Mathematics Solutions Solutions for Class 7 Math Chapter 17 - Miscellaneous Problems : Set 2

Answer:

Amount deposited = Rs 15000
Rate of interest = 9%
Simple interest = Rs 5400
Let the number of years be n. 
SI = $\frac{\mathrm{PRT}}{100}$
$$\begin{gathered} \Rightarrow 5400=\frac{15000\times 9\times n}{100} \\ \Rightarrow \frac{5400\times 100}{15000\times 9}=n \\ \Rightarrow n=4 \end{gathered}$$
Thus, the number of years for which she had deposited the money is 4 years. 

Answer:

Answer:

Nasruddin invested Rs 40,000
Mahesh invested Rs 60,000
Profit = 30%
Profit earned by Nasruddin = $30\% \mathrm{of} \mathrm{Rs} 40,000=\frac{30}{100}\times 40,000=12,000$
Profit earned by Mahesh = $30\% \mathrm{of} \mathrm{Rs} 60,000=\frac{30}{100}\times 60,000=18,000$
Thus, Nasruddin got Rs 12,000 profit and Mahesh earned Rs 18,000 as profit. 
 

Answer:

Diameter = 5.6 cm
Radius = $\frac{\mathrm{diamete}r}{2}=\frac{5.6}{2}=2.8 \mathrm{cm}$
Circumference = $2\mathrm{\pi }r=2\times \frac{22}{7}\times 2.8=17.6 \mathrm{cm}$
Thus, the circumference of the circle is 17.6 cm. 

Answer:

(i) (2a – 3b)²
Using the identity ${\left(x-y\right)}^2=x^2-2xy+y^2$
${\left(2a-3b\right)}^2={\left(2a\right)}^2-2\times 2a\times 3b+{\left(3b\right)}^2=4a^2-12ab+9b^2$

(ii) (10 + y)² 
Using the identity: ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} ={\left(10\right)}^2+2\times 10\times y+y^2 \\ =100+20y+y^2 \end{gathered}$$

(iii) ${\left(\frac{p}{3}+\frac{q}{4}\right)}^2$
Using the identity: ${\left(a+b\right)}^2=a^2+2ab+b^2$
$$\begin{gathered} ={\left(\frac{p}{3}\right)}^2+2\times \frac{p}{3}\times \frac{q}{4}+{\left(\frac{q}{4}\right)}^2 \\ =\frac{p^2}{9}+\frac{pq}{6}+\frac{q^2}{16} \end{gathered}$$

(iv) ${\left(y-\frac{3}{y}\right)}^2$
Using the identity ${\left(a-b\right)}^2=a^2-2ab+b^2$
$$\begin{gathered} =y^2-2\times y\times \frac{3}{y}+{\left(\frac{3}{y}\right)}^2 \\ =y^2-6+\frac{9}{y^2} \end{gathered}$$
 

Answer:

We use the formula: $\left(x-a\right)\left(x+a\right)=x^2-a^2$
(i) (x – 5)(x + 5)
$\left(x^2-5^2\right)=x^2-25$

(ii) (2a – 13)(2a + 13)
$$\begin{gathered} ={\left(2a\right)}^2-{13}^2 \\ =4a^2-169 \end{gathered}$$

(iii) (4z – 5y)(4z + 5y)
$$\begin{gathered} ={\left(4z\right)}^2-{\left(5y\right)}^2 \\ =16z^2-25y^2 \end{gathered}$$

(iv) (2t – 5)(2t + 5)
$$\begin{gathered} ={\left(2t\right)}^2-5^2 \\ =4t^2-25 \end{gathered}$$

Answer:

Diameter = 1.05 m
Circumference of the wheel = $2\mathrm{\pi }r$
= $2\times \frac{22}{7}\times \left(\frac{1.05}{2}\right)=3.3$m
In 1000 rotations, the distance covered = 3.3 m × 1000 = 3300 m = 3.3 km

Answer:

Length of the garden = 40 m
Area = 1000 sq m
$$\begin{gathered} \Rightarrow lb=1000 \\ \Rightarrow 40b=1000 \\ \Rightarrow \Rightarrow b=25 \mathrm{m} \end{gathered}$$
Perimeter = $2\left(l+b\right)=2\left(40+25\right)=130 \mathrm{m}$
For 1 round of fencing leaving the entrance of 4 m, the length of wire required = 130 m $-$ 4 m = 126 m
For 3 such rounds of fencing, $3\times 126 \mathrm{m}=378 \mathrm{m}$ of wire required. 
Rate of fencing 1 m = Rs 250
Rate of fencing 378 m = $378\times 250=\mathrm{Rs} 94500$
Thus, Rs 94500 is required for fencing the garden. 

Answer:

Given here is a right angled triangle. So, we can apply the Pythagoras theorem.
$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2 \\ \Rightarrow {20}^2+{21}^2={\mathrm{AC}}^2 \\ \Rightarrow {\mathrm{AC}}^2=400+441=841 \\ \Rightarrow \mathrm{AC}=29 \end{gathered}$$
Thus, the length of hypotenuse is 29 units. 
Perimeter of ∆ABC = AB + BC + CA = 20 + 21 + 29 = 70 units.

Answer:

Edge of the cube = 8 cm
Total surface area = $6a^2=6\times {\left(8\right)}^2=384 {\mathrm{cm}}^2$
Thus, the total surface area of the cube is 384 cm²

Answer:

365y⁴z³ – 146y²z^{4
$\mathrm{Taking} 73y^2{z}^3 \mathrm{common}$, 
$=73y^2{z}^3\left(5y^2-2z\right)$}

Answer:

Average = 36
$$\begin{gathered} \Rightarrow \frac{33+34+35+x+37+38+39}{7}=36 \\ \Rightarrow \frac{216+x}{7}=36 \\ \Rightarrow 216+x=252 \\ \Rightarrow x=36 \end{gathered}$$
Hence, the correct answer is option (iv). 

Answer:

(61² – 51² ) 
$$\begin{gathered} =\left(61-51\right)\left(61+51\right) \left(\because \left(a^2-b^2\right)=\left(a+b\right)\left(a+b\right)\right) \\ =10\times 112 \\ =1120 \end{gathered}$$
Hence, the correct answer is option (i). 

Answer:

Rs 2600 are divided among Sameer and Smita.
Ratio = 8 : 5
Total = 8 + 5 = 13
Sameer's share = $\frac{8}{13}\times 2600=1600$
Smita's share = $\frac{5}{13}\times 2600=1000$
Thus, Sameer's share is Rs 1600 and Smita's share is Rs 1000.
Hence, the correct answer is option (iv).