RD Chapter 19 Arithmetic Progressions Ex 19.7 Solutions
Question - 1 : - A man saved ₹ 16500 in ten years. In each year after the first he saved ₹ 100/- more than he did in the preceding year. How much did he saved in the first year?
Answer - 1 : -
Given: A mansaved ₹16500 in ten years
Let ₹ x be his savingsin the first year
His savings increasedby ₹ 100 every year.
So,
A.P will be x, 100 +x, 200 + x………………..
Where, x is first termand
Common difference, d =100 + x – x = 100
We know, Sn isthe sum of n terms of an A.P
By using the formula,
Sn =n/2 [2a + (n – 1)d]
where, a is firstterm, d is common difference and n is number of terms in an A.P.
Given:
Sn =16500 and n = 10
S10 =10/2 [2x + (10 – 1)100]
16500 = 5{2x + 9(100)}
16500 = 5(2x + 900)
16500 = 10x + 4500
-10x = 4500 – 16500
–10x = –12000
x = 12000/10
= 1200
Hence, his saving inthe first year is ₹ 1200.
Question - 2 : - A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.
Answer - 2 : -
Given:
First year savings is₹ 32
Second year savings is₹ 36
In this process heincreases his savings by ₹ 4 every year
Then,
A.P. will be 32, 36,40,………
Where, 32 is firstterm and common difference, d = 36 – 32 = 4
We know, Sn isthe sum of n terms of an A.P
By using the formula,
Sn =n/2 [2a + (n – 1)d]
where, a is firstterm, d is common difference and n is number of terms in an A.P.
Given:
Sn =200, a = 32, d = 4
Sn =n/2 [2a + (n – 1)d]
200 = n/2 [2(32) +(n-1)4]
200 = n/2 [64 + 4n –4]
400 = n [60 + 4n]
400 = 4n [15 + n]
400/4 = n [15 + n]
100 = 15n + n2
n2 +15n – 100 = 0
n2 +20n – 5n – 100 = 0
n (n + 20) – 5 (n +20) = 0
(n + 20) – 5 (n + 20)= 0
(n + 20) (n – 5) = 0
n = -20 or 5
n = 5 [Since, n is apositive integer]
Hence, the manrequires 5 days to save ₹ 200
Question - 3 : - A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the instalment.
Answer - 3 : -
Given:
40 annual instalmentswhich form an arithmetic series.
Let the firstinstalment be ‘a’
S40 =3600, n = 40
By using the formula,
Sn =n/2 [2a + (n – 1)d]
3600 = 40/2 [2a +(40-1)d]
3600 = 20 [2a + 39d]
3600/20 = 2a + 39d
2a + 39d – 180 = 0 …..(i)
Given:
Sum of first 30 termsis paid and one third of debt is unpaid.
So, paid amount = 2/3× 3600 = ₹ 2400
Sn =2400, n = 30
By using the formula,
Sn =n/2 [2a + (n – 1)d]
2400 = 30/2 [2a +(30-1)d]
2400 = 15 [2a + 29d]
2400/15 = 2a + 29d
2a + 29d -160 = 0 ….(ii)
Now, let us solveequation (i) and (ii) by substitution method, we get
2a + 39d = 180
2a = 180 – 39d … (iii)
Substitute the valueof 2a in equation (ii)
2a + 29d – 160 = 0
180 – 39d + 29d – 160= 0
20 – 10d = 0
10d = 20
d = 20/10
= 2
Substitute the valueof d in equation (iii)
2a = 180 – 39d
2a = 180 – 39(2)
2a = 180 – 78
2a = 102
a = 102/2
= 51
Hence, value of firstinstallment ‘a’ is ₹ 51
Question - 4 : - A manufacturer of the radio sets produced 600 units in the third year and 700 units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find
(і) the production in the first year
(іі) the total product in the 7 years and
(ііі) the product in the 10th year.
Answer - 4 : -
Given:
600 and 700 radio setsunits are produced in third and seventh year respectively.
a3 =600 and a7 = 700
(і) The production in thefirst year
We need to find theproduction in the first year.
Let first yearproduction be ‘a’
So the AP formed is,a, a+x, a+2x, ….
By using the formula,
an = a+ (n-1)d
a3 = a+ (3-1)d
600 = a + 2d …. (i)
a7 = a+ (7-1)d
700 = a + 6d
a = 700 – 6d…. (ii)
Substitute value of ain (i) we get,
600 = a + 2d
600 = 700 – 6d + 2d
700 – 600 = 4d
100 = 4d
d = 100/4
= 25
Now substitute valueof d in (ii) we get,
a = 700 – 6d
= 700 – 6(25)
= 700 – 150
= 550
∴ The production in thefirst year, ‘a’ is 550
(іі) the total product inthe 7 years
We need to find thetotal product in 7 years i.e. is S7
By using the formula,
Sn =n/2 [2a + (n-1)d]
n = 7, a = 550, d = 25
S7 =7/2 [2(550) + (7-1)25]
= 7/2 [1100 + 150]
= 7/2 [1250]
= 7 [625]
= 4375
∴ The total product inthe 7 years is 4375.
(ііі) the product in the 10th year.
We need to find theproduct in the 10th year i.e. a10
By using the formula,
an = a+ (n-1)d
n = 10, a = 550, d =25
a10 =550 + (10-1)25
= 550 + (9)25
= 550 + 225
= 775
∴ The product in the 10th yearis 775.
Question - 5 : - There are 25 trees at equal distances of 5 meters in a line with a well, the distance of well from the nearest tree being 10 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.
Answer - 5 : -
Given: totaltrees are 25 and equal distance between two adjacent trees are 5 meters
We need to find thetotal distance the gardener will cover.
As gardener is comingback to well after watering every tree:
Distance covered bygardener to water 1st tree and return back to the initialposition is 10m + 10m = 20m
Now, distance betweenadjacent trees is 5m.
Distance covered byhim to water 2nd tree and return back to the initial positionis 15m + 15m = 30m
Distance covered bythe gardener to water 3rd tree return back to the initialposition is 20m + 20m = 40m
Hence distance coveredby the gardener to water the trees are in A.P
A.P. is 20, 30, 40………upto 25 terms
Here, first term, a =20, common difference, d = 30 – 20 = 10, n = 25
We need to find S25 whichwill be the total distance covered by gardener to water 25 trees.
So by using theformula,
Sn =n/2 [2a + (n – 1)d]
S25 =25/2 [2(20) + (25-1)10]
= 25/2 [40 + (24)10]
= 25/2 [40 + 240]
= 25/2 [280]
= 25 [140]
= 3500
∴ The total distancecovered by gardener to water trees all 25 trees is 3500m.
Question - 6 : - A man is employed to count ₹ 10710. He counts at the rate of ₹ 180 per minute for half an hour. After this he counts at the rate of ₹ 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
Answer - 6 : -
Given: Amount tobe counted is ₹ 10710
We need tofind time taken by man to count the entire amount.
He counts at the rateof ₹ 180 per minute for half an hour or 30 minutes.
So, Amount to becounted in an hour = 180 × 30 = ₹ 5400
Amount left = 10710 –5400 = ₹ 5310
Sn =5310
After an hour, rate ofcounting is decreasing at ₹ 3 per minute. This rate will form an A.P.
A.P. is 177, 174,171,……
Here a = 177 and d =174 – 177 = –3
By using the formula,
Sn =n/2 [2a + (n – 1)d]
5310 = n/2 [2(177) +(n-1) (-3)]
5310 = n/2 [354 -3n +3]
5310 × 2 = n [357 –3n]
10620 = 357n – 3n2
10620 = 3n(119 – n)
10620/3 = n(119 – n)
3540 = 119n – n2
n2 –119n + 3540 = 0
n2 –59n – 60n + 3540 = 0
n(n – 59) – 60(n – 59)= 0
(n – 59) (n – 60) = 0
n = 59 or 60
We shall considervalue of n = 59. Since, at 60th min he will count ₹ 0
∴ The total time takenby him to count the entire amount = 30 + 59 = 89 minutes.
Question - 7 : - A piece of equipment cost a certain factory ₹ 600,000. If it depreciates in value 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?
Answer - 7 : -
Given: A piece ofequipment cost a certain factory is ₹ 600,000
We need to find thevalue of the equipment at the end of 10 years.
The price of equipmentdepreciates 15%, 13.5%, 12% in 1st, 2nd, 3rd yearand so on.
So the A.P. will be15, 13.5, 12,…………… up to 10 terms
Here, a = 15, d = 13.5– 15 = –1.5, n = 10
By using the formula,
Sn =n/2 [2a + (n – 1)d]
S10 =10/2 [2(15) + (10-1) (-1.5)]
= 5 [30 + 9(-1.5)]
= 5 [30 – 13.5]
= 5 [16.5]
= 82.5
The value of equipmentat the end of 10 years is = [100 – Depreciation %]/100 × cost
= [100 – 82.5]/100 ×600000
= 175/10 × 6000
= 175 × 600
= 105000
∴ The value ofequipment at the end of 10 years is ₹ 105000.
Question - 8 : - A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?
Answer - 8 : -
Given: Price ofthe tractor is ₹12000.
We need tofind the total cost of the tractor if he buys it in installments.
Total price = ₹ 12000
Paid amount = ₹ 6000
Unpaid amount = ₹12000 – 6000 = ₹ 6000
He pays remaining ₹6000 in ‘n’ number of installments of ₹ 500 each.
So, n = 6000/500 = 12
Cost incurred by himto pay remaining 6000 is
The AP will be:
(500 + 12% of 6000) +(500 + 12% of 5500) + … up to 12 terms
500 × 12 + 12% of(6000 + 5500 + … up to 12 terms)
By using the formula,
Sn =n/2 [2a + (n – 1)d]
n = 12, a = 6000, d =-500
S12 =500×12 + 12/100 × 12/2 [2(6000) + (12-1) (-500)]
= 6000 + 72/100 [12000+ 11 (-500)]
= 6000 + 72/100 [12000– 5500]
= 6000 + 72/100 [6500]
= 6000 + 4680
= 10680
Total cost = 6000 +10680
= 16680
∴ The total cost of thetractor if he buys it in installment is ₹ 16680.