Question -11
A and B each have certain number of mangoes. A says to B,” if you give 30 of your mangoes, I will have twice as many as left with you.’’ B replies,’’ if you give me10, I will have thrice as many as you left with you.’’ How many mangoes does each have?
Solution
A:34, B:62
Question -12
8 Men and 12 Women can finish a piece of work in 10 days while 6 men and 8 women can finish it in 14 days. Find the time taken by one man alone and that by one woman alone to finish the work.
Solution
Suppose that one man alone can finish the work in x days and one woman alone
can finish it in y days. Then,
One man’s one day’s work =1/x; One woman’s one day’s work =1/y
∴ Eight men’s one day’s work =8/x; 12 women’s one day’s work =12/y
Since 8 men and 12 women can finish the work in 10 days ⇒10(8/x+12/y)=1
⇒80/x+120/y =1 ..(i) Again, 6 men and 8 women can finish the work
in 14 days. ∴14(6/x+8/y)=1 ⇒84/x + 112/y =1 (ii) Putting 1/x=u and 1/y=v in
equations (i) and (ii), we get
80u+120v−1=0
84u+112v−1=0 ; By using cross-multiplication method, we have
u = (-120+112)/(8960-10080) = -8/-1120 = 1/140 ; ⇒ 1/x = 1/140 ; ⇒ x = 140
v = (-84+80)/-1120 = -4/-1120 = 1/280 ; ⇒ 1/y =1/ 280 ; ⇒ y = 280
Thus, one man alone can finish the work in 140 days and one woman alone
can finish the work in 280 days.
Question -13
On selling a T.V. at 5 % gain and a fridge at 10 % gain, a shop keeper gains Rs.2000. But if he sells the T.V. at 10 % gain and the fridge at 5 % loss , he gains Rs.1500 on the transaction. Find the actual prices of T.V. and fridge.
Solution
Let, the price of a TV be Rs.x and that of a fridge be Rs.y
Then, we have:
5x/100 +10y/100 =2000
⇒ 5x+10y=200000; ⇒x+2y=40000 ;
⇒x=40000−2y….(i)
Now, 10x/100 − 5y/100 =1500
⇒10x−5y=150000
⇒2x−y=30000
⇒2(40000−2y)−y=30000 from equation (i)
⇒80000−4y−y=30000
⇒5y=80000−30000
⇒5y=50000
⇒y=Rs10000
∴x=Rs40000−Rs20000
⇒x=Rs.20000
∴ Price of a T.V. =Rs.20000; Price of a fridge =Rs.10000
Question -14
A shopkeeper sells a saree at a profit of 8% and a sweater at a discount of 10%, thereby getting a sum rs 1008.If he had sold the saree at a profit of 10% and the sweater at a discount of 8% he would have got rs 1028.Find the cost of the saree and the list price (price before discount) of the sweater.
Solution
Let the cost price of a saree is x and the list price of a sweater is y.
As per question 108% of x + 90% of y =1008 which solving gets
6x+5y=5600 and 110% of x +92% of y =1028 which gives by solving
55x+46y=51400 solving the equations we get x=600 and y =400.
Question -15
The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and the breadth
is increased by 3 units. The area is increased by 67 square units if length is increased by 3 units and
breadth is increased by 2 units. find the perimeter of the rectangle.
Solution
Let length of given rectangle be x and breadth be y ,
then area of rectangle will be xy .
According to the first condition we have
(x-5)(y+3) = xy -9
or, 3x – 5y= 6 ………(1)
According to the second condition, we have
(x+3) (y+2) = xy – 67
or, 2x +5y = 61 ……..(2)
Multiplying equation (1) by 3 and equation (2) by 5
and then adding,
9 x – 15 y = 18
10x + 15y = 305
x = 323/19
x = 17
Substituting this value of x in equation (1),
3(17) -5y =6
5y =51-6
y = 45/5
y = 9
Hence, perimeter = 2(x+y)
= 2(17+9)
= 52 units
Question -16
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 ₹2530.
Also one reserved first class ticket and one reserved first class half ticket from A to B costs
₹3810.Find the full first class fare from station A to B and also the reservation charges for a
ticket.
Solution
Let the cost of full fare be ₹ 𝑥 and the cost of half first class fare be ₹ 𝑥/ 2 ,
respectively and reservation charges be ₹ 𝑦 per ticket.
Case I
The cost of one reserved first class ticket from the stations A to B =₹ 2530
x + y = 2530 … ( i )
Case II
The cost of one reserved first class ticket and one reserved first class half ticket from
stations A to B = ₹ 3810
⇒ 𝑥 + 𝑦 + 𝑥/ 2 + 𝑦 = 3810
⇒ 𝑥 + 𝑥/ 2 + 𝑦 + 𝑦 = 3810
⇒ 3𝑥/ 2 + 2 𝑦 = 3810
Multiplying throughout by 2, we get
⇒ 3𝑥 + 4𝑦 = 7620 . . . (𝑖𝑖)
Now, multiplying Eq. (i) by 4 and then subtracting from Eq. (ii), we get
3 𝑥 + 4 𝑦 − 4𝑥 − 4𝑦 = 7620 − 10120
− 𝑥 = − 2500 ⇒ 𝑥 = 2500
On putting the value of x in Eq. (i), we get
2500+y=2530 ⇒ 𝑦 = 30 Hence, full first-class fare from stations A to B is ₹ 2500 and
the reservation for a ticket is ₹ 30
Question -17
The taxi charges in a city consist of a fixed charge together with the charge for the distance
covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the
charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution
Let the fixed charge be x
& the variable charge be y.
Given that charge paid for 10 km is Rs 105 and for 15 km is Rs 155
So,
Fixed charge +10×(Charge per km)=Rs105
x+10y=105⇒x=105−10y (1)
And for 15 kilometers,
x+15y=155⇒x=155−15y (2)
From (1) and (2)
105−10y=155−15y⇒y=10
putting value of y in equation (1)
x=105−10×10
5
x=5
For travelling 25 km, he has to pay
=x+25y
(5 + 25 × 10)
= Rs.255.
Question -18
A cyclist, after riding a certain distance, stopped for half an hour to repair his bicycle, after which he completes the whole journey of 50 km at half speed in 4 hours. If the breakdown had occurred 15 km farther off, he would have done the whole journey in 3 hours. Find where the breakdown occurred and
his original speed.
Solution
Let the cycle breakdown occurred at distance x km and speed of the cycle is y km/hr.
As per given questions , We have ,
x / y + ( 50 – x ) / ( y / 2 ) = 4 ————————————–(1)
( x + 15 ) / y + ( 50 – ( x + 15 ) ) / ( y / 2 ) = 3 ————————————-(2)
On solving above equations, we get , x = 40 and y = 15.
Therefore, the cycle breakdown occurred at distance 40 km and original speed of
the cycle is 15km/hr
Question -19
The perimeter of a rectangular garden is 20 m. If the length is 4 m more than the breadth, find the length and breadth of the garden.
Solution
Let the length of the garden be x m. Therefore, breadth of garden = (x -4)
m.
Since, perimeter is 20 m, so
2[x + (x – 4)] = 20
⇒ 2(2x – 4) = 20
⇒ 2x – 4 = 10
⇒ 2x = 10 + 4 = 14
⇒ x = 7
Hence, length = 7 m and breadth = 7 – 4 = 3 m.
Alternatively, you can solve the problem using two variables.
Let the length of garden = x m
Width of garden = y m
Therefore, x = y + 4 … (i)
Also, perimeter is 20 m, therefore
2(x + y) = 20
⇒ x + y = 10 … (i)
Solving (i) and (ii), we get x = 7, y = 3
Hence, length = 7 m and breadth = 3 m
Question -20
A and B each have certain number of oranges. A says to B, “if you give me 10 of your oranges, I will have twice the number of oranges left with you.” B replies, “if you give me 10 of your oranges, I will have the same number of oranges as left with you.” Find the number of oranges with A and B separately.
Solution
Suppose A has x oranges and B has y oranges.
According to the given conditions, we have
𝑥 + 10 = 2(𝑦 − 10) ⇒ 𝑥 − 2𝑦 + 30 = 0 (𝑖)
and 𝑦 + 10 = 𝑥 − 10 ⇒ 𝑥 − 𝑦 − 20 = 0 (𝑖𝑖)
Subtracting equation (ii) from equation (i), we get
− 𝑦 + 50 = 0 ⇒ 𝑦 = 50
Putting y = 50 in equation (i), we get x = 70
Hence, A has 70 oranges and B has 50 oranges
SELF ASSESSMENT
Q1. Determine algebraically and graphically the vertices of the triangle formed by
the lines 3x-y=3,2x-3y=2 and x+2y=8.
Q2. A train covered a certain distance at a uniform speed. If the train would have
been 6km/h faster, it would have taken 4 hours less than the scheduled time.
And if the train were slower by 6km/h, it would have taken 6 hours more than
the scheduled time. Find the length of the journey
Q3. After covering a distance of 30 km with a uniform speed there is some defect in
a train engine and therefore, its speed is reduced to 4/5 of its original speed.
Consequently, the train reaches its destination late by 45 minutes. Had it
happened after covering 18 kilometers more, the train would have reached 9
minutes earlier. Find the speed of the train and distance of the journey.
Q4. Raman usually go to a dry fruit shop with his mother. He observes the
following two situations.
On 1st day: The cost of 2 kg of almonds and 1 kg of cashew was Rs 1600.
On 2nd day: The cost of 4 kg of almonds and 2 kg of cashew was Rs 3000.
Denoting the cost of 1 kg almonds by Rs x and cost of 1 kg cashew by Rs
y, answer the following questions.
(i) Represent algebraically the situation of day-I.
(a) x + 2y = 1000 (b) 2x + y = 1600 (c) x – 2y = 1000 (d) 2x – y = 1000
(ii) Represent algebraically the situation of day- II.
(a) 2x + y= 1500 (b) 2x- y= 1500 (c) x + 2y=1500 (d) 2x + y = 750
(iii) The linear equation represented by day-I, intersect the x axis at
(a) (0,800) (b) (0,-800) (c) (800,0) (d) (-800,0)
(iv) The linear equation represented by day-II, intersect the y-axis at
(a) (1500,0) (b) (0, -1500) (c) (-1500,0) (d) (0,1500)
(v) Linear equations represented by day-I and day -II situations, are
(a) non parallel (b) parallel
(c) intersect at one point (d) overlapping each other.
Q5. A company has a locker in which valuable documents are kept. The passcode is
a four-digit number of the form xyyx. The Chief Executive Officer (CEO) and the
Vice President (VP) of the company have each been given one clue. On solving
BOTH clues, the passcode that opens the locker can be found.
CEO’s clue: When twice the ones digit is subtracted from the tens digit, the
result is 1.
VP’s clue: Three more than the tens digit is thrice the ones digit.
Find the passcode that opens the locker.
Q6. The total cost of snowden ice cream parlour is divided into fixed cost(x) and
variable cost(y). Fixed cost is the cost that the ice cream parlour has to incur
even at zero level of production and variable cost is the cost that will be
directly proportional to each unit of ice cream sold. The parlour launched a
new flavour of ice cream and wanted to find the fixed and variable cost
associated with it. They found that their total cost for that flavour was Rs
27500 after selling 150 units and Rs 32500 after selling 250 units. Answer the
following questions based on the given information.
Q7. The numerator of a fraction is 5 less than the denominator. If the numerator
increased by 1 and the denominator increased by 4 , then they are in the ratio
1:3 . Determine the fraction.
Q8. 3 years ago Rekha was 3 years older than thrice as old as her nephew. 5 years
from now, she will be 6 years more than twice as old as her nephew .Find the
age of Rekha after 10 years?
