Arithmetic Progression Exe-10E Concise Class-10 Maths Solution Ch-10. In this article you would learn solving word problems in A.P. We Provide Step by Step Solutions / Answer of questions for Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Arithmetic Progression Exe-10E Concise Class-10 Maths Solution Ch-10
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-10 | Arithmetic Progression |
| Writer | R.K. Bansal |
| Exe-10E | Word Problems in A.P |
| Edition | 2025-2026 |
How to Solve Word Problems in A.P.
Arithmetic Progression Exe-10E Concise Class-10 Maths Solution Ch-10
Que-1: Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km h-1. The second car goes at a speed of 8 km h-1 in the first hour and thereafter increasing the speed by 0.5 km h-1 each succeeding hour. After how many hours will the two cars meet ?
Ans: Let the two cars meet after n hours.
That means the two cars travel the same distance in n hours.
Distance travelled by the 1st car in n hours = 10 × n km
Distance travelled by the 2nd car in n hours = n/2[2×8+(n-1)×0.5]km
10×n=n/2[2×8+(n-1)×0.5]
20 = 16 + 0.5n – 0.5
0.5n = 4.5
n = 9
Thus, the two cars will meet after 9 hours
Que-2: A sum of Rs 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is Rs 20 less than its preceding prize; find the value of each of the prizes.
Ans: Total amount of Prize Sn = Rs. 700
Let the value of the first prize be Rs. a
Number of prizes = n = 7
Let the value of first prize be Rs. a
Depreciation in next prize = Rs. 20
We have,
Sn=n/2[2a+(n-1)d]
700=7/2[2a+6(-20)]
700=7/2[2a=120]
1400 = 14a – 840
14a = 2240
a = 160
Value of 1st Prize = Rs. 160
Value of 2nd Prize = Rs. (160 – 20) Rs. 140
Value of 3rd Prize = Rs. (140 – 20) = Rs. 120
Value of 4th Prize = Rs. (120 – 20) = Rs. 100
Value of 5th Prize = Rs. (100 – 20) Rs. 80
Value of 6th Prize = Rs. (80 – 20) = Rs. 60
Value of 7th Prize = Rs. (60 – 20) = Rs. 40.
Que-3: An article can be bought by paying Rs 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs 3,000 and every other installment is Rs 100 less than the previous one, find :
(i) amount of installment paid in the 9th month
(ii) total amount paid in the installment scheme.
Ans: Total price of an article = Rs 28000
No. of installments (n) = 12
First installment (a) = Rs 3000
Depreciation in installment = d = –100
i Amount of installment paid in the 9th month
= t9
= a + 8d
= 3000 + 8 × (–100)
= 3000 – 800
= Rs. 2200
ii. Total amount paid in the installment scheme
= S12
= 12/2[2×3000+11×(-100)]
= 6[6000 – 1100]
= 6 × 4900
= Rs. 29,400.
Que-4: A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Ans: Since the production increases uniformly by a fixed number every year, the sequence formed by the production in different years in an A.P.
Let the production in the first year = a
Common difference = number of units by which the production increases every year = d
We have,
t3 = 600
a + 2d = 600 …(1)
t7 = 700
a + 6d = 700 …(2)
Subtracting (1) from (2), we get
4d = 100
d = 25
a + 2 × 25 = 600
a = 550
i. The production in the first year = 550 Tv sets
ii. The production in the 10th year = t10
= 550 + 9 × 25
= 775 Tv sets
iii. The total production in 7 years = S7
= 7/2[2×550+6×25]
= 7/2[1100+150]
= 7/2×1250
= 4375 Tv sets
Que-5: Mrs. Gupta repays her total loan of Rs 1.18,000 by paying installments every month. If the installment for the first month is Rs 1,000 and it increases by RS 100 every month, what amount will she pay as the 30th installment of loan? What amount of loan she still has to pay after the 30th installment?
Ans: Total loan to be paid by Mrs. Gupta = Rs 118000
Installment for the first month (a) = Rs 1000
Increase in installment every month = d = Rs. 100
30th installment = t30
= a + 29d
= 1000 + 29 × 100
= 1000 + 2900
= Rs. 3900
Now, amount paid in 30 installments = S30
= 30/2[2×1000+29×100]
= 15[2000 + 2900]
= 15 × 4900
= Rs. 73,500
∴ Amount of loan to be paid after the 30th installments
= Rs. (1,18,000 – 73,500)
= Rs. 44,500
—: Arithmetic Progression Exe-10E Concise Class-10 Maths Ch-10 Solving Word Problems :—
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