subtracting 5 from both sides, we get
⇒ 3x + 5 - 5 = 50 - 5
⇒ 3x = 45
Dividing both sides by 3, we get
$\frac{3\text{x}}3 = \frac{45}3$
x = 15
Thus, the required number is 15.

View full question & answer→

Question 183 Marks

The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

Answer

Let the smaller number be ‘x’.So, the larger number = x + 7.
According to question:
6x + (x + 7) = 77
6x + x + 7 = 77
7x + 7 = 77
Subtracting 7 from both sides, we get
7x + 7 - 7 = 77 - 7
7x = 70
Dividing both sides by 7, we get
$\frac{7\text{x}}{7}=\frac{70}{7}$
$\text{x}=10$
Thus, the smaller number = x = 10, and the larger number = x + 7 = 10 + 7 = 17.
The two required numbers are 10 and 17.

View full question & answer→

Question 193 Marks

Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.

Answer

Let the first number be ‘x’.
Hence, the second number = x + 1 and the third number = x + 2.
⇒ Sum of first and second numbers = (x) + (x + 1).
According to question:
(x) + (x + 1) = 15 + (x + 2)
⇒ 2x + 1 = 17 + x
Transposing x to L.H.S. and 1 to R.H.S., we get
⇒ 2x - x = 17 - 1
⇒ x = 16
So, first number = x = 16, second number = x + 1 = 16 + 1 = 17 and third number = x + 2 = 16 + 2 = 18
Thus, the required consecutive natural numbers are 16, 17 and 18.

View full question & answer→

Question 203 Marks

A bag contains 25 paise and 50 paise coins whose total value is Rs. 30. If the number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.

Answer

Let the number of 50 paise coins = ‘x’.
So, the money value contribution of 50 paise coins = 0.5x.
The number of 25 paise coins = ‘4x’.
The money value contribution of 25 paise coins = 0.25(4x) = x.
According to the question:
0.5x + x = 30
⇒ 1.5x = 30
Dividing both sides by 1.5, we get
$\Rightarrow\frac{1.5\text{x}}{1.5}=\frac{30}{1.5}$
$\Rightarrow\text{x}=20$
Thus, the number of 50 paise coins = ‘x’ = 20, and the number of 25 paise coins = ‘4x’ = 4 (20) = 80.

View full question & answer→

Question 213 Marks

If 5 is subtracted from three times a number, the result is 16. Find the number.

Answer

Let the required number be ‘x’.
Then, 5 subtracted from 3 times x = 3x - 5.
⇒ 3x - 5 = 16
Adding 5 to both sides, we get
= 3x - 5 + 5 = 16 + 5
⇒ 3x = 21
Dividing both sides by 3, we get
$\frac{3\text{x}}{3}=\frac{21}{3}$
$\text{x}=7$
Thus, the required number is 7.

View full question & answer→

Question 223 Marks

3(x + 2) = 15

Answer

3 (x + 2) = 15
Dividing both sides by 3, we get
$\frac{3(\text{x}+2)}{3}=\frac{15}{3}$
(x + 2) = 5
Subtracting 2 from both sides, we get
x + 2 - 2 = 5 - 2
x = 3
Verification:
Substituting x = 3 in L.H.S., we get
L.H.S. = 3 (x + 2) = 3(3 + 2) = 3(5) = 15, and R.H.S. = 15
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 233 Marks

Shikha is 3 years younger to her brother Ravish. If the sum of their ages is 37 years, what are their present ages?

Answer

Let the present age of Shikha = ‘x’ years.So, the present age of Shikha’s brother Ravish = (x + 3) years.
So, sum of their ages = x + (x + 3)
⇒ x + (x + 3) = 37
⇒ 2x + 3 = 37
Subtracting 3 from both sides, we get
⇒ 2x + 3 - 3 = 37 - 3
⇒ 2x = 34
Dividing both sides by 2, we get
$\Rightarrow\frac{\text{2x}}{2}=\frac{34}{2}$
$\Rightarrow\text{x}=17$
So, the present age of Shikha = 17 years, and the present age of Ravish = x + 3 = 17 + 3 = 20 years.

View full question & answer→

Question 243 Marks

Mrs. Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as old as Nilu. Find their present ages.

Answer

Let the present age of Nilu = ‘x’ years
Therefore, the present age of Nilu’s mother, Mrs. Jain = (x + 27) years
So, after 8 years,
Nilu’s age = (x + 8), and Mrs. Jain’s age = (x + 27 + 8) = (x + 35) years
⇒ x + 35 = 2(x + 8)
Expanding the brackets, we get
⇒ x + 35 = 2x + 16
Transposing x to R.H.S. and 16 to L.H.S., we get
⇒ 35 - 16 = 2x - x
⇒ x = 19
So, the present age of Nilu = x = 19 years, and the present age of Nilu’s mother = x + 27 = 19 + 27 = 46 years.

View full question & answer→

Question 253 Marks

$\frac{\text{x}}{2}=0$

Answer

$\frac{\text{x}}{2}=0$
Multiplying both sides by 2, we get
$\Rightarrow\frac{\text{x}}{2}\times2=0\times2$
$\Rightarrow\text{x}=0$
Verification:
Substituting x = 0 in L.H.S., we get
L.H.S. = $\frac02$ = 0 and R.H.S. = 0
L.H.S. = 0 and R.H.S. = 0
L.H.S. = R.H.S.

View full question & answer→

Question 263 Marks

Solve the following equations by trial and error method:
x - 7 = 10

Answer

x - 7 = 10
Here, L.H.S. = x - 7 and R.H.S. = 10.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
9	9 - 7 = 2	10	No
10	10 - 7 = 3	10	No
11	11 - 7 = 4	10	No
12	12 - 7 = 5	10	No
13	13 - 7 = 6	10	No
14	14 - 7 = 7	10	No
15	15 - 7 = 8	10	No
16	16 - 7 = 9	10	No
17	17 - 7 = 10	10	Yes

Therefore, if x = 17, L.H.S. = R.H.S.
Hence, x = 17 is the solution to this equation.

View full question & answer→

Question 273 Marks

The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16. How old are the three children?

Answer

Let the age of the girl = ‘x’ years.So, the age of her younger sister = (x - 4) years.
Thus, the age of the brother = (x - 4 - 4) years = (x - 8) years.
According to question:
(x - 4) + (x - 8) = 16
x + x - 4 - 8 = 16
2x - 12 = 16
Adding 12 to both sides, we get
2x - 12 + 12 = 16 + 12
2x = 28
Dividing both sides by 2, we get
$\frac{2\text{x}}{2}=\frac{28}{2}$
x = 14
Thus, the age of the girl = x = 14 years, the age of the younger sister = x - 4 = 14 - 4 = 10 years, and the age of the younger brother = x - 8 = 14 - 8 = 6 years.

View full question & answer→

Question 283 Marks

Solve the following equations by trial and error method:
$\frac{15}{\text{x}}=3$

Answer

$\frac{15}{\text{x}}=3$ Here, $\text{L.H.S.}=\frac{15}{\text{x}}$ and R.H.S. = 3.
Since R.H.S. is a natural number, $\frac{15}{\text{x}}$ must also be a natural number, so we must substitute values of x that are factors of 15.

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
$1$	$\frac{15}{1}=15$	$3$	No
$3$	$\frac{15}{3}=5$	$3$	No
$5$	$\frac{15}{5}=3$	$3$	Yes

Therefore, if x = 5, L.H.S. = R.H.S.
Hence, x = 5 is the solution to this equation.

View full question & answer→

Question 293 Marks

Solve the following equations by trial and error method:
$\frac{\text{x}}{18}=20$

Answer

$\frac{\text{x}}{18}=20$
Here, $\text{L.H.S.}=\frac{\text{x}}{18}$ and R.H.S. = 20.
Since R.H.S. is a natural number, $\frac{\text{x}}{18}$ must also be a natural number, so we must substitute values of x that are multiples of 18.

X	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
$324$	$\frac{324}{18}=18$	$20$	No
$342$	$\frac{342}{18}=19$	$20$	No
$360$	$\frac{360}{18}=20$	$20$	Yes

Therefore, if x = 360, L.H.S. = R.H.S.
Hence, x = 360 is the solution to this equation.

View full question & answer→

Question 303 Marks

Solve the following equations by trial and error method:
2x + 4 = 3x

Answer

2x + 4 = 3x
Here, L.H.S. = 2x + 4 and R.H.S. = 3x

x	L.H.S.	R.H.S.	Is L.H.S. = R.H.S.
1	2(1) + 4 = 6	3(1) = 3	No
2	2(2) + 4 = 8	3(2) = 6	No
3	2(3) + 4 =10	3(3) = 9	No
4	2(4) + 5 = 12	3(4) = 12	Yes

Therefore, if x = 4, L.H.S. = R.H.S.
Hence, x = 4 is the solution to this equation.

View full question & answer→

Question 313 Marks

3(x + 6) = 24

Answer

3(x + 6) = 24
Dividing both sides by 3, we get
$\frac{3(\text{x}+6)}{3}=\frac{24}{3}$
(x + 6) = 8
Subtracting 6 from both sides, we get
x + 6 - 6 = 8 - 6
x = 2
Verification:
Substituting x = 2 in L.H.S., we get
L.H.S. = 3(x + 6) = 3(2 + 6) = 24, and R.H.S. = 24
L.H.S. = R.H.S.
Hence, verified.

View full question & answer→

Question 323 Marks

Find the number which when multiplied by 7 is increased by 78.

Answer

Let the required number be ‘x’.
Thus, when multiplied by 7, it gives 7x, and x increases by 78.
7x = x + 78
Transposing x to L.H.S., we get
7x - x = 78
6x = 78
Dividing both sides by 6, we get
$\frac{6\text{x}}{6}=\frac{78}{6}$
$\text{x}=13$
Thus, the required number is 13.

View full question & answer→

Maths 3 Mark Question

Take a timed test