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Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.

 

Solution:
Let bananas in lot A = x
Bananas in lot B = y
According to question:
\left ( \frac{x}{3} \right )\left ( 2 \right )+y\left ( 1 \right )= 400
2x + 3y = 400 × 3
2x + 3y = 1200 … (1)
x(1) + \left ( \frac{y}{5} \right )(4) = 460
5x + 4y = 460 × 5
5x + 4y = 2300 … (2)
Multiply equation (1) by 5 and eq. (2) by 2
10x + 15y = 6000                    
10x + 8y = 4600
–         –         –
7y = 1400

y= \frac{1400}{7}= 200
Put y = 200 in eq. (1)
2x + 3(200) = 1200
2x + 600 = 1200
2x = 600
x = 300

Total number of bananas = x + y = 300 + 200 = 500

 

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Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme

Solution:
Let money invested in A and B are x, y respectively.
So, according to question.
0.08x + 0.09y = 1860                 … (1)

x\times \frac{9}{100}+y\times \frac{8}{100}= 1880
0.09x + 0.08y = 1880              … (2)
Multiply equation (1) by 9 and (2) by 8
0.72x + 0.81y = 16740
0.72x + 0.64y = 15040
–             –              –
0.17y = 1700

y = \frac{1700}{0\cdot 7}= 10000
Put y = 10000 in eq. (1)
0.08x + 900 = 1860
0.08x = 960
x = \frac{960}{8}\times 100 = 12000
x = 12000
He invested Rs. 12000 in A and Rs. 10000 in B.

 

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A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, there by, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028. Find the cost price of the saree and the list price (price before discount) of the sweater.

 

Solution:
Let the cost price of saree = Rs. x
Let the cost price of sweater = Rs. y
Saree with 8% profit =  (1+\frac{8}{100})x=\frac{108x}{100}
Sweater with 10% discount = (1-\frac{10}{100})y = \frac{90y}{100}
Saree with 10% profit = x + x\times \frac{10}{100}= \frac{110x}{100}

Sweater with 8% discount = y –  y\times \frac{8}{100}= \frac{92y}{100}

According to question:

\frac{108x}{100}+ \frac{92y}{100}= 1008
108x + 90y = 100800         
6x + 5y = 5600 … (1)             (Divide by 18)

\frac{110x}{100}+ \frac{92y}{100}= 1028
110x + 92y = 102800
55x + 56y = 51400 … (2)          (Divide by 2)
Multiply equation (1) by 46 and eq. (2) by  5
276 x + 230 y = 257600
275 x + 230 y = 257000
–              –             –
x = 600
Put x = 600 in eq. (1)
6(600) + 5y = 5600
5y = 5600 –3600
5y = 2000
y = 400
Price of saree = Rs. 600
Price of sweater = Rs. 400
 

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A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the station A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs Rs 3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket.

Solution:
Let the full ticket cost = y Rs.
Let reservation charge = x Rs.
According to questions
x + y = 2530                            … (1)

(x + y) + (x + y/2) = 3810       …(2)

Here x + y is for full ticket and x + y/2 is for half.
Solve equation (1) and (2)
x + y + x + \frac{y}{2} = 3810

2x + y + \frac{y}{2} = 3810
4x + 2y + y = 3810 × 2
4x + 3y = 7620                        … (3)
Multiply equation (1) by 3
3x + 3y = 7590
4x + 3y = 7620
–      –        –
–x = – 30
Put x = 30 in eq. (1)
30 + y = 2530
y = 2500
Hence Reservation charge  = 30Rs.
Full First class charge = 2500 Rs.

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A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.

Solution:
Let the first digit = x
Let the ten’s digit = y
Therefore the number = 10y + x
As per the question
8(x + y) –5 = 10y + x
8x + 8y – 10y – x = 5
7x – 2y = 5 … (1)
16(y – x) + 3 = 10y + x
16y – 16x – 10y – x = – 3
–17x + 6y = –3
17x – 6y = 3 … (2)
Multiply equation (1) by 3
21x – 6y = 15
17x – 6y = 3
–        +     –
4x = 12
x = 3
Put x = 3 in eq. (1)
7(3) –2y = 5
–2y = 5 –21

y = \frac{16}{2} = 8
Hence the number is = 10y + x = 10(8) + 3 = 83
 

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A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

Solution:
Let speed of boat in still water = x km/h
Let speed of stream = y km/h
Speed of boat in upstream = (x - y) km/h
Speed of boat in downstream = (x + y) km/h
We know that time =\frac{distance}{speed}

So, \frac{30}{\left ( x-y \right )}+\frac{28}{x+y}= 7… (1)

\frac{21}{\left ( x-y \right )}+\frac{21}{x+y}= 5 … (2)

Put \frac{1}{\left ( x-y \right )}= u,\frac{1}{\left ( x+y \right )}= v in (1), (2)

30u + 28v = 7 … (3)
21u + 21v = 5 … (4)
Solve equation (3), (4)
Multiply equation (3) by 21 and (4) by 30
630u + 588 v = 147
630u + 630v = 150
_-______-____-________
+42v = +3

v= \frac{3}{42}= \frac{1}{14}

Put v= \frac{1}{14} in eq. (4)

21u+21\left ( \frac{1}{14} \right )= 5

21u= 5-\frac{3}{2}

21u= \frac{10-3}{2}

21u = \frac{7}{2}
u= \frac{7}{2\times 21}\Rightarrow u= \frac{1}{6}
We put u= \frac{1}{x-y}\, and\, v= \frac{1}{x+y}

So,

\frac{1}{x-y}= \frac{1}{6}            |           \frac{1}{x+y}= \frac{1}{14}
x-y=6..(5)                          x+y=14...(6)

By adding equation (5) and (6)
2x = 20
x = 10
Put x = 10 in eq. (6)
10 + y = 14
y = 4
Hence speed of the boat in still water = 10 km/hr.
Speed of the stream = 4 km/hr.

 

 

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A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Solution:

According to question,
Person’s speed with upstream = (5 – x) km/h
Person’s speed with downstream = (5 + x) km/h

Time with upstream = \frac{40}{5-x}\left ( Using\, t= \frac{D}{S} \right )

Time with downstream = \frac{40}{5+x}\left ( Using\, t= \frac{D}{S} \right )

\frac{40}{5-x}= \frac{3\times 40}{5+x}
(5 + x)40 = 120 (5 – x)
200 + 40x = 600 - 120 x
160x = 400
x = 2.5
Speed of stream = 2.5 km/h

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Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus.On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

Solution:
We know that

Speed = \frac{Distance}{Time}
Time =  Distance / Speed
Let the speed of rickshaw is x and of bus is y.
So, according to question

\frac{2}{x}+\frac{\left ( 14-2 \right )}{y}= \frac{1}{2}Hr

\frac{2}{x}+\frac{12}{y}= \frac{1}{2}\cdots (1)

 \frac{4}{x}+\frac{10}{y}= \frac{39}{60}\, Hour\cdots (2)

Solve eq. (1) and (2)

\left ( \frac{2}{x}+\frac{12}{y} = \frac{1}{2}\right )\times 2

\frac{4}{x}+\frac{24}{y}= 1

\frac{4}{x}+\frac{10}{y}= \frac{39}{60}
–     –      –

\frac{24}{y}-\frac{10}{y}=1- \frac{39}{60}

\frac{24-10}{y}= \frac{60-39}{60}

\frac{14}{y}= \frac{21}{60}
21y = 840

y= \frac{840}{21}= 40
Put y = 40 in eq. (1)

\frac{2}{x}+\frac{12}{40}= \frac{1}{2}

\frac{2}{x}= \frac{20-12}{40}

\frac{2}{x}= \frac{8}{40}
40 = 4x
x = 10
Speed of rickshaw = 10 km/h
Speed of bus = 40 km/h

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Determine, algebraically, the vertices of the triangle formed by the lines 3x – y =  3, 2x – 3y = 2 and x + 2y = 8

Solution:
Determine, algebraically, the vertices of the triangle formed by the lines
3x – y =  3 … (1)
2x – 3y = 2 … (2)
x + 2y = 8 ... (3)

For vertices of A, solve equation (1) and (3)
3x – y = 3
x + 2y = 8 × 3
3x + 6y = 24
3x + y = –3
7y = 21
y = 3
Put y = 3 in eq. (1)
3x – 3 = 3
3x = 6
x = 2
point A is (2, 3)
For vertices of B solve equation (1) and (2)
3x – y = 3 × 3
2x – 3y = 2
9x – 3y = 9
–    +    –
–7x = –7
x = 1
Put x = 1 in equation (1)
3(1) – y = 3
3 – 3 = y
y = 0
point B is (1, 0)
For vertices of C solve (2) and (3)
2x – 3y = 2
x + 2y = 8 × 2
2x + 4y = 16
2x – 3y = 2
–    +     –
7y = 14
y = 2
Put y = 2 in eq. (2)
2x – 3(2) = 2
2x = 2 + 6

x = \frac{8}{2} = 4
Point C (4, 2)
Hence vertices of triangle ABC are A (2, 3), B (1, 0) and C (4, 2)

 

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The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

Solution:
Let the cost of a pen is x and the cost of a pencil box is y.
As per the question
4x + 4y = 100 … (1)
3x = y + 15 … (2)
Divide equation (1) by 4.
x + y = 25    ………..(3)
By adding equation (1) and (3) we get
4x = 40
x = 10
Put x = 10 in eq. (1)
40 + 4y = 100
4y = 100 – 40
y = 60/4 = 15
x = 10, y = 15
Cost of pen = Rs. 10
Cost of pencil box = Rs.15

 

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