CBSE Class 7 Mathematics Question Paper Set 29 with Solutions

Answer: (b) 3x. Three times a number means the number is multiplied by 3, so it is written 3 × x = 3x.

Teacher's Note:
1. In letter-number expressions, a number written next to a letter means multiplication.
2. Do not confuse 3 + x (three more than x) with 3x (three times x).

Answer: (a) \(\dfrac{1}{5}\). Divide the numerator and denominator by their HCF 3: \(\dfrac{3}{15}=\dfrac{3\div 3}{15\div 3}=\dfrac{1}{5}\). Option (c) is not in simplest form, and option (b) is negative.

Teacher's Note:
1. To reach the simplest form, divide the top and bottom by their highest common factor.
2. The sign of a fraction does not change when you reduce it.

Answer: (b) 9ab, 7ba. Like terms have exactly the same letter factors with the same powers. Since ba = ab, 9ab and 7ba are like terms. In (a) the powers of x differ and in (c) the letters differ.

Teacher's Note:
1. Compare only the letter parts, never the numerical coefficients, when checking for like terms.
2. The order of letters in a product does not matter, so ab and ba are the same.

Answer: (b) −2x²y + 2xy².
(2xy² + xy) − (xy + 2x²y) = 2xy² + xy − xy − 2x²y = 2xy² − 2x²y.

Teacher's Note:
1. Change the sign of every term of the expression being subtracted before combining.
2. Only like terms can be added or cancelled, so xy cancels but xy² and x²y stay separate.

Answer: False. Putting x = 3 gives LHS = 3 + 3 = 6, which is not equal to RHS = 0. The solution of x + 3 = 0 is x = −3.

Teacher's Note:
1. To check a solution, substitute the value and see whether LHS equals RHS.
2. Remember that the solution can be a negative number.

Answer: False. The angles of a triangle must add up to 180°, but 105° + 80° + 70° = 255°, which is more than 180°.

Teacher's Note:
1. Always add the three given angles first and check that the total is 180°.
2. A triangle can have at most one angle of 90° or more.

Answer: First part:
1. 5m(3 − m) = 15m − 5m².
2. Sum = 15m − 5m² + 6m² − 13m = (−5m² + 6m²) + (15m − 13m).
3. = m² + 2m
OR
1. (5mn + 2m) − (8m − 6mn) = 5mn + 2m − 8m + 6mn.
2. = (5mn + 6mn) + (2m − 8m)
3. = 11mn − 6m

Teacher's Note:
1. Open the bracket first by multiplying, then group the like terms.
2. When subtracting, the sign of every term in the second expression flips, so −(−6mn) becomes +6mn.

Answer: 1. LCM of 3 and 2 is 6.
2. \(\dfrac{2}{3}=\dfrac{4}{6}\) and \(\dfrac{5}{2}=\dfrac{15}{6}\).
3. Since 15 > 4, \(\dfrac{5}{2}\) is greater. (Also \(\dfrac{2}{3}<1\) while \(\dfrac{5}{2}=2\dfrac{1}{2}>1\).)

Teacher's Note:
1. Make the denominators equal using the LCM, then compare the numerators.
2. A quick check against 1 often decides the answer without any working.

Answer: Multiply numerators and denominators:
\(\dfrac{3\times 2}{11\times 5}=\)\(\dfrac{6}{55}\). No common factor of 6 and 55 exists, so this is in simplest form.

Teacher's Note:
1. For multiplication of fractions, there is no need to find a common denominator.
2. Cancel common factors before or after multiplying to give the answer in simplest form.

Answer: 1. \(\dfrac{2b}{3}-5=3\)
2. Add 5 to both sides: \(\dfrac{2b}{3}=8\).
3. Multiply both sides by 3: 2b = 24.
4. Divide both sides by 2: b = 12.
Check: \(\dfrac{2\times 12}{3}-5=8-5=3\), which equals the RHS.

Teacher's Note:
1. Do the same operation on both sides and undo the operations one at a time.
2. Always substitute your answer back into the equation to be sure.

Answer: Steps of construction:
1. Draw a line segment BC = 6.5 cm using a ruler.
2. With B as centre and radius 6.5 cm, draw an arc above BC.
3. With C as centre and the same radius 6.5 cm, draw another arc cutting the first arc at A.
4. Join AB and AC.
ΔABC is the required equilateral triangle, with AB = BC = CA = 6.5 cm and each angle 60° (this can be checked with a protractor).

Teacher's Note:
1. Keep the compass opening fixed at 6.5 cm for both arcs so all three sides are equal.
2. Draw the arcs lightly but do not erase them, as they show your construction.

Answer: 1. Sum: (2y² + 3yz) + (−y² − yz − z²) + (yz + 2z²)
y² terms: 2y² − y² = y²
yz terms: 3yz − yz + yz = 3yz
z² terms: −z² + 2z² = z²
Sum = y² + 3yz + z².
2. Subtract 2y² + yz: (y² + 3yz + z²) − (2y² + yz) = y² + 3yz + z² − 2y² − yz.
3. = −y² + 2yz + z²

Teacher's Note:
1. Arrange the three expressions so that like terms are in the same column, then add column by column.
2. A common error is forgetting to change the sign of yz when subtracting 2y² + yz.

Answer: 1. Take east as positive and west as negative. Total displacement = \(\dfrac{2}{3}-1\dfrac{5}{7}\).
2. \(1\dfrac{5}{7}=\dfrac{12}{7}\).
3. LCM of 3 and 7 is 21: \(\dfrac{2}{3}=\dfrac{14}{21}\) and \(\dfrac{12}{7}=\dfrac{36}{21}\).
4. \(\dfrac{14}{21}-\dfrac{36}{21}=-\dfrac{22}{21}=-1\dfrac{1}{21}\).
5. Sam is \(1\dfrac{1}{21}\) km to the west of P.

Teacher's Note:
1. Choosing a positive direction and a negative direction turns a story problem into a simple subtraction.
2. Convert the mixed number to an improper fraction before finding the common denominator.

Answer: Here the angle sum 30° + 100° = 130° < 180°, so the triangle exists.
Steps of construction:
1. Draw XY = 6 cm.
2. At X, use a protractor to draw a ray XP making an angle of 30° with XY.
3. At Y, draw a ray YQ making an angle of 100° with YX, on the same side of XY as XP.
4. The rays XP and YQ meet at the point Z.
5. ΔXYZ is the required triangle. (Check: ∠XZY = 180° − 130° = 50°.)

Teacher's Note:
1. This is an angle-side-angle (ASA) construction, so the given side must lie between the two given angles.
2. Draw both rays on the same side of XY, or they will never meet.

Answer: 1. The term 6t has the coefficient 6.
2. The factors of 56t³ are 56 × t × t × t. (Writing 56 as 2 × 2 × 2 × 7, the factors are 2 × 2 × 2 × 7 × t × t × t.)

Teacher's Note:
1. The coefficient is the number multiplied with the letters in a term.
2. The power t³ means t is a factor three times.

Answer: No, Riya is not correct.
1. Like terms must have the same letter with the same power.
2. The terms 56t³, 12t² and 6t all have different powers of t, and 16s² and 2s have different powers of s.
3. 106 is a constant with no letter.
Hence the expression has no pair of like terms.

Teacher's Note:
1. Check each term for the same letter and the same power, not just the same letter.
2. Terms such as 12t² and 16s² are unlike because their letters are different.