Linear growth and decay

Correct answer: B. 38 cm

Given \(H(t)=18+4t\). Substituting \(t=5\), \(H(5)=18+4(5)=18+20=38\) cm. Hence, 38 cm is correct. Getting 32 cm would mean calculating the growth term \(4t\) incorrectly. Exam tip: To find a function value, substitute the given value for the variable and then simplify.

Correct answer: A. 72 L

Given \(W(t)=120-6t\). Substituting \(t=8\), \(W(8)=120-6(8)=120-48=72\) L. Therefore, 72 L is correct. 48 L is only the amount of water removed, not the water remaining. Exam tip: substitute the value of \(t\) first, then multiply and subtract.

Correct answer: A. \(P(n)=40-3n\)

In \(P(n)=40-3n\), the power of \(n\) is 1 and its coefficient is \(-3\), so the quantity decreases by 3 for every one-unit increase. \(40-3n^2\) is not linear. Exam tip: linear decay has degree 1 with a negative variable coefficient.

Correct answer: A. \(12-3t\)

In \(12-3t\), the power of \(t\) is 1, so the change is linear. The coefficient \(-3\) means the quantity decreases by 3 per unit time. \(12+3t\) shows growth. Exam tip: check the sign of the slope.

Correct answer: C. 6

To make the amount ₹500, put S(m)=500: 200+50m=500. Thus, 50m=300, so m=6. Therefore, the amount becomes ₹500 after 6 months. After 5 months, the amount is only ₹450, so it is a close but incorrect option. Exam tip: Set the given linear expression equal to the required amount and solve for m.

Correct answer: B. 5 घंटे

For the battery to be at 60%, put \(B(h)=60\) in the model: \(100-8h=60\). Thus, \(8h=40\), so \(h=5\) hours. At 4 hours, the battery would be \(100-8(4)=68\%\), so that option is not correct. Exam tip: In linear decay questions, equate the function to the required target value first, then solve for the variable.

Correct answer: C. \(y=35+6x\)

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the change per step. Here, the initial value is 35 and the growth is 6, so the rule is \(y=35+6x\). In option B, 6 is subtracted, so it represents decay. Exam tip: put \(x=0\) to identify the initial value.

Correct answer: D. \(y=140-10x\)

A linear rule has the form \(y=a+bx\), where \(a\) is the initial value and \(b\) is the change per step. Here, the initial value is 140 and the value decreases by 10 each step, so the change is \(-10\). Therefore, the rule is \(y=140-10x\). Option A represents an increase of 10 per step. Exam tip: words such as “decrease” or “decay” indicate a negative slope.

Correct answer: A. \(y=12-2x\)

In \(y=12-2x\), the coefficient of x is \(-2\), so y falls by 2 units for every 1-unit rise in x. Option B represents growth. Exam tip: linear decay has x to power 1 with a negative coefficient.

Correct answer: B. \(y=90-7x\)

In option B, the coefficient of \(x\) is \(-7\). So, when \(x\) increases by 1, \(y\) decreases by 7 each time; this is constant-rate linear decay. Relations involving \(x^2\) or \(1/x\) are not linear. Exam tip: a negative slope indicates decay.

Correct answer: B. 20

Given A(t)=a+4t and A(3)=32, substitute t=3: a+4×3=32. Thus, a+12=32, so a=20. If a were 24, then A(3)=24+12=36, not the given value 32. Exam tip: when a function value is given, substitute the corresponding value of t directly into the function.

Correct answer: D. 80

Given \(R(t)=b-6t\). Substituting \(t=5\) gives \(R(5)=b-6(5)=b-30\). Since \(R(5)=50\), \(b-30=50\), so \(b=80\). Option 50 is the value of \(R(5)\), not the value of \(b\). Exam tip: When a function value is given, first substitute the stated input into the formula and form an equation.

Correct answer: A. \(120-8t\)

In \(120-8t\), the highest power of \(t\) is 1 and its coefficient is \(-8\), so the quantity decreases by 8 per time unit. \(120+8t\) shows growth. Exam tip: linear decay has a negative coefficient of the variable.

Correct answer: A. \(P(t)=12+3t\)

In \(P(t)=12+3t\), the coefficient of \(t\) is \(+3\), so the quantity increases by 3 for every unit of time. Option B is linear but shows decay because its coefficient is negative. Exam tip: for linear growth, the power of \(t\) must be 1 and its coefficient must be positive.

Correct answer: C. (y=12+5x)

For linear growth the coefficient must be positive and the initial value must be (12). (y=12+5x) satisfies both.

Correct answer: B. \(y=60-5t\)

Initial value means value at \(t=0\). For \(y=60-5t\):
\[y(0)=60-5\cdot 0=60\]
Also the coefficient of \(t\) is \(-5\), which is negative, so the quantity decreases with time (linear decay).

Correct answer: C. 45

Given \(M(t)=m+5t\). Substituting \(t=2\), we get \(M(2)=m+10=25\), so \(m=15\). Now, for \(t=6\), \(M(6)=15+5\times6=45\). Hence, 45 is correct. The nearby option 40 can result from incorrectly counting the growth; from \(t=2\) to \(t=6\), the increase is \(5\times4=20\). Exam tip: first use the given condition to find the unknown constant \(m\), then substitute the required value.

Correct answer: A. 13

Given \(N(t)=n-3t\). Substituting \(t=4\), \(N(4)=n-12=28\), so \(n=40\). Now, for \(t=9\), \(N(9)=40-3\times9=40-27=13\). Hence, 13 is correct. A value such as 16 can result from using the rate of decrease or the time value incorrectly. Exam tip: first use the given condition to find the unknown constant \(n\), then substitute the required value of \(t\).

Correct answer: B. 5

At \(x=3\), the value of \(y\) is 35. Substituting these values in \(y=20+kx\) gives \(35=20+3k\). Hence, \(3k=15\), so \(k=5\). Option 15 is the value of \(3k\), not of \(k\). Exam tip: In a linear expression, substitute the given values of \(x\) and \(y\) first, then isolate the unknown coefficient.

Correct answer: B. 6

Given \(x=4\) and \(y=76\), substitute these values in \(y=100-kx\): \(76=100-4k\). Hence, \(4k=24\), so \(k=6\). Option 24 is the value of \(4k\), not of \(k\). Exam tip: In a linear expression, substitute the given values of \(x\) and \(y\) first, then isolate the unknown term.

Correct answer: B. The value of B(t) is greater

Substituting t=5 gives A(5)=10+3(5)=25 and B(5)=25+5=30. Since 30 is greater than 25, B(t) has the greater value. The “both values are equal” option is incorrect because the values differ by 5. Exam tip: To compare linear models at a given t, substitute that value into both expressions before comparing.

Correct answer: B. \(B(5)>A(5)\)

Substituting \(t=5\), we get \(A(5)=10+3(5)=25\) and \(B(5)=25+5=30\). Since \(30>25\), \(B(5)>A(5)\) is correct. \(A(5)=B(5)\) is incorrect because the values differ by 5. Exam tip: Substitute the given value of \(t\) into each expression separately before comparing them.

Correct answer: C. Both values are equal

Substituting \(t=5\), we get \(C(5)=80-4(5)=60\) and \(D(5)=70-2(5)=60\). Hence, both models give the same value. Options A and B are incorrect because neither value is larger than the other. Exam tip: Substitute the given value of \(t\) into each expression before comparing them.

Correct answer: B. \(t=8\)

Given \(T(t)=30+2t\) and \(T(t)=46\), set \(30+2t=46\). This gives \(2t=16\), so \(t=8\). Therefore, \(t=8\) is correct. For example, substituting \(t=6\) gives \(T(t)=42\), not 46. Exam tip: equate the function to the required target value first, then solve the linear equation for \(t\).

Correct answer: C. \(t=9\)

Given \(U(t)=64-4t\), set \(U(t)=28\): \(64-4t=28\). Thus, \(4t=36\), so \(t=9\). For example, at \(t=8\), \(U(8)=32\), not 28. Exam tip: substitute your final value back into the original expression to verify it.