CBSE Class 7 Mathematics Algebraic Expressions Assignment Set 03

Question 1. Find the value of \( m \) if the value of \( 2x^2 + x - m \) equal 5, when \( x = 0 \)
Answer: Given expression is \( 2x^2 + x - m \). We are given that this expression equals 5 when \( x = 0 \). Substitute \( x = 0 \) into the expression: \( 2(0)^2 + 0 - m = 5 \)
\( 2(0) + 0 - m = 5 \)
\( 0 - m = 5 \)
\( -m = 5 \)
\( \implies m = -5 \) Hence, the value of \( m \) is \( -5 \).
In simple words: Put 0 in place of x. That leaves -m = 5, which means m = -5.
Exam Tip: Keep the minus sign attached to \( m \) during substitution so you do not accidentally write \( +5 \).

Question 2. If \( x = 2 \) and \( y = 5 \), find the value of \( x^2 + xy - 2 \)
Answer: Given that \( x = 2 \) and \( y = 5 \). Substitute these values into the given expression: \( x^2 + xy - 2 = (2)^2 + (2)(5) - 2 \)
\( = 4 + 10 - 2 \)
\( = 14 - 2 \)
\( = 12 \) Thus, the required value is 12.
In simple words: Put 2 for x and 5 for y. Work out the powers and multiplication first, then add and subtract.

Exam Tip: Always evaluate powers and products before performing addition and subtraction.

Question 3. Simplify and find the value when \( a = -1 \), \( b = 2 \), \( c = -3 \)
(i) \( a^2 + c^2 + b - 5 \)
(ii) \( a^2 + 3(b^2 - 5) + ac \)
Answer:
(i) Given expression: \( a^2 + c^2 + b - 5 \) Substitute \( a = -1 \), \( b = 2 \), and \( c = -3 \): \( = (-1)^2 + (-3)^2 + 2 - 5 \)
\( = 1 + 9 + 2 - 5 \)
\( = 12 - 5 \)
\( = 7 \)

(ii) Given expression: \( a^2 + 3(b^2 - 5) + ac \) Substitute \( a = -1 \), \( b = 2 \), and \( c = -3 \): \( = (-1)^2 + 3((2)^2 - 5) + (-1)(-3) \)
\( = 1 + 3(4 - 5) + 3 \)
\( = 1 + 3(-1) + 3 \)
\( = 1 - 3 + 3 \)
\( = 1 \)
In simple words: Replace the letters with the given numbers. Remember that multiplying two negative numbers gives a positive result.

Exam Tip: Be extra careful with negative signs - squaring any negative number always gives a positive answer.

Question 4. If \( a = 1 \), \( b = 2 \), \( c = -3 \), find the value of \( a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)
Answer: Given values are \( a = 1 \), \( b = 2 \), and \( c = -3 \). Substitute these values into the expression: \( a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)
\( = (1)^2 + (2)^2 + (-3)^2 + 2(1)(2) + 2(2)(-3) + 2(-3)(1) \)
\( = 1 + 4 + 9 + 4 + (-12) + (-6) \)
\( = 18 - 12 - 6 \)
\( = 18 - 18 \)
\( = 0 \) Hence, the value of the expression is 0 (Zero).
In simple words: Put in the values for each letter. The positive and negative terms cancel each other out completely to give 0.

Exam Tip: Watch the signs carefully when multiplying terms like \( 2(2)(-3) \) to ensure you write \( -12 \), not \( +12 \).

Question 5. If \( p = -5 \), find the value of \( p^2 + 10p + 25 \)
Answer: Given that \( p = -5 \). Substitute \( p = -5 \) into the given expression: \( p^2 + 10p + 25 = (-5)^2 + 10(-5) + 25 \)
\( = 25 - 50 + 25 \)
\( = 50 - 50 \)
\( = 0 \) Therefore, the value of the expression is 0 (Zero).
In simple words: Replace p with -5. Calculate each part to get 25 - 50 + 25, which equals 0.

Exam Tip: You can verify this result quickly because \( p^2 + 10p + 25 \) is the expansion of \( (p + 5)^2 \), which equals \( (-5 + 5)^2 = 0 \).

Question 6. Simplify the expression and find its value for \( a = 5 \), \( b = -3 \): \( 3(a^2 - ab) + 5 + ab \)
Answer: First, expand the bracket and combine like terms: \( 3(a^2 - ab) + 5 + ab = 3a^2 - 3ab + 5 + ab \)
\( = 3a^2 - 2ab + 5 \)
Now substitute \( a = 5 \) and \( b = -3 \) into the simplified expression: \( = 3(5)^2 - 2(5)(-3) + 5 \)
\( = 3(25) - (-30) + 5 \)
\( = 75 + 30 + 5 \)
\( = 110 \) Thus, the value of the expression is 110.
In simple words: Open the bracket first and combine the like terms. Then put the numbers in place of the letters to find the answer.

Exam Tip: Always simplify the expression before plugging in numbers - it reduces the number of steps and avoids calculation errors.

Question 7. What should be the value of \( x \) if the value of \( 2y^2 + y - x \) equals to 15 for \( y = -1 \)
Answer: Given that the value of \( 2y^2 + y - x \) equals 15 when \( y = -1 \). Set up the equation: \( 2y^2 + y - x = 15 \) Substitute \( y = -1 \): \( 2(-1)^2 + (-1) - x = 15 \)
\( 2(1) - 1 - x = 15 \)
\( 2 - 1 - x = 15 \)
\( 1 - x = 15 \)
\( -x = 15 - 1 \)
\( -x = 14 \)
\( \implies x = -14 \) Therefore, the value of \( x \) is \( -14 \).
In simple words: Put -1 in place of y. Simplify the numbers on the left, then solve the equation to find x = -14.

Exam Tip: When you reach \( -x = 14 \), remember to multiply both sides by \( -1 \) to find the value of positive \( x \).

Question 8. For the expression \( (3n + 2) \), find its 5th, 90th term
Answer: Given expression for the \( n^{\text{th}} \) term is \( 3n + 2 \).
To find the 5th term, put \( n = 5 \): \( 3(5) + 2 = 15 + 2 = 17 \)
To find the 90th term, put \( n = 90 \): \( 3(90) + 2 = 270 + 2 = 272 \)
Hence, the 5th term is 17 and the 90th term is 272.
In simple words: Put n = 5 to get the 5th term, and put n = 90 to get the 90th term.

Exam Tip: The position of the term gives the value of \( n \); state both requested terms clearly in your final step.

Question 9. If the expression is \( n^2 + 5 \), what will be its 100th term
Answer: The given expression for the \( n^{\text{th}} \) term is \( n^2 + 5 \). To find the 100th term, substitute \( n = 100 \): \( (100)^2 + 5 = 10000 + 5 \)
\( = 10005 \) Therefore, the 100th term is 10005.
In simple words: Multiply 100 by itself to get 10000, then add 5 to get 10005.

Exam Tip: Count your zeros carefully when calculating \( 100^2 \); it is 10000 (four zeros), not 1000.

Question 10. Find the 50th term of expression \( \frac{n(n + 1)}{2} \)
Answer: The given expression is \( \frac{n(n + 1)}{2} \). Substitute \( n = 50 \) to find the 50th term: \( \frac{50(50 + 1)}{2} = \frac{50(51)}{2} \)
\( = 25 \times 51 \)
\( = 1275 \) Thus, the 50th term of the expression is 1275.
In simple words: Put 50 in place of n. Divide 50 by 2 to get 25, then multiply 25 by 51 to get 1275.

Exam Tip: Simplify by dividing 50 by 2 before multiplying with 51 - this keeps your multiplication small and easy.