Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Maths Solutions

Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Foundation Mathematics Solutions. We provide step by step Solutions of lesson/ Chapter-9 Percentage for ICSE Class-7 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Maths Solutions

Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Foundation Maths Solutions

Board ICSE
Publications Goyal brothers Prakashan
Subject Maths
Class 7th
Chapter-9 Unitary Method
Writer RS Aggrawal
Book Name Foundation
Topics Solution of Exe-9C
Academic Session 2024 – 2025

Word Problems on Percentage Change

Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Foundation Mathematics Solutions

Question-1. Increase:

(i) 375 by 4%

Ans-(i) Increase = (4/100) * 375 = 15

so, increased value = (375 + 15) = 390.

(ii) 500 by 3.4%

Ans-(ii) Increase = (3.4/100) * 500 = 17

so, increased value = (500 + 17) = 517.

(iii) 70 by 140%

Ans-(iii) Increase = (140/100) * 70 = 98

so, increased value = (70 + 28) = 168.

(iv) 48 by 12½%

Ans-(iv) Increase = (25/ 2 * 100) * 48 = 6

so, increased value = (48 + 6) = 54.

(v) 90 by 2%

Ans-(v) Increase = (2/100) *  90 = 1.8

so, increased value = (90 + 1.8) = 91.8

Question-2. Decrease:

(i) 70 by 40%

Ans-(i) Decrease = (40/100) * 70 = 28

so, decreased value = (70 – 28) = 42

(ii) 340 by 35%

Ans-(ii) Decrease = (35/100) * 340 =119

so, decreased value = (340 – 119) = 221.

(iii) 65 by 4%

Ans-(iii) Decrease = (4/100) * 65 = 2.6

so, decreased value = (65 – 2.6) = 62.4

(iv) 36 by 16(2/3)%

Ans-(iv) Decrease = (50 / 3 * 100) * 36 = 6

so, decreased value = (36 – 6) = 30.

(v) 260 by 1.5%

Ans-(v) Decrease = (1.5/100) * 260 = 3.9

so, decreased value = (260 – 3.9) = 256.1

Question-3. Find a number :

(i) Which when increased by 10% becomes 66.

Ans-(i) Let the number = x

10 % of x = x * 10/100 = x/10

Increasing the number by 10% = x+(x/10) = 11x/10

now, according to the question

11x/10 = 66

11x = 660

x = 60.

(ii) Which when increased by 120% becomes 77.

Ans-(ii) let the number be x

now, according to the question

x + (120x/100) = 77

x + 6x/5 = 77

5x + 6x = 385

11x = 385

x = 35.

(iii) Which when increased by 2.5% becomes 246.

Ans-(iii) let the number be x

now, x + 2.5% of x = 246

x + 25x/1000 = 246

41x/40 = 246

x = 246 * (40/41)

x = 6 * 40

Hence  = 240.

Question-4. Find a number:

(i) Which when decreased by 35% becomes 52.

Ans-(i) let the number be x

So according to the question,

x – 35% of x = 52

x – (35/100)x = 52

(100x – 35x)/100 = 52

65x = 5200

x = 5200/65

x = 80.

(ii) Which when decreased by 8% becomes 115.

Ans-(ii) let the number is x

now according to question,

x – 8% of x = 115

x – (8x/100) = 115

92x/100 = 115

x = 11500/92

x = 500/4

Hence = 125.

(i) Which when decreased by 3.4% becomes 483.

Ans-(iii) let the number is x

now according to question,

(x – 3.4) % of x = 483

(x – 34/1000) * x = 483

(1000x – 34) / 1000  = 483

966x / 1000 = 483

x = (483 * 1000) / 966

x = 500.

Question-5:  By what number must a given number be multiplied to increase it by 12%?

Answer- Let the number be x

It is increased by 12%

New number = x+(12/100) * x = (112 / 100) * x = (28/25) x

The given number should be multiplied by (28/25) to increase it by 12%.

Question-6:  By what number must a given number be multiplied to decrease it by 30%?

Answer- Let the number be x

It is multiplied with y and x is decreased by 30%

So,

x * y = x – (30x /100)  ……… eq 1

Now , from eq. 1 we get

y = (100x – 30x) / 100x   ……… eq 2

On solving we get-

y= 7/10

So, the no. must be multiplied with (7/10) to reduce it by 30%.

Question-7:  The price of a fan increases from Rs 3260 to Rs 3749. Find the increase per cent in its price.

Answer- Initial price of fan = Rs. 3260

Updated price of fan = Rs. 3749

Increase in fan price = Rs. 3749−Rs. 3260 = Rs. 489

Increase percent = (489/3260) * 100 = 15%.

Question-8: The monthly salary of Mr Rakesh is Rs 32500. After deducting the provident fund, he gets Rs 29900 per month. What per cent of the salary is deducted as provided fund?

Answer – The monthly salary of mr. Rakesh = Rs 32500

After reducing provident fund = Rs 29900

The total reduced salary = Monthly salary – after reduced monthly salary

= 32500 – 29900

=Rs 2600

Percentage of reduced salary = [(2600/32500 )* 100]% = 8%.

Question-9: A car was purchased last years for Rs 415000. Now, its value is Rs 356900. At what rate is the car depreciating?

Answer-  Depreciation value = 415000 – 356900 = Rs 58100

Now Depreciation % = Depreciation value/CP * 100

= 58100/415000 * 100

= 0.14 * 100

Hence  14 %.

Hence, car is depreciating at 14% per annum.

Question-10:  On decreasing the price of a car by 6%, its value becomes Rs 249100. What was the original price of the car?

Answer-  Let the initial price of Car to be x rupees

Now if we decrease 6 percent of x then the amount will be 229100

6% of x = x * (6/100) = 3x/50

Now deduct 3x/100 from the initial price to get the new price of car-

x – (3x/100) = x – (3x/50) = (50x – 3x) / 50 = 47x/50

Now, 47x/50 is the new price.

According to the question the new price is equivalent to 24910.

So, 47x/50 = 249100

47x = 249100 * 50

x = 249100 * 50/47

x = 5300 * 50,    Hence x = Rs 265000.

Question-11: On increasing the salary of a man by 12%, his salary in increased by Rs 2316. What was his original salary?

Answer- Let his original salary be x

Given that On increasing a man’s salary by 12%, his salary increases by Rs 2316

12% of x = 2316

12x/100 = 2316

12x = 2316 * 100

12x = 231600

x = 19300.

Question-12: The salary of Gopal was increased by 10% and then the increased salary was decreased by 10%. Find the net increase or decrease per cent in his original salary.

Answer-12  let salary= Rs 100

increased salary = Rs10

new salary= Rs (100+10) = 110

decrease salary =110×10/100 =11

new salary = 110-11=99

overall decrease = 1%.

Question-13: After deducting 4% of a bill, the amount still to be paid is Rs 1488. How much was the original bill?

Answer- 100% – 4% = 96%

96% of x is 1488

96/100*x=1488

x = 1488*100/96

x = Rs 1550.

Question-14:  The weight of a boy was 40 kg. But, it was wrongly measured as 42 kg. Find the error per cent.

Answer- Actually weight = 40kg

error weight = 42kg

difference = (40 kg – 40 kg) = 2kg

error % will be-

= 2/40×100%

= 5%.

— : end of Percentage Class 7 RS Aggarwal Exe-9C Goyal Brothers ICSE Foundation Maths Solutions :–

Return to: – ICSE Class -7 RS Aggarwal Goyal Brothers Math Solutions

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