Class 7 Maths Chapter 10 HOTS Questions - Algebraic Expressions

Q1: If P = 2x² – 5x + 2, Q = 5x² + 6x – 3 and R = 3x² – x – 1. Find the value of 2P – Q + 3R.
Sol: 2P – Q + 3R = 2(2x² – 5x + 2) – (5x² + 6x – 3) + 3(3x² – x – 1)
= 4x² – 10x + 4 – 5x² – 6x + 3 + 9x² – 3x – 3
= 4x² – 5x² + 9x² – 10x – 6x – 3x + 4 + 3 – 3
= 8x² – 19x + 4
Required expression.

Q2: If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
Sol: A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
= -2x – 13 – 3x + 6 – 2x + 7
= -2x – 3x – 2x – 13 + 6 + 7
= -7x
Since A + B + C = kx
-7x = kx
Thus, k = -7

Q3: Rohan’s mother gave him ₹ 3xy² and his father gave him ₹ 5(xy² + 2). Out of this total money he spent ₹ (10 – 3xy²) on his birthday party. How much money is left with him? [NCERT Exemplar]

Sol: Money give by Rohan’s mother = ₹ 3xy²
Money given by his father = ₹ 5(xy² + 2)
Total money given to him = ₹ 3xy² + ₹ 5 (xy² + 2)
= ₹ [3xy² + 5(xy² + 2)]
= ₹ (3xy² + 5xy² + 10)
= ₹ (8xy² + 10).
Money spent by him = ₹ (10 – 3xy)²
Money left with him = ₹ (8xy² + 10) – ₹ (10 – 3xy²)
= ₹ (8xy² + 10 – 10 + 3x²y)
= ₹ (11xy²)
Hence, the required money = ₹ 11xy²   

Q4: Add: 5x−8y+2z,3z−4y−2x,6y−z−x and 3x−2z−3y.

Sol: Let us add the given expressions 5x−8y+2z,3z−4y−2x,6y−z−x and 3x−2z−3y as shown below:
(5x−8y+2z)+(3z−4y−2x)+(6y−z−x)+(3x−2z−3y)
=5x−8y+2z+3z−4y−2x+6y−z−x+3x−2z−3y
=(5x−2x−x+3x)+(−8y−4y+6y−3y)+(2z+3z−z−2z)(Combiningliketerms)
=(5−2−1+3)x+(−8−4+6−3)y+(2+3−1−2)z
=5x−9y+2z
Hence, (5x−8y+2z)+(3z−4y−2x)+(6y−z−x)+(3x−2z−3y)=5x−9y+2z.

Q5: Add: (4x² −7xy+4y² −3),(5+6y² −8xy+x²) and (6−2xy+2x² −5y²).
Sol: Let us add the given expressions 4x² −7xy+4y² −3,5+6y² −8xy+x² and 6−2xy+2x² −5y² as shown below:
(4x² −7xy+4y² −3)+(5+6y² −8xy+x²)+(6−2xy+2x² −5y²)
=4x² −7xy+4y² −3+5+6y² −8xy+x² +6−2xy+2x² −5y² 
=(4x² +x² +2x²)+(−7xy−8xy−2xy)+(4y² +6y² −5y²)+(−3+5+6)(Combiningliketerms)
=(4+1+2)x² +(−7−8−2)xy+(4+6−5)y2 +8
=7x² −17xy+5y² +8
Hence, (4x² −7xy+4y² −3)+(5+6y² −8xy+x²)+(6−2xy+2x² −5y²)
=7x² −17xy+5y² +8.

Q6: Subtract4a+3b from 2a+2b−4.
Sol: Let us subtract the given expression 4a+3b from 2a+2b−4 as shown below:
(2a+2b−4)−(4a+3b)
=2a+2b−4−4a−3b
=(2a−4a)+(2b−3b)−4(Combiningliketerms)
=(2−4)a+(2−3)b−4
=−2a−b−4
Hence, (2a+2b−4)−(4a+3b)=−2a−b−4.

Q7: Simplify 2(3b−5a)−7[9−62−5(a−6)]
Sol:
=2(3b−5a)−7[9−62−5(a−6)]
=6b–10a–(63–434–35(a–6))
=6b–10a–(63–434–35a–210)
=6b–10a–63+434+35a+210
=25a+6b+581.

Q8: Add: 6p+4q−r+2r−5p−6,11q−7p+2r−1 and 2q−3r+4.
Sol: Let us add the given expressions 6p+4q−r+2r−5p−6,11q−7p+2r−1 and 2q−3r+4 as shown below:
(6p+4q−r)+(2r−5p−6)+(11q−7p+2r−1)+(2q−3r+4)
=6p+4q−r+2r−5p−6+11q−7p+2r−1+2q−3r+4
=(6p−5p−7p)+(4q+11q+2q)+(−r+2r+2r−3r)+(−6−1+4)(Combiningliketerms)
=(6−5−7)p+(4+11+2)q+(−1+2+2−3)r−3
=−6p+17q−3
Hence, (6p+4q−r)+(2r−5p−6)+(11q−7p+2r−1)+(2q−3r+4)=−6p+17q−3.