an amount of 50000 is out into three investments at the rate of 6%, 7%, 8% respectively. the total annual income is 3580. if the combined income from the first two investments is 70 more than the income from the third. find the amount invested in each type of investment separately. Determinants ka question hai.
Answers
Answer:
Let investments be x, y, z.
x=
100
6×5000
=300
y=
100
7×5000
=350
z=
100
8×5000
=400
Now,
6x+7y+8z=100 ………(1)
x+y+z=358 ……(2)
x+y−z=70 ………(3)
Δx=B
⎣
⎢
⎢
⎡
6
1
1
7
1
1
8
1
−1
⎦
⎥
⎥
⎤
⎣
⎢
⎢
⎡
x
y
z
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
100
358
70
⎦
⎥
⎥
⎤
=Δ=
∣
∣
∣
∣
∣
∣
∣
∣
6
1
1
7
1
1
8
1
−1
∣
∣
∣
∣
∣
∣
∣
∣
=(−6+8+7)−(8+6−7)
=9−7=2
Δ
1
=
∣
∣
∣
∣
∣
∣
∣
∣
100
358
70
7
1
1
8
1
−1
∣
∣
∣
∣
∣
∣
∣
∣
=(−100+358×8+490)−(560+100−7×358)
=−70+15×358
=5300
Δ
2
=
∣
∣
∣
∣
∣
∣
∣
∣
6
1
1
100
358
70
8
1
−1
∣
∣
∣
∣
∣
∣
∣
∣
=(−6×358+100−560)−(8×358+420−100)
=−14×358+660−320
=−4672
Δ
3
=
∣
∣
∣
∣
∣
∣
∣
∣
6
1
1
7
1
1
100
358
70
∣
∣
∣
∣
∣
∣
∣
∣
=(1420+100+7×358)−(100+6×358+490)
=288
x=
Δ
Δ
1
=
2
5300
=2650
y=
Δ
Δ
2
=
2
−4672
=2336
z=
Δ
Δ
3
=
2
288
=144
Amount of each investment is Rs. 2650, Rs. 2336, Rs. 144
Answer:
Let investments be x, y, z.
x=1006×5000=300
y=1007×5000=350
z=1008×5000=400
Now,
6x+7y+8z=100 ………(1)
x+y+z=358 ……(2)
x+y−z=70 ………(3)
Δx=B
⎣⎢⎢⎡61171181−1⎦⎥⎥⎤⎣⎢⎢⎡xyz⎦⎥⎥⎤=⎣⎢⎢⎡10035870⎦⎥⎥⎤
=Δ=∣∣∣∣∣∣∣∣6
=(−6+8+7)−(8+6−7)
=9−7=2
Δ1=∣∣∣∣∣∣∣∣1003587071181−1∣∣∣∣∣∣∣∣
=(−100+358×8+490)−(560+100−7×358)
=−70+15×358
=5300
Δ2=∣∣∣∣∣∣∣∣6111003587081−1∣∣∣∣∣∣∣∣
=(−6×358+100−560)−(8×358+420−100)
=−14×358
=−4672
Δ3=∣∣∣∣∣∣∣∣61171110035870∣∣∣∣∣∣∣∣
=(1420+100+7×358)−(100+6×358+490)
=288
x=ΔΔ1=25300=2650
y=ΔΔ2=2−4672=2336
z=ΔΔ3=2288=144
Amount of each investment is Rs. 2650, Rs. 2336, Rs. 144.