Mathematics Solutions Solutions for Class 7 Maths Chapter 17 Miscellaneous Problems : Set 2 are provided here with simple step-by-step explanations. These solutions for Miscellaneous Problems : Set 2 are extremely popular among Class 7 students for Maths Miscellaneous Problems : Set 2 Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 17 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

Page No 100:

Question 1:

Angela deposited 15000 rupees in a bank at a rate of 9 p.c.p.a. She got simple interest amounting to 5400 rupees. For how many years had she deposited the amount?

Answer:

Amount deposited = Rs 15000
Rate of interest = 9%
Simple interest = Rs 5400
Let the number of years be n
SI = PRT100
5400=15000×9×n1005400×10015000×9=nn=4
Thus, the number of years for which she had deposited the money is 4 years. 

Page No 100:

Question 2:

Ten men take 4 days to complete the task of tarring a road. How many days would 8 men take?

Answer:

Number of men Number of days
10 4
8 x

More the men lesser the number of days required to finish the work. 
So, the Number of men and the number of days required to complete the work are inversely proportional to each other. 
108=x4x=10×48x=5
Thus, when 8 men are involved so, 5 days will be required. 

Page No 100:

Question 3:

Nasruddin and Mahesh invested Rs 40,000 and Rs 60,000 respectively to start a business. They made a profit of 30%. How much profit did each of them make?

Answer:

Nasruddin invested Rs 40,000
Mahesh invested Rs 60,000
Profit = 30%
Profit earned by Nasruddin = 30% of Rs 40,000=30100×40,000=12,000
Profit earned by Mahesh = 30% of Rs 60,000=30100×60,000=18,000
Thus, Nasruddin got Rs 12,000 profit and Mahesh earned Rs 18,000 as profit. 
 

Page No 100:

Question 4:

The diameter of a circle is 5.6 cm. Find its circumference.

Answer:

Diameter = 5.6 cm
Radius = diameter2=5.62=2.8 cm
Circumference = 2πr=2×227×2.8=17.6 cm
Thus, the circumference of the circle is 17.6 cm. 

Page No 100:

Question 5:

Expand.
(i) (2a – 3b)2

(ii) (10 + y)2

(iii) p3+q42

(iv) y-3y2

Answer:

(i) (2a – 3b)2
Using the identity x-y2=x2-2xy+y2
2a-3b2=2a2-2×2a×3b+3b2=4a2-12ab+9b2

(ii) (10 + y)
Using the identity: a+b2=a2+2ab+b2
=102+2×10×y+y2=100+20y+y2

(iii) p3+q42
Using the identity: a+b2=a2+2ab+b2
=p32+2×p3×q4+q42=p29+pq6+q216

(iv) y-3y2
Using the identity a-b2=a2-2ab+b2
=y2-2×y×3y+3y2=y2-6+9y2
 

Page No 100:

Question 6:

Use a formula to multiply.
(i) (x – 5)(x + 5)
(ii) (2a – 13)(2a + 13)
(iii) (4z – 5y)(4z + 5y)
(iv) (2t – 5)(2t + 5)

Answer:

We use the formula: x-ax+a=x2-a2
(i) (x – 5)(x + 5)
x2-52=x2-25

(ii) (2a – 13)(2a + 13)
=2a2-132=4a2-169

(iii) (4z – 5y)(4z + 5y)
=4z2-5y2=16z2-25y2

(iv) (2t – 5)(2t + 5)
=2t2-52=4t2-25

Page No 100:

Question 7:

The diameter of the wheel of a cart is 1.05 m. How much distance will the cart cover in 1000 rotations of the wheel?

Answer:

Diameter = 1.05 m
Circumference of the wheel = 2πr
2×227×1.052=3.3 m
In 1000 rotations, the distance covered = 3.3 m × 1000 = 3300 m = 3.3 km

Page No 100:

Question 8:

The area of a rectangular garden of length 40 m, is 1000 sqm. Find the breadth of the garden and its perimeter. The garden is to be enclosed by 3 rounds of fencing, leaving an entrance of 4 m. Find the cost of fencing the garden at a rate of 250 rupees per metre.

Answer:

Length of the garden = 40 m
Area = 1000 sq m
lb=100040b=1000b=25 m
Perimeter = 2l+b=240+25=130 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, 3×126 m=378 m of wire required. 
Rate of fencing 1 m = Rs 250
Rate of fencing 378 m = 378×250=Rs 94500
Thus, Rs 94500 is required for fencing the garden. 

Page No 100:

Question 9:

 From the given figure, find the length of hypotenuse AC and the perimeter of ∆ABC.

Answer:

Given here is a right angled triangle. So, we can apply the Pythagoras theorem.
AB2+BC2=AC2202+212=AC2AC2=400+441=841AC=29 
Thus, the length of hypotenuse is 29 units. 
Perimeter of ∆ABC = AB + BC + CA = 20 + 21 + 29 = 70 units.

Page No 100:

Question 10:

If the edge of a cube is 8 cm long, find its total surface area.

Answer:

Edge of the cube = 8 cm
Total surface area = 6a2=6×82=384 cm2
Thus, the total surface area of the cube is 384 cm2

Page No 100:

Question 11:

Factorise.  365y4z3 – 146y2z4

Answer:

365y4z3 – 146y2z4
Taking 73y2z3 common
=73y2z35y2-2z

Page No 100:

Question 1:

Choose the right answers from the options given for each of the following questions.
If the average of the numbers 33, 34, 35, x , 37, 38, 39 is 36, what is the value of x ?
(i) 40
(ii) 32
(iii) 42
(iv) 36

Answer:

Average = 36
33+34+35+x+37+38+397=36216+x7=36216+x=252x=36
Hence, the correct answer is option (iv). 

Page No 100:

Question 2:

Choose the right answers from the options given for each of the following questions.
The difference of the squares, (612 – 512 ) is equal to .................. .
(i) 1120
(ii) 1230
(iii) 1240
(iv) 1250

Answer:

(612 – 512 ) 
=61-5161+51                    a2-b2=a+ba+b=10×112=1120
Hence, the correct answer is option (i). 

Page No 100:

Question 3:

Choose the right answers from the options given for each of the following questions.
If 2600 rupees are divided between Sameer and Smita in the proportion 8 : 5, the share of each is ............... and ............... respectively .
(i) Rs 1500, Rs 1100
(ii) Rs 1300, Rs 900
(iii) Rs 800, Rs 500
(iv) Rs 1600, Rs 1000

Answer:

Rs 2600 are divided among Sameer and Smita.
Ratio = 8 : 5
Total = 8 + 5 = 13
Sameer's share = 813×2600=1600
Smita's share = 513×2600=1000
Thus, Sameer's share is Rs 1600 and Smita's share is Rs 1000.
Hence, the correct answer is option (iv). 



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