Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 7 Finding the Unknown MCQ improves accuracy in objective exams.
MCQ on Finding the Unknown Class 7
Finding the Unknown MCQ Class 7
Class 7 Maths Finding the Unknown MCQ
Question 1.
A weighing scale is balanced when two identical boxes and a 4-unit weight are placed on one pan, and one identical box with a 12-unit weight is placed on the other pan. What is the weight of one box?
(a) 6
(b) 4
(c) 8
(d) 12
Solution:
(c) 8
Let the weight of one box be x units. Then,
Total weight on one pan = (2x + 4) units
Total weight on other pan = (x + 12) units
As weighing scale is balanced, weights on boththe pans are equal.
∴ 2x + 4 = x + 12
⇒ 2x – x = 12 – 4 ⇒ x = 8
Hence, the weight of one box is 8 units.
Question 2.
A sequence of arrangements uses 4 sticks in the first step and increases by 3 sticks in every next step. Which expression gives the number of sticks in step n?
(a) 3n
(b) 3n + 1
(c) 3n + 2
(d) 3n – 1
Solution:
(b) 3n + 1
Number of sticks in:
1st step = 4
2nd step = 4 + 3 × 1
3rd step = 4 + 3 × 2
4th step = 4 + 3 × 3
.
.
.
.
.
nth step = 4 + 3 × (n – 1) = 4 + 3n – 3
= 3n + 1
Hence, the number of sticks in step n is 3n + 1.
Question 3.
The value of m so that 0.6 m – (-0.8 m + 0.4) = 0.2 – 0.5 m, is:
(a) \(\frac{7}{19}\)
(b) \(\frac{6}{19}\)
(c) \(\frac{9}{19}\)
(d) \(\frac{8}{19}\)
Solution:
(b) \(\frac{6}{19}\)
Given, 0.6m – (- 0.8 m + 0.4) = 0.2 – 0.5m
⇒ 0.6m + 0.8m – 0.4 = 0.2 – 0.5m
⇒ 1.4m – 0.4 = 0.2 – 0.5m
⇒ 1.4m + 0.5m = 0.2 + 0.4
⇒ 1.9m = 0.6
⇒ m = \(\frac{0.6}{1.9}\) ⇒ m = \(\frac{6}{19}\)
Question 4.
Which of the following is correct about x = x?
(a) The equation has no solution.
(b) The equation has exactly one solution.
(c) The equation has infinitely many solutions.
(d) The equation is incorrect.
Solution:
(c) The equation has infinitely many solutions.
We have, x = x
It means it is true (LHS = RHS) for all values of the variable. Hence, it has infinitely many solutions.
Question 5.
It is given that \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\), then the value of \(\frac{45}{56} \times \frac{14}{9} \times 5\) is:
(a) \(\frac{3150}{504}\)
(b) \(\frac{2850}{564}\)
(c) \(\frac{3150}{564}\)
(d) \(\frac{2850}{504}\)
Solution:
(a) \(\frac{3150}{504}\)
Given, \(\frac{45}{56} \times \frac{14}{9} \times \frac{7}{3} \times 5=\frac{22050}{1512}\)
⇒ \(\frac{45}{56} \times \frac{14}{9} \times 5=\frac{22050}{1512} \times \frac{3}{7}=\frac{3150}{504}\)
Question 6.
If 3x – \(\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\), then the value of x which satisfies the given equation is:
(a) Natural number greater than 7
(b) Natural number between 5 and 7
(c) Natural number less than 2
(d) Natural number between 2 and 4
Solution:
(d) Natural number between 2 and 4
Given, \(3 x-\frac{4}{5}=\frac{5 x}{3}+\frac{16}{5}\)
⇒ \(3 x-\frac{5 x}{3}=\frac{16}{5}+\frac{4}{5}\) ⇒ \(\frac{9 x-5 x}{3}=\frac{16+4}{5}\)
⇒ \(\frac{4 x}{3}=\frac{20}{5}\) ⇒ \(\frac{4 x}{3}=4\)
⇒ x = \(\frac{4 \times 3}{4}\) ⇒ x = 3
Question 7.
The sum of two consecutive even numbers is 70. The greater one is:
(a) 32
(b) 34
(c) 36
(d) 38
Solution:
(c) 36
Let x and x + 2 be two consecutive even numbers.
According to question, we have
x + (x + 2) = 70
⇒ 2x + 2 = 70 ⇒ 2x = 70 – 2
⇒ 2x = 68 ⇒ x = \(\frac{68}{2}\) = 34
So, smaller even number = 34 and greater even number = 34 + 2 = 36
Question 8.
Rajini has three balls, A, B and C. A is twice as heavy as B, B weighs one third of C. The heaviest ball is:
(a) A
(b) B
(c) C
(d) A and C
Solution:
(c) C
Let the weight of ball C be x units. Then,
Weights of ball B and ball A are \(\frac{x}{3}\) units and \(\frac{2x}{3}\) units respectively. Clearly, x > \(\frac{2x}{3}\) > \(\frac{x}{3}\)
[As weight cannot be negative, x > 0]
Thus, the heaviest ball is C.
Question 9.
Which of the following equations represents the statement “twice of a number is 25 more than one-third of the number”?
(a) \(2 x=\frac{x}{3}+25\)
(b) \(2 x+25=\frac{x}{3}\)
(c) \(2 x=\frac{25+x}{3}\)
(d) \(3 x=25+\frac{x}{2}\)
Solution:
(a) \(2 x=\frac{x}{3}+25\)
Let the number be x. Then,
Twice of the number = 2x and one-third of the number = \(\frac{x}{3}\)
According to question, we have 2x = \(\frac{x}{3}\) + 25
Question 10.
Adding 7 to the thrice of a whole number gives 34. The whole number is:
(a) 13
(b) 9
(c) 14
(d) 11
Solution:
(b) 9
Let the whole number be x. Then, thrice of the whole number = 3x
According to question, we have
3x + 7 = 34 ⇒ 3x = 34 – 7
⇒ 3x = 27 ⇒ x = \(\frac{27}{3}\) = 9
Question 11.
Which of the following equations have t = – 3 as a solution?
(i) 2t + 7 = 1
(ii) 4(t + 5) = 8
(iii) 2t + \(2\left(t+\frac{1}{4}\right)=\frac{-5}{2}\) + t
(iv) -2 = \(\frac{3}{5}\) – 7t
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (i) and (ii)
(i) LHS = 2t + 7 = 2 × (-3) + 7 = -6 + 7
= 1 = RHS
∴ t = – 3 is the solution of equation 2t + 7 = 1.
(ii) LHS = 4(t + 5) = 4(-3 + 5) = 4 × 2 = 8 = RHS
∴ t = – 3 is the solution of equation
4 (t + 5) = 8.
(iii) LHS = 2t + 2\(\left(t+\frac{1}{4}\right)\)
= 2 x (-3) + 2 \(\left(-3+\frac{1}{4}\right)\)
= -6 + \(2\left(\frac{-11}{4}\right)\) = -6 – \(\frac{11}{2}\)
= \(-\frac{23}{2}\)
RHS = \(\frac{3}{5}-7 t=\frac{3}{5}-7 \times(-3)\)
= \(\frac{3}{5}+21=\frac{3+105}{5}=\frac{108}{5}\) ≠ -2 = LHS
Question 12.
“48 is divided into 2 parts such that three times the greater part is 10 less than four times the smaller part”.
Identify the correct statements about the given situation.
(i) The smaller part is 16.
(ii) The smaller part is 22.
(iii) The greater part is 26.
(iv) The greater part is 32.
Choose the correct option from the following:
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Solution:
(c) (ii) and (iii)
Let the greater part be x. Then, the smaller part = 48 – x
According to given situation, we have
3x = 4(48 – x) – 10 ⇒ 3x = 4 × 48 – 4x – 10
⇒ 3x = 192 – 4x – 10 ⇒ 3x + 4x = 192 – 10
⇒ 7x = 182 ⇒ x = \(\frac{182}{7}\) = 26
So, the greater part is 26 and the smaller part is 48 – 26 = 22.
Thus, statements (ii) and (iii) are correct.
Question 13.
The sum of three consecutive odd numbers is 63.
Identify the correct statements about the given situation.
(i) Smallest odd number is 17.
(ii) Largest odd number is 23.
(iii) Smallest odd number is 19.
(iv) Largest odd number is 21.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii)
We know, difference between two consecutive odd numbers is 2.
Let x, x + 2 and x + 4 be the three consecutive odd numbers.
∴ x + (x + 2) + (x + 4) = 63
[Given, sum of three consecutive odd numbers is 63.]
⇒ 3x + 6 = 63 ⇒ 3x = 63 – 6
⇒ 3x = 57 ⇒ x = \(\frac{57}{3}\) = 19
So, the three consecutive odd numbers are 19, 21 and 23.
Thus, the smallest odd number is 19 and the largest odd number is 23.
Finding the Unknown Class 7 Assertion and Reason Questions
The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Question 1
(A): Only addition can be done on both sides of an equation, not subtraction.
(R): An equation remains balanced when we perform the same operation on both sides.
Solution:
(d) A is false but R is true.
An equation remains balanced when we perform the same operation (addition, subtraction, multiplication, division) on both sides.
∴ Assertion (A) is false, but Reason (R) is true.
Question 2.
(A): Two complementary angles differ by 8° can be represented by equation x – (90° – x) = 8°.
(R): Sum of complementary angles is 180°.
Solution:
(c) A is true but R is false.
We know, sum of complementary angles is 90°.
Given, two complementary angles differ by 8°.
Let x be one of the two complementary angles. Then, the other angle = 90° – x
∴ x – (90° – x) = 8° is the required equation.
Thus, Assertion (A) is true, but Reason (R) is false.
Question 3.
(A): For the given triangle, y + 2 (y + 15°) = 180°.
(R): In an isosceles triangle, angles opposite to equal sides are equal.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
The given triangle is isosceles.
We know, angles opposite to equal sides of an isosceles triangle are equal.
∴ First angle = y, second angle = third angle
= y + 15°
Now, y + (y + 15°) + (y + 15°) = 180°
[The sum of all the angles of a triangle is 180°.]
⇒ y + 2(y + 15°) = 180°
∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Finding the Unknown Class 7 Fill in the Blanks
Question 1.
If the solution is known, the number of equations that can be built are ________ .
Solution: infinite
If the solution is known, the number of equations that can be built are infinite.
Question 2.
Fill in the blanks with integers:
(i) 11 (11 + ____ ) – 98 = 100
(ii) – 3 × (-9 + _____) = 42
Solution: 7, -5
(i) Let y be the required value to fill the blank.
∴ 11(11 + y) – 98 = 100
⇒ 11(11 + y) = 100 + 98
[Adding 98 to both sides]
⇒ 11(11 + y) = 198
⇒ (11 + y) = 198 ÷ 11
[Dividing both sides by 11]
⇒ 11 + y= 18
⇒ y = 18 – 11 = 7
So, 11(11 + 7) – 98 = 100
(ii) Let y be the required value to fill the blank.
∴ -3 × (- 9 + y) = 42
⇒ (- 9 + y) = 42 ÷ (-3)
[Dividing both sides by – 3]
⇒ – 9 + y = -14
⇒ y = – 14 + 9 [Adding 9 to both sides]
⇒ y = – 5
So, – 3 × (- 9 + (- 5)) = 42
Question 3.
In an equation, there is always an ________ sign.
Solution: equals
In an equation, there is always an equals sign.
Question 4.
The value of the variable which satisfies a equation is called a ________ to the equation.
Solution: solution
The value of the variable which satisfies an equation is called a solution to the equation.
Question 5.
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}\) = ______ .
Solution: \(\frac{x}{6}\) + 3
Let x be the number. Then,
One-fifth of the number = \(\frac{x}{5}\) and one-sixth of the number = \(\frac{x}{6}\)
\(\frac{x}{5}=\frac{x}{6}\) + 3
The equation representing “one-fifth of a number is 3 more than one-sixth of the same number” is \(\frac{x}{5}=\frac{x}{6}\) + 3.
Question 6.
If \(\frac{x}{7}\) of a number is 60. the number is _______ .
Solution: 105
Let the number be x.
∴ \(\frac{4}{7}\) × x = 60
⇒ x = 60 × \(\frac{7}{4}\) = 15 × 7 = 105
If \(\frac{4}{7}\) of a number is 60, the number is 105.